Second Edition
BALLISTICS
Theory and Design
of Guns and Ammunition
© 2014 by Taylor & Francis Group, LLC
© 2014 by Taylor & Francis Group, LLC
Second Edition
BALLISTICS
Theory and Design
of Guns and Ammunition
Donald E. Carlucci and Sidney S. Jacobson
Boca Raton London New York
CRC Press is an imprint of the
Taylor & Francis Group, an informa business
© 2014 by Taylor & Francis Group, LLC
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accuracy of the text or exercises in this book. This book’s use or discussion of MATLAB® software or related products
does not constitute endorsement or sponsorship by The MathWorks of a particular pedagogical approach or particular
use of the MATLAB® software.
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Version Date: 20130627
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© 2014 by Taylor & Francis Group, LLC
MATLAB® is a trademark of The MathWorks, Inc. and is used with permission. The MathWorks does not warrant the
accuracy of the text or exercises in this book. This book’s use or discussion of MATLAB® software or related products
does not constitute endorsement or sponsorship by The MathWorks of a particular pedagogical approach or particular
use of the MATLAB® software.
CRC Press
Taylor & Francis Group
6000 Broken Sound Parkway NW, Suite 300
Boca Raton, FL 33487-2742
© 2014 by Taylor & Francis Group, LLC
CRC Press is an imprint of Taylor & Francis Group, an Informa business
No claim to original U.S. Government works
Printed on acid-free paper
Version Date: 20130627
International Standard Book Number-13: 978-1-4665-6437-4 (Hardback)
This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been
made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright
holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this
form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may
rectify in any future reprint.
Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the
publishers.
For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://
www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923,
978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For
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Library of Congress Cataloging-in-Publication Data
Carlucci, Donald E.
Ballistics : theory and design of guns and ammunition / Authors, Donald E. Carlucci, Sidney S.
Jacobson. -- 2nd edition.
p. cm.
Includes bibliographical references and index.
ISBN 978-1-4665-6437-4
1. Ballistics. I. Jacobson, Sidney S. II. Title.
UF820.C28 2014
623’.51--dc23
Visit the Taylor & Francis Web site at
http://www.taylorandfrancis.com
and the CRC Press Web site at
http://www.crcpress.com
© 2014 by Taylor & Francis Group, LLC
2013014363
To Peg C., Sandy J., and our families,
without whose patience and support
we could not have brought
this work to completion
© 2014 by Taylor & Francis Group, LLC
© 2014 by Taylor & Francis Group, LLC
Contents
Preface to the Second Edition ...................................................................................................... xi
Preface to the First Edition......................................................................................................... xiii
Acknowledgments ........................................................................................................................xv
Authors ........................................................................................................................................ xvii
Part I
Interior Ballistics
1. Introductory Concepts ...........................................................................................................3
1.1 Ballistic Disciplines.......................................................................................................4
1.2 Terminology ...................................................................................................................4
1.3 Units and Symbols ........................................................................................................5
2. Physical Foundation of Interior Ballistics .........................................................................7
2.1 Ideal Gas Law ................................................................................................................7
2.2 Other Gas Laws ........................................................................................................... 14
2.3 Thermophysics and Thermochemistry.................................................................... 15
2.4 Thermodynamics ........................................................................................................ 21
2.5 Combustion .................................................................................................................. 26
2.6 Solid Propellant Combustion .................................................................................... 38
2.7 Fluid Mechanics .......................................................................................................... 48
References ............................................................................................................................... 68
3. Analytic and Computational Ballistics ............................................................................ 69
3.1 Computational Goal ................................................................................................... 69
3.2 Lagrange Gradient ...................................................................................................... 70
3.3 Chambrage Gradient ................................................................................................ 100
3.4 Numerical Methods in Interior Ballistics .............................................................. 102
3.5 Sensitivities and Efficiencies ................................................................................... 106
References ............................................................................................................................. 109
4. Ammunition Design Practice ........................................................................................... 111
4.1 Stress and Strain........................................................................................................ 111
4.2 Failure Criteria .......................................................................................................... 115
4.3 Ammunition Types ................................................................................................... 119
4.4 Propellant Ignition .................................................................................................... 119
4.5 The Gun Chamber .................................................................................................... 120
4.6 Propellant Charge Construction ............................................................................. 121
4.7 Propellant Geometry ................................................................................................ 122
4.8 Cartridge Case Design ............................................................................................. 123
4.9 Projectile Design ....................................................................................................... 126
4.10 Shell Structural Analysis ......................................................................................... 126
4.11 Buttress Thread Design ............................................................................................ 148
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© 2014 by Taylor & Francis Group, LLC
Contents
viii
4.12 Sabot Design .............................................................................................................. 156
References ............................................................................................................................. 165
Further Readings ................................................................................................................. 165
5. Weapon Design Practice .................................................................................................... 167
5.1 Fatigue and Endurance ............................................................................................ 167
5.2 Tube Design ............................................................................................................... 169
5.3 Gun Dynamics........................................................................................................... 175
5.4 Muzzle Devices and Associated Phenomena ....................................................... 181
Gun Dynamics Nomenclature ........................................................................................... 189
References ............................................................................................................................. 189
Further Readings ................................................................................................................. 190
Part II
Exterior Ballistics
6. Introductory Concepts ....................................................................................................... 193
References ............................................................................................................................. 203
Further Reading ................................................................................................................... 203
7. Dynamics Review ............................................................................................................... 205
Reference ............................................................................................................................... 221
Further Readings ................................................................................................................. 221
8. Trajectories ...........................................................................................................................223
8.1 Vacuum Trajectory ....................................................................................................223
8.2 Simple Air Trajectory (Flat Fire).............................................................................. 231
8.3 Wind Effects on a Simple Air Trajectory ............................................................... 248
8.4 Generalized Point Mass Trajectory......................................................................... 261
8.5 Six Degree-of-Freedom (6-DOF) Trajectory........................................................... 270
8.6 Modified Point Mass Trajectory .............................................................................. 290
8.7 Probability of First Round Hit................................................................................. 301
References .............................................................................................................................305
Further Readings .................................................................................................................305
9. Linearized Aeroballistics .................................................................................................. 307
9.1 Linearized Pitching and Yawing Motions ............................................................309
9.2 Gyroscopic and Dynamic Stabilities ...................................................................... 320
9.3 Yaw of Repose............................................................................................................ 328
9.4 Roll Resonance .......................................................................................................... 329
References ............................................................................................................................. 331
10. Mass Asymmetries ............................................................................................................. 333
References ............................................................................................................................. 335
11. Lateral Throwoff ................................................................................................................. 337
11.1 Static Imbalance ........................................................................................................340
11.2 Dynamic Imbalance..................................................................................................342
References .............................................................................................................................346
© 2014 by Taylor & Francis Group, LLC
Contents
ix
12. Swerve Motion .................................................................................................................... 347
12.1 Aerodynamic Jump................................................................................................... 347
12.2 Epicyclic Swerve ........................................................................................................ 351
12.3 Drift ............................................................................................................................. 352
Reference ............................................................................................................................... 353
13. Nonlinear Aeroballistics ................................................................................................... 355
13.1 Nonlinear Forces and Moments.............................................................................. 355
13.2 Bilinear and Trilinear Moments.............................................................................. 358
References ............................................................................................................................. 361
Part III
Terminal Ballistics
14. Introductory Concepts ....................................................................................................... 365
15. Penetration Theories .......................................................................................................... 369
15.1 Penetration and Perforation of Metals ................................................................... 369
15.2 Penetration and Perforation of Concrete ............................................................... 391
15.3 Penetration and Perforation of Soils ....................................................................... 399
15.4 Penetration and Perforation of Ceramics ............................................................... 406
15.5 Penetration and Perforation of Composites........................................................... 414
References ............................................................................................................................. 416
16. Shock Physics ...................................................................................................................... 419
16.1 Shock Hugoniots ....................................................................................................... 419
16.2 Rarefaction Waves ..................................................................................................... 437
16.3 Stress Waves in Solids .............................................................................................. 460
16.4 Detonation Physics....................................................................................................480
16.5 Explosives Equations of State .................................................................................. 499
16.5.1 JWL Equation of State .................................................................................500
16.5.2 JWLB Equation of State ............................................................................... 501
16.5.3 Analytic Cylinder Model ............................................................................ 501
References .............................................................................................................................505
Further Readings ................................................................................................................. 506
17. Introduction to Explosive Effects .................................................................................... 507
17.1 Gurney Method ......................................................................................................... 507
17.2 Taylor Angles ............................................................................................................. 511
17.3 Mott Formula ............................................................................................................. 516
References ............................................................................................................................. 522
Further Reading ................................................................................................................... 522
18. Shaped Charges .................................................................................................................. 523
18.1 Shaped Charge Jet Formation ................................................................................. 525
18.2 Shaped Charge Jet Penetration................................................................................534
References .............................................................................................................................546
Further Reading ...................................................................................................................546
© 2014 by Taylor & Francis Group, LLC
x
Contents
19. Wound Ballistics ................................................................................................................. 547
References ............................................................................................................................. 557
Further Reading ................................................................................................................... 557
Appendix A ................................................................................................................................. 559
Appendix B.................................................................................................................................. 569
Index ............................................................................................................................................. 575
© 2014 by Taylor & Francis Group, LLC
Preface to the Second Edition
When we were asked to create a second edition to Ballistics: Theory and Design of Guns and
Ammunition, we were a bit concerned about what could or should be added or changed.
One thing that was certain was that the numerous errors that we became aware of through
teaching and student observations had to be corrected. For that, we are indebted to the students and readers who have taken the time and effort to point these out. It was certainly a
humbling experience. Hopefully we got them all this time. With the errors corrected, the
task of deciding what we could put in to add more value to the book while maintaining an
introductory treatment rose to the forefront.
A great deal of work went into this edition that might not be apparent on the surface. We
added a large number of new problems and updated the solutions manual accordingly.
In fact, the solutions manual increased in size from 247 pages to 543 pages! One might
notice that the new problems do not have answers in the text. This was intentionally
done to allow instructors to assign these problems as tests or graded homework. We also
added some Mathcad® codes to the CRC website to assist our readers with solution of the
problems. These are available at http://www.crcpress.com/product/isbn/9781466564374.
The more obvious changes include an increase in topical areas. We decided to add
a section to Chapter 8 that would, in a general sense, discuss the topic of probability
of first round hit for direct fire weapons. This treatment is general and certainly does
not even attempt to reflect the state of the art since that data are not appropriate for a
general audience. Another area of major improvement is the additional sections added
to Chapter 16, which discuss explosive equations of state. Dr. Ernie Baker, our professional colleague and recognized expert in the field of explosives effects and insensitive
munition design, added this section. We are extremely grateful to him for this. His
schedule was incredibly busy, yet he found the time to assist us. Finally, there was
a significant update to Chapter 19 on wound ballistics. During any course on terminal ballistics, it seems that this topic area is the one people seem most interested in.
Granted, the treatment is still general, and, as is common in our experience, any statement made about the wounding process will be hotly debated, we decided to put more
flesh on the bones of this section. Hopefully it is of help, keeping in mind the introductory nature of the treatment.
Michael S. L. Hollis is the creator of the cover art, which depicts a typical kinetic
energy penetrator in the midst of discarding its sabot. The sabot consists of the three
“petals,” each with bow shocks forming ahead of them, peeling away from the subprojectile. The flames emanating from the rear of the subprojectile represent the tracer that
allows the firer to visualize its trajectory downrange. These projectiles are generally
fired at high Mach numbers and are used in weapons with bore diameters that range
from 20–120 mm.
We are indebted to Jonathan Plant of CRC Press, Taylor & Francis Group, for encouraging
(i.e., gently prodding) us to undertake the work of revision and facilitating its publication.
xi
© 2014 by Taylor & Francis Group, LLC
xii
Preface to the Second Edition
Disclaimer
The design, fabrication, and use of guns, ammunition, and explosives are, by their very
nature, dangerous. The techniques, theories, and procedures developed in this book
should not be utilized by anyone without the proper training and certifications. In the
checking and editing of these techniques, theories, and procedures, every effort has been
made to identify potential hazardous steps, and safety precautions have been inserted
where appropriate. However, these techniques, theories, and procedures must be exercised
at one’s own risk. The authors and the publisher, its subsidiaries and distributors, assume
no liability and make no guarantees or warranties, express or implied, for the accuracy of
the contents of this book or the use of information, methods, or products described within.
In no event shall the authors, the publisher, its subsidiaries or distributors be liable for
any damages and expense resulting from the use of information, methods, or products
described in this book.
MATLAB® is a registered trademark of The MathWorks, Inc. For product information,
please contact:
The MathWorks, Inc.
3 Apple Hill Drive
Natick, MA 01760-2098 USA
Tel: 508-647-7000
Fax: 508-647-7001
E-mail: info@mathworks.com
Web: www.mathworks.com
© 2014 by Taylor & Francis Group, LLC
Preface to the First Edition
This book is an outgrowth of a graduate course taught by the authors for the Stevens
Institute of Technology at the Picatinny Arsenal in New Jersey. Engineers and scientists
at the arsenal have long felt the need for an armature of the basic physics, chemistry, electronics, and practice on which to flesh out their design tasks as they go about fulfilling
the needs and requirements of the military services for armaments. The Stevens Institute
has had a close association with the arsenal for several decades, providing graduate programs and advanced degrees to many of the engineers and scientists employed there. It is
intended that this book be used as a text for future courses and as a reference work in the
day-to-day business of weapons development.
Ballistics as a human endeavor has a very long history. From the earliest developments
of gunpowder in China more than a millennium ago, there has been an intense need felt
by weapon developers to know how and why a gun works, how to predict its output in
terms of the velocity and range of the projectiles it launched, how best to design these projectiles to survive the launch, fly to the target and perform the functions of lethality, and
the destructions intended.
The discipline over the centuries has divided itself into three natural regimes: Interior
ballistics or what happens when the propellant is ignited behind the projectile until the
surprisingly short time later when the projectile emerges from the gun; exterior ballistics
or what happens to the projectile after it emerges and flies to the target and how to get it to
fly there reproducibly shot after shot; and terminal ballistics or once it is in the vicinity of
the target, how to extract the performance from the projectile for which the entire process
was intended, usually lethality or destruction.
Ballisticians, those deeply involved in the science of ballistics, tend to specialize in only
one of the regimes. Gun and projectile designers, however, must become proficient in all
the regimes if they are to successfully field weapons that satisfy the military needs and
requirements. The plan of this book is bilateral: first, an unfolding of the theory of each
regime in a graduated ascent of complexity, so that a novice engineer gets an early feeling
for the subject and its nomenclature and is then brought into a deeper understanding of the
material; second, an explanation of the design practice in each regime. Most knowledge of
weapon design has been transmitted by a type of apprenticeship with experienced designers sharing their learning with newer engineers. It is for these engineers that this work is
intended, with the hope that it will make their jobs easier and their designs superior.
xiii
© 2014 by Taylor & Francis Group, LLC
© 2014 by Taylor & Francis Group, LLC
Acknowledgments
In the four years of development of this book, many people have contributed in a variety
of ways. This has probably been the most difficult section that we had to write for fear of
neglecting key people.
We are a generation apart in our careers, mentors, and experience, but not in our enthusiasm for the subject. As the elder, Sidney S. Jacobson must acknowledge several people
who inspired his enthusiasm beginning in the 1950s: Robert Schwartz, designer of the 280
mm atomic shell who also wrote the first set of ammunition design notes; Alfred A. Loeb,
aeroballistician and friend; and Ralph F. Campoli, a peerless ammunition designer and
longtime friend.
Donald E. Carlucci would particularly like to thank for encouragement and mentorship Michael P. Devine, William DeMassi, James Pritchard, Howard Brunvoll, Vincent
Marchese, Robert Reisman, Dr. Daniel Pillasch, Donald Rybarczyk, Dale Kompelien,
Anthony Fabiano, Carmine Spinelli, Stephen Pearcy, Ami Frydman, Walter Koenig,
Dr. Peter Plostins, and William R. Smith, all of whom have contributed to his professional
development and thus to the ultimate publication of this book.
Both of us must acknowledge as mentor, role model, and friend, the late Victor Lindner,
an acknowledged leader in the world of weaponry, whose career spanned both of ours.
On the inspirational as well as technical side of the ledger, our heartfelt thanks go to Paul
Cooper and Dr. John Zukas, whose gentle prodding to get the project moving and encouragement throughout has been unflagging. From a technical standpoint, we would like to thank
Dan Pangburn for allowing us to incorporate his method of buttress thread calculation,
Dr. Bryan Cheeseman for his comments and help on the composites and ceramics sections,
all of the reviewers of the book, Dr. Costas Chassapis, and Dr. Siva Thangam for their support while the material was being developed and taught as courses for Stevens Institute of
Technology, Mark Minisi, Stanley DeFisher, Shawn Spickert-Fulton, Miroslav Tesla, Patricia
Van Dyke, Dr. Wei-Jen Su, Yin Chen, John Thomas, Dr. Bill Drysdale, Dr. Bill Walters, Igbal
Mehmedagic, and Julio Vega who, as teachers, students, friends, and co-workers, have either
contributed analyses, checked problems, or suggested corrections to the manuscript.
We would like to thank the following persons who very kindly assisted in pointing out errors in the text: Dr. Ernie Baker, James Boatright, Sean Brandt, Liam Buckley,
Dominic Cimorelli, Paul Cooper, Michael Colonnello, Charles T. Freund, John Geaney,
Nicholas Grossman, Jonathan Jablonski, Dennis Mackin, Mark Minisi, N. Rus Payne,
Stephen Recchia, Lindsay Roberts, Oscar Ruiz, Shawn Spickert-Fulton, Robert Terhune
and Caitlin Weaver.
Additionally, we would like to thank Jonathan Plant and Richard Tressider for their
help and encouragement, patience, and commitment to the publication of both the first
and second editions of the book. With regard to the first edition, we wish to thank
Sathyanarayanamoorthy Sridharan for actually seeing the edition through publication.
Similarly, we wish to thank Arun Kumar Aranganathan for seeing this second edition
through publication.
Only we are responsible for any errors of commission or omission.
xv
© 2014 by Taylor & Francis Group, LLC
© 2014 by Taylor & Francis Group, LLC
Authors
Donald E. Carlucci has been an engineer at the U.S. Army Armament, Research,
Development and Engineering Center, Picatinny Arsenal, since May 1989. He is currently
the U.S. Army Senior Scientist for Computational Structural Modeling based at Picatinny.
He was formerly chief of the Analysis and Evaluation Technology Division, Fuze and
Precision Munitions Technology Directorate responsible for the modeling and evaluation
of cannon-launched munitions programs at Picatinny, and chief scientist for the XM982
Excalibur guided projectile. Dr. Carlucci also has formerly held the position of development program officer (chief engineer) for sense and destroy armor (SADARM). Before his
employment at Picatinny, he was a design engineer for Titanium Industries, located in
Fairfield, New Jersey.
Dr. Carlucci has held positions as chief engineer, quality assurance manager, and purchasing manager for Hoyt Corporation, located in Englewood, New Jersey. He is a licensed
professional engineer in the states of New Jersey and New York and holds a doctor of philosophy in mechanical engineering (2002) and a master of engineering (mechanical) (1995)
degree from the Stevens Institute of Technology, Hoboken, New Jersey. In 1987, he received
his bachelor of science degree in mechanical engineering from the New Jersey Institute of
Technology, Newark, New Jersey.
Dr. Carlucci is an adjunct professor of mechanical engineering at the Stevens Institute of
Technology where he teaches graduate classes on interior, exterior, and terminal ballistics
as well as undergraduate classes on engineering design.
Sidney S. Jacobson was a researcher, designer, and developer of ammunition and weapons at the U.S. Army’s Picatinny Arsenal in New Jersey for 35 years. He rose from junior
engineer through eight professional levels in research and development laboratories to
become associate director for R&D at the arsenal. His specialty for most of his career was
in the development of large caliber tank munitions and cannons. Many of these weapons,
such as the long rod, kinetic energy penetrators (APFSDS rounds), and the shaped charge,
cannon-fired munitions (HEAT rounds), have become standard equipment in the U.S.
Army. For these efforts and successes he earned several awards from the army including,
in 1983, the Department of the Army Meritorious Civilian Service Medal. In 1972, he was
awarded an Arsenal Educational Fellowship to study continuum mechanics at Princeton
University where he received his second MS degree (1974). He earned a master of science
in applied mechanics from Stevens Institute of Technology (1958) and a bachelor of arts in
mathematics from Brooklyn College (1951).
He retired in 1986 but maintains his interest in the field through teaching, consulting, and
lecturing. He holds two patents and was a licensed professional engineer in New Jersey.
xvii
© 2014 by Taylor & Francis Group, LLC
© 2014 by Taylor & Francis Group, LLC
Part I
Interior Ballistics
© 2014 by Taylor & Francis Group, LLC
© 2014 by Taylor & Francis Group, LLC
1
Introductory Concepts
The subject of ballistics has been studied for centuries by people at every level of academic achievement. Some of the world’s greatest mathematicians and physicists such as
Newton, Lagrange, Bernoulli, and others solved problems in mathematics and mechanics
that either directly or indirectly were applied to the various ballistic disciplines. At the
other end of the academic scale, there are individuals such as James Paris Lee (inventor
of the Lee-Enfield rifle) who developed his first weapon (not the famous Lee-Enfield) at
age 12 with no formal education.
The dominant characteristic of any of the ballistic disciplines is the “push–pull” relationship of experiment and analysis. It is a rare event, even as of this writing, when an
individual can design a ballistic component or device, either digitally or on paper, and
have it function “as designed” in the field. Some form of testing is always required and
consequent tweaking of the design. This inseparable linkage between design and test is
due to three things: the stochastic nature of ballistic events, the infinite number of conditions into which a gun–projectile–charge combination can be introduced, and the lack of
understanding of the phenomena.
The stochastic behavior that dominates all of the ballistic disciplines stems from the
tremendous number of parameters that affect muzzle velocity, initial yaw, flight behavior,
etc. These parameters can be as basic as how or when the propellant was produced to
what was the actual diameter of the projectile measured to 0.0001 in. Even though, individually, we believe that we understand the effect of each parameter, when all parameters
are brought together the problem becomes intractable. Because of this parameter overload
condition, the behavior is assumed to be stochastic.
The number of battlefield and test conditions that a gun–projectile–charge combination
can be subjected to is truly infinite. For safety and performance estimates, the U.S. Army is
often criticized for demanding test conditions, which could not possibly occur. While this
may be true, it is simply a means of over-testing a design to assure that the weapon system
is safe and reliable when the time comes to use it. This philosophy stems from the fact that
you cannot test every condition and also because soldiers are an ingenious bunch and will
invent new ways to employ a system beyond its design envelope.
Lack of understanding of the phenomena may seem rather strong wording even though
there are instances where this is literally true. In most cases, we know that parameters are
present that affect the design. We also know how they should affect the design. Some of
these parameters cannot be tested because there is some other, more fundamental variable
that affects the test setup to a far greater degree.
The overall effect of ballistic uncertainty, as described earlier, is that it will be very
unusual for you to see the words “always” or “never” when describing ballistic phenomena in this work.
3
© 2014 by Taylor & Francis Group, LLC
4
Ballistics: Theory and Design of Guns and Ammunition
1.1 Ballistic Disciplines
The field of ballistics can be broadly classified into three major disciplines: interior ballistics, exterior ballistics, and terminal ballistics. In some instances, a fourth category named
intermediate ballistics has been used.
Interior ballistics deals with the interaction of the gun, projectile, and propelling charge
before emergence of the projectile from the muzzle of the gun. This category would include
the ignition process of the propellant, the burning of propellant in the chamber, pressurization of the chamber, the first-motion event of the projectile, engraving of any rotating
band and obturation of the chamber, in-bore dynamics of the projectile, and tube dynamics during the firing cycle.
Intermediate ballistics is sometimes lumped together with interior ballistics, but has
come into its own category of late. Intermediate ballistics deals with the initial motion of
the projectile as it is exiting the muzzle of the tube. This generally includes initial tip-off,
tube and projectile jump, muzzle device effects (such as flash suppression and muzzle
brake venting), and sabot discard.
Exterior ballistics encompasses the period from when the projectile has left the muzzle
until impact with the target. One can see the overlap here with intermediate ballistics. In
general, all that the exterior ballistician is required to know is the muzzle velocity and
tip-off and spin rates from the interior ballistician, and the physical properties (shape and
mass distribution) from the projectile designer. In exterior ballistics, one generally is concerned with projectile dynamics and stability, the predicted flight path and time of flight,
and angle, velocity and location of impact. More often, now than in previous years, the
exterior ballistician (usually called an aero-ballistician) is also responsible for designing or
analyzing guidance algorithms carried onboard the projectiles.
Terminal ballistics covers all aspects of events that occur when the projectile reaches the
target. This means penetration mechanics, behind armor effects, fragment spray patterns
and associated lethality, blast overpressure, nonlethal effects, and effects on living tissue.
This last topic is becoming more and more important because of the great interest in lessthan-lethal armaments and, indeed, it has been categorized into its own discipline known
as wound ballistics.
1.2 Terminology
Throughout this work we will be using the word “gun” in its generic sense. A gun can be
loosely defined as a one-stroke internal combustion engine. In this case, the projectile is
the piston and the propellant is the air–fuel mixture. Guns themselves can be classified in
four broad categories: a “true” gun, a howitzer, a mortar, and a recoilless rifle.
A true gun is a direct-fire weapon that predominantly fires a projectile along a relatively
flat trajectory. Later on we will decide what is truly flat and what is not. Notice the word
“predominantly” crept in here. A gun, say on a battleship, can fire at a high trajectory
sometimes. It is just usually used in the direct-fire mode. A gun can be further classified as rifled or smooth bore, depending upon its primary ammunition. Guns exhibit a
relatively high muzzle velocity commensurate with their direct-fire mission. Examples of
guns include tank cannon, machine guns, and rifles.
© 2014 by Taylor & Francis Group, LLC
Introductory Concepts
5
A howitzer is an indirect-fire weapon that predominantly fires projectiles along a curved
trajectory in an attempt to obtain improved lethal effects at well-emplaced targets. Again,
howitzers can and have been used in a direct-fire role; it is simply not one at which they
normally excel.
A mortar is a tube that is usually man-portable used to fire at extremely high trajectories to provide direct and indirect support to the infantry. Mortars generally have much
shorter ranges than howitzers and cannot fire a flat trajectory at all.
A recoilless rifle is a gun designed with very little weight. They are usually mounted on
light vehicles or man emplaced. They are used where there is insufficient mass to counteract the recoil forces of a projectile firing. This is accomplished by venting the high-pressure
gas out of a rear nozzle in the breech of the weapon in such a way as to counter the normal
recoil force.
A large listing of terminology unique to the field of ballistics is included in the glossary
in Appendix A.
1.3 Units and Symbols
The equations included in the text may be used with any system of units. That being said,
one must be careful of the units chosen. The literature that encompasses the ballistic field
uses every possible system and is very confusing for the initiate engineer. The U.S. practice
of mixing the International System of Units (SI), United States Customary System (USCS),
and Centimeter–Gram–Seconds (CGS) units is extremely challenging for even the most
seasoned veteran of these calculations. Because of this an emphasis has been placed on the
units in the worked-out examples and cautions are placed liberally in the text.
Intensive and extensive properties (where applicable) are denoted by lowercase and
uppercase symbols, respectively. In some instances, it is required to use the intensive
properties on a molar basis. These will be denoted by an overscore tilde. In all cases, the
reader is advised to always be sure of the units.
© 2014 by Taylor & Francis Group, LLC
© 2014 by Taylor & Francis Group, LLC
2
Physical Foundation of Interior Ballistics
2.1 Ideal Gas Law
The fundamental means of exchanging the stored chemical energy of a propellant into
the kinetic energy of the projectile is through the generation of gas and the accompanying
pressure rise. We shall proceed in a disciplined approach, whereby we introduce concepts
at their simplest level and then add the complications associated with the real world.
Every material exists in some physical state of either solid, liquid, or gas. There are several variables that we can directly measure and some that we cannot but that are related
to one another through some functional relationship. This functional relationship varies
from substance to substance and is known as an equation of state.
Thermodynamically, the number of independent properties required to define the state
of a substance is given by the so-called state postulate, which is described in Ref. [1]. For all
of the substances examined in this text, we shall assume they behave in a simple manner.
This essentially means that the equilibrium state of all of our substances can be defined by
specification of two independent, intrinsic properties. In this sense, an intrinsic property
is a property that is characteristic of (in other words, governed by) molecular behavior.
The ideal gas law is essentially a combination of three relationships [2]. Charles’s law
states that volume of a gas is directly proportional to its temperature. Avogadro’s principle states that the volume of a gas is directly proportional to the number of moles of
gas present. Boyle’s law states that volume is inversely proportional to pressure. If we
combine these three relationships, we arrive at the famous ideal gas law, which states in
extensive form:
pvɶ = NℜT
(2.1)
where
p is the pressure of the gas
ṽ is the molar specific volume
N is the number of moles of the gas
ℜ is the universal gas constant
T is the absolute temperature
The units of Equation 2.1 are not always convenient to work with. For this reason, the form
of the ideal gas law that we shall use most often in this text is
pv = RT
(2.2)
7
© 2014 by Taylor & Francis Group, LLC
8
Ballistics: Theory and Design of Guns and Ammunition
In this case
p is the pressure of the gas
v is the specific volume (in mass units as we are used to)
R is the specific gas constant, unique to each gas
T is again the absolute temperature
The specific gas constant can be determined from the universal gas constant by dividing
the latter by the molar mass:
R=
ℜ
M
(2.3)
where M is the molar mass of the gas (e.g., 15.994 lbm/lb-mol for oxygen). There are many
other variants of the ideal gas law, which differ only in units. The other two versions that
we occasionally utilize are
pV = mg RT
(2.4)
p = ρ RT
(2.5)
and
In these equations
V (nonitalicized) is the volume the gas occupies
mg is the mass of the gas
ρ is the gas density
One should always check units when using these equations.
The pressure in a vessel filled with gas is caused by innumerable collisions of the gas
molecules on the walls of the vessel [2]. The more tightly packed the molecules are, the
more collisions occur—the higher the pressure is. Similarly, temperature excites the gas
molecules so that they move faster, collide more—thus also increasing pressure. It is these
collisions, among other things, that must be handled somehow by our equation of state.
The ideal gas law relies upon the fact that the gas molecules are very far apart relative
to one another [3]. If the molecules linger in the neighborhood of one another they will be
influenced by strong intermolecular forces, which can either attract or repel them from one
another. Thus, the ideal gas law ignores this effect. The ideal gas law further assumes intermolecular collisions occur completely elastically (i.e., like billiard balls). These assumptions must be kept in mind when using the ideal gas law. We shall soon see that under
the pressures and temperatures in a gun that these assumptions are invalid. Nevertheless,
they provide us with a point of departure and a useful stepping-stone for our studies.
To use the ideal gas law to determine the state of the gas in a gun, we need to invoke classic
thermodynamic relationships. The second law of thermodynamics can be stated as follows:
Q = ∆U + W + losses
(2.6)
where
Q is the energy added to the system
∆U is the change in internal energy
W is the work done on the system, and the losses term contains all of the energy that cannot
be recovered if, say, we pushed the projectile back to its starting position in the gun tube
© 2014 by Taylor & Francis Group, LLC
Physical Foundation of Interior Ballistics
9
Our sign convention shall be that a Q will be positive when energy is added to the system,
∆U will be positive if the internal energy of the system is increased, and W will be positive
if work is done on the system. Losses always remove energy from the system.
If we tailor Equation 2.6 to a gun launch situation, then Q would be the energy released
by burning our propellant, ∆U would be the change in internal energy of the propellant,
and W would be the work done on the projectile.
Let us further define the work term in the classical sense. It is typical of a first year engineering curriculum to define the work as follows:
∫
W = F ⋅ dx
(2.7)
In Equation 2.7, work is defined as a scalar that results from the vector dot product of force,
F with the distance over which the force acts, also a vector, dx (note that all vectors are
characterized by bold type in this book). If we restrict our analysis to a gun system, we
can see that, given pressure acting on the base of a projectile, it only has one direction to
travel due to the constraints of the gun tube. If we imagine that this gun tube is perfectly
straight (it never is) and we align a coordinate system with the axis of the tube, then the
displacement vector, dx, must be aligned with force vector, F (i.e., the angle between F and
dx is zero therefore the cosine of the angle between them is unity); and our relation for a
dot, or more formally, the scalar product of these two vectors gives us
F ⋅ dx = F ⋅ dx ⋅ cos(0) = Fdx
(2.8)
Our work definition for this case is then
∫
W = Fdx
(2.9)
This relationship for work has to be refined somewhat to fulfill our needs. We will need
to put the force acting on the projectile in terms of the pressure and sometimes would like
the volume to be included in the equation. If we look at the ideal gas equation of state in
the form of Equation 2.4, we do not see a force in there but we do see a pressure term and
a volume term.
We know from the mechanics of materials [4] that
F = pA
(2.10)
This has not been written in vector form so as to keep things simple (we will write it differently later). Equation 2.10 states that the resultant force, F, on a body is equal to the average
pressure, p, on that body times the area, A, over which the pressure acts. So we can rewrite
Equation 2.9 using this result as
W=
∫ pA dx
(2.11)
We now need to get volume in there somehow. We shall use the fact that, except for the
chamber of a gun (and a few notable exceptions with the bore), the area over which the
pressure acts is constant and equal to the bore cross-sectional area that we have defined
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
10
as A earlier. The area of the rifling grooves does contribute here if the tube is rifled, but
let us assume a nice smooth cylindrical bore for now. If A is the cross-sectional area and
dx is a differential element of length, then the differential element of volume, dV, can be
defined as
dV = Adx
(2.12)
We can now write Equation 2.11 in terms of pressure and volume as
W=
∫ p dV
(2.13)
You may recall this form of the definition of work from thermodynamics [5].
We now have two equations and a definition at our disposal as a pedagogical device that
can help illustrate the energy exchange mechanism in a gun. The equations are an ideal
gas equation of state Equation 2.4 and the second law of thermodynamics, Equation 2.6;
and the definition of how we defined work in Equation 2.13.
Let us imagine that we have a simple gun as depicted in Figure 2.1.
We shall assume that we have somehow placed a mass, mg, of a gas that behaves according to the ideal gas equation of state in the tube and compressed it, adiabatically, using the
projectile and no leakage has occurred. We shall further assume that there is no friction
between the projectile and the tube wall. Thus, in the situation depicted by Figure 2.1,
we have an ideal gas trapped between the projectile and the breech, compressed to some
pressure, p, at some absolute temperature, T. We shall further assume that the projectile of
mass, mp, is somehow held at position x = 0 and no gas or energy can escape. In this situation, the volume the gas occupies, which we shall call the chamber volume, Vc, is given by
Vc =
πd 2
l
4
(2.14)
What we have done essentially is compressed the projectile against an imaginary spring
(the gas), which now has a potential energy associated with it. From a thermodynamic
standpoint, we can reduce Equation 2.6 to
0 = ∆U + W
(2.15)
Recapping, we note that Q = 0 because there was no heat lost through the tube wall (adiabatic compression) and there is no propellant per se that will burn to generate heat. The
losses were zero because we have no friction.
mp
l
d
mg
x
FIGURE 2.1
Simple gun system.
© 2014 by Taylor & Francis Group, LLC
L
Physical Foundation of Interior Ballistics
11
Now that everything is set, we need to release our projectile and see what happens. If we
substitute Equation 2.4 into Equation 2.13, we can write
∫
W = mg RT
dV
V
(2.16)
This equation now shows how much work is being done on the projectile as a function of
the volume. It is noteworthy here that we are assuming the gas that is actually pushing on
the projectile is massless. By this we mean that no energy is being applied to accelerate the
mass of the gas. We will remove this assumption later in our studies. What we do not like
about Equation 2.16 is that temperature still appears as a variable.
By our earlier assumptions, we stated that the process was frictionless and adiabatic.
Recall, again from thermodynamics, that this actually defines an isentropic process [1]. For
a closed system (one with constant mass), it can be shown [6] that the absolute temperature,
T, of our system is related to the initial temperature of the gas, Ti, through
V
T = Ti c
V
(γ −1)
(2.17)
where
V is the volume at a given time t
Vc is the initial chamber volume
γ is the specific heat ratio of the gas (defined later)
If we substitute Equation 2.17 into Equation 2.16, we can write
V
(γ −1)
c
W = mg RTi V
∫V
−γ
dV
(2.18)
Vc
This equation is easy to work with because we know most of the terms on the RHS (righthand side) when we set up our pedagogical gun. We know the mass, mg, of the gas. We
know R and γ because we picked which gas it was. We know the initial temperature of the
gas and we know the chamber volume.
Now that we did all of this work with volumes, we want to convert these back to distances. A typical output desired by ballisticians is the pressure versus travel (i.e., distance)
curve. This plot helps the gun designer determine where to make his tube thick and where
he can get away with thinning the wall. If we again recognize that our gun has a constant
inner diameter, we can use Equation 2.14 to write Equation 2.18 as
L
W = mg RTil
(γ −1)
∫ (l + x )
−γ
dx
(2.19)
0
If we perform this integration, we obtain
W=
© 2014 by Taylor & Francis Group, LLC
mg RTil(γ −1)
(l + L)(1−γ ) − l(1−γ )
(1 − γ )
(2.20)
Ballistics: Theory and Design of Guns and Ammunition
12
We need to recall from dynamics that the kinetic energy of the projectile can be written as
K.E.projectile =
1
mpVm2
2
(2.21)
where Vm is the muzzle velocity. If we assume that all of the energy of the gas is converted
with no losses into kinetic energy of the projectile, then we can use Equation 2.15 to state that
(2.22)
K.E.projectile = W
We can make use of Equations 2.20 and 2.21 to write this as
mg RTil(γ −1)
1
[(l + L)(1−γ ) − l(1−γ ) ]
mpVm2 =
2
(1 − γ )
(2.23)
This is an important result as it relates muzzle velocity to the properties and amount of the
gas used, the mass of the projectile, and includes the effect of tube length. We can use this
equation to estimate muzzle velocity. So a convenient form of this equation is
Vm = 2
mg RTil(γ −1)
[(l + L)(1−γ ) − l(1−γ ) ]
mp (1 − γ )
(2.24)
In some instances, we would like to use these relationships to determine the state of the
gas or velocity of the projectile at some point in the tube other than the muzzle. If this is
the case, the procedure would be as follows:
1. Solve for the work term up to the position of interest, xproj, using
xproj
W (xproj ) = mg RTil
(γ −1)
∫ (l + x )
−γ
dx
(2.25)
0
2. Determine the volume at the position of interest using
V( xproj ) =
πd 2
(l + xproj )
4
(2.26)
3. Determine the gas temperature at this position from Equation 2.17
4. Determine pressure from the ideal gas Equation 2.4
This procedure is relatively straightforward.
If, as an example, we look at an idealized 155 mm compressed air gun and assume the
following parameters
Projectile weight = 100 lbm
Initial pressure = 45 MPa (approximately 6500 psi)
Tube length = 6 m
From Figures 2.2 through 2.4, we can depict the results of a calculation for temperature,
pressure, and velocity versus distance for this idealized situation.
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Physical Foundation of Interior Ballistics
13
350
T (K)
300
Temperature (K)
250
200
150
100
50
0
0
1
2
3
4
Distance (m)
5
6
7
FIGURE 2.2
Temperature versus distance in an ideal gas gun.
50,000,000
p (Pa)
45,000,000
40,000,000
Pressure (Pa)
35,000,000
30,000,000
25,000,000
20,000,000
15,000,000
10,000,000
5,000,000
0
0
1
2
3
4
Distance (m)
5
6
7
FIGURE 2.3
Pressure versus distance in an ideal gas gun.
Problem 1
Assume we have a quantity of 10 g of 11.1% nitrated nitrocellulose (C6H8N2O9) and it is
heated to a temperature of 1000 K assuming it changes from solid to gas somehow without changing chemical composition. If the process takes place in an expulsion cup with a
volume of 10 in.3, assuming ideal gas behavior, what will the final pressure be in pounds
per square inch?
lbf
Answer: p = 292 2
in.
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
14
250
Velocity (m/s)
200
150
100
50
V (m/s)
0
0
1
2
3
4
Distance (m)
5
6
7
FIGURE 2.4
Velocity versus distance in an ideal gas gun.
2.2 Other Gas Laws
There are many times when ideal gas behavior is insufficient to model real gases. This is
certainly true under the pressures and temperatures of gun launch. Although there are
many models that attempt to account for the deviation of real gases from ideal or perfect
behavior [2,3], we shall examine only two, the simplest of which we shall use.
Ideal gas behavior is approached when the distance between molecules (known as the
mean free path) is large. Thus, molecules do not collide or interact with one another very
often. Temperature is a measure of the internal energy of the gas. Thus, when the temperature is high, the molecules are moving around faster and have more of an opportunity
to interact with one another. Pressure is a result of how closely the molecules are packed
together, thus a higher pressure tends to put the molecules in close proximity. It is for these
reasons that we cannot normally use the ideal gas law in gun launch applications.
The Noble–Abel equation of state is given by
p( V − mg b) = mg RT
(2.27)
where
p is the pressure of the gas
V is the volume the gas occupies
mg is the mass of the gas
R is the specific gas constant
T is the absolute temperature
b is the co-volume of the gas
The co-volume of the gas has been described as a parameter that takes into account the
physical size of the molecules and any intermolecular forces created by their proximity
© 2014 by Taylor & Francis Group, LLC
Physical Foundation of Interior Ballistics
15
to one another. Think of it as not having physical meaning but as simply a number that
allows for a better fit to observed experimental data. The units of the co-volume are cubic
length per mass unit. Usually, the gas co-volume is provided in the literature but an estimation tool has been provided by Corner [7], which will not be repeated here since actual
data exist.
Occasionally, the Noble–Abel equation of state is insufficient to suit our needs. At these
times, it is typical to use a Van der Waals equation of state given by
p=
ɶ
RT
a′
−
vɶ − b′ vɶ2
(2.28)
In this case
p is again the pressure of the gas
ṽ is the molar specific volume
R̃ is the molar specific gas constant, unique to each gas
T is again the absolute temperature
a' and b' are constants particular to the gas
The Noble–Abel equation of state is the basis for nearly all of our work in this text, therefore Equation 2.27 is very important. At times, we may write it a little differently but you
will always be reminded of where it originated.
Problem 2
Perform the same calculation as in Problem 1, but use the Noble–Abel equation of state and
assume the co-volume to be 32.0 in.3/lbm
lbf
Answer: p = 314.2 2
in.
Problem 3
A hypothetical “air mortar” is to be made out of a tennis ball can using a tennis ball as
the projectile. The can has a 2–1/2” inside diameter and is 8” long. If a tennis ball of the
same diameter weighs 2 oz. and initially rests against the rear of the can, to what air
pressure must one pressurize the can to in order to achieve a 30 ft/s launch of the tennis
ball? Assume that the tennis ball can be held against this pressure until released, that it
perfectly obturates and also assume an isentropic process and ideal gas behavior with
γ = 1.4 for air.
2.3 Thermophysics and Thermochemistry
The main energy exchange process of conventional interior ballistics is through combustion. Once ignited, the chemical energy of the propellant is released through an oxidation
reaction. This energy release will be in the form of heat, which, in turn, increases the
pressure in the volume behind the projectile (i.e., in a combustion chamber). The pressure
exerts a force on the projectile, which accelerates it to the desired velocity.
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Ballistics: Theory and Design of Guns and Ammunition
In general, combustion requires three main ingredients to commence: a fuel, an oxygen source, and heat. In a common combustion reaction, such as an internal combustion
engine like the one in your car, oxygen is supplied to the reaction independently of the
fuel. The heat in this case is generated by a spark ignition and the burning of the air–fuel
combination that ensues.
A gun chamber has very little room for oxygen once it is stuffed with propellant. It
is important to note that, for other reasons, there is always free volume in the chamber
(called ullage)—we will explain this later. For now, we should understand that although
there is some oxygen in the chamber, the amount is insufficient to completely combust the
propellant. It is for this reason that propellants are formulated to contain both the fuel and
the oxidizer. In general, the propellant burning is an under-oxidized reaction. This has
some implications as the propellant gases leave the muzzle—again, we shall discuss this
in more detail later.
This brief introduction should make clear the reason to examine thermochemistry, thermophysics, and combustion phenomena. To proceed, we shall first define each field of
study. The definitions of Ref. [8] shall be used here to describe the first two topics as they
are extremely straightforward and clear. Thermophysics is defined as the quantification of
changes in a substance’s energy state caused by changes in the physical state of the material. An example of this would be the determination of the amount of energy required to
vaporize water in your teapot. Thermochemistry is then the quantification of changes in a
substance’s energy state caused by changes in the chemical composition of the material’s
molecules. An example of this would be the energy required to dissociate (break up) water
molecules into hydrogen and oxygen. Combustion is defined in Ref. [1] as the quantification of the energy associated with oxidizer–fuel reactions. Thus, combustion is a natural
outgrowth of thermophysics and thermochemistry.
Now that we have categorized these three fields of study, we shall attack them in a somewhat jumbled order. The reason for this is that, from our perspective, we really need not
distinguish between any of them and all of them appear in our gun launch physics. It is
also important to realize that whether the energy change comes from a chemical reaction
or a phase change from solid to gas, as long as we can calculate the extent of the energy
change, we can perform a valuable analysis.
Energy to all intents and purposes consists of two types: potential and kinetic. Potential
energy can be considered as stored energy. There are many ways to store energy. We can
store energy by compressing a steel bar or spring, by lifting a mass to a higher elevation
in the earth’s gravitational field, and by chemically preparing a compound that, whether
by combustion or chemical reaction, will release energy. Each of these forms of potential
energy: elastic strain; gravitational potential and chemical potential energy, has a different
method of storing and releasing the energy but they are all potential energies. There are
other forms of potential energy but we need not deal with them in this context.
Kinetic energy is the energy of a mass in motion. It can be observed in objects that are
in translational motion or in rotational motion. To extract some or all of this energy, it is
necessary to slow or stop the moving mass that has the kinetic energy. The energy in a
spinning flywheel is an example of rotational kinetic energy.
The field of thermodynamics is the study of energy transformations. It quantifies the
balance of energy between kinetic and potential. In thermodynamics, it is common to see
two energy transformation mechanisms: heat and work.
Heat transfer is essentially an exchange of energy through molecular motion. As we
shall soon see, molecules of a substance are always in motion. The faster they are in
motion, the hotter the substance is. These molecules can influence other molecules when
© 2014 by Taylor & Francis Group, LLC
Physical Foundation of Interior Ballistics
17
they are placed in contact with them, thus giving up some of their energy and increasing
the energy of the contacted substance. Temperature is a sensible measure of an object’s
internal energy.
Work is a means of increasing an object’s energy by application of a force through a distance. This method of energy transfer can create either potential energy, as in compressing
a spring, or kinetic energy as applied to a free, rigid mass. While the equations for heat
transfer can be the subject of entire texts (e.g., [9]), work can be defined through the vector
equation
W = F ⋅ dx
(2.29)
where
W is the work done on or by the system
F is the force vector
dx is the vector distance through which the force acts, known as the displacement vector
We must note that this is a vector equation. The work term is a scalar because the dot
product of two vectors results in a scalar. Because of the dot product term, the sign of W
is dependent upon the cosine of the angle between F and dx. Recall the definition of a dot
product as
A ⋅ B = AB cos θ
(2.30)
where A and B are the scalar magnitudes of the vectors A and B (Figure 2.5). If we use
Equation 2.30 with the variables of Equation 2.29, this tells us that if the angle between the
force vector and the displacement vector is between 0° and 90° or 270° and 0°, the work is
positive, i.e., it is work performed on the system. If, however the angle is between 90° and
270°, the work is negative, and therefore work performed by the system.
Internal energy, U, of a substance can be considered a form of potential energy. Some authors
[5] categorize the internal energy separately from potential and kinetic energies. This can
clearly be done in general, but for the application of gun launch it seems proper to group it as
a potential energy. The internal energy of a substance is manifested in the molecular motions
within that substance. These motions generally are translational or vibrational in nature. The
molecules of a substance are attracted to and repelled by one another and are in some degree
of translational motion. Additionally, the attractive or repulsive forces within a molecule itself
allow us to use an analogy of springs holding the atoms together. Imagine a structure of a
water molecule, for instance as depicted in Figure 2.6. If the oxygen and hydrogen atoms are
assumed to be steel balls and the molecular bond springs, we could pick this molecule up,
A
θ
B
FIGURE 2.5
Depiction of two vectors for scalar product definition.
© 2014 by Taylor & Francis Group, LLC
18
Ballistics: Theory and Design of Guns and Ammunition
Oxygen
atom
Hydrogen
atoms
FIGURE 2.6
Model of a water molecule.
hold the oxygen atom, and shake it. If the springs were really stiff in bending and much less
so in tension or compression, we would see the hydrogen atoms oscillating in and out at some
frequency. The greater the frequency, the more energy we would need to put into the system.
Even though the springs are stiff in bending, it does not mean that they cannot bend. This just
takes more energy. Like springs, we can store energy in the molecules this way.
This simple model of a molecule is a crude but useful approximation. Imagine now that
we put our model on a frictionless surface, like an ice hockey rink. If we hit the molecule
in a random way, we will excite these vibrational modes as well as create translational and
rotational motion. Now, if we fill the ice hockey rink with models … well, you get the idea.
As stated previously, the level of this interaction (collisions) must be represented somehow.
The metric used is internal energy with the level of activity defined as zero at the temperature known as absolute zero (0° on the Kelvin or Rankine scales).
The internal energy also includes the energy required to maintain a particular phase
of the material such as solid, liquid, or gas. Additionally, certain phases associated with
molecular structure such as face-centered cubic (FCC), body-centered cubic (BCC), etc., are
accounted for in the internal energy.
Quite often we shall see internal energy and what is commonly known as “pdV” work terms
together in our energy balance equations. The term is called pdV work because it is special
and separate from work generated by, say, a paddle wheel moving fluid around. This work
term arises from pressure pushing on a given volume. If the volume changes by an infinitesimal amount, dV, we essentially have force acting through a distance. To prove this to yourself,
look at the units. Because we see these terms together so often, it is convenient for us to group
them into one term, which we will call enthalpy, H. Mathematically, the enthalpy is defined as
H = U + pV
(2.31)
Notice here that we have removed the differential from the work term. The reason for this
is that, considering both enthalpy and internal energy, we are concerned with changes in
H and U. Therefore, the differential appears when we write the entire equation in differential form as
dH = dU + pdV
(2.32)
For proof of this result, refer to any thermodynamics text (e.g., [1,5]). An example of the
difference between internal energy and enthalpy is the rigid container or piston container.
Consider a rigid container that has some amount of gas in it. Assume the container is sealed
© 2014 by Taylor & Francis Group, LLC
Physical Foundation of Interior Ballistics
19
so that matter cannot enter or leave. Let us also assume that the container will allow energy
to be transferred to and from the gas. If we transfer heat (energy) to the gas, the temperature
will rise as will the pressure. Since the volume of the container is fixed, no work can be done;
thus all of the energy added to the gas is internal energy. From Equation 2.32, we see that
in this case the change in enthalpy would be exactly equal to the change of internal energy.
Now we assume that, instead of our container being rigid, the roof of the container is
a sealed yet moveable piston. In this case, once again matter cannot escape; however, the
volume is able to change. Now the only thing holding up the roof is the pressure of the gas
acting to just counteract the weight of the roof itself. Let us add the same amount of heat
that we added to the original, rigid, container. In this case, the temperature of the gas will
increase (but less than before) and the volume will increase because the piston is moveable
and the pressure must remain constant and just sufficient to counteract the weight of the
roof. In this instance, the enthalpy would be greater than the internal energy because it
includes the work done in lifting the piston.
When a substance changes form, chemically or physically, energy is either absorbed or
released. The method that we use to quantify this energy change is through heats of formation and the like. Though called a “heat,” what is really implied is an enthalpy change.
We shall proceed through these different enthalpy changes, attempting to list some of
the more common ones. For greater detail, the reader is encouraged to consult thermodynamics texts in addition to the descriptions provided in Ref. [8]. Specific values for text
problems will be given as needed. It is not the intent of the authors to tabulate the different
energy parameters of different materials.
When a substance is formed, atomic bonds in the constituent molecules are destroyed
and then recreated (at least this is a clean way to think of it from a bookkeeping perspective). The energy absorbed or generated by this process is commonly called the heat of
reaction, ∆H r0 . The Δ reminds us that we always are concerned with changes in enthalpy
from a particular reference state (usually standardized as 25°C and 1 atm). The “0” superscript is a convenient reminder that this is from a reference state of 1 atm. As the subscript,
sometimes we see “298” meaning 298 K. Though 298 K and 25°C are the same value, one
must always be wary of the reference state chosen by a particular author.
The heat of formation, ∆H f0, is the energy required to form a particular substance from
its individual component atoms. The heats of formation are the building blocks that determine the heat of reaction. Any elemental substance in its stable configuration at standard
conditions has a heat of formation equal to zero at that state. For instance, diatomic nitrogen, N2, has ∆H f0 = 0 at 25°C and 1 atm. We will provide an example of the heat of formation calculation in a later section.
Now that with the aforementioned quantities defined, we can write an equation for the
heat of reaction
∆H r0 =
∑
products
∆H f0 −
∑
∆H f0
(2.33)
reactants
Equation 2.33 states that the heat of reaction for a given substance is equal to the sum of the
heats of formation of the final products of the reaction that created the substance minus the
sum of the heats of formation of the materials that had to be reacted together to create the
new substance. This is further reinforcement of the definition of the heat of reaction. Recall
that we stated the atomic bonds of the molecules were destroyed and then remade. This is
essentially what Equation 2.33 is saying. The energy it took to create each of the reactants
has to be accounted for and then the energy it takes to create the new substances from the
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
20
constituents is calculated—energy is conserved. If the heat of reaction is a negative number, heat is liberated by the reaction otherwise it is absorbed.
When a compound is specifically combusted with sufficient oxygen to attain its most
oxidized state, the heat of reaction has a special name: the heat of combustion. The heat of
combustion is identified by the symbol ∆H c0. The heat of combustion is typically what is
obtained when propellant is burned in a closed bomb. The equation for the heat of combustion mirrors that of the heat of reaction, the only difference being as noted earlier:
∑
∆H c0 =
∆H f0 −
fully oxidized
products
∑
∆H f0
(2.34)
reactants
The heats of detonation and explosion have meanings that seem to be reversed. The heat of
detonation is the heat of reaction taken when detonation products are formed from an explosive compound during a detonation event. The formula for the heat of detonation is given by
∆Hd0 =
∑
∑
∆H f0 −
detonation
products
∆H f0
(2.35)
original
explosive
What is termed the heat of explosion is the amount of energy released when a propellant
or explosive is burned (not detonated) and is given by
0
∆H exp
=
∑
∆H f0 −
burning
products
∑
∆H f0
(2.36)
original
propellants
The heat of afterburn is another type of heat of reaction that occurs often in propellants
and explosives. Because the composition of propellants and explosives usually force an
under-oxidized reaction, the reaction products will often combine with the oxygen present
in the air outside the gun or explosive device, given sufficient temperature and pressure.
This secondary reaction results in a second pressure wave or blast and a fireball. The heat
of afterburn can be described mathematically as
0
∆H AB
=
∑
fully oxidized
products
∆H c0 −
∑
∆Hd0
(2.37)
remaining
detonation
products
Not all energy changes involve chemical reactions. We mentioned earlier that changes in
physical state and structure require energy. When a solid melts to form a liquid or a liquid
solidifies, we call the energy required, the latent heat of fusion, λf. These values are tabulated in any chemistry book or thermodynamics text. Some authors use different symbols
so one must, as always, be careful.
In a similar vein, the energy required to vaporize a liquid to a gas or condense a gas to a
liquid is known as the latent heat of vaporization and given by the symbol λfg.
If a material changes the structure of its atoms, say from BCC to FCC, the energy is
known as the heat of transition, λt.
There are many other types of material transitions that require energy. The types
described earlier cover the needs of this work.
© 2014 by Taylor & Francis Group, LLC
Physical Foundation of Interior Ballistics
21
2.4 Thermodynamics
The combustion process that occurs in a gun is a thermodynamic process. The term thermodynamics is a bit misleading because it implies that the dynamics of the combustion
process is examined. This is not quite true. Classical thermodynamics is based on the
examination of the various processes through equilibrium states. This is somewhat akin
to frames of a motion picture. We examine the state of the system before some event and
we usually examine it at some point, later in time, we are interested in.
Some of the concepts of thermodynamics were introduced in earlier sections, work and
energy being the major ones. Here we shall look in detail at two ways of describing thermodynamic systems to proceed with our study.
We shall define energy for an arbitrary system as
E=U+
1
mV 2 + mgz
2
(2.38)
Equation 2.38 is our extensive form of the definition of the system energy, E. In this equation, U is the internal energy, m is the system mass, V is the system velocity, g is a gravitational constant, and z is some height above a reference datum. The second and third
terms on the RHS of the equation are the kinetic and potential energies, respectively. If we
examine this equation, it is easy to see why some authors group the internal energy as a
separate energy type. However, in the case of a gun launch, the potential energy term is
insignificant. This focuses us on the transfer of energy between internal and kinetic.
We sometimes write Equation 2.38 in its intensive form as
1
e = u + V 2 + gz
2
(2.39)
Recall from our earlier discussions that an intensive property is the associated extensive
property divided by mass.
We shall now examine the first law of thermodynamics as it is applied to two different
types of systems: a fixed mass of material and a fixed volume of space through which
material flows. The first type of analysis, where the material is a fixed mass, is known
as a Lagrangian approach, while the fixed or control volume (CV) approach is known as
Eulerian. Both are important from a ballistic analysis standpoint and are prevalent in interior, exterior, and terminal ballistic studies.
For a fixed mass of material, undergoing some thermodynamic process, the first law of
thermodynamics can be written as
Q1− 2 + W1− 2 = ∆E1− 2
(2.40)
where
Q is the heat or energy added to the system
W is the work performed on or by the system
∆E is the change in the energy state of the material
The subscript 1–2 simply lets us know that the process began at some state 1 and ends at
some state 2. The signs on the terms are very important. We assume a positive change in
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
22
energy comes about through adding heat to the system and doing work on the system.
Thus, work performed on the system is positive and heat added is also positive. Different
thermodynamics texts write the first law slightly different, but if you understand that the
net result of work on the system or heat transfer to the system is to increase its energy, then
few mistakes will be made.
An interesting observation of Equation 2.40 is that the energy state change has an infinite number of paths that lead to the same result. For instance, if we wanted to add 24 kJ
of energy to some arbitrary system, we could do it by adding 12 kJ of heat and performing
12 kJ of work on the system. We could obtain the same result by adding 36 kJ of heat and
extracting 12 kJ of work from the system. The possibilities are limitless. This reinforces our
assertion that thermodynamics is really only concerned with end states.
Caution is warranted at this point. Equation 2.40 does not say how the energy, once
added to the system, is partitioned between potential (internal) energy or kinetic energy.
This reveals something. Heat and work are added to or removed from the system at the
system boundaries while the distribution of energy between internal or kinetic energy is
done within the system.
We shall now write out Equation 2.40 explicitly for a Lagrangian system
1
1
Q1− 2 + W1− 2 = m u2 + V22 − u1 + V12
2
2
(2.41)
Here we have neglected the gravitational potential energy terms and used the intensive
form of the energy, multiplied by the system mass. As previously stated, many times we
would like to use enthalpies instead of internal energies. If this is the case, we can rewrite
Equation 2.41 using our relationship between the two from Equation 2.40. We shall use the
intensive form of Equation 2.40 to yield
1
1
Q1− 2 + W1− 2 = m h2 − pv2 + V22 − h1 − pv1 + V12
2
2
(2.42)
Here we note that h is the specific enthalpy and v is the specific volume.
We shall now examine the first law of thermodynamics in the Eulerian frame of reference. Recall that in the Eulerian frame, we chose a CV (real or imaginary) and observed
how the energy within the volume changes based upon the energy carried into or out
of it by any entering or exiting substance as well as any heat or work done at the system
boundaries. It is convenient for us to write the first law in terms of the time rate of change
of energy, heat, and work. We start by writing Equation 2.40 as a rate equation
dQ dW dE
+
=
dt
dt
dt
(2.43)
Qɺ + Wɺ = Eɺ
(2.44)
or
Here the dots over the heat and work terms indicate the time rate of change of the variable.
Proper thermodynamics terminology would require us to use the “δ” instead of “d” in
© 2014 by Taylor & Francis Group, LLC
Physical Foundation of Interior Ballistics
23
Equation 2.43 because of path dependency considerations, but for our purposes we shall
ignore this fact. The reader is advised to consult any thermodynamics text for a better
understanding of the difference.
The substitutions that were performed to arrive at Equation 2.41 are not as straightforward in this case. Because we have material entering and leaving the CV, we can imagine
that this material can enter or leave with a different pressure and density as it interacts
with our fixed CV. Because of this, we must account for the energy used to make these
changes. Alternatively, one can envision the material coming in at a higher pressure or
density and wanting to push our imaginary CV outward, but since we fixed our CV it cannot. The energy from this must go somewhere so it works on the fluid in and around our
CV. Mathematically, this results in the energy term in Equation 2.44 having to include a
pv term. This is sometimes known as flow work [10]. With this in mind, Equation 2.44 can
be written as
ɺ out (eout + pout vout ) − m
ɺ in (ein + pin vin )
Qɺ + Wɺ = m
(2.45)
.
Here, by multiplying the intensive properties by the mass flow rate, m, we have the rate
of change of the energy terms. We have also arbitrarily assumed one inlet and one outlet.
If more inlets or outlets in our CV were present and they had different mass flow rates or
pressures, we would have to consider each with a term identical to our outlet or inlet terms
given earlier. We now can make the substitution for our energy terms to yield
1 2
1 2
ɺ
ɺ out uout + Vout
Qɺ + Wɺ = m
+ pout vout − m
Vout + pinvin
in uin +
2
2
(2.46)
In this case, we have also assumed a uniform velocity over the inlets and outlets. With one
inlet and outlet, the mass flow in must equal the mass flow out so we can write Equation
2.46 as
1 2
1 2
ɺ uout + Vout
+ pinvin
Qɺ + Wɺ = m
+ pout vout − uin + Vout
2
2
(2.47)
Substitution of enthalpy into the aforementioned equation puts it into a compact form:
1 2
1 2
ɺ hout + Vout
Qɺ + Wɺ = m
− hin + Vout
2
2
(2.48)
In many fluid dynamics texts, there are wonderful examples of how these equations
are used with multiple inlets and outlets [11]. You may be asking yourself how useful
are these equations if we only use one inlet or outlet? The answer is that they are very
useful. Except for flows through muzzle devices or through internal ports like bore
evacuators and ports for automatic weapons, a gun is a right circular tube that contains
the propellant gas. Any flow field analysis we perform on the moving gases will have
just one inlet (toward the breech) and one outlet (toward the projectile). Thus, as we
develop our equations later for in-bore motion, we can use these simple equations in the
aforementioned form.
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
24
As a review, we have two equations that state the first law of thermodynamics. For a
fixed mass of material (Lagrangian frame), we have
1
1
Q1− 2 + W1− 2 = m h2 − pv2 + V22 − h1 − pv1 + V12
2
2
(2.42)
and for a fixed volume that material can flow in and out of (Eulerian frame)
1 2
1 2
ɺ hout + Vout
Qɺ + Wɺ = m
− hin + Vout
2
2
(2.48)
These equations have been repeated here because of their critical importance to our work.
In many instances, we will find that we require a relationship between internal energy
or enthalpy and temperature. If we have a gas that is not reacting and intermolecular forces
are small enough to ignore, we can consider the gas to be thermally perfect [12]. The implications of this are that internal energy and enthalpy are functions of the temperature alone.
With this model, we can write expressions for internal energy and enthalpy as follows:
du = cvdT
(2.49)
dh = cpdT
(2.50)
where
cv is the specific heat at constant volume
cp is the specific heat at constant pressure
Normally, cp and cv vary with temperature. In many practical cases, this variation is small
and we can further assume that the gas is calorically perfect, which results in the aforementioned equations being written as
u = c vT
(2.51)
h = cpT
(2.52)
For a thermally or calorically perfect gas (not a reacting gas), there is a relationship between
cp, cv, and R. If we define γ as the ratio of specific heats where
γ =
cp
cv
(2.53)
then we can write the aforementioned relationships as
© 2014 by Taylor & Francis Group, LLC
cp − c v = R
(2.54)
cp =
γR
γ −1
(2.55)
cv =
R
γ −1
(2.56)
Physical Foundation of Interior Ballistics
25
The second law of thermodynamics defines the concept of entropy for us [1]. We know
from the second law of thermodynamics that
Tds = du + pdv
(2.57)
or, if we insert the definition of enthalpy
Tds = dh − vdp
(2.58)
If we evaluate Equations 2.57 and 2.58 under the assumptions of a calorically perfect gas,
we obtain
p
T
s2 − s1 = cp ln 2 − R ln 2
T1
p1
(2.59)
T
v
s2 − s1 = cv ln 2 + R ln 2
T
1
v1
(2.60)
In these expressions, the subscripts “1” and “2” indicate the initial and final states of the
substance, respectively. An isentropic process is a process in which there is no entropy
change. This is also known as a reversible process. In a real system, entropy must always
increase or, at best, stay constant. Many processes have slight enough entropy increases
as to be considered isentropic. Isentropic processes also are excellent to examine as theoretical limits on real processes. If we examine Equations 2.59 and 2.60 under an isentropic
assumption, we see that the left-hand side (LHS) is zero in both. This has implications that
allow us to write (for an isentropic process)
γ
p2 ρ 2 v2
= =
p1 ρ1 v1
−γ
γ
T γ −1
= 2
T1
(2.61)
Problem 4
The M898 SADARM projectile weighs 102.5 lb. The projectile was fired from a 56 caliber, 155 mm weapon and a pressure–time trace was obtained. The area under the pressure–time curve was (after converting the time to distance) calculated to be 231,482 psi-m.
Calculate the muzzle energy of the projectile in megajoules. Assume the bore area to be
29.83 in.2
Answer: E = 30.7 [MJ].
Problem 5
An 8 in. Mk. 14 Mod. 2 Navy cannon is used at NSWC Dahlgren, VA for “canister” firings.
These firings are used to gun harden electronics that are carried in an 8 in. projectile. The
projectile used weighs 260 lb. The measured muzzle velocity is around 2800 ft/s. Calculate
the muzzle energy of the projectile in megajoules. Assume the bore area to be 51.53 in.2 The
rifled length of the tube (distance of projectile travel) is 373.65 in.
Answer: E ≈ 43 [MJ].
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
26
2.5 Combustion
As stated in the previous two sections, combustion is the process through which the
energy of the solid propellant is converted to useful work. The purpose of this section
is to quantify the oxidation reaction. The tactic we shall employ is to examine the more
common, everyday combustion processes that combine (relatively) simple fuels with air to
produce work. In this way, we shall, hopefully, bring to mind the combustion thermodynamics that has been taught at an undergraduate level and perhaps has been forgotten or
not exercised since it was first learned.
If we utilize the concept of a fixed CV, we can imagine a combustion chamber as depicted
in Figure 2.7. In this CV, we can envision a mass of fuel entering as well as some mass of
air. The two are then combusted with one another and the gaseous products leave as a
mixture. We can write the first law of thermodynamics for this system then as in Equation
2.46, which we shall repeat here with subscripts that reflect Figure 2.7:
1 2
1 2
1 2
ɺ
ɺ products hproducts + Vproducts
ɺ fuel hfuel + Vfuel
Qɺ + Wɺ = m
− mair hair + Vair − m
2
2
2
(2.62)
In Equation 2.62, we can see how the heat and energy generated are affected by the amount
of mass flow, the enthalpies, and the velocities of the fuel, the oxidizer (air in this case), and
the product gases. We require some means of determining the energy converted through
the chemical reaction. We achieve this through the balancing of the chemical reaction. We
shall return to Equation 2.62 once we have discussed chemical reactions.
One of the most important compounds in the study of combustion is air. We shall adopt
a convention that is standard in many thermodynamics texts [5,13,14] that models air as
21% diatomic oxygen (O2) and 79% diatomic nitrogen (N2). This means that every mole of
oxygen carries with it 3.76 moles of nitrogen. This relationship comes about because
moles N 2
0.79
mole air = 3.76 moles N 2
mole O
moles O 2
2
0.21
mole air
(2.63)
As can be seen in Appendix B.1, the molecular weight for our simple model of air is
28.97 kg/kg-mol.
•
mair
•
•
mfuel
FIGURE 2.7
Fixed control volume (CV) combustion chamber.
© 2014 by Taylor & Francis Group, LLC
mproducts
Physical Foundation of Interior Ballistics
27
The balancing of a chemical reaction determines what the molecular composition of the
combustion products will be and furthermore helps us to quantify the amount of energy
absorbed or released. If energy is absorbed in a chemical reaction, in other words, if we
had to add energy to force the reaction to completion, the reaction is said to be endothermic. If heat is liberated, the reaction is said to be exothermic [15].
A reaction can be said to be theoretically or stoichiometrically balanced if the reaction
goes to completion and there is no excess oxygen in the products [1]. We shall define a
complete reaction as one in which all of the oxygen combines first with all of the hydrogen to form steam and then with all the carbon to form carbon dioxide. Oxygen has a
greater affinity for combining with hydrogen than with carbon [1]. The only time that
carbon monoxide (CO) will be formed is if there is insufficient oxygen. We must keep in
mind that in any real reaction there will usually be some amounts of carbon monoxide
and other compounds such as nitric oxide (NO) in the combustion products. We shall
return to this issue later. For the time being, we shall assume that the only reaction
products in the stoichiometric reaction are CO2 and H2O. The balancing of these chemical reactions is an important part of our study of the combustion process, which we shall
now examine.
We shall use two convenient forms of chemical equations: a molar-based equation
and a mass-based equation. In the molar-based equation, we shall usually combust one
mole of fuel with some amount of air. The result may be multiplied by the number of
moles of fuel actually burned to obtain a final answer. When the mass-based equation
is employed, we generally use one mass unit of fuel (lbm or kg) and some amount of air,
again multiplying the solution by whatever the actual mass of fuel happens to be. The
techniques just described are applicable to a system where the mass is fixed. The same
equations can be used with mass or molar flow rates if the system happens to be a steady
flow or open system.
It is informative to balance the chemical reactions in the context of everyday systems
that combust a fuel with air. Usually, this fuel is a hydrocarbon composition. The stoichiometric amount of air required would be enough so that all of the carbon combusts with
sufficient oxygen to form CO2 and all of the hydrogen combusts to form water or steam.
If we had a hydrocarbon fuel of chemical composition CxHy, we would like to find the
number of moles, a, of air required to completely combust the fuel and we would write the
balanced chemical reaction as
C x H y + a(O 2 + 3.76N 2 ) → xCO 2 +
y
H 2O + 3.76 aN 2
2
(2.64)
We could solve for a to yield
a= x+
y
4
(2.65)
As an example, let us say we have one mole of benzene (C6H6) that we would like to burn
in air. The balanced, stoichiometric equation would be found by first determining a from
Equation 2.65
a = 6+
© 2014 by Taylor & Francis Group, LLC
6
= 7.5
4
(2.66)
Ballistics: Theory and Design of Guns and Ammunition
28
Now the balanced equation is found using Equation 2.64
C6H6 + 7.5(O 2 + 3.76N 2 ) → 6CO 2 + 3H 2O + 28.2N 2
(2.67)
This is an example of a stoichiometrically balanced equation using a molar basis. There are
times when a particular fuel is burned with too much air (over oxidized) or too little air
(under oxidized). The latter is usually the case with propellants in the chamber of a gun.
When a fuel is over oxidized, we usually categorize it by stating how much excess air is
included in the reaction. For instance, 50% excess air used in the reaction of Equation 2.67
would alter the balanced equation to be written as
C6H6 + (1.5)(7.5)(O 2 + 3.76N 2 ) → 6CO 2 + 3H 2O + 3.75O 2 + 42.3N 2
(2.68)
If the fuel were burned with 50% deficient air, we would have
C6H6 + (0.5)(7.5)(O 2 + 3.76N 2 ) → 4.5CO + 3H 2O + 1.5C + 14.1N 2
(2.69)
In this case, we have used the rules set forth earlier where steam is formed first then carbon monoxide. At this point, all of the oxygen has been used up so solid carbon is formed.
From this simple example, you can see that the amount of air used in the combustion is
critical to determination of the products.
We can now define an air–fuel ratio as the ratio mass of air combusted to the mass of fuel
combusted. This is given mathematically by
A−F =
ɺ
mair
m
= air
ɺ fuel
mfuel m
(2.70)
If we continue using our three examples, we could find the mass fuel ratio for each of the
reactions defined in Equations 2.67 through 2.69. If we note here that the molar mass of
Benzene is 78.11 lbm/lb-mol and the molar mass of air is 28.97 lbm/lb-mol, we have for the
stoichiometric reaction
A − FStoich
lbm
(7.5) [mol air ]( 4.76)(28.97 )
lb-mol = 13.24 lbm air = 13.24
=
lbm
lbm C6 H6
(1) [mol C6 H6 ](78.11)
lb-mol
(2.71)
For the reaction with 50% excess air, we have
A − F50%excess
lbm
(1.5)(7.5) [mol air ]( 4.76)(28.97 )
lb-mol = 19.85 lbm air = 19.85
=
lbm
lbm C6 H6
(1) [mol C6 H6 ](78.11)
lb-mol
© 2014 by Taylor & Francis Group, LLC
(2.72)
Physical Foundation of Interior Ballistics
29
For the reaction with 50% deficient air, we have
A − F50%deficient
lbm
(0.5)(7.5) [ mol air ] ( 4.76)(28.97 )
lb-mol = 6.61 lbm air
=
lbm
lbm C6 H6
(1) [ mol C6 H6 ] (78.11)
lb-mol
= 6.61
(2.73)
Now that we have introduced the process of chemical equation balancing and some of the
mathematics required, we must quantify the energy released (or absorbed) by the chemical reaction. We have already introduced the concept of enthalpy as well as defined the
enthalpy of formation. We shall pause here to examine how a heat of formation is obtained.
We shall consider carbon dioxide for our example. If we have a combustion chamber
in which we react pure oxygen with solid carbon, we can put the two substances into the
container at 25°C and start the reaction somehow. The balanced equation on a molar basis
would be
C(s) + O 2 → CO 2
(2.74)
The first law of thermodynamics states that
Q + W = N products hproducts − N reactants hreactants
(2.75)
Here we have used specific values so that everything is on a molar basis. Since the container
is rigid, there is no work performed on or by the system, thus Equation 2.75 reduces to
Q = N products hproducts − N reactants hreactants
(2.76)
If we were to perform this experiment, we would find that the container would get hot.
Theoretically, we could extract this heat from the container until the temperature returned to
25°C; if we were to do this, we would find that 393,546 kJ/kg-mol of energy would have been
produced. Examination of Appendix B.1 reveals that this is exactly the value of the heat of formation of carbon dioxide recalling that a negative value denotes heat given off by the reaction.
The enthalpy of a substance allows us to quantify the energy state of a material. The
enthalpy of formation was defined as the energy required to form a particular composition from its basic elements resulting in the compound as a product at some reference temperature and pressure (we shall use 25°C or 298 K and 1 atm as this reference condition).
If we were to take this compound and arbitrarily increase its temperature or pressure by
some amount and if there were no phase change or change in composition, we will have
increased its enthalpy. If we restrict our analysis to an ideal gas, it can be shown [1] that
the enthalpy is a function of temperature only. With this, we can write for a composition
hT = hf0 + ∆h298 →T
where
hT is the enthalpy of the material at temperature T,
hf0 is the enthalpy of formation
∆h298→T is the change in enthalpy from the reference state to the temperature, T
© 2014 by Taylor & Francis Group, LLC
(2.77)
Ballistics: Theory and Design of Guns and Ammunition
30
We define ∆h298→T as
0
∆h298 →T = h (T ) − ( h298
)
(2.78)
Tables of enthalpies are located in Appendix B at the end of the book. As an example, consider carbon monoxide at 2000 K. The enthalpy of this compound using Appendices B.1
and B.2 would be
kJ
kJ
kJ
hCO2000 K = −110, 541
+ 56, 737
= −53, 804
kg-mol
kg-mol
kg-mol
(2.79)
Now that we have worked with enthalpies a bit, we can begin to apply what we have
learned. We shall look at an example of these principles applied first to a closed bomb
where there is no work performed and then to a gun where there is.
For a closed bomb, we shall tailor Equation 2.42 to our needs. If we consider a closed
vessel, we realize that there is no velocity into or out of the CV, and there is no work performed on or by the system. This allows us to write Equation 2.42 as
Q1− 2 = m[( h2 − pv2 ) − ( h1 − pv1 )] = m(u2 − u1 )
(2.80)
If we write this equation on a molar basis as limit to ideal gas behavior, we can state that
Q=
∑ N (h
i
i
prod
− RuTprod ) −
∑ N (h
i
reac
− RuTreac )
(2.81)
i
This relationship is important because it tells us that the heat given off by the closed bomb
is affected by the enthalpy change of the chemical reaction and the temperature of the
products.
We shall examine a pressure vessel containing 0.001 kg of methane (CH4) and 0.002 kg of
air. The enthalpy of formation for methane is −74,850 kJ/kg-mol and its molecular weight
is 16.04 kg/kg-mol. The reaction will begin at 298 K and we shall remove enough heat from
the vessel that the final temperature becomes 1500 K. We would like to determine how
much heat is given off.
We need to balance the chemical reaction on a molar basis, so we shall determine how
many moles of methane and air we have in the container. For methane, we have
N CH4 =
(0.001) kg CH4
= 6.23 × 10 −5 kg-mol CH4
kg
(16.04)
kg-mol
(2.82)
For the air, we have
N air =
(0.002) kg air
= 6.90 × 10 −5 kg-mol air
kg
(28.97 )
kg-mol
© 2014 by Taylor & Francis Group, LLC
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Physical Foundation of Interior Ballistics
31
Our balanced reaction is then
(6.23 × 10 −5 )CH 4 + (6.90 × 10 −5 ) ( (0.21)O 2 + (0.79)N 2 )
→ (2.9 × 10 −5 )H 2O + (9.56 × 10 −5 )H 2 + (6.23 × 10 −5 )C(s) + (5.45 × 10 −5 )N 2
(2.84)
We shall examine the reactants first. For the methane, we have
(
N CH4 hf0 + ∆h298 →T − RuTCH4
)
kJ
kJ
+ 0 − (8.314)
= (6.23 × 10 −5 ) [kg-mol] −74, 850
(298) [K]
kg-mol
kg-mol ⋅ K
(
)
N CH4 hf0 + ∆h298 →T − RuTCH4 = −4.82 [kJ]
For the oxygen and nitrogen, we have
kJ
(298) [K]
N O2 hf0 + ∆h298 →T − RuTO2 = (1.45 × 10 −5 ) [kg-mol] 0 + 0 − (8.314)
kg-mol ⋅ K
(
)
(
)
N O2 hf0 + ∆h298 →T − RuTO2 = −0.036 [kJ]
kJ
N N2 hf0 + ∆h298→T − RuTN2 = (0.79)(6.90 × 10−5 )[kg-mol] 0 + 0 − (8.314)
(298) [K]
⋅
kg-mol
K
(
)
(
)
N N2 hf0 + ∆h298→T − RuTN2 = −0.135 [kJ]
The enthalpies of the reactants are therefore
∑ N (h
i
reac
)
− RuTreac = −4.82 [kJ] − 0.036 [kJ] − 0.135 [kJ] = −4.99 [kJ]
i
For the products we have (using the tables in the appendix)
(
N H2 O hf0 + ∆h298 →T − RuTH2 O
)
kJ
= (2.9 × 10 −5 ) [kg-mol] −241, 845 + 48, 181 − (8.314)
(1, 500) [K]
kg-mol ⋅ K
(
)
N H2 O hf0 + ∆h298 →T − RuTH2 O = −5.98 [kJ]
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
32
(
N H2 hf0 + ∆h298 →T − RuTH2
)
kJ
(1, 500) [K]
= (9.56 × 10 −5 ) [kg-mol] 0 + 36, 307 − (8.314)
kg-mol ⋅ K
(
)
N H2 hf0 + ∆h298 →T − RuTH2 = +2.28 [kJ]
(
N C hf0 + ∆h298 →T − RuTC
)
kJ
kJ
− (8.314)
= (6.23 × 10 −5 ) [kg-mol] 0 + 23, 253
(1, 500) [K]
kg-mol ⋅ K
kg-mol
(
)
N C hf0 + ∆h298 →T − RuTC = 0.67 [kJ]
(
N N2 hf0 + ∆h298 →T − RuTN2
)
kJ
kJ
− (8.314)
= (5.45 × 10 −5 ) [kg-mol] 0 + 38, 404
(1, 500) [K]
kg-mol
kg-mol ⋅ K
(
)
N N2 hf0 + ∆h298 →T − RuTN2 = 1.41 [kJ]
The enthalpies of the products are then given by
∑ N (h
i
prod
− RuTprod ) = −5.98 [kJ] + 2.28 [kJ] + 0.67 [kJ] + 1.41 [kJ] = −1.62 [kJ]
i
The heat given off by the reaction is then calculated through Equation 2.81 as
Q = (−1.62) [kJ] − (−4.99) [kJ] = +3.37 [kJ]
(2.85)
This illustrates the process of calculating the amount of energy given off by a closed bomb
reaction as well as the effect of temperature on the reaction products. In this case, energy
has to be added to get the reaction to go to completion. It must be noted that had we
decided to lower the temperature of the products, energy would eventually have to have
been removed. This will be examined as a problem at the end of the chapter.
If we apply the same principles to a gun launch, we can determine the amount of energy
imparted to the projectile and in so doing, obtain a feeling for the process of energy conversion between propellant chemical energy and projectile kinetic energy.
Unlike the fixed boundary examined in the closed-bomb problem, a gun launch involves
a boundary that is moving (the base of the projectile). This problem is similar to a piston of
an internal combustion engine that undergoes one stroke. We have defined work earlier as
a form of energy and if we assume all of the energy of the propellant goes into heating of
© 2014 by Taylor & Francis Group, LLC
Physical Foundation of Interior Ballistics
33
the gaseous products, kinetic energy of the projectile, and a loss term (including friction,
swelling of the gun tube, etc.), we can write the first law of thermodynamics as given in
Equation 2.75. Rewriting this by assuming the velocity of the seated projectile is zero, we
obtain our thermodynamic equation for a gun launch as
Q+
1
mV 2 =
2
∑ N (h
i
i
prod
)−
∑ N (h
i
reac
) + losses
(2.86)
i
We have neglected potential energy changes here because they are usually quite small
relative to the other terms. We shall examine an example in the form of a potato gun to
illustrate the use of Equation 2.86 and the other methods of this chapter.
A potato gun is a device that people use to project potatoes at targets. These devices can
be very dangerous to the operator as well as the target. We would like to calculate the muzzle velocity of a half-pound potato projectile used in a particular gun. This gun is made of
2 in. diameter PVC pipe (a very good insulator). The projectile rests on a stop when loaded
through the muzzle so that there is a 6 in. long chamber. The device in question was injected
with 0.005 oz (mass) of lighter fluid as a gas (n-butane—C4H10 (g) hf0 = −124, 733 kJ/kg-mol ,
n = 58.123 kg/kg-mol) to fire the potato. We shall assume the potato obturates perfectly and
that there is no bore friction. The travel of the potato in the gun tube is 24 in. The weapon
is fired under standard conditions of 77°F and 14.7 psi. Assume the reactants and the products both exist at these conditions. We would like to determine the velocity of the potato at
the completion of combustion in feet per second assuming no losses.
The chamber was 6 in. long and 2 in. in diameter, so our chamber volume is
Vi = Al = π
(2)2
[in.2 ](6) [in.] = 18.85 [in.3 ]
4
(2.87)
The air weighs 28.97 lbm/lb-mol and if we assume ideal gas behavior, the density of air is
calculated from
pv = RT → ρ =
p
RT
(2.88)
lbf
lbm
(14.7 ) 2 (28.97 )
lbm
in.
lb-mol
ρ=
= 0.0000428 3
ft-lbf
in.
in.
(12) (537 ) [R ]
(1545)
lb-mol − R
ft
So the amount of air we actually have is
lbm
mair = ρ Vi = (0.0000428) 3 (18.85) [in.3 ] = 0.0008068 [lbm]
in.
The amount of fuel was given in ounces
lbm
mfuel = (0.005) [oz](0.0625)
= 0.0003125 [lbm]
oz
© 2014 by Taylor & Francis Group, LLC
(2.89)
Ballistics: Theory and Design of Guns and Ammunition
34
For the actual combustion, we need to use our mass information and convert it to molar
values, recognizing that the molar mass is the same whether it is kg/kg-mol or lbm/lb-mol.
For the fuel and air, we have
N fuel =
mfuel
= (0.0003125) [lbm]
nfuel
1
= 0.0000054 [lb-mol]
lbm
(58.123)
lb-mol
N air =
mair
= (0.0008068) [lbm]
nair
1
= 0.00000278 [lb-mol]
lbm
(28.97 )
lb-mol
(2.90)
(2.91)
For each lb-mol of air, we know that 1/4.76 lb-mol of it is oxygen so we have
N O2 =
1
(0.0000278) [lb-mol] = 0.0000058 [lb-mol]
4.76
N N2 =
3.76
(0.0000278) [lb-mol] = 0.0000220 [lb-mol]
4.76
Now we can write our combustion equation as
(0.0000054)C 4H10 ( g ) + (0.0000058)O 2 + (0.0000220)N 2 →
(0.0000116)H 2O + (0.0000216)C + (0.0000154)H 2 + (0.0000220)N 2
To determine the muzzle velocity, we start with our first law of thermodynamics equation, simplified by the fact that there is no heat transfer and no shaft work. Then the energy
of the fuel–air mixture equals the work done on the projectile plus the energy of the products of combustion.
H R = H p + Wp
(2.92)
Let us look at the internal energies for each of the reactants
Reactant
Enthalpy of Formation (kJ/kg-mol)
Enthalpy of Formation (in.-lbf/lb-mol)
C4H10 (g)
O2
N2
−124,733
0
0
−500,728,155
0
0
The conversion used here is as follows:
BTU
kJ
ft-lbf
in.
in.-lbf
lb-mol
( x)
(12) → 4014.4 x
(0.4299) kJ (778..16)
BTU
ft
lb-mol
kg-mol
kg-mol
© 2014 by Taylor & Francis Group, LLC
(2.93)
Physical Foundation of Interior Ballistics
35
For the products, we have
Product
Enthalpy of Formation (kJ/kg-mol)
Enthalpy of Formation (in.-lbf/lb-mol)
H2O (g)
N2
C2
H2
−241,845
0
0
0
−970,862,568
0
0
0
We will rearrange our first law equation as follows:
Wp = H R − H p
We calculate HR first
(
)
(
)
(
H R = N C4 H10 hf0 + ∆h298 →T + N O2 hf0 + ∆h298 →T + N N2 hf0 + ∆h298 →T
)
Plugging in the numbers we have, we get
in.-lbf
H R = (0.0000054) [lb-mol](−500, 728 , 155 + 0)
lb-mol
in.-lbf
+ (0.0000058) [lb-mol](0 + 0)
lb-mol
in.-lbf
+ (0.0000220) [lb-mol](0 + 0)
lb-mol
H R = −2704 [in.-lbf ]
We calculate Hp in a similar manner
(
)
(
)
(
)
(
H p = N H2 O hf0 + ∆h298 →T + N H2 hf0 + ∆h298 →T + N N2 hf0 + ∆h298 →T + N C hf0 + ∆h298 →T
in.-lbf
H p = (0.0000116) [lb-mol](−970, 862, 568 + 0)
lb-mol
in.-lbf
+ (0.0000154) [lb-mol](0 + 0)
lb-mol
in.-lbf
+ (0.0000220) [lb-mol](0 + 0)
lb-mol
in.-lbf
+ (0.0000216) [lb-mol](0 + 0)
lb-mol
H p = −11, 262 [in.-lbf ]
© 2014 by Taylor & Francis Group, LLC
)
Ballistics: Theory and Design of Guns and Ammunition
36
Then the work done on the projectile is
Wp = −2, 704 [in.-lbf ] − (−11, 262) [in.-lbf ] = 8, 558 [in.-lbf ]
Since this work equals the muzzle energy of the projectile
Wp =
1
mV 2 = 8, 558 [in.-lbf ]
2
Therefore,
V=
2Wp
=
m
lbm-ft
(2)(8558) [in.-lbf ](32.2)
2
lbf-s = 303 ft
s
in.
(0.5) [lbm](12)
ft
Wow! That is pretty fast but we used a lot of butane, assumed the products return to
ambient conditions quickly, and neglected things. Also note that the length of the tube
did not come into play. We would definitely have to account for this as we shall later see.
One important parameter in determining the amount of energy transferred to the projectile
is the temperature of the product gases. As you can see from our example, an increase in the
temperature of the product gases will result in a decrease in the projectile velocity because
Hp goes up. Typically, we can assume the product gases exit at a temperature between 0.6T0
and 0.7T0, where T0 is the adiabatic flame temperature of the product gases [7]. The adiabatic
flame temperature of a gas is the temperature that is achieved if the gases burn to completion
in the absence of any heat transfer or work being performed [1]. The calculation of the adiabatic flame temperature is relatively straightforward but requires iteration. This is beyond
the scope of this text but the reader is referred to the references at the end of this chapter for a
complete description of the procedure. In addition, there are several commercially available
codes (including some that come with the purchase of textbooks now, for instance [13]). To
achieve our objectives, the temperature of the reaction products will always be given.
Problem 6
Calculate the A–F ratio for the combustion of the following fuels. Calculate the ratio with
both theoretical air and 10% excess air.
1. Benzene (C6H6)
Answer: 13.24 and 14.56
2. n-Butane (C4H10)
Answer: 15.42 and 16.96
3. Ethyl alcohol (C2H5OH)
Answer: 8.98 and 9.88
Problem 7
Let us examine a pressure vessel identical to the example problem in the text containing
0.001 kg of methane (CH4) and 0.002 kg of air. The enthalpy of formation for methane is
−74,850 kJ/kg-mol and its molecular weight is 16.04 kg/kg-mol. The reaction will begin at 298 K
and we shall remove enough heat from the vessel that the final temperature becomes 1000 K.
© 2014 by Taylor & Francis Group, LLC
Physical Foundation of Interior Ballistics
37
1. Determine the maximum heat given off.
Answer: Q = +0.606 [kJ].
2. Compare the result in (1) with the example problem in this chapter.
Answer: This situation removes 2.764 kJ more energy than the example.
Problem 8
A really interesting person takes the tennis ball mortar we built in problem 3 and modifies it—squirting in and igniting 0.003 oz. of acetylene gas (C2H2(g)). If we assume the
combustion kinetics are fast enough such that the energy release occurs before the ball
can move, we want to determine the muzzle velocity of the tennis ball. Proceed along the
following steps:
a. Balance the stoichiometric reaction equation for acetylene.
b. Balance the actual equation neglecting the volume the acetylene occupies in the
chamber. Assume the air initially in the chamber is at 14.7 psia and 77°F.
c. Determine the increase in internal energy of the gas.
d. Assuming the gas is calorically perfect (∆U = mgcv∆T) and that cv = 0.33 BTU/lbm-R
for the mixture, determine the increase in temperature of the gas.
you will have to do (c) and (d) by iteration, first assuming a final reaction temperature, carrying out the calculation for ∆U and seeing if the ∆T you get matches—if not
iterate again – once you get an answer within say 10% that is good enough.
No t e :
e. Based on the result of (d), determine the initial pressure on the tennis ball assuming the specific gas constant of the products is R = 80 ft-lbf/lbm-R.
f. Use the result of (e) and possibly your results from problem 2 to determine the
muzzle velocity of the tennis ball. Assume γ = 1.4
g. Determine the temperature of the gases at shot exit.
For acetylene, n = 26.038 lbm/lbmol, ∆h0f = +97,477 BTU/lbmol.
Problem 9
A potato is stuffed into the 3” diameter exhaust pipe of a car that is not running too well.
1 g of incompletely combusted combustion products (assume gaseous Heptane) mixes
with a stoichiometric amount of air behind the potato and ignites. If the potato is wedged
4” into the exhaust (i.e., it has 4” of travel) and weighs 0.25 lbm, and assuming the combustion takes place before the potato moves, determine the theoretical maximum “muzzle”
velocity of the potato. Also calculate the muzzle velocity assuming an isentropic expansion. For gas expansion purposes you can assume the volume available initially behind the
potato is equal to the volume between the point of obturation and the end of the exhaust
and assume a “smeared” specific heat ratio of 1.3 for the product gases. Also assume the
combustion begins at 500 K and completes at 1500 K. Assume the total enthalpy at 500 K
for n-Heptane (C7H16) is −120,000 kJ/kg-mol.
Problem 10
Assume we have a quantity of 29 lb of 11.1% nitrated nitrocellulose (C6H8N2O9) and it
is placed in an empty chamber of a gun at 77°F and 14.7 psia. The chamber is 1160 in.3
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
38
in volume. The propellant density is 0.060 lbm/in3. If the air in the chamber is NOT
neglected and assuming the volume is fixed:
a. Write the balanced equation for this combustion (assume the oxygen goes preferentially into CO2 instead CO this time—you will find the difference later).
b. Using the tables in the textbook, estimate the adiabatic flame temperature of the
resultant gas (Hint: recall the definition of adiabatic flame temperature).
2.6 Solid Propellant Combustion
Now that we have examined the background of the thermochemistry and thermodynamics
of combustion, we shall see how this applies to the behavior of a burning solid propellant.
We shall endeavor, in this section, to come up with definitions and relationships that will
allow us to define the state of the propellant behind a projectile at any given time. The process we will use is somewhat simplified because the real situation behind a moving projectile
is generally a two phase, reacting flow field. Some of our assumptions, though not necessarily valid in the purest sense, are good enough to predict bulk behavior of the propelling gas.
In the previous sections, we have discussed how energy is evolved by the propellant. We
saw that thermodynamic properties were not dynamic at all, merely means of accounting
for energy knowing the initial and end states and making assumptions on the process
between them. This section will allow us to add in some time dependency to the equations
to somewhat understand the rates at which combustion is occurring.
Solid propellants are generally nitrocellulose compounds that are manufactured by
nitrating through immersion in acid. The details of this process for various materials can
be examined in detail in Refs. [7,8,16–18]. This material is then chopped and worked into a
doughy substance and pushed though dies to form various shapes. The material then has
solvents removed and it is dried. When this process is complete, the propellant has the
consistency of uncooked (i.e., hard and somewhat brittle) pasta. Though this statement is
general, there are, as always, exceptions.
The burning of solid propellant is a surface phenomenon. The rate of gas evolution is dependent upon the amount of surface area of the propellant. Because of this, the shape that the
propellant takes is extremely important. Burning is the mechanism of transforming the solid
propellant to a gas. The burn rate of a propellant is highly dependent upon the pressure at
which the burning reaction takes place. Essentially, the greater the pressure, the faster the propellant burns. These two behavioral observations tell us that if we can control the geometry and
confinement of a given propellant, we can, to a large degree, control the rate of gas evolution.
We shall examine a single propellant grain to gain an understanding of how the geometry affects the rate of evolution of gas. Consider a long cylinder of solid propellant that is
commonly referred to as a grain. If the cylinder were long enough, we could see that most
of the surface area would be located along the circumference and length. In other words,
we can neglect the two small surface areas that comprise the ends. This is illustrated in
Figure 2.8. If we neglect the burning of the end surfaces, it allows us to examine the geometry through simple mathematical relationships.
As our grain begins to burn, solid material will be evolved into gas. Thus, we can imagine the solid surfaces shrinking toward the centerline of the grain. If we examine our
grain from the end looking down its axis, we would see a circular section as depicted in
© 2014 by Taylor & Francis Group, LLC
Physical Foundation of Interior Ballistics
39
This is where most of the surface area
is located
FIGURE 2.8
Long cylindrical propellant grain.
d
FIGURE 2.9
Propellant grain cross section.
Figure 2.9. We could then write an expression for the surface area of our grain as a function of its diameter and length:
A(t) = πd(t)l
(2.94)
In this expression, A(t) is the surface area of the grain, d(t) is the diameter, and l is the length.
We have denoted the surface area and diameter as functions of time to remind us of our
assumption of no burning at the ends of the grain. After some time, t, the grain surface will
have regressed such that our diameter has decreased. This is depicted in Figure 2.10. This
graphically shows us that at time t1 the grain clearly has more surface area than at time t2;
therefore, as burning progresses, the rate of evolution of gas slows down. This is commonly
called regressive burning.
Propellant geometry is characterized by a quantity known as the web thickness or simply the web. The symbol used for the web is D. The web is the smallest thickness of the initial propellant grain. In the case of our cylindrical grain, it would be the initial diameter.
d(t1)
d(t2)
FIGURE 2.10
Propellant grain cross section at two times.
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
40
In the interior ballistics analysis of a gun system, we need to track how much gas
is evolved and also how much solid is remaining. This is important because we have
seen that all of our equations of state are dependent upon volume as well as pressure
and temperature, and these, in turn, affect the burning rate. The amount of solid propellant remaining is tracked through use of the web fraction, f. The web fraction is the
fraction of web remaining at a given time, t. Through use of this web fraction, we can
write an expression for the amount of propellant remaining at any time as a function
of the web:
d(t) = fD
(2.95)
This is illustrated for a grain with a single perforation (known colloquially as a perf) in
Figure 2.11. It is important to note here that for a single perf grain, the web is defined as the
outside radius minus the inside radius. This sometimes is confusing for new ballisticians
since we use D as the web thickness. Also one can see from the figure that an advantage
of a single perf grain is that it burns from both the inside out and the outside in, thus
decreasing the surface on the outside while increasing the surface on the inside—known
as neutral burning behavior.
Use of the web fraction is convenient because, mathematically, it is a function that varies
from unity to zero. The manner in which it varies may be somewhat complex, but at least
the end states are well defined. An example plot of web fraction versus time is shown in
Figure 2.12. In this figure, tB is the time at which all of the propellants have evolved into
gas—the burnout time.
fD
D
FIGURE 2.11
Burning of a single perforated propellant grain.
f
1
t
f = fraction of remaining web
f = f (t)
FIGURE 2.12
Fraction of remaining web versus time.
© 2014 by Taylor & Francis Group, LLC
f (0) = 1
f (tB) = 0
tB
Physical Foundation of Interior Ballistics
41
Many times, we are interested more in the volume of the propellant that has evolved
into gas rather than the fraction of the web remaining. It should be clear that the two
quantities are related since the gas had to come from the solid material and conservation of mass states that we can neither destroy nor create mass. This is handled through
use of the fraction of propellant burnt, ϕ. Since ϕ is a function of f and f is a function of
time, we see that ϕ must be also a function of time. Since propellant geometries can be
fairly complicated, ϕ can be a rather complicated function of f. For simple shapes, this
relationship is straightforward. For instance, a single perforated grain has the functional
relationship that
φ (t) = 1 − f (t)
(2.96)
Most shapes can be simplified to express ϕ as a quadratic function of f through use of a
shape function, θ:
φ (t) = [1 − f (t)][1 + θ f (t)]
(2.97)
This expression allows us to cover almost any simple geometry, the most notable exception
being a sphere. Figure 2.13 depicts how variation in the shape function affects the relationship between f and ϕ.
With the formulations earlier, we have been able to mathematically define the effect
of propellant geometry on the rate of gas evolution. The second important parameter
in this generation of gas was stated to be the effect of pressure on burning. Whenever
a propellant burns, say in a fixed volume, two competing processes are happening: the
volume into which the gaseous propellant is moving is increasing because there is less
solid material—this decreases the pressure, and the more and more propellant gas is
being pushed into a confined space—this increases the pressure. The rate at which the
Form functions for various θ and spherical grains
1.0
0.9
0.8
0.7
(t)
0.6
0.5
θ = –1
0.4
θ = –0.4
0.3
θ=0
0.2
θ = 0.4
θ=1
0.1
Spherical grain
0.0
1.0
0.9
0.8
0.7
0.6
0.5
f(t)
FIGURE 2.13
Effect of different values of θ on ϕ and f.
© 2014 by Taylor & Francis Group, LLC
0.4
0.3
0.2
0.1
0.0
Ballistics: Theory and Design of Guns and Ammunition
42
surface area decreases affects this relationship. The simplest model for the relationship
between burn rate and pressure is given by
D
df
= − β pB (t)
dt
(2.98)
where
D(df/dt) is the time rate of change of the web (i.e., the burning rate)
β is a burn rate coefficient
pB is the pressure (we will discuss the subscript later)
The negative sign comes about because the amount of propellant would be increasing if
D(df/dt) were to result in a positive number. This simple relationship facilitates our analysis of propellant behavior in a gun. Other relationships can more accurately describe propellant behavior, but their complexity is such that computer codes must be used to obtain
answers with them. Two very common burn relationships are
df
= − β [ pB (t)]α
dt
(2.99)
df
= − β [ pB (t) − P1 ]
dt
(2.100)
D
D
Equation 2.99 is by far the most commonly used in computer codes. Caution must be exercised when using burn rate data from the literature as the units will be an indicator of the
proper burn rate form of the governing equation. If we examine the units of D df/dt, we
see that they are in terms of [length]/[time]. This type of data are usually obtained from
a strand burner. A strand burner is a device that can accurately measure the rate of linear
burning in a propellant. Reference [19] contains an excellent diagram of a strand burner.
If we consider a pressure vessel so thick as to be rigid and the amount of propellant so
small such that we can neglect its contribution to the volume, we can describe the burning of
the propellant as a constant volume process. This is the essence of closed-bomb testing. We
can further assume that this pressure vessel can be isolated thermally and the gas behavior
is ideal. In this case, our closed bomb, with internal volume, V, would resemble Figure 2.14.
V
FIGURE 2.14
Diagram of a closed bomb.
© 2014 by Taylor & Francis Group, LLC
Physical Foundation of Interior Ballistics
43
Since we assumed ideal gas behavior, we can write an expression for pressure as a function
of volume and temperature
pB V = mg RT
(2.101)
where
pB is the pressure
V is the volume
mg is the mass of the gas
R is the specific gas constant
T is the temperature
When we place our solid propellant into the closed bomb, it has an initial weight that we
would call c. So initially, we can write
c = ρ Vsolid
(2.102)
where
ρ is the density of the solid propellant
Vsolid is its volume
If we now assume that the propellant is cylindrical, we can write its volume as the product
of its cross-sectional area and its length. The initial diameter is the web for a cylindrical
grain, so
Vcyl.grain = π
D2
l
4
(2.103)
[d(t)]2
l
4
(2.104)
This volume at any time, t, can be expressed as
V(t)cyl.grain = π
Because mass is conserved, the amount of solid propellant burned is equal to the amount
of gas generated. This is an important concept. If we started with 1 lbm of propellant and
completely burned it, we would be left with 1 lbm of gas. Based on this, we can write for
the mass of the gas as
mg (t) = ρ Vcyl.grain − V(t)cyl.grain = ρ
π
l D2 − [d(t)]2
4
{
}
(2.105)
We discussed the fraction of propellant burnt, ϕ, earlier. We are now in a position to formally define it as
φ (t) ≡
© 2014 by Taylor & Francis Group, LLC
mg (t)
c
(2.106)
Ballistics: Theory and Design of Guns and Ammunition
44
If we substitute Equations 2.102, 2.103, and 2.105 into Equation 2.106, we obtain
φ (t) =
ρ
π
l D2 − [d(t)]2
[d(t)]2
4
= 1 −
π
D2
ρ lD2
4
{
}
(2.107)
Now we insert Equation 2.95 into Equation 2.107 to yield
2
fD
2
φ (t) = 1 −
= 1 − f = (1 − f )(1 + f )
D
(2.108)
Comparing this expression (derived for a cylindrical grain) to Equation 2.97 shows that
the shape factor θ = 1 for a cylindrical grain. Also, by comparison to Equation 2.96 we see
that the shape factor θ = 0 for a single perforated grain. Essentially, any shape factor can
be derived using this same procedure. So up to this point, we have determined that the
shape factor
θ = 0 for single perforated grains
θ = 1 for cylindrical grains
An interesting thing has happened. We started this section attempting to find a relationship for the mass of gas evolved from the solid propellant and we have come around to
finding the relationship between ϕ and f again. The key procedure here is now to rearrange Equation 2.106:
mg (t) = cφ (t)
(2.109)
This is the relationship that governs the amount of gas evolved from the burning propellant. It looks rather simple, but consider that ϕ is a function of f and t, and f is a function of
pB and t. We shall return to this later.
The burning propellant in our closed bomb must generate pressure. To take this further,
we need to rearrange Equation 2.98 into
pB (t) = −
D df
β dt
(2.110)
In this expression, we know that D is the initial web and therefore a constant, and we shall
assume that β is a constant (β actually increases somewhat with pressure).
Because we want to work with masses of substances, f is not a convenient variable. We shall
use a relationship to express it in terms of ϕ. At this point, caution must be exercised. Recall
that the relationship between ϕ and f varies with propellant geometry. We shall proceed using
our cylindrical grain relationship (Equation 2.108). Rearranging Equation 2.108, we obtain
f (t) = 1 − φ (t)
(2.111)
If we differentiate this relationship with respect to time, we obtain
df
1
dφ
=−
dt
2 1 − φ (t) dt
© 2014 by Taylor & Francis Group, LLC
(2.112)
Physical Foundation of Interior Ballistics
45
This form allows us to rewrite Equation 2.110 as
pB (t) = −
dφ
D
2β 1 − φ (t) dt
(2.113)
We now have all the expressions we need to bring this together. We have an equation of
state
pB (t)V = mg (t)RT (t)
(2.114)
We have an expression for conservation of mass (relationship between mg and ϕ)
mg (t) = cφ (t)
(2.115)
and we have an expression that relates the amount of pressure generated to the amount of
propellant burnt (burn rate equation)
pB (t) = −
dφ
D
2β 1 − φ (t) dt
(2.116)
All these expressions are in terms of constants we know beforehand or f, ϕ, and T.
To describe the temperature of gas, we need to define a parameter used often in interior ballistics, the propellant force, λ. Propellant force is a constant that is defined as the
amount of energy released from a propellant under adiabatic conditions. In other words,
it is the most energy one can obtain by burning a propellant. Mathematically, we express
it as
λ ≡ RT0
(2.117)
where
R is the specific gas constant
T0 is the adiabatic flame temperature of the gas
This constant has units of energy per unit mass. Sometimes, T0 is referred to as the
uncooled explosion temperature. In our development, we shall assume that all gases are
evolved at the adiabatic flame temperature. There are many theories that describe combustion. Introductory treatments are provided in Refs. [20,21], but all of the references in the
end of this section cover the topic to some degree. References [22–24] treat the topic in great
detail. If we utilize this reactive assumption, we can rewrite Equation 2.114 using Equation
2.117 to give us
pB (t)V = λ mg (t)
(2.118)
Now we can combine Equations 2.118 and 2.116 to yield (for a cylindrical grain)
λ mg (t)
D
dφ
=−
V
2β 1 − φ (t) dt
© 2014 by Taylor & Francis Group, LLC
(2.119)
Ballistics: Theory and Design of Guns and Ammunition
46
We then substitute Equation 2.115 into the aforementioned expression, resulting in
dφ
λ cφ (t)
D
=−
V
2β 1 − φ (t) dt
(2.120)
1
dφ
2 β λc
=−
DV
φ (t) 1 − φ (t) dt
(2.121)
This can be rearranged to yield
This is a separable, first-order, nonlinear, differential equation, which can be written in
integral form as
1
∫
0
tB
2 β λc
dφ
dt
=−
DV
φ (t) 1 − φ (t)
∫
(2.122)
0
The solution of which is
1
1−φ − 1
2βλc tB
t
ln
=−
1 − φ + 1
DV 0
0
(2.123)
This expression is somewhat problematic because of its singular behavior at ϕ = 1 and ϕ = 0.
The equation was approximated numerically to yield
tB ≈ 350
DV
β λc
(2.124)
In this case, the solution to this expression was problematic; however, in many cases, it can
be evaluated more readily. The techniques that will follow are much simpler from a hand
calculation standpoint.
Even though the closed bomb may seem academic, it is actually quite a useful device for
determining propellant parameters. If we consider Equations 2.110 and 2.113, we see that
since we know the initial web, D, and we can measure the pressure, the only thing missing
is β and ϕ or f. Equation 2.114 tells us that if we measure pB and T and know V, we can get ϕ
or f. Thus, the closed bomb is useful for determining the burn rate coefficient, β.
Problem 11
M1 propellant is measured in a closed bomb. Its adiabatic flame temperature is 3906°F.
Its molar mass is 22.065 lbm/lb-mol. What is the effective mean force constant in
ft-lbf/lbm?
ft-lbf
Answer: λ = 305, 709
lbm
© 2014 by Taylor & Francis Group, LLC
Physical Foundation of Interior Ballistics
47
Problem 12
M15 propellant was tested in a strand burner to determine the linear burning rate. The
average pressure evolved was 10,000 psi. If the burning exponent, α, was known to be 0.693
and the pressure coefficient, β, was known to be 0.00330 in./s/psi0.693. Determine the average linear burning rate, B in inch per second.
in.
Answer: B( p) = 1.952
s
Problem 13
Derive the functional form of ϕ in terms of f for a flake propellant. Assume cylindrical
geometry.
Hint: Flake propellant consists of grains that have thicknesses much smaller than any
other characteristic dimension.
Answer: ϕ(t) = 1 − f
Problem 14
An M60 projectile is to be fired from a 105 mm M204 Howitzer. The propellant used in
this semi-fixed piece of ammunition is 5.5 lbm of M1 propellant. M1 propellant consists of
single perforated grains (θ = 0) with a web thickness of 0.0165 in., if the average pressure
(over the launch of this projectile) developed in the weapon is 20,455 psi. Calculate the
average burning rate coefficient in in.3/lbf-s if the burn rate is (we use a negative sign in
the burn rate to make the form come out right later):
df
= −185.9 [s −1 ]
dt
in.3
Answer: β = 1.50 × 10 −4
lbf-s
Problem 15
β is actually a function of pressure and temperature (it is really given in tables at 25°F at
this value). For simplification (and illustration), we will assume it is constant. Given this
assumption, calculate the functional form of the web fraction, f from Problem 14.
β pavg
Answer: f = 1 −
t
D
Problem 16
Given the data provided in Problems 14 and 15, determine the proper form of the fraction
of charge burnt.
Answer: ϕ(t) = 185.9t
Problem 17
You are asked to characterize a commercial propellant. In order to do this, you take one
grain of the propellant and place it in a closed bomb of 0.5 in.3 volume, initially evacuated.
You have a temperature and pressure sensor in the device. After 0.063 s you decide that the
propellant has fully combusted. You read the data—pressure was measured to be 3.706 psi
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
48
(this is not a big value but it was only one small grain of propellant) but it looks as though
the temperature sensor is broken. The initial propellant grain weighed 0.003189 grains and
it was 0.1 in. long by 0.01 in. diameter. Based on this data only—
a. Estimate the propellant force, λ, in ft-lbf/lbm.
b. Estimate the linear burn rate coefficient, β in in/s/psi.
c. List all assumptions and explain why you believe these estimates are too high or
too low (certain assumptions may make the estimates high while others make
them low)—There are at least four assumptions used here.
2.7 Fluid Mechanics
The entire field of ballistics is steeped in the principles of fluid mechanics. The flow of
propellant gases in the gun tube, the flow of the propellant gases through a muzzle brake
upon shot exit, the flow of the air around the projectile in flight, and even, as we shall see,
the flow of target material during a penetration event can many times be modeled as a
fluid. This section is devoted to a basic treatment of fluid mechanics principles. Some of
these we will use very soon, others will be used at a later time. All of them are important
in the study of ballistics.
A fluid differs from a solid in its behavior when placed in shear. In general, fluids can
support little or no shear loads or tensile stress. Fluids are generally characterized by their
behavior under shear stress. Because a fluid will, in general, flow readily under a shear
stress, this behavior is normally plotted in a graph of rate of deformation versus shear
stress as depicted in Figure 2.15.
A fluid is considered to exhibit Newtonian behavior if there is a linear relationship
between shear stress and rate of deformation. A fluid is non-Newtonian otherwise. Some
fluids such as an ideal plastic or a thixotropic material actually do exhibit a yield stress.
Newtonian
Rate of deformation
Ideal fluid
Non-Newtonian
Ideal plastic
Thixotropic
Yield stress
FIGURE 2.15
Rate of deformation versus shear stress.
© 2014 by Taylor & Francis Group, LLC
Shear stress
Physical Foundation of Interior Ballistics
49
In the case of an ideal plastic, after a certain yield stress is achieved, the material exhibits
a linear relationship between stress and deformation rate. A thixotropic material exhibits a nonlinear relationship after yield stress is reached. An ideal fluid is one where the
material will flow and continue to accelerate regardless of the amount of shear stress
applied.
Many of the fluids we will deal with are Newtonian. Mathematically, the relationship
between applied shear stress and deformation rate is given by
τ =µ
∂u
∂y
(2.125)
where
τ is the applied shear stress
μ is the dynamic viscosity of the fluid
𝜕u/𝜕y is the deformation gradient (change in velocity with respect to a spatial coordinate)
The ratio of the dynamic viscosity to the fluid density occurs so often that it is customary
to define a kinematic viscosity as
ν=
µ
ρ
(2.126)
where
ν is the kinematic viscosity
ρ is the density of the fluid
In the section on thermodynamics, we introduced the concept of a Lagrangian or control
mass approach and an Eulerian or control volume approach to solving transport problems.
In examination of a fluid’s behavior, we need to develop both of these techniques. Our plan
of attack will be to develop these equations in a CV and provide equations to change the
reference frame afterwards. For a more complete treatment, the reader is referred to Refs.
[11,12,25–28].
The basis for our development of the following equations are the tenets that (a) mass
must be conserved and (b) Newton’s second law must hold true. Newton’s second law can
be written as
∑ F = dt (mV)
d
(2.127)
where
ΣF is the vector sum of all the forces acting on a body (or blob of fluid or CV)
m is the mass of the body
V is the vector velocity of the body
It is important to note that throughout this work, V is volume (a scalar), V is velocity (as a
scalar quantity), and V is velocity (as a vector quantity).
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
50
Vt
V
CV
Vt
Vn
V
Vn
FIGURE 2.16
Depiction of normal and tangential velocity components with respect to an arbitrary CV.
Since CVs can be oriented in an arbitrary manner, it is important to understand that
only that component of velocity normal to the control surface (CS) (i.e., the boundary of the
CV) transports material or energy into the CV. If we examine Figure 2.16 where we have
broken the velocity vectors into normal and tangential components (denoted Vn and Vt,
respectively), we can clearly see why this is so.
Consider an arbitrary property, N, of a substance. We would like to see how this property is transported into and out of a CV. If we define an intensive property, η, such that
N
m
or N = η m
∫ ηρdV + ∫
ηρ V ⋅ dA +
η=
(2.128)
Then we can write
dN ∂
=
dt ∂t
CV
outflow area
∫
ηρ V ⋅ dA
(2.129)
inflow area
This equation defines how a property of interest is transported into and out of the CV. If
we look at each of the terms, we see that this is an intuitively satisfying equation. The term
on the LHS is the time rate of change (decrease) of any property of the CV over a time of
interest. The first term on the RHS tells us how much of that property is stored in the CV
over this time. The second term on the RHS tells us how much material has left the
CV, while the third term tells us how much material has entered.
Now wait a minute! If we look at the signs on the second and third terms, they seem to
be incorrect—should not the stuff leaving have a negative sign and the stuff entering have
a positive sign? The answer to this is yes, but Equation 2.129 is written correctly. The key to
this seemingly inconsistent sign convention lies in the fact that the dot product in the second term is positive when we define the area as a vector that points outward and is normal
to the surface. Similarly, the inflow term will always lead to a negative number since the
velocity vector points inward and the area vector points outward.
We shall now examine the flow of propellant gases in a suitable CV located somewhere
behind a projectile at an instant in time. This will serve to foster understanding of the CV
approach.
Consider a CV in a gun tube located somewhere behind a moving projectile as depicted
in Figure 2.17. There will be a velocity associated with the propelling gases (we will see
this later) such that the gases are flowing in one side and out the other, but no gases flow
through the walls.
The ends of this cylindrical CV are designated as CSs. The inlet side is CS1 and the outlet
side is CS2. If we would like to write an equation for how mass is transferred into or out of
© 2014 by Taylor & Francis Group, LLC
Physical Foundation of Interior Ballistics
CS1
51
CS2
No flow through tube walls
FIGURE 2.17
Typical gun tube CV (2D representation).
this CV, we set N, the flow variable in Equation 2.129, equal to m, the property of interest.
When this is done, Equation 2.128 tells us that
η=
N m
= =1
m m
(2.130)
So for this case, we can write
dm ∂
=
dt ∂t
∫ ρdV + ∫
CV
ρ V ⋅ dA +
outflow area
∫
ρ V ⋅ dA
(2.131)
inflow area
We know mass can neither be created nor destroyed, so dm/dt = 0, then we arrive at what
is commonly called the equation of conservation of mass, or the continuity equation. In a
general form, it is given as
∂
∂t
∫ ρdV + ∫
CV
ρ V ⋅ dA +
outflow area
∫
ρ V ⋅ dA = 0
(2.132)
inflow area
The first term on the LHS states how the mass in the CV is changing with time. The second term is the amount of mass exiting the CV and the third term is the amount of mass
entering the CV.
The flow inside a gun tube is never steady or uniform. Nevertheless, it is informative
to look at this expression using these two assumptions to gain some physical insight into
the nature of the terms. The steady flow assumption means that there is no increase or
decrease in material flow into or out of our CV. This implies that the first term is zero. So
for the special case of steady flow, we have
∫
outflow area
ρ V ⋅ dA +
∫
ρ V ⋅ dA = 0
(2.133)
inflow area
Simply put, this equation states that what comes into the CV equals what goes out of the CV.
Uniform flow is a special case where fluid viscosity effects are neglected. This results in
a constant velocity across the CSs. In essence, the velocity at the wall of the gun tube is the
same as the velocity on the centerline of the tube. We will discuss this and its implications
in more detail later.
When we apply this assumption to Equation 2.133 and note that V · dA is negative at CS1
(because the vectors have opposite directions) and positive at CS2, we obtain the following
simple relationship:
ρ1V1 A1 = ρ 2V2 A2 = mɺ
.
Thus, under the steady flow assumption, the mass flow rate, m, is constant.
© 2014 by Taylor & Francis Group, LLC
(2.134)
Ballistics: Theory and Design of Guns and Ammunition
52
We shall now examine the use of momentum, mV, as our flow variable. Use of Equation
2.128 with this flow variable yields
η=
N mV
=
=V
m
m
(2.135)
Now we can include this into Equation 2.129 to obtain
d(mV ) ∂
=
dt
∂t
∫ VρdV + ∫
CV
Vρ V ⋅ dA +
outflow area
∫
Vρ V ⋅ dA
(2.136)
inflow area
Through Newton’s second law, we know that the term on the LHS (time rate of change of
momentum) equals the forces on the system. The first term on the RHS is the change in
the system’s momentum through storage in the CV. The second and third terms are the
momentum leaving and momentum entering the CV, respectively. It is again informative
to examine the steady flow case, which reduces our equation to
∫
F=
Vρ V ⋅ dA +
outflow area
∫
Vρ V ⋅ dA
(2.137)
inflow area
Here we have replaced the time rate of change of momentum term with the force. Once
again we shall use the uniform flow assumption to facilitate our understanding of this equation. Consider the same gun tube CV as earlier, drawn slightly differently in Figure 2.18.
As discussed earlier, the velocity and area scalar products result in a negative sign on the
inflow and a positive sign on the outflow side. With this uniform flow assumption (recall
we also included a steady flow assumption to reduce the equation to the form of Equation
2.137), our Equation 2.137 would become
F = ρ 2V2V2 A2 − ρ1V1V1 A1
(2.138)
Note that this is still a vector equation with the vectors V1 and V2 determining the direction of F. If we had already worked out or it was obvious what direction the resultant force
would be in, then we could write
F = ρ 2V22 A2 − ρ 1V12 A1
(2.139)
Equation 2.138 only tells us part of the story. It tells us the inertial reaction of the CV to the
forces arising from a fluid passing through it. There are two types of forces that occur on
the LHS in response to or independent of this, body forces and surface tractions.
V2
V1
A1
CS1
A2
CS2
No flow through tube walls
FIGURE 2.18
Typical gun tube CV (3D representation).
© 2014 by Taylor & Francis Group, LLC
Physical Foundation of Interior Ballistics
53
p2
p1
A1
A2
CS2
CS1
FIGURE 2.19
CV with no viscous forces acting.
Body forces are those that act through the bulk of the material (i.e., directly affecting
every molecule). Examples of this are gravitational loads, electromagnetic loads, etc. It is
customary to write these loads on a unit mass basis to be consistent with the rest of the
equation. In many cases, these are small and are neglected.
Surface tractions are forces that act on the CS. These forces tend to be large and can be
categorized into normal forces and shear forces. As the name implies, normal forces act
normal to the CS. Pressure is the most common normal force. Because pressure cannot be
negative, it always acts opposite to the surface area vector.
Shear stresses are a result of the fluid’s propensity to stick to a solid (or other fluid) surface. The fluid viscosity, as defined earlier, is a measure of the intensity of these stresses.
Shear stresses always act opposite to the direction of flow and along the CS. If a fluid is
modeled as inviscid, there can be no shear stresses.
Picking up from Equation 2.138, if we model a flow as steady with no viscosity, there will
still be pressure forces present. This is depicted in Figure 2.19.
Since pressure forces always act opposite to the area vector, it is customary to define the
pressure forces as
Fp = −
∫
outflow area
pdA −
∫
pdA
(2.140)
inflow area
In Equation 2.140, the signs of the area vectors would define the direction of the force.
Before we establish a CV with viscous forces acting, it is instructive to describe these
viscous forces and their effect on the flow field. As previously established, viscosity is a
property of a fluid. The greater the viscosity of a fluid is, the more difficult it is to shear
the material. If the viscosity is high enough or the flow velocity low enough, a fluid will
exhibit what is known as laminar flow. Laminar flow is a very orderly shearing of the fluid
from a solid surface where the fluid sticks to the boundary. In a tube or pipe, after some
entrance length required for the flow to establish itself, the fluid will achieve a parabolic
velocity distribution as depicted in Figure 2.20.
V
Tube wall
FIGURE 2.20
Laminar velocity profile in a tube.
© 2014 by Taylor & Francis Group, LLC
54
Ballistics: Theory and Design of Guns and Ammunition
V
Tube wall
FIGURE 2.21
Uniform velocity profile in a tube.
The laminar profile in Figure 2.20 is in stark contrast to the uniform profile that we had
assumed in our previous discussions depicted in Figure 2.21. If the flow velocity is high
enough or the viscosity low enough, the flow will transition from laminar flow to what
is known as turbulent flow. Turbulent flow is characterized by a large number of eddies
that swirl around in the flow. These eddies are important in that they tend to distribute
momentum, energy, and matter throughout the fluid resulting in better mixing and very
different transport properties. Many more flows are turbulent than laminar. The dimensionless parameter that governs this behavior is known as the Reynolds number and is
given by
Re =
ρ Vd Vd
=
µ
ν
(2.141)
where
Re is the Reynolds number and is dimensionless
ρ is the fluid density
V is the fluid velocity
d is a relevant characteristic length of the system (an internal diameter of a pipe, a length
of a projectile, etc.)
μ and ν are the dynamic and kinematic viscosities of the fluid, respectively
If the Reynolds number is high enough, the flow will be turbulent. This demarcation is,
in general, a range of values that also depends whether the flow is an internal one (such
as the gas flow in a gun tube) or an external one (such as the flow about a projectile). The
velocity profile of a turbulent flow is depicted in Figure 2.22. Here we can see that the
effect of the eddies is to distribute the momentum, resulting in a profile that is flatter and
more akin to our inviscid flow model of Figure 2.21.
If we now return to our discussion on the surface tractions, we can discern that the effect
of fluid viscosity is to create a shear stress at the boundary between the fluid inside a gun
V
Tube wall
FIGURE 2.22
Turbulent velocity profile in a tube.
© 2014 by Taylor & Francis Group, LLC
Physical Foundation of Interior Ballistics
55
p2
A2
p1
A1
τw
CS1
CS2
FIGURE 2.23
Surface tractions on a gun tube CV.
tube and the solid tube itself (i.e., on our CS). If we consider the diagram of Figure 2.19,
we can redraw this figure to include the effect of shear stresses as depicted in Figure 2.23.
Since the shear stress, τw, acts all over the area of our CV, we can add a term in for this into
Equation 2.140 to obtain an expression for all of the surface forces as follows:
Fsurface = −
∫
∫
pdA −
outflow area
∫
pdA −
inflow area
τ wdA
(2.142)
surface area
We can insert this into our expression for the conservation of momentum Equation 2.137
to obtain, for steady flow
−
∫
∫
pdA −
outflow area
pdA −
inflow area
∫
∫
τ wdA =
surface area
Vρ V ⋅ dA +
outflow area
∫
Vρ V ⋅ dA
inflow area
(2.143)
or, in a more general sense,
∫
−
pdA −
outflow area
=
∂
∂t
∫
pdA −
∫
τ wdA
inflow area
surface area
∫ VρdV + ∫
Vρ V ⋅ dA +
CV
outflow area
∫
Vρ V ⋅ dA
(2.144)
inflow area
The next transport property we shall examine is that of energy. In Sections 2.4 and 2.5, this
was discussed to a degree. The objective of this section is to demonstrate that we can use
the same transport Equation 2.129 to come up with the energy equations we have used
earlier. We start by recognizing that our transport variable is energy, E. With this in mind,
Equation 2.128 can be rewritten as
η=
E
=e
m
(2.145)
Recall that lower case letters are intensive properties. Then we can write
dE ∂
=
dt ∂t
∫ eρdV + ∫
CV
eρ V ⋅ dA +
outflow area
∫
eρ V ⋅ dA
(2.146)
inflow area
This states that the change in energy of a system is equal to the change in energy stored in
the system minus that which is advected away plus that which is advected into the system.
Recall from Equation 5.6 that
dQ dW dE
+
=
dt
dt
dt
© 2014 by Taylor & Francis Group, LLC
(2.147)
Ballistics: Theory and Design of Guns and Ammunition
56
From our definition of work, we know that
W=
∫ pdV
(2.148)
But volume is nothing more than a length times an area. This allows us to write
∫ px ⋅ dA
W=
(2.149)
If we take the derivative of this expression with respect to time assuming pressure is an
average value over the time increment, we can write
dW
=
dt
dx
∫ p dt ⋅ dA = ∫ pV ⋅ dA
(2.150)
There are many types of work terms. The aforementioned term happens to be called pdV
work or pressure work. The other types of work, such as shaft work, are usually not present in a gun launch so we shall neglect them. Insertion of Equation 2.150 into Equation
2.146 and rearranging yields
dQ ∂
=
dt ∂t
p
p
e + ρ V ⋅ dA
e + ρ V ⋅ dA +
ρ
ρ
inflow area
outflow area
∫
∫ eρdV + ∫
CV
(2.151)
In the section on thermodynamics, we defined the specific energy through Equation 2.39.
If we insert this definition into the aforementioned expression, we obtain
dQ ∂
=
dt ∂t
∫
eρ dV +
CV
p
V2
+ u + ρ V ⋅ dA
gz +
2
ρ
CS
∫
(2.152)
Here we have combined the last two terms on the RHS of Equation 2.151 with the understanding that the integral of the last term in Equation 2.152, being an integral over the
entire CS, accounts for the difference between inflow and outflow. It is informative to look
at this equation with respect to a gun launch. The term on the LHS represents the transfer
of heat to or from the system. The first term on the RHS represents the change in stored
energy of the system (such as energy released by propellant combustion). The last term on
the RHS is the change in energy of the system. Since gravitational potential energy, the
product gz, is small relative to the other energy terms, it is usually neglected allowing us
to rewrite the expression as
dQ ∂
=
dt ∂t
∫
CV
eρ dV +
V2
p
+ u + ρ V ⋅ dA
ρ
2
CS
∫
(2.153)
Earlier in this section, we introduced the common practice of characterizing a fluid based on its
behavior under shear stress. This allowed us to come up with a relationship between applied
© 2014 by Taylor & Francis Group, LLC
Physical Foundation of Interior Ballistics
57
shear stress and deformation rate. Another distinction has to be made between fluids with
respect to the density. If the density is considered constant in a fluid or solid that we model,
we call this material incompressible. If the density varies, we must analyze the problem with
the assumption of compressible material. This has many ramifications. The most significant
ramification is that if the material is incompressible, then the energy equation is decoupled
from the momentum equation and we can solve them independently [25]. This makes problem solving much simpler. We do not have this luxury when the density varies significantly.
In fluid flows, such as those that we shall study later, a dimensionless parameter known
as the Mach number, Ma, is used as a measure to determine the effect of compressibility,
among its other uses. The Mach number is given by
Ma =
V
a
(2.154)
where
V is some characteristic velocity in the material
a is the speed of sound in the material
In general, if the Mach number is below 0.3, the deviation from incompressible flow is
small so the assumption of incompressibility leads to an acceptably small error [16]. In an
ideal gas, the speed of sound is given by the relation
a = γ RT
(2.155)
where
γ is the specific heat ratio
R is the specific gas constant
T is the absolute temperature (i.e., in degrees Rankine or Kelvin)
The speed of sound in any material is formally defined as
a=
∂p
∂ρ
(2.156)
s
That is to say that the speed of sound in a material is equal to the square root of the partial derivative of pressure with respect to density evaluated with constant entropy. The
interested reader is referred to any of Refs. [15,16,25] for the detailed proof of this equation.
The speed of sound is essentially the fastest speed at which a disturbance can be propagated by molecular interaction. If a disturbance is created that is strong enough, a shock
will form. This shock must always move faster than the speed of sound in the material.
We will discuss this in detail later.
In the study of compressible flows, it is common practice to utilize stagnation values
in many of our calculations. Stagnation values are the values of the enthalpy, pressure,
temperature, and density that are achieved by adiabatically slowing a flow down to zero
velocity. The assumption of adiabatic behavior is warranted in many of the situations we
will examine, particularly in exterior ballistics. The stagnation enthalpy is given by
1
h0 = h + V 2
2
© 2014 by Taylor & Francis Group, LLC
(2.157)
Ballistics: Theory and Design of Guns and Ammunition
58
In this and the following equations, the subscript “0” indicates the stagnation value, V is
the velocity of the flowing fluid, and the values without the subscript are the static value,
in the case of Equation 2.157, h is the static enthalpy. Equation 2.157 holds for any material.
If the material is an ideal gas, we can define the stagnation temperature, pressure, and
density as
T0 = h +
1 V2
2 cp
or
γ −1
T0
= 1+
Ma 2
2
T
(2.158)
γ
p0
γ −1
γ −1
Ma 2
= 1+
p
2
(2.159)
1
ρ0 γ − 1
γ −1
= 1+
Ma 2
ρ
2
(2.160)
In each of these cases, thermodynamic relations have been used for an ideal gas
(Equation 2.61).
Shock waves are formed in materials when disturbances of sufficient strength propagate
through the medium. “Sufficient” strength is a term that we throw about rather loosely
to describe conditions where shocks are formed—it can be cast in terms of flow velocities
or pressures (the two are linked as we shall see). Shocks can be classified as normal or
oblique, depending upon the direction of material flow into them. They can also be analyzed as steady or transient. In general, shocks can take curved and rather complex shapes,
but the simple analytical tools we have allow us to look at them only under simplified
geometries. More complex geometries require the assistance of a computer.
We shall only examine normal shocks in this brief review and direct the reader to Ref.
[16] for the handling of oblique shocks. The best way to examine the behavior of a shock
is to look at a shock tube. This simple device will allow us to introduce all of the material
necessary for the introductory study of ballistics and set the stage for later work when we
discuss stress waves in solids.
Before we look at a shock tube, we need to discuss the principle of superposition as
applied to shock waves. Consider two shocks as depicted in Figure 2.24. One of these cases
is a stationary shock where we could consider ourselves “riding on the wave,” while in the
other case, we can consider ourselves to be sitting on the ground watching the shock pass
by. If, in both cases, the shock were passing into a stagnant medium, we would see some
important correlations. The passage of a shock wave always induces motion that follows
the wave. Consider the situation where we are sitting on the ground, the air about us is
V2
Gas motion
downstream
V1
up > 0
Fluid in motion
behind shock
Gas motion
upstream
(a)
FIGURE 2.24
(a) Stationary and (b) moving shock waves.
© 2014 by Taylor & Francis Group, LLC
x
(b)
U
Stagnant gas
ahead of shock
x
Physical Foundation of Interior Ballistics
59
stagnant and all of a sudden a shock passed by us just as is shown in Figure 2.24b. If the
shock were moving at velocity, U, we would feel an induced motion, a wind, immediately
afterwards moving at velocity up in the same direction that the shock was moving. If we
experienced this same situation but instead were riding on the shock, we would feel a wind
of velocity U coming toward our face. This would be analogous to the situation in Figure
2.24a. In this situation, the velocity V1 would be equal to U. Note the direction of the velocity
vectors in the figure. The velocity vector of magnitude V2 is moving away from the wave.
The figure is drawn correctly, but in the case that was just described, based on superposition, since U is larger than up (and it always is). If we were riding on the wave, we would
see material leaving us at velocity (U − up). When we examine a shock wave in the frame
of Figure 2.24b, we are said to be using an Eulerian frame of reference. If we analyze the
very same situation as shown in Figure 2.24a, we are using a Lagrangian reference frame.
The difference between Lagrangian and Eulerian reference frames is important because
we sometimes prefer to solve a problem in one frame or the other because the mathematics
are simpler. As long as the reference frame motion is accounted for, solving in one frame
or the other leads to the same answer.
We shall now use the Lagrangian approach to examine the governing equations for a
stationary normal shock wave. Consider the situation in Figure 2.25 where a shock wave
is moving to the left at velocity, U. Since we would like to examine the behavior of this
shock, we will put ourselves in a reference frame attached to the shock itself. We form a CV
enclosing the shock only. We observe, while riding on this shock, that fluid enters the CV
at velocity U and leaves at velocity u2. We can write the conservation of mass, momentum,
and energy equations for this system as follows:
Conservation of mass (continuity equation)
ρ1U = ρ 2u2
(2.161)
p1 + ρ1U 2 = p2 + ρ 2u22
(2.162)
1
1
h1 + U 2 = h2 + u22 = h01 = h02 = h0 = constant
2
2
(2.163)
Conservation of momentum
Conservation of energy
We see from the last equation that across a shock wave the stagnation enthalpy must
remain constant. This falls out directly from the fact that we assumed the shock wave
FIGURE 2.25
Stationary shock wave.
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
60
FIGURE 2.26
Stationary shock wave moving into a stagnant fluid.
was adiabatic. These equations are coupled through a material model such as the ideal
gas equation of state (relates p, V, and T) and the calorically perfect assumption (relates
h to T). If we consider the special case where the shock under examination is moving
into a stagnant fluid as depicted in Figure 2.26, we can write the aforementioned three
equations as
ρ1U = ρ 2 (U − up )
(2.164)
p1 + ρ1U 2 = p2 + ρ 2 (U − up )2
(2.165)
1
1
h1 + U 2 = h2 + (U − up )2 = h0 = constant
2
2
(2.166)
The conservation of mass, momentum, and energy equations can be combined as detailed
in Refs. [10,16] to yield the Rankine–Hugoniot relationship. This relationship determines
how the energy changes across a normal shock wave. It is very important and will appear
again when the terminal ballistics material is discussed. It can be written in terms of total
specific energy, e, or if some of the energy components are negligible, it can be written in
terms of enthalpy, h. At this stage, we will use the latter expression, but we shall switch
when we discuss shock in the terminal ballistics section. Writing the Rankine–Hugoniot
relationship in terms of enthalpy, we have
h2 − h1 =
1
1
1
( p2 − p1 )
−
2
ρ
ρ
2
1
(2.167)
The strength of a shock is normally assessed by the change in pressure across it. In other
words, its strength is given by the ratio p2/p1. If we assume the material through which this
shock is propagating is an ideal gas, Equations 2.164 through 2.166 can be combined with
the relationships provided in Equation 2.61 to yield expressions that relate all of the values
ahead of the shock to values after the passage of the shock. The details of this are available
in Ref. [16]. These expressions are as follows:
γ + 1 p2
+
T2 p2 γ − 1 p1
=
T1 p1
γ + 1 p2
1+
γ − 1 p1
© 2014 by Taylor & Francis Group, LLC
(2.168)
Physical Foundation of Interior Ballistics
ρ1
=
ρ2
61
γ + 1 p2
γ − 1 p1
γ + 1 p2
+
γ − 1 p1
1+
(2.169)
The real power of these equations lies in the fact that with just the strength of the shock
known we can determine all of the other items of interest. In the aforementioned equations, we have seen that given the pressure ratio (i.e., the strength) of the shock, we know
the temperature behind the wave and the increase in density across the wave. We can also
determine the wave speed, U, and induced velocity, up, through
U = a1
γ + 1 p2
− 1 + 1
2γ p1
(2.170)
2γ
ρ1 a1 p2
γ +1
up = U 1 − = − 1
ρ 2 γ p1
γ − 1 + p2
γ + 1 p1
(2.171)
If we change reference frames to one in which we are stationary and the shock is moving,
then the assumption of constant stagnation enthalpy, h0, is no longer valid. The reason is
best illustrated by an example. Consider the gas ahead of the shock wave. It was initially
motionless so h1 = h01. After the wave passes, we know that the temperature must increase
so h2 > h1. Additionally, the gas is now moving at velocity, up, so that we can see
h1 = h01 < h02 = h2 +
1 2
up
2
(2.172)
It is by this very same logic that the stagnation pressure, temperature, and density must
also increase.
We have discussed some governing equations but let us break for a moment to discuss
why a gas shocks up. If we examine Equation 2.170 closely, we see that a higher pressure
causes a faster motion of the wave. If we imagine a shock wave as depicted in Figure 2.27
moving to the right, we can pick out three points that we shall follow for some time. Point A
is essentially the beginning of the pressure increase and at the un-shocked initial pressure.
p
C
C
B
C
B
A
B
A
A
t, x
FIGURE 2.27
Formation of a shock wave.
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
62
Point B is at some pressure in between the peak pressure of the shock and the initial
pressure of the material into which the shock is propagating. Point C is at the peak shock
pressure. From Equation 2.170, we see that the local velocity of point B must be greater
than point A and also that the local velocity of point C is greater still. This means that at
some time, t, these points must converge thereby forming a step discontinuity in pressure. This step discontinuity is the way we model the shock—there is actually a very
small distance over which a shock will develop so that the pressure increase is rapid, but
continuous. With this information, we see that compression shocks are the only admissible shocks. Later we will introduce rarefactions that are the converse of shocks. Since
the pressure decreases in a rarefaction wave, the wave will tend to spread out over time
and distance.
Now that we have the governing equations, we shall examine the behavior of a shock
wave in a shock tube. A shock tube is a device as depicted in Figure 2.28 that contains two
regions of gas. These regions are separated by a diaphragm that can be burst very quickly
and contain one gas at high pressure and another at lower pressure. The gases could be
different (thus all of their properties as well) as can their temperatures. Below the graphic
of the shock tube is a pressure versus distance plot showing that the pressure in region
4 (the high-pressure region) is greater than that of region 1 and the diaphragm divides
the two regions. If the diaphragm is burst, then a shock will propagate into the lower
pressure region, increasing the pressure, and a rarefaction wave (to be discussed later)
will propagate into the high-pressure region, decreasing the pressure. If we examine the
shock tube after some very short time, t, the situation will appear as shown in Figure 2.29
with the corresponding pressure–distance profile. One of the most interesting aspects of
compressible fluid flow is that if we know what the initial states of the ideal gases in the
p4, T4, a4, γ4
p1, T1, a1, γ1
4
Diaphragm
p4
p
1
p1
FIGURE 2.28
The shock tube in its initial state. (From Anderson, J.D., Modern Compressible Flow with Historical Perspective, 3rd
edn., McGraw-Hill, New York, 2003. With permission.)
up
4
2
3
U
1
Diaphragm burst
p
p4
p3 = p2
p1
x
FIGURE 2.29
The shock tube after some short time, t. (From Anderson, J.D., Modern Compressible Flow with Historical Perspective,
3rd edn., McGraw-Hill, New York, 2003. With permission.)
© 2014 by Taylor & Francis Group, LLC
Physical Foundation of Interior Ballistics
63
shock tube are, we can predict the pressures and temperatures of the unsteady motion
afterwards by Equations 2.161 through 2.171. In fact, we can predict the pressure behind
the initial shock from
−2γ 4
( γ4 − 1)
a p
(γ 4 − 1) 1 2 − 1
a
p
p 4 p2
4 1
=
1 −
p1 p1
p2
2γ 1 2γ 1 + (γ 1 + 1) − 1
p1
(2.173)
Equation 2.173 needs to be solved for the initial shock strength, p2/p1, but afterwards
Equations 2.161 through 2.171 can be used directly to calculate the parameters of interest.
The details of this derivation can be found in Ref. [16].
You can see from Figures 2.28 and 2.29 that the shock tube is not infinite in extent. At
some point, the shock produced by the bursting of the diaphragm will reach the right end
of the tube. When this occurs, the condition at the wall is such that no flow through it is
possible. Consider that all the fluid behind the shock wave is moving with induced velocity, up, toward the wall. Clearly, this situation is at odds with the wall-imposed boundary
condition of zero velocity. Nature handles this issue by creating a shock wave of strength UR
that propagates back into the fluid that is heading toward the wall at velocity up. Notice that
we have used a velocity here to define the strength of the shock—it should be clear by now
that if we know either the velocity of the shock or the pressure ratio, we can find the other.
The net effect of this reflected shock is that it stagnates the fluid between it and the fixed
end of the shock tube as depicted in Figure 2.30. In this figure, we shall assume the tube is
extremely long on the rarefaction side so we do not have to discuss rarefaction reflections,
yet. If we look at our conservation Equations 2.161 through 2.163 and consider that the shock
wave sees material coming into it at velocity UR + up, we can write equations for the reflected
shock that are analogous to Equations 2.164 through 2.166 for the incident wave. These are
ρ 2 (U R + up ) = ρ 5U R
(2.174)
p2 + ρ 2 (U R + up )2 = p5 + ρ 5U R2
(2.175)
1
1
h2 + (U R + up )2 = h5 + U R2
2
2
(2.176)
up
4
3
2
UR
5
u5 = 0
p4
p
p3 = p2
p5
x
FIGURE 2.30
The shock tube after a reflection of the incident wave.
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
64
A simple method for determination of the speed of the reflected shock is to first determine
the Mach number of the incident pulse, Mas, through
Mas =
U
a1
(2.177)
The relationship between the incident shock velocity and the reflected velocity is derived
in Ref. [16] and given by
MaR
Mas
2(γ − 1)
( Mas2 − 1) γ + 1 2
=
1+
2
2
2
MaR − 1 Mas − 1
Mas
(γ + 1)
(2.178)
Here MaR is the Mach number of the reflected shock that can be converted to a velocity
through use of
MaR =
U R + up
a2
(2.179)
In our discussions on shock waves throughout the terminal ballistics sections, we will make
use of time–distance diagrams, so-called t–x plots. It is prudent to introduce them here as
reinforcement of the shock wave discussion. An t–x plot places distance on the abscissa and
time on the ordinate. Because of this placement, which is opposite to normal function versus time plots, we need to adjust some of our logic that we are used to. For instance, slopes
of straight lines on these diagrams are reciprocal velocities. If we consider the situation in
Figure 2.30 and draw an t–x plot for it with the origin starting from the initial diaphragm
location, we would have a plot as depicted in Figure 2.31. We shall examine the shocks in this
diagram first. If we assume that the incident shock forms immediately (this is not really true,
as we learned earlier, but close enough for our purposes), it propagates toward the wall that is
located at point x2 in our figure. If we wanted to determine what the velocity distribution was
t
u5 = 0
t3
Reflected
shock,
slope = 1/UR
Particle path,
slope = 1/up
t2
x = 0 is
diaphragm
location
Wall
t1
Incident shock,
slope = 1/U
u1 = 0
x1
x3
x2
x
FIGURE 2.31
t–x plot for the reflection of a shock wave. (From Anderson, J.D., Modern Compressible Flow with Historical
Perspective, 3rd edn., McGraw-Hill, New York, 2003. With permission.)
© 2014 by Taylor & Francis Group, LLC
Physical Foundation of Interior Ballistics
65
in this device at any time, t, we would examine a horizontal line in the figure. For instance, if
we examined the situation at time, t1, we would see that the material in the unshaded region
up to point x1 would have a velocity up and everything between x1 and x2 (the wall) would
have zero velocity. Once the incident shock reflects off the wall, a new shock of velocity UR
propagates back into the fluid. This is depicted by the upper line in the diagram. Note that the
slope is greater on this reflected shock, indicative of a lower velocity than the incident wave.
The material in the shaded region behind this wave has been stagnated to zero velocity. We
can use an t–x diagram to determine how a particle moves over time. Consider a particle
located initially at location x1. It remains stationary until the shock wave passes by at time t1,
as indicated by a vertical line. At time t1, the incident shock passes it and induces a velocity,
up to the particle. When the particle moves at velocity, up, it will trace out a line on the diagram that has a slope of 1/up. While this particle is moving at velocity up, the shock interacts
with the wall and reflects at time t2. While the reflected shock is approaching, the observed
particle has no idea anything is about to happen and continues to move at velocity up until
the reflected shock passes by at time t3. This passage of the reflected shock stagnates the particle to zero velocity and its motion (or lack thereof) traces out a vertical line. A final point of
interest regarding t–x plots is that we can actually see the compression of the material. If we
consider all of the material initially between points x1 and x2, we see that, after the passage of
the shock and its reflection, it has all been compressed to the region between x3 and x2. With
this information, the basis for our future discussions using t–x plots is established.
A rarefaction wave, sometimes known as an expansion or relief wave, is the means by
which nature handles a sudden drop in pressure. As we stated earlier, compression waves
(also known as condensations) eventually coalesce into shocks that are analyzed as step
discontinuities in pressure. This coalescence was brought about by the fact that the local
velocity increases with increasing pressure. In a rarefaction, the opposite is true. A rarefaction increases over time because the pressure at the head of the wave is greater than that
at the tail of the wave. In the case of our shock tube, the head of the rarefaction will propagate at the local speed of sound in the material (a4 in Figure 2.29), while the tail will propagate
at velocity (u3 − a3) that is equal to (up − a2). This is depicted schematically in Figure 2.32.
Tail slope = 1/UT = 1/(u3–a3)
t
3
u = u3
a = a3
x = 0 is
diaphragm
location
Particle path,
slope = 1/u3 = 1/up
Head slope = 1/UH = 1/(u4–a4) = –1/a4
3
4
u = u4 = 0
a = a4
4
x
FIGURE 2.32
(See color insert.) t–x plot for a rarefaction wave.
© 2014 by Taylor & Francis Group, LLC
66
Ballistics: Theory and Design of Guns and Ammunition
Throughout the rarefaction wave, the velocity continuously decreases between these two
values. Because of this continuous decrease in velocity, it is common to model the decrease
as a series of wavelets. The more wavelets we include, the smoother the curve. If we use
Figure 2.32 to trace a particle path after the bursting of the diaphragm, we see that the particle would not move until the head of the rarefaction wave passed by it. After the passage
of the head of the wave, the velocity would continuously increase until passage of the tail
of the wave, after which it would be moving at velocity up. The length of the rarefaction
can be determined at any time by scribing a horizontal line through the diagram. If we do
this at two points in time on the diagram, we can see how the length of the wave increases.
What is depicted in Figure 2.32 is a simple, centered rarefaction wave. A wave is considered simple if all of the characteristics (the rays emanating from the origin) are straight.
Reflections of a rarefaction are somewhat more complicated than that of a shock. The
reflection of the head of the rarefaction wave must pass through the characteristics of the
rest of the wave being both affected by as well as affecting them. The result is that the
characteristics tend to bend making the calculations somewhat more complex. We will
handle this in a simplified fashion later, but the interested reader is directed to Ref. [16] for
an outstanding treatment for handling these situations.
We now have sufficient information to handle the fluid mechanics of interior and exterior ballistics. We shall treat the formation of shocks and rarefactions as necessary in the
terminal ballistics section.
Problem 18
The principle behind a muzzle brake on a gun is to utilize some of the forward momentum of the propelling gases to reduce the recoil on the carriage. In the simple model below
(Figure 2.33), the brake is assumed to be a flat plate with the jet of gases impinging upon
it. If the jet diameter is 105 mm and the velocity and density of the gas (assume air) are
750 m/s and 0.457 kg/m3, find the force on the weapon in Newtons assuming the gases
are directed 90° to the tube and the flow is steady.
Answer: −2225.9 [N]
Problem 19
Some engineer gets the idea that if deflecting the muzzle gases to the side is a good idea,
then deflecting it rearward would be better (until of course an angry gun crew gets hold
of him). If the jet diameter is again 105 mm and the velocity and density of the gas (again
assume air) are 750 m/s and 0.457 kg/m3, find the force on the weapon in Newtons assuming the gases are directed 150° to the tube and the flow is steady (Figure 2.34).
Answer: −4153.5 [N]
F
FIGURE 2.33
Normal deflection of flow through a muzzle brake.
© 2014 by Taylor & Francis Group, LLC
Physical Foundation of Interior Ballistics
67
F
30°
(typ.)
FIGURE 2.34
Rearward deflection of flow through a muzzle brake.
Problem 20
Consider a shock tube that is 6 ft long with a diaphragm at the center. Air is contained in
both sections (γ = 1.4). The pressure in the high-pressure region is 2000 psi. The pressure in
the low pressure region is 14.7 psi. The temperature in both sections is initially 68°F. When
the diaphragm is burst, determine the following:
1. The velocity that the shock wave propagates into the low–pressure region.
Answer: 2798 [ft/s]
2. The induced velocity behind the wave.
Answer: 1946 [ft/s]
3. The velocity of a wave reflected normally off the wall (relative to the laboratory).
Answer: 1232 [ft/s]
4. The temperature behind the incident wave.
Answer: 657 [°F]
5. Draw an t–x diagram of the event. Include the path of a particle located 2 ft from
the diaphragm.
Problem 21
An explosion generates a shock wave in still air. Assume we are far enough from the initial explosion that we can model the wave as a one-dimensional shock. Assume that the
pressure generated by the explosion was 10,000 psi and the ambient atmospheric pressure, density, and temperature are 14.7 psi, 0.06 lbm/ft3 and 68°F, respectively. Determine
1. The static pressure behind the wave (assume γ = 1.4 and since we are far away
from the effects of the explosion assume a1/a4 ≈ 0.5)
Answer: p2 = 376.6 [psi]
2. The velocity that the wave propagates in still air
Answer: U = 5294 [ft/s]
3. The induced velocity that a building would see after the wave passes
Answer: up = 4212 [ft/s]
4. The velocity of a wave reflected normally off a building
Answer: UR = 1921 [ft/s]
© 2014 by Taylor & Francis Group, LLC
68
Ballistics: Theory and Design of Guns and Ammunition
References
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2. Jones, L. and Atkins, P., Chemistry, Molecules, Matter, and Change, 4th edn., W.H. Freeman and
Co., New York, 2003.
3. Masterson, W., Slowinski, E., and Stanitski, C., Chemical Principles, 6th edn., Saunders College
Publishing, Philadelphia, PA, 1985.
4. Beer, F.P. and Johnson, R., Mechanics of Materials, 2nd edn., McGraw-Hill, New York, 1992.
5. Van Wylen, G.J. and Sonntag, R.E., Fundamentals of Classical Thermodynamics, 3rd edn., John
Wiley and Sons, New York, 1986.
6. Fermi, E., Thermodynamics, Dover Publications, New York, 1956.
7. Corner, J., Theory of the Interior Ballistics of Guns, John Wiley and Sons, New York, 1950.
8. Cooper, P.W., Explosives Engineering, Wiley-VCH Inc., New York, 1996.
9. Sucec, J., Heat Transfer, William C. Brown Publishers, Dubuque, IA, 1985.
10. Kays, W.M. and Crawford, M.E., Convective Heat and Mass Transfer, 3rd edn., McGraw-Hill,
New York, 1993.
11. Fox, R.W. and McDonald, A.T., Introduction to Fluid Mechanics, 4th edn., John Wiley and Sons,
New York, 1992.
12. Anderson, J.D., Modern Compressible Flow with Historical Perspective, 3rd edn., McGraw-Hill,
New York, 2003.
13. Cengel, Y.A. and Boles, M.A., Thermodynamics and Engineering Approach, 4th edn., McGraw-Hill,
New York, 2002.
14. Moran, M.J. and Shapiro, H.N., Fundamentals of Engineering Thermodynamics, 5th edn., John
Wiley and Sons, New York, 2004.
15. Masterson, W.L., Slowinski, E.J., and Stanitski, C.L., Chemical Principles, 5th edn., Saunders
College Publishing, Philadelphia, PA, 1981.
16. Cooper, P.W. and Kurowski, S.R., Introduction to the Technology of Explosives, Wiley-VCH, New
York, 1996.
17. Hayes, T.J., Elements of Ordnance, John Wiley and Sons, New York, 1938.
18. Eringen, A.C., Liebowitz, H., Koh, S.L., and Crowley, J.M., Eds., Mechanics and Chemistry of Solid
Propellants, Proceedings of the Fourth Symposium on Naval Structural Mechanics, Pergamon Press,
London, U.K., 1965.
19. Kubota, N., Propellants and Explosives, Wiley-VCH, New York, 2002.
20. Turns, S.R., An Introduction to Combustion, 2nd edn., McGraw-Hill, New York, 2000.
21. Borman, G.L. and Ragland, K.W., Combustion Engineering, WCB-McGraw-Hill, New York, 1998.
22. Yang, V., Brill, T.B., and Ren, W.-Z., Solid propellant chemistry, combustion, and motor interior
ballistics, Progress in Astronautics and Aeronautics, Vol. 185, American Institute of Astronautics
and Aeronautics, Reston, VA, 2000.
23. Kuhl, A.L., Leyer, J.C., Borisov, A.A., and Sirignano, W.A., Dynamics of deflagrations and reactive systems flames, Progress in Astronautics and Aeronautics, Vol. 131, American Institute of
Astronautics and Aeronautics, Washington, DC, 1989.
24. Kuhl, A.L., Leyer, J.C., Borisov, A.A., and Sirignano, W.A., Dynamics of gaseous combustion, Progress in Astronautics and Aeronautics, Vol. 151, American Institute of Astronautics and
Aeronautics, Washington, DC, 1993.
25. White, F.M., Fluid Mechanics, 5th edn., McGraw-Hill, New York, 2003.
26. Panton, R.L., Incompressible Flow, 2nd edn., John Wiley and Sons, New York, 1995.
27. Currie, I.G., Fundamental Mechanics of Fluids, 2nd edn., McGraw-Hill, New York, 1993.
28. White, F.M., Viscous Fluid Flow, 3rd edn., McGraw-Hill, New York, 2006.
© 2014 by Taylor & Francis Group, LLC
3
Analytic and Computational Ballistics
Chapter 2 has provided us with the necessary background to discuss procedures that calculate the behavior of projectiles and propellant in the gun tube. The chapter had to be
brief because detailed treatment of any one of the subjects could be (and are) collected into
complete texts in their own right. The reader is directed to the references at the end of the
chapter if a more complete background in the individual subject is felt to be necessary.
Much like other introductory texts on difficult subjects, this chapter shall begin with
fundamental treatments that will allow the reader to perform meaningful calculations of
interior ballistic problems. This simplified treatment will, by its very nature, not provide
exact answers but answers that are reasonable from an engineering viewpoint. As will be
discussed, more exact methods require a varying degree of computer assets.
3.1 Computational Goal
The interior ballistician is charged with devising a propellant charge that will deliver the
projectile of interest to the gun muzzle intact, with the desired muzzle velocity, with no damage to the weapon from excess pressure, and with high probability that successive charges
propelling the same projectiles will produce the same results. To do this, the ballistician
must be able to predict a priori what the charge will do, i.e., what pressures will both the gun
and the projectile experience during travel down the bore and what the velocity and acceleration profile would be during the travel to the muzzle. Over the centuries, ballisticians,
including some quite eminent mathematicians and physicists, have devised computational
schemes that can be used to make such predictions. We intend to explore a few of these analytic tools in sufficient depth so that the physics and mathematics become clear to the user,
who would then also be able to discern reasonable answers from patently erroneous ones.
It is important to understand how predictions of pressure and velocity are verified
experimentally in real guns. Such understanding has led to the development of pressure ratios that allow the gun and projectile designers to know what pressures are acting on the gun and on the projectile at locations that practical instrumentation has some
difficulty capturing. Pressure is most readily measured at the base of the gun chamber,
where the gas flow is minimal or nonexistent. When pressure taps are introduced along
the bore to take measurements while the projectile is traveling and the gases are flowing,
it has been found that turbulent flow and shock waves make such measurements difficult to interpret. Copper crusher gauges are used in which small copper cylinders are
crushed to a barrel shape in the gauge by the applied pressure and the distortion of the
cylinders measured. These gauges are placed in the base of the charge and recovered after
firing. Distortion is checked against a calibration chart and the pressure is quickly read.
Of course, pressure measured in this way is representative only of the maximum pressure
sensed by the gauge, which gives no indication of its profile in time or in travel.
69
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
70
Even such a primitive measurement was and still is of use; because the designer
would know the maximum pressure, the projectile and gun would have to contend
with an indication that piezo-type pressure gauges are functioning properly. These
gauges are still widely used to check the pressure consistency of already developed
charges. Knowledge of how that copper pressure was related to pressures at other locations during the travel was a great advance. When the pressure ratios were devised that
related chamber pressure to the pressure at the base of the projectile during its travel
down the bore, these were greatly appreciated by the designers. Even better was the
introduction of electronic piezo gauges installed through the breech that allowed the
measurement of pressure over time so that a pressure–time profile could be available.
The study of a few of the computational theories that develop these ratios follows in
succeeding sections.
3.2 Lagrange Gradient
To determine the time-dependent motion of the projectile, we need to make some assumptions about the behavior of the gas pushing it out of the gun. These assumptions will
involve the pressure, mass, and density distribution of the gas (see Figure 3.1). We shall
continue to use x as the distance from the projectile base position at the seating location to
its position at all later times with the time derivative defined as
dx
= xɺ = V
dt
(3.1)
xpmax
p
pmax
pmuz
x=L
pB
p
FIGURE 3.1
Pressure–distance relationship in a typical gun firing.
© 2014 by Taylor & Francis Group, LLC
pS
x
Analytic and Computational Ballistics
71
We will first assume that the gas density is uniform in the volume behind the projectile
at time t. We can then write, for any time, t, that
ρ = ρ ( xg , t )
(3.2)
In this equation, xg is the x-location of the gas mass center behind the projectile. We shall
also assume that there is no spatial gradient in density at any time, thus
∂ρ
=0
∂xg t
(3.3)
We can also write the continuity equation for a compressible fluid as
∂ρ
∂
ρVx g = 0
+
∂t ∂xg
(
)
(3.4)
We can expand the continuity equation (3.4) as
∂Vx g
∂ρ ∂ρ
Vx g + ρ
+
=0
∂t ∂xg
∂xg
(3.5)
Inserting our assumption of the absence of a spatial density gradient allows us to simplify
this expression to
∂Vx g
∂ρ
+ρ
=0
∂t
∂xg
(3.6)
Now because we stated that the density was not a function of x, we can remove the partial
derivative notation from the temporal term and rearrange to yield
∂Vx g
1 dρ
=−
ρ dt
∂xg
(3.7)
Assume at this point that the solid propellant in the charge has all turned to gas, then
what was initially a solid propellant of charge weight, c, is now a gas of identical weight, c.
So the gas density is this weight divided by the volume the gas occupies, or
ρ (t)|c =
c
V(t)
(3.8)
where the subscript “c” refers to conditions after the charge has burned out, i.e., all the
solid has evolved into gas. If the base of the projectile has moved a distance, x, and the bore
area is A, then the volume behind the projectile containing gas is
V(t) = Ax(t)
© 2014 by Taylor & Francis Group, LLC
(3.9)
72
Ballistics: Theory and Design of Guns and Ammunition
If we insert Equation 3.9 into Equation 3.8 and then take the derivative with respect to
time, the result can be simplified to Equation 3.10:
1 dx ∂Vx g
=
x dt
∂xg
(3.10)
Note that there is a difference here between x and xg:
x is the location of the base of the projectile
xg is the location of the mass center of the gas
If we integrate Equation 3.10 with respect to xg and use the boundary conditions of Vxg = 0
when xg = 0, then we get
xg dx
= Vxg ( xg )
x dt
(3.11)
Now, since x is the position of the base of the projectile at time t we see that dx/dt is the
velocity of the projectile at time t, so we can write
V Vxg
=
x
xg
(3.12)
This implies that the gas particle velocity varies linearly from the breech face to the projectile base, and is a fundamental tenet of the Lagrange* approximation. We can describe the
kinetic energy of the gas stream as
KEg =
1
mgVx2g
2
(3.13)
But, as described earlier, the mass of the gas is its density times the volume it occupies at
time t, therefore
x
KEg =
1
∫ 2 ρ AV
2
xg
dxg
(3.14)
0
Moving the spatially constant terms, ρA/2, outside the integral and performing the integration gives us
ρ A V 2 xg3
KEg =
2 x2 3
x
=
0
1
ρ AxV 2
6
(3.15)
But we know from our earlier work that
ρ Ax = c
(3.16)
So we can write
KEg =
1 2
cV
6
* Joseph-Louis Lagrange, 1736–1813, Italian/French mathematician.
© 2014 by Taylor & Francis Group, LLC
(3.17)
Analytic and Computational Ballistics
73
The total kinetic energy of the system (neglecting recoil) is
KE g =
3
ρ A V 2 xg
2 x2 3
x
1
cV
6
=
0
(3.18)
But the kinetic energy of the projectile is
KEshot =
1
wpV 2
2
(3.19)
where wp is the projectile mass.
So the Lagrange approximation for kinetic energy is
KEtot =
1
1
1
c
wpV 2 + cV 2 = wp + V 2
2
6
2
3
(3.20)
In this development the volume of gas is assumed to be a cylinder of cross-sectional
area A. In reality, it is not; while the bore is cylindrical, the chamber is not. Chamber
diameters can be much greater than bore diameters. To account for this, an effort to modify the Lagrange gradient approximations has been performed [1]. This will be explored
subsequently. The changes from the Lagrange gradient will be found to be small but not
insignificant and the so-called chambrage gradient will be explained in Section 3.3 and
incorporated in the discussion of numerical methods in Section 3.4.
We can describe the linear momentum of the gas stream as
Mom g = mgVxg
(3.21)
But, again, the mass of the gas is its density times the volume it occupies at time t, therefore
x
∫
Mom g = ρ AVxg dxg
(3.22)
0
We can use our continuity relationship in Equation 3.11 to write
x
x
xg dx
xg
Mom g =ρ A
dxg = ρ A x V dxg
x
t
d
0
0
∫
∫
(3.23)
1
ρ AxV
2
(3.24)
Performing the integration gives us
Mom g = ρ A
© 2014 by Taylor & Francis Group, LLC
V xg2
x 2
x
=
0
Ballistics: Theory and Design of Guns and Ammunition
74
If we recall Equation 3.16, we can write
Mom g =
1
cV
2
(3.25)
The total linear momentum of the system (neglecting the weapon) is
Mom tot = Mom shot + Mom g
(3.26)
The linear momentum of the projectile is
Mom shot = wpV
(3.27)
So the Lagrange approximation for linear momentum is
Mom tot = wpV +
1
c
cV = wp + V
2
2
(3.28)
Because we are looking for the parameters, we can readily measure breech pressure and
muzzle velocity, and we must develop predictive equations for them, i.e., equations for
pressure in terms of charge parameters and equations of motion of the projectile. To do
this, we adopt a Lagrangian approach to track the motion of a particle of gas. What follows
is a derivation for the equation of motion for an element of gas. For a rigorous, complete
treatment, see any text on fluid mechanics, (e.g. [2]).
For differentiation that tracks a fluid element (the Lagrangian approach), the following
differential operator (called the substantial derivative or material derivative) is used:
D
∂
∂
∂
∂
= +u +v
+w
Dt ∂t
∂x
∂y
∂z
(3.29)
where u, v, and w are the velocity components in the x, y, and z directions, respectively.
If we consider a one-dimensional flow operating on the velocity Vxg ( x) (here Vxg is the
axial velocity and replaces u given earlier)
DVxg ∂Vxg
∂Vxg
=
+ Vxg
∂t
∂x
Dt
(3.30)
In vector notation, the gradient of a function is
∇=i
∂
∂
∂
+j +k
∂x
∂y
∂z
(3.31)
Force is the time rate of change of momentum
F=
© 2014 by Taylor & Francis Group, LLC
∂
( mv )
∂t
(3.32)
Analytic and Computational Ballistics
75
It can be shown using Gauss’s theorem [3] that the rate of change of linear momentum of
the fluid inside a surface S in changing to surface S’ in time, dt, is
dv
∫ ρ dt dV
(3.33)
V
From the equations of motion for an inviscid fluid we know that the total force equals
the pressure on the boundary element integrated over the boundary plus the body force F
integrated over the mass in S, or
∫ pndS +∫ FρdV = −∫ ∇pdV +∫ FρdV
V
S
V
(3.34)
V
Because by Gauss’s theorem
∫ pndS = − ∫ ∇pdV
(3.35)
V
S
Setting the RHS of Equation 3.35 equal to Equation 3.33, we get
dv
∫ Fρ − ∇p − ρ dt dV = 0
(3.36)
V
Since V is chosen arbitrarily, the sum in brackets must equal zero
Fρ − ∇p − ρ
dv
=0
dt
(3.37)
In the absence of a body force F, we can rewrite this as
dv
1
= − ∇p
dt
ρ
(3.38)
We can write Equation 3.38 as follows for one-dimensional flow and negligible body forces
∂Vxg
∂Vxg
dp
+ Vxg
= −ρ
∂xg
∂
dxg
t
(3.39)
Note here that we have used the substantial derivative for the velocity of the gas stream.
If we insert the relationship for the gas stream velocity we obtained through the continuity equation (3.11) into Equation 3.39, we can write
∂ xg dx
∂Vxg
dp
= −ρ
+ Vxg
∂
∂xg
t
x
t
dxg
d
© 2014 by Taylor & Francis Group, LLC
(3.40)
Ballistics: Theory and Design of Guns and Ammunition
76
or
∂ xg dx xg dx ∂ xg dx
dp
= −ρ
+
dxg
∂t x dt x dt ∂xg x dt
(3.41)
We can combine terms in Equation 3.41 as follows:
xg dx 2 xg d 2 x xg dx 2
dp
+
= − ρ − 2
+
x dt 2 x 2 dt
dxg
x dt
(3.42)
Simplifying the expression gives us
xg d 2 x
dp
= −ρ
x dt 2
dxg
or
xg
dp
= −ρ
xɺɺ
x
dxg
(3.43)
If we use our relationship between density and charge weight in Equation 3.8, we can write
cxg
dp
=−
xɺɺ
Ax 2
dxg
(3.44)
We can integrate this expression with respect to the gas mass center as
xg
∫
0
xg
dp
c
dxg = −
xɺɺ xgdxg
dxg
Ax 2
∫
(3.45)
0
Performing the integration yields
p=−
cxg2
xɺɺ + constant
2 Ax 2
(3.46)
Let us now define
pS = pressure at the projectile base
pB = pressure at the breech
p = mean pressure in volume behind projectile
pR = pressure resisting projectile motion (force/bore area)
We will develop the equations of motion both with a resistive force in the bore (such as
friction and the air being compressed in front of the projectile) and neglecting the resistance. If we write Newton’s second law for a projectile being acted upon by propellant
gases, we have
wxɺɺ = ApS
© 2014 by Taylor & Francis Group, LLC
(3.47)
Analytic and Computational Ballistics
77
Writing this in terms of the acceleration we get
xɺɺ =
A
pS
w
(3.48)
where w is the projectile mass. Since the base of our projectile is at location x and the local pressure on the base is pS, we can substitute these values into Equation 3.46 for xg and p to obtain
pS = −
c
xɺɺ + constant
2A
(3.49)
Keep in mind that this is a local condition that we applied to the gas in the vicinity of the
base (that gas’s mass center is approximately at x). We can rearrange Equation 3.49 to yield
our constant of integration:
c
xɺɺ
2A
(3.50)
c A
c
pS = 1 +
p
2A w
2w
(3.51)
constant = pS +
If we use Equation 3.48, we obtain
constant = pS +
Inserting this constant back into our Equation 3.46 gives us
p=−
cxg2 A
cxg2
c
c
ɺɺ
=
−
+
+
x
p
1
S
pS + 1 +
pS
2 Ax 2
2w
2 Ax 2 w
2w
(3.52)
or
xg2 c
p = pS + pS 1 − 2
x 2w
(3.53)
This equation relates the pressure at the base of the projectile to that at the location of the
gas mass center. By similar logic, at the breech, xg = 0, and the pressure, p = pB, so we can
substitute the values into Equation 3.53 to obtain a relationship between the breech pressure and the pressure at the projectile base
pB = pS + pS
c
c
= pS 1 +
2w
2w
(3.54)
The space-mean pressure is formally defined as
x
1
p=
pdxg
x
∫
(3.55)
0
If we insert Equation 3.53 into this equation, we get
x
p=
1
x
∫
0
© 2014 by Taylor & Francis Group, LLC
xg2 c
pS + pS 1 − 2
dxg
x 2w
(3.56)
Ballistics: Theory and Design of Guns and Ammunition
78
Solving this integral, inserting the limits of integration, and simplifying yields
x
1
c
c xg3
p = pS xg + pS
xg − pS
2w
2w 3 x 2 0
x
(3.57)
Inserting the limits of integration gives us
p = pS + pS
1
c
c
− pS
2w 3 2w
(3.58)
Simplifying we get
c
p = pS 1 +
3
w
(3.59)
This equation relates the space-mean pressure to the base pressure acting on the projectile.
We now have equations that relate breech pressure to base pressure (Equation 3.54) and
space-mean pressure to base pressure (Equation 3.59). What is missing is a relationship
between breech pressure and space-mean pressure. We can arrive at the desired result by
dividing Equation 3.59 by Equation 3.54, simplifying to yield
c
pS 1 +
p
3w
=
c
pB
pS 1 +
w
2
(3.60)
For easier manipulation, it is sometimes desirable to expand Equation 3.60 in a Taylor
series, which, neglecting higher order terms, would be
p
c
= 1−
+⋯
pB
6w
(3.61)
To account for the effects of bore resistance, we again write Newton’s second law for a
projectile being acted upon by propellant gases and bore friction as
w1xɺɺ = A( pS − pR )
(3.62)
Here we have used w1 to represent the mass of the projectile (you will see why later) and
have included a resistive pressure, pR, that fights the gas pressure. Note that the resistive
pressure is simply the resistive force divided by the bore cross-sectional area so that the
terms in the aforementioned equation can be conveniently grouped—it is not actually a
pressure at all. Writing this in terms of the acceleration we get
xɺɺ =
© 2014 by Taylor & Francis Group, LLC
A
( pS − pR )
w1
(3.63)
Analytic and Computational Ballistics
79
Again, since the base of our projectile is at location x and the local pressure on the base is
pS, we can substitute these values into Equation 3.46 for xg and p to obtain
pS = −
c
xɺɺ + constant
2A
(3.64)
Remember that this is a local condition that we applied to the gas in the vicinity of the base
where the gas’s mass center is approximately at x.
Following the same procedure that we used to arrive at a general expression for pressure, but now with bore resistance, we rearrange Equation 3.64 to find the constant of
integration, and with simplification arrive at
c
xɺɺ
2A
(3.65)
c A
c
c
( pS − pR ) = 1 +
pS − 2w pR
w
2 A w1
2
1
1
(3.66)
constant = pS +
If we use Equation 3.63, we obtain
constant = pS +
Inserting this constant back into Equation 3.64 gives us
cxg2
c
c
xɺɺ + 1 +
pS −
pR
2Ax 2
2w1
2w1
(3.67)
cxg2 A
c
c
pR
( pS − pR ) + 1 +
pS −
2
2 Ax w1
2w
2w1
(3.68)
p=−
or
p=−
or
p = pS + ( pS − pR )
xg2
c
1− 2
x
2w1
(3.69)
which relates the pressure at the base of the projectile to the pressure at the gas mass center, but with the effect of bore friction included. Having this general equation we can again
proceed as we did earlier to find equations that relate breech to base pressure, space-mean
to base pressure, and space-mean to breech pressure for the bore friction case. These are
c
c
pS −
pR
2w1
2w1
(3.70)
c
c
p = pS 1 +
− pR
3w1
3 w1
(3.71)
pB = pS +
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
80
p
1+ 1 − R
pS
p
=
pB
p
1+ 1 − R
pS
c
3w1
c
2w1
(3.72)
If we plot breech, space-mean, and base pressure versus x, the position of the projectile
base, we shall see that a gradient of pressure exists in which the breech pressure is always
the greatest and the base pressure is always the smallest. This is the so-called Lagrange
gradient and is fundamental to our modeling of the propellant gas. There are instances
where this gradient is reversed and this usually means that we have a problem—a socalled negative delta-p. This is indicative of a fragmented propellant charge caused by
poor ignition. A charge designed to move with the accelerating projectile, the traveling
charge, is a notable exception.
We are essentially prepared now to treat the F in the equation F = ma, which is in its
simplest form, the base pressure times the base area. We now need to determine what
generates the pressure, what the acceleration of the projectile will be, and how the acceleration and the ever-increasing volume behind the projectile affect the pressure. To do this,
we shall review the equations from our initial discussions of propellant burning as well
as revisiting our notation before moving on to combining everything into the equations of
motion of the projectile.
We have previously defined the following quantities and shall simply list them here for
..
ease of reference. The first quantity is the projectile’s acceleration, x . The pressure acting
on the base of the projectile is the stimulus that causes the acceleration
pS (t) = pressure at the base of the projectile at time t.
We usually measure pressure at the breech of the weapon and it is this pressure
that we are determining when we examine the burning of the propellant. We need
to constantly refer this breech pressure to the base pressure. We do this by invoking
the Lagrange gradient assumption, keeping in mind that we begin by neglecting bore
resistance
c
pB = pS 1 +
2w
(3.73)
We can write Newton’s second law for the force on the projectile base as
wxɺɺ = pS A
(3.74)
If we substitute our Lagrange gradient into this equation to put it in terms of the breech
pressure and the projectile velocity, we can write
w
© 2014 by Taylor & Francis Group, LLC
pB
dV
=
A
c
dt
1
+
2w
(3.75)
Analytic and Computational Ballistics
81
If we want to include losses, w can be replaced by w1, an effective projectile mass that
can be thought of as an added mass due to the combination of resistance of bore friction,
engraving by the rifling, resistance due to compression of the air ahead of the shot, etc.
Then we have
c dV
= pB A
w1 +
2
dt
(3.76)
The burning of the propellant generates the pressure that pushes on the projectile. Let us
now recall the equation that relates the amount of propellant turned to gas
φ = (1 − f ) (1 + θ f )
(3.77)
Also recall that the rate of gas evolution (burning) is a function of the pressure
D
df
= − β p ≈ β pB
dt
(3.78)
In our earlier study of solid propellant combustion, we developed an equation of state
for the gas that related ϕ to the pressure and the distance the projectile traveled
cλφ
pB ( x + l ) =
A
c
1 + 2w
1
1+ c
3w1
(3.79)
Finally, we have our equation of motion for the projectile
c dV
= pB A
w1 +
2
dt
(3.80)
whose initial conditions are x = 0, V = 0, f = 1 at t = 0.
These equations may be manipulated to determine the parameters of interest as functions of the fraction of the remaining web f = f(t)
x = projectile travel
V = projectile velocity
pB = breech pressure
If we combine Equations 3.79 and 3.80, eliminating the breech pressure between them,
we can write
−
© 2014 by Taylor & Francis Group, LLC
c dV
D df w1
=
1+
β dt A
2w1 dt
(3.81)
Ballistics: Theory and Design of Guns and Ammunition
82
We can rearrange this to get the equation in terms of the projectile acceleration
DA 1 df dV
=
−
β w1 1 + c dt dt
2w1
(3.82)
This can be integrated resulting in
V=−
AD
f + constant
c
β w1 1 +
2w1
(3.83)
If we insert the initial conditions that V = 0 when f = 1, Equation 3.83 yields
constant =
AD
c
β w1 1 +
2w1
(3.84)
This gives us
V (t) =
AD
(1 − f (t))
c
β w1 1 +
2w1
(3.85)
From the aforementioned equation, we can rearrange Equation 3.82 as follows:
dV
=−
dt
df
AD
c dt
β w1 1 +
2w1
(3.86)
We can now substitute our relationship between velocity and fraction of web remaining
(Equation 3.86) into our projectile equation of motion (Equation 3.80), algebraically simplifying it and inserting the relationship for base pressure (Equation 3.79) to yield
c
1+
cλφ
D df
2w1
−
=
β dt A( x + l) 1 + c
3w1
(3.87)
This may be rearranged to obtain
c
1+
df
cλφβ
2w1
=−
AD ( x + l) 1 + c
dt
3w1
© 2014 by Taylor & Francis Group, LLC
(3.88)
Analytic and Computational Ballistics
83
Using the chain rule transformation between distance and time
df df dx
df
=
=V
dt dx dt
dx
(3.89)
df
1 df
=
dx V dt
(3.90)
This can be written as
Now let us substitute Equations 3.85 and 3.88 into Equation 3.90, simplify the result and
yield
2
c
1
+
df
w1cλφβ 2
2w1
=− 2 2
dx
A D ( x + l)(1 − f ) 1 + c
w
3 1
(3.91)
To examine the rate of change of f, the fraction of web remaining, with the travel distance, x, we take the reciprocal of Equation 3.91
c
1+
A D (1 − f )
dx
3w1 ( x + l)
=−
2
df
w1cλφβ 2
c
1 +
2w1
2
2
(3.92)
Here l is an initial chamber length, to be described subsequently.
By inserting the relationship between ϕ and f, from Equation 3.77 we get
c
1+
dx
AD
3w1 ( x + l)
=−
2
df
w1cλβ 2
c (1 + θ f )
1+
2w1
2
2
(3.93)
Equation 3.93 is cumbersome and following Corner [4] we find that we can define a
dimensionless central ballistic parameter, M, that is a function of the gun, the charge, and
the projectile, i.e., the system
c
1+
AD
3w1
M=
2
w1cλβ 2
c
1+
2w1
2
© 2014 by Taylor & Francis Group, LLC
2
(3.94)
Ballistics: Theory and Design of Guns and Ammunition
84
This simplifies our distance–web fraction relationship to
dx
( x + l)
= −M
df
(1 + θ f )
(3.95)
The dimensionless nature of M can be shown if we note that c and w1 are mass units. We
can also write the units of the burning rate coefficient as
D df
L2T
[β ] =
⇒
[
β
]
=
M
pB dt
(3.96)
The units of the propellant force, λ, are
2
energy ML L L
[λ ] =
=
× = = [velocity ]2
2
M T
mass T
(3.97)
Using these in our definition of the central ballistic parameter, we can show
L6
[ M] =
= [0]
2
2
2
L
T
L
M×M
M
T
(3.98)
thereby demonstrating that M is dimensionless. Equation 3.95, repeated here,
dx
( x + l)
= −M
df
(1 + θ f )
(3.95)
shows how M relates the burning of the propellant, f, with the expansion of the volume
represented by x, the travel. A similar concept appears in all interior ballistic theories.
We are now in a position to compute the parameters that the interior ballistician really
seeks, the projectile’s velocity and the concomitant instantaneous breech pressure for each
point along its travel down the tube. If we wish to know the pressure on the base of the
projectile or the space-mean pressure in the volume behind the projectile, we need only
apply the appropriate Lagrange approximation to the breech pressure. This is an extraordinary result. By simply understanding the amount of propellant burnt and some gun or
propellant or projectile data, we have determined everything we need to know about the
interior ballistics.
We can now take the distance–web fraction relationship and integrate it directly. But we
must examine two distinct cases for θ, the form factor of the grain. One where θ ≠ 0 and
one where θ = 0. Let us separate the variables in Equation 3.95 to obtain
df
dx
= −M
( x + l)
(1 + θ f )
© 2014 by Taylor & Francis Group, LLC
(3.99)
Analytic and Computational Ballistics
85
Then we can write for θ ≠ 0
x
∫
0
f
df
dx
= −M
( x + l)
(1 + θ f )
∫
(3.100)
0
or, for θ = 0
f
x
∫
0
dx
= − M df
( x + l)
∫
(3.101)
0
Evaluation of the integral equation (3.100) for θ ≠ 0 gives us
ln( x + l) = −
M
−M
ln(1 + θ f ) + ln(K ) = ln K (1 + θ f ) θ
θ
(3.102)
Solving for K with the initial conditions, f = 1 at x = 0 we get
M
K = l(1 + θ ) θ
(3.103)
This constant, when inserted in the original Equation 3.102, gives us
M
1+θ θ
x + l = l
1+θ f
(3.104)
In a similar fashion, we can evaluate Equation 3.101 to give us the distance-remaining web
fraction relation for θ = 0
x + l = le M ( 1 − f )
(3.105)
We now know how the web fraction, f, varies with distance, and have, incidentally,
shown the algebraic simplification inherent in the central ballistic parameter, M. We can
now pursue a relationship between pressure and web fraction. If we look at Equation 3.88,
we see the quotient on the RHS and note that this occurs frequently. We define it as our
Lagrange ratio, RL, another simplification.
c
2w1
RL =
c
1+
3w1
1+
(3.106)
This will allow us to rewrite Equation 3.79 in simpler form as
pB ( x + l ) =
© 2014 by Taylor & Francis Group, LLC
cλφ
RL
A
(3.107)
Ballistics: Theory and Design of Guns and Ammunition
86
We will make an assumption that the chamber and bore diameters are the same and
relate the volume behind the projectile to a fictitious chamber length, l. (We will correct
this subsequently when we examine the chambrage gradient.)
Vi = Al = U −
c
δ
(3.108)
In this expression, U is the empty chamber volume and c/δ is the volume occupied by the
solid propellant charge.
We continue by substituting Equations 3.108 and 3.77 into Equation 3.107 and rearranging to give our relationship between the breech pressure and the fraction of remaining
web for θ ≠ 0:
M
λ cRL
1 −θ f θ
pB =
(1 − f ) (1 + θ f )
Vi
1 −θ
for θ ≠ 0
(3.109)
We can also proceed in similar fashion for θ = 0 by substituting Equations 3.105 and 3.77
into Equation 3.107 to find the relationship between the breech pressure and the fraction
of remaining web:
pB =
λ cRL
(1 − f ) (1 + θ f )exp − M(1 − f ) for θ = 0
Vi
(3.110)
Summarizing, we now have the definition of the central ballistic parameter (Equation 3.94)
and equations that relate velocity as a function of remaining web (Equation 3.85) and travel
as a function of remaining web for different form functions (Equations 3.104 and 3.105)
as well as breech pressure as a function of remaining web for different form functions
(Equations 3.109 and 3.110). With these we can now integrate the governing equations and
find solutions for velocity at peak pressure, at all-burnt point of travel, and at muzzle exit.
Equations 3.109 and 3.110 are somewhat cumbersome to work with, so we shall define a
parameter, Q, as follows:
c
1+
λc
2w1 λ c
Q=
RL
=
c Vi
Vi
1
+
3w1
(3.111)
Then we can rewrite Equation 3.109 in a more compact way
M
1+θ f θ
pB = Q(1 − f ) (1 + θ f )
1+θ
(3.112)
The maximum or peak pressure attained is then found by taking the first derivative of
pB with respect to f and setting it equal to zero
M
M +1
dpB
M
= Q (1 − f )
+ 1 θ (1 + θ f ) θ − (1 + θ f ) θ = 0
df
θ
© 2014 by Taylor & Francis Group, LLC
(3.113)
Analytic and Computational Ballistics
87
Let us solve Equation 3.113 for f. By introducing the subscript “m” to denote maximum,
we obtain the product of two terms
(1 + θ f m ) ( M + θ )(1 − f m ) − (1 + θ f m ) = 0
(3.114)
Solving this we have two choices here, either
(1 + θ f m ) = 0 or
( M + θ )(1 − f m ) − (1 + θ f m ) = 0
(3.115)
The first would only be admitted for the special case of θ = −M, thus, our criteria for
determination of fm is
( M + θ )(1 − f m ) − (1 + θ f m ) = 0
(3.116)
and
fm =
M +θ − 1
M + 2θ
(3.117)
Equation 3.117 works for all values of θ. If we want to determine ϕm, the fraction of propellant burnt at peak pressure, we call on our relationship between f and ϕ, Equation 3.77.
Here we have denoted peak values with the subscript “m”:
φ = (1 − f )(1 + θ f )
(3.118)
Substitution of Equation 3.117 into the aforementioned equation yields
M +θ − 1
M + θ − 1
φm = 1 −
1 + θ
M + 2θ
M + 2θ
(3.119)
This, when simplified, gives
φm =
(1 + θ ) [ M + θ + θ ( M + θ )] (1 + θ ) [( M + θ )(1 + θ )]
=
[ M + 2θ ]2
[ M + 2θ ]2
(3.120)
or the following (valid for all θ):
φm =
( M + θ )(1 + θ )2
[ M + 2θ ]2
(3.121)
In designing a gun (and for other reasons), it is desirable to know where a projectile is in
its travel down bore when the pressure is at a maximum. This involves substitution of
Equation 3.117 into Equation 3.104 and for the case where θ ≠ 0 this yields
M
θ
1+θ
xm + l = l
1+θ M +θ − 1
M + 2θ
© 2014 by Taylor & Francis Group, LLC
(3.122)
Ballistics: Theory and Design of Guns and Ammunition
88
Simplifying, we finally get
M
( M + 2θ ) θ
xm + l = l
(M + θ )
for θ ≠ 0
(3.123)
For the case where θ = 0, we substitute Equation 3.117 into Equation 3.105, which we
rewrite as follows:
xm + l = l exp M(1 − f m )
(3.124)
M + θ − 1
xm + l = l exp M 1 −
M + 2θ
(3.125)
On substitution we get
Simplifying this result and substituting θ = 0 into it gives us
xm + l = le for θ = 0
(3.126)
for a zero form factor.
Knowing now the position of the peak pressure in the bore, we can then ask what the
breech pressure would be at this point. We can insert the value we have for the fraction
of remaining web at peak pressure, fm, back into the breech pressure equation for θ ≠ 0
(Equation 3.109)
M
1 + θ fm θ
pBm = Q(1 − f m )(1 + θ f m )
1+θ
(3.127)
With considerable algebraic simplification including substituting the values for Q, and the
Lagrange ratio, RL, for the case θ ≠ 0, we finally arrive at
pB m
c
1+
λc
2
w
1
=
Vi 1 + c
3w1
(1 + θ )2 ( M + θ ) Mθ +1
M +2
( M + 2θ ) θ
(3.128)
Following a similar procedure, we now insert the value we have for the fraction of remaining web at peak pressure, fm, into the breech pressure equation for θ = 0 (Equation 3.110)
pBm = Q(1 − f m )exp − M(1 − f m )
(3.129)
Then substituting for Q and RL and simplifying, we see that we have characterized the
breech pressure at the instant peak pressure is achieved down bore:
pB m
© 2014 by Taylor & Francis Group, LLC
c
1+
λc
2w1
=
Vi 1 + c
3w1
1
M exp[1]
(3.130)
Analytic and Computational Ballistics
89
Determining the breech pressure and travel when the solid grains have been completely
consumed is also of considerable interest. We shall use the subscript “c” to represent
charge burnout. If the charge is designed properly, it will burnout somewhere in the bore
that allows us to extract most energy from the propellant and reduces the muzzle blast.
Recall from our previous discussions that at t = 0, x = 0, f = 1, and ϕ = 0 but at all burnt
(subscript “c”), t = tc, x = xc, f = 0, and ϕ = 1. If we substitute f = 0 in Equations 3.109 and 3.110,
we obtain the breech pressure at the instant of charge burnout
c
M
1+
λc
2w1 1 θ
pB c =
c 1 + θ
Vi
1 + 3w
1
for θ ≠ 0
(3.131)
c
1+
λ c
2w1
exp[ − M] for θ = 0
pB c =
c
Vi
1
+
3w1
(3.132)
and
The travel of the projectile at burnout is a data point we usually want to know because
if this distance turns out to be longer than the barrel length, then the charge is not completely burnt when the projectile exits. If we substitute f = 0 in Equations 3.104 and 3.105,
we obtain the position of the projectile at the instant of charge burnout
M
xc + l = l(1 + θ ) θ
for θ ≠ 0
(3.133)
xc + l = lexp[M] for θ = 0
(3.134)
and
It is a good idea to use these equations first to see whether the propellant burns out in
the tube with the parameters we have designed into the grain. Still-burning grains leaving
the tube signify a poorly designed charge. For completeness, however, if charge burnout
happens outside the bore, the pressure at the breech location when the projectile leaves the
muzzle may be calculated by evaluating f at the muzzle through Equation 3.104 or 3.105
and using this value to calculate pB from Equation 3.109 or 3.110. The muzzle velocity could
then be obtained from Equation 3.85.
If charge burnout is, as desired, in the bore, recall that there is still a net force (pressure)
pushing on the projectile. A simple means of calculating this pressure is to assume that
the process occurs so quickly that it is essentially adiabatic and that the gas behaves as an
ideal gas. With these assumptions and the initial conditions that the pressure is pBc and
the distance is xc, we have a closed form solution to the problem. It is vitally important to
note that the expansion of the gas after charge burnout is neither adiabatic nor isentropic,
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
90
however the result is usually within about 5% with respect to pressure. The isentropic
relationships for an ideal gas are
p ρ
= =
p0 ρ 0
γ
1
v
1
v0
γ
v0 γ v −γ
= =
v v0
(3.135)
This equation relates pressure to specific volume in a general way, but we need to involve
the projectile travel as well. We can express the volume behind the projectile as a function
of distance as
V( x) = ( x + l)A
(3.136)
Then the specific volume of the gas is this value divided by the mass of the gas, which we
know is still c after burnout. Thus, we can write Equation 3.136 in its intensive form as
v( x) = ( x + l)
A
c
(3.137)
Furthermore, we can specialize this to the point at which the charge burns out and write
v( xc ) = ( xc + l)
A
c
(3.138)
We can now tailor Equation 3.135 to our needs by substituting the conditions at burnout as
our reference conditions
p( x) v( x)
=
pc
v( xc )
−γ
x+l
=
xc + l
−γ
(3.139)
This condition occurs often so we define
r( x ) =
x+l
xc + l
(3.140)
which can be written in more compact form as
p( x )
= r ( x )− γ
pc
(3.141)
A sketch of this situation is depicted in Figure 3.2.
These extensive preparations have finally brought us to the goal of interior ballistics
and the design of a gun system—imparting a desired velocity to a projectile and being
able to repeat that process at will. We have developed the means for predicting how the
© 2014 by Taylor & Francis Group, LLC
Analytic and Computational Ballistics
91
x
xc
l
Chamber
Propellant burnout
FIGURE 3.2
Position of projectile at charge burnout.
propellant burns over time, how the breech, space-mean, and base pressures vary with
time, and where the projectile moves to in relation to these pressures. Now we will focus
on the velocity of the projectile during this ballistic cycle. Recall that the kinetic energy of
the projectile plus the gas losses was written as
KEtot =
1
1
1
c
wpV 2 + cV 2 = wp + V 2
2
6
2
3
(3.20)
The work done on the projectile and the gas from charge burnout to the point of interest
(usually muzzle exit) is
x
∫
W = A p dx
(3.142)
xc
Combining Equations 3.20 and 3.142 and inserting Equation 3.139 yields
x
−γ
1
c 2
x+l
2
w1 + V ( x) − V ( xc ) = Apc
dx
2
3
xc + l
x
∫
(3.143)
c
We must keep in mind that we are using space-mean pressure here because the work is
being done on both the projectile and the gas. We can use any of breech, space-mean, and
base pressure (with the appropriate relationship) because we know each in terms of the
others. Integrating and rearranging we get
x
1
1
c 2
2
( x + l )− γ + 1
w1 + V ( x) − V ( xc ) = Apc
2
3
(1 − γ )(xc + l)−γ
x
(3.144)
c
Evaluation of the limits of integration yields
1
1
c 2
2
( x + l)1−γ − ( xc + l)1−γ
w1 + V ( x) − V ( xc ) = Apc
2
3
(1 − γ )(xc + l)−γ
© 2014 by Taylor & Francis Group, LLC
(3.145)
Ballistics: Theory and Design of Guns and Ammunition
92
Rearranging and inserting Equation 3.132 into the aforementioned equation, we get a
velocity relationship after burnout for θ = 0
V 2 ( x ) − V 2 ( xc ) =
( x + l)1−γ
2 Aλ ce − M ( xc + l)1−γ
− 1
−γ
Vi (1 − γ )( xc + l) (w1 + 3c ) ( xc + l)1−γ
(3.146)
But recall that the volume Vi = Al and if we define
1− γ
2 x + l
Φ=
− 1
(1 − γ ) xc + l
(3.147)
We can write
V 2 ( x ) − V 2 ( xc ) =
λ c( xc + l)exp[ − M]
Φ
c
l w1 +
3
(3.148)
Here again we revert to the cases of the form factor being zero or not zero and examine the
former first. Recall Equation 3.105. For conditions after charge burnout, there is no remaining web (f = 0), so we can write
x c + l = le M
for θ = 0 and
f =0
(3.149)
Rearranging this and substituting it into Equation 3.149 yields
V 2 ( x ) − V 2 ( xc ) =
λc
c
w1 +
3
Φ
(3.150)
This allows us to calculate the velocity of a projectile after burnout of the web for θ = 0. The
case for nonzero θ requires further examination and manipulation. In Equation 3.85, we had
a general expression for velocity as a function of remaining web. After burnout this becomes
V ( xc ) =
AD
(3.151)
c
β w1 +
2
Since we are working with kinetic energy, squaring this gives
V 2 ( xc ) =
A 2D 2
c
β w1 +
2
2
(3.152)
2
We defined the central ballistic parameter, M, in Equation 3.94, and can rearrange it for our
purposes into the form
cλ M
A 2D 2
=
2
c
c
w1 + β 2 w1 +
3
2
© 2014 by Taylor & Francis Group, LLC
(3.153)
Analytic and Computational Ballistics
93
When this is compared with Equation 3.152, we conclude that
V 2 ( xc ) =
λ cM
c
w1 +
3
(3.154)
This is an important result—it says that just by knowing the physical parameters of
the weapon, projectile, and charge one can predict the projectile velocity at charge burnout. With this result and continuing the examination for nonzero θ, we can then say that
Equation 3.150 is valid for any θ. If we solve Equation 3.150 for the velocity at any point,
V(x), insert Equation 3.154, and rearrange the terms, we get
V 2 ( x) =
cλ
( M + Φ ) for θ ≠ 0
c
w1 +
3
(3.155)
This result along with our earlier work allows us now to determine projectile velocity
at all points in the gun for charge grains of all form factors both before and after burnout.
We have been through many derivations that have led us to the essentials of interior ballistics—breech pressure and velocity in terms of projectile travel. These results, furthermore,
are in closed form, accessible to computation by hand calculator. Specialized pressures, spacemean pressure and projectile base pressure, may be computed from the breech pressure data
using the Lagrange approximations. Projectile design and gun design proceed from these
equations. In the following sections, we shall discuss refinements to the Lagrange formulation
with an emphasis on the use of modern computer programs that take the drudgery out of hand
calculation and provide the ability to iterate solutions for small changes in the parameters.
Problem 1
You are asked to analyze the pressure of a charge zero (igniter) firing in an M31 boom for a
120 mm mortar projectile. You decide to examine it as a closed bomb first. Assume we have
59 g of M48 propellant (properties given later). The volume of the closed bomb is 5.822 in.3.
The propellant grains are balls (roughly spherical) with a diameter (web) of 0.049 in.
M48 propellant properties
lbm
Density ρ = 0.056 3
in.
Ratio of specific heats γ = 1.21
in.3
Co-volume b = 26.72
lbm
Isochoric flame temperature T0 = 3720°F
Burn rate exponent α = 0.9145
in.
Average burn rate coefficient β = 0.0095
s ⋅ psi
Burn rate D
df
in.
= 40.341
dt
s
ft-lbf
Force constant λ = 391, 000
lbm
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
94
1. Come up with the equation for the web fraction, f, as a function of time.
Answer: f = (1 − 823.29t) [%]
2. For a sphere, the fraction of propellant burnt has the functional form ϕ = 1 − f 3,
write this in terms of time and ϕ.
Answer: ϕ = 2,470t − 2,033,419t2 + 558,031,251t3
3. Determine how long it will take the propellant to burn halfway through and all
the way through.
Answer: Time to burn through halfway is 0.6 ms
4. Using the Noble–Abel equation of state, determine the pressure in the vessel when
half of the propellant is burnt and when all of the propellant is burnt. Note that this
cannot usually occur as the propellant is a charge zero firing that is vented into the
main ullage volume behind the mortar bomb (significantly greater volume).
lbf
Answer: p = 258, 746 2
in.
Problem 2
If we use the Lagrange approximation in examination of a 155 mm projectile launch, what
is the average pressure in the volume behind a 102 lbm projectile if the breech pressure is
55,000 psi? The propelling charge weighs 28 lbm.
lbf
Answer: p = 52, 787 2
in.
Problem 3
A 120 mm projectile is to be examined while in the bore of a tank cannon at a time 4 ms
from shot start. Over this time period, the projectile has acquired an average velocity of 1000 ft/s. The propellant grain (M15) is single perf (θ = 0) with a 0.034 in. initial
web. The co-volume of the propellant is 31.17 in.3/lbm. The density of the propellant is
0.06 lbm/in.3. If the projectile weighs 50.4 lbm, the propellant weighs 12.25 lbm and the
chamber volume is 330 in.3. At this time, 0.02 in. of the web remains. The propellant
force is 337,000 ft-lbf/lbm. Determine the breech pressure in the weapon. Be careful
with the units!
lbf
Answer: pB = 21, 784 2
in.
Problem 4
The Paris gun was a monstrous 210 mm weapon designed by Germany during the First
World War to bombard Paris from some 70 miles away. It was unique in that it fired the
first exo-atmospheric projectile ever designed. The weapon had a chamber volume of
15,866 in.3. Very little of the projectile protrudes into the chamber after it seats (so ignore
the volume the base occupies). The length of travel for the projectile from shot start to
shot exit is 1182 in. The projectile weighs 234 lb. The propelling charge weighs 430.2 lb.
The propellant used was specially designed and was similar to U.S. M26 propellant. It
consisted of 64%–68% NC, 25%–29% NG with 7% Centralite (symmetrical diethyl diphenylurea C17H20N2O), and some other additives. The propellant was single perforated with
a web thickness described later. Assume the propellant has the following properties.
© 2014 by Taylor & Francis Group, LLC
Analytic and Computational Ballistics
95
(Note that these are the authors’ guesses—a better estimate of the properties can be
found in Ref. [5].)
Adiabatic flame temperature T0 = 2881 K
Specific heat ratio γ = 1.237
Co-volume b = 1.06 cm3/g
Density of solid propellant ρ = 1.62 g/cm3
Propellant burn rate coefficient β = 0.0707 (cm/s)/(MPa)
Web thickness D = 0.217 in.
Propellant force λ = 1019 J/g
1. Using the aforementioned data, determine (a) the projectile base pressure in psi,
(b) velocity in ft/s, and (c) distance down the bore of the weapon in inches for peak
pressure.
lbf
Answer: (a) pSmax = 31, 548 2
in.
ft
(b) Vpmax = 2880
s
(c) x pmax = 270.1 [in.]
2. Determine the pressure in psi at a point 3 in. behind the projectile base when the
charge burns out.
lbf
Answer: px− 3 = 28, 637 2
in.
3. Assuming the gas behaves according to the Noble–Abel equation of state, determine the muzzle velocity of the projectile in ft/s.
ft
Answer: V = 5791
s
Problem 5
A British 14 in. Mark VII gun has a chamber volume of 22,000 in.3. A 5 in. of the projectile
protrude into the chamber after it seats. The length of travel for the projectile from shot
start to shot exit is 515.68 in. The weapon has a uniform twist of 1 in 30. The projectile
weighs 1590 lb. The propelling charge weighs 338.25 lb. The propellant used is called “SC”
and consists of 49.5% NC (12.2% nitrated), 41.5% NG with 9% Centralite. Assume SC propellant has the following properties:
Adiabatic flame temperature T0 = 3090 K
Specific heat ratio γ = 1.248
Co-volume b = 26.5 in.3/lbm
Density of solid propellant ρ = 0.0567 lbm/in.3
Propellant burn rate β = 0.000331 (in./s)/(psi)
Web thickness D = 0.25 in.
Specific molecular weight n = 0.04262 lb-mol/lbm
1. Determine the force constant, λ in ft-lbf/lbm.
2. Determine the central ballistic parameter for this gun–projectile combination.
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
96
3. Using the aforementioned data, determine the projectile base pressure, velocity,
and distance down the bore of the weapon for both peak pressure and charge
burnout assuming the grain is a cylindrical propellant (θ = 1).
Answers:
ft-lbf
λ = 366, 246
lbm
M = 1.933
lbf
pBmax = 43, 281 2
in.
lbf
pBc = 26, 350 2
in.
lbf
psmax = 39, 120 2
in.
lbf
psc = 23, 817 2
in.
ft
Vpmax = 1082
s
ft
Vc = 2128
s
x pmax = 75.6 [in.]
xc = 293.7 [in.]
Problem 6
Verify Equation 3.150 is valid for any θ.
Problem 7
You are asked to design a gun to propel a fragment for an explosive initiation test at
3000 ft/s. The diameter of the chamber and bore is to be 0.50 in. You have on hand some
M10 flake propellant with the ballistic properties given later. Because of space limitations
the device (chamber and bore) cannot exceed 5 ft in length. The fragment plus sabot weighs
0.1 lbm. Assume the propellant has the following properties:
Adiabatic flame temperature
Specific heat ratio
Co-volume
Density of solid propellant
Propellant burn rate coefficient
Web thickness
Propellant force
© 2014 by Taylor & Francis Group, LLC
T0 = 3000 K
γ = 1.2342
b = 27.76 in.3/lbm
δ = 0.0602 lbm/in.3
β = 0.00002468 (in./s)/(psi)
D = 0.011 in.
λ = 339,000 ft-lbf/lbm
Analytic and Computational Ballistics
97
(a) Using the aforementioned data, determine the bore and chamber length for the
weapon as well as the amount of propellant required—be careful to leave some air
space in the chamber.
(b) Once the system is established in (a), determine the central ballistic parameter,
value of peak breech pressure, and location of the projectile when peak pressure
occurs.
Problem 8
A test of an experimental propellant grain yields a parametric relation for the fraction of
web remaining of
f (t) = C1t 2 + C2t + 1
In the equation above, C1 and C2 are constants. The burn rate model for this propellant was
determined to be
D
df
= − β pαB
dt
Determine the expression for the velocity of the projectile as a function of time for a given
gun-projectile system. Describe the assumptions used and why they are relevant.
Problem 9
A black powder charge is to be designed to throw a 3 in. diameter firework canister
that weighs 2 lbs at 100 ± 2 ft/s. The tube is 3 in. in diameter and the chamber where
the propellant charge sits is 6 in. long. The projectile travel distance can be up to 18 in.
long but you may size the tube length. The propellant grains are spherical and obey
the relation
φ (t) = 1 − f 3
(1)
The average diameter of the propellant grains (they vary quite a bit and they are not exactly
spherical) is 0.02 in. The tube can handle a maximum pressure of 2000 psi.
Assume the linear burn rate for the black powder is 2.22 in./s measured at 3,329 psi and
the force constant is 105,000 ft · lbf/lbm. Assume a specific heat ratio of 1.21 and a solid
density of 0.06 lbm/in.3. Listing all assumptions
a.
b.
c.
d.
Develop the proper equations for the motion of the projectile.
Size the charge so that the system will function as required—make sure it fits too!
Determine the peak pressure in psi and location of peak pressure in inches.
Determine the location of charge burnout in inches and the velocity at burnout in ft/s. Please note that if your design allows unburnt propellant to exit
the tube please state that and calculate muzzle exit pressure and velocity in
this case.
e. Determine the muzzle velocity in ft/s.
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
98
Problem 10
A 7 perf grain is a very common geometry in weapons. The geometry is such that the web
between the outer diameter of the grain and the I.D. of the outer perforations and the web
between the perforations themselves is equal. Determine, up to the point of slivering, the
following:
a. The equation for ϕ(t) for any perforation size and any web
b. If possible obtain an estimate for the value of θ assuming the perforations are ¼ of
the web thickness, if not, show why not
End effects may be neglected.
Problem 11
A Japanese 18.1 in. Type 94 gun was the largest weapon ever mounted on a warship. The
gun had a chamber volume of 41,496 in.3. An estimated 8 in. of the projectile protrude into
the chamber after it seats. The length of travel for the projectile from shot start to shot exit
is 806.3 in. The weapon has a uniform twist of 1 in 28. The type 91 AP projectile weighs
2,998 lbs. The propelling charge weighs 794 lbs. The propellant used is called “DC1” and
consists of 51.8% NC (11.85% nitrated), 41.0% NG with 4.5% Centralite (symmetrical diethyl
diphenylurea C17H20N2O), 2.0% orthotolyl urethane (added as an improvement to the centralite) and 0.7% mineral matter (salts for wear and flash reduction) [6]. Assume DC1 propellant has the following properties:
Adiabatic flame temperature
Specific heat ratio
Co-volume
Density of solid propellant
Propellant burn rate
Web thickness
Propellant force
T0 = 3000 K
γ = 1.23
b = 27.0 in.3/lbm
δ = 0.059 lbm/in.3
β = 0.000300 (in./s)/(psi)
D = 0.184 in.
λ = 284,000 ft lbf/lbm
a. Determine the central ballistic parameter for this gun/projectile combination.
b. Using the aforementioned data, determine the projectile base pressure, velocity,
and distance down the bore of the weapon for both peak pressure and charge
burnout assuming the grain is single perforated propellant.
c. Determine the muzzle velocity of the weapon and the pressure acting on the projectile at muzzle exit.
Problem 12
A British “2 Pounder” (so-called because the projectile weighed about two pounds) was
their main anti-tank weapon for the first two and a half years of the Second World War.
It fired a 40 mm projectile that weighed 1.94 lbm. The gun had a chamber volume of 23.0
in3. Since the shot had a flat base, when crimped to the cartridge case none of the projectile
protrudes into the chamber. The length of travel for the projectile from shot start to shot
exit is 70 in. The weapon has a uniform right-hand twist of 1 in 30. The propelling charge
© 2014 by Taylor & Francis Group, LLC
Analytic and Computational Ballistics
99
weighs 0.583 lbs. The propellant used was single perforated cordite. Assume the cordite
propellant has the following properties:
Adiabatic flame temperature
Specific heat ratio
Co-volume
Density of solid propellant
Propellant burn rate
Web thickness
Propellant force
T0 = 2442 K
γ = 1.21
b = 31.32 in.3/lbm
δ = 0.059 lbm/in.3
β = 0.00024 (in./s)/(psi)
D = 0.0197 in.
λ = 318,000 ft lbf/lbm
a. Determine the central ballistic parameter for this gun/projectile combination.
b. Using the aforementioned data, determine the projectile base pressure, velocity,
and distance down the bore of the weapon for both peak pressure and charge
burnout.
c. Determine the muzzle velocity of the weapon and the pressure acting on the projectile at muzzle exit.
d. Create a pressure–travel and velocity–travel curve for this system. Annotate the
location of charge burnout on the pressure–travel curve.
Problem 13
Develop the equations required to model a propellant/charge/gun system where the propellant behaves according to a βpα burn rate. Do this for both θ = 0 and θ ≠ 0. Describe, in
bulletized form, how you would solve these equations numerically.
Hint: Start with Equations 3.80 and 3.77 through 3.79 making sure that you alter
Equation 3.78 to the new burn rate.
Problem 14
Write a code using any software you want to solve Problem 13. Only write the code so it
solves the interior ballistics problem up to charge burnout. Check the code by setting α = 1
and show that you get answers close to that obtained in Problem 12.
Problem 15
Use your code developed in Problem 14 to obtain a solution to the same problem assuming
black powder is the propellant. Only use your code to take the problem to the all-burnt
point. Assume the properties of black powder are as follows:
Adiabatic flame temperature
Specific heat ratio
Co-volume
Density of solid propellant
Propellant burn rate
Web thickness
Propellant force
© 2014 by Taylor & Francis Group, LLC
T0 ≈ 2300 K
γ ≈ 1.22
b ≈ 31 in.3/lbm
δ = 0.060 lbm/in.3
β = 0.04044 (in./s)/(psi0.511)
D = 0.0197 in.
λ = 105,000 ft lbf/lbm
100
Ballistics: Theory and Design of Guns and Ammunition
Problem 16
A Japanese (designed by the British firm of Vickers) 14 in./45 cannon is to be examined.
It fired a 14 in. (36 cm) projectile that weighed 1485 lbm. The gun had a chamber volume
of 17,996 in.3 [6]. Assume 4 in. of the projectile protrudes into the chamber. The length of
travel for the projectile from shot start to shot exit is 540.8 in. [6]. The weapon has a uniform
right-hand twist of 1 in 28. The propelling charge has four increments, where each weighs
78.45 lbs. The propellant used was DC that consisted of 64.8% NC, 30% NG, 4.5% centralite,
and 0.7% mineral matter [6]. Assume the propellant geometry is such that θ = 0.1. Assume
the DC propellant has the following properties:
Adiabatic flame temperature
Specific heat ratio
Co-volume
Density of solid propellant
Propellant burn rate
Web thickness
Propellant force
T0 = 3200 K
γ = 1.23
b = 27.0 in.3/lbm
δ = 0.059 lbm/in.3
β = 0.000298 (in./s)/(psi)
D = 0.165 in.
λ = 365,000 ft lbf/lbm
The weapon was “zoned” to fire using 2, 3, and 4 bags that were called, “weak,” “reduced,”
and “full” [7]. For each of these charge configurations
a. Determine the central ballistic parameter for this gun/projectile/propellant
combination.
b. Using the aforementioned data, determine the projectile breech pressure for both
peak pressure and charge burnout.
c. Using the aforementioned data, determine the projectile base pressure, velocity,
and distance down the bore of the weapon for both peak pressure and charge
burnout.
d. Determine the muzzle velocity of the weapon and the pressure acting on the projectile at muzzle exit.
e. Plot the pressure versus distance based on the aforementioned results at the
instant of peak pressure and muzzle exit.
3.3 Chambrage Gradient
In our derivation of the Lagrange gradient approximations, we assumed that the chamber
of the gun was simply an extension of the bore. The volume of the chamber was converted
to a cylinder of bore diameter and the tube was lengthened appropriately behind the projectile. In doing this, we neglected the effects of short, larger diameter chambers (the definition of chambrage is the ratio of the diameter of the chamber to the bore inner diameter)
and all calculations that are functions of distance from the breech, x–xs, are inaccurate in
the distance term. If we account for these differences by deriving a chambrage gradient,
we find that the two methods yield similar but close answers. Nevertheless, one should
understand how the answers relate to each other and to the real problem. Fredrick W.
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Analytic and Computational Ballistics
101
Robbins of the Army Research Laboratory, who has allowed us to base this section on his
excellent work, derived the chambrage gradient formulation that follows.
The formulation of the chambrage gradient follows much the same pattern that was used
in the development of the Lagrange gradient. It leads, however, to an algorithm that is best
applied with the aid of a computer. Small increments of time (hence distance) are chosen
and computations of pressure (breech, mean, and base), velocity, acceleration, and distance
traveled are made for the end point of the interval. The calculation is then repeated for the
next increment of time. This is done until the projectile exits the bore. A representation of
the situation is shown in Figure 3.3 for a chosen time step.
The definitions of the terms used in Figure 3.3 are shown in Figure 3.4.
In Robbins’ derivation, certain integrals called J integral factors are developed and must
be computed. They are
x0
J 1 ( x0 ) =
V(x)
∫ A(x) dx
(3.156)
0
J 1 ( xs ) = J 1 ( x0 ) +
1
Ab
Ab
2
V( x0 )( xs − x0 ) + 2 ( xs − x0 )
(3.157)
2
J 2 ( xs ) =
[ V(x0 ) + Ab (xs − x0 )]
(3.158)
Ab2
Ab
Vs
A(x)
V(x)
x, V(x)
xs , V(xs)
FIGURE 3.3
Chamber with large chambrage.
Velocity of propellant gas
Velocity of the projectile
at position, x measured from
at the time of interest
the breech at the time of interest
V(x) = Ab
VS
V(xS)
Cross-sectional area of the bore
V(x)
A(x)
Cross-sectional area of the weapon
at position, x at the time of interest
Volume behind the projectile base
at the time of interest
FIGURE 3.4
Definitions of terms used in chambrage gradient development.
© 2014 by Taylor & Francis Group, LLC
Volume at position, x
at the time of interest
Ballistics: Theory and Design of Guns and Ammunition
102
J 3 ( xs ) = J 3 ( x0 ) + Ab J1( x0 )( xs − x0 ) +
V( x0 )
A
( x s − x 0 )2 + b ( x s − x 0 )3
2
6
V( x0 ) + Ab ( xs − x0 )3 − [ V( x0 )]
J 4 ( xs ) = J 4 ( x0 ) +
3 Ab2
(3.159)
3
(3.160)
The acceleration at any point, a, appears in one of our algorithm factors explicitly
a(t) = a1(t) + a2 (t)ps
(3.161)
where
a1(t) =
AbVs2 cAb presist
+
2
mp
[ V(xs )] V(xs )
cAb
a2 (t) = −
cAb2
mp [ V( xs )]
2
(3.162)
(3.163)
Another factor required in the algorithm is b(t) derived as
b(t) = −
cAb2Vs2
2 [ V( xs )]
3
(3.164)
The way the algorithm is used is (roughly) as follows:
At each time step
•
•
•
•
•
•
The breech pressure is calculated from the burning rate equations
J1 through J4 are calculated
a(t) and b(t) are calculated
The projectile acceleration, velocity, and distance down the bore are calculated
The volume behind the projectile is updated
The process moves to the next time step
This gradient, while only slightly more accurate than the Lagrange gradient in the computed distance from the breech, is used in some modern interior ballistic computer codes.
3.4 Numerical Methods in Interior Ballistics
In this section, we shall briefly discuss methods for solving the interior ballistics problem
through use of computational tools. In recent decades, computational capabilities have
increased at an astronomical rate. One of the most famous early uses of the computer to
© 2014 by Taylor & Francis Group, LLC
Analytic and Computational Ballistics
103
solve the exterior ballistics problem (firing tables) was the use of the electronic numerical integrator and computer (ENIAC) machine during and immediately after the Second
World War. In this case, the computer was used to solve tedious exterior ballistics problems in rapid order.
In the field of interior ballistics, the computer revolution has given the individual ballistician the tools (although some commercial packages can be expensive) to solve extremely
complicated interior ballistics problems and optimize a system quickly. The complexity of
these tools is driven by the physics that are incorporated in the particular code. We shall
discuss some general categories of software, their uses, and their limitations.
Many interior ballistics codes are of the zero-dimensional variety. In these types of
codes, the density of the propellant gas (as stipulated by the Lagrange approximation) is
considered constant in the volume between the breech and the projectile. The Lagrange
pressure gradient is assumed to be in effect and results in a nice, always well-behaved
launch. These codes are extremely useful for predictive applications because they run fast.
One of the features of these codes that make them so useful is that we can easily include
and track burn characteristics of multiple propellant types (both geometry and chemical
composition). This allows us to tailor the burn characteristics so that a particular pressure–distance distribution is achieved while maintaining a particular muzzle velocity.
Another excellent feature of this type of code is that heat transfer to the weapon can be
accounted for in the energy balance. This provides a more realistic muzzle velocity than if
it is neglected and can be of great value to the gun designer. Friction and blow-by effects
can be fudged in and burn rate parameters varied to replicate actual tests. Additionally,
the effects of the regression of all surfaces (recall that we neglected end effects in our hand
calculation methods) can be simulated and accounted for. Zero-dimensional codes can
also include the effects of inhibitors on the propellant grains as well as highly nonlinear
pressure–burn rate relationships. Since zero-dimensional codes track the pressure, it is
simple enough to use them to develop recoil models as well. All in all, zero-dimensional
codes are probably the most effective tools at the disposal of the interior ballistician for
basic ballistics design work. Once a set of experiments have been conducted to validate
these codes, their accuracy is excellent.
A quasi-one-dimensional code is one in which the density of the propellant gas behind
the projectile is a known function of some other variable. An example of this would be a
zero-dimensional code that incorporated the chambrage gradient. Essentially, beyond the
ability to track the effect of variable chamber or bore area on the density, the limitations
and benefits of this type of code are the same as discussed in the zero-dimensional section.
A one-dimensional interior ballistics code allows density to vary based on the physical equations and conservation laws in the axial direction only. Thus, at a given cross
section, the density is considered constant throughout the radial direction. These codes
are very good at predicting pressure waves and therefore can estimate the pressure differential along the volume behind the projectile. The benefit of this is that, since propellant generally burns faster under higher pressure, the local burn rate and therefore the
amount of gas evolved can be tracked. This allows the user to see pressure waves develop
and propagate. The disadvantage is that, since the code can only track pressure waves in
the axial direction, unless the charge fully fills the volume behind the projectile, it is difficult to completely match the physics of the firing. This occurs because the presence of
solids and gases in the chamber is generally not uniform—the solids are usually at the
bottom of the chamber. This affects the gas dynamics. Solids will also be entrained by
the gas flow down the bore and some modeling of their motion has to be accomplished
(or ignored). In most cases, the propellant bed is assumed to be a monolithic mass that
© 2014 by Taylor & Francis Group, LLC
104
Ballistics: Theory and Design of Guns and Ammunition
regresses and stretches as the propellant is burned. These codes are usually very good but
the user should completely understand the assumptions on how the propellant is allowed
to move before using them.
A two-dimensional model is one where the density can vary in the radial direction
as well. These models are better at predicting pressure waves but take somewhat longer
time to run than one-dimensional models. Pressure can be tracked in the radial direction and the propellant motion included. The same issues with propellant motion are
present as they were in the one-dimensional models though it is possible to track propellant motion.
A three-dimensional model has it all. Because of this they usually take an excruciatingly long time to set up and run. This time constraint makes them generally reserved for
failure investigations rather than predictive simulations. Individual propellant grains can
regress and be tracked and one can imagine the difficulty with this in the sense of model
validation. With suitable stress and failure models, grain fracture can also be examined. If
erosion models are incorporated, the effect of gas wash on propellant burn rate can even
be included. One has to ask oneself if all of this is really necessary. In some cases, these
models are crucial, in other cases, they are certainly overkill. The usefulness of this type
of model is still somewhat limited by computer speed, but as computers become faster the
limitation will change to a lack of accurate physical models for motion, surface regression,
propellant and gun tube erosion, grain fracture, etc. These issues are certainly solvable,
but finding a proponent who will fund the research is difficult.
Now that we have described the general types of models, it is important to explain
their use further. In general, all of them are used in a similar manner. We shall use the
zero-dimensional model as an example and leave the rest to the reader’s imagination
(and budget restrictions). Typically, a propellant formulation and geometry is chosen
as a point of departure given that we have a preliminary gun design and a projectile to
work with. This propellant is then further developed in terms of geometry or chemical
composition. Some zero-dimensional codes are provided with optimization subroutines
so that particular characteristics of the ballistic cycle can be achieved. The pressure–
time, acceleration–time, and pressure–distance curves are examined and, if suitable,
some experimental charges are made up. The configuration is then fired and the results
checked against the code. These results then can be used to adjust burn rates and resistive characteristics, and the model can be used to predict all future firings and design
iterations.
A particular example of the power of these codes is their usefulness in assessing the
interior ballistics of systems that vary widely in matters of scale, e.g., in mass of projectile,
diameter of bore, and muzzle velocity. In the 1960s, ballisticians J. Frankle and M. Baer
at the Ballistics Research Laboratories at Aberdeen, Maryland [8] and others elsewhere
devised codes largely based on Corner’s zero-dimensional analysis that we described in
detail in Section 3.2. Among these the Frankle–Baer simulation, still in use today, which
examined and expanded on the basic energy equation,
Energy released by burning propellant
= change in internal energy of the gases + work done on the projectile + secondary losses
or
Q = ∆U + W + Losses
© 2014 by Taylor & Francis Group, LLC
(3.165)
Analytic and Computational Ballistics
105
developed equations of state of the propellant gases based on more recent thermodynamic
theories and refined the losses term from new experimental data. This led to more refined
ratios for breech, mean, and shot base pressures, and more accurate equations of motion
for the projectile.
To examine the effects of scale we computed the relevant pressure ratios for three widely
different gun–projectile combinations. We show these combinations and the resultant
ratio values in Tables 3.1 through 3.5. What is noteworthy is the applicability of the theory
over the range of size, projectile mass, and propellant type and volume. Notice also the
closeness of the pressure ratios for each projectile between the Corner and the Frankle–
Baer simulations.
TABLE 3.1
Inputs for Comparison of Corner and Frankle–Baer
Parameter
Expression or
Value (J. Corner)
M735
M1
M193
M735
M1
M193
Expression or Value
(Frankle–Baer)
Charge weight
c
13.125
9.000
4.020 × 10 −3
c
13.125
9.000
4.020 × 10 −3
Projectile weight
w
12.78
31.97
−
3
7.86 × 10
wp
12.78
31.97
−
3
7.86 × 10
Propellant type
—
M30
M1
Ball
—
M30
M1
Ball
TABLE 3.2
Burn Characteristic Inputs for Numerical Comparison of Corner and Frankle–Baer
Parameter
Expression or
Value (J. Corner)
M735
M1
M193
Expression or Value
(Frankle–Baer)
M735
M1
M193
Propellant impetus
(force)
λ
3.64 × 10 5
5
3.05 × 10
3.32 × 10 5
λ
3.64 × 10 5
5
3.05 × 10
3.32 × 10 5
Specific heat ratio
γ
1.2385
1.2592
1.26
γ
1.2385
1.2592
1.26
Polytropic index
1
γ −1
© 2014 by Taylor & Francis Group, LLC
n
Ballistics: Theory and Design of Guns and Ammunition
106
TABLE 3.3
Pressure Gradient Calculations for Numerical Comparison
of Corner and Frankle–Baer
Parameter
Expression or Value
(Frankle–Baer)
Expression or Value (J. Corner)
p
pS
1+
1
δ
n/a
1
ab
n/a
2n + 3 2(n + 1)
+
c / wp
δ
p
pB
1 c
1 − 6 w
1 c
1 −
δ wp
pB
pS
1 c
1 + 2 w
−( n+1)
(1 − ab )
c
3w
1+
1 c
δ wp
1 + c1 β n
1
1 + α n
2n + 3
1 + c1n
n+1
(1 − ab )
TABLE 3.4
Specific Frankle–Baer Computations for the M735, M1, and M193 Projectiles
Projectile
M735 (105 mm KE)
M1 (105 mm HE)
M193 (5.56 mm ball)
α
β
c1
1
δ
0.56
0.63
0.60
1.07
1.01
1.03
1.05
1.02
1.04
0.333
0.322
0.315
ε=
c
wp
1.027
0.282
0.511
1
ab
(1−ab)n+1
11.002
37.811
22.325
0.683
0.876
0.800
TABLE 3.5
Specific Gradient Comparison of Corner and Frankle–Baer for the M735, M1, and M193 Projectiles
p
pB
p
pS
Projectile
M735 (105 mm KE)
M1 (105 mm HE)
M193 (5.56 mm ball)
1+
1 c
2w
Corner
Frankle–Baer
Corner
Frankle–Baer
Corner
Frankle–Baer
1.342
1.094
1.170
1.342
1.091
1.161
0.829
0.953
0.915
0.917
0.956
0.929
1.514 (1.619)
1.141 (1.148)
1.256
1.464 (1.463)
1.142 (1.141)
1.250 (1.250)
3.5 Sensitivities and Efficiencies
Having explored the detailed development of theories of interior ballistic events, we will
now probe the outcome of varying some of the parameters that are under the control of the
charge designer. To do this, we will be referring back to definitions and equations developed under Section 3.2.
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Analytic and Computational Ballistics
107
A useful quantity for our analysis is the dimensionless central ballistic parameter, M
c
1+
AD
3w1
M=
2
w1cλβ 2
c
1+
2w1
2
2
(3.166)
Of particular importance in this are the variables D and β, the original web dimension
and the burning rate coefficient, respectively. If we examine Equation 3.167
pB m
c
1+
λc
2
w
1
=
Vi 1 + c
3w1
1
M exp[1]
(3.167)
We can see that, at least for the case of θ = 0, M is in the denominator and as the ratio D/β
decreases, M decreases and from Equation 3.167, the peak pressure, pB, increases. That is,
if the original web size is decreased, the peak pressure will increase. This is a parameter
much under the control of the designer.
Referring again to Equation 3.166, we see that if the charge mass (weight), c, is increased,
then M decreases (the cs in the gradient term largely cancel out and c in the first term
denominator governs). In Equation 3.167, c appears in the numerator and M in the denominator causing pB, the peak pressure, to again rise.
Let us now examine the shift in location of xm, the point in travel where the peak pressure exists:
M
( M + 2θ ) θ
xm + l = l
(M + θ )
(3.168)
Equation 3.168 relates xm to M. If the ratio D/β or the charge mass, c, decreases, then M
decreases and consequently xm is reduced (it moves toward the breech). This kind of shift
is important in gun design since wall thickness and center of mass are important considerations for weapon mounting.
The sensitivity of muzzle velocity, V, to changes in web size or charge weight can be seen
in Equation 3.169:
V 2 ( x ) − V 2 ( xc ) =
λ c( xc + l)exp[ − M]
Φ
c
l w1 +
3
(3.169)
where M is the governing term and is decreased as we showed earlier, if charge mass is
increased or web size reduced. Because M has a negative exponent in the equation, its
reduction drives an increase in V.
Finally, the influence of travel on muzzle velocity can be shown to be quite weak. The
computation is complex and will not be shown here. But, e.g., by doubling the travel, velocity increases only by a factor of about a tenth, hardly worth the effort in the real world.
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
108
pB
m
p
FIGURE 3.5
Average and maximum breech pressure for a typical gun firing.
There are two measures of efficiency that are of interest to the interior ballistician: piezometric or pressure efficiency and ballistic or energy efficiency.
Piezometric efficiency, εp, is the ratio of the average pressure during the entire ballistic
cycle to the peak pressure during the cycle:
εp =
p
pB m
(3.170)
An illustration of the space-mean pressure and maximum breech pressure is provided
as Figure 3.5.
Increasing εp implies that the muzzle pressure will be high (usually an undesirable
trait), and that the charge burnout point will move toward the muzzle (hopefully never
outside the muzzle). High piezometric efficiency usually means poor regularity, i.e.,
round-to-round muzzle velocity repeatability is poor (an undesirable trait). For powerful,
high-velocity cannons, this efficiency is usually in the 50%–60% range. Other cannons are
lower. High piezometric efficiency also implies that the expansion ratio, the ratio of total
gun volume to chamber volume, will be low: powerful guns have large chambers and
consume lots of propellant.
Ballistic efficiency, ε b, is defined as the ratio of the kinetic energy of the projectile as it
exits the muzzle to the total potential energy of the propellant charge:
εb =
muzzle KE
=
propellant PE
wV 2 (γ − 1)wV 2
=
λc
2λ c
γ −1
1
2
(3.171)
because the potential energy is defined as
propellant PE =
RT0
(γ − 1)
and λ = RT0
(3.172)
Increasing ε b tends to shift the all-burnt position toward the breech and increases the
expansion ratio. Reducing the central ballistic parameter, M, by going to a smaller web will
also increase ε b. The ballistic efficiency of most guns is approximately 0.33.
© 2014 by Taylor & Francis Group, LLC
Analytic and Computational Ballistics
109
References
1.
2.
3.
4.
5.
Robbins, F., Interior Ballistics Course Notes, Self Published, Aberdeen, MD, 2002.
Panton, R.L., Incompressible Flow, 2nd edn., John Wiley & Sons, New York, 1995.
Currie, I.G., Fundamental Mechanics of Fluids, 2nd edn., McGraw-Hill, New York, 1993.
Corner, J., Theory of the Interior Ballistics of Guns, John Wiley & Sons, New York, 1950.
Bull, G.V., Murphy, C.H., Paris Kanonen—The Paris Guns (Wilhelmgeschutze) and Project HARP,
Verlag, E.S. Mittler & Sohn GmbH, Herford und Bonn, 1988.
6. Campbell, J., Naval Weapons of World War Two, Naval Institute Press, Annapolis, MD, 1985.
7. Jordan, J. (Ed.), Warship 2012, Conway Publishing, London, U.K., 2012.
8. Frankle, J.M., Interior Ballistics of High Velocity Guns Experimental Program, Phase I, BRL
Memorandum Report 1879, U.S. Army Ballistic Research Laboratory, Aberdeen Proving
Ground, MD, November 1967.
© 2014 by Taylor & Francis Group, LLC
© 2014 by Taylor & Francis Group, LLC
4
Ammunition Design Practice
Chapter 3 provided us with the information necessary to determine the forces acting on
the projectile and gun. This chapter endeavors to describe techniques necessary for the
projectile or weapon designer to be successful. Sections 4.1 and 4.2 describe topics in the
field of mechanics of materials. This material will form the basis by which we will evaluate designs. Sections 4.3 through 4.8 apply these concepts to the design of projectiles and
guns. This chapter ends with practices and techniques used to design modern ammunition that must be fired from a gun.
4.1 Stress and Strain
Before proceeding with our examination of design practices, a discussion of the fundamentals of the general state of stress in materials is in order. Consider an arbitrary cube
of material under load as depicted in Figure 4.1. The state of stress can be completely
defined by six stress components: σx, σy, σz, τxy, τyz, and τzx. Here, we have used a Cartesian
coordinate system where the normal stresses are denoted by σ and the shear stresses are
denoted by τ.
The first subscript represents the plane in which the stress acts (defined by its normal
vector) while the second subscript indicates the direction of action. These components
form the stress tensor, which is actually a 3 × 3 matrix of nine elements; except that we
have assumed that τxy = τyx, τzy = τyz, and τxz = τzx. When written as a tensor, the state of stress
in a material is defined as
σ x
σ = τ xy
τ zx
τ xy
σy
τ yz
τ zx
τ yz
σ z
(4.1)
It can be shown that the coordinate system in which we measure the stresses can be rotated
so that the shear stresses vanish. The three remaining stresses are normal stresses, known
as the principal stresses, and are denoted as σ 1, σ 2, and σ 3.
These stresses are important because, regardless of what coordinate system we view
the component in, the stress state is uniquely determined. Also, in some materials, these
stresses are associated with failure and fracture.
These points are sometimes shown graphically through use of Mohr’s circle. The determination of the principal stresses will be discussed later in this section.
111
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Ballistics: Theory and Design of Guns and Ammunition
112
σy
σz
τyx
τyz
τxy
τzy
σx
σx
τzx
τxz
σz
σy
FIGURE 4.1
Cartesian stress components.
It is also very important to understand this when we try to examine the stress levels in a
part experimentally with a strain gage. Stress is a point function defined by force per unit
area expressed as
σ=
F
A
(4.2)
where
σ is the stress
F is a force
A is the cross-sectional area of the component
The same equation also holds if we use the symbol τ signifying a shear stress.
When we examine a structure, we normally are given the loads that are imposed on it.
We then either choose a material or evaluate a given material to see how it will behave
under the applied loads. This process requires us to convert the external loads to stress.
These stresses will cause movement of the material in the form of either stretching (tension) or compression. This movement is the actual displacement of the material. There is
an intermediate analytical step between these two where we need to define the strain of
the material. The strain in the material is defined as the change in length of a part over its
initial, unstressed length. Mathematically, this is expressed as
ε=
∆l
l
(4.3)
We require a relationship between stress and strain to evaluate material behavior under
a load. The link between stress and strain is called a stress–strain relationship. The most
common and simplest stress–strain relationship is that for a linear-elastic material. This is
known as Hooke’s law and is given for small deformations and uniaxial loading by
ε=
© 2014 by Taylor & Francis Group, LLC
σ
E
(4.4)
Ammunition Design Practice
113
where E is the modulus of elasticity, sometimes known as Young’s modulus. In a linearelastic material, any loading and unloading of the structure occurs along a curve in stress–
strain space that has a slope equal to the modulus of elasticity. Under the assumption of
general loading, material will be “pulled in” in the transverse directions as it is stretched
longitudinally. The ratio of lateral strain to axial strain is denoted as ν and called Poisson’s
ratio and is given for an isotropic material as
ν =−
εy
ε
=− z
εx
εx
(4.5)
This assumption of general loading changes our Hooke’s law relation as follows:
εx =
σ x νσ y νσ z
−
−
E
E
E
(4.6)
εy = −
νσ x σ y νσ z
+
−
E
E
E
(4.7)
εz = −
νσ x νσ y σ z
−
+
E
E
E
(4.8)
While we have defined ε to represent longitudinal strain in a material, a different type of
strain can be examined—shear strain. Shear strain, γ, is defined as the angular deviation
of a material from its original, undeformed shape. Shear strain is given by its own version
of Hooke’s law as
γ =
τ
G
(4.9)
where G is known as the shear modulus of the material.
In an isotropic material, E, ν, and G are not independent. The relationship that links
them is
G=
E
2(1 + v)
(4.10)
When we perform hand calculations, it is customary to convert the loads to stresses, then
the stresses to strains, and finally strains to deformations. The process is somewhat different (i.e., reversed) in a finite element analysis (FEA).
The determination of the principal stresses is important in several failure criteria. When
a part is being examined experimentally during a gun launch, it is customary to utilize a
strain gage. A strain gage measures the change in a parts length using the fact that resistance increases in a conductor as it is stretched. Strain gages are not always placed along
the directions in which it is desired to compute stress however. Since strain gages only
measure in-plane stress, it is common to transform this two-dimensional measurement
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Ballistics: Theory and Design of Guns and Ammunition
114
σy
y
y´
σy΄
τxy
x´
θ
τx΄y΄
σx΄
σx
x
FIGURE 4.2
Transformation of stress components.
into a desired in-plane direction. To transform stress from the strain gage coordinate system to the desired coordinate system, we use the following equations:
σ x′ = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
(4.11)
σ y ′ = σ x sin 2 θ + σ y cos 2 θ − 2τ xy sin θ cos θ
(4.12)
τ x′y ′ = τ xy (cos 2 θ − sin 2 θ ) + (σ y − σ x )sin θ cos θ
(4.13)
In each of these equations, the primed variables are those in the desired direction and
the unprimed variables are those measured by the strain gages. This is depicted in
Figure 4.2.
Rotation of coordinate systems in three dimensions is covered in excellent detail in Ref.
[1]. It was stated earlier that a rotation can be made such that the shear stresses vanish
and this results in what are known as principal stresses [2]. To determine the values of the
principal stresses, we determine the stress invariants through solution of the eigenvalue
problem. The three stress invariants are given by
I1 = σ x + σ y + σ z
(4.14)
2
2
I2 = σ xσ y + σ yσ z + σ zσ x − τ xy
− τ yz
− τ z2x
(4.15)
2
2
2
I 3 = σ xσ yσ z − σ xτ yz
− σ yτ zx
− σ zτ xy
+ 2τ xyτ yzτ zx
(4.16)
Once these invariants are obtained, the principal stresses are obtained through
I1 2 2
I1 − 3 I 2 cos φ
+
3 3
(4.17)
σ2 =
I1 2 2
2π
I1 − 3 I 2 cos φ +
+
3 3
3
(4.18)
σ3 =
I1 2 2
4π
I1 − 3 I 2 cos φ +
+
3 3
3
(4.19)
σ1 =
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Ammunition Design Practice
115
In Equations 4.17 through 4.19, the quantity 𝜙 is calculated through
φ=
2I 3 − 9I I + 27 I
1
1 2
3
cos −1 1
3
3
2 ( I 2 − 3I ) 2
1
2
(4.20)
Now we have all of the basic information necessary to discuss failure criteria. Limits of
space prevent a more in-depth treatment of this topic. The reader is referred to the references at the end of this chapter for a more detailed treatment.
Problem 1
For the state of stress below, find the principal stresses and the maximum shear stress.
20
[σ ] = 15
0
15
4
0
0
0 [MPa]
−9
Answer: τmax = 19 [Mpa]
4.2 Failure Criteria
When embarking on the design of a particular projectile component, we must initially
determine certain characteristics of the material contemplated for the design: Will we
use a metal or a plastic? Does it have a distinct yield point? Is it brittle or very ductile?
Such determinations will govern which criteria we use when we calculate the stresses
that will cause failure of the component. There are three commonly used criteria for
yield or failure: von Mises, which is also known as the maximum distortion energy
criterion; Tresca, which is known as the maximum shear stress criterion; and Coulomb,
which uses a maximum normal stress criterion. Other materials may require unique
failure criteria, for example, composites or nonisotropic metals may require Tsai–Wu or
Tsai–Hill criteria.
The von Mises or maximum distortion energy criterion is used typically when the
component is to be made of metal. It assumes that the energy required to change the
shape of the material is what causes yielding and that a hydrostatic state of stress will
not result in failure. The materials for which it is used should have a distinct yield point.
Our convention shall follow that of structural engineers in which we shall assume tensile
stress to be positive. By this criterion, we assume that the distortion of the material will
precipitate the failure. We shall order the stresses with 1 as largest to 3 being smallest
and state the following:
(σ 1 − σ 2 )2 + (σ 2 − σ 3 )2 + (σ 3 − σ 1 )2 = constant
© 2014 by Taylor & Francis Group, LLC
(4.21)
Ballistics: Theory and Design of Guns and Ammunition
116
σ2
If stress state falls in this
region, the component
is OK
+σY
–σY
+σY
–σY
σ1
This is a 2D
representation of an
ellipsoid which also
includes the σ3 (out of the
plane of the paper)
direction
FIGURE 4.3
von Mises failure surface.
We set this constant equal to 2σ Y2 or 6K 2. Here σ Y is the yield stress in simple tension and K
is the yield stress in pure shear. This implies that
1 2
σY = K2
3
(4.22)
or
K=
2 σY
σY
= 1.155
3 2
2
(4.23)
σ Y is also known as the equivalent stress and either σ Y or K can be found experimentally.
In σ 1–σ 2–σ 3 space, the criterion is represented by an ellipsoidal surface whose inner region
symbolizes stress states that are safe (nondistorting). This is shown two dimensionally in
Figure 4.3.
The Tresca or maximum shear stress criterion is used when the material is known to
have great ductility. It assumes the failure mechanism is by slippage along shear planes
generated by the shear stress in the material. This assumption says that the material will
not fail unless the shear stress it is experiencing is greater than that exhibited by a tensile
test specimen of the same material at its failure point. Again we assume that tensile stress
is positive and order the stresses with 1 the largest to 3 the smallest and state the following:
(σ 1 − σ 3 )
= constant
2
(4.24)
We set this constant equal to σ Y2 or K. Here σ Y is the yield stress in simple tension and K is
the yield stress in pure shear. This implies that for a component not to exhibit failure
(σ 1 − σ 3 ) < σ Y
(4.25)
σ Y = 2K
(4.26)
and
© 2014 by Taylor & Francis Group, LLC
Ammunition Design Practice
117
σ2
If stress state falls in this
region, the component
is OK
+σY
–σY
σ1
+σY
–σY
This is a two-dimensional
representation of a
polyhedron which also
includes the σ3 (out of the
plane of the paper)
direction
FIGURE 4.4
Tresca failure surface.
as well as
(σ 1 − σ 2 ) < σ Y , (σ 1 − σ 3 ) < σ Y , and (σ 2 − σ 3 ) < σ Y
(4.27)
Once again either σ Y or K can be found experimentally. In σ 1–σ 2–σ 3 space, this is represented by a polyhedral surface whose inner region includes all stress states that are safe
(nonfailing). This is shown as a 2D sketch in Figure 4.4.
The Tresca criterion is slightly more conservative if used for metals than von Mises. The
Tresca polyhedron is contained within (circumscribed by) the ellipsoid of von Mises.
The third failure criterion we will examine is the Coulomb or maximum normal stress
criterion. Here, we assume that the normal stress in the material will precipitate the failure. Tensile stress is again assumed to be in the positive direction and stresses from 1–3
are again in order of decreasing magnitude. In this criterion, we require that for a material
that does not exhibit failure
σ 1,σ 2 ,σ 3 < σ U
(4.28)
That is, all of the principal stresses must be less than the ultimate stress, σ U, in the material in that particular direction. Recall that we use this for brittle materials where there
is no yield point or yielding behavior. The failure surface is a rectangular polyhedron
whose edges are the ultimate stresses in each principal direction. Stress levels within
the polyhedron will not cause failure. A two-dimensional representation is depicted in
Figure 4.5. Even though this Figure is shown as a square, in many materials the compressive strength is much greater than the tensile strength, resulting in different limits
and thereby changing the appearance (and sometimes resulting in a name change as
well to a Mohr–Coulomb criteria) of the failure surface. In this instance, the failure
surface would look like Figure 4.6. In the Mohr–Coulomb failure criteria, a greater
compressive normal stress allows the material to carry more load. This is caused by the
locking of slip planes akin to the sliding friction of a block causing greater resistance
when the block gets heavier (i.e., an increase in normal stress on the slip plane). When
this is applicable, our criteria result in an equation for the failure surface as follows:
max[|τ |−λ ⋅ (σ )] = σ E
© 2014 by Taylor & Francis Group, LLC
(4.29)
Ballistics: Theory and Design of Guns and Ammunition
118
If stress state falls in this
region, the component
is OK
σ2
σUT
σUC
σUT
Here σUC is the
ultimate stress in
compression and σUT
is the ultimate stress
in tension
σ1
σUC
FIGURE 4.5
Coulomb failure surface.
σ2
If stress state falls in this
region, the component is OK
This is a two-dimensional representation
of a polyhedron which also includes the
σ3 (out of the plane of the paper) direction
σYT
σYC
σYT
σ1
Here σYC is the yield stress
in compression and σYT is
the yield stress in tension
σ1 + σ2 = 0
σYC
FIGURE 4.6
Mohr–Coulomb failure surface.
This equation results in a greater stress to failure due to the internal friction coefficient, λ.
Since compressive strength is negative and λ is a positive quantity, the equivalent failure
stress, σ E, is greater with greater normal stress, σ.
Occasionally, it will be essential that we combine two or more of these criteria due to a
change in material behavior. We shall describe this in due course.
Problem 2
A component has principal stress values of 20,000, 56,000, and −220,000 psi (note that
negative means compressive stress), if the yield strength in a simple tension test of the
material was found to be 180,000 psi, will the part survive based on the von Mises failure
criteria?
Answer: No the part will fail.
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4.3 Ammunition Types
Just as weapons are categorized by their usage as guns (low angle, line of sight, direct fire),
howitzers (high angle, beyond line of sight, indirect fire), or mortars (very high angle, short
range, indirect fire), the munitions for them are also categorized, not by use, but by their
construction or assembly methods. Ammunition can be fixed, separable, or separate loaded.
Fixed ammunition, usually called a cartridge, consists of a container for the propellant
charge, called the cartridge case, that is firmly attached to the projectile by crimping or cement
and that remains in the weapon after firing and is ejected near it or is consumed during firing,
and the projectile that flies downrange to the target. The charge, priming, and ignition system
are assembled inside the case and are not alterable. This type of ammunition is characteristically used in tank, antiaircraft, aircraft weapons, and in most small arms (rifles and pistols).
Separable ammunition (also called semi-fixed ammunition) also consists of the cartridge
case and projectile, but the case is not attached firmly to the projectile and can be removed
in the field to adjust the charge, which can be changed incrementally. This type of ammunition was used in older howitzers and is still used in shotguns.
Separate-loaded ammunition (sometimes called separated ammunition) consists of the
projectile, which is loaded first into the weapon, the propellant charge loaded next, and
finally the primer and igniter loaded last. The charge, which is supplied to the weapon
site, is in bagged increments and is altered, along with the quadrant elevation of the
weapon, to vary the range. The primer is usually loaded into the weapon’s breechblock.
The block is self-sealing and assumes this function, which in fixed ammunition is done
by the cartridge case. Ammunition of this type is used in howitzers and large naval guns.
Mortar ammunition is essentially of the separated type. The charge is incremental to
help vary the range by altering the muzzle velocity. The charge increments are held in
place on the projectile body by clips or holders. Increments may be added or deleted in
the field by the gunner. Priming is done through an integral attachment to the projectile
(a boom). Primer initiation is by a firing pin in the weapon that strikes an initiator in the
boom at the termination of the fall of the projectile as it is dropped down the tube from the
muzzle end. Trigger firing is also possible in some weapon designs.
The practical design of fixed ammunition cartridges, which is what we will mainly dwell
on, encompasses the design of the propellant charge ignition system, the construction of
the main body of the propellant charge, the design of the projectile body itself, including its shape and mass distribution, and its obturation and stabilization components. The
design must also incorporate into the projectile the ancillary systems necessary for its
intended functioning, for example, fuzes, expulsion charges, explosive trains, and in modern projectiles, guidance, and control.
4.4 Propellant Ignition
Energetic devices that are combined in a specific manner into the ignition train accomplish the initiation of combustion of the propellant charge. The first of these elements is
the highly sensitive, detonator cup filled with material such as lead azide, which itself
receives an energy pulse from a trigger mechanism that delivers the pulse in the form of
a mechanical (spring actuated) or electrical (hot wire or laser) impulse. The sensitive mix
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120
is detonated by the impulse and flashes into the main, less sensitive ignition charge. This
material is contained in the primer head.
Secondary ignition takes place in the main primer body, where the ignition material
known as the primer charge is stored. This material has traditionally been fine-grained
black powder, which is known to have certain undesirable properties such as hygroscopicity. Attempts have been made to replace black powder, but it still remains the chief secondary ignition material. Two basic forms of primer charge are used in large caliber munitions:
flat base-pad igniter charges are used with separate-loaded bagged propellant charges and
in fixed, stick propellant charges; central core or bayonet-type primer bodies are used in
most fixed, loose, granular propellant charges.
The design goal of all ignition systems is to provide rapid but smooth ignition of the
main propellant charge avoiding at all cost pressure surges or spikes. Such surges can
crush individual grains or sticks causing large, uncontrolled increases in burning surfaces
and uncontrolled burning of the main charge. Symptoms of such burning are negative
delta pressure (−∆p) waves, that is, negative gradients of pressure along the length of the
chamber. One cause of pressure surges are the so-called blind primers, where vent holes
are missing along the length of the primer body tube. The pressure build-up in the tube
can rupture it causing asymmetric ignition and a −∆p.
Other caveats are to avoid overly sensitive detonator mixes and to provide gas flow
space in the main propellant charge. Ignition and burning are surface phenomena and too
tightly packed charges do not provide the necessary surfaces.
4.5 The Gun Chamber
To the rear of the long cylindrical portion of the gun (the bore) is the chamber, shown in Figure
4.7. The tapers shown facilitate the removal rearward of the spent cartridge case that hugs the
chamber wall. During the firing cycle, the case swells because of the internal pressure and
firmly contacts the chamber wall sealing the gases from exiting rearward. When the pressure
decays, a properly designed case comes away from the wall and the tapers insure that it does
Tapers greatly exaggerated
Bottom of groove
Top of land
Dc
D
Forcing cone
Shoulder
Rear face of tube
FIGURE 4.7
Chamber geometry.
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121
not stick in the chamber. When a fixed round of ammunition is loaded into the chamber, the
rear face of the tube provides the stop and seat for the rim of the case. During the expansion
of the pressurized case, the forcing cone of the chamber forms the seal for the hot gases by the
extrusion and engraving of the rotating band in a rifled bore or the extrusion of the obturating
band in a smooth bore. The ratio Dc/D is known as the chambrage, an important characteristic
of the design. Large values of the chambrage tend to cause turbulent flow of the gases as they
enter the bore. Such turbulence contributes to the erosion of the bore surfaces.
The gun designer is caught in a curious bind: for a desired volume of propellant, a large
chambrage provides a shorter cartridge length, frequently a highly desirable parameter in the
tight confines of a turret, for example; on the other hand, large chambrage values subject the
bore to more erosion. Some of this difficulty has been overcome by the use of erosion reducing coolants. It has been found that much of the erosive wear in high-performance guns and
howitzers can be ameliorated by the introduction of a cool liquid, gaseous, or particulate layer
between the hot propellant gases and the bore. Materials such as titanium dioxide, wax, talc,
or silicone oil have proven efficacious. If these materials are assembled in the body of the propellant charge so that the gas flow keeps the coolant at the bore wall, a substantial decrease of
erosion results. This is called laminar flow and is observed in low chambrage guns. Thus, a
compromise may have to be made in the chambrage to reduce the turbulent flow.
4.6 Propellant Charge Construction
In fixed cartridges, the most common practice is to fill a metallic cartridge case with perforated
granular propellant grains around a bayonet-type primer that has already been inserted in
the case. The grains commonly have seven perforations for progressive burning. In high-performance rounds, vibrating the case to help settle the grains maximizes the loading density
of the charge. Tank munitions are often loaded with perforated stick propellant. The sticks
are bundled and carefully laid up around the boom and fin components that intrude into
the depth of the case. Supplementary granular propellant is occasionally added to the stick
bundles to further boost the charge mass and increase the progressivity of burning. Rocket
grain configurations with complex star and slit perforations have been tried as well as 19-perf
grains to raise the burning rate, but these are difficult to make and are not standard.
Howitzer (separate-loaded) charges are made up of bagged increments that are ignited
by the last increment loaded, the base-pad igniter. A primer in the breechblock sets off the
igniter. In these and in the fixed ammunition charges, coolants are strategically emplaced
to promote erosion resistance. With bagged propelling charges, since there is no cartridge
case present, it is extremely important that all of the material be combusted. Great care is
taken in selecting materials—silk was used for many years in the Navy—to assure that
there are no burning embers left in the weapon after it was fired. It is typical for a howitzer
crewman to look down the bore and shout “bore clear” during firing operations. If burning materials are present and a fresh charge is inserted into the bore, the propellant may
ignite and cause serious injury to the gun crew. This has been termed “cook-off.”
There have been extensive efforts to take advantage of the convenience of stowage, low
cost, and inherent safety of liquid bipropellants (LP). However, severe operational and
performance problems have prevented their adoption. These problems have centered on
combustion instability that manifests itself in destructive, unpredictable pressure peaks, particularly in bulk-loaded systems. Attempts to get around these so-called Taylor instabilities
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122
have had some success with regenerative pressurized systems that atomize the pumped-in
liquids, ignite this cloud, and avoid the pressure wave unpredictability of an ignited bulk
of liquid. This concept, even though it has shown promise, still may not be able to overcome
the poor low-temperature properties of the liquid propellants. They show marked increases
in viscosity at low temperatures causing severe flow and pumping problems.
Two other concepts of gun propulsion should be mentioned. These are the use of electromagnetically generated force to propel a projectile down a gun and the idea of using a
low molecular weight gas to propel the projectile—the light gas gun. At the time of this
writing neither concept has shown the ability to progress beyond the laboratory stage to
a fieldable weapon, although light gas guns are in common use in laboratories to reach
velocities with small projectiles approaching meteorite entry speeds.
4.7 Propellant Geometry
The geometry of the propellant grain is one of the parameters available to the interior ballistician to tailor the pressure curve in the gun. Production of gas from a grain depends on
the evolution of the total surface of the grain as the burning proceeds. If the surface area
increases with time, the grain is considered progressive. If the total surface remains constant over time, the grain is neutral, and if the surface decreases with time the grain is considered regressive. The perforations in the grain affect the surface area and therefore the
burning characteristics. In cylindrical grains, the number of perforations is usually one of
the numbers in the sequence: 1, 7, 19, and 37. The largest number in use in the United States
is 19, and this is rarely found because of the difficulty of manufacture. The various types
of grains are shown in Figure 4.8. The web, D, that is the smallest thickness of propellant
between any two surfaces is one of the major parameters in interior ballistic computations.
D
fD
D
Single perf. grain—neutral
fD
Seven perf. grain—progressive
W
D
D << 1
W
4
Strip or flake—regressive
Ball—regressive
FIGURE 4.8
Typical propellant grain geometries.
© 2014 by Taylor & Francis Group, LLC
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4.8 Cartridge Case Design
The design of a metallic cartridge case must fulfill four basic roles: the case must seal or
obturate the gun breech so that gases do not stream backward out of the gun; it must serve
as a protective container for the propellant charge; it must act as a structural member of the
cartridge assembly to allow for vigorous handling during shipping, stowage, and loading
into the chamber; and it must be easily extractable from the chamber after the round is
fired. Metallic cases have been used for much more than a hundred years and the design
practices are well established to fulfill these roles. Yet difficulties still arise in the extraction of the case after firing—it can stick in the chamber, rendering the weapon useless
until it is removed. The case by itself cannot sustain the gun pressure and is intended to be
supported by the chamber walls. Yet, the case must be designed with sufficient clearance
to permit loading and ramming. The analysis of sticking that follows must be part of the
design engineer’s overall task before a new weapon can be fielded.
It is possible, through use of some relatively simple equations, to determine if a cartridge
case will expand enough to stick in the chamber of the weapon after firing. Graphically, we
can depict this as shown in Figure 4.9. In this figure, we see the effect when a case with a
low yield strength is loaded to the same levels as a good case. The expansion and contraction of the gun tube itself must be taken into account when the cartridge case is designed.
This condition can be approximated using a bilinear, kinematic hardening model where
the stress–strain curves of the case material are modeled, as depicted in Figure 4.10.
The first step in this procedure is to model the gun tube. In this case, we assume that the
material is perfectly elastic—which will be the case for any properly designed tube—and
we can determine the radial expansion through [2]
utube =
a′
(1 −ν )( p1a′2 − p2b 2 ) + (1 + ν )b 2 ( p1 − p2 )
Etube (b 2 − a′2 )
(4.30)
In this equation (which has been tailored from a previous formula for a thick-walled
cylinder because the point we are interested in is on the inside radius of the tube wall),
Hoop stress in case wall
Clearance between case and
chamber before firing
Elastic expansion of the
chamber on firing
Yp Normal case
Yp Low yield case
Expansion (strain)
Low yield case interferes
with chamber by this
amount
FIGURE 4.9
Stress–strain diagram of a normal case and one with low yield strength.
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
124
Hoop stress in case wall
Clearance between case and
chamber before firing
Elastic expansion of the
chamber on firing
Tangent
modulus
Modulus of
elasticity
Modulus of
elasticity
Expansion (strain)
Low yield case interferes
with chamber by this
amount
FIGURE 4.10
Stress–strain diagram of a normal case and one with low yield strength modeled as bilinear kinematic hardening materials.
a′ is the inner radius of the chamber, b is the outer radius of the gun tube, p1 is the internal pressure, p2 is the external pressure (usually conservatively taken as 0), ν is Poisson’s
ratio for the tube material, and Etube is the modulus of elasticity.
We now calculate the stress, strain, and displacement of the case through use of the thinwall cylinder equations [3]
a 2 p1
Ecase h
(4.31)
σ θθ =
ap1
h
(4.32)
εθθ =
σ θθ
Ecase
(4.33)
ucase =
In these equations, ucase is the radial expansion of the case, σ θθ is the hoop stress in the case,
εθθ is the hoop strain, a is the outside radius of the case, and h is the case wall thickness.
Now, the gun tube will stop the case from expanding further once contact is made so the
maximum expansion of the case will be as follows:
rcase + ucasemax = rtube + utube = rcase + aεθθmax
(4.34)
Because we know the pressure and the tube dimensions and therefore the value of utube,
we can calculate εθθmax. We can then use this value to calculate the stress in the case at the
maximum expansion:
εθθmax − ε Y =
σ θθmax − σ Y
Ecase-tangent
(4.35)
In this expression, the subscript Y indicates yield values and Ecase-tangent is the tangent
modulus of the cartridge case material. Once we determine the stress at the maximum
© 2014 by Taylor & Francis Group, LLC
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125
expansion, we need to recall that a material which has yielded will retract along its original elastic modulus. Thus, we can write
ε return =
σ θθmax
Ecase
(4.36)
Now the residual strain in the case is given by
ε residual = εθθmax − ε return
(4.37)
We can then find the permanent radial displacement through
uresidual = aε residual
(4.38)
If we now add uresidual to the original radius of the case, a, we can see that if
uresidual + a ≥ a′, the case will stick
(4.39)
uresidual + a < a′, the case will not stick
(4.40)
or if
Over the last 20 years, the metallic case of drawn brass, extruded steel, or spirally wrapped
steel has been replaced in certain systems by a fully combustible or consumable case. These
cases are manufactured of felted nitrocellulose and usually consist of a base and sidewall
that are assembled with cement, filled with granular or stick propellant, and attached to
the projectile with clamps and cement. Since the cases are consumed completely, they do
not seal the breech. With these munitions, a self-sealing breech must be designed for the
weapon. For guns with nonsealing breeches that are already fielded for use with conventional metallic cartridge cases, a case that has a metallic stub and a combustible sidewall has
been devised to take advantage of the small volume of the ejected stub in the confines of a
tank turret, e.g., and the overall reduction in cost and weight of the round. While systems
with the combustible case have been fielded, the success of this development has not been
complete. Occasional problems with incomplete combustion of a case that leaves smoldering residue capable of igniting the next loaded round (cook-off) have required scavenging
systems for the chamber to be installed. The inherent structural weakness of nitrocellulose
has also posed problems of case attachment and handling. Yet the obvious advantages of the
combustible case have kept the concept in the weapon designer’s toolbox for possible use.
Problem 3
A design for a 105 mm weapon is being considered. The chamber is stated to withstand the
desired 35,000 psi and is essentially a steel cylinder of 4.5 in. ID and 7 in. OD (Etube = 30 × 106
psi, ν = 0.3). We have decided to use brass with an OD of 4.490 in. If we use a bilinear, kinematic hardening model where the brass has a modulus of elasticity of 15 × 106 psi, a local
tangent modulus of 12.5 × 106 psi, a yield stress of 15,000 psi (yield occurs in this material at
ε = 0.001), and an ultimate tensile strength of 45,000 psi, with the information given, what
is the radial clearance between the case and the chamber after firing neglecting thermal
effects?
Answer: Approximately 0.004 in. radial clearance
© 2014 by Taylor & Francis Group, LLC
126
Ballistics: Theory and Design of Guns and Ammunition
4.9 Projectile Design
While propulsion systems are fairly straightforward in design because their intended use
is simple, projectiles vary widely in use and as a consequence their designs are complex
and demanding. The propulsion system must get the projectile through the launch environment with consistent muzzle velocities, but without undue stress to the gun or the
projectile. The projectile, on the other hand, must withstand the forces of launch, be efficient, consistent and precise in its flight environment, and deliver its intended utility at the
target. We will explore only projectile design for launch in this section, reserving design
for flight and terminal effects until later.
Projectiles may be classified into two general types: cargo carriers and pure kinetic
energy deliverers. The cargo carriers include shells that deliver high explosives (HE), submunitions and mines, pyrotechnics, smart munitions, and other specialized lethal systems, e.g., shaped charges (HEAT) and explosively formed penetrators (EFP) shells. The
kinetic energy delivery systems, used chiefly for the attack of armor, are monobloc steel
shot (AP), saboted, long-rod, heavy metal penetrators (APFSDS), and older types of spinstabilized, saboted (APDS) projectiles.
The stresses induced into a projectile during launch are chiefly due to the acceleration
that the gases impart to it. The cargo carriers are shells whose stresses are due to relatively
low accelerations and which, except for the tank cannon fired HEAT shell, achieve only
moderate muzzle velocities. We will therefore explore the kinds of stresses and failures
inherent in shell-like structures under load in Section 4.10. Kinetic energy munitions, on
the other hand, are subject to extremely high accelerations and have high muzzle velocities. For these types, we will explore the driving mechanism stresses and other aspects of
these designs.
The gamut of topics in projectile design is almost unlimited. However, several suggest
themselves because of their general applicability or timely interest. Shell design is a ubiquitous problem and will be explored in depth in Section 4.10. The use of buttress threads
is so common in projectile and gun design that it warrants its own in Section 4.11. Sabot
design is more specialized as are the problems of kinetic energy rods and their buttress
driving grooves. These will be explored in Section 4.12.
Modern projectiles employ a variety of electronic and electromechanical devices for fuzing, target detection, and guidance and control. This relatively new engineering discipline
called “gun hardening” deals with designing these devices to survive the harsh environment of gun launch.
4.10 Shell Structural Analysis
Most cargo-carrying projectiles, whether fin- or spin-stabilized, are designed with cargo
bodies in the shape of an axisymmetric cylindrical shell. Because the loads on these cylinders are the result of spin and acceleration of the shells and their contents, the stresses
encountered are highly variable along and through their walls. These stresses will be
examined as will the consequences of failure criteria.
© 2014 by Taylor & Francis Group, LLC
Ammunition Design Practice
127
The symbols and definitions of the constants and variables of shell loading are
tabulated next:
A—Bore area of the gun
a—Linear acceleration
d—Diameter of bore (across lands), diameter of assumed shear circle in base of shell
di—Inside diameter (ID) of projectile
do—Outside diameter (OD) of projectile
Fb—Maximum force on base of projectile and rotating band
FT—Maximum tangential force on projectile wall
FTR—Hoop tension (force) in wall of projectile resulting from rotation of the shell
FT′ —Tangential force at section of shell
f′—Setback force
g—Acceleration due to gravity
h—Total depth of filler from nose
h′—Total depth of filler from nose end of cavity to section under consideration
Izz —Polar moment of inertia
I ′zz —Polar moment of inertia of metal parts forward of section when section is ahead of
rotating band and aft of it when section is aft of the rotating band
n—Twist of the rifling
pb—Maximum propellant pressure
ph—Filler pressure due to setback
prot—Filler equivalent pressure due to rotation, includes wall inertia
ri—Inside radius of projectile
ro—Outside radius of projectile
S—Compressive strength of the rotating band
S1—Longitudinal stress
S2—Tangential stress
S3—Radial stress
τ—Shear stress
σ Y—Static yield stress in tension
T—Torque applied to the projectile
t—Base thickness, wall thickness
V—Muzzle velocity
w—Total projectile weight
w′—Weight of metal parts forward of section under consideration
w′f —Weight of filler forward of section under consideration
α—Angular acceleration
ρm—Density of projectile material
ρf—Density of filler material
ω—Angular velocity
r b—Radius of band seat
pband—Band pressure
We distinguish between thin-walled and thick-walled cylinders in this analysis so that
the designer may run quick, ballpark estimates of the stress levels encountered. In practice, FEA is usually conducted on the components, but as emphasized earlier, the designer
should have a good idea of the bounds of the answer before beginning the FEA.
© 2014 by Taylor & Francis Group, LLC
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128
We begin with a review of basic mechanics of materials as applied to cylinders. If a cylinder is subjected to an axial load and does not buckle, the axial stress can be determined from
S1 = −
FAxial
FAxial
=−
A
π ro2 − ri1
(
)
(4.41)
The stress–strain relationships for a cylinder is as follows:
ε rr =
1
[σ rr −ν (σ θθ + σ zz )]
E
(4.42)
εθθ =
1
[σ θθ −ν (σ rr + σ zz )]
E
(4.43)
ε zz =
1
[σ zz −ν (σ θθ + σ rr )]
E
(4.44)
where
ν is Poisson’s ratio
σrr is the radial stress
σ θθ is the transverse (hoop) stress
σzz is the axial stress
E is Young’s modulus
If a cylinder is subjected to a torsional load, it will twist. We typically assume that this
deformation is small and plane sections remain plane. Thus, when we apply a torque, T, to
a cylinder of length, L, with shear modulus, G, and polar moment of inertia, J, the structure
will rotate through an angle ϕ (in radians):
φ=
TL
JG
(4.45)
For a hollow cylinder,
J=
(
1
π ro4 − ri4
2
)
(4.46)
For a material that behaves according to Hooke’s law,
G=
E
2(1 + ν )
(4.47)
Such a material under pure torsion will only exhibit shear stress according to
τθ z =
© 2014 by Taylor & Francis Group, LLC
Tr
J
(4.48)
Ammunition Design Practice
129
S2
r
S1
t
FIGURE 4.11
Thin-wall cylinder geometry.
While the thick-wall cylinder analysis, which we describe next, is an exact solution, a quick
way to assess the major stresses if the wall thickness is less than 10% of the cylinder radius
is to assume that the stresses in the radial direction, S3, are negligible.
Thus, we examine only the meridional or longitudinal and the circumferential or hoop
stresses. We define S1 as the longitudinal stress, S2 as the hoop stress, and p as the pressure
depicted in Figure 4.11.
If the cylinder has closed ends, then internal pressure can cause a longitudinal stress
S1 = σ zz =
pr
2t
(4.49)
otherwise S1 = 0. Internal (or external) pressure always causes hoop stress
S2 = σ θθ =
pr
t
(4.50)
In practical shell design, we always perform a thick-wall cylinder analysis assuming that
the stresses in the radial direction are significant enough to be considered. Thus, we must
examine longitudinal, hoop, and radial stresses. We again define S1 = σzz = longitudinal
stress, S2 = σ θθ = hoop stress, S3 = σrr = radial stress, and p = pressure. This is depicted in
Figure 4.12.
The following solutions are known as the Lamé formulas and assume open ends, which
implies S1 = 0 if no axial loads are present. If axial loads are present, they must be accounted
for. Internal (or external) pressure always causes hoop stress. (Note that the subscripts “o”
and “i” refer to the outer and inner surfaces, respectively.)
S2 = σ θθ =
(
2
ri2ro2 ( po − pi )
1
2
p
r
−
p
r
−
i
i
o
o
r2
ro2 − ri2
)
S3
S2
ro
r
ri
FIGURE 4.12
Thick-wall cylinder geometry.
© 2014 by Taylor & Francis Group, LLC
S1
(4.51)
Ballistics: Theory and Design of Guns and Ammunition
130
with a maximum at r = ri. The radial stress can be calculated from
S3 = σ rr =
(
2
r 2r 2 ( p − pi )
1
p r − po ro2 + i o o2
2 i i
r
r − ri
2
o
)
(4.52)
with a maximum again at the inner surface r = ri, and equal to S3 = −pi.
Initially, we will analyze the state of stress caused by the centrifugal loading induced by
the rotation of a projectile in a rifled gun tube. In a spin-stabilized projectile, besides the
longitudinal loads induced by the acceleration through the tube, the rotation of the projectile, which is dependent upon the axial velocity and the twist of the rifling in the tube,
induces stresses in the walls. The twist of the rifling is usually measured in revolutions
per caliber of travel (i.e., a twist of 1 in 20 means the projectile makes one revolution in
20 calibers of travel [n = 20]). The units of n are calibers per revolution. If we multiply n by
the diameter, d, we get units of length per revolution:
length
caliber length
n
×d
= nd
revolution
caliber
revolution
(4.53)
Since there are 2π radians per revolution, the angular velocity a projectile has attained is
defined as
radians length
2π V revolution time radians
=
= [t −1 ]
ω=
=
length
time
nd
revolu
ution
(4.54)
The centrifugal force directed radially outward on an element of material at radius r is
Fc = mar =
w 2
rω
g
(4.55)
In the tangential direction, the inertial forces on an element of material can be determined
from
Ft = mat =
w
rα
g
(4.56)
We can determine the centrifugal force on the cylinder wall caused by spinning the cylinder in the absence of other loads by integrating Equation 4.55 from the inner diameter
to the outer diameter. To do this, we consider the differential element as depicted in
Figure 4.13. From this diagram, we see that the mass of an infinitesimal annular ring of
material is
dm =
© 2014 by Taylor & Francis Group, LLC
dw
= ρ dV = ρ l 2π rdr
g
(4.57)
Ammunition Design Practice
131
l
Angular velocity, ω
Density
ρ
r
dr
ri
ro
Projectile model
FIGURE 4.13
Differential thickness element geometry.
Inserting Equation 4.57 into Equation 4.55 yields
dFc = ar dm = ρ l 2π r 2ω 2dr
(4.58)
which, when integrated from the inner to the outer radius, gives
ro
FcWALL = 2πρ lω
2
∫
r 2 dr =
ri
2πρ lω 2 3
ro − ri3
3
(
)
(4.59)
This is the radial force on the wall due to the inertia of the wall material only. If the projectile is filled with material, we need to account for this filler as well. Thus, if we integrate
from the centerline to the inner radius of the projectile wall, we obtain
ri
∫
dFcFILL = 2πρFILL lω 2 r 2dr =
0
2πρFILL lω 2 3
ri
3
( )
(4.60)
The total force acting on the projectile wall due to spin is then
Fc = FcWALL + FcFILL =
2π lω 2
ρ ro3 − ri3 + ρFILL ri3
3
(
)
( )
(4.61)
For stress computations, we require an internal pressure; thus, we need to convert the centrifugal forces to an equivalent internal pressure. If we assume that our centrifugal forces are acting
on the interior of the shell, pushing radially outward, the area for our equivalent pressure is
Arad = 2π ril
© 2014 by Taylor & Francis Group, LLC
(4.62)
Ballistics: Theory and Design of Guns and Ammunition
132
Thus, our equivalent pressure can be written as
prot =
ω2
Fc
ρ ro3 − ri3 + ρFILL ri3
=
Arad 3ri
(
)
( )
(4.63)
In Equation 4.56, we determined the tangential force arising from the angular acceleration.
If we perform a similar analysis to that which developed Equation 4.61, we will obtain an
expression for the torque as follows:
T = MWALL + MFILL =
(
)
( )
1
πα l ρ ro4 − ri4 + ρFILL ri4
2
(4.64)
The derivation of this is left as an exercise for the interested reader and is included as a
problem at the end of the chapter.
The formulas for calculating the tangential and radial stresses at radial location, r, in a
rotating cylinder where ro > 10(ro − ri) can be given as
2 2
3 + ν 2 2 ri ro 1 + 3ν 2
r
σ θθ = ρω 2
ri + ro + 2 −
3 +ν
r
8
(4.65)
2 2
3 + ν 2 2 ri ro
2
σ rr = ρω 2
ri + ro + 2 − r
8
r
(4.66)
We are reminded that the longitudinal stress (assuming the structure does not buckle)
is simply the axial acceleration multiplied by the weight of all of the material forward
of the location of interest divided by the shell cross-sectional area—we will discuss this
presently. These formulas were developed for the centrifugal loading of a spinning projectile by forces that act during both in-gun setback and flight. The axial load on a
projectile, however, is for the most part only present during acceleration in the tube, is a
function of time, and occurs whether the projectile is spinning or not. Beyond this, there
is also an applied torque due to the angular acceleration, which is applied through the
rotating band or slip obturator. The setback load and (if spinning) the centrifugal and
torsional loads must all be superimposed on the projectile to determine its state of stress.
The axial force on the projectile during firing is given by
F = ps A
(4.67)
where ps is the pressure acting on the base of the projectile defined by the Lagrange
approximation
1
ps = pB
c
1+
2w
© 2014 by Taylor & Francis Group, LLC
(4.68)
Ammunition Design Practice
133
The D’Alembert force is the force due to acceleration that exactly equals this pressure force
a=
ps Ag
w
(4.69)
At any axial position, the force on the cross-sectional area can be shown to be proportional
to the weight of material forward of the section:
f′ =
w′
ps A
w
(4.70)
To calculate the force (or really the pressure) in the filler material, we usually resort to a
hydrostatic model
ph = ρ ha = ρ h
ps Ag
w
(4.71)
where
ρ is the density of the filler
h is the filler head height
ph is the hydrostatic pressure that is developed
In a spin-stabilized projectile, the angular acceleration, α, is proportional to the linear
acceleration, a, where
α = Ka
(4.72)
Then,
α =K
ps Ag
w
(4.73)
where K has units of length−1 and is dependent upon the twist, n (in calibers of travel per
turn), and the bore diameter, d, thus
K = tan θ =
2π
nd
(4.74)
where θ is the angle between the circumferential twist distance and the axial distance
traveled. From Equations 4.73 and 4.74, we get
α=
2π ps Ag
nd w
(4.75)
If we define a tangential force applied to the rotating band of the projectile as FT, then the
torque on the projectile is
T = FT
© 2014 by Taylor & Francis Group, LLC
d
2
(4.76)
134
Ballistics: Theory and Design of Guns and Ammunition
We know that the torque is equal to the product of the polar moment of inertia of the projectile and its angular acceleration
T = I zzα
(4.77)
Solving for the angular acceleration in terms of the tangential force, we get
α=
FT d
I zz 2
(4.78)
Inserting this into Equation 4.56 and solving for FT yields
FT =
π 2I zz ps
nw
(4.79)
Since
A =π
d2
4
(4.80)
The force that is applied by the rifling to the rotating band is transmitted through the
structure to regions both forward and aft of the rotating band. These forces are proportional to the moment of inertia of the sections ahead of or behind the application of the
torque load, I ′zz. We assume that this force acts over a mean diameter of the outer and inner
wall surfaces of the shell and then we get
FT′ =
16π I ′zz ps A
n(do + di )2 w
(4.81)
Because the rotating band is intended to act as a gas seal (obturator) as well as the rotational driver, designs typically exhibit a diameter over the band that is slightly larger
than the groove diameter of the weapon. The engraving action of the gun lands and the
interference fit in the grooves causes a plastic flow of the band, resulting in a pressure
on the band seat as well as a developed reaction in the gun wall. This pressure can be
greater than the gas base pressure on the projectile. Measurements of this pressure have
been obtained by strain gaging of the gun tube and computing the stress at the weapon’s
inner diameter. The pressure required to cause this stress is called the interface pressure.
It has been shown that cannelures or circumferential grooves cut into the band surface
reduce this pressure substantially by allowing room for band material to flow rather than
being loaded in a quasi-hydrostatic condition. This is depicted in Figure 4.14. The composition/material of the rotating band can have a dramatic effect upon the behavior of
the projectile in the tube as well as tube wear. An excellent example of this relationship
is contained in Ref. [4].
We have the forces on the projectile structure but now must translate these into
stresses that allow us to determine how much design margin is present. Once determined, these stresses are then linked to well-established failure criteria to determine
the failure point of the material. Since projectiles may be made of a variety of materials,
© 2014 by Taylor & Francis Group, LLC
Ammunition Design Practice
135
pband
ps
Cannelures
FIGURE 4.14
Rotating band pressure.
r
z
Point A
j=1
j=4
j=2
j=3
FIGURE 4.15
Stress locations in an M1 high explosive (HE) projectile.
specialized criteria may have to be used on each material. This full procedure is somewhat complicated and beyond the scope of this book, but we will attempt to describe
the basics through an examination of a simple M1, high-explosive projectile structure
depicted in Figure 4.15.
Assume a thick-walled cylinder as shown for stress calculations where
S1j —Longitudinal stress at the jth location
S2j —Hoop stress at the jth location
S3j —Radial stress at the jth location
τ 11—Longitudinal shear at the base
τ 2j —Torsional (shear) stress at the jth location
It is helpful to recap here all of the loads on an element of projectile wall material at a
generalized location (such as point A) in the diagram. This element of material is
• Compressed in the axial direction due to the axial acceleration
• Loaded in tension in the hoop direction because of the wall mass being pulled
radially outward due to the spin
• Loaded in tension in the hoop direction because of the filler material moving outward due to the setback load and the spin
• Loaded in shear due to the rotating band accelerating the projectile in an angular
direction
• Loaded in shear due to the greater stress in the outer wall than on the inside wall
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
136
pband
ps
FIGURE 4.16
Load conditions for an M1 HE projectile.
Note that when including mass forward of a particular section we must include all mass
transmitting loads to the section, e.g., fuze, bushings, and cups. The pressures applied to
our model of the M1 projectile are shown in Figure 4.16.
Now let us examine specific locations of interest along the shell where experience tells us
failures might occur. For convenience, these have been tabulated in Table 4.1 and tailored
to each individual location with the symbol, source load, and type of stress noted.
At location 1, these are the formulas used to calculate stresses due to the setback of filler
on base, the moments caused thereby, and by gas pressure on base:
S11 = − ph
(4.82)
S21 = 0
(4.83)
S31 =
ro2
( ps − ph )
t2
3r 3
o
S′31 = − ps
2 ro3 − ri3
(
)
(4.84)
r 3 + 2r 3
i
+ ph o
2 ro3 − ri3
(
)
(4.85)
TABLE 4.1
Typical Stresses in a High Explosive (HE) Projectile and Their Sources
Type of Stress
Symbol
Compressive load on base
Radial stress on base at centerline
S11
S31
Radial stress on base at centerline
′
S31
Hoop stress at rear of the band
S22
Radial stress at ends of band and
maximum ID
Longitudinal stress at ends of band and
maximum ID
Hoop stress at forward end of band and
maximum ID
Shear stress through thickness t
S32, S33, S34
τ11
Torsional shear in projectile wall
τ22, τ23, τ24
© 2014 by Taylor & Francis Group, LLC
S12, S13, S14
S23, S24
Source of Load
Setback of filler
Moments of filler setback and base
pressures (flat base)
Moments of filler setback and base
pressures (round base)
Setback of filler, rotation, and
external pressure (band and gas)
Rotation of projectile, filler setback,
and filler rotation
Setback of metal parts in wall (filler
contribution usually neglected)
Filler pressure and rotation of wall
Moments of filler setback and base
pressures (round base)
Setback of filler, rotation, and
external pressure (band and gas)
Ammunition Design Practice
137
Equation 4.82 is the axial component stress. We can see that it is just driven by the reaction
of the fill and shell to the axial acceleration. Since this is a centerline location, by definition
there is no hoop stress, which is defined by Equation 4.83. Equation 4.84 specifies the radial
stress assuming the base is flat faced. This comes about from the difference in the base
pressure reacting against the internal forces and attempting to push the center of the base
into the fill. Equation 4.85 is the radial stress equation assuming the base is a rounded bottom (i.e., with the concave portion enclosing the fill). We can see from this equation that the
stresses are much lower as it carries the load more efficiently than a flat bottom shell. The
drawback is that a base of this type requires a skirted boat tail which is more expensive to
manufacture but saves considerable weight.
Moving to location 2, these are the stresses due to setback of filler, filler rotation, wall
rotation, and band pressure:
S12 = −
w′ + w′f ps A
w π ro2 − ri2
(
)
( ph + prot ) ro2
+
ro2 − ri2
(
)
(4.86)
r2 + r2
r2
S22 = ( ph + prot ) o2 i2 − pband 2 b 2
ro − ri
rb − ri
(4.87)
S32 = −( ph + prot )
(4.88)
At this location, we see that the axial stress defined by Equation 4.86 has two parts. The first
term on the RHS is the inertia of all the fill and shell material ahead of this location. The second term is the axial stress caused by the internal pressure of the fill expanding. In Equation
4.87, the first term on the RHS is the contribution of spin to the hoop stress and the second
term is the restoring force caused by the gun tube pushing in on the rotating band. Equation
4.88 is simply the radial stress caused by the rotation and compression of the fill and wall.
Further forward on the shell at location 3, the stresses due to setback of filler, filler rotation, wall rotation, and band pressure have identical formulas to location 2 but with, of
course, different values of the variables due to the lower hydrostatic pressure component:
S13 = −
w′ + w′f ps A
w π ro2 − ri2
(
)
( ph + prot ) ro2
+
ro2 − ri2
(
)
(4.89)
r2 + r2
r2
S23 = ( ph + prot ) o2 i2 − pband 2 b 2
ro − ri
rb − ri
(4.90)
S33 = −( ph + prot )
(4.91)
Finally at location 4, near the nose of the shell, the stresses due to setback of filler, filler
rotation, and wall rotation are as follows:
S14 = −
© 2014 by Taylor & Francis Group, LLC
w′ + w′f ps A
w π ro2 − ri2
(
)
( ph + prot ) ro2
+
ro2 − ri2
(
)
(4.92)
Ballistics: Theory and Design of Guns and Ammunition
138
r2 + r2
S24 = ( ph + prot ) o2 i2
ro − ri
(4.93)
S34 = −( ph + prot )
(4.94)
At each location, one must be certain to use the proper head height of filler and the proper
inner and outer radii of the shell.
We must also account for the shear stresses which are most severe at location 1. For simplicity, we will assume a flat base and calculate the shear stress due to wall torsion.
Wherever these calculations are done on the shell, the proper Izz and the proper inner
and outer diameters must be used:
τ 11 =
τ 22 , τ 23 , τ 24 =
( ps − ph )πri2 ( ps − ph )ri
=
2πrt
2t
i
(4.95)
ps A
64I ′zz
FT′
=
3
π 2
( do − di2 ) n(do + di ) (do − di ) w
4
(4.96)
A typical loading of the shell using known weights, pressures, and acceleration is shown
in Figure 4.17 and Table 4.2.
Location of interest—
on inside wall
6 in.
Y CS00
Z X
Section properties at or ahead of 6-in. location:
m6 = 17.15 lbm
Mass
do 6 = 4.10 in.
OD
di6 = 2.95 in.
ID
h6 = 8.189 in.
Head height
Izz6 = 41.15 lbm-in.2
Moment of inertia
Polar MOI
J6 = 108.3 in.4
FIGURE 4.17
Location of interest on a 105 mm M1 projectile.
TABLE 4.2
Typical Values for Use in an HE Projectile Design
Component
Weight (lbm)
Fuze
Body
Rotating band
Filler (TNT)
Total
© 2014 by Taylor & Francis Group, LLC
2.1
34.0
0.4
5.5
42.0
Loads
Breech pressure (psi)
Spin rate—maximum p (Hz)
Base pressure (psi)
Acceleration (g’s)
Angular acceleration (rad/s2)
38,400
82.4
37,150
11,873
348,600
Ammunition Design Practice
139
The common practice currently used in projectile design is to dispense with the hand
calculations and go right to a FEA. While this is usually very accurate and saves a good
deal of time, there are instances when one would like to check the answers through a hand
calculation. Let us examine one location on this 105 mm M1 HE projectile fired from an
M2A2 cannon at 145°F.
Projectile data:
Shell material: HF-1 steel
•
•
•
•
•
•
Density—0.283 lbm/in.3
Projectile OD—4.10 in.
Projectile ID (average)—2.95 in.
Projectile effective (including friction) mass (fuzed)—42 lb
Projectile base intrusion into cartridge case—85.43 in.3
Izz —80.24 lbm-in.2
Fill material: TNT
•
•
•
•
•
Density—0.036 lbm/in.3
Total length of explosive column—13.44 in.
Izz —5.17 lbm-in.2
Average fill cross-sectional area—6.49 in.2
Fill surface area—124 in.2
The M1 projectile fired from our cannon is depicted in Figure 4.17. The properties of the
section ahead of the location of interest are provided in Figure 4.17. We shall determine the
stress tensor at the location shown. We shall assume the projectile obturates perfectly and
that there is no friction between the projectile and the tube.
To begin, we should always draw a free-body diagram of an infinitesimal element at the
point of interest.
Let us look at the hoop direction first. We shall use Equation 4.51:
σ θθ =
(
2
r 2r 2 ( p − pi )
1
p r − po ro2 − i o o2
2 i i
r
r − ri
2
o
)
(4.97)
σθθ
σθθ
In this case, r = ri and po = 0 so we can write
σ θθ =
© 2014 by Taylor & Francis Group, LLC
(
(
)
1
pi ri2 + ro2
ro2 − ri2
)
(4.98)
Ballistics: Theory and Design of Guns and Ammunition
140
The internal pressure is found through our equivalent pressure technique mentioned earlier:
prot =
ωp2max
ρ ro3 − ri3 + ρfill ri3
3ri
(
(4.99)
2
2
prot
)
rev
rad
(82.4)2
(2π )2
s
rev
=
in.
lbm-ft
(3)(1.475) [in.](12) (32.2)
2
ft
lbf-s
lbm
lbm
× (0.283) 3 (2.05)3 − (1.475)3 [in.3 ] + (0.036) 3 (1.475)3 [in.3 ]
in.
in.
lbf
prot = 258 2
in.
For the hydrostatic component of the equivalent pressure, we know that
ph = ρfill ap max h6
(4.100)
lbm
ft
(0.036) 3 (382,300) 2 (8.189) [in.]
in.
s
ph =
lbm-ft
(32.2)
2
lbf-s
lbf
ph = 3500 2
in.
The equivalent internal pressure is then
peq = prot + ph
lbf
peq = pi = 3758 2
in.
The hoop stress is then
σ θθ =
2
2
lbf 4.10 2.95
2
(
)
+
3758
[in. ]
2
2
2
4.10 2.95
in. 2 2
2
[in. ]
−
2
2
1
lbf
σ θθ = 11, 830 2
in.
© 2014 by Taylor & Francis Group, LLC
(4.101)
Ammunition Design Practice
141
Now let us look at the axial stress. This is the stress at the point due to two things: the axial
inertia of all the material ahead of the cut setting back and the effective internal pressure
caused by the rotation of the projectile and the hydrostatic compression of the fill material:
σ zz =
(p r
(r
2
i i
2
o
− po ro2
−r
2
i
)
)−
FAxial
π ro2 − ri2
(
)
(4.102)
σzz
σzz
We shall use the radii given in the problem statement. We put negative sign in the aforementioned equation to denote compressive stress because only the axial component loads
the inner wall in compression. The force acting on the section of interest due to setback is
given by
FAxial = m6 apmax
FAxial
ft
(382,300) 2 (17.15) [lbm]
s
= 203,600 [lbf ]
=
lbm-ft
(32.2)
2
lbf-s
Using this result, we have
2
σ zz
lbf 2.95
2
(3758) 2
[in. ]
(203,600) [lbf ]
in.
2
−
=
2
2
π
4.10 2.95
( 4.10)2 − (2.95)2 [in.2 ]
2
[in. ] 4
−
2 2
lbf
σ zz = −27 ,940 2
in.
© 2014 by Taylor & Francis Group, LLC
(4.103)
Ballistics: Theory and Design of Guns and Ammunition
142
Many times we neglect the first term in the aforementioned equation for conservatism. In
the radial direction, we only have our equivalent pressure pushing radially outward and
our location of interest is on the ID, so
σ rr = − peq
(4.104)
lbf
σ rr = −3758 2
in.
σrr
σrr
The angular acceleration will generate a torque through the rotating band that results in a
shear stress in the plane normal to the axis of the projectile.
τzθ
τzθ
The torque on the projectile is also the opposite of the torque on the gun tube and comes
directly from Equation 4.77:
T6 = I zz6 α pmax
(4.105)
The moments of inertia were provided and we must use the angular acceleration calculated at peak pressure provided earlier. Now the torque comes about through
2
rad 1 lbf-s 1 ft
T6 = ( 41.15) [lbm-in.2 ](348,600) 2
s 32.2 lbm-ft 12 in.
T6 = 37 ,130 [lbf-in.]
The in-plane shear stress is given by
τ=
Tr
J
Then we have
τ zθ
2.95
(37 ,130) [lbf-in.]
[in.]
lbf
2
=
= 506 2
4
(108.3) [in. ]
in.
© 2014 by Taylor & Francis Group, LLC
(4.106)
Ammunition Design Practice
143
The shear stress caused by the rotation is generated by the shell trying to spin up the
explosive fill. The torque on the explosive fill is determined through
Tfill = I zzfill α pmax
(4.107)
2
rad 1 lbf-s 1 ft
Tfill = (5.17 ) [lbm-in.2 ] (348,600) 2
s 32.2 lbm-ft 12 in.
Tfill = 4644 [lbf-in.]
This generates a force at the internal radius of
Ffill =
Ffill =
Tfill
ri
(4.108)
( 4644) [lbf-in.]
= 3162 [lbf ]
2.95
[in.]
2
Smearing this over the entire internal surface area gives us
τ rθ =
(3162) [lbf ]
lbf
= 25.5 2
2
(124) [in. ]
in.
The axial shear is approximated as a worst case by calculating the hydrostatic pressure at
the bottom of the explosive column, transforming it into a force, and smearing that force
over the entire internal cavity area. We know the entire explosive column height is
h = 13.44 [in.]
Then the peak hydrostatic pressure of the fill is
ph = ρfill apmax h
(4.109)
lbm
ft
(0.036) 3 (382,300) 2 (13.44) [in.]
in.
s
ph =
lbm-ft
(32.2)
2
lbf-s
lbf
ph = 5744 2
in.
Calculating this pressure over the average cross-sectional area of the projectile, we obtain
Fbase = ph Aavgfill
lbf
FAxial = (5744) 2
in.
(6.49) [in.2 ] = 37 ,280 [lbf ]
© 2014 by Taylor & Francis Group, LLC
(4.110)
Ballistics: Theory and Design of Guns and Ammunition
144
Now this force smeared over the interior surface area will yield the stress
τ rz =
(37 ,280) [lbf ]
− Faxial
lbf
=−
= −300 2
Afill
(124) [in.2 ]
in.
(4.111)
We can now write our stress tensor
σ rr
σ = τ rθ
τ rz
τ rθ
σ θθ
τθ z
τ rz −3,758
τ θ z = 25.5
σ zz −300
25.5
11, 830
506
−300
lbf
506 2
in.
−27 ,940
It must be noted that these equations assumed that there were no other forces acting
on the projectile. For instance, in some projectiles with poorly designed rotating bands,
leaking of the propellant gases (known as blow-by) causes the exterior of the projectile
to be pressurized. This load must be considered because it has been known to collapse
projectiles in development. Another point is that, while it is common to check a projectile at peak acceleration, the spin rate at this location is not a maximum. Maximum spin
occurs at the exit of the muzzle of the weapon where the velocity is the highest. It is
always good practice to check a projectile for maximum spin with no axial acceleration
to simulate this.
Problem 4
A high-explosive projectile is to be designed for a 155 mm cannon using a 12 in. thick steel
wall with TNT as the filler material. Assume the shell and filler are a cylinder 0.75 m in
length. It is to be capable of surviving a worn-tube torsional impulse (angular acceleration)
of 440,000 rad/s2.
1. Derive the expression to calculate the torque on the projectile that achieves this
acceleration if the torque is applied at the OD of the shell.
2. Calculate the value of the torque assuming the density of steel is 0.283 lbm/in.3
and TNT is 0.060 lbm/in.3
Hint: Start from FT = maT
Answer:
(
)
1. T = MWALL + MFILL = 12 πα l ρ ro4 − ri4 + ρFILL ri4 , 2. T = 796,600 [lbf-in.]
Problem 5
To participate in a failure investigation of an explosive, someone asks you to look at their
design of a cylinder that was supposed to hold the explosive during a 155 mm Howitzer
launch. Assume the explosive sticks completely to the interior wall. The firing conditions
at the time of the failure were as follows:
© 2014 by Taylor & Francis Group, LLC
Ammunition Design Practice
145
6.092 in.
1 in.
1 in.
10 in.
4 in.
Axial acceleration = 10,000 g
Angular acceleration = 300,000 rad/s2
Angular velocity = 100 Hz
The projectile was as shown above:
The wall is AISI 4140
lbf
E = 30 × 106 2
in.
ν = 0.29
lbm
ρ = 0.283 3
in.
The explosive is composition B
g
ρfill = 0.71 3
cm
Write the stress tensor for a point on the inside diameter, 4 in. from the base
Answer:
σ rr
σ = τ rθ
τ rz
τ rθ
σ θθ
τθ z
© 2014 by Taylor & Francis Group, LLC
τ rz −2265
τ θ z = 21
σ zz −266
21
5564
−2864
−266
lbf
−2864 2
in.
−19,307
Ballistics: Theory and Design of Guns and Ammunition
146
Problem 6
A 155 mm projectile is fired from a tube with a 1 in 20 twist. Its muzzle velocity is 1000 m/s.
What is the spin rate at the muzzle in Hz?
Answer: 322.6 [Hz]
Problem 7
It is requested that a brass slip ring be constructed for a spin test fixture to allow electrical signals to be passed (although real noisy) to some instrumentation. The design
requirements are for the ring to have an ID of 4 in., a length of 2 in., and be capable of
supporting itself during a 150 Hz spin test. How thick does the ring have to be? The
properties of brass are as follows: yield strength of 15,000 psi and density of approximately 0.32 lbm/in.3
Answer: 1/4 in. thickness will work but it can be thinner
Problem 8
A 155 mm projectile is to be designed with a rotating band designed to discard as the projectile leaves the muzzle of the weapon. The rotating band is fixed to the projectile (so it
transmits the proper torque to the projectile) with splines that prevent rotational motion
relative to the projectile while allowing the band to expand in the radial direction for
proper discard. This function must occur at the highest as well as the lowest spin rates.
The two extreme muzzle velocities are 250 and 800 m/s, respectively, with corresponding
peak axial accelerations of 2000 and 15,500 g, respectively. The projectile mass is 98 lbm
and the axial moment of inertia is 41 lbm in2. Geometry constraints require the band to
have an engraved outside diameter of 6.2 in. and the outer diameter of the band seat (band
inside diameter) is 5.5 in. The band is 2.5 in. long. The 48 rifling lands are 0.05 in. high and
are 0.2 in. wide. Consider both a copper and a soft iron band with the properties provided
next and determine whether the bands will
(a) Withstand the shear at peak angular acceleration.
(b) Break up upon muzzle exit.
(c) If one (or both) designs fail to work properly, what can be done with the analysis
and/or design to make it work? Is there anything we must be careful of?
Assume the following: The weapon has a 1 in 20 twist. At peak acceleration, only shear
on the rotating band need be considered. Ignore the increase in shear area caused by the
rifling helix. Ignore the stress concentration developed by the engraving for discard calculations. Ignore the effect of the splines. For conservatism, on muzzle exit ignore the mass
of material above the rifling marks (i.e., use an OD of 6.1 in. for the band).
The properties for the copper and steel are as follows:
Ultimate tensile strength
Shear strength
Density
Poisson’s ratio
Copper
Soft Iron
35,000 psi
19,250 psi
0.316 lbm/in.3
0.28
40,000 psi
22,000 psi
0.263 lbm/in.3
0.34
Problem 9
An experimental 40 mm gun has an average chamber inner diameter 60 mm. The weapon
is expected to develop a maximum breech pressure of 35,000 psi. If we would like the
© 2014 by Taylor & Francis Group, LLC
Ammunition Design Practice
147
weapon to withstand 10,000 cycles at this pressure and given the properties of the steel
given later, determine the outside diameter of the chamber. Without proper design experience an interference fit can sometimes be catastrophic. If we were instead to design this
chamber out of two tubes, each at half of this thickness but with the outer tube compressing the inner tube by 0.002 in. diametrally, what is the maximum pressure the design will
accommodate and still function for the 10,000 cycles?
Assume AISI 4340 steel with a yield strength (SY) of 100,000 psi. The endurance stress
(S’n) for 4340 is 0.875SY for the amount of cycles desired. Assume the following factors
from our cyclic loading discussion: CR = 0.93, CG = 0.95 and CS = 0.99. Assume the chamber is open ended. The modulus of elasticity and Poisson’s ratio are 30 × 106 psi and 0.3,
respectively.
Problem 10
You are to design a fragment throwing gun system for another organization to be used in
fragment impact testing. The gun is to throw a 22 g, 0.500 in. diameter, cylindrical fragment at 2300 m/s. Your design must use a brass cartridge case to assist with obturation of
the breech. Other assumptions and information are
1. You do not need to design the breech—assume it will hold the cartridge case in
properly (in reality we can always add more threads to the design).
2. Even though there will be a slight taper on the chamber (which must be larger
than the bore diameter for seating purposes), assume, for calculation purposes,
that the chamber is cylindrical at its maximum diameter.
3. The tube is to be steel and assume that the yield strength is 60,000 psi (this accounts
for the effect of cyclic loading). The modulus of elasticity is 29 × 106 psi. Poisson’s
ratio is 0.29.
4. Assume the propellant is either cylindrical or single perforated (and state your
assumption).
5. Choose from the following propellants:
Propellant
Linearized
Burn Rate,
β (in./s/psi)
Solid
Density, δ
(lbm/in.3)
Adiabatic
Flame
Temperature,
T0 (°R)
Propellant
Force, λ
(ft-lbf/lbm)
Specific
Heat
Ratio, γ
IMR
M12
Bullseye
Red dot
Navy pyro
0.000132
0.000137
0.000316
0.000153
0.000135
0.0602
0.0600
0.0590
0.0593
0.0566
5103
5393
6804
5774
4477
327,000
362,000
425,000
375,000
321,000
1.2413
1.2326
1.2523
1.2400
1.2454
6. Assume the cartridge case is brass and use a bilinear kinematic hardening model
where the brass has a modulus of elasticity of 15 × 106 psi, a local tangent modulus of 13 × 106 psi, and a yield stress of 16,000 psi (yield occurs in this material at
ε = 0.002).
7. Weight is not a major concern; however, you should make the design light enough
to be moved using reasonable test range equipment.
© 2014 by Taylor & Francis Group, LLC
148
Ballistics: Theory and Design of Guns and Ammunition
The design is to proceed as follows (not necessarily in the order given):
A. Interior ballistics design
a. Size the chamber length and diameter.
b. Determine the amount of propellant needed based on your choice of the
aforementioned propellants and propellant geometry (make sure it fits in the
chamber).
c. Determine a web thickness for the propellant.
d. Determine the length of the gun.
e. Determine V, pB and x for the projectile at peak pressure.
f. Determine Vc, pBc and xc for the projectile at charge burnout.
g. Determine the muzzle velocity of the projectile.
B. Gun tube design
a. Based on the calculations of part (A), develop a pressure–distance curve to use
as criteria for your gun design
b. Determine the outside diameter of the gun tube. To keep the design light as
possible, use the design rules provided in the text and taper the tube toward
the muzzle. If needed, over the chamber, you may shrink fit cylinders to build
up a composite tube.
c. Determine the weight of your gun and comment on if it is reasonable.
C. Cartridge case design
a. Determine a thickness and tolerance for your cartridge case.
b. Determine the outside diameter and tolerance for the cartridge case.
c. Decide on a tolerance for your chamber inside diameter.
Note that for these calculations show that the case may be easily extracted at the limits of
the tolerance.
It is important that you write down all of your assumptions. It is also highly likely that as
you proceed further along with your design you may come upon a situation that requires
you to revisit an assumption you made earlier—this is to be expected and it is part of the
design process.
4.11 Buttress Thread Design
There are a variety of instances where a buttress thread form is the desired means of
transmitting loads between mating components. In some instances, the thread form is not
the usual continuous spiral associated with a normal thread but a series of discontinuous grooves that exhibit the cross-sectional form of the buttress. In this section, we will
discuss a true thread with lead-ins and partial thread shapes, but we will assume that the
basic analysis will apply to buttress grooves as well.
Buttress threads are designed to maximize the load-carrying capability in one direction of a threaded joint. There are many variations on such threads, but on ammunition
© 2014 by Taylor & Francis Group, LLC
Ammunition Design Practice
149
Pitch of the thread
Pressure flank
7°
45°
Load carrying
or shearing
FIGURE 4.18
Depiction of a standard buttress thread.
components we predominantly use threads with a pressure flank angle (described later) of
7° as shown in Figure 4.18. Thread callouts on drawings usually appear, e.g.,
2.750-4UNC-2A LH Buttress
The meanings of these callouts are as follows:
•
•
•
•
First number is the major diameter of the thread (here it is in inches)
Second number is how many threads per inch
The letters are the thread form callout (UNC = Unified National Coarse)
The last number is the class of fit of the thread related to clearances in the engagement (3 is the tightest fit, 1 the loosest)
• The last letter determines whether the thread is male (A) or female (B)
(mnemonic − A = Adam = male)
• LH means left handed (there will be no callout if the threads are right-hand twist
or if the thread is a groove and not a continuous spiral)
Thread nomenclature of relevance is as follows:
• Major diameter is the largest diameter of the thread form
• Minor diameter is the smallest diameter of the thread form
• Pitch diameter is the diameter where there is 1/2 metal and 1/2 air
We use buttress threads for several reasons: most important is to improve the directional
loading characteristics of the thread; also to allow for a more repeatable, controllable
shear during an expulsion event, i.e., if we want the thread to intentionally and controllably fail allowing separation of the components; and to prevent thread slip in joints with
fine threads or threads on thin shell walls. If thread slip occurs, the threads can either
dilate or contract elastically and the joint can pop apart with little or no apparent damage
to the threads.
When we design for strength, we typically calculate the strength based on the shear
area at the pitch diameter in the weaker material. This, of course, translates to half the
length of engagement of the threads. This is acceptable because we usually use conservative properties and add a safety factor to account for material variations and tolerances.
© 2014 by Taylor & Francis Group, LLC
150
Ballistics: Theory and Design of Guns and Ammunition
Element at
point A
r
Compressive stress
from F
F
Neutral
axis
di
σ=
Mc
I
Tensile stress
Mc
from σ =
I
FIGURE 4.19
Depiction of a standard buttress thread.
We must always base our calculations on the weaker material if the design is to be robust.
When designing to actually fail the threads, however, we need to be more exact in our
analysis and take everything such as actual material property variation and tolerancing
into account or our answers will be wrong.
We will proceed in this analysis in meticulous detail, initially, as a cantilevered beam
subjected to compressive and tensile stresses caused by contact forces and bending
moments. This technique was first developed during the U.S. Army’s sense and destroy
armor (SADARM) program by Dan Pangburn of Aerojet Corporation [5] and has been
used by the U.S. Army.
We consider the thread form as a short, tapered, cantilever beam and assume that failure
will occur as a result of a combination of stresses and that combined bending and compressive stress precipitate the failure. This is depicted in Figure 4.19. If we examine this
figure, we see that the distributed force, F, causes our beam to bend in the classical sense
with the loaded flank in tension and the unloaded flank in compression about the neutral
axis. We have separated an element of material out from point A in the Figure. The freebody diagram of this element shows that the bending of the beam puts it in tension, while
the loading on the pressure flank puts it in compression. It is this combined load that will
cause failure of the material.
If we were analyzing this in a finite element code, the bending and compression would
cause combined stresses and the part would fail by one of the failure criteria that were
discussed earlier. However, in this case, we will use the maximum shear criteria to check
for failure at some radius in the thread and will also check the load at which failure occurs
with the von Mises criteria at the thread roots, di on the male thread and do on the female
thread. These are the diameters of the loading (i.e., the mating thread contact areas) as
depicted in Figure 4.20.
For simplicity, we shall call the male thread the “bolt” (subscript 1) and the female thread
the “nut” (subscript 2). The loading is further described by Figure 4.21. In this figure, the
radius, r, is the plane at which the threads will shear.
If we assume the contact is frictionless, the average normal stress is simply the total axial
force, F, divided by the projected area, A. We have assumed that the normal stress is constant over the contact area. This gives us a negative value because the stress is compressive.
Figure 4.22 shows the configuration where the normal force has been termed F4 and the
thread area is A4. Since an axial loading is what shears the threads, we need to project the
© 2014 by Taylor & Francis Group, LLC
Ammunition Design Practice
151
do
Bolt (1)
r
F
di
The location where these
stresses are the greatest
is here along the contact
surface
FIGURE 4.20
Definition of load radii.
Bolt (1)
r = Shear radius
do = Outer diameter
di = Inner diameter
r
do
F2
t2
F1
Nut (2)
di
t1
1
2
FIGURE 4.21
Loading diagram of buttress threads.
F
F4
A4
1
A
FIGURE 4.22
Loading of a thread surface.
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
152
components of this force along the axis of the projectile (i.e., rotate through the angle, ϕ1).
This allows us to express the stress as
−F
σN = 4 =
A4
F
F
cos φ1
= − = σv
A
A
cos φ1
−
(4.112)
where σ N and σv are the normal and axial stresses, respectively. By substituting the area,
A, we get
σv = −
F
π 2 2
do − di
4
(
)
(4.113)
If we assume that failure takes place at a radius, r, yet to be determined, the bearing force
on the external thread (bolt) that produces bending in the thread is
d 2
F1 = −πσ v o − r 2
2
(4.114)
Similarly, the force that produces bending in the internal thread (nut) is
2
d
F2 = −πσ v r 2 − i
2
(4.115)
Now the pitch diameter is defined as the location where the thickness of the thread is onehalf the thread pitch. Since thread failure occurs at an assumed radius, r, we need to define
the thicknesses of both the male and female threads at this location.
First, recall that the thread pitch is p and then define dpf as the internal (female) thread
pitch diameter and dpm as the external (male) thread pitch diameter. Then t1 and t2 from our
earlier diagram can be expressed as follows:
t1 =
dpm
p
−r −
2
2
(tan φ1 + tan φ2 )
(4.116)
t2 =
p dpf
−
− r (tan φ1 + tan φ2 )
2 2
(4.117)
Then the bending stress can be calculated from simple beam theory as
t
M
3M
Mc
2
σ=
=
=
1
3
I
rt 2
π
(2π r )t
12
© 2014 by Taylor & Francis Group, LLC
(4.118)
Ammunition Design Practice
153
where c is the distance from the point of interest, r, to the neutral (bending) axis and I is
the area moment of inertia of the cross section. The bending stress in the external (male)
thread is then
d
3 F1 o − r
2
σ1 =
2π rt12
(4.119)
Similarly, we can show that the bending stress in the internal (female) thread is
d
3 F2 r − i
2
σ2 =
2π rt22
(4.120)
In considering the failure criteria, we shall assume that the maximum shear stress in the
material must not exceed 0.6 times the material strength in a tensile test. We will use the
yield strength as this material strength because at that point in failure the geometry of
the part is changing. Experience has shown that once this begins to happen the part is in
the process of failing anyway and will not recover.
In a state of combined loading, the maximum shear stress can be found from
1
τ max = |σ max + σ min |
2
(4.121)
This averaging can be shown to be
τ max =
σ −σN
= 0.6Y
2
(4.122)
Here we are reminded that σ N and σv are compressive therefore negative numbers and Y is
the yield stress in tension. The equivalent stress at failure in the male thread is then
Y1 =
σ1 −σ v
1.2
(4.123)
Y2 =
σ2 −σv
1.2
(4.124)
and in the female thread it is
In these equations, Y1 and Y2 are the yield stress in the male and female threads, respectively.
We will now combine Equations 4.123 and 4.119 as well as Equations 4.124 and 4.120 to
eliminate σ 1 and σ 2, respectively. This yields
1
do − r σ
Y1 = 1.25F1 2 2 − v
π rt1
1.2
© 2014 by Taylor & Francis Group, LLC
(4.125)
Ballistics: Theory and Design of Guns and Ammunition
154
and
Y2 = 1.25F2
1
di
2 − σv
2
1.2
π rt2
r−
(4.126)
We now combine Equations 4.125 and 4.116 as well as Equations 4.126 and 4.117 to eliminate the thicknesses, t1 and t2, respectively. This yields
1
do − r
σv
2
Y1 = 1.25F1
2 −
1
.2
1
1
π r p − r − dpm (tan φ1 + tan φ2 )
2
2
(4.127)
and
1
di
σv
2
Y2 = 1.25F2
2 −
.2
1
1
1
π r p − dpf − r (tan φ1 + tan φ2 )
2
2
r−
(4.128)
We will now insert Equation 4.114 into Equation 4.127 and Equation 4.115 into Equation
4.128 to eliminate F1 and F2, respectively. This yields
(
Y1 = −0.3125σ v d − 4r
2
o
2
)
1
do − r
σv
2
2 −
1.2
1
1
r p − r − dpm (tan φ1 + tan φ2 )
2
2
(4.129)
and
1
di
σv
2
Y2 = −0.3125σ v ( 4r − d )
2 −
1.2
1
1
r p − dpf − r (tan φ1 + tan φ2 )
2
2
2
r−
2
i
(4.130)
Now we must solve Equations 4.129 and 4.130 in terms of σv. The first of these is
σv =
−Y1
G3 + G2 + G1 + G0 +
1
1.2
(4.131)
where
G3 =
0.15625do3
r(0.5 p − r tan φ1 − r tan φ2 + 0.5dpm tan φ1 + 0.5dpm tan φ2 )2
© 2014 by Taylor & Francis Group, LLC
(4.132)
Ammunition Design Practice
155
G2 =
−0.3125do2
(0.5 p − r tan φ1 − r tan φ2 + 0.5dpm tan φ1 + 0.5dpm tan φ2 )2
(4.133)
G1 =
−0.625rdo
(0.5 p − r tan φ1 − r tan φ2 + 0.5dpm tan φ1 + 0.5dpm tan φ2 )2
(4.134)
G0 =
1.25r 2
(0.5 p − r tan φ1 − r tan φ2 + 0.5dpm tan φ1 + 0.5dpm tan φ2 )2
(4.135)
The second equation is
σv =
−Y2
1
H 3 + H 2 + H1 + H 0 +
1.2
(4.136)
where
H3 =
0.15625di3
r(0.5 p + r tan φ1 + r tan φ2 − 0.5dpf tan φ1 − 0.5dpf tan φ2 )2
(4.137)
H2 =
−0.3125di2
(0.5 p + r tan φ1 + r tan φ2 − 0.5dpf tan φ1 − 0.5dpf tan φ2 )2
(4.138)
H1 =
−0.625rdi
(0.5 p + r tan φ1 + r tan φ2 − 0.5dpf tan φ1 − 0.5dpf tan φ2 )2
(4.139)
H0 =
1.25r 2
(0.5 p + r tan φ1 + r tan φ2 − 0.5dpf tan φ1 − 0.5dpf tan φ2 )2
(4.140)
We now solve Equation 4.113 for F and we get
F=
π
σ v do2 − di2
4
(
)
(4.141)
Substitution of Equation 4.131 for σv yields (for a full thread on the bolt)
F=
π 2
−Y1
do − di2
1
4
G3 + G2 + G1 + G0 +
1.2
(
)
(4.142)
We perform a similar operation with Equation 4.136 giving us (for a full thread on the nut)
F=
© 2014 by Taylor & Francis Group, LLC
π 2
−Y2
do − di2
1
4
H 3 + H 2 + H1 + H 0 +
1.2
(
)
(4.143)
156
Ballistics: Theory and Design of Guns and Ammunition
Equations 4.142 and 4.143 now contain only two unknowns, r and F. The procedure now
involves solving both Equations 4.142 and 4.143 and plotting the force, F versus r. The lowest value in either equation is then the force (and location) at which the joint will fail. It is
recommended that these solutions be performed with the aid of a computerized numerical
calculation program such as MathCAD.
Partial threads can have a significant effect on the failure strength of a joint. If the joint
were designed to survive, it is generally best to ignore the additional strength afforded by
partial threads and base the design margin on the calculation method earlier. When a joint
is designed to fail, however, the lead in and run out must be accounted for unless sufficient
margin is available in the expulsion system such that two additional threads may be added
to the calculation, yet still allow the joint to be overcome with ease.
4.12 Sabot Design
Sabots (French for wooden shoe) are used in both rifled and smoothbore guns to allow a
standard weapon to fire a high density, streamlined sub-projectile whose diameter is much
smaller than the bore, at a velocity higher than would normally be possible if the gun were
sized to the sub-projectile’s diameter. Discarding sabots have been in general use since the
Second World War and are still popular (in fact, an artist’s rendition of a discarding sabot
is illustrated on the cover of this book). They are called “discarding sabots” since they are
shed from the sub-projectile at the muzzle allowing it to fly unencumbered to the target.
As stated previously, velocity is proportional to the square root of the pressure achieved
in the tube, the area of the bore, the length of travel, and inversely proportional to the
square root of the projectile weight. In mathematical terms,
V~
pAL
wp
(4.144)
We can see that if the area over which the pressure is applied is much greater than the area
presented at the rear of the sub-projectile, a larger force would be applied to accelerate it
than if it were fired at the same pressure from a bore of its own diameter. Furthermore,
decreasing the launch weight of the as-fired assembly also increases the velocity. Therefore,
we must design as light a sabot as feasible so that we can maintain a very dense, small
diameter sub-projectile (usually an armor penetrator). The combination of the full bore area,
a dense, streamlined sub-projectile, and a lightweight sabot has the overall effect of generating unusually high velocities, a characteristic essential for kinetic energy armor penetration.
There are many requirements for a successful sabot:
• It must seal the propellant gases behind the projectile (obturate)
• It must support the sub-projectile during travel in the bore to provide stable motion
(called providing a suitable wheelbase)
• It must transfer the pressure load from the propellant gases to the sub-projectile
• It must completely discard at the muzzle of the weapon without interfering with
the flight of the sub-projectile
© 2014 by Taylor & Francis Group, LLC
Ammunition Design Practice
Rotating band
157
Sabot
Sub-projectile
FIGURE 4.23
Simplified diagram of an armor-piercing, discarding-sabot (APDS) projectile.
• The discarded sabot parts must also fall reliably within a danger area in front of
the weapon so as not to injure troops nearby
• It must be minimally parasitic, i.e., it must be as light as possible and remove as
little energy from the sub-projectile as possible
These are formidable requirements that necessitate great ingenuity on the part of the
designers.
The problem has been solved in a variety of ways. In the 1950s, designers, chiefly British,
used cup- or pot-type sabots to launch armor-penetrating, discarding-sabot (APDS) subprojectiles (Figure 4.23). The guns from which these munitions were fired were rifled to
launch conventional full caliber, spin-stabilized rounds and so the sub-projectiles of the
APDS rounds were spin-stabilized too. Such armor-defeating munitions were highly effective against the tank armor of the times, and pot-type, saboted, kinetic energy penetrators
were adopted in tank cannon around the world.
Tank armor changed in the 1960s and became more difficult to penetrate with the tungsten carbide cores of the sub-projectiles in use. Initially, incremental changes were made
in the material of the core (sintered tungsten was used instead of sintered tungsten carbide), but it was eventually realized that longer, smaller diameter, high-density penetrators
were the answer. There are physical limits to the degree of sub-calibering practical in spinstabilized projectiles: the spin required for flight stability increases as the square of the
ratio of bore to sub-projectile for conventionally shaped projectiles and it becomes nearly
impossible to spin-stabilize very long projectiles. Rifling twists were increased to attempt
to accommodate the APDS rounds; in one case, a 1:12 twist was tried when the normal twist
would have been 1:40. In the end, APDS designs were abandoned in favor of very long, finstabilized penetrators (APFSDS) that used a radically different type of sabot (Figure 4.24).
Obturator
Fins
Sabot
Sub-projectile
FIGURE 4.24
Simplified diagram of an armor-piercing, fin-stabilized, discarding-sabot (APFSDS) projectile.
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
158
laft
lsabot
lfwd
ps
ps
a
T—Shear traction
FIGURE 4.25
Free-body diagram for rings and rods.
The guns too were changed to smoothbores; although to preserve older weapons in use,
designers learned how to make fin-stabilized munitions firable in rifled guns as well.
The basic type of sabot used with long-rod, fin-stabilized penetrators is the ring with its
subvarieties: base pull, double ramp, and saddle sabots. Whereas pot sabots were essentially discarded rearward as a unit, ring sabots are segmented into three or more sections
and discard radially outward at the muzzle to clear the fins that are larger in diameter
than the rod. The finned sub-projectile is frequently imparted with a slow spin to average out unavoidable manufacturing asymmetries during flight that could cause trajectory
drift. This type of munition is now in the arsenals of all nations.
The design of the ring sabot begins with the stress analysis of the shear traction between
the sabot inner diameter and the penetrator outer diameter. This analysis is crucial for
determining the mass of the ring and thus the parasitic weight of the sabot. We will follow
the work of Drysdale [6] throughout this development. The essential parameters of the
computation are shown in Figure 4.25.
From this free-body diagram, we can infer that
T = ps ( A − Ap ) − msabot a
(4.145)
A reasonable estimate for the masses where the symbols are as follows:
msabot =
1
msub-projectile
2
(4.146)
where
T is the total shear traction force
A is the bore area
Ap is the area of the penetrator cross section
msabot is the mass of the sabot
msub-projectile is the mass of the sub-projectile
a is the projectile acceleration
ps is the pressure on the base of the shot (note that the net pressure on the fins is zero)
σ 1 is the axial stress on the penetrator
Because the sabot needs to be as light as possible, the material is usually much weaker
than the penetrator; thus, the sabot length depends mostly on the sabot material. If the
© 2014 by Taylor & Francis Group, LLC
Ammunition Design Practice
159
penetrator were weaker for some reason, the sabot length would depend upon that material. Thus, we can write for the surface traction
Tallow =
π
dplsabotτ allow
2
(4.147)
where
dp is the diameter of the penetrator or sub-projectile
Tallow is the allowable traction force
τallow is the maximum shear stress allowed in the weaker material
The shear traction is usually transmitted through matching grooves or threads. Analysis
of these surfaces can be rather complicated but is similar to standard or buttress thread
design practice. Given no actual data on the allowable shear stress in the material, we can
use the following formulas based on the Tresca or the von Mises yield criteria:
σe
2
(4.148)
1.155σ e
= 0.577σ e
2
(4.149)
τ allow =
By the Tresca criteria or
τ allow =
by the von Mises criteria. In both of these expressions, σe is the equivalent stress as discussed in Section 4.2. Thus, the allowable surface traction can be stated as
Tallow = Kπ dplsabotσ e
(4.150)
where K is either 0.25 or 0.2885 dependent upon the failure criteria.
If we substitute Equation 4.150 into Equation 4.145, we can solve for the proper sabot
length
Kπ dplsabotσ e = ps ( A − Ap ) − msabot a
(4.151)
and
lsabot =
ps A
m a
− 1 Ap − sabot
Kπ dpσ e Ap
Kπ dpσ e
(4.152)
π 2
dp
4
(4.153)
m a
ps dp A
− 1 − sabot
4Kσ e Ap
Kπ dpσ e
(4.154)
But
Ap =
Then,
lsabot =
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
160
Now, by our earlier assumption (Equation 4.146)
ps A = ma = (msabot + msub-projectile )a = 3msabot a
(4.155)
Then,
lsabot =
ps dp A
ps a
− 1 −
4Kσ e Ap
3Kπ dpσ e
(4.156)
Multiplying and dividing the second RHS term by Ap and simplifying, we get
lsabot =
ps dp A
ps dp A
− 1 −
Kσ e Ap
4Kσ e Ap
12
(4.157)
More generally, if the mass of the sabot is not half of the sub-projectile mass, then we must
use Equation 4.154 to determine the proper length.
The shape of ring sabots evolved over time from quite heavy designs to highly efficient
ones. Early sabots were saddle shaped (Figure 4.26). These had points of high shear stress
concentrations near the ends.
These sabots had an excellent wheel base (the distance between the forward and aft
bourrelets), which prevented balloting in the tube and provided good accuracy. The parasitic weight, however, was high and sufficiently high muzzle velocities were not attained.
Single- and double-ramp sabots have come into use because of the favorable weight
reduction that can be obtained with this design. They utilize gun pressure to help clamp
the sabot to the penetrator and have the added advantage of maintaining an almost constant shear stress between the sabot and the penetrator. The double-ramp sabot is shown
in Figure 4.27.
Detailed studies have shown that a higher order (nonlinear) curved ramp yields a constant shear stress under load. The method of solution for finding the best shape of the sabot
taper depends on a free-body analysis of the sabot and the penetrator. Figures 4.28 and
4.29 show differential elements of the sub-projectile and the sabot, respectively.
If we examine Figure 4.28, we see that the axial forces consist of the net internal stress,
(dσzp/dz)∆z; the inertial resistance to acceleration, ρpVpa; and the shear stress imparted by
Saddle
ps
Penetrator
σ1
τ
Shear
stress
Axial distance
FIGURE 4.26
Shear stress variation in a saddle-type sabot.
© 2014 by Taylor & Francis Group, LLC
Ammunition Design Practice
161
Double-ramp sabot
ps
Penetrator
σ1
τ
Shear
stress
Axial distance
FIGURE 4.27
Shear stress variation in a double ramp-type sabot.
r
Δz
τ
RP
σzp +
σzp
dσzp
dz
Δz
FIGURE 4.28
Differential element in a sub-projectile showing forces acting.
ps
r
RS(z)
σsp +
σzs
dσzs
dz
Δz
τ
Δz
FIGURE 4.29
Differential element in a sabot showing forces acting.
the sabot, τ. Similarly, on the sabot, we have the net internal stress, (dσzs/dz)∆z; the inertial
resistance to acceleration, ρsVs a; the shear stress imparted by the sub-projectile, τ; and the
component of pressure in the axial (z) direction. We proceed by initially finding an expression for the volume of the sabot free body. Details of this derivation are found in Ref. [6].
The incremental volume of the sabot can be shown as follows:
Vs = π Rs2 ( z) − Rp2 ∆z
© 2014 by Taylor & Francis Group, LLC
(4.158)
Ballistics: Theory and Design of Guns and Ammunition
162
We then sum the forces on the sabot in the axial direction
psπ Rs2 ( z + ∆z) − Rp2 ∆z − σ zsπ Rs2 ( z) − Rp2
dσ zs
∆z π Rs2 ( z + ∆ z)Rp2 − ρs Vs a − 2π Rpτ∆z = 0
+ σ zs +
dz
(4.159)
After collection of terms and simplification, we get
( ps + σ zs )
dRs2 dσ zs
+
− ρs a Rs2 ( z) − Rp2 − 2Rpτ = 0
dz dz
(4.160)
Note here that Rs and σzs are functions of z.
Next we find σzp assuming it is linear in z through the expression
σ zp =
(
)
F
1
1
ρ pπ Rp2 a − 2π Rpτ ∆z + σ 1
( ρ p Vp a − 2π Rpτ∆z) + σ 1 =
=
A π Rp2
π Rp2
(4.161)
or
2τ
σ zp = ρ p a −
∆z + σ 1
Rp
(4.162)
where σ 1 is the axial stress in the penetrator as depicted earlier. Now we need to relate σzp
to σzs by applying the assumption of strain compatibility, i.e., the strain in the sabot equals
the strain in the penetrator.
We then use the appropriate elastic moduli and Poisson’s ratio in Hooke’s law to relate
the penetrator stresses to those in the sabot
ε zs =
1
1
[σ zs −ν s (σ rs + σ θ s )] = ε zp = E σ zp −ν p (σ rp + σ θ p )
Es
p
(4.163)
Es
σ zp −ν p (σ rp + σ θ p ) + ν s (σ rs + σ θ s )
Ep
(4.164)
Thus,
σ zs =
If we ignore the bimetallic nature of the components and assume that
σ rp + σ θ p = σ rs + σ θ s = −2 ps
(4.165)
then Equation 4.164 becomes
σ zs =
© 2014 by Taylor & Francis Group, LLC
Es
(σ zp + 2ν p ps ) − 2ν s ps
Ep
(4.166)
Ammunition Design Practice
163
r
Sabot profile
Conical approximation
Penetrator OD
Reduction of thickness, R(z) by
Increasing Es /Ep
Decreasing τ
Increasing σθ
Increasing ps
z
FIGURE 4.30
Sabot radial profile. (Based on analysis from Drysdale, W.H., Design of kinetic energy projectiles for structural integrity, Technical Report ARBRL-TR-02365, U.S. Army Ballistic Research Laboratory, Aberdeen, MD,
September 1981.)
These assumptions allow integration of the differential equation for R(z) producing the
profile in Figure 4.30 (solid curve).
Two of the basic types of sabots are shown in Figures 4.26 and 4.27. The double ramp also
incorporates a front air scoop to facilitate discard in the air stream as well as providing
additional bourrelets riding surface in the tube.
A great deal of work on the effect of sabot design parameters has been accomplished
at the U.S. Army Research Laboratory (formerly the Ballistics Research Laboratory) and
Picatinny Arsenal. A treatment of the effect of sabot stiffness on how clean a projectile
launch is can be found in Ref. [7].
Problem 11
You are to design a rifled 20 mm gun system for an antivehicle application. The gun is
to throw a 0.25 lbm, 0.5 in. diameter, cylindrical sub-projectile at 4500 ft/s. A sabot will
be placed on the outside of the sub-projectile. The sabot can be any material you like and
can find properties for. The sabot shall be made of three pieces, which are at least 3 in. in
length. Your design must use a brass cartridge case to assist with obturation of the breech.
Other assumptions and information are
1. You do not need to design the breech—assume it will hold the cartridge case in
properly (in reality we can always add more threads to the design).
2. Even though there will be a slight taper on the chamber (which must be larger
than the bore diameter for seating purposes), assume, for calculation purposes,
that the chamber is cylindrical at its maximum diameter.
3. The tube is to be steel and assume that the yield strength is 60,000 psi (this accounts
for the effect of cyclic loading). The modulus of elasticity is 29 × 106 psi. Poisson’s
ratio is 0.29.
4. Assume the propellant is either cylindrical or single perforated (and state your
assumption)
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
164
5. Choose from the following propellants:
Propellant
Linearized
Burn Rate,
β (in./s/psi)
Solid
Density, δ
(lbm/in.3)
Adiabatic
Flame
Temperature,
T0 (°R)
Propellant
Force, λ
(ft-lbf/lbm)
Specific Heat
Ratio, γ
IMR
M12
Bullseye
Red Dot
Navy Pyro
0.000132
0.000137
0.000316
0.000153
0.000135
0.0602
0.0600
0.0590
0.0593
0.0566
5103
5393
6804
5774
4477
327,000
362,000
425,000
375,000
321,000
1.2413
1.2326
1.2523
1.2400
1.2454
6. Assume the cartridge case is brass and use a bilinear kinematic hardening model
where the brass has a modulus of elasticity of 15 × 106 psi, a local tangent modulus of 13 × 106 psi, and a yield stress of 16,000 psi (yield occurs in this material at
ε = 0.002)
7. The weapon shall be as light as possible.
The design is to proceed as follows (not necessarily in the order given):
a. Interior ballistics design
i. Size the chamber length and diameter.
ii. Determine the amount of propellant needed based on your choice of the
aforementioned propellants and propellant geometry (make sure it fits in
the chamber).
iii. Determine a web thickness for the propellant.
iv. Determine the length of the gun.
v. Determine V, pB, and x for the projectile at peak pressure.
vi. Determine Vc, pBc, and xc for the projectile at charge burnout.
vii. Determine the muzzle velocity of the projectile.
b. Gun tube design
i. Based on the calculations of part (a), develop a pressure–distance curve to
use as criteria for your gun design.
ii. Determine the outside diameter of the gun tube. To keep the design light
as possible, use the design rules provided in the text and taper the tube
toward the muzzle. If needed, over the chamber, you may shrink fit cylinders to build up a composite tube.
iii. Determine the weight of your gun and comment on if it is reasonable.
c. Cartridge case design
i. Determine a thickness and tolerance for your cartridge case.
ii. Determine the outside diameter and tolerance for the cartridge case.
iii. Decide on a tolerance for your chamber inside diameter.
Note that for these calculations you must show that the case may be easily
extracted at the limits of the tolerance.
© 2014 by Taylor & Francis Group, LLC
Ammunition Design Practice
165
d. Sabot design
i. Based on a twist rate of 1 in 30, calculate the thickness of a restraining
band, 1” wide located over the petal C.G., required to reliably break at shot
exit. This band can be considered as a thin cylinder evenly loaded by the
three sabot petals (i.e., no moment). “Reliable” breakage can be considered
as when the hoop stress is 1.25 times the allowable hoop stress. Neglect the
dynamic pressure cause by the forward velocity of the projectile.
It is important that you write down all of your assumptions. It is also highly likely that as
you proceed further along with your design you may come upon a situation that requires
you to revisit an assumption you made earlier—this is to be expected and it is part of the
design process.
References
1. Budynas, R.G., Advanced Strength and Applied Stress Analysis, 2nd edn, McGraw-Hill, New York,
1999.
2. Boresi, A.P., Schmidt, R.J., and Sidebottom, O.M., Advanced Mechanics of Materials, 5th edn., John
Wiley & Sons, New York, 1993.
3. Beer, F.P., Johnston, E.R., and DeWolf, J.T., Mechanics of Materials, 4th edn., McGraw-Hill,
New York, 2006.
4. Montgomery, R.S., Interaction of copper containing rotating band metal with gun bores at the
environment present in a gun tube, Report AD-780-759, Watervliet Arsenal, New York, June 1974.
5. Pangburn, D., Personal communications with author December 1995 to March 2004.
6. Drysdale, W.H., Design of kinetic energy projectiles for structural integrity, Technical Report
ARBRL-TR-02365, U.S. Army Ballistic Research Laboratory, Aberdeen, MD, September 1981.
7. Plostins, P., Clemins, I., Bornstein, J., and Diebler, J.E., The effect of sabot front borerider stiffness on the launch dynamics of fin-stabilized kinetic energy ammunition, BRL-TR-3047, U.S.
Army Ballistic Research Laboratory, Aberdeen, MD, October 1989.
Further Readings
Barber, J.R., Intermediate Mechanics of Materials, McGraw-Hill, New York, 2001.
Ugural, A.C. and Fenster, S.K., Advanced Strength and Applied Elasticity, 3rd edn., Prentice Hall, Upper
Saddle River, NJ, 1995.
© 2014 by Taylor & Francis Group, LLC
© 2014 by Taylor & Francis Group, LLC
Part II
Exterior Ballistics
© 2014 by Taylor & Francis Group, LLC
© 2014 by Taylor & Francis Group, LLC
6
Introductory Concepts
When the projectile has left the environment of the gun, including the effects of the exiting propellant gases that momentarily surround it, it enters the realm of the exterior ballistician. Here, it is subject to the force of the pressure of the atmosphere that it is flying
through, the force induced by its spin, and the force due to the acceleration of gravity.
The projectile in flight is no longer constrained in lateral motions by the walls of the gun
and, as a free body, can develop motions that are complex and occasionally inimical to the
intent of its user and embarrassingly, to its designer. The study of these motions and the
progress of the projectile to its target are the subject of this part of the book.
We will begin with the simplest case, consideration of the projectile as a point mass
flying in a vacuum with only the force due to gravity acting on it. Then we will proceed
to introduce the force due to the pressure of the air, but still considering the projectile as
a mass concentrated at a point. Finally, we will consider the projectile as a three-dimensional body acted upon by the air, its spin, and gravity. In the final sections of this part
of the text, we shall examine the complex motions arising from the coupling of projectile
dynamics and aero-mechanical forces. Our object will be to examine the conditions necessary for a precise, predictable, satisfactory trajectory enabling the projectile to fulfill its
terminal ballistic utility.
Since this text is intended to have a broad scope, some of the material is not derived in
detail. The reader is encouraged to seek the more detailed treatments in the references
noted throughout each section.
Many of the principles and terms concerned with fluid mechanics required for the
understanding of interior ballistics were introduced in Section 2.7. These principles will
be extended in this section with a view toward an exterior ballistician—commonly called
an aero-ballistician.
We shall first examine the elements of a trajectory as depicted in Figure 6.1. These terms
are commonly used in the military by gunners and researchers alike. Although most of
the symbols and terms in this figure are self-explanatory, some require comment. First is
the so-called map range. This is the range to the target that the gunner would see if he or
she were to plan firing using a map. The base of the trajectory is quite important and is
defined as being level in a plane with the firing point. Gunners of large caliber weapons
and mortars take great care in assuring that the sights on the weapon are leveled in the
direction depicted as well as the plane out of the paper.
Since larger ordnance fires over extensive ranges, it is common to assume that the origin
of the trajectory is coincident with the ground beneath the artillery piece. The line of site
and angle of site (yes, they are spelled that way in much of the literature) are what the gunner uses to aim at the target. As you can see, they only assist in determination of the
pointing of the weapon and the relative height of the target.
An important feature of this diagram is the line of departure. You have probably noticed
that it is not collinear with the elevation of the weapon (i.e., where the bore is pointed).
The reality is that a projectile almost never leaves the bore of a gun aligned with the
bore—we shall discuss this in detail later. For now, we will simply state that this is due to
193
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
194
Vertical jump
Quadrant
elevation of
departure
Origin
e
f
eo
pa
de
n
Li
Quadrant
elevation
r
rtu
n
Trajectory
atio
lev
fe
eo
Lin
Maximum ordinate
Line of fall
Angle of lift
Base of trajectory
Level point
Angle of site
Line
Angle of elevation
of sit
e
Point of impact
Map range
FIGURE 6.1
Elements of a trajectory.
the dynamics of the projectile and gun as well as aerodynamic effects. It should be noted
that Figure 6.1 is drawn as two dimensional. The out-of-plane angular position of the projectile at muzzle exit is known as lateral or azimuthal jump. This will combine vectorially
with the vertical jump that is depicted to give a resultant jump vector.
The angle of lift and line of fall are defined for the level point; however, it is common to
see these used at the target even though officially these quantities at the target are called
angle of impact and line of impact (sometimes shot line).
The aerodynamics and ballistics literature are quite diverse and terminology is far from
consistent. This has particular significance in the coordinate systems used to define the
equations of motion. In this text, we shall use the coordinate system of Ref. [1] as depicted
in Figure 6.2. The primary difference between this scheme and those of, say, Refs. [2–5], is
that the y-axis is deemed to be positive pointing up, with the z-axis as positive to the right
as opposed to the z-axis down and y-axis to the right. This makes sense to the authors with
up being a more intuitive positive direction. The only issues (and some people consider
them significant) with this scheme are that first, the nice right-handed naming convention
y
Positive yaw
x
Positive roll
z
Positive pitch
FIGURE 6.2
Definition of projectile coordinates.
© 2014 by Taylor & Francis Group, LLC
Introductory Concepts
195
x
B
α
V sin αt
αt
O
A
β
C
V cos β
V cos β sin α
x
V sin β
αt
V
V
Trajectory
FIGURE 6.3
Generalized yaw of a projectile.
of the aerodynamic coefficients is disturbed (as we shall see later x–y–z corresponds to
l–n–m not l–m–n as one would normally like); second, what would normally be a positive
rotation in the y-direction (i.e., nose left) is defined as negative—we shall handle this when
we define the associated equations.
We shall now define some terminology and, more importantly, the forces, moments, and
associated coefficients that are used throughout this part of the text. It is important for the
reader to recognize that these force, moment, and coefficient definitions are by no means
an all-inclusive collection. Occurrences of additional forces or moments at times require
additional definitions—e.g., control deflections. We shall adhere to the broad scope of this
text by including only what is necessary for a basic understanding of ballistics.
We mentioned the yaw and pitch of the projectile earlier in this section. The projectile
geometry in an arbitrary state of yaw is depicted in Figure 6.3. This illustration shows the
projectile yawed and pitched to some angle, αt, relative to the velocity vector. The illustration also shows the trajectory, which is defined as the curve traced out by the velocity vector. Thus, the velocity vector is everywhere tangent to the trajectory curve. The inset shows
the decomposition of the angle between the projectile axis of symmetry, x (OB), and the
velocity vector, V (OA). We first measure the sideslip angle, β (∠AOC), and then measure
the yaw angle, α (∠COB), from the axis of symmetry, x, to side OC = V cos β. The side BC of
right triangle OBC then has a value of V cos β sin α. The resulting angle ∠AOB is defined
as the total yaw angle, αt, and in triangle AOB where side AB = V sin α t . It should be noted
that triangle ABC with sides V sin αt, V cos β sin α, and V sin β is not a right triangle.
Most projectiles have at least trigonal symmetry. This is symmetry about three planes
through the projectile long axis, 120° apart. Because of symmetry, it is common to vectorially combine the yaw and pitch of the projectile into one term, which we simply call total
yaw, αt. All of our coefficients will be based on this total yaw. Later, when we discuss
advanced topics, it will be necessary to once again separate them.
An examination of Figure 6.3 shows that we can relate the total yaw to α and β through
sin α t = sin 2 β + cos 2 β sin 2 α
(6.1)
The drag on a projectile is the force exerted on it by the medium through which it is moving, usually air. Since the drag is generated by the motion of the projectile through the air,
it is naturally directed opposite to the velocity vector as illustrated in Figure 6.4.
There are, in general, two types of drag: pressure drag and skin friction drag. This is
because nature can only act on the surface area of the projectile in two ways: normal to the
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
196
x
Drag force
αt
V
Trajectory
FIGURE 6.4
Drag of a projectile.
surface and along it. A third type of drag called wave drag is a form of pressure drag generated by a shock wave formed when the local velocity along the surface of the projectile
reaches Mach 1. We will discuss drag in further detail later, but in all cases it is convenient
to lump the effect of the drag into one coefficient called the drag coefficient. The drag force
is defined in terms of this drag coefficient as
1
1
Drag force = FD = − ρ SCD VV = ρV 2SCD
2
2
(6.2)
Equation 6.2 shows two forms of the defining expression for drag force, vector, and scalar. We shall define all of our forces and moments in this way because; although we will
initially examine the scalar forms, it will be necessary later to use the vector forms. For
now, knowing that the drag force is opposite to the velocity vector and that its scalar magnitude, as depicted on the far RHS in Equation 6.2, is sufficient.
Like many of the coefficients we shall discuss, the drag coefficient can be a complicated
function of the yaw angle. In a more general form, we can write the drag coefficient as the
sum of a linear part and a yaw-dependent term:
CD = CD 0 + CD δ 2 δ 2
(6.3)
δ = sin α t
(6.4)
where δ is the total yaw defined as
The first term on the RHS is the linear part of the drag coefficient, known as the zeroyaw drag coefficient, while the second term is known as the yaw drag coefficient. The
reason that there is no intermediate term is that, for a symmetric body, the drag has to be
the same whether the body is angled at, say +5° or −5°. This is discussed more elegantly
in Ref. [2].
We shall see later that the drag coefficient varies with Mach number in a complex manner.
Dynamic pressure is a quantity defined as 1/2ρV2, where ρ is the density of a fluid that
an object is immersed in and V is the velocity of the fluid relative to the object. It is simply
the physical reaction of the fluid when trying to force an object through it and occurs so
often that it has been given its own name. This dynamic pressure is multiplied by a reference area, S. It is always important to know what reference area is used in the definition
of the coefficients. In every case we shall examine, this reference area is based on the projectile circular cross section. Also, as we shall soon see, moments require a length scale as
well. In all of these instances, we shall use the projectile diameter as the reference length.
© 2014 by Taylor & Francis Group, LLC
Introductory Concepts
197
x, p
Mlp
αt
V
Trajectory
FIGURE 6.5
Spin damping of a projectile.
When a projectile spins in a medium, the viscous interaction of the medium and the
projectile surface is such that the projectile will spin down throughout the flight. This phenomenon is accounted for by a moment applied to the projectile called the spin-damping
moment. It is defined as
Spin-damping moment = M1p =
1
pd
ρV 2Sd C1p
2
V
(6.5)
This moment is directed opposite to the spin vector, p, of the projectile as depicted in
Figure 6.5, and the tendency is for the projectile to spin down, thus there is no negative
sign in Equation 6.5 because the vector handles the decay. One needs to note that the figure
is drawn for a right-hand twist. If a left-hand twist were involved, the spin vector, p, and
the spin-damping moment vector would be reversed.
Some projectiles have fins or jets that impart a roll torque to the projectile such that the
spin rate increases. This rolling moment is depicted in Figure 6.6 and defined through
Rolling moment = M1δ =
1
ρV 2Sdδ FC1δ
2
(6.6)
In this expression, δ F is a cant angle provided to the fins to generate the lift required to
sustain rotation.
Lift is defined as the aerodynamic force that acts orthogonal to the velocity vector. This
is depicted in Figure 6.7. The lift force can be defined in both scalar and vector notations as
Lift force = FL =
1
1
ρ SCLα [V × (x × V )] = ρV 2SCLα δ
2
2
(6.7)
x
M1δ
αt
V
Trajectory
FIGURE 6.6
Roll moment of a projectile.
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
198
x
αt
FL
V
Trajectory
FIGURE 6.7
Lift vector of a projectile.
The lift force coefficient can be written in its nonlinear form as
CLα = CLα 0 + CLα 2 δ 2
(6.8)
With a symmetric projectile, we must note that if there is no angle of attack (i.e., δ = 0) there
is no lift. This is obvious even for the linear case since δ appears in Equation 6.7. Some
authors prefer to work in coordinates other than those we are utilizing here. In those cases,
expressions such as Cx and CN are used for drag and lift, respectively. In these cases, it is
important that proper transformations are used to change the coefficients. An example of
this is provided in Ref. [1].
At this point, we must discuss two quantities known as center of pressure (CP) and
center of gravity (CG) (sometimes called center of mass). The CG is the location on the projectile where all of the mass can be concentrated so that for an analysis the gravitational
vector will operate at this point. The CP is the point through which a vector can be drawn,
i.e., the resultant of all infinitesimal pressure forces acting on the projectile. For most projectiles that are spin-stabilized, the CP is ahead of the CG and the reverse is true with finor drag-stabilized projectiles. Figure 6.8 is an illustration of this.
The separation of the CP and CG gives rise to an overturning moment in all projectiles
(Figure 6.9). As we shall see later, this moment is destabilizing for spin-stabilized projectiles (which is why they must be spun) and stabilizing for fin-stabilized projectiles. The
overturning moment (sometimes called the pitching moment) is defined as
Overturning moment = Mα =
1
1
ρ SdVCMα (V × x) = ρV 2SdCMα δ
2
2
CP
CG
(center of mass)
FIGURE 6.8
Center of gravity (CG) and center of pressure (CP) illustrated.
© 2014 by Taylor & Francis Group, LLC
(6.9)
Introductory Concepts
199
x
αt
Mα
V
Trajectory
FIGURE 6.9
Overturning moment vector of a projectile.
We can see from Equation 6.9 that this moment is a function of the angle of attack and
because of the cross-product, a positive overturning moment (nose up) is oriented along
the positive z-axis.
The overturning moment coefficient can be written in a nonlinear form similar to the lift
and drag forces as
CMα = CMα 0 + CMα 2 δ 2
(6.10)
When a body of circular cross section is immersed in a flow-field perpendicular to its
axis and is spun about its axis, a force known as the Magnus force is developed [6]. This
force comes about because on one side of the body the free stream velocity of the flow
is added to the velocity of the surface, while on the other side the free stream velocity is
reduced by the surface velocity. On the basis of Bernoulli’s equation (Equation 6.11), we see
that along the body surface streamline the pressure must be higher on the side with the
lower velocity [7]:
p 1 2
+ V + z = constant
ρ 2
(6.11)
This results in a side force on the body as illustrated in Figure 6.10.
This might not seem like a big deal because a projectile almost never flies sideways, but
if we consider a projectile in a crosswind, or, more importantly, one that is yawed, we see
that this side component can contribute somewhat to the aerodynamic loading. For all
Angular velocity, ω
Upper surface velocity = rω – V∞
Free stream velocity, V∞
Body radius, r
Lower surface velocity = rω + V∞
FIGURE 6.10
Magnus effect on a projectile.
© 2014 by Taylor & Francis Group, LLC
Body will move in this
direction—Magnus force
direction
Ballistics: Theory and Design of Guns and Ammunition
200
x
αt
FNpα
V
MMpα
Trajectory
FIGURE 6.11
Magnus force and moment on a projectile.
practical purposes, however, if a projectile is not yawed in flight, then there is no Magnus
force. The Magnus force is defined for our purposes as
Magnus force = FNpα =
1
1
ωd
2 ωd
ρ SV
CN pα δ
CN pα (V × x) = ρV S
2
2
V
V
(6.12)
The Magnus force coefficient can be written in a nonlinear form in the same manner as
Equation 6.10, which we will not repeat (Figure 6.11).
In many cases, the Magnus force is small and is usually neglected with respect to the
other forces acting on the projectile. In contrast, the moment developed because of this
force is considerable. We define the Magnus moment as
Magnus moment = M Mpα =
1
1
ωd
ωd
2
ρVSd
CM pα [x × (V × x)] = ρV Sd
CM pα δ
2
2
V
V
(6.13)
The Magnus moment contributes significantly to the stability of the projectile and will be
discussed in detail later. The Magnus moment coefficient can be written as a nonlinear
term in the same way as all our other coefficients.
The CP for the lift force and the CP for the Magnus force are usually not the same, thus
the moments will act through differing moment arms. The reason for this is the different
physics that give rise to the different phenomena. These change during flight as well since
a projectile’s yaw changes as it moves downrange.
Pitch damping is the tendency of a projectile to cease its pitching motion due to air
resistance. It is usually more difficult to visualize for someone new to the field. It is
relatively simple to think about a right circular cylinder mounted in a fixture with its
spin axis held by a frictionless bearing on each end. If we spin the projectile, it will
slow down because of the sticking of the fluid to the surface and the resultant viscous
action (remember the bearings are magically frictionless). If we mount the projectile
such that the bearing is transverse to the long axis and spin it, we will still have the
viscous action slowing the projectile down; however, this will be overwhelmed by the pressure forces that retard the motion and the projectile will spin down much faster. This
combination of forces is called pitch damping. For projectiles, we can define the pitch
damping force as
Pitch damping forces = F Nq +αɺ =
© 2014 by Taylor & Francis Group, LLC
1
1
dx dl
dx
−
ρVSd
CNq + ρVSdCNαɺ
2
2
dt dt
dt
(6.14)
Introductory Concepts
201
Motion with
q = sin ωt
α=0
Motion with
q=0
α = sin ωt
FIGURE 6.12
.
Pictorial description of q and α.
or in scalar terms,
Pitch damping forces = FNq +αɺ =
q d
1
αɺ d
ρV 2S t CNq + t CNαɺ
2
V
V
(6.15)
In Equation 6.16, we have defined the total pitching motion, qt, and the total rate of change
of angle of attack, α̇t, as
qt = q2 + r 2
and αɺ t =
dα t
dt
(6.16)
We note here that this pitch damping comes about through two motions. The first motion
is brought about through the pitching rate q, while the second is developed because of the
resistance to the changing angle of attack. This is described in eloquent detail in Ref. [5].
The simplest way of depicting this is to assume a sinusoidal motion of a projectile along its
flight path. With this assumption, Figure 6.12 shows what motions would result if q only
.
was present and contrasts this with motion if α only were present.
.
It is generally difficult to separate q and α in experimental flight data. For this reason, the
two coefficients are almost always written as a sum and recorded in the literature as such.
With assumptions on the yawing motion of the projectile and the practice of combining
coefficients, as described previously in Equations 6.14 and 6.15, they can be combined as
detailed in Ref. [1] into
Pitch damping force = FNq +αɺ =
=
1
dx
ρVSd(CNq + CNαɺ )
2
dt
1
qd
ρV 2Sd t (CNq + CNαɺ )
2
V
(6.17)
The pitch damping force is, like the Magnus force, generally neglected because it is small
with respect to the other forces such as lift and drag. The moment caused by this pitch damping is frequently significant (Figure 6.13). It can be described mathematically as follows:
Pitch damping moment = M Mq +αɺ
=
dx
dl
1
dx
1
2
ρVSd 2 x ×
− x ×
CMq + ρVSd C M αɺ x ×
dt
dt
2
dt
2
© 2014 by Taylor & Francis Group, LLC
(6.18)
Ballistics: Theory and Design of Guns and Ammunition
202
x
FNq+α
qt
αt
V, l
Trajectory
MMq+α
FIGURE 6.13
Pitch damping force and moment on a projectile.
In scalar form, we can write
Pitch damping moment = M Mq +αɺ =
q d
1
αɺ d
ρV 2Sd t C Mq + t CMαɺ
2
V
V
(6.19)
1
dx
ρVSd 2 (CMq + CMαɺ ) x ×
2
dt
(6.20)
1
qd
ρV 2Sd t [CMq + CMαɺ ]
2
V
(6.21)
These can be simplified as per Ref. [1] into
Pitch damping moment = M Mq+αɺ =
Pitch damping moment = M Mq +αɺ =
At certain times and in some special cases, there are other combinations of forces and
moments and therefore additional coefficients require attention. We will not go any further here as this text is meant to be most general.
We now have the basic terms defined that we shall use in our study of exterior ballistics.
Problem 1
In a test range, a 0.50 caliber M33 ball projectile has a velocity vector which is at an angle
of 10° to the horizontal (assume zero azimuth) with a velocity of 3013 ft/s. The initial pitch
and yaw angles are 1.030° and 1.263°, respectively. The initial rotational rate of change of
the axial unit vector (dx/dt) is provided later. If the projectile has the coefficients below at
this particular instant, draw the situation and determine the following:
a.
b.
c.
d.
e.
f.
Velocity vector (ft/s)
Projectile axial unit vector (x)
Drag force vector (lbf)
Spin damping moment vector (lbf-in)
Overturning moment vector (lbf-in)
Magnus moment vector (lbf-in)
Assume the weapon has a right-hand twist.
© 2014 by Taylor & Francis Group, LLC
Introductory Concepts
203
Projectile information:
CD = 0.2938
(CMq + CMαɺ ) = −5.5
I P = 7.85 [g cm 2 ]
CMα = 2.88
(CNq + CNαɺ ) = 0.004
I T = 74.5 [g cm 2 ]
CLα = 2.69
CMpα = 0.05
m = 42.02 [g]
lbm
ρ = 0.0751 3
ft
rad
ω = 15, 404
s
CNpα = −0.01
Clp = −0.003
dx
rad
= {0.405e1 − 1.963e2 − 0.981e3 }
dt
s
Please supply all answers in an inertial coordinate system labeled 1, 2, 3 with 1 being along
the downrange direction and 3 being to the right side. Treat all missing coefficients as
equal to zero. It is very important that you DRAW the situation. This will have a great deal
of influence in obtaining the correct answer
References
1. McCoy, R.L., Modern Exterior Ballistics, Schiffer Military History, Atglen, PA, 1999.
2. Murphy, C.H., Free flight motion of symmetric missiles, Ballistics Research Laboratory Report
No. 1216, Aberdeen Proving Ground, MD, 1963.
3. McShane, E.J., Kelley, J.L., and Reno, F.V., Exterior Ballistics, University of Denver Press, Denver,
CO, 1953.
4. Nicolaides, J.D., On the free flight motion of missiles having slight configurational asymmetries, Ballistics Research Laboratory Report No. 858, Aberdeen Proving Ground, MD, 1953.
5. Nielsen, J.N., Missile Aerodynamics, AIAA reprint, American Institute of Aeronautics and
Astronautics, Reston, VA, 1988.
6. White, F.M., Fluid Mechanics, 5th edn., McGraw-Hill, New York, 2003.
7. Fox, R.W. and McDonald, A.T., Introduction to Fluid Mechanics, 4th edn., John Wiley & Sons,
New York, 1992.
Further Reading
Bull, G.V. and Murphy, C.H., Paris Kanonen—The Paris Guns (Wilhelmgeschutze) and Project HARP,
Verlag E.S. Mittler & Sohn GmbH, Herford und Bonn, Herford, Germany, 1988.
© 2014 by Taylor & Francis Group, LLC
© 2014 by Taylor & Francis Group, LLC
7
Dynamics Review
Throughout the study of exterior ballistics, dynamics play a great role in the flight of the
projectile. The Coriolis effect in long-range trajectories or the drag changes due to the precessional and nutational motion of the projectile are just two examples of the effect of projectile
body dynamics on flight. We will find that at least a cursory review of dynamics is essential
to the understanding of projectile motion. Analyzing dynamics of projectile flight is best
approached through the use of vectors and we will begin our review with their study.
A vector is defined as a quantity having a magnitude and a direction. Two vectors are
considered equal if both their magnitude and direction are identical. However, this does
not mean that they have to originate at the same point, i.e., a translation has no effect on
whether vectors are equal. A scalar is simply a numerical quantity (a magnitude). When
a scalar and a vector are multiplied (in any order) they form a vector. Thus, we can define
any vector as a scalar magnitude multiplied by a vector of unit length (a unit vector) in the
proper direction (Figure 7.1).
A = Ae A
(7.1)
A vector can be written as the sum of its scalar magnitude in each individual coordinate
direction times a unit vector in that particular direction.
A = Axi + Ay j + Az k
(7.2)
The magnitude of the vector is defined as
A =|A|= Ax2 + Ay2 + Az2
(7.3)
Vectors may be added together in any order by summing the individual components in
each direction. This is the commutative property:
A + B = B + A = ( Ax + Bx )i + ( Ay + By )j + ( Az + Bz )k
(7.4)
The following is also true when adding more than one vector together:
(A + B) + C = A + (B + C)
(7.5)
Equation 7.5 represents the associative property of vectors. In all of the above expressions,
note that i, j, and k are the unit vectors in the x, y, and z coordinate directions, respectively.
Multiplication of vectors can occur in two different ways—each applicable to particular
situations. Consider two vectors A and B shown in Figure 7.2, we define the scalar product
or dot product as
A ⋅ B = A ⋅ B ⋅ cosθ
(7.6)
205
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1
A
A
eA
FIGURE 7.1
Vector and associated unit vector.
B
θ
A
FIGURE 7.2
Vector pair illustrated.
Both the commutative and associative laws of multiplication apply to the dot product.
A ⋅B = B⋅A
(7.7)
(A + B) ⋅ C = A ⋅ C + B ⋅ C
(7.8)
The dot product of two vectors is given by
A ⋅ B = ( Axi + Ay j + Az k) ⋅ (Bxi + By j + Bz k)
(7.9)
This equation when expanded is
A ⋅ B = Ax Bxi ⋅ i + Ax Byi ⋅ j + Ax Bzi ⋅ k + Ay Bx j ⋅ i + Ay By j ⋅ j + Ay Bz j ⋅ k + AzBx k ⋅ i
+ AzBy k ⋅ j + AzBz k ⋅ k
(7.10)
However, since the unit vectors are orthogonal, and the dot product of two orthogonal
vectors is identically zero while the dot product of parallel vectors is unity as follows from
i ⋅ i = j ⋅ j = k ⋅ k = 1 ⋅ 1 ⋅ cos(0 ) = 1
i ⋅ j = j ⋅ i = j ⋅ k = k ⋅ j = i ⋅ k = k ⋅ i = 1 ⋅ 1 ⋅ cos(90 ) = 0
Therefore, we can write
A ⋅ B = Ax Bx + Ay By + AzBz
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(7.11)
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207
B
en
θ
A
FIGURE 7.3
Vector cross product normal unit vector.
The second type of vector multiplication is the vector or cross product, which is defined as
A × B = A ⋅ B ⋅ sinθ en
(7.12)
Here en is a unit vector normal to the plane made by vectors A and B. This is depicted in
Figure 7.3. The cross product does not obey the commutative property because
A × B = −B × A
(7.13)
The distributive property, however, does apply to the cross product. Thus,
(A + B) × C = A × C + B × C
(7.14)
The cross product of two vectors is given by
A × B = ( Axi + Ay j + Az k) × (Bxi + By j + Bz k)
(7.15)
When expanded, Equation 7.15 can be written as
A × B = Ax Bxi × i + Ax Byi × j + Ax Bzi × k + Ay Bx j × i + Ay By j × j + Ay Bz j × k
+ AzBx k × i + AzBy k × j + AzBz k × k
(7.16)
Since the unit vectors are orthogonal,
i × i = j × j = k × k = 1 ⋅ 1 ⋅ sin(0 )en = 0 and i × j = 1 ⋅ 1 ⋅ sin(90 ) = en
But, since we have a right-handed coordinate system, by the right-hand rule, the normal
to i and j is the unit vector k, thus i × j = k. We can also invoke Equation 7.13 to get j × i =
−i × j = −k. We can carry this logic further to show that j × k = i or k × j = −i and i × k = −j
or k × i = j. Thus, we can rewrite Equation 7.16 as
A × B = ( Ay Bz − AzBy )i + ( AzBx − AxBz )j + ( AxBy − Ay Bx )k
(7.17)
This Equation 7.17 is the following determinant expanded by its minors:
i
A × B = Ax
Bx
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j
Ay
By
k
Az
Bz
(7.18)
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208
A + ΔA
ΔA
A
FIGURE 7.4
Vector sum illustrated.
We will proceed next to the calculus of vectors. Let us consider a vector, A, dependent
upon a scalar variable, u, as shown in Figure 7.4. Then A + ∆A corresponds to u + ∆u and
we can write for its derivative
dA
∆A
= lim
du ∆u →0 ∆u
(7.19)
d(A + B) dA dB
=
+
du
du du
(7.20)
Differentiation is distributive so that
The chain rule also applies for scalar and vector products so that
dg
d
dA
( gA) =
A+ g
du
du
du
(7.21)
d
dA
dB
(A ⋅ B) =
⋅B + A ⋅
du
du
du
(7.22)
d
dA
dB
(A × B) =
×B + A×
du
du
du
(7.23)
Consider a vector A dependent upon time, t. If we take its derivative with respect to time,
we get
dAy
dA dAx
dAz
di
dj
dk
i+
j+
k + Ax
=
+ Ay
+ Az
dt
dt
dt
dt
dt
dt
dt
(7.24)
If the coordinate system is inertial (i.e., it does not move), we can write
dAy
dA dAx
dAz
i+
j+
k
=
dt
dt
dt
dt
(7.25)
If the coordinate system is moving (like on a rotating earth), the rate of change terms for
the unit vectors cannot be neglected. This gives rise to what we call “Coriolis terms” as we
shall discuss later.
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209
y
P
r
O
x
z
FIGURE 7.5
Position vector.
We will now examine the kinematics of a particle. Kinematics is the study of the motion
of particles and rigid bodies without regard to the forces which generate the motion.
Particle kinematics assumes that a point can represent the body. The rotations of the particle itself are neglected making this a three degree of freedom (DOF) model. If we have
the inertial reference frames x, y, and z, the position of a particle, P, is defined by a position
vector, r, drawn from the origin to the particle as is shown in Figure 7.5.
If the particle, P, moves along a trajectory, T, its instantaneous velocity is always in a
direction tangent to the trajectory and its magnitude is the speed at which it moves along
the curve. Thus, the tip of this vector, r, traces out the trajectory (Figure 7.6) and the velocity, v, is defined as the time rate of change of r, written as
v=
dr
dt
(7.26)
If we were to take the velocity vector at every instant of time and fix its tail to the origin of
an inertial coordinate system, then the curve traced out by its tip would be called a hodograph (Figure 7.7) and the velocity of the tip would be the time rate of change of velocity or
the acceleration, a. Thus, we can write
a=
dv d 2r
=
dt dt 2
(7.27)
y
v
T
P
O
r
x
z
FIGURE 7.6
Trajectory curve.
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y
a
v
O
x
z
FIGURE 7.7
Hodograph.
er
eθ
r
θ
O
FIGURE 7.8
Differentiation of a vector through use of tangential and radial unit vectors.
Now, if we examine the particle as moving in two dimensions only, we can break its motion
up into two components, one parallel to and one perpendicular to the position vector, r
(Figure 7.8). The position vector written in this coordinate system is given by
r = re r
(7.28)
So from our definition for the velocity in Equation 7.26, we get
v=
dr dr
de
=
er + r r
dt dt
dt
(7.29)
Since er is a unit vector (its magnitude is a constant = 1) the only thing that changes with
time is its direction.
This introduces the concept of curvilinear motion with radial coordinates, (r, θ). The
direction is defined by the angle, θ. For a small change in the angle, θ, we can write
der
∆e
= lim r
dt ∆t →0 ∆t
(7.30)
However, we can observe that for small angles
∆er =|er |sin(∆θ ) = (1)sin(∆θ ) ≈ ∆θ
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(7.31)
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211
Δe
r
er
er + Δer
eθ
Δθ
FIGURE 7.9
Rotation of the radial unit vector.
We also see from Figure 7.9 that ∆er acts in the eθ direction thus,
∆er ≈ ∆θ eθ
(7.32)
Then, returning to Equation 7.30 we can write
∆θ dθ
der
= eθ lim
=
eθ
t
∆
0
→
∆t
dt
dt
(7.33)
Now, we can insert Equation 7.33 into Equation 7.29 to get the desired relation for the velocity.
v=
dθ
dr dr
=
er + r
eθ
dt dt
dt
(7.34)
The first term on the RHS of Equation 7.34 is the radial velocity, the second term is the
tangential velocity sometimes denoted as vr and vθ, respectively. The magnitude of the
velocity is given by
2
dr dθ
v =|v|= vr2 + vθ2 = + r
dt dt
2
(7.35)
To obtain the acceleration in curvilinear coordinates, we need to take the time derivative
of Equation 7.34 as follows:
a=
d 2θ
dθ deθ
dv d dr
dθ d 2r
dr der dr dθ
+
= er + r
eθ = 2 er +
eθ + r 2 eθ + r
dt
dt dt
dt dt dt
dt dt
dt dt dt dt
(7.36)
We have already solved for the derivative of er with respect to time; now, in a similar manner, we will find the derivative of the tangential component. Again, since eθ is a unit vector,
the only thing that changes with time is its direction. This direction is again defined by the
angle, θ, so for a small change in the angle, θ, we can write
deθ
∆e
= lim θ
∆
t
→
0
∆t
dt
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(7.37)
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212
Δeθ
eθ
er
eθ + Δeθ
Δθ
FIGURE 7.10
Rotation of the tangential unit vector.
But we see again that for small angles
∆eθ =|eθ|sin(∆θ ) = (1)sin(∆θ ) ≈ ∆θ
(7.38)
which acts in the negative er direction as depicted in Figure 7.10. If
∆eθ ≈ −∆θ er
(7.39)
∆θ
deθ
dθ
er
= −er lim
=−
∆t → 0 ∆t
dt
dt
(7.40)
then we can write,
Insertion of Equations 7.33 and 7.40 into Equation 7.36 yields
2
a=
d2r
dr dθ
dr dθ
d 2θ
dθ
r
e
+
e
+
e
+
eθ − r
θ
θ
r
er
2
2
dt
dt dt
dt dt
dt
dt
(7.41)
Rearranging and combining like terms gives us
2
d 2r
d 2θ
dr dθ
dθ
+
a = 2 −r
e
r
r 2 + 2
eθ
dt dt
dt
dt
dt
Each of these terms has a specific name and meaning in the dynamics of a body
d2r
= Radial acceleration
dt 2
2
dθ
r
= Centripetal acceleration
dt
d 2θ
= Angular acceleration
d 2t
2
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dr dθ
= Coriolis acceleration
dt dt
(7.42)
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213
y΄
y
B
rB/A
rB
O
x΄
Body-fixed coordinate
A
rA
system (translates with the
body and the body
does not rotate)
x
Inertial coordinate system
(fixed in space)
rA = Position vector of point A
rB = Position vector of point B
rB/A = Relative position vector of
point B with respect to point A
vA = Velocity of point A
vB = Velocity of point B
vB/A = Relative velocity of point B
with respect to point A
aA = Acceleration of point A
aB = Acceleration of point B
aB/A = Acceleration of point B with
respect to point A
FIGURE 7.11
Definition of vectors associated with rigid body translational motion.
To move on in our study, we need to examine the planar kinematics of a rigid body. First,
we will examine a pure translation where we have a body-fixed coordinate system moving relative to our inertial coordinate system. Here we note from Figure 7.11 that by vector
addition, we obtain
rB = rA + rB/A
(7.43)
To determine the velocity of point B, which is under a pure translation, we have to differentiate Equation 7.43 to get
vB =
drB drA drB/A
=
+
dt
dt
dt
(7.44)
We know, however, that since this is a pure translation (no rotation) drB/A/dt = 0 and
drA/dt = vA. Thus, for a pure translation,
vB = v A
(7.45)
If we differentiate Equation 7.45, we get the acceleration of a point during a pure translation as
aB =
dv B dv A
=
= aA
dt
dt
(7.46)
We will now examine the rotation of a body fixed in space. Let us define the angular velocity, ω, as the time rate of change of angular position, θ, thus
ω=
dθ
dt
(7.47)
Let us further define the angular acceleration, α, as the time rate of change of angular
velocity, or
α=
dω d 2θ
= 2
dt
dt
(7.48)
The angular velocity, ω, and the angular acceleration, α, are depicted in Figure 7.12. We will
now look at the rotation in terms of the vector kinematic equations for point, P. We first
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α, ω, dθ
ω
α
O
O
r P
dθ
r
P
θ
dθ
θ
FIGURE 7.12
Example of rigid body rotation. On the left is the body rotating in space. On the right is a view of this same body
looking down the axis toward O.
ω
ω
O
r
v
O
P
r
P
rP
E
FIGURE 7.13
Rigid body rotation. On the left is the body rotating at angular velocity, ω. On the right is a view of this same
body looking down the axis toward O.
examine the velocity whose direction we specify by the right-hand rule. Then let us define
the position of point, P, by the position vector, r, as shown in Figure 7.13. Now we can write
the velocity, v, in terms of radial and circumferential components as we discussed earlier.
Thus, from Equation 7.34 we have
v=
dθ
dr dr
=
er + r
eθ
dt dt
dt
(7.49)
But, since this is a rigid body, dr/dt = 0, so we get
v=r
dθ
eθ = rωeθ
dt
(7.50)
If we were instead to draw the position vector from a more general location such as point
E in Figure 7.13, we see that
r = rP sin φ
(7.51)
If we substitute Equation 7.51 into Equation 7.50, we see a form we have derived earlier.
v = rP ω sin φ eθ
(7.52)
This can be written in vector form if we invoke Equation 7.12. Thus, we have
v = ω × rP
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(7.53)
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ω, α
ω, α
O
r
an
at
P
O
an
at
r
rP
P
E
FIGURE 7.14
Rigid body rotation with acceleration. On the left is the body rotating at angular velocity, ω, and accelerating
with angular acceleration, α. On the right is a view of this same body looking down the axis toward O.
We now will examine the acceleration whose direction is once more specified by the righthand rule. We shall define the position of point, P, by the position vector, r, as shown in
Figure 7.14. We can then write the acceleration, a, in terms of radial and circumferential
components as discussed earlier.
We need to recall Equation 7.42 and note that dr/dt = 0. This leaves us with
dθ 2
d 2θ
a = −r
er + r 2
dt
dt
eθ
(7.54)
Here we need to note that the first term is negative because it acts in the negative radial
direction. Now, from our previous definitions we can rewrite Equation 7.54 as
a = (−rω2 )er + (rα )eθ
(7.55)
We can define the normal and tangential components of acceleration as
a n = −rω2er
and a t = rα eθ
(7.56)
where
an is the normal (centrifugal) acceleration
at is the tangential acceleration
We can differentiate Equation 7.53 to obtain the more general result
a=
dr
dv dω
=
× rP + ω P
dt
dt
dt
(7.57)
If we insert Equations 7.48 and 7.53 into Equation 7.57, we get the general vector form for
the acceleration of a rigid body rotating in an inertial coordinate system.
a = α × rP + ω × (ω × rP )
(7.58)
To move closer toward a more general treatment, we shall now derive the kinematic equations for plane motion of a rigid body using a translating coordinate system (the body is
free to rotate). We can break down any planar motion of the rigid body into a translation
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y΄
rA = Position vector of point A
rB = Position vector of point B
rB/A = Relative position vector of
point B with respect to point A
rB/A
rB
vA = Velocity of point A
x΄
vB = Velocity of point B
A
vB/A = Relative velocity of point B
Translating
coordinate
rA
with respect to point A
system (translates
aA = Acceleration of point A
with
the
body)
x
O
aB = Acceleration of point B
Inertial coordinate system
aB/A = Acceleration of point B with
(fixed in space)
respect to point A
y
B
FIGURE 7.15
Definition of vectors associated with rigid body translational and rotational motion.
and a rotation about some point. Let us choose point A in Figure 7.15 to be a location about
which the body rotates. Equation 7.43 is still valid, but dr/dt no longer equals zero.
Thus, from Equation 7.44, we have
vB =
drB drA drB/A
=
+
dt
dt
dt
(7.59)
Not only is drB/A/dt ≠ 0 but also, since we chose point A as one about which rotation will
take place, drA/dt is a pure translation and drB/A/dt is a pure rotation about point A. Thus,
we can write Equation 7.59 in terms of the velocities as
v B = v A + v B/A
(7.60)
We saw earlier that for a pure rotation, we can write the velocity in the form of Equation
7.53. Thus, we have
v B/A = ω × rB/A
(7.61)
If we substitute Equation 7.61 into Equation 7.60, we obtain the vector equation for planar
motion of a rigid body in which our coordinate system translates with a point in the body
but does not rotate with the body.
v B = v A + ω × rB/A
(7.62)
To obtain the acceleration of point B, we need to differentiate Equation 7.60, thus
aB =
dv B dv A dv B/A
=
+
dt
dt
dt
(7.63)
Let us examine this equation term by term. The first term is straightforward, showing that
the acceleration of the translation is simply the linear acceleration of our chosen reference
point.
dv A
= aA
dt
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The second term is differentiated as follows:
dv B/A d
dω
dr
= (ω × rB/A ) =
× rB/A + ω × B/A
dt
dt
dt
dt
(7.65)
We again need to call upon Equations 7.48 and 7.53 to get Equation 7.65 into a more general
form.
dv B/A
= α × rB/A + ω × (ω × rB/A )
dt
(7.66)
Inserting Equations 7.64 and 7.66 into Equation 7.63 yields the kinematic equation for the
acceleration of a rigid body in a coordinate system that translates with the body. The coordinate system in this case moves with the body but does not rotate allowing the body to
rotate relative to the moving coordinate system.
a B = a A + α × rB/A + ω × (ω × rB/A )
(7.67)
We shall now derive the kinematic equations for plane motion of a rigid body using a
translating and rotating coordinate system (the body is capable of movement in both coordinate systems). We choose point A in Figure 7.16 to be a location from which we want
to measure the motion of the body. At the instant considered, point A has a position, rA,
a velocity, vA, and an acceleration a A, while the x–y axes (and the body) are rotating with
angular velocity, ω, and accelerating with angular acceleration, α.
Equation 7.43 is still valid for determination of the position vectors. For the velocity of
point B, we shall use the form
vB = v A +
drB/A
dt
(7.68)
If we examine our earlier derivations under the curvilinear motion and replace er with i
and eθ with j (i and j represent the unit vectors in our x–y coordinate system), we obtain
the following relations:
di dθ
=
j = ωj
dt dt
(7.69)
dj dθ
=
(−i) = −ωi
dt dt
(7.70)
y΄
y
B
rB /A
x΄
rB
A
rA
Coordinate system
(translates and rotates
independent the body)
x
O
Inertial coordinate system
(fixed in space)
FIGURE 7.16
Definition of vectors associated with rigid body translational and rotational motion including a rotating local
coordinate system.
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y
B
rB/A
x
A
ω
FIGURE 7.17
Body rotating in moving coordinate system.
Using the definition of the cross product and noting they are orthogonal, Equations 7.69
and 7.70 can be rewritten as
di
= ω×i
dt
(7.71)
dj
= ω× j
dt
(7.72)
If we look at the body and our moving coordinate system as shown in Figure 7.17, we see
that if the body translates and rotates relative to our x–y axes we can write, in light of
Equation 7.62
drB/A
= (v B/A )xyz + ω × rB/A
dt
(7.73)
Substituting this into Equation 7.68 yields the general relation for the velocity of a point in
a rigid body as seen from an arbitrary coordinate system.
v B = v A + (v B/A )xyz + ω × rB/A
(7.74)
To obtain the acceleration of our point B, we need to differentiate Equation 7.74 with respect
to time, thus
aB =
drB/A
dv A d(v B/A )xyz dω
+
+
× rB/A + ω ×
dt
dt
dt
dt
(7.75)
Let us again move term by term through Equation 7.75. The first term is
dv A
= aA
dt
(7.76)
Since our point A does not rotate, the acceleration of its translation is simply this linear
acceleration.
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The second term is differentiated by first breaking up vB/A into its components along the
x and y axes of our moving frame. Since we are looking only at motion in the x–y plane,
the z component is nonexistent.
(v B/A )xyz = (v B/A )x i + (v B/A )y j
(7.77)
This allows us to write the second term as
d(vB/A ) xyz d(vB/A ) x
d(vB/A ) y
di
dj
=
i+
j + (vB/A ) x
+ (vB/A ) y
dt
dt
dt
dt
dt
(7.78)
The first pair of terms in Equation 7.78 are the acceleration components of point B relative to point A as seen by an observer moving with the coordinate system at point A. The
second pair of terms of Equation 7.78 can be rewritten as the cross product of the angular
velocity of the x–y coordinate system and the velocity vector of point B relative to point A.
So we can write
d(v B/A )xyz
= (a B/A )xyz + ω × (v B/A )xyz
dt
(7.79)
Returning now to Equation 7.75, in the third term, we simply rewrite the term dω/dt as α.
Finally, for the last term of Equation 7.75, we use Equation 7.73 to obtain
a B = a A + (a B/A )xyz + ω × (v B/A )xyz + α × rB/A + ω × (v B/A )xyz + ω × (ω × rB/A )
(7.80)
After a slight rearrangement and combination of like terms, we have the general kinematic
equation for acceleration of point B
a B = a A + α × rB/A + ω × (ω × rB/A ) + 2ω × (v B/A )xyz + (a B/A )xyz
(7.81)
It is important to review each of the terms in Equations 7.74 and 7.81. First, let us review the
generalized velocity and acceleration equations we derived.
v B = v A + (v B/A )xyz + ω × rB/A
(7.74)
a B = a A + α × rB/A + ω × (ω × rB/A ) + 2ω × (v B/A )xyz + (a B/A )xyz
(7.81)
The terms in these equations have meanings as tabulated in Table 7.1.
As one can imagine the addition of the third dimension in these equations adds significant complexity to the expressions although the basic principles remain the same. The
interested reader is referred to Ref. [1] or any similar text on dynamics to familiarize themselves with the three-dimensional uses of these equations.
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TABLE 7.1
Vector Terms Used in Equations 7.74 and 7.81
Variable
rB/A
vA
vB
(vB/A)xyz
aA
aB
(aB/A)xyz
ω
α
Definition
Relative position vector of point B with respect to point A
Velocity of point A in the inertial coordinate system
Velocity of point B in the inertial coordinate system
Relative velocity of point B with respect to point A in the xyz coordinate system
Acceleration of point A in the inertial coordinate system
Acceleration of point B in the inertial coordinate system
Acceleration of point B with respect to point A in the xyz coordinate system
Angular velocity of the xyz coordinate system measured from the inertial coordinate system
Angular acceleration of the xyz coordinate system measured from the inertial coordinate system
Problem 1
A 155 mm projectile is in flight at its maximum ordinate. At this instant in time, the nose
of the projectile is pointing along (and spinning about) the unit vector:
x = (0.998e1 + 0.030e2 + 0.056e3 )
The projectile velocity vector is
ft
V = (1199e1 + 0e2 + 49e3 )
s
In both of these cases, e1, e2, and e3 are unit vectors in the x, y, and z planes, respectively.
Also at this location the air density, spin rate, and projectile mass are as follows:
lbm
rev
ρ = 0.052 3 , p = ω = 150
, and m = 100 [lbm]
ft
s
The projectile characteristics are assumed to be
CD = 0.29
CMα = 3.0 (CMq + CM αɺ ) = −10.2
CLα = 2.12 (CNq + CN αɺ ) = 0.002
CNpα = −0.010 = −10.2 CMpα = 0.51
Clp = −0.015
Please answer the following questions:
1. Draw the situation.
2. Determine the drag force vector.
Answer: FD = (−68.39e1 − 2.80e3) [lbf]
© 2014 by Taylor & Francis Group, LLC
Dynamics Review
221
3. Determine the lift force vector.
Answer: FL = (−0.310e1 + 14.980e2 + 7.600e3) [lbf]
4. Determine the overturning moment vector.
Answer: MM = (−0.441e1 − 5.468e2 + 10.783e3) [ft-lbf]
5. Determine the magnus moment vector.
Answer: MMpα = (0.043e1 − 0.732e2 − 0.369e3 ) [ft-lbf]
Reference
1. Greenwood, D.T., Principles of Dynamics, 2nd edn., Prentice Hall, Englewood Cliffs, NJ, 1988.
Further Readings
Beer, F.P. and Johnston, E.R., Vector Mechanics for Engineers—Statics and Dynamics, 7th edn., McGraw-Hill,
New York, 2004.
Colley, S.J., Vector Calculus, Prentice Hall, Upper Saddle River, NJ, 1998.
Hibbeler, R.C., Engineering Mechanics—Statics and Dynamics, 7th edn., Prentice Hall, Englewood
Cliffs, NJ, 1995.
O’Neil, P.V., Advanced Engineering Mathematics, 2nd edn., Wadsworth Publishing Co., Belmont, CA,
1986.
© 2014 by Taylor & Francis Group, LLC
© 2014 by Taylor & Francis Group, LLC
8
Trajectories
Now that the basics of the terminology and the dynamic equations have been presented,
we shall begin to look at their uses in the form of prediction of trajectories. The aero-ballistician is usually faced with one of two problems: “If I want to hit a target at position x,
to what elevation (and perhaps with how much propelling charge) do I have to elevate the
weapon?” or “My weapon is elevated to elevation x and I expect muzzle velocity y—where
is the projectile going to end up?”
To approach this in a logical and easily understandable fashion, we shall begin with a great
many simplifying assumptions, relieving these as we progress. Each section builds upon the
previous one so that we recommend even seasoned veterans progress in numerical order.
Initially, we will only look at the effect that gravity imposes on the projectile, a vacuum trajectory, so that even the air is removed from our area of concern, thus neutralizing the fluid
mechanics for a while. As we progress, we shall add in the atmosphere but neglect dynamics,
atmospheric perturbations, and earth rotation. One by one we shall continually step up the
complexity until finally we shall introduce the full six degree-of-freedom (6 DOF) equations.
One might initially think that these simplified models have no practical use, but are merely
educational stepping stones. Nothing could be further from the truth. In many instances,
some of the complications only slightly affect the solution and a ballistician is well placed
to assume them away. Some of these common situations will be pointed out as they arise.
8.1 Vacuum Trajectory
In this section, we will make two broad assumptions: First, that the projectile mass is concentrated at a point (which allows us to neglect body dynamics affected by mass distribution),
and second, that the only force acting on the projectile is that due to the acceleration of
gravity (this allows us to neglect the rather complicated fluid dynamic effects when a solid
body moves through a fluid). With these assumptions, the two governing differential
equations of motion are
mxɺɺ = 0
(8.1)
yɺɺ = − g
(8.2)
The solutions to these equations, found by integrating with respect to time, are
x = V0t cos φ0
y = V0t sin φ0 −
1 2
gt
2
(8.3)
(8.4)
223
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
224
y
V0 = 800 m/s
V
mg
0= ?
x
R = 20,000 m
FIGURE 8.1
Vacuum trajectory.
A sketch of this simplified trajectory is seen in Figure 8.1. Notice that unlike the generalized trajectory shown previously in Figure 6.1 the terminal point is on the line y = 0 and
the entire trajectory is in the x–y plane.
But from Equations 8.3 and 8.4, we can write
y
gt
= tan φ0 −
2V0 cos φ0
x
(8.5)
By solving Equation 8.3 for time, t, we get
t=
x
V0 cos φ0
(8.6)
We can then write Equation 8.5 putting y in terms of x only. Therefore,
y = x tan φ0 −
gx 2
2V02 cos 2 φ0
(8.7)
This equation is in the form of a parabola in x and y coordinates, the path the projectile will
follow in a vacuum. Solving for the range, x, when y = 0 gives
x(2V02 cos φ0 sin φ0 − gx) = 0
(8.8)
This says that either x = 0 (the trivial solution) or
x=
V2
2V02
cos φ0 sin φ0 = 0 sin 2φ0
g
g
(8.9)
Because the trajectory is a parabola, maximum range is attained at
π
4
(8.10)
π
=1
2
(8.11)
φ0 =
since
sin
© 2014 by Taylor & Francis Group, LLC
Trajectories
225
When we substitute this into Equation 8.9, the maximum range can be found to be
xmax =
V02
g
(8.12)
The maximum ordinate of the trajectory is at half the maximum range and is
ymax =
V02
4g
(8.13)
If we differentiate Equation 8.9 with respect to ϕ 0 and set this equal to 0, we can prove
Equation 8.10 as shown next:
dxmax 2V02
=
cos 2φ0 = 0
dφ0
g
(8.14)
This gives the launch angle for maximum range in a vacuum as π/4.
Except for the maximum range, there are two quadrant elevations (QE) that will allow a
projectile to impact at a given distance. We will designate the second QE with a caret, “∧.”
Its existence is due to the identity
sin φ = sin(180 − φ )
(8.15)
sin 2φ0 = sin 2(90° − φ0 )
(8.16)
1
gR
φˆ0 = 90° − φ0 = 90° − sin −1 2
2
V0
(8.17)
Then,
Thus,
where x has been replaced by the range, R. The maximum ordinate is achieved when the
y-component of the velocity is 0. By differentiating Equation 8.4 with respect to time and
setting the result equal to 0, we get
V0 sin φ0 − gts = 0
(8.18)
or
ts =
V0 sin φ
g
(8.19)
Substituting this into Equation 8.4 gives
2
V sin φ0 1 V0 sin φ0
1 V02 sin 2 φ0
=
−
y s = V0 sin φ0 0
g
2
g
g
g
2
© 2014 by Taylor & Francis Group, LLC
(8.20)
Ballistics: Theory and Design of Guns and Ammunition
226
If we note that at impact the y-coordinate is zero, we can find the time of flight to impact
with Equation 8.4
0 = V0tI sin φ0 −
1 2
gtI
2
(8.21)
or
tI =
2V0 sin φ0
g
(8.22)
This is double the time to the maximum ordinate and the trajectory in a vacuum is symmetrical about this ordinate. Further evidence of the symmetry may be seen by examining
the angle of fall. If we differentiate Equation 8.7 with respect to x and substitute the value
of x we found at impact in Equation 8.9 in the differentiated result, we see that
dy
sin 2φ0
= tan φ0 −
cos 2 φ0
dx I
(8.23)
sin 2φ0 = 2 sin φ0 cos φ0
(8.24)
But
Therefore,
tan φI = tan φ0 − 2
sin φ0
= − tan φ0
cos φ0
(8.25)
Thus, in a vacuum trajectory, the projectile impacts at the mirror image of the angle it had
when it was launched.
For any given launch velocity, V0, maximum range in a vacuum is achieved with an
initial launch angle of 45°. To reach any range shorter than the maximum, there are two
launch elevations, one greater than 45° and the other less, a high angle and a low angle of
fire. The trajectory envelope is a curve that bounds all possible trajectories that attempt to
reach all ranges from zero to the maximum range possible for the given launch velocity [1].
We shall now mathematically describe this curve.
We know from Equation 8.7 that
y = x tan φ0 −
gx 2
2V02 cos 2 φ0
(8.7)
y = x tan φ0 −
gx 2
sec 2 φ0
2V02
(8.26)
This can also be written as
© 2014 by Taylor & Francis Group, LLC
Trajectories
227
If we make use of the trigonometric identity sec2 ϕ = 1 + tan2 ϕ, we can, with substitution
and manipulation, write
tan 2 φ0 −
2V02
2V 2
tan φ0 + 02 y + 1 = 0
gx
gx
(8.27)
Equation 8.27 is quadratic in ϕ 0 and as such, when solved, yields two roots that correspond
to the two elevations that achieve the same range as discussed earlier.
The exceptions to this are when the range is zero or the range is maximum. These conditions yield a repeated root. The other instances a repeated root occurs are whenever the
trajectory touches the trajectory envelope. This occurs only once at any given elevation. If
the roots of this equation are complex conjugates, the range in question cannot be achieved
with the given muzzle velocity. We can solve for all of the double roots to obtain the equation of the trajectory envelope.
We proceed by first completing the square in Equation 8.27 noting that
2
V2
2V 2
V2
tan φ0 − 0 tan φ0 + 0 = tan φ0 − 0
gx
gx
gx
2
2
(
)
(8.28)
2
By adding and subtracting a term, V02 gx , to Equation 8.27, we complete the square of a
part of the equation and can operate on the remainder of it:
2
2
V 2 V 2 2V 2
2V 2
tan φ0 − 0 tan φ0 + 0 − 0 + 02 y + 1 = 0
gx
gx
gx gx
2
(8.29)
Breaking apart Equation 8.29 into two terms and setting each equal to zero gives us from
Equation 8.28
2
tan 2 φ0 −
2
V2
2V02
V2
tan φ0 + 0 = tan φ0 − 0 = 0
gx
gx
gx
(8.30)
and
2
V2
2V02
y − 0 +1 = 0
2
gx
gx
(8.31)
The double root in Equation 8.30 occurs when
tan φ0 =
V02
gxe
(8.32)
where xe = x on the envelope curve. We can also pursue the equation for the envelope curve
more directly
2
V2
2V02
ye − 0 + 1 = 0
2
gxe
gxe
© 2014 by Taylor & Francis Group, LLC
(8.33)
Ballistics: Theory and Design of Guns and Ammunition
228
2000
1800
1600
Altitude (ft)
1400
1200
1000
800
600
400
200
0
0
500
1000
1500
2000
2500
Range (ft)
3000
3500
4000
4500
FIGURE 8.2
Trajectory envelope.
where ye is the y-coordinate on the envelope curve. Equation 8.33 can be further rearranged to yield the final equation of the trajectory envelope:
ye =
1 V02 gxe2
−
2 g 2V02
(8.34)
A typical trajectory envelope is illustrated in Figure 8.2.
To move to a different subject in the study of the vacuum trajectory, when the trajectory of
the projectile is relatively flat, certain simplifying assumptions may be made, which allow the
equations of motion to be solved with greater ease. In particular, if we rewrite Equation 8.7 as
y = x tan φ0 −
gx 2
sec 2 φ0
2V02
(8.35)
we now take its derivative with respect to ϕ 0, we get
gx
dy
gx 2
= x sec 2 φ0 − 2 tan φ0 sec 2 φ0 = x 1 − 2 tan φ0 sec 2 φ0
V
dφ0
V0
0
(8.36)
Now because
sec 2 φ0 = 1 + tan 2 φ0
and if tan2 ϕ 0 ≪ 1 then sec2 ϕ 0 ≈ 1. This occurs when ϕ 0 < 5°. This is the requirement for
what is commonly called the flat fire approximation to be valid. We can then translate
Equations 8.35 and 8.36 into
y ≈ x tan φ0 −
© 2014 by Taylor & Francis Group, LLC
gx 2
2V02
(8.37)
Trajectories
229
and
dy
gx
≈ x 1 − 2 tan φ0
dφ0
V0
Equation 8.38 can be even further simplified for short ranges if −
dy
≈x
dφ0
(8.38)
gx
tan φ0 ≪ 1 then
V02
(8.39)
This is sometimes known as the rigid trajectory because the trajectory appears to rotate
rigidly with the elevation angle. The vertical error that arises from use of the flat fire
approximation in a vacuum trajectory is
εy =
gx 2
tan 2 φ0
2V02
(8.40)
which, as is readily seen, states that as the range or launch angle increases, the error
increases. Flat fire is characteristic of the engagements experienced with high-powered,
high-velocity tank cannons where initial launch angles for direct-fire ranges of several
kilometers are less than 5°. Elevation changes to correct fire are measured in fractions of a
degree (known as mils) as well. One mil is equal to 1/6400 of a circle.
Problem 1
A target is located at 20 km. A projectile muzzle velocity is 800 m/s, assuming a vacuum
trajectory, at what quadrant elevation (QE) should one set the weapon to hit the target?
Answer: ϕ 0 = 158.7 [mil].
Problem 2
The enemy in the aforementioned problem is very smart and has located his unit on the
reverse slope of a hill that is 3,000 m in height with its peak located 18,000 m from your
firing position. Assuming that the target is at the same level as you (just behind the hill),
determine a firing solution (QE, if there is one) to hit him assuming a vacuum trajectory.
Answer: It can be hit.—You find the initial QE.
Problem 3
The U.S. pattern 1917 (M1917) “Enfield” rifle was the most numerous rifle used by our troops
in the First World War. It was an easier rifle to manufacture than the M1903 “Springfield” (even
though the Springfield was officially the U.S. Army’s service rifle) and the troops liked its accuracy better. In fact, the famous Sergeant Alvin York was actually armed with an Enfield, not
a Springfield as is commonly believed, when he single-handedly captured over 100 German
soldiers in the Argonne Forest in 1918. The pattern 1917 used the standard M1 30-’06 cartridge
in U.S. service. The bullet had a mass of 174 grains (a grain is a common unit of measure in
© 2014 by Taylor & Francis Group, LLC
230
Ballistics: Theory and Design of Guns and Ammunition
small arms ammunition and is defined as 1/7000 of a lbm) and a diameter of 0.308 in. This
cartridge–rifle combination has a muzzle velocity of 2800 ft/s. Assuming a vacuum trajectory:
1. Determine the angle in degrees to set the sights on the rifle (i.e., the QE) if the target is level with the firer and at 200 yards range.
Answer: ϕ 0 = 0.0705°.
2. If the target is at the same horizontal range but 20 yards higher, and the firer does
not adjust the sights, how much higher or lower will the bullet strike?
Answer: ymiss = −0.0125 [in.].
Problem 4
You are asked to create a rough safety fan for a maximum range test at Yuma Proving
Ground. The test consists of a U.S. M198 155 mm howitzer firing an M549 projectile at maximum charge with rocket off. The projectile weighs 96 lbm. The muzzle velocity is 880 m/s.
1. Using a vacuum trajectory, calculate and plot the trajectory envelope for the test.
Answer: R = 78,940 [m].
2. Determine the longest time of flight of the projectile.
Answer: tI = 179.4 [s].
Problem 5
It is desired to use a 105 mm battery to set up a line of illumination candles over an enemy
defensive position. The enemy is positioned in depth of about 2 km. Assuming that four weapons are available and the candle array is to be placed such that the closest candle is 9 km away
from the gun position (assume all guns occupy the same point in space) and there is one candle
every 500 m (i.e., four candles total at 9.0, 9.5, 10.0, and 10.5 km), determine the QE for each
weapon, the time set on each fuze, and the time and order of fire of each gun such that all of
the expulsion events occur simultaneously at an altitude of 750 m over the heads of the enemy.
Assume a muzzle velocity of 600 m/s and a projectile mass of 30 lbm. Use a vacuum trajectory
for the purposes of this problem. (Note that this would NEVER be acceptable in actual practice.)
Could the same result be achieved with less than 4 guns? Show why or why not with a
calculation.
Problem 6
The fire control problem: During the age of battleships, it was essential that a fire control computer be installed on the ships. From 1915 onwards, these were mechanical devices by which
one could put in their own ship’s course and speed as well as the target ship’s estimated
course and speed. These data and the expected muzzle velocity of the projectile (as well as
time of flight) allowed the guns to be aimed where the enemy ship would be when the shells
landed. To get a feel for the magnitude of the problem, we are going to examine it in a greatly
simplified form assuming a vacuum trajectory. Given the data that follow, and assuming a
vacuum trajectory, provide a firing angle off the bow, elevation and timing to fire each of four
guns so that a pattern is created to hit if the target veers 10° to the port or starboard of its present course. A hit can be assumed to occur if the shell lands on the point where the enemy ship
will actually be or if it lands in its “danger space.” Because of the trajectory of the shells, a hit
will occur if the trajectory passes over the target and lands within 100 yards behind it—the
ship creates a “shadow” or danger space. An example pattern might look like the one drawn
in the figure but feel free to create your own if it meets the aforementioned criteria. The shells
© 2014 by Taylor & Francis Group, LLC
Trajectories
231
should all be fired at the same time. Assume your speed estimate of the target is exact. Also
assume the four guns are mounted in two pairs so that you only have two azimuths to work
with but the elevations can be varied independently. The target is about 400 ft long so some
error in azimuth is acceptable—but the error should be quantified. Plot the impact points and
target position at the time of impact. Remember that your ship is moving!
The weapons are British 12” mark IX naval guns with a muzzle velocity of 2800 ft/s and
a maximum elevation of 20°.
20 knots
25°
4,000 yards
15 knots
12,000 yards
8.2 Simple Air trajectory (Flat Fire)
As we progress in our study of exterior ballistics, we now introduce the concept of drag by
substituting air for the medium through which our point mass projectile flies. We do this
so that projectile dynamics do not enter yet into the equations of motion. We are essentially still dealing with a spherical, nonrotating cannon ball. Furthermore, to simplify the
mathematics, we will insist on a flat fire trajectory, with launch angles below about 6°.
A flat fire trajectory is depicted in Figure 8.3. The methods and equations we will develop
were used in the 1950s for direct-fire calculations over relatively short ranges [1].
We begin with Newton’s second law in an inertial reference frame and use vector representations (boldface) where appropriate.
F = ma
m
(8.41)
dV
= ΣF + m g
dt
(8.42)
V0
y
Vy 0
0
Vx 0
V0
j
V
0
i
FIGURE 8.3
Flat fire trajectory.
© 2014 by Taylor & Francis Group, LLC
mg
x
Ballistics: Theory and Design of Guns and Ammunition
232
where
m is the projectile mass
V is the projectile velocity vector
t is the time
dv
a=
= Vector acceleration
dt
∑ F is the vector sum of all aerodynamic forces
g is the vector acceleration due to gravity
The inertial reference frame allows us to neglect the Coriolis acceleration, which is the
result of the earth’s rotation. Since there is no angle of yaw, the lift and drag forces due to
yaw and the Magnus force due to spin are also negligibly small. These will be discussed in
detail later. Thus, only the projectile drag forces (base, wave, and skin-friction) are working to slow the projectile down and gravity is pulling it toward the earth. The aerodynamic drag force acting on the projectile is then given by
1
1
FD = − ρ SCD VV = ρV 2SCD
2
2
(8.43)
where CD is the drag coefficient, introduced earlier. Another coefficient in common use in
ballistics is the ballistic coefficient, C, which is defined as
C=
m
d2
(8.44)
where m and d are the mass and diameter of the projectile. As a matter of convenience, we
also define
ρ SCD ρπ CD
=
(8.45)
Cˆ D* =
2m
8 C
This allows us to save a little energy in typing since this combination of parameters
appears so often. It is known as a starred coefficient [2]. Equation 8.45 stems from the fact
that S, the frontal area of the projectile, is
S=
π d2
4
(8.46)
We can combine Equations 8.42 and 8.45 and divide by the mass to get an expression for
the time rate of change of velocity (acceleration)
dV
1
=−
ρ SCD VV + g
dt
2m
(8.47)
The negative sign was placed in front of the force (Equation 8.43) because the drag always
opposes the velocity vector (otherwise, it is called thrust). We can separate the velocity,
acceleration, and gravitational vectors into components along the coordinate axes so that
they will be convenient to work with:
dV
= Vxi + V y j + V z k + (Vxi + Vy j + Vz k )
dt
© 2014 by Taylor & Francis Group, LLC
(8.48)
Trajectories
233
and
g = −gj
(8.49)
But because we are in an inertial frame, i̇ = j̇ = k̇ = 0 and therefore
dV
= Vxi + V y j + V z k
dt
(8.50)
If we break Equation 8.47 into its components, we get three coupled, ordinary, nonlinear
differential equations
V x = −Cˆ D* VVx
V y = −Cˆ D* VVy − g
V z = −Cˆ D* VVz
(8.51)
(8.52)
(8.53)
The equation that couples Equations 8.51 through 8.53 is
V = Vx2 + Vy2 + Vz2
(8.54)
We can linearize these equations by making a few assumptions. First, let us assume that
there is no crosswind, so Vz = 0, and if we further constrain the ratio of the vertical velocity
to the horizontal velocity to |Vy/Vx| = tan ϕ < 0.1, then V and Vx are within 0.5% of each
other, and we have constrained the launch and fall angles to be less than 5.7°, the angles
introduced in the preceding section for the flat fire approximation.
So with the assumptions that V = Vx and Vz = 0, we can develop Equations 8.51 through
8.53 into
V x = −Cˆ D* Vx2
(8.55)
V y = −Cˆ D* VxVy − g
(8.56)
V z = 0
(8.57)
These differential equations use time as the independent variable. It is often convenient
to use distance as the independent variable. By making a common transformation of variables to allow distance along the trajectory to be the independent variable instead of time,
we can improve our ability to work with these expressions. Performing the transformation
results in equations of the form
VxVx′ = −Cˆ D* Vx2
(8.58)
VxVy′ = −Cˆ D* VxVy − g
(8.59)
where the prime denotes differentiation with respect to distance.
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
234
By dividing both equations by Vx we obtain Equations 8.60 and 8.61 that use the downrange distance, x, as the independent variable. For these equations, an analytic solution
does exist:
Vx′ = −Cˆ D* Vx
(8.60)
g
Vy′ = −Cˆ D* Vy −
Vx
(8.61)
Equation 8.60 can be integrated by separation of variables as
x
Vx = Vx0 exp − Cˆ D* dx1
0
∫
(8.62)
In this equation and future equations, we use a variable xi or ti as a dummy variable
of integration. Equation 8.61 can also be solved by quadrature methods since it is of
the form
dVy ˆ *
g
+ CDVy = −
Vx
dx
(8.63)
Equation 8.61 can be solved for Vy for initial conditions at x = 0, t = 0, and Vy = Vy0 as
x
ˆ
*
Vy = exp − C D dx1 Vy 0 −
0
∫
x
∫
0
x2
g
ˆ
*
V exp C D dx1 dx2
x
0
∫
(8.64)
If we take the ratio of the x and y velocity components, we can obtain the relation for the
angle the velocity vector makes with the horizontal. This can be shown to be
1
tan φ = tan φ0 −
Vx
x
∫
0
g
V
x
x2
Cˆ *D dx1 dx2
exp
0
∫
(8.65)
where ϕ 0 is the initial launch angle.
To complete our study of the flat fire trajectory, we need to find the elements of it, i.e., the
x and y values along it, from launch to termination. To do this, we must integrate over time
the velocities we have found in Equations 8.62 and 8.64, which we had earlier transformed
into distance variables. We know by definition
t
∫
t
∫
y = Vydt and x = Vxdt
0
© 2014 by Taylor & Francis Group, LLC
0
(8.66)
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235
Substituting Equations 8.62 and 8.64 into each of the equations of 8.66 in turn and performing the integrations, we can show that with the initial conditions of x = 0, t = 0, and y = y0
x
x
x2
g
*
ˆ
ˆ
*
y = t exp C Ddx1 Vy0 − exp C D dx1 dx2 + y0
Vx
0
0
0
∫
∫
∫
(8.67)
Now we can find t from Vx = dx/dt. Separating the variables and substituting Equation 8.62
for Vx, we get
x2
exp Cˆ D* dx1 dx2
0
0
x
t=
1
Vx0
∫
∫
(8.68)
Through a somewhat tedious set of algebraic substitutions and manipulations that are
contained in Ref. [1], we can arrive at our desired equation in x and y; the launch angle, ϕ 0;
the dummy range variables x1, x2, and x3; the initial launch velocity, Vx0; the initial ordinate, y0;
and the drag coefficient, ĈD* :
y = y0 + x tan φ0 −
gx 2 2
2Vx20 x 2
x x3
∫∫
0 0
x2
exp 2 Cˆ *D dx1 dx2dx3
0
∫
(8.69)
The disadvantage of these equations is that the variation of drag coefficient has to be simple to
evaluate the integrals. Since the drag coefficient does not vary in a simple manner with Mach
number, this makes the analytic solutions inaccurate and difficult to accomplish. Figure 8.4
0.48
0.46
0.44
0.42
0.40
0.38
0.36
0.34
0.32
0.30
0.28
0.26
0.24
0.22
0.20
0.18
0.16
0.14
0.12
0.10
0.08
0.06
0.04
0.02
0.00
Plot area
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
2.6
2.8
3.0
3.2
3.4
3.6
3.8
4.0
4.2
4.4
4.6
4.8
5.0
5.2
5.4
5.6
5.8
6.0
6.2
6.4
6.6
6.8
7.0
7.2
7.4
7.6
7.8
Zero yaw drag coefficient
Zero yaw drag
Mach
FIGURE 8.4
Drag coefficient versus Mach number for a typical projectile.
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Ballistics: Theory and Design of Guns and Ammunition
236
depicts a typical drag curve that varies with Mach number. One can see from this figure that
there is no simple analytic solution to this variation. With computer power nowadays we usually solve or approximate the exact solutions numerically, doing the quadratures by breaking
the area under the curve into quadrilaterals and summing the areas.
To integrate these equations analytically, we will examine three forms of the drag
coefficient:
1. Constant CD that is useful for the subsonic flight regime, M < 1
2. CD inversely proportional to the Mach number that is characteristic of the highsupersonic flight regime, M ≫ 1
3. CD inversely proportional to the square root of the Mach number that is useful in
the low-supersonic flight regime, M ≥ 1
First, we will examine the case of a constant drag coefficient. If we examine Figure 8.4,
we can see that this would be a useful approximation for our projectile behavior if the
launch velocity was, say, between Mach 0.8 and 0. We assume that the drag force varies
with the square of the velocity (the drag coefficient was the drag force divided by the
2
1
dynamic pressure, 2 ρV ) and we set the drag coefficient equal to a constant, K1. We shall
use terminology consistent with Ref. [1] so that
ĈD* =
ρS
ρS
CD =
K 1 = k1
2m
2m
(8.70)
which we then substitute in Equation 8.62:
x
Vx = Vx0 exp −k1 dx1 = Vx0 exp(− k1x)
0
∫
(8.71)
We can find t by substituting Equation 8.70 into Equation 8.68 to give
x2
1
exp k1dx1 dx2 =
Vx0
0
0
x
1
t=
Vx0
∫
∫
x
∫ exp(k x )dx
1 2
2
(8.72)
0
or
t=
1
1
(exp[k1x] − exp[0]) =
(exp[k1x] − 1)
Vx0 k1
Vx0 k1
(8.73)
Noting that from Equation 8.62
x
V
exp − Cˆ *D dx1 = x
Vx0
0
∫
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x
V
x
and exp Cˆ D* dx1 = 0
Vx
0
∫
(8.74)
Trajectories
237
and also noting that (Vy0 Vx0 ) = tan φ0 , we can show through manipulation [1] that
gt
Vy = Vx tan φ0 −
V
x0
Vx0 k1t
1+
2
(8.75)
To find the angle of fall, ϕ, as a function of range, x, and the instantaneous velocity at x, Vx,
we solve Equation 8.71 for k1:
k1 =
1 Vx0
ln
x Vx
(8.76)
We now substitute Equation 8.76 into Equation 8.73, for t. Taking the result and recalling that
tan ϕ = Vy/Vx for any x, we use this new equation for t and transform Equation 8.65 into
tan φ = tan φ0 −
gt
Vx0
Vx0 t 1 Vx0
ln
1 +
2 x Vx
(8.77)
Finally, to find the altitude, y, at any point along the trajectory as a function of the range
and the velocity at that range, we transform Equation 8.65 with the constant drag coefficient, k1, use the new equation for t that we derived earlier, and after manipulation
arrive at
g x
1
y = y0 + x tan φ0 −
2 Vx0 Vx0
ln
Vx
2
2
1 Vx
Vx
Vx
0 − 1 + 0 − 1 − ln 0
Vx
Vx
2 Vx
(8.78)
A constant drag coefficient is useful when analyzing low-subsonic projectiles since most
of them have nearly constant drag coefficients. Also, projectiles at hypersonic speeds
(usually described as a Mach number greater than five) can be analyzed with this
assumption (look again at Figure 8.4). Essentially, we are linearizing the problem when
we do this.
Our next effort will be to examine a nonconstant drag coefficient, one varying as the
inverse of the Mach number. In this case, we assume that the drag force varies linearly
with the velocity (because the drag coefficient is the drag force divided by the dynamic
pressure, 12 ρV 2 , and when we divide by the Mach number, we essentially divide by the
velocity times a constant). Now we set the drag coefficient equal to K 2/M, then
K2
M
(8.79)
ρS
ρS K2
CD =
2m
2m M
(8.80)
CD =
and
ĈD* =
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Ballistics: Theory and Design of Guns and Ammunition
238
Recall that the Mach number is V/a where a is the speed of sound in air. Then for our flat
fire approximation, we can define a constant k2 such that
ρS
K2a
2m
(8.81)
ρ S K 2 a k2
=
2m Vx
Vx
(8.82)
k2 =
Then,
ĈD* =
From Equations 8.55 and 8.56, we see that
Vɺ x = −k 2Vx
(8.83)
Vɺ y = −k 2Vy − g
(8.84)
Vx′ = −Cˆ D* Vx
(8.85)
and
Also from Equation 8.60, we see that
Using these three equations and proceeding in the same fashion as we did with the constant CD, we can derive equations for the x- and y-velocities; the time of flight to any range, x;
the angle of fall, ϕ, and the trajectory ordinate at any range. These equations are (details
in Ref. [1])
Vx = Vx0 exp(−k 2t) (in terms of t)
g
g
Vy = Vy0 + exp(−k 2t) −
k2
k2
(in terms of t)
x
Vx0
ln
Vx0 Vx
t=
Vx
1−
Vx0
gx
tan φ = tan φ0 + 2
Vx0
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Vx0
1− V
x
1 − Vx
Vx0
(8.86)
(8.87)
(8.88)
(8.89)
Trajectories
239
gt 2
y = y0 + x tan φ0 −
2 ln Vx0
Vx
Vx
1− V
x0
1−
Vx
ln 0
Vx
(8.90)
These relations, for CD proportional to 1/M, are useful in the analysis of high-supersonic
projectiles such as kinetic energy armor penetrators where 2.5 < M < ∼5.
For the case where the drag coefficient varies as the M , we assume that the drag varies
with velocity to the 3/2 power and we set the drag coefficient equal to K 3/ M , then
ρS K3
Cˆ D* =
2m M
(8.91)
Since
M=
Vx
a
(8.92)
we can define a new constant as
ρS
K3 a
2m
k3 =
(8.93)
We can then write
ρS
a
k
Cˆ D* =
K3
= 3
2m
Vx
Vx
(8.94)
Proceeding as we did in the earlier two cases, we can derive Vx, Vy, t, ϕ, and the ordinate, y.
The details of the derivations are again available in Ref. [1]:
Vx =
Vy = −
(k
3
Vx0 t + 2
)
)
(
t=
x
Vx0
(8.95)
(in terms of t)
4Vy0
k 32Vx0 t 3
+ 2k 3 Vx0 t 2 + 4t +
Vx0 t + 2 3
k 3 Vx0 t + 2
g
3
(k
4Vx0
Vx0
Vx
)
2
(in terms of t)
(8.96)
(8.97)
gt
Vx0
1 Vx
Vx0
+ 1
0 +
Vx
3 Vx
(8.98)
y = y0 + x tan φ0 −
1 2 1
Vx
gt 1 + 2
2
Vx0
3
(8.99)
tan φ = tan φ0 −
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240
Ballistics: Theory and Design of Guns and Ammunition
These last Equations 8.95 through 8.99 are useful for flight in the low- to moderatesupersonic regime, 1 < M < 2.5.
In summary, we have derived the equations of motion assuming a flat fire trajectory. We
use them when the angle of departure and angle of fall are both below 5.7°. We have solved
them with three drag assumptions:
1. A constant drag coefficient that is useful in the subsonic and hypersonic regimes
and can be used over short distances in all Mach regimes.
2. A drag coefficient inversely proportional to the Mach number that is useful in the
high-supersonic regime.
3. A drag coefficient inversely proportional to the square root of the Mach number
that is useful in the low-supersonic flight regime.
Problem 7
The French infantry rifle model 1886 called the Lebel was their standard weapon from
1886 into First World War and even saw limited use in the Second World War. You
can see this 51 in. long monster in any movie involving the French Foreign Legion. It
used an 8 mm cartridge called the balle D with a bullet mass of 198 grains and a diameter of 0.319 in. This cartridge–rifle combination has a muzzle velocity of 2296 ft/s.
Assuming flat fire with K 3 = 0.5 and using standard sea level met data (ρ = 0.0751 lbm/ft3,
a = 1120 ft/s)
1. Create a table containing range (yards), impact velocity (ft/s), time of flight (s), initial QE angle (min), and angle at impact (min) in 200 yard increments out to 1000
yards.
2. If an infantryman is looking at a target at 2000 yards, what angle will the sight
have relative to the tube assuming they used standard met in the design?
Answer: About 10.3°.
3. Comment on the validity of this method with respect to (2).
Problem 8
British 0.303 in. ball ammunition is to be fired in an Mk.1 Maxim machine gun. The bullets’ mass is 175 grains. When used in this weapon, it has a muzzle velocity of 1820 ft/s.
Assuming flat fire with K3 = 0.5 and using standard sea level met data (ρ = 0.0751 lbm/ft3,
a = 1120 ft/s)
1. Create a table containing range (yards), impact velocity (ft/s), time of flight (s), initial QE angle (min), and angle at impact (min) in 200 yard increments out to 1000
yards.
2. The weapon was used by British units assigned to bolster the Italians in the Alps
during the First World War (Italy came in on the Allied side because they wanted
the Tyrol region from Austria more than they wanted the Nice region from France).
At an altitude of 3000 ft, how much higher or lower will a bullet fired from this
weapon impact a level target if the sights are set using the sea level conditions
given earlier and the target is at 600 yards? At this altitude assume the density and
temperature of the atmosphere are ρ = 0.0551 lbm/ft3 and T = 20°F.
Answer: y = 3.078 [ft] (too high).
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241
Problem 9
The main armament in the Italian M13-40 during the Second World War was a 47 mm/
32-caliber weapon designed and built by the Ansaldo Arms company. The most effective
antitank projectile it carried was an APBC (Armor-Piercing, Ballistic Capped) round, which
had a muzzle velocity of 2060 ft/s. With this particular projectile–weapon combination, the
assumption of constant drag coefficient seems to yield reasonable results. The k1 value for this
case is 0.00025 [1/m]. Using the flat fire, point mass trajectory create a table of range (yards),
velocity (ft/s), initial QE (min), and impact QE (min) out to 1000 yards in 200 yard increments.
Problem 10
A U.S. 37 mm projectile is fired with a muzzle velocity of 2600 [ft/s]. The projectile
weighs 1.61 lbm. Assuming K 2 = 0.841 [unitless] and using standard sea level met data
(ρ = 0.0751 lbm/ft3, a = 1120 ft/s, R = 1716 [ft · lbf/slug · R])
1. Determine the drag coefficient CD and drag force on the projectile if the projectile
is fired in still air.
Answer: FD = 33.04 [lbf].
2. Create a table containing range (yards), impact velocity (ft/s), time of flight (s), initial
QE angle (min), and angle at impact (min) in 100 yard increments out to 800 yards.
3. If this weapon is used at an increased altitude and assuming the density and temperature of the atmosphere are ρ = 0.060 lbm/ft3 and T = 30°F, how much higher or
lower will the weapon have to be aimed to hit a target at 800 yards?
Answer: The weapon must be aimed 0.28 mil or 0.98 min lower.
Problem 11
You are asked to create a rough safety fan for at maximum range test at Yuma Proving
Ground. The test consists of an experimental 155 mm howitzer projectile at a severe overpressure charge. The projectile weighs 106 lbm. The muzzle velocity is 980 m/s.
a. Using a vacuum trajectory calculate and plot the trajectory envelope for the test.
b. Determine the longest time of flight of the projectile.
If we assume an average constant drag coefficient (on the way up) of CD = 0.5/M and the
projectile were fired vertically, what would the maximum ordinate be? Assume standard
sea level met data (ρ = 0.0751 lbm/ft3, a = 1120 ft/s, R = 1716 [ft · lbf/slug · R])
Hint: Look at the flat fire assumptions and re-derive the equation of motion assuming
there never is an x-velocity.
Problem 12
The drag of a sphere below Mach number of 0.5 is well approximated by CD = AM2 + BM +
C, where A = 0.0262, B = 0.0456 and C = 0.4666, and M is the Mach number. If we would
like to analyze the motion of a cannon ball fired from a Demi-Culverin circa 1646, which
fired a spherical shot weighing 9 lbm and of 4.5 in. diameter. Please develop the equation of motion for Vx (only) as a function of downrange distance based on the flat fire
assumptions. Then using standard sea level met data (ρ = 0.0751 lbm/ft3, a = 1120 ft/s,
R = 1716[ft ⋅ lbf/slug ⋅ R ]), determine Vx at a range of 1800 yards (the reported maximum
range of the weapon). Assume a muzzle velocity of 550 ft/s.
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Ballistics: Theory and Design of Guns and Ammunition
Problem 13
A French 240 mm projectile is fired from a model 1873 cannon with a muzzle velocity of
440 [m/s]. The projectile mass is 144 kg. Assuming K1 = 0.55 [unitless] and using standard
sea level met data (ρ = 0.0751 lbm/ft3, a = 1120 ft/s, R = 1716 [ft · lbf/slug · R])
a. Create a table containing range (m), impact velocity (m/s), time of flight (s), initial
quadrant elevation angle (degrees), and angle at impact (degrees) in 600 m increments out to 7800 m.
b. Experiments were conducted in France in 1884 to determine the effect of rotating
and locations on the range of this projectile. These experiments were reported on
by Breger in Notes on the Construction of Ordnance No. 27, 10 June 1884. The farther
rearward the rotating band was placed, the greater the projectile yaw at muzzle.
The same QE that provided a range of 7800 m actually only made 7688 m with a
(assume constant) yaw angle of 4°. Assume that the 7800 m range was achieved
with zero initial yaw. With this information calculate the yaw drag coefficient, CDδ 2.
Problem 14
It is desired to develop a close protection system using a 0.50 caliber machine gun. The
muzzle velocity of the weapon is 2950 ft/s. The projectile is an M8 bullet with a diameter of
0.50 in. and a mass of 0.09257 lbm. Since the projectile interception mission is to occur relatively close to the firing platform, we can assume the projectile behaves according to the flat
fire assumption with a constant drag coefficient of K1 = 0.45 [unitless]. Assuming the flat fire
assumption is valid for the trajectory and the intercept point is to be 20 ft above the ground,
develop a firing table for the intercept out to 300 yards in 50 yard increments. Assume standard sea level met data (ρ = 0.0751 lbm/ft3, a = 1,120 ft/s, R = 1716 [ft · lbf/slug · R])
Create a table containing range (yards), impact velocity (ft/s), time of flight (s), and initial
quadrant elevation angle (degrees) in 50 yard increments out to 300 yards. Comment on
the accuracy of your answer.
Problem 15
If we were to assume that the flat fire conditions were to hold in a coordinate system that were
elevated to an angle θ, re-derive the differential equations in time and space coordinates—
DO NOT solve them, just put them in the form of Equations 8.60 and 8.61. Also write the
transformation equations between the normal and slant velocity (Vn and Vs) and Vx and Vy.
Problem 16
One of the problems with hit-to-kill close protection systems is the ability to accurately
point the weapon and fire in a timely fashion at a very small target. Assume that we must
impact a sphere 4 in. in diameter at the ranges developed in problem 14. For each range,
assuming perfect timing as well as perfect projectile tracking, determine the allowable
tolerance in QE to impact the target.
Problem 17
Let us now assume that the pointing against the incoming round in problem 14 is absolutely perfect. Determine the tolerance in lock time (officially the time from pulling the
trigger to weapon firing—but we will assume it is to muzzle exit) to hit the target at the
conditions of problem 14. Assume that the incoming projectile is moving at 300 ft/s. How
does this change if the velocity estimate is ±20 ft/s?
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243
Problem 18
The main armament of the last pre-war U.S. Heavy Cruisers (known as the “tin-clads”)
was an 8 in./55 cal weapon. The effective range of this weapon was 30,000 yards at
an elevation of 40°43′. During the Second World War, there were many night actions
in the Pacific where these weapons were used at an extremely short range (less than
10,000 yards). You are asked to create a firing table for this weapon at the short ranges.
The muzzle velocity of the weapon/projectile/propellant combination is 2500 ft/s. The
projectile is an APC (Armor Piercing, Capped) with a diameter of 8 in. and a mass of
335 lbm. Since the range is short, we can assume the projectile behaves according to the
flat fire assumption with a drag coefficient inversely proportional to the Mach number
of K 2 = 0.62 [unitless] (note that this is not really a great fit for this projectile). Assuming
the flat fire assumption is valid for the trajectory, develop a firing table for the system to
10,000 yards in 1,000 yard increments. Assume standard sea level met data (ρ = 0.0751
lbm/ft3, a = 1,120 ft/s, R = 1716 [ft · lbf/slug · R]).
Create a table containing range (yards), impact velocity (ft/s), time of flight (s), and initial
quadrant elevation angle (degrees) in 1,000 yard increments out to 10,000 yards.
Problem 19
The analog fire control computers installed on board ships during the Second World War
were amazing devices. The inputs required were course and speed of the firing ship, estimated range to the target and course and speed of the target. Inaccuracies in the target course
and speed estimates were compensated for by generating a “ladder”; this was a shell pattern
that was a linear array using as many guns as were available in one “salvo.” The ideal result
was that the target would end up directly in the middle of the “salvo” and be “straddled.”
Because of the relatively flat projection of the shells, being straddled usually guaranteed that
the target was hit by at least one projectile. The shorter the range to a target, the larger the
“danger space” offered and the better the chance of a hit. In this problem you assume the
role of the fire control computer. The weapons available are nine 8”/55 caliber guns with
the ballistic performance from problem 18. Your ship is moving due north at 30 knots (one
knot is one nautical mile per hour or 2000 yards per 3600 s). At the instant of fire, the enemy
ship is dead ahead of your ship traveling at 35 knots on course 090 (see the figure) at a range of
8000 yards. Ignore the effect of the launch platform motion on the drag (only) of the projectile.
± 30º
enemy ship
8,000 yds
θ
your ship
a. Determine the firing solution assuming both ships continue straight ahead (QE and
relative angle to the bow of your ship) for one shell to impact the enemy. (Because
of the flat trajectory, it is good to aim so the shell falls a little behind the enemy
ship)—Hint: Remember the projectiles are leaving from a ship that is moving!
b. Perform the same calculation if the enemy turns 30° toward the firing ship and
30° away from the firing ship.
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Ballistics: Theory and Design of Guns and Ammunition
244
c. Using a method of your choosing, examine the sensitivity of the fire control problem to (i) an incorrect firing ship speed, (ii) an incorrect target ship course, and
(iii) an incorrect target ship speed. Quantify their relative importance.
Problem 20
The U.S. 7.62 mm Ball M80 (projectile diameter = 0.308 in., mass m = 147 grains) is fired
in a test range. Based on data given next, estimate the coefficient CD. Assume the projectile is fired with a muzzle velocity of 2810 ft/s, under standard sea level met conditions
(ρ = 0.0751 lbm/ft3, a = 1120 ft/s). Justify your answer by explaining why you chose the
appropriate drag model. Validate your answer with an appropriate calculation.
Range (yards)
V0 (ft/s)
Vx (ft/s)
2810
2810
2810
1960
1765
1580
400
500
600
Problem 21
Many times all of the data we need for a projectile are not provided to us and we have to
extract the information from different sources. You are given the following information
about a British 2 pounder projectile [3]:
Projectile diameter: 40 mm
Projectile weight: 2.375 lbm
Muzzle velocity: 2600 ft/s
Armor penetration as a function of distance
Distance (yards)
thickness Perforated (mm)
100
500
1000
1500
55
47
37
27
In terminal ballistics, we will find that, based on some work by Zener and Holloman in
1942, the penetration of this type of projectile is proportional to the velocity as follows:
V ∼t
d
m
where
t is the target thickness
d is the projectile diameter
m is the projectile mass
With only this information at your disposal
a. Determine the best drag model for this projectile
b. Generate the proper coefficient from the data
c. Create a table of range (yards), velocity (ft/s), time of flight (s), launch angle (minutes),
impact angle (minutes) if the projectile is fired with no wind at each position
Please justify your answer.
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245
Problem 22
A Hornady 0.308” diameter 208 grain Amax bullet is to be analyzed. Based on data collected by Litz [4], it has the following drag characteristics:
V [ft/s]
1500
2000
2500
3000
CD
0.354
0.306
0.274
0.250
If it is fired with a muzzle velocity of 2780 ft/s, assuming standard sea level met data
(ρ = 0.0751 lbm/ft3, a = 1120 ft/s)
a. Determine a proper drag model to use over this range. Justify your answer by
comparing numbers from all three drag models that were introduced.
b. Assuming K3 = 0.409, create a table of range (yards), velocity (ft/s), time of flight (s),
launch angle (min), impact angle (min), and drift (yards) if the projectile is fired
with no wind every 100 yards out to 600 yards.
c. Determine the deflection in inches assuming the projectile experiences a
headwind of 20 ft/s for the entire flight. Tabulate every 100 yards out to 600
yards.
d. Determine the deflection in inches assuming the projectile experiences a crosswind (left or right—your choice) of 20 ft/s for the entire flight. Tabulate every 100
yards out to 600 yards.
Problem 23
Many precision shooters develop specific propellant loads for competition shooting
for a given propellant type, projectile, barrel, cartridge and primer combination. The
strategy for optimizing a propellant loading is to incrementally increase the propelling charge with the weapon aimed at the same location and determine the charge at
which the projectiles “group” the most. This method, called a Creighton Audette ladder test, finds the so-called “sweet spot” of the gun. The projectiles “group” due to a
combination of barrel motion owing to recoil and vibration. To illustrate this point, let
us assume that we have a perfectly rigid (i.e., no vibration) 22 in. barrel that fires the
projectile of problem 22. Assume a range of 300 yards. Assume that the projectile velocity during barrel transit can be approximated as ½ of the muzzle velocity. Also assume
that the recoil is fairly constant and causes a constant upward rotation of the muzzle of
3.3 rad/s. For the following 20 velocities, determine the muzzle velocity (which would
correspond to a propellant load) that is in the “sweet spot.” Please plot impact height
versus muzzle velocity.
Hint: The muzzle rise will affect the initial launch angle but the flatter trajectory will compensate to some degree. Although not done in practice, for this calculation, start at the high
muzzle velocity and track the bullet impact heights as you lower the velocity.
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246
V0 [ft/s]
2780
2775
2770
2765
2760
2755
2750
2745
2740
2735
2730
2725
2720
2715
2710
2705
2700
2695
2690
2685
Problem 24
Normally on a fin-stabilized projectile the spin damping due to the body is much smaller
than that due to the fins. A 155 mm projectile weighs 101 lbm and is designed so that it leaves
the muzzle of the weapon at 600 m/s and spinning at 220 Hz. After 0.5 s fins are deployed.
At this instant in time the spin rate is 197 Hz and the velocity is 500 m/s. After another 0.5 s
the projectile has achieved its steady state spin rate of 12 Hz and is at a velocity of 400 m/s.
The changes in polar moment associated with the fin deployment are provided next (be sure
to think about which one is before and which is after). Determine the following:
a. The drag coefficient of the both flight configurations
b. The spin damping coefficient for the body (only) and the fins (only)
c. The roll coefficient (Clδ) of the fins assuming a 1° cant
Assume
lbm
m
ρ = 0.076 3 a ≈ 330 I PA = 10.1[lbm ⋅ ft 2 ] I PB = 10.8 [lbm ⋅ ft 2 ]
ft
s
It is important in this problem, since it is pretty open ended, that you list all of your
assumptions and present your answer in a manner appropriate to those assumptions.
Problem 25
In a test range, a 0.50 caliber M33 ball projectile is fired at an elevation of 10° with a muzzle velocity of 3013 ft/s. The initial pitch and yaw angles are 1.030° and 1.263°, respectively.
The initial pitch and yaw rates are 2 rad/s nose down and 1 rad/s nose left, respectively. If
the projectile has the coefficients below at this particular instant, write the acceleration vector
and the angular momentum vector.
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Please ignore the Coriolis acceleration and assume the weapon has a right-hand twist.
Projectile information:
CD = 0.2938
(C
Mq
+ CMαɺ = −5.5
)
I P = 7.85 [g-cm 2 ]
CMα = 2.88
(C
Nq
+ CNαɺ = 0.004
)
I T = 74.5 [g-cm 2 ]
CLα = 2.69
CMpα = 0.05
m = 42.02 [g]
CNpα = −0.01
lbm
ρ = 0.0751 3
ft
rad
p = 15, 404
s
Please supply all answers in an inertial coordinate system labeled 1, 2, 3 with 1 being along
the downrange direction and 3 being to the right side. Treat all missing coefficients as
equal to zero. It is very important that you DRAW the situation.
Problem 26
The projectile given in problem 25 is fired off of a fast attack boat chasing down some
pirates. All of the conditions in problem 25 are identical except for the mounting of the
weapon and its movement. The weapon is mounted on the starboard (right) side of the boat.
At the instant of firing the boat is moving at 40 knots in a straight line. The boat is rolling
at 6 Hz. in a counterclockwise direction as viewed from behind. The boat is pitching into
a swell so that the bow is dropping at 3 Hz. The gunner is slewing the weapon toward the
bow at 1.5 rad/s while simultaneously lowering the muzzle at 2 rad/s. Assuming, at the
instant of muzzle exit, the weapon is pointed directly to starboard and has the same elevation as in the test firing (i.e., 10°), find the values for the acceleration and angular momentum
as was done in problem 25. Comment on the results. For a proper comparison, use the same
coordinate system as in problem 25 but now with the 1 direction pointing to starboard and
the 3 direction pointing to the stern of the boat. Once again the drawings are important.
Problem 27
In a test range, a modified 105 mm M1 projectile is fired at an elevation of 7° with a muzzle
velocity of 1022 ft/s. The initial pitch and yaw angles are 1.0° and 1.5°, respectively. The
initial pitch and yaw rates are 3 rad/s nose down and 2 rad/s nose left, respectively. If the
projectile has the coefficients below at this particular instant, write the acceleration vector
and the angular momentum vector.
Please ignore the Coriolis acceleration and assume the weapon has a right-hand twist.
Projectile information:
CDδ 2 = 4.20
(C
(C
CMα = 4.30
C Mpα = −0.892
CLα = 1.65
Clp = −0.028
CNpα = −0.55
lbm
ρ = 0.0751 3
ft
CD = 0.131
© 2014 by Taylor & Francis Group, LLC
)
)≈0
Mq
+ CMαɺ = −8.7
Nq
+ C Nα
I P = 0.547 [lbm-ft 2 ]
I T = 5.377 [lbm-ft 2 ]
m = 32.1[lbm]
rad
p = 932
s
Ballistics: Theory and Design of Guns and Ammunition
248
Please supply all answers in an inertial coordinate system labeled 1, 2, 3 with 1 being along
the downrange direction and 3 being to the right side. Treat all missing coefficients as
equal to zero. It is very important that you DRAW the situation. This will have a great deal
of influence in obtaining the correct answer
Problem 28
A modified 105 mm M1 projectile is fired downward at an angle of −45° from the horizontal from an aircraft moving horizontally at 200 knots with a muzzle velocity of 1022 ft/s.
The initial pitch and yaw angles are 1.0° and 1.5°, respectively. The initial pitch and yaw
rates are 3.5 rad/s nose down and 2.5 rad/s nose left, respectively.
a. If the projectile is fired off the right side of the aircraft and has the coefficients
below at this particular instant, write the acceleration vector and the angular
momentum vector.
b. Write the acceleration vector and the angular momentum vector assuming everything is the same except now the projectile is fired off the left side of the aircraft.
c. Comment on the differences between parts (a) and (b).
Please ignore the Coriolis acceleration and assume the weapon has a right-hand twist.
Projectile information:
CDδ 2 = 4.20
(C
(C
CMα = 4.30
C Mpα = −0.892
CLα = 1.65
Clp = −0.028
CNpα = −0.55
lbm
ρ = 0.0751 3
ft
CD = 0.131
)
)≈0
Mq
+ CMαɺ = −8.7
Nq
+ C Nα
I P = 0.547 [lbm-ft 2 ]
I T = 5.377 [lbm-ft 2 ]
m = 32.1[lbm]
rad
p = 932
s
Please supply all answers in an inertial coordinate system labeled 1, 2, 3 with 1 being along
the downrange direction and 3 being to the right side of the gun looking from the breech
(be careful as this will change between parts (a) and (b)). Treat all missing coefficients as
equal to zero.
8.3 Wind Effects on a Simple Air Trajectory
We continue with our study of a point mass projectile model by adding a further complication to its flat fire trajectory—a crosswind or a range wind, as dynamic atmospheric
phenomena. In the basic equations, we have neglected any change in air density with a
change in altitude since the effect is small. We also have assumed the equations could be
solved in closed form. We want to be able to solve them with winds that are both constant
and variable along the flight path.
© 2014 by Taylor & Francis Group, LLC
Trajectories
249
We begin with a modified version of Equation 8.47 using vector notation
dV
ρ SCD ɶ
=−
V (V − W) + g
dt
2m
(8.100)
dV
ɶ (V − W) + g
= −Cˆ D* V
dt
(8.101)
or
where
m is the projectile mass
W is the wind velocity vector
V is the projectile velocity vector
g is the vector acceleration due to gravity
t is the time
ρ is the air density
dV
a=
is the vector acceleration
dt
S is the projectile reference area
ρ SCD
ĈD* =
2m
CD is the dimensionless drag coefficient
In the aforementioned equations, we have replaced the velocity vector, V, by the vector
(V − W) because drag measurements are made relative to the air stream not relative to the
ground. We have also replaced the scalar velocity (the speed) with
ɶ =| V − W |
V
This is the scalar difference of the projectile and wind velocities. A diagram of the problem
is shown in Figure 8.5.
Height
y
Wy
Wx
Wz
FD
V0
j
on
cti
fle
i
Range
De
k
mg
0
z
FIGURE 8.5
Coordinate system for projectile launch including wind effects.
© 2014 by Taylor & Francis Group, LLC
V
x
Ballistics: Theory and Design of Guns and Ammunition
250
We can resolve V, W, and g into components along the coordinate axes as follows:
V = Vxi + Vy j + Vz k
(8.102)
W = Wxi + Wy j + wz k
(8.103)
g = −g j
(8.104)
Note that
ɶ 2 =|V − W|⋅|V − W|
V
and
|V| = Vx2 + Vy2 + Vz2
This leads us to
|V − W|⋅|V − W|= (Vx − Wx )2 + (Vy − Wy )2 + (Vz − Wz )2
Then,
ɶ = (Vx − Wx )2 + (Vy − Wy )2 + (Vz − Wz )2
V
(8.105)
If we insert Equations 8.102 through 8.104 into Equation 8.101, we get
dV
ɶ (Vx − Wx )]i + [ − Cˆ *D V
ɶ (Vy − Wy ) − g]j + [− Cˆ *D V
ɶ (Vz − Wz )]k
= [− Cˆ *D V
dt
We can separate this vector equation into its three scalar components:
dVx
ɶ (Vx − Wx )
= −Cˆ D*V
Vɺ x =
dt
(8.106)
dVy
ɶ (Vy − Wy ) − g
= −Cˆ D* V
Vɺ y =
dt
(8.107)
dVz
ɶ (Vz − Wz )
= −Cˆ D* V
Vɺ z =
dt
(8.108)
Equations 8.106 through 8.108 are the exact equations for a point mass trajectory of a projectile acted upon by gravity, wind, and aerodynamic drag. They are first-order, nonlinear,
coupled, ordinary differential equations that are coupled through Equation 8.105.
© 2014 by Taylor & Francis Group, LLC
Trajectories
251
The nonlinearity, as previously discussed, creates difficulties when we attempt to solve
these expressions analytically. We can only solve the exact equations using numerical
methods. This will necessitate making the simplifying assumption of flat fire, which will
allow us to solve them in closed form. We can alter Equation 8.105 by multiplying by the
fraction (Vx − Wx)/(Vx − Wx) = 1 and then simplifying to get
ɶ = (Vx − Wx ) 1 + ε y2 + ε z2
V
(8.109)
where
εy =
(Vy − Wy )
(Vx − Wx )
(8.110)
εz =
(Vz − Wz )
(Vx − Wx )
(8.111)
and
Using the binomial expansion of the form
1
1
1 + ζ = 1 + ζ − ζ 2 + ⋯
2
8
we can operate on the radical of Equation 8.109 arriving at
ɶ = (Vx − Wx ) 1 + 1 ( ε y2 + ε z2 ) − 1 ( ε y2 + ε z2 )2 + ⋯
V
2
8
(8.112)
Because a projectile’s velocity is usually much greater than winds of even hurricane force,
we can assume that
|Wx |, |Wy |, and |Wz |≪ Vx
(8.113)
and
ε y2
and ε z2 ≪ 1
(8.114)
If we look at the first inequality of Equation 8.114 and consider the assumptions of Equation
8.113, we find that
εy =
Vy Wy Vy
−
≈
= tan φ ≪ 1
Vx Wx Vx
This was the approximation developed around a similar binomial expansion in Section
8.2. If we recall that this relation restricted us to Vy/Vx < 0.1, which, by squaring, results in
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
252
the requirement that winds be at least two orders of magnitude smaller than the velocity, Vx,
and this is easily the case. The second inequality of Equation 8.114 is also satisfied if Wz and
Vz are comparable in size from
εz =
(Vz − Wz ) (Vz − Wz )
=
≪1
(Vx − Wx )
(Vx )
All this results in Ṽ and (Vx − Wx) being within about 1% of each other. So if Ṽ ≈ (Vx − Wx),
we can rewrite Equations 8.106 through 8.108 as
dVx
= −Cˆ D* (Vx − Wx )2
Vɺ x =
dt
(8.115)
dVy
= −Cˆ D* (Vx − Wx )(Vy − Wy ) − g
Vɺ y =
dt
(8.116)
dVz
= −Cˆ D* (Vx − Wx )(Vz − Wz )
Vɺ z =
dt
(8.117)
Updrafts and downdrafts are usually so small (and usually have the same effect as a crosswind for reasons we shall later describe) that we neglect them completely. Thus, we shall
set Wy equal to zero from now on. We will now look first at the effect where only a crosswind is present (i.e., where Wx = Wy = 0) and then examine the effect of a headwind or
tailwind. If we make this substitution into Equations 8.115 through 8.117, we obtain
dVx
= −Cˆ D* Vx2
Vɺ x =
dt
(8.118)
dVy
= −Cˆ D* VxVy − g
Vɺ y =
dt
(8.119)
dVz
= −Cˆ D* Vx (Vz − Wz )
Vɺ z =
dt
(8.120)
Equations 8.118 and 8.119 are identical to Equations 8.55 and 8.56 from our earlier work in
the zero wind case. If we now change from time to space variables, as we did in the zero
wind case, and recall that dt/dx = 1/(dx′/dt), then we arrive at the equations as follows:
© 2014 by Taylor & Francis Group, LLC
Vx′ = −Cˆ D* Vx
(8.121)
g
Vy′ = −Cˆ D* Vy −
Vx
(8.122)
Vz′ = −Cˆ D* (Vz − Wz )
(8.123)
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253
Once again, the prime symbol represents differentiation with respect to x, and Equations
8.121 and 8.122 are identical to those developed for the zero wind case. Now we have
already solved differential Equations 8.121 and 8.122 under their previous guise with the
result of
x
Vx = Vx0 exp − Cˆ D* dx1
0
∫
x
Vy = exp − Cˆ *D dx1 Vy0 −
0
∫
x
∫
0
g
V
x
(8.124)
x2
Cˆ *D dx1 dx2
exp
0
∫
(8.125)
Equation 8.123 is somewhat more difficult to solve. It is a first-order, linear differential
equation of the form y΄ + P(x)y = Q, whose solution, after the necessary integrations and
substitution of initial conditions that at x = 0, Vz = 0, is
x
x
x
Vz = exp − Cˆ *D dx1 Cˆ *D Wz exp Cˆ *D dx1 dx2
0
0
0
∫
∫
∫
(8.126)
From Equation 8.124, we see that the exponential is Vx /Vx0, and this can be inserted directly
into Equation 8.126. Also if we assume that Wz is a constant, it can be removed from the
integral to give
Vz =
We can integrate
∫
Cˆ D* exp
0
x
x
x
Vx
Wz Cˆ D* exp Cˆ D* dx1 dx2
Vx0
0
0
∫
∫
(8.127)
x
∫ Cˆ * dx dx by parts in Equation 8.127 to yield
D
1
2
0
x
x
x
Cˆ D* exp Cˆ D* dx1 dx2 = exp Cˆ D* dx1 Cˆ D* dx2 −
0
0
0
0
x
∫
∫
∫
∫
x
x
x
Cˆ D* dx2 Cˆ D* exp Cˆ D* dx1 dx1
0
0
∫∫
0
∫
(8.128)
The integral of the last term can be solved through a series of substitutions and evaluations
at the limits to yield
x
x
ˆ
ˆ
*
*
CD exp CDdx1 = exp Cˆ D* dx1 − 1
0
0
0
x
∫
© 2014 by Taylor & Francis Group, LLC
∫
∫
(8.129)
Ballistics: Theory and Design of Guns and Ammunition
254
If this is inserted into Equation 8.127, the result is
Vz =
x
Vx
Wz exp CD* dx1 − 1
Vx0
0
∫
(8.130)
We can further manipulate Equation 8.130 by inserting the value of the exponential from
Equation 8.124. In doing so, we get
Vz =
Vx
V
Vx
Wz 0 − 1 = Wz 1 − x
V
Vx0
V
x0
x
(8.131)
Since 0 < Vx < Vx0 , Vz always has to be less than the wind speed Wz. Thus, Wz is an upper
bound on Vz.
If we examine the deflection due to a constant crosswind, we can write
t
t
t
1
V
z = Vzdt = Wz 1 − x dt = Wz t 0 −
V
V
x0
x0
0
0
∫
∫
x
z = Wz t −
V
x0
Vxdt
0
t
∫
(8.132)
Equation 8.132 is known as the lag rule for predicting crosswind effects. It is an exact
solution for a constant crosswind. The quantity in the brackets is known as the lag time
because a projectile in a real atmosphere would take longer to reach the same range than
one fired in a vacuum.
Another interesting point is seen from examination of Equation 8.131. If Vx is always
equal to the initial x velocity, no matter how hard the wind blows, the projectile will not
be affected. Thus, a rocket motor that maintains the initial x velocity could make the projectile insensitive to wind, a concept called automet. Note also that if the thrust is greater
than the initial velocity, the projectile will actually move into the wind.
We consider next the effect of a variable crosswind. A simple way to model this effect
on a projectile is to superimpose solutions for constant crosswinds over incremental distances and piece the resultant trajectory together. This technique of superposition works
only with linear phenomena. However, since Equation 8.132 is linear in x and t, we can
apply this method. An alternative approach would be to apply Equation 8.132 in a piecewise fashion using the information from the previous calculation in the subsequent one.
To do this, we shall rewrite Equation 8.132 as a difference equation
∆x
∆zi = wzi ∆ti − i
V
xi0
(8.133)
where
∆zi is the distance traveled in the z-direction from time i − 1 to the time i
∆xi is the distance traveled in the x-direction from time i − 1 to the time i
∆ti is the time between time i − 1 to the time i
wzi is the constant crosswind acting on the projectile between time i − 1 and time i
Vxi0 is the x-velocity at time i − 1
© 2014 by Taylor & Francis Group, LLC
Trajectories
255
We can rewrite Equation 8.133 as
( x − xi − 1 )
zi − zi −1 = wzi (ti − ti −1 ) − i
Vxi 0
(8.134)
To use this method, one must first tabulate t, Vx, and x as described earlier and then perform the calculation for z at each interval. With some modifications, a forward difference
technique can also be used. These tedious calculations are best done with a computer
program for small intervals of time.
We will now examine the effects of a constant range wind, both head-on and a tailwind.
We do this by comparing the effects to a flat fire, no-wind flight and will determine the
effects on time of flight, impact, and velocity at impact.
We make the initial assumption that there is no crosswind, i.e., Wy = Wz = 0, and insert
this into Equations 8.115 through 8.117, the component differential equations for a pointmass, flat fire trajectory:
dVx
= −Cˆ D* (Vx − Wx )2
Vɺ x =
dt
(8.135)
dVy
= −Cˆ D* (Vx − Wx )Vy − g
Vɺ y =
dt
(8.136)
dVz
= −Cˆ D* (Vx − Wx )Vz
Vɺ z =
dt
(8.137)
Because there is no crosswind, Equation 8.137 reduces to
Vz = 0
By a change of time to space variables and various algebraic manipulations, we can change
Equation 8.135 to
W
Vx′ + Cˆ D* Vx = Cˆ D* Wx 2 − x
Vx
(8.138)
Similarly, we do the same to Equation 8.136 and arrive at a distance equation in a
y-variable only
W
Vy′ + Cˆ D* 1 − x
Vx
−g
Vy = V
x
(8.139)
Recall from our earlier discussion that the wind speed is about two orders of magnitude
smaller than the projectile velocity, so mathematically we can express this condition as
Wx
≤ 0.01
Vx
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
256
We can then rewrite Equations 8.138 and 8.139 allowing them to be equalities as follows:
Vx′ + Cˆ D* Vx = 2Cˆ D* Wx
(8.140)
g
Vy′ + Cˆ D* Vy = −
Vx
(8.141)
and
We can then solve these equations for Vx and Vy.
As we saw in the earlier solutions for the constant crosswind, with appropriate integrations, algebraic manipulation, and the insertion of the initial condition that at x = 0, Vx = Vx0,
we see that
x
x
x
x
ˆ
ˆ
ˆ
*
*
*
Vx = exp − C D dx1 2 C D Wx exp C D dx1 dx2 + Vx0 exp − Cˆ D* dx1
0
0
0
0
∫
∫
∫
∫
(8.142)
From Equation 8.129, recall that
x
x
ˆ
ˆ
*
*
CD exp CDdx1 = exp Cˆ D* dx1 − 1.
0
0
0
x
∫
∫
∫
Using this fact and by substituting it into Equation 8.142, factoring the result, and considering that Wx is constant, we arrive at
x
x
Vx = Vx0 exp − Cˆ D* dx1 + 2Wx 1 − exp − Cˆ D* dx1
0
0
∫
∫
(8.143)
The first term on the RHS of Equation 8.143 is simply the velocity decay caused by drag of
the projectile. The second term is the effect of the range wind on it. If we examine Equation
8.124, which was an analysis for a firing in the absence of range wind, the first term of
Equation 8.143 represents Vx, the x-velocity with no wind. The second term, when we substitute for the exponential, then represents the effect of the range wind on the flight. Thus,
we can see the range wind effects shown as the variable of interest with a tilde (∼) in the
following:
ɶx = Vx + 2Wx 1 − Vx
V
Vx0
(8.144)
This equation shows that at any time, t, a tailwind (i.e., one blowing in the positive x-direction)
has the effect of increasing the velocity (relative to the ground), while the opposite is true
of a headwind. This is important because if we had a table of velocities versus range for the
no-wind case, we could then tabulate the effect of range wind.
© 2014 by Taylor & Francis Group, LLC
Trajectories
257
If we now look at the y-velocity, we can operate on Equation 8.141 with the initial condition that at x = 0, Vy = Vy0. This provides us with the solution of the space variable equation
x
x
x
x
g
*
*
ˆ
ˆ
exp C D dx1 dx2 + Vy0 exp − Cˆ D* dx1
Vy = − exp − C D dx1
Vx
0
0
0
0
∫
∫
∫
∫
(8.145)
At this point, we can introduce the range wind by inserting Equation 8.143 for Vx
arriving at
x
x
Vy = − g exp − Cˆ *D dx1
0
0
∫
∫
+ Vy0 exp −
∫
x
exp Cˆ *D dx1
0
x
Vx0 exp − Cˆ *D dx1 + 2Wx 1 − exp −
0
∫
∫
∫
*
ˆ
C D dx1
0
dx2
x
Cˆ *D dx1
0
x
(8.146)
This rather complicated integral can be simplified somewhat; however, another approach
[1] to the problem that makes use of the no-wind method used previously in Equation 8.144
simplifies things even further. This is seen as
x
x
x
x
Vy0
g
*
*
ˆ*
ˆ
ˆ
ɶ
Vy = − exp − C D dx1
C
V
exp
D dx1 dx 2 +
x0 exp − C D dx1
ɶx
V
Vx0
0
0
0
0
∫
∫
∫
∫
or since
Vy0
Vx0
= tan φ0
x
x
x
1
*
ˆ
ɶ
Cˆ *D dx1 dx2 + Vx tan φ0
Vy = − g exp − C D dx1
exp
ɶx
V
0
0
0
∫
(8.147)
∫
∫
(8.148)
Further use of Ṽx and some algebraic manipulation gives
x
ɶy = Vx tan φ0 − gVx
V
1
∫ V Vɶ dx
0
x
2
(8.149)
x
Recalling Equation 8.144, we can rewrite the denominator of the integral as
2
ɶx = Vx2 + 2Wx Vx − Vx = Vx2 1 + 2Wx − 2Wx
VxV
Vx0
Vx
Vx0
© 2014 by Taylor & Francis Group, LLC
(8.150)
Ballistics: Theory and Design of Guns and Ammunition
258
If we again use the fact that the wind velocity is at least two orders of magnitude
smaller than the projectile velocity, the last two terms in the product on the RHS
vanish, leaving
ɶx ≈ Vx2
VxV
(8.151)
Then we can rewrite Equation 8.149 as
x
ɶy = Vx tan φ0 − gVx
V
1
∫V
0
2
x
dx2
(8.152)
This equation has exactly the same form as the flat fire equation for Vy. Hence, we can say
that for a flat fire trajectory, with a small range wind compared to the projectile velocity,
the vertical component of velocity is not appreciably affected.
We can now turn our attention to the time of flight of a projectile with a constant range
wind by first defining an average downrange velocity following the procedure by McCoy
[1] as
x
ɶx = 1 V
ɶxdx1
V
avg
x
∫
(8.153)
0
For Ṽx, we substitute Equation 8.144 giving
ɶx = 1
V
avg
x
x
Vx
dx1
x0
∫ V + 2W − 2W V
x
x
0
x
Performing the integration on the second term of the integrand and rewriting, we get
x
ɶx = 1 2Wx x + Vxdx1 − 2 Wx
V
avg
x
Vx0
0
∫
Vxdx1
0
x
∫
(8.154)
Rearranging Equation 8.154 and knowing that the velocity averaging also applies to the
no-wind case, i.e.,
x
Vxavg =
1
Vxdx1
x
∫
(8.155)
0
we get
ɶx = Vx + 2Wx 1 − Vxavg
V
avg
avg
Vx0
© 2014 by Taylor & Francis Group, LLC
(8.156)
Trajectories
259
The time of flight can be expressed as the range divided by the average velocity for either
the case of no range wind or with range wind included. Thus, we can write
R
Vxavg
(8.157)
ɶt = R
ɶx
V
avg
(8.158)
t=
or
By taking the reciprocal of Equation 8.158, performing judicious substitutions, gathering
terms, and finally taking the reciprocal of the result, we can write
t
ɶt =
t
1
1 + 2Wx −
R Vx0
(8.159)
This shows, as we would expect, that a tailwind (Wx positive) reduces the time of flight
while a headwind (Wx negative) increases it.
Let us summarize what we have done for crosswinds and range winds. We modified the
flat fire equations to account for crosswind and range wind. We use them when the angle of
departure and angle of fall are both below 5.7°. We solved the crosswind equations assuming constant and variable crosswinds and introduced the classic lag rule. With variable
crosswinds, we saw it is fairly accurate to piece the trajectory together using locally constant
values for the crosswind. We have solved the range wind equations assuming only constant
range wind. We could treat variable range wind in a manner similar to variable crosswinds,
but the difference in results is usually not worth the added effort. For range wind, we used
a solution technique that compared the velocities, positions, and time to the no-wind case.
Problem 29
A U.S. 37-mm AP projectile is fired with a muzzle velocity of 2600 [ft/s]. The projectile
weighs 1.61 lbm. Assuming flat fire with K 2 = 0.841 [unitless] and using standard sea level
met data (ρ = 0.0751 lbm/ft3, a = 1120 ft/s)
1. Create a table containing range (yards), impact velocity (ft/s), time of flight (TOF) (s),
initial QE angle (min), and angle at impact (min) in 200 yard increments out to
1000 yards assuming no-wind effects.
Answer: At 1000 yards, V = 1837 [ft/s].
2. Determine the deflection of the projectile with a 20-mi/h crosswind blowing from
left to right as viewed from behind the weapon.
Answer: At 1000 yards, z = 6.217 [ft].
3. Determine the impact velocity, change in TOF, and how high the projectile will hit
if fired at the same QE’s with a 20-mi/h tailwind and no crosswind.
Answer: The projectile will hit 1.402 in. higher than expected.
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Ballistics: Theory and Design of Guns and Ammunition
260
Problem 30
A British 12-in. projectile has a K3 of 0.8 and a weight of 850 lbm. If it is fired at an initial
QE of 130 mil with a muzzle velocity of 2800 ft/s
1. Create a table of range (yards), altitude (yards), velocity (ft/s), time of flight (s),
inclination angle (degrees), and drift (yards) if the projectile is fired with no wind.
2. Repeat part (1) if the projectile is fired with a headwind of 25 ft/s for the first
3000 yards of flight and a crosswind (left or right—your choice) of 35 ft/s for the
remainder of the flight. Tabulate every 1000 yards with the impact location as the
last entry in the table.
Problem 31
The U.S. 0.30 caliber Ball M2 (projectile diameter = 0.308in.) was the standard infantry rifle
cartridge in WWII. Based on data collected by McCoy [1], it is a 150 grain, flat-based Spitzer
shape with a K3 of 0.491 from Mach 1.2 to 3. If it is fired with a muzzle velocity of 2780 ft/s,
assuming standard sea level met data (ρ = 0.0751 lbm/ft3, a = 1120 ft/s):
a. Create a table of range (yards), velocity (ft/s), time of flight (s), launch angle
(minutes), impact angle (min), and drift (yards) if the projectile is fired with no
wind every 100 yards out to 600 yards.
b. Determine the deflection in inches assuming the projectile experiences a headwind of 20 ft/s for the entire flight. Tabulate every 100 yards out to 600 yards.
c. Determine the deflection in inches assuming the projectile experiences a crosswind (left or right—your choice) of 20 ft/s for the entire flight. Tabulate every
100 yards out to 600 yards.
Problem 32
Precision shooters are always in search of “tight groupings.” That is, the grouping of the
impact points of the projectiles at a given range. We shall examine the rifle and projectile
combination given in problem 31 and determine the effect each of several parameters has
on the precision. Vary each of the following parameters individually by 1% (up and down)
from problem 31 and determine the miss distance in inches for a 200 yard range. Please
carry answers to five significant figures as a baseline of comparison:
a.
b.
c.
d.
e.
f.
g.
h.
i.
Muzzle velocity
Projectile mass
Drag coefficient (CD)
Air density
Air temperature (in Rankine)
Launch angle of departure (we will assume this is due to weapon or shooter motion)
Headwind (0 ± 0.2 ft/s)
Crosswind (0 ± 0.2 ft/s)
Choose any two of these and vary them together—what is the result? Is the answer
simply a linear superposition of the two individual errors? Why or Why not? Is
this true for all of the parameters? Can you draw any conclusions from looking at
all of the results?
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261
8.4 Generalized Point Mass Trajectory
In keeping with our practice of introducing ever-increasing complexity into our theory, we
will now remove most of the restrictions of the earlier work. We will examine the effects
of an unrestricted launch angle and make the high-angle fire of mortars and howitzers
amenable to trajectory analysis. We still reserve for later study the effects on flight of a
three-dimensional body whose shape, physical properties, and motions add a significant
level of complexity to trajectory analysis.
The aerodynamic behavior of a projectile can be examined from three separate viewpoints: motion affected only by the acceleration of gravity and the initial velocity (vacuum
trajectory); motion affected by gravity, initial velocity, and aerodynamic drag (point mass
trajectory); motion affected by the projectile’s shape, physical properties, and dynamics
(which actually manifests itself as changing drag) as well as gravity and launch conditions. We will concentrate on the second viewpoint in this section.
In the equations that follow, we are assuming that the projectile is still a cannon ball with
all of its mass concentrated at one point. This allows us to continue to neglect the rigid body
kinematics that would be present in a distributed mass. However, we shall include wind
effects and earth-rotational effects, and therefore three-dimensional motion. As stated, flat
fire restrictions are removed so that the analysis is applicable to all launch angles.
We begin with the same set of equations of motion, except for the addition of a term for
the Coriolis force, mΛ:
F = ma
m
dV
=
dt
∑ F + m g + mΛ
(8.160)
(8.161a)
where
m is the projectile mass
V is the velocity vector
t is the time
dV
a=
is the vector acceleration
dt
ΣF is the vector sum of all aerodynamic forces
g is the vector acceleration due to gravity
Λ is the vector Coriolis acceleration due to rotation of the earth
We also recall from our earlier work with wind effects
dV
ɶ (V − W) + g
= −Cˆ D* V
dt
(8.162)
ˆ * = ρ SC /2m. In the aforementioned equations,
where W is the wind velocity vector and C
D
D
we have replaced the velocity vector V by the vector (V − W) because drag measurements
are made relative to the air stream, not relative to the ground. We have also again replaced
the scalar velocity (the speed) with Ṽ = |V − W|, which is the scalar difference of the projectile and wind velocities. A diagram of this is shown in Figure 8.6.
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Ballistics: Theory and Design of Guns and Ammunition
262
Wy
Height
y
Wx
Wz
FD
V0
j
mg
V
De
fle
c ti
on
0
i
k
Range
x
z
FIGURE 8.6
Generalized point mass trajectory.
Without repeating the entire procedure, it can be shown that we may separate Equation
8.162 into individual components to obtain the differential equations for a point mass:
dVx
ɶ (Vx − Wx )
= − Cˆ *D V
Vɺ x =
dt
(8.163)
dVy
ɶ (Vy − Wy ) − g
= − Cˆ *D V
Vɺ y =
dt
(8.164)
dVz
ɶ (Vz − Wz )
= − Cˆ *D V
Vɺ z =
dt
(8.165)
ɶ = (Vx − Wx )2 + (Vy − Wy )2 + (Vz − Wz )2
V
(8.166)
The scalar velocity, Ṽ, is again
In all of the aforementioned equations, the wind velocity is variable and is considered
positive when it blows in the positive direction of one of the coordinate axes. Equations
8.163 through 8.165 are nonlinear, coupled differential equations which are the exact solution to Newton’s laws governing the motion of a projectile affected by wind, gravity, and
aerodynamic drag. These equations are coupled through Equation 8.166. Now, as we did in
our discussion of flat fire, we would like to evaluate Equations 8.163 through 8.165 by using
the downrange distance, x, as the independent variable. To do this, we simply note that for
each of the time derivatives, we can write
dVx
dVx dVx dt dx
=
= Vx
= VxVx′
Vɺ x =
dt
dt dx dt
dx
(8.167)
Vɺ y = VxVy′
(8.168)
Vɺ z = VxVz′
(8.169)
And similarly
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Trajectories
263
We can now write the three equations of motion with x as the independent variable as follows:
Vx′ =
Vy′ =
ɶ
V
1 ɺ
Vx = − Cˆ *D
Vx
Vx
ɶ
V
1 ɺ
Vy = − Cˆ *D
Vx
Vx
Vz′ =
(Vx − Wx )
(8.170)
g
(Vy − Wy ) −
Vx
(8.171)
ɶ
V
1 ɺ
Vz = − Cˆ *D (Vz − Wz )
Vx
Vx
(8.172)
As we noted earlier, the vertical component of the wind, Wy, is usually extremely small and
will be neglected in further treatment. Further, as we mentioned, these equations are impossible to solve in closed form and we must resort to numerical methods for their solution.
Without the restrictions of flat fire, projectiles fired at high angles of departure may
traverse the atmosphere to great altitudes. In their flight, they encounter air temperatures
and pressures that constantly change. These changes must be accounted for in the numerical computations to adequately solve the trajectory. Hence, knowledge of the standard
atmosphere must serve as input to the calculations. There are two standards in common
use: Army Standard Metrology and the International Civil Aviation Organization (ICAO)
atmosphere. ICAO atmosphere is the most used of the two. Temperature and pressure
versus altitude are shown for the ICAO model in Figure 8.7. These atmospheric models are
usually incorporated into ballistics codes.
Now, to become familiar with the physics of the Coriolis acceleration which was brought
to its final form by Gaspard de Coriolis in 1835, we will study the effects of the earth’s
rotation on a flat fire, vacuum trajectory example. We do this because we will be able to
see these effects without resorting to a computer for calculation. The effect is really due
to the fact that the firing point and target are located on the rotating earth, thus when the
projectile lands, the earth has rotated through an angle and has thus moved the target.
Figure 8.8 shows the geometry of the earth, the latitude of the firing site, and the orientation of the axes.
The Coriolis acceleration is defined as
60
60
50
50
Altitude (km)
Altitude (km)
2ω × ( v Β/ Α )xyz = 2Ω × ( v )xyz
40
30
20
40
30
20
10
10
0
(a)
(8.173)
–60 –40 –20 0 20
Temperature (°C)
0
(b)
40
80
Pressure (kPa)
120
FIGURE 8.7
International Civil Aviation Organization (ICAO) models for (a) atmospheric temperature and (b) pressure.
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Ballistics: Theory and Design of Guns and Ammunition
264
Ω
x
N
y
AZ
x
z
L
..
z
FIGURE 8.8
Angles used for Coriolis acceleration calculations. Picture on the right represents a map of the corresponding
area on the globe.
We have written Equation 8.173 in this way because the angular velocity we are considering is that of the earth and our projectile velocity is relative to our firing position (and
therefore the earth), which moves with the x–y–z coordinate system. For this equation to
be useful to us, we have to write the earth’s angular velocity, Ω, in terms of our x–y–z coordinate system. We will see that this acceleration is independent of the projectile weight but
dependent upon its velocity. From Figure 8.8, we see that we can readily define Ω in terms
of our moving coordinate system as
Ω = Ω cos L cos AZi + Ω sin Lj − Ω cos L sin AZk
(8.174)
If we also note that v is defined as
(v )xyz = Vxi + Vy j + Vz k
(8.175)
Then inserting Equations 8.174 and 8.175 into Equation 8.173 gives us
2Ω × ( v )xyz
(Vz sin L + Vy cos L sin AZ)i
= 2Ω (−Vz cos L cos AZ − Vx cos L sin AZ))j
(Vy cos L cos AZ − Vx sin L)k
(8.176)
We will write the Coriolis acceleration in terms of a D’Alembert force (i.e., the negative of
what we have in Equation 8.176), so we shall define the Coriolis term in our equation of
motion (Equation 8.161) as
Λ = −2Ω × ( v )xyz
© 2014 by Taylor & Francis Group, LLC
(−Vy cos L sin AZ − Vz sin L)i
= 2Ω (Vx cos L cos AZ + Vz cos L cos AZ)j
(Vx sin L − Vy cos L cos AZ)k
(8.177)
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265
Here we need to define the following variables:
Λ is the vector Coriolis acceleration
Ω is the angular velocity of the earth about its polar axis = 0.00007292 (rad/s)
L is the latitude of the firing site, positive in the northern hemisphere, negative in the
southern
AZ is the azimuth angle of fire, measured clockwise from north
Vx, Vy, Vz is the velocity in the x, y, z directions, respectively, positive along the positive coordinate axes
Now that we have defined some terminology, we shall examine the effect that the Coriolis
acceleration has on a vacuum trajectory. While this is stretching the vacuum trajectory
much beyond its usefulness in ballistics, we remind the reader that the purpose is to demonstrate the physics that result from Coriolis effects. We begin by recalling Equation 8.161
m
dV
= ΣF + m g + mΛ
dt
(8.161)
Now, since this is a vacuum trajectory, the force term on the RHS is zero, and we can divide
by the mass, m, to obtain the vector equation for a vacuum trajectory
dV
=g+Λ
dt
(8.178)
Rewriting Equation 8.178 in terms of its vector components gives (note that the “g” term
appears only in Equation 8.180)
dVx
= 2Ω(−Vy cos L sin AZ − Vz sin L)
dt
(8.179)
dVy
= 2Ω(−Vx cos L sin AZ + Vz cos L cos AZ) − g
dt
(8.180)
dVz
= 2Ω(Vx sin L − Vy cos L cos AZ)
dt
(8.181)
We shall now provide examples of the effect. These examples are based on the work of
McCoy and can also be found in his work [1]. Let us consider first a purely vertical firing
(i.e., Vx = Vz = 0). One may, initially, consider this a trivial example, but for test purposes we
occasionally do fire vertically. And, by the way, as we will see, what goes up does not come
straight down. Let us also choose due east as positive x, so AZ = 90°. With these assumptions, Equations 8.179 through 8.181 become
© 2014 by Taylor & Francis Group, LLC
dVx
= −2Ω
ΩVy cos L
dt
(8.182)
dVy
= −g
dt
(8.183)
dVz
=0
dt
(8.184)
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266
These equations are well behaved and no longer coupled, so we can solve them independently. We shall integrate Equation 8.182 by first rewriting it, then integrating it:
dy
dVx
= −2Ω
cos L
dt
dt
(8.185)
Vx = −2Ωy cos L + C
(8.186)
To determine C, we know that at y = y0, Vx = 0 so we can write
Vx = −2Ωy cos L + 2Ωy0 cos L = −2Ω cos L( y − y0 )
(8.187)
If you recall our coordinate system, this means a projectile fired straight up will drift to the
west and one fired (or dropped) straight down will drift to the east. Now we will integrate
Equation 8.183 to get
Vy = − gt + C
(8.188)
Again, solving for the constant by inserting the initial conditions that at t = 0, Vy = Vy0 ,
we get
Vy = Vy0 − gt
(8.189)
Now we shall rewrite and integrate Equation 8.189 a second time making use of the fact
that at t = 0, y = y0, to obtain
y = Vy0 t −
1 2
gt + y0
2
(8.190)
We can now insert Equation 8.190 into Equation 8.187 and rewrite it as
dx
1
1
= −2Ω cos L y0 + Vy0 t − gt 2 − y0 = −2Ω cos L Vy0 t − gt 2
dt
2
2
(8.191)
This can be integrated using the initial conditions that at t = 0, x = 0 to give
1
x = −Ω cos L Vy0 t 2 − gt 3
3
(8.192)
Let us now look at the special case of a bomb dropped from a given height with Vy0 = 0 and
let y = 0. If we know the altitude from which we are dropping the bomb, we can determine
its time of flight from Equation 8.190, thus
y0 =
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1 2
gt
2
(8.193)
Trajectories
267
or
t=
2 y0
g
(8.194)
If we insert this into Equation 8.192, we get
2y
1
x = gΩ cos L 0
3
g
3/ 2
(8.195)
This says that since we are on the positive x-axis, the bomb will drift to the east. This drift
would be greatest at the equator and zero at the poles.
Another example that uses the vacuum trajectory analysis is a projectile that is fired
vertically upward with velocity, Vy0 . We can find the time to apogee from Equation 8.189
knowing that at apogee, Vy = 0:
t=
Vy0
g
(8.196)
If we insert this value of t into Equation 8.192, we get
2
x = − Ωgt 3 cos L
3
(8.197)
The time to apogee can be put in terms of the height at apogee, ys, through Equation 8.190
ys =
1 2
gt
2
(8.198)
ts =
2 ys
g
(8.199)
Therefore, the time to apogee is
And therefore the Coriolis-caused displacement at apogee along the x-axis is found by
inserting Equation 8.199 into Equation 8.197 giving
2 ys3
4
xs = − Ω
cos L
g
3
(8.200)
Lastly, we can apply the Coriolis analysis to the vacuum trajectory, flat fire situation and
determine a correction for the acceleration in that case. We begin by making the usual
assumptions for the flat fire trajectory of
Vy ≪ Vx
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and Vz ≪ Vx
(8.201)
Ballistics: Theory and Design of Guns and Ammunition
268
We substitute these into Equations 8.179 through 8.181 yielding
dVx
≈0
dt
(8.202)
dVy
≈ 2ΩVx cos L sin AZ − g
dt
(8.203)
dVz
≈ 2ΩVx sin L
dt
(8.204)
Solution of Equation 8.202 with the initial conditions of Vx = Vx0 at t = 0 yields
Vx ≈ Vx0
(8.205)
Solution of Equation 8.203 after insertion ofV Equation 8.205 with the initial conditions of
Vy = Vy0 at t = 0 and integrating yields
2ΩVx0
Vy ≈ Vy0 − gt 1 −
cos L sin AZ
g
(8.206)
Solution of Equation 8.204 with the initial conditions of Vz = 0 at t = 0 yields after insertion
of Equation 8.205 and integrating
Vz ≈ 2ΩVx0 t sin L
(8.207)
If we now integrate Equations 8.205 through 8.207 subject to x = 0, y = y0, and z = 0 at t = 0
to get the displacements in the x, y, and z directions, we get
x ≈ Vx0 t
y ≈ y0 + Vy0 t −
gt 2
2
2ΩVx0
1 −
g
z = ΩVx0 t 2 sin L
(8.208)
cos L sin AZ
(8.209)
(8.210)
If we want to parameterize Equations 8.209 and 8.210 in terms of the downrange distance, x,
we can rewrite Equation 8.208 as
t≈
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x
Vx0
(8.211)
Trajectories
269
We can then insert this value of time into Equations 8.209 and 8.210 to obtain
y ≈ y0 +
Vy0
Vx0
x−
gx 2 2ΩVx0
1−
cos L sin AZ
2
2Vx0 g
(8.212)
and
z≈
Ωx 2
sin L
Vx0
(8.213)
Equation 8.212 was rearranged in this form (including the substitution of tan ϕ 0) for comparison with Equation 8.37 also modified to include a y0:
gx 2
2V02
(8.37)
gx 2 2ΩVx0
1−
cos L sin AZ
2
2Vx0 g
(8.214)
y ≈ y0 + x tan φ0 −
y ≈ y0 + x tan φ0 −
From this comparison, we see that the incorporation of the Coriolis acceleration in the flat
fire vacuum trajectory manifests itself in a modification to the gravitational term. Thus, as
defined in Ref. [1], we can define a Coriolis factor, fC, as
2ΩVx0
f C = 1 −
cos L sin AZ
g
(8.215)
and we could rewrite Equation 8.214 as
y ≈ y0 + x tan φ0 − f C
gx 2
2Vx20
(8.216)
If we look closely at Equation 8.214, we note several things: the value of cos L is anywhere
between 0 and 1 for all possible latitudes; thus, if we were firing due north or due south,
there would be no effect on the vertical component of impact; if we fired due east (AZ = 90°),
the Coriolis effect essentially weakens the gravity term and the bullet would hit high; a
due west firing would strike low; and the maximum effect on gravity is to alter it by 1.8%.
Since sin L fluctuates between +1 and −1, the drift, the z-component, will vary right or left
depending on the hemisphere where the firing occurs.
Now that the physics of the Coriolis effect are understood, the only difference when
applied to the nonvacuum point mass trajectory is the fact that the velocity is changing
with time due to drag. This is best handled numerically and will not be covered here.
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Ballistics: Theory and Design of Guns and Ammunition
270
y
V0 = 2800 ft/s
V
mg
0 = 130
mils
x
R = ?, z = ?
FIGURE 8.9
Graphical representation of long-range fire for problem 33.
In summary, for the generalized point mass trajectory, we included drag but ignored the
projectile’s dynamic effects on drag. We described the origins of the Coriolis acceleration
acting on a projectile. The physics was demonstrated through the vacuum trajectory and
further examined with the flat fire assumptions. Incorporation of the Coriolis acceleration
into the generalized point mass assumption is only affected by the variation of velocity
over the trajectory and best handled numerically.
Problem 33
A projectile fired from a British 12 in. Mark IX naval gun had a muzzle velocity of 2800 ft/s
and was fired at a QE of 130 mil (Figure 8.9). Assuming a vacuum trajectory, at what deflection would the shot hit the ground?
Assume the firing is taking place at 50° south latitude and the round is being fired due
north.
Answer: z = −75.8 [ft].
8.5 Six Degree-of-Freedom (6-DOF) Trajectory
In keeping with our plan of increasing the complexity of our analyses to approach more
closely the physical realities of projectile flight, we will now consider the projectile as a
distributed mass. Since projectiles are relatively stiff structures, a 6-DOF model can adequately represent its position and attitude at any time. Each degree of freedom is tied to a
coordinate necessary to completely describe the position of a body.
While this model is necessarily more complex than anything we have studied so far, the
underlying physical principles remain the same. In the following work, we will use vectors (boldfaced, nonitalicized letters) in many of the derivations. We continue to do this
because of the brevity and elegance of the notation.
In the equations that follow, we assume that the projectile is a rigid body of finite length
with its mass distributed based on its geometry. This allows us to account for the effect of
projectile attitude on drag and also allows the full dynamics to come into play. We shall
use direction cosines with respect to the projectile axis of symmetry (and thus a coordinate system with unit vectors i, j, k that translates with the CG but does not rotate and
remains aligned with the projectile axis) as opposed to Eulerian angles (angles that are
measured relative to the inertial coordinate system). This is illustrated in Figure 8.10.
© 2014 by Taylor & Francis Group, LLC
Trajectories
271
i
y, 2
V
at
j
k
mg
x, 1
z, 3
FIGURE 8.10
Coordinate system for 6-DOF model.
Once again we restate the equations of motion, which for generality includes a term for
rocket propulsion of the projectile. However, because this force is usually assumed to be
aligned with the projectile’s longitudinal axis, its effect on the motions we will study is
uncoupled from the other motions and may be added in afterward. Consequently, we will
ignore it in our further work:
F = ma
m
dV
=
dt
(8.217)
∑ F + m g + mΛ + ∑ R
T
(8.218)
where
m is the projectile mass
V is the projectile velocity vector
t is the time
dV
a=
is the vector acceleration
dt
ΣF is the vector sum of all aerodynamic forces
g is the vector acceleration due to gravity
Λ is the vector Coriolis acceleration due to rotation of the earth
ΣRT is the vector sum of all rocket thrust forces (to be ignored)
We can also write the equation for the conservation of angular momentum as
dH
=
dt
∑M + ∑R
M
where
H is the vector angular momentum of the projectile
ΣM is the vector sum of all aerodynamic moments
ΣR M is the vector sum of all rocket thrust moments (to be ignored)
© 2014 by Taylor & Francis Group, LLC
(8.219)
272
Ballistics: Theory and Design of Guns and Ammunition
Because the projectile is assumed to be symmetric, every axis transverse to the longitudinal axis through the CG is a principal axis of inertia. The longitudinal axis itself is also, of
course, a principal axis of inertia. The definition of the inertia tensor, which we will use, is
I xx
I = −I yx
−I zx
−I xy
I yy
−I zy
−I xz
−I yz
I zz
(8.220)
Here the diagonal terms are called the moments of inertia and the off-diagonal terms are
called the products of inertia. We know that there is a rotation that can be applied to this
tensor such that the off-diagonal elements go to zero. In this orientation, the axes are said
to be principal axes of inertia and the tensor is written as
I x
I = 0
0
0
Iy
0
0
0
I z
(8.221)
In our coordinate system, we shall define the unit vectors, i, j, and k so that they all lie along
the projectile’s principal axes. Because of this unique situation, the total angular momentum of the projectile can be expressed as the sum of two vectors: the angular momentum
about i and the angular momentum about any axis perpendicular to i through the CG.
Since the i-axis is what we usually call the polar axis, we will denote the polar moment of
inertia as IP. With the symmetry of the projectile, the other moments of inertia about axes
perpendicular to i are known as the transverse moments of inertia, Iy = Iz = IT. We can then
rewrite the inertia tensor as
IP
I= 0
0
0
IT
0
0
0
I T
(8.222)
If a projectile is spinning at spin rate, p, the angular momentum about the polar axis is
defined as
HP = I P pi
(8.223)
The angular momentum about any transverse axis is defined as
di
HT = I T i ×
dt
(8.224)
With this, we can write the total momentum vector as
di
H = I P pi + I T i ×
dt
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(8.225)
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273
By defining a specific angular momentum, h = H/IT, we can write
h=
I P p di
i + i×
IT
dt
(8.226)
If we take the derivative of Equation 8.226 with respect to time, we get
I p di di di d 2i
dh I P
= pɺ i + P
+ × + i×
I T dt dt dt dt 2
dt I T
(8.227)
Since the cross-product of a vector with itself is zero, we get
I p di d 2i
dh I P
= pɺ i + P
+ i×
I T dt dt 2
dt I T
(8.228)
In anticipation of a later need, we shall take the dot product and cross-product of the vector h
with the unit vector i to get
I p di
I p
h ⋅ i = P i + i × ⋅ i = P
d
I
t
IT
T
(8.229)
I p di
di
h × i = P i + i × × i =
d
d
I
t
t
T
(8.230)
In Equations 8.229 and 8.230, we have used the orthogonality properties of vectors as follows:
i⋅i = 1
i⋅j = i⋅k = 0
i×i = 0
i × j = k → (i × j) × i = (k) × i = j ∴ (i × j) × i = j
We will now examine all of the forces and then the moments acting on the projectile and
combine them into Equations 8.218 and 8.219. We have discussed all of these items in
Chapter 6, so we shall simply refresh their meanings briefly and move on. The first force acting on the projectile is the drag force, which acts opposite to the velocity vector, so we have
1
Drag Force = FD = − ρ SCD VV
2
(8.231)
The second force is the lift force, which we modified for our coordinate system as follows:
1
Lift Force = FL = − ρ SCLα [V × (i × V )]
2
© 2014 by Taylor & Francis Group, LLC
(8.232)
Ballistics: Theory and Design of Guns and Ammunition
274
This equation contains a vector triple product in it that we replace with the relationship
from vector algebra
A × (B × C) = (A ⋅ C)B − (A ⋅ B)C
(8.233)
which, for the product in Equation 8.232, can be written as
V × (i × V ) = V 2i − (V ⋅ i)V
When inserted into Equation 8.232, we have
Lift force = FL =
1
ρ SCLα [V 2i − (V ⋅ i)V]
2
(8.234)
The next force is the Magnus force, brought on by the spin or roll of the projectile and
taken from Equation 6.13
Magnus force = FM =
1
pd
ρ SV CNpα (V × i)
2
V
(8.235)
However, from Equation 8.229, we know that p = (IT/IP)(h · i) and, from vector algebra, V × i =
−i × V. Then, we can manipulate Equation 8.235 to the form
1
I
Magnus force = FM = − ρ SdCNpα T
2
IP
(h ⋅ i)(i × V )
(8.236)
Next we need to include the pitch damping force from Equation 6.15 where we will write
v′ as the unit vector along the velocity vector:
Pitch damping force =
1
1
di
di dv′
ρVSd CNq + ρVSdCNαɺ −
2
2
dt
dt dt
(8.237)
If we assume dv′/dt = di/dt (this means that the rate at which the velocity vector is rotating to follow the curve of the trajectory is much smaller than the rate at which the axis
of the projectile is moving) and we include Equation 8.230, we get a relation similar to
Equation 6.18
Pitch damping force =
1
di
di 1
ρVSdCNq + ρVSdCNαɺ
2
dt
dt 2
(8.238)
1
ρVSd(CNq + CNαɺ )(h × i)
2
(8.239)
or by Equation 8.230
Pitch damping force =
© 2014 by Taylor & Francis Group, LLC
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275
With all our forces now expressed in terms of our defined coefficients, we can divide Equation
8.218, omitting the rocket motor, by the projectile mass and inserting the coefficients to give
ρ SdCNpα I T
dV
ρVSCD
ρ SCLα 2
=−
V+
[V i − (V ⋅ i)V] −
dt
2m
2m
2m I P
+
ρVSd ( CNq + CNαɺ )
2m
(h ⋅ i)(i × V )
(h × i) + g + Λ
(8.240)
We will now examine the moments involved in Equation 8.219, the first of which is the spin
damping moment written as
1
pd
ρV 2Sd Clp i
2
V
Spin damping moment = MS =
(8.241)
However, if we again insert Equation 8.229 into the aforementioned equation, we get
1
I
ρVSd 2Clp T (h ⋅ i)i
2
IP
Spin damping moment = MS =
(8.242)
The rolling moment comes from Equation 6.7 and is
Rolling moment = MR =
1
ρV 2Sdδ FClδ i
2
(8.243)
The overturning moment can be written from Equation 6.10 as
Overturning moment = Mα =
1
ρ SdVCMα (V × i)
2
(8.244)
The Magnus moment can be written from Equation 6.14 and, by using the relations of
Equations 8.233 and 8.236, we get
Magnus moment = Mpα =
1
I
ρ Sd 2CMpα T (h ⋅ i)[V − (V ⋅ i)i]
2
IP
(8.245)
We can obtain the pitch damping moment by rewriting Equation 6.19 as well as using the
relation of Equation 8.229 to get
Pitch damping moment = Mq =
(
)
1
ρVSd 2 CMq + CMαɺ [h − (h ⋅ i)i]
2
(8.246)
We can now place all of these relations into Equation 8.219, again omitting the rocket term,
to yield
dH
= MS + MR + Mα + Mpα + Mq
dt
© 2014 by Taylor & Francis Group, LLC
(8.247)
Ballistics: Theory and Design of Guns and Ammunition
276
Equation 8.247 can be changed to a more desirable form by dividing by IT, which yields
dh MS MR Mα Mpα Mq
=
+
+
+
+
IT
IT
IT
IT
IT
dt
(8.248)
This, in turn, may be rewritten by inserting the various moment equations derived
earlier as
ρ V 2Sdδ FClδ
dh ρ VSd 2Clp
ρVSdCMα
=
(h ⋅ i)i +
(V × i)
i+
dt
2I P
2I T
2I T
+
ρVSd 2 (CMq + CMαɺ )
ρ Sd 2CMpα
(h ⋅ i)[V − (V ⋅ i)i] +
[h − (h ⋅ i)i]
2I P
2I T
(8.249)
Note that the equations of motion are highly coupled to one another and the reason we
call the model a 6 DOF is readily apparent. When we break the equations up into their
individual components, we have six equations and six unknowns (x, y, z, p, α, and β). Let
us recall that the x, y, z axes are axes fixed to the earth, independent of the projectile,
while i is the unit vector along the axis of symmetry of the projectile and has components along the x, y, z earth axes and p, α, and β are the spin rate, pitch, and yaw angles,
respectively. For convenience, clarity, and to facilitate analysis, we will relabel the x, y, z
unit vectors (normally i, j, k) as e1, e2, and e 3, respectively, letting the subscripts denote
the x, y, z axes in that order (see Figure 8.10). Then, in terms of components in the earthfixed system,
h = h1e1 + h2e2 + h3e3
(8.250)
i = i1e1 + i2e2 + i3e3
(8.251)
V = V1e1 + V2e2 + V3e3
(8.252)
W = W1e1 + W2e2 + W3e3
(8.253)
v = V − W = (V1 − W1 )e1 + (V2 − W2 )e2 + (V3 − W3 )e3
(8.254)
v1 = (V1 − W1 ) v2 = (V2 − W2 ) v3 = (V3 − W3 )
(8.255)
v = v12 + v22 + v32
(8.256)
We shall also define
and further defining
and
© 2014 by Taylor & Francis Group, LLC
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277
We can insert the definition for v in place of V in Equations 8.245 and 8.249 to yield
ρ SdCNpα I T
dV
ρ vSCD
ρ SCLα 2
=−
v+
[v i − (v ⋅ i)v] −
dt
2m
2m
2m I P
+
(h ⋅ i)(i × v )
ρ vSd(CNq + CNαɺ )
(h × i) + g + Λ
2m
(8.257)
and
ρ v 2Sdδ FClδ
ρ vSdCMα
dh ρ vSd 2Clp
=
i+
( v × i)
(h ⋅ i)i +
dt
2I P
2I T
2I T
+
ρ Sd 2CMpα
ρ vSd 2 (CMq + CMαɺ )
(h ⋅ i) [ v − (v ⋅ i)i ] +
[h − (h ⋅ i)i]
2I P
2I T
(8.258)
Our goal is to examine Equations 8.257 and 8.258 and break each into three equations, one
for each coordinate direction (actually for the acceleration in each coordinate direction).
But before we attempt to break Equations 8.257 and 8.258 into their components, it will be
best to solve for some of the vector quantities that occur in them. Beginning with the second term of Equation 8.257 we can, with appropriate vector multiplication, obtain
(
)
(
)
(v ⋅ i)v = v12i1 + v1v2i2 + v1v3i3 e1 + v1v2i1 + v22i2 + v2v3i3 e2
(
)
+ v1v3i1 + v2v3i2 + v32i3 e3
(8.259)
Another useful relation is that
(v ⋅ i) = (v1e1 + v2e2 + v3e3 ) ⋅ (i1e1 + i2e2 + i3e3 )
(8.260)
(v ⋅ i) = v1i1 + v2i2 + v3i3
(8.261)
or
However, we can show that
cos α t =
(v ⋅ i) v1i1 + v2i2 + v3i3
=
v
v
(8.262)
The next relations in Equation 8.257 are
(h ⋅ i) = h1i1 + h2i2 + h3i3
(8.263)
e1 e2 e3
(i × v ) = i1 i2 i3 = (i2v3 − i3v2 )e1 + (i3v1 − i1v3 )e2 + (i1v2 − i2v1 )e3
v1 v2 v3
(8.264)
and
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
278
Now, also in Equation 8.257 is the term
(h ⋅ i)(i × v ) = ( h1i1 + h2i2 + h3i3 )[(i2v3 − i3v2 )e1 + (i3v1 − i1v3 )e2 + (i1v2 − i2v1 )e3 ]
(8.265)
But with the fact that (h · i) = IPp/IT as shown earlier, then we can write
(h ⋅ i)(i × v ) =
I p
I p
IP p
(i2v3 − i3v2 )e1 + P (i3v1 − i1v3 )e2 + P (i1v2 − i2v1 )e3
IT
IT
IT
(8.266)
Another relation we have to deal with is
e1 e2 e3
(h × i) = h1 h2 h3 = ( h2i3 − h3i2 )e1 + ( h3i1 − h1i3 )e2 + ( h1i2 − h2i1 )e3
i1 i2 i3
(8.267)
The next relation we generate with the help of Equation 8.229 is
(h ⋅ i)i =
I p
I p
IP p
i1e1 + P i2e2 + P i3e3
IT
IT
IT
(8.268)
In Equation 8.258, we need
e1 e2 e3
(v × i) = v1 v2 v3 = (i3v2 − i2v3 )e1 + (i1v3 − i3v1 )e2 + (i2v1 − i1v2 )e3
i1 i2 i3
(8.269)
And finally in Equation 8.258
(v ⋅ i)i = (v1i1 + v2i2 + v3i3 )(i1e1 + i2e2 + i3e3 )
(8.270)
or
(
)
(
)
(v ⋅ i)i = v1i12 + v2i1i2 + v3i1i3 e1 + v1i1i2 + v2i22 + v3i2i3 e2
(
)
+ v1i1i3 + v2i2i3 + v3i32 e3
(8.271)
Let us now look at Equation 8.257 with all of the vector quantities broken into their
components
dV1
dV2
dV3
ρ vSCD
(v1e1 + v2e2 + v3e3 )
e1 +
e2 +
e3 = −
dt
dt
dt
2m
ρ SCLα 2
[v i1e1 + v 2i2e2 + v 2i3e3 − v cos α t (v1e1 + v2e2 + v3e3 )]
+
2m
ρ SdCNpα I T I P p
−
[(v3i2 − v2i3 )]e1 + (v1i3 − v3i1 )e2 + (v2i1 − v1i2 )e3 ]
2m I P I T
+
ρ vSd ( CNq + CNαɺ )
[( h2i3 − h3i2 )e1 + ( h3i1 − h1i3 )e2 + ( h1i2 − h2i1 )e3 ]
2m
+ g1e1 + g 2e2 + g 3e3 + Λ 1e1 + Λ 2e2 + Λ 3e3
© 2014 by Taylor & Francis Group, LLC
(8.272)
Trajectories
279
Similarly, let us perform the same operation on Equation 8.258
ρ vSd 2Clp I P p
dh1
dh
dh
e1 + 2 e2 + 3 e3 =
I (i1e1 + i2e2 + i3e3 )
dt
dt
dt
2I P
T
+
ρ v 2Sdδ FClδ
ρ vSdCMα
(i1e1 + i2e2 + i3e3 ) +
[(v2i3 − v3i2 )e1 + (v3i1 − v1i3 )e2 + (v1i2 − v2i1 )e3 ]
2I T
2I T
+
ρ Sd 2CMpα I P p
I [(v1e1 + v2e2 + v3e3 ) − v cos α t (i1e1 + i2e2 + i3e3 )]
2I P
T
+
ρ vSd 2 ( CMq + CMαɺ )
IP p
(i1e1 + i2e2 + i3e3 )
( h1e1 + h2e2 + h3e3 ) −
2I T
IT
(8.273)
We will first operate on Equation 8.272 by collecting all of the terms with the unit vectors e1,
then e2 and e3, and by putting them into the equations for linear and angular momentum:
ρ SdCNpα p
pSCLα 2
dV1
ρ vSCD
v i1 − vv1 cos α t −
v1 +
=−
(v3i2 − v2i3 )
dt
2m
2m
2m
+
ρ vSd ( CNq + CNαɺ )
2m
( h2i3 − h3i2 ) + g1 + Λ1
(8.274)
ρ SdCNpα p
dV2
ρ vSCD
ρ SCLα 2
v i2 − vv2 cos α t −
v2 +
=−
(v1i3 − v3i1 )
dt
2m
2m
2m
+
ρ vSd ( CNq + CNαɺ )
2m
( h3i1 − h1i3 ) + g 2 + Λ 2
(8.275)
ρ SdCNpα p
dV3
ρ vSCD
ρ SCLα 2
v i3 − vv3 cos α t −
v3 +
=−
(v2i1 − v1i2 )
dt
2m
2m
2m
+
ρ vSd ( CNq + CNαɺ )
2m
( h1i2 − h2i1 ) + g 3 + Λ 3
(8.276)
Next is Equation 8.273, where the same procedure will be followed:
dh1 ρ vSd 2ClP p
ρ v 2Sdδ FClδ
ρ vSdCMα
i1 +
i1 +
=
(v2i3 − v3i2 )
dt
2I T
2I T
2I T
ρ vSd 2 ( CMq + CMαɺ )
ρ Sd 2CMpα p
IP p
[v1 − vi1 cos α t ] +
+
h1 −
i1
2I T
2I T
IT
© 2014 by Taylor & Francis Group, LLC
(8.277)
Ballistics: Theory and Design of Guns and Ammunition
280
dh2 ρ vSd 2ClP p
ρ v 2Sdδ FClδ
ρ vSdCMα
i2 +
i2 +
=
(v3i1 − v1i3 )
dt
2I T
2I T
2I T
+
ρ vSd 2 ( CMq + CMαɺ )
ρ Sd 2CMpα p
IP p
[v2 − vi2 cos α t ] +
h2 −
i2
2I T
2I T
IT
(8.278)
dh3 ρ vSd 2ClP p
ρ v 2Sdδ FClδ
ρ vSdCMα
i3 +
i3 +
=
(v1i2 − v2i1 )
dt
2I T
2I T
2I T
ρ vSd 2 ( CMq + CMαɺ )
ρ Sd 2CMpα p
IP p
[v3 − vi3 cos α t ] +
+
h3 −
i3
2I T
2I T
IT
(8.279)
We can simplify Equations 8.274 through 8.279 considerably by defining the following
coefficients:
ɶ D = ρ vSCD
C
2m
2
ɶl = ρ vSd ClP p
C
P
2I T
ɶL = ρ SCLα
C
α
2m
ρ
ɶ N = SdCNpα p
C
pα
2m
2
ɶl = ρ v Sdδ FCl δ
C
δ
2I T
ɶ M = ρ vSdCMα
C
α
2I T
ɶN =
C
q
2
ɶ M = ρ Sd CMpα p
C
pα
2I T
ρ vSd ( CNq + CNαɺ )
2m
ρ vSd ( CMq + CMα )
2
ɶM =
C
q
2I T
With these coefficients, we can write Equations 8.274 through 8.276 in a more compact form:
dV1
ɶ Dv1 + C
ɶL (v 2i1 − vv1 cos α t ) − C
ɶ N (v3i2 − v2i3 )
= −C
pα
α
dt
ɶ N ( h2i3 − h3i2 ) + g1 + Λ1
+C
q
(8.280)
dV2
ɶ D v2 + C
ɶL (v 2i2 − vv2 cos α t ) − C
ɶ N (v1i3 − v3i1 )
= −C
pα
α
dt
ɶ N ( h3i1 − h1i3 ) + g 2 + Λ 2
+C
q
(8.281)
dV3
ɶ D v3 + C
ɶL (v 2i3 − vv3 cos α t ) − C
ɶ N (v2i1 − v1i2 )
= −C
pα
α
dt
ɶ N ( h1i2 − h2i1 ) + g 3 + Λ 3
+C
q
© 2014 by Taylor & Francis Group, LLC
(8.282)
Trajectories
281
We can do the same with Equations 8.277 through 8.279:
(
)
IP p
h1 −
i1
IT
(8.283)
(
)
IP p
h2 −
i2
IT
(8.284)
dh3
ɶl + C
ɶl i3 + C
ɶ M (v1i2 − v2i1 ) + C
ɶ M (v3 − vi3 cos α t ) + C
ɶ M h3 − I P p i3
= C
α
q
δ
P
pα
I
dt
T
(8.285)
dh1
ɶl + C
ɶl i1 + C
ɶ M (v2i3 − v3i2 ) + C
ɶ M (v1 − vi1 cos α t ) + C
ɶM
= C
α
q
δ
P
pα
dt
dh2
ɶl + C
ɶl i2 + C
ɶ M (v3i1 − v1i3 ) + C
ɶ M (v2 − vi2 cos α t ) + C
ɶM
= C
α
q
δ
P
pα
dt
(
)
Now that we have the six, coupled, equations for our six accelerations, we would like to
determine the position of the projectile in space and time. We do this by creating a vector, X,
to the center of mass of the projectile. If we note that X = [xe1 + ye2 + ze3] in the earth-fixed
coordinate system, then we can break the individual components into
t
∫
x = x0 + V1dt
(8.286)
0
t
∫
y = y0 + V2dt
(8.287)
0
∫
t
z = z0 + V3dt
(8.288)
0
Recognize that when firing a long-range weapon we usually do so with grid coordinates
on a map of the earth. A map is, in theory, created by peeling the geometry off a globe.
Thus, the coordinates and distances are correct in the downrange and cross-range directions (x and z). However, the altitude, y, has to be corrected for the curvature of the earth.
This is depicted with the applicable equations in Figure 8.11. A similar rotation occurs with
the gravity vector as depicted in Figure 8.12.
With this relationship, we can write the projectile position vector in earth coordinates as
x 2
x2
x
[e e3
E ≈ [E1e1 + E2e2 + E3e3 ] = xe1 + y +
e2 + ze3 = y +
1
R
2
2
z R
e3 ]
(8.289)
where R is the average radius of the earth, taken to be 6,951,844 yards or 6,356,766 m.
The use of earth coordinates is recommended at ranges beyond about 2000 yards
(at 2000 yards there is a 10.36-in. difference in height [1]). Furthermore, the acceleration
of gravity varies with altitude (and, in fact, latitude and longitude as well) and we need
to consider this.
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
282
θ
2
x = Rθ
θ E
2
2R sin θ
2
θ
R
E2 cos θ ≈ E2
2R sin2
E2 ≈
E2cos θ = 2R sin2 θ
2
θ2
θ
≈ 2R
4
2
Rθ2 = R x2
2
2 R2
E2 ≈
x2
2R
FIGURE 8.11
Altitude error over long trajectories.
θ
θ
g
g
g cos θ = g cos 2 θ
2
= g 1−2sin2
x = Rθ
θ
g=g
R
g sin θ ≈ gθ = g x
R
y
x
e
e + g 1−2
R 1
R 2
θ
x2
=g 1− 2
2
2R
x
R
y
g=g 1
1−2
R
0
[ e1
e2 e3 ]
g = 32.174 [1 – 0.0026 cos (2L)] ft2
s
FIGURE 8.12
Rotation of the gravity vector due to earth curvature and associated equations.
To complete the equations of motion, we must consider the form of the Coriolis acceleration vector. We have discussed this extensively previously so we shall simply write the
components of this vector as
Λ1 = 2Ω(− V2 cos L sin AZ − V3 sin L)
(8.290)
Λ 2 = 2Ω(V1 cos L sin AZ + V3 cos L cos AZ)
(8.291)
Λ 3 = 2Ω(V1 sin L − V2 cos L cos AZ)
(8.292)
Or as a vector
Λ1
Λ = Λ 2 e1
Λ 3
e2
−V2 cos L sin AZ − V3 sin L
e3 = 2Ω V1 cos L sin AZ + V3 cos L cos AZ e1
V1 sin L − V2 cos L cos AZ
© 2014 by Taylor & Francis Group, LLC
e2
e3
(8.293)
Trajectories
283
We now have the differential equations of motion but need initial conditions to solve them.
Let us examine the projectile at the instant of muzzle exit without worrying about how it
attained its state of motion there (this is the job of the interior ballistician). We shall define
the initial tube angle in azimuth and elevation as θ 0 and ϕ 0, respectively. Then our initial
velocity vector can be defined as
Λ1
cos φ0 cos θ 0
V0 = Λ 2 [e1 e2 e3 ] = V0 sinφ0 cos θ 0 [e1 e2 e3 ]
Λ 3
sinφ0
(8.294)
And, if we also take the wind into account, we have
V10 − W10
v10
v 0 = V0 − W0 = v20 [e1 e2 e3 ] = V20 − W20 [e1 e2 e3 ]
V30 − W30
v30
(8.295)
Here the usual relationships for these vectors apply. These are
V0 = V120 + V220 + V320
(8.296)
v0 = v120 + v220 + v320
(8.297)
The initial orientations of the body-fixed unit vectors in the earth-fixed system are
cos((φ0 + α 0 )cos(θ 0 + β 0 )
i10
i0 = i10 e1 + i20 e2 + i30 e3 = i20 [e1 e2 e3 ] = sin(φ0 + α 0 )cos(θ 0 + β 0 ) [e1 e2 e3 ]
i30
sin(θ 0 + β 0 )
(8.298)
− cos 2 (θ 0 + β 0 ) sin(φ0 + α 0 )cos(φ0 + α 0 )
j10
2
1
j0 = j10 e1 + j20 e2 + j30 e3 = j20 [e] =
cos (θ 0 + β 0 )cos 2 (φ0 + α 0 ) + sin 2 (θ 0 + β 0 ) [e1 e2 e3 ]
Q
j30
− sin(θ 0 + β 0 )cos(θ 0 + β 0 )sin(φ0 + α 0 )
(8.299)
− sin(θ 0 + β 0 )
k10
1
0
k 0 = k10 e1 + k 20 e2 + k 30 e3 = k 20 [e] =
[e1
Q
k 30
cos(θ 0 + β 0 )cos(φ 0 + α 0 )
e2
e3 ]
(8.300)
In the aforementioned equations, α 0 and β 0 are the initial pitch and yaw angles, respectively, of the projectile. Thus, they add directly to the weapon azimuth and elevation
angles. The quantity Q we define following Ref. [1] as
Q = sin 2 (θ 0 + β 0 ) + cos 2 (θ 0 + β 0 )cos 2 (φ0 + α 0 )
© 2014 by Taylor & Francis Group, LLC
(8.301)
Ballistics: Theory and Design of Guns and Ammunition
284
If we now consider the rotation, (ω)ijk, of the projectile about its axis of symmetry (thus relative to the i-j-k triad) and we define an arbitrary initial projectile rotation as
(ω0 )ijk = ωi0 i0 + ω j0 j0 + ωk0 k 0
(8.302)
Here this initial angular velocity is dependent upon the initial orientation of the unit vector, i0. Then, the initial velocity of the unit vector can be written as follows:
i0
j0
di0
= (ω0 )ijk × i0 = ωi0 ω j0
dt
1
0
iɺ10
ωk0 = [ωk0 j0 − ω j0 k 0 ] = iɺ20 [e]
ɺ
0
i30
k0
ωk0 j10 − ω j0 k10
= ωk0 j20 − ω j0 k 20 [e1 e2 e3 ]
ωk j3 − ω j k 3
0
0
0 0
(8.303)
Note that Equation 8.303 is a tensor equation. Tensors are higher order vectors but can be
treated the same. If we insert the results of Equations 8.299 and 8.300 into the aforementioned equation, we get
1
= ω j0 si n(θ 0 + β 0 ) − ωk0 cos 2 (θ 0 + β 0 )sin(φ0 + α 0 )cos(φ0 + α 0 )]
iɺ10 =
Q
(8.304)
1
= ωk0 cos 2 (θ 0 + β 0 )cos 2 (φ0 + α 0 ) + ωk0 sin 2 (θ 0 + β 0 )
iɺ20 =
Q
(8.305)
1
−ω j0 cos(θ 0 + β 0 )cos(φ0 + α 0 ) − ωk0 sin (θ 0 + β 0 )cos(θ 0 + β 0 )siin(φ0 + α 0 )
iɺ30 =
Q
(8.306)
Continuing with our statement of the initial conditions, a positive pitch rotates the nose
of the projectile upward and a positive yaw rotates the nose to the left as viewed from the
rear. The initial value of the modified angular momentum vector is given by
I P p0
di
i0 + i0 × 0
dt
IT
(8.307)
di0 ɺ
= i10 e1 + iɺ20 e2 + iɺ30 e3
dt
(8.308)
h0 =
We can rewrite di0/dt as
© 2014 by Taylor & Francis Group, LLC
Trajectories
285
which then allows us to write
e1
i0 ×
di0
= i10
dt
iɺ10
e2
e3
i20
(
)
(
)
(
)
i30 = i20 iɺ30 − i30 iɺ20 e1 + i30 iɺ10 − i10 iɺ30 e2 + i10 iɺ20 − i20 iɺ10 e3
(8.309)
iɺ20 iɺ30
We can then incorporate Equation 8.309 into Equation 8.307 to yield
I P p0
ɺ
ɺ
I i1 0 + i20 i30 − i30 i20
T
h10
I p
h0 = h20 [e] = P 0 i2 0 + i30 iɺ10 − i10 iɺ30 [e1 e2 e3 ]
IT
h30
I P p0 i + i iɺ − i iɺ
I T 3 0 10 20 20 10
(8.310)
Here the initial value of the spin rate p0 is determined by the axial velocity and the twist
rate, n (in calibers per revolution), of the weapon through p0 = 2πV0/nd. We have thus completed all of the initial conditions necessary to perform the calculation.
As we will discuss in a later section, the projectile’s motion can be characterized as epicyclic. The tip of a vector drawn from the CG of the projectile to the nose will trace out a
curve that contains two cyclic modes: a fast mode, known as nutation, and a slow mode,
known as precession. If the round is stable, these modes will eventually damp down to
near zero, leaving only some movement because of nonlinear forces and moments. We
shall explore this more later.
Some other terms come up in the succeeding sections that require definitions. Since they
are essential to the understanding of trajectories, we will define them now.
A projectile’s yaw of repose is the yaw created by the action of gravity on the projectile
as it attempts to follow its trajectory curve. As stated earlier, the nose of the projectile is
usually above the trajectory. There is then a net aerodynamic force through the CP which
wants to rotate the nose up. With a right-hand spinning projectile, this results in a yaw of
the nose to the right. This is called the yaw of repose.
Failure to trail is a situation that arises when the base of the projectile does not follow the
nose (it flies base first after apogee). This is depicted in Figure 8.13.
The trail angle is the quadrant elevation angle (particular to a gun, projectile, and charge
combination) above which the projectile will not turn over and will fail to trail.
We can summarize this section by saying that for a rigid projectile the 6-DOF model is
as accurate as one can get to the trajectories. If the model yields an inaccurate answer, the
problem is usually a wrong assumption in the metrology, initial conditions, or projectile
mass properties. Lastly, the only practical method of solving these equations is by numerical methods and, with the speed of computers today, the codes run very efficiently and
quickly. This last statement makes it difficult to generate meaningful problems for the
interested reader. We have endeavored to create useful exercises by stipulating a large
number of conditions and requiring the reader to examine the accelerations of the projectile at a point in space.
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
286
i
y, 2
V
αt
j
k
mg
mg
V
x, 1
z, 3
FIGURE 8.13
A projectile that has failed to trail.
Problem 34
A British bomber is flying at a speed of 200 mph in still air. If the 0.303 in. machine
guns are fired sideways, calculate the axial acceleration vector and the angular
acceleration vector acting on the projectile through use of the 6-DOF equations if the
projectile is
1. Fired to the right
Answer:
ft
a = [ − 1376e1 + 5.89e2 − 1184e3 ] 2
s
dh
rad
= [39.85e1 − 35, 922e2 − 2.45e3 ] 2
dt
s
2. Fired to the left
Answer:
ft
a = [ − 1376e1 + 70.29e2 − 1184e3 ] 2
s
dh
rad
= [39.85e1 + 35, 922e2 + 2.45e3 ] 2
dt
s
3. Discuss the effect of the angular momentum on the projectile nose (which way
does it tip?)
Please ignore the Coriolis acceleration, assume there is no yaw at muzzle exit and assume
a muzzle velocity of 2440 ft/s, the weapon has a right-hand twist.
© 2014 by Taylor & Francis Group, LLC
Trajectories
287
Projectile information:
C D 0 = 0.35
C
2
Dδ
= 3.46
C M α = 2.36
(C
(C
)
) = 0.003
Mq
+ C M αɺ = −16.2
I P = 0.00026[lbm-in.2 ]
Nq
+ C N αɺ
I T = 0.00258[lbm-in.2 ]
C M P α = 0.02
m = 0.025[lbm]
lbm
ρ = 0.060 3
ft
rev
p = 2033
s
CL α = 2.81
CN Pα = −0.67
Please supply all answers in an inertial coordinate system labeled 1, 2, and 3 with 1 being
along the aircraft flight path and 3 being off the right side of the plane. Treat all missing
coefficients as equal to zero.
Problem 35
One of the interesting aspects of the forces acting on a projectile occurs as the projectile leaves
an aircraft sideways. This problem is encountered all the time in the AC-130 gunship. Let us
examine a 105 mm HE projectile being fired into a city from both the top of a building and from
the AC-130 in flight. The velocity of the projectile is 1510 ft/s. With the information provided
1. Calculate the total acceleration vector for both cases
2. Comment on the differences
Positional information:
33.5° north latitude
Azimuth of velocity vector: 80° True
Angle of velocity vector to horizontal: −10°
Wind is calm
α = +2° (nose up), β = −1.5° (nose to the left looking downrange)
The projectile nose is rotating to the right of the velocity vector at 0.5 rad/s
The aircraft is flying at 300 mph to the north
Projectile information:
C D2 = 8.0
(C
(C
C M α = 3.80
C M p α = 0.05
m = 32.1[lbm]
lbm
ρ = 0.060 3
ft
rev
p = 200
s
C D 0 = 0.39
δ
)
) = 0.005
Mq
+ C M αɺ = −6.5
I P = 0.547 [lbm-ft 2 ]
Nq
+ C N αɺ
I T = 5.377 [lbm-ft 2 ]
CL α = 1.9
CN p α = −0.01
Please supply all answers in an inertial coordinate system labeled 1, 2, and 3 with 1 being
due north and 3 being due east.
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
288
Problem 36
The Paris gun was built by Germany in the First World War to shell Paris from 75 miles
away. The weapon was a 210 mm diameter bore with the shells pre-engraved to compensate for wear of the tube. During firing of this weapon, all things such as wind effects,
Coriolis, etc., had to be accounted for (they really could have used a good 6-DOF model
and a computer). Write the acceleration vector for this projectile at an instant in its trajectory when the velocity (relative to the ground) is 2500 ft/s and the following conditions
apply (please note that there is “no” rocket motor):
Positional information:
48.75° north latitude
Azimuth of velocity vector: 300° True
Angle of velocity vector to horizontal: +10°
Wind is blowing at 20 mph due south and horizontal
α = 1°, β = 1.5°
The projectile nose is rotating up at 2 rad/s
Projectile information:
C M α =3.50
(C
(C
CL α = 2.50
C M pα = 0.55
m = 220 [lbm]
CN pα = − 0.02
lbm
ρ = 0.060 3
ft
rev
p = 150
s
C D = 0.28
)
) = 0.005
Mq
+ C M αɺ = −16.5
I P = 19.13[lbm-ft 2 ]
Nq
+ C N αɺ
I T = 66.40[lbm-ft 2 ]
Note that the aforementioned numbers are guesses at the projectile’s characteristics; they do not
represent the real projectile’s performance as no data are available from any source researched.
Please supply all answers in an inertial coordinate system labeled 1, 2, and 3 with 1 being
due west and 3 being due north.
Answer:
ft
Linear acceleration vector is a = { − 85e1 − 33e2 − 35e3 } 2
s
Angular acceleration vector is
dh
rad
= { − 14e1 − 42e2 + 38e3 } 2
dt
s
Problem 37
An AC-130 is flying at a speed of 200 miles per hour in still air. If the 105 mm weapon is fired
sideways, calculate the axial acceleration vector and the angular acceleration vector acting on
the projectile through use of the 6-DOF equations if the projectile is fired to the left. Discuss
the effect of the angular momentum on the projectile nose (which way does it want to tip?).
Please ignore the Coriolis acceleration, assume there is no yaw at muzzle exit and assume
a muzzle velocity of 1500 ft/s, the weapon has a right-hand twist.
© 2014 by Taylor & Francis Group, LLC
Trajectories
289
Projectile information:
C D2 = 8.0
(C
(C
C M α = 3.80
C M p α = 0.05
m = 32.1[lbm]
lbm
ρ = 0.060 3
ft
rev
p = 220
s
C D 0 = 0.39
δ
)
) = 0.005
Mq
+ C M αɺ = −6.5
I P = 0.547 [lbm-ft 2 ]
Nq
+ C N αɺ
I T = 5.377 [lbm-ft 2 ]
CL α = 1.9
CN p α = −0.01
Please supply all answers in an inertial coordinate system labeled 1, 2, 3 with 1 being along
the aircraft flight path and 3 being off the right side of the plane. Treat all missing coefficients as equal to zero.
Problem 38
An F-86 Saber jet flying with a speed of 600 mph. The aircraft is pulling up out of a dive
and the pilot is experiencing 2.5 g′s. The pilot fires his 0.50 caliber machine guns at a
ground target during the pull out. At the instant the plane makes an angle of 30° to the
ground, a projectile leaves the muzzle of the weapon with a muzzle velocity of 1800 ft/s,
3° to the right of the bore axis and is rotating to the left at 2 rad/s in the plane of the gun,
relative to the gun. Also the projectile is pitched up 1.5° and yawed left 1° to the line of fire.
Ignoring Coriolis effects, write the vectors that completely define the initial conditions of
the projectile so that you could use them in a 6-DOF model. DO NOT solve the equations
of motion—just write the initial conditions. Assume coordinate directions as follows:
1. Horizontal direction in which the aircraft is flying
2. Straight up
3. To the right as viewed from the rear of the aircraft
Problem 39
In the study of ballistics, we normally neglect Magnus forces as small and only the
moments generated by these forces are significant. In low-velocity projectiles, however,
Magnus forces may be considerable. Consider a baseball thrown with spin toward a batter.
The ball weighs 5 oz and is 2.9in. in diameter. The distance from the pitcher’s mound to
home plate is 60′ 6”. Assume the ball is thrown at 70 miles/h and that there is no wind. The
release point is 4 ft above the ground with an upward angle of 4°. We need to determine
how much spin is required to move the ball 2 ft to the left as viewed from the pitcher.
Please perform the following calculations:
a. Assuming constant drag and Magnus force coefficients, develop the equations of
motion for the ball assuming flat fire is valid and the spin axis remains vertical for
the entire flight.
b. Determine the spin rate and direction to cause the desired motion.
c. Determine the final velocity, time of flight, and height of the ball.
d. Determine if the assumption of flat fire really was valid.
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
290
Please list all of your assumptions. The following coefficients may be assumed:
lbm
CD = 0.52 ρ = 0.076 3 CNpα = 0.09
ft
8.6 Modified Point Mass Trajectory
The 6-DOF model’s equations, when fully developed, described an epicyclical motion with
fast (nutational) and slow (precessional) modes that (hopefully) would damp out early on,
allowing the projectile to assume a yaw of repose for the remainder of the flight. This yaw
of repose, which remains nearly constant, we assume will account for most of the drag
induced by the yawing of the projectile. If we can simplify the 6-DOF model, which is
computationally expensive to run, by accounting only for the yaw of repose, we could get
a model that will allow the projectile to drift the proper amount and still be quite accurate.
Once again, following McCoy [1], we will make the point mass assumption in the equations that follow. Recall that because of this assumption the projectile is essentially represented as a cannon ball with all of its mass concentrated at one point. We shall then add
some details, which will account for the yawing of the projectile, by assuming the projectile yaw is relatively constant or varies little with time compared to the steady state yaw
angle. This assumption is usually valid except in high-angle fire situations.
We begin with the usual equations of motion and Newton’s second law
F = ma
(8.311)
particularized as
m
∑ F + mg + mΛ
dV
=
dt
(8.312)
Here the variables are the same as we described in the 6-DOF section. In the aforementioned equations, we replace the velocity vector V by the vector (V − W) because drag
measurements are made relative to the air stream not relative to the ground. We will also
again replace the scalar velocity (the speed) with the difference between the projectile and
wind velocities:
ɶ =|V − W|
v = V−W →V
(8.313)
The diagram of the problem is shown in Figure 8.14.
From our work on 6-DOF model, recall Equations 8.257 and 8.258, which we rewrite here
neglecting the pitch damping and rocket forces:
ρ SdC Npα I T
dV
ρ vSCD
ρ SCLα 2
=−
v+
[v i − (v ⋅ i)v] −
dt
2m
2m
2m I P
© 2014 by Taylor & Francis Group, LLC
(h ⋅ i)(i × v ) + g + Λ
(8.314)
Trajectories
291
y
Wy
Height
Wz
Wx
FD
V0
j
on
c ti
De
f le
k
V
mg
0
Range
i
x
z
FIGURE 8.14
Modified point mass trajectory.
2
dh ρ vSd Clp
ρ v 2Sdδ FCl δ
ρ vSdC Mα
(h ⋅ i)i +
i+
(v × i)
=
dt
2I P
2I T
2I T
+
ρ vSd 2 ( CMq + C Mαɺ )
ρ Sd 2C Mpα
(h ⋅ i)[v − (v ⋅ i)i] +
[h − (h ⋅ i)i]
2I P
2I P
(8.315)
Recall also from 6-DOF model (Equation 8.228) that introduces the polar and transverse
moments of inertia
I p di d 2i
dh I P
= pɺ i + P
+ i×
I T dt dt 2
dt I T
(8.316)
Equations 8.314 and 8.315 may be simplified by introducing the tilde (∼) coefficients:
ɶ D = ρ vSCD
C
2m
2
ɶl = ρ vSd ClP p
C
P
2I T
ɶL = ρ SCLα
C
α
2m
ρ
ɶ N = SdCNpα p
C
pα
2m
2
ɶl = ρ v Sdδ FCl δ
C
δ
2I T
ɶ M = ρ vSdCMα
C
α
2I T
ɶN =
C
q
2
ɶ M = ρ Sd CMpα p
C
pα
2I T
ρ vSd ( CNq + CNαɺ )
2m
ρ vSd ( CMq + CMα )
2
ɶM =
C
q
2I T
Using this notation, the modified equations are written as
dV
ɶD v + C
ɶL [v 2i − (v ⋅ i)v] − C
ɶN IT
= −C
pα
α
dt
IP
(
)
(h ⋅ i)(i × v) + g + Λ
dh
ɶl + C
ɶl i + C
ɶ M (v × i) + C
ɶ M (h ⋅ i)[v − (v ⋅ i)i] + C
ɶ M [h − (h ⋅ i)i]
= C
p
δ
α
pα
q
dt
© 2014 by Taylor & Francis Group, LLC
(8.317)
(8.318)
Ballistics: Theory and Design of Guns and Ammunition
292
or alternatively
dV
ɶD v + C
ɶL [v × (i × v)] − C
ɶ N (v × i) + g + Λ
=−C
α
pα
dt
(
)
(8.319)
dh
ɶl + C
ɶl i + C
ɶ M (v × i) + C
ɶ M (h ⋅ i)[v − (v ⋅ i)i] + C
ɶ M [h − (h ⋅ i)i]
= C
q
P
δ
α
pα
dt
(8.320)
We have shown earlier that since the unit vector, i, is always perpendicular to its derivative
di/dt, the dot product of i and di/dt is identically zero. We shall combine Equations 8.316
and 8.320 to yield
IP
I di d 2i
ɶ
ɶ
ɶ
ɶ
ɶ
pɺ i + P p
+ i×
= ClP + Clδ i + CMα (v × i) + CMpα [i × (v × i)] + CMq
IT
I T dt dt 2
(
)
di
i × dt
(8.321)
We will now take the dot product of i with Equation 8.321 to yield
(
)
(
IP
ɶl + C
ɶl → dp = I T C
ɶl + C
ɶl
pɺ = C
p
p
δ
δ
IT
dt I P
)
(8.322)
Here use has been made of the facts that a cross-product results in a vector that is orthogonal to both of the original vectors and that the dot product of orthogonal vectors is identically zero. These relationships are written in mathematical terms here:
di
= i ⋅ (ω × i) = 0
dt
i ⋅ i × (v × i) = 0
i⋅
di
i⋅i× = 0
dt
i ⋅ (i × v) = i ⋅ (v × i) = 0
Equation 8.322 has an important consequence; for a rotationally symmetric projectile,
the spin is decoupled from the yawing motion. Now, if we substitute Equation 8.322 into
Equation 8.321, we get
(Cɶ
lp
2
ɶl i + I P p di + i × d i
+C
δ
I T dt dt 2
)
ɶ M [i × (v × i)] + C
ɶ M i × di
ɶl + C
ɶl i + C
ɶ M (v × i) + C
= C
pα
q
δ
α
p
dt
(
)
(8.323)
or
I P di
d 2i ɶ
ɶ
ɶ
p
+ i × 2 = C
Mα (v × i ) + CM pα [i × (v × i)] + CMq
I T dt
dt
di
i × dt
(8.324)
With Equations 8.319, 8.322, and 8.324, we have merely restated our 6-DOF model. Murphy
[2] formulated the differential equation of motion as a second-order equation in terms
© 2014 by Taylor & Francis Group, LLC
Trajectories
293
j
l
sin αt
αt
i
cos αt
FIGURE 8.15
Projectile axial unit vector, l, illustrated.
of complex variables and solved it. The particular solution was the (relatively) constant
yaw of repose, and the complimentary solution was the transient epicyclic motion. In the
modified point mass approach, we extract the particular solution and ignore the transient
motion, instead concentrating on the yaw of repose, the drift, and the effect of the yaw
drag. We assume that the epicyclic pitching and yawing motion are negligible everywhere
along the trajectory, i.e., in many instances, reasonable, since it should damp early in the
trajectory and thus contributes little to the drift. We proceed by defining another unit vector triad in the same sense as our i-j-k triad. Instead of it being aligned with the geometric
axis of the projectile, we align it with the velocity vector and utilize l-m-n as the principal
directions. We can then define l (Figure 8.15) as
l=
v
v
(8.325)
and formally define our vector yaw of repose as
α R = l × (i × l)
(8.326)
But we know that, where αt is the total angle of attack,
l × (i × l) = (1)2i − (l ⋅ i)l = i − (1) (1) cos α t l
(8.327)
α R = l × (i × l) = i − (cos α t )l
(8.328)
so that in terms of αt
For simplicity [1,2], if we choose the plane that l lies in to be the plane that j lies in as well,
we can write Equation 8.328 as
α R = i − cos α t (cos α ti + sin α t j) = (1 − cos 2α t ) i + sin α t cos α t j
then
α R = (1 − cos 2α t )2 + cos 2α t sin 2α t =
© 2014 by Taylor & Francis Group, LLC
sin 4α t + cos 2α t sin 2α t
Ballistics: Theory and Design of Guns and Ammunition
294
and
α R = sinα t sin 2α t + cos 2α t = sinα t
We shall now differentiate Equation 8.328 with respect to time:
dα R
di
dl
=
− (cos α t )
+ sin α t l
dt
dt
dt
(8.329)
We have made the assumption early in this analysis that the yaw of repose is relatively
constant, thus (dαR/dt) ≈ 0. We also note that for a small yaw angle sin αR ≈ 0 ≪ cos αR. If
we incorporate these approximations into Equation 8.329, we get
di
dl
= (cos α t )
dt
dt
(8.330)
Taking the time derivative of Equation 8.330 yields
d 2i
d2l
dl
α
cos
=
(
)
− (sin α t )
t
2
2
dt
dt
dt
(8.331)
But the small angle approximation still applies so that
d 2i
d2l
= (cos α t ) 2
2
dt
dt
(8.332)
We can also solve Equation 8.328 for i to get
i = α R + (cos α t )l
(8.333)
Since v and l are parallel and cross-products of parallel vectors are zero, we can write
i × v = [α R + (cos α t ) l] × v = α R × v + (cos α t ) l × v = α R × v
(8.334)
v × (i × v ) = v × (α R × v) = v 2 α R − (v ⋅ α R ) v = v 2α R
(8.335)
and
Also in a similar fashion operating on Equation 8.334, we can write
v × i = v × [α R + (cos α t ) l] = v × α R + v × (cos α t ) l = v × α R
(8.336)
We can also show that
i × (v × i) = v cos α tα R + v(sin 2 α t )l
© 2014 by Taylor & Francis Group, LLC
(8.337)
Trajectories
295
We now have relations in Equations 8.330, 8.332, and 8.333 for i, di/dt, and d2i/dt 2, respectively, and can substitute them into Equations 8.319 and 8.324 to eliminate i. We shall start
with Equation 8.319, and also noting that
v × α R = (vl) × α R = v(l × α R )
(8.338)
dV
ɶ Dv + C
ɶ L v 2α R + C
ɶ N v(l × α R ) + g + Λ
= −C
α
pα
dt
(8.339)
we get
It is also worth noting that
dV
d
dV
dl ɺ
dl
=
l+V
Vl =
= Vl + V
dt
dt
dt
dt
dt
(8.340)
We will now attack each term of Equation 8.324, but first we define
γ = (l ⋅ i) = cos α t
(8.341)
Then this and the succeeding relations follow as
dl
dl
I P di I P
I
≈ p cos α t
=γ P p
p
dt
I T dt I T
I T dt
(8.342)
d 2i
d2l 2 d2l
i × 2 = γ αR × 2 + γ l × 2
dt
dt
dt
(8.343)
dl 2 dl
di
i × dt = γ α R × dt + γ l × dt
(8.344)
Combining the terms of Equation 8.324, we get
γ
d2l ɶ
I P dl
d2l
2
ɶ
p + γ αR × 2 + γ 2 l × 2 = C
Mα v( l × α R ) + CM pα [vγα R + v sin α t l]
I T dt
d
t
d
t
ɶ M γ α R × dl + γ 2 l × dl
+C
q
dt
dt
(8.345)
At this point we continue with our simplifying assumptions and neglect the Coriolis term
in comparison with the gravitational acceleration term and also neglect the sin2 αt in comparison to γ. Thus, we can rewrite Equations 8.339 (including the relation of Equation 8.340)
and 8.345 as
dl
ɶD v + C
ɶL v 2α R + C
ɶ N v(l × α R ) + g
Vɺ l + V
= −C
α
pα
dt
© 2014 by Taylor & Francis Group, LLC
(8.346)
Ballistics: Theory and Design of Guns and Ammunition
296
and
γ
d 2l ɶ
I P dl
d 2l
ɶ
ɶ
p + γ αR × 2 + γ 2 l × 2 = C
Mα v( l × α R ) + CM pα vγα R + CM q
I T dt
dt
dt
dl 2 dl
γ α R × dt + γ l × dt
(8.347)
We shall now take the vector cross-product of l with Equations 8.346 and 8.347 and observe
how each term behaves. First the LHS
dl
dl
dl
= 0 +V l ×
l × Vɺ l + V = l × Vɺ l + l × V
dt
dt
dt
(8.348)
Then each term on the RHS of Equation 8.346
(
ɶ D v ) = l × (−C
ɶ Dvl) = −C
ɶ Dv(l × l) = 0
l × (−C
(8.349)
ɶL v 2α R = C
ɶL v 2 (l × α R )
l×C
α
α
(8.350)
)
ɶ N v (l × α R ) = −C
ɶ N v[l × (l × α R )] = C
ɶ N v[l × (α R × l)]
l × −C
pα
pα
pα
ɶ N v[α R − (l ⋅ α R )l]
=C
pα
(8.351)
l× g = l× g
(8.352)
and
Continuing on Equation 8.347, first the LHS
l×γ
I P dl
I dl
p
= γ P p l×
I T dt
I T dt
d 2l
d 2l
l × γ αR × 2 = γ l × αR × 2 = γ
dt
dt
d 2 l
d2l
l ⋅ 2 α R − ( l ⋅ α R ) 2
dt
dt
d2l
d 2 l d 2 l
d2l
l × γ 2 l × 2 = γ 2 l × l × 2 = γ 2 l ⋅ 2 l − 2
dt
dt dt
dt
(8.353)
(8.354)
(8.355)
Then each term on the RHS of Equation 8.347
ɶ M v(l × α R ) = C
ɶ M v[l × (l × α R )] = −C
ɶ M v[α R − (l ⋅ α R )l]
l×C
α
α
α
(8.356)
ɶ M vγα R = C
ɶ M vγ (l × α R )
l×C
pα
pα
(8.357)
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297
and continuing
ɶ M γ α R × dl = C
ɶ M γ l × α R × dl
l×C
q
q
d
dt
t
ɶ M γ α R × dl = C
ɶM γ
l×C
q
q
dt
dl
dl
dl ɶ
l ⋅ dt α R − (l ⋅ α R ) dt = CMq γ (l ⋅ α R ) dt
(8.358)
(8.359)
and finally
ɶ M γ 2 dl
ɶ M γ 2 l × dl = C
ɶ M γ 2 l × l × dl = C
ɶ M γ 2 0 − (l ⋅ l) dl = − C
l×C
q
q
q
q
dt
d
t
t
d
dt
(8.360)
We will now insert Equations 8.348 through 8.352 into Equation 8.346 to get
dl ɶ 2
ɶ
V l× = C
Lα v ( l × α R ) − CN pα v[α R − ( l ⋅ α R )l] + (l × g)
dt
(8.361)
We then do the same with Equations 8.353 through 8.360, inserting them into Equation
8.347 yielding
γ
d 2 l
d 2 l d 2 l
d2l
I P dl
p l × + γ l ⋅ 2 α R − ( l ⋅ α R ) 2 + γ 2 l ⋅ 2 l − 2
dt
I T dt
dt
dt dt
ɶ M γ (l ⋅ α R ) dl + γ 2 dl
ɶ M v[α R − (l ⋅ α R )l] + C
ɶ M vγ (l × α R ) − C
= −C
α
pα
q
dt
dt
(8.362)
We now have a pair of equations that essentially are comprised of two vector variables: the
yaw of repose, αR, and the vector, l × αR. Cumbersome as it may seem, this is a linear system
that can be solved readily through the use of matrices and their determinants to yield αR in
terms of l and the other scalars and coefficients. The solution for αR after much manipulation is
ɶ M vγ V l × dl − (l × g)
−C
pα
dt
2
2
ɶL v 2γ 2 l ⋅ d l l − d l + C
ɶL v 2γ I P p l × dl + C
ɶL C
ɶ M v 2γ 2 dl
+C
α
q
α
α
2
2
dt
I T dt
dt dt
αR =
2
ɶN C
ɶ M v 2γ − C
ɶL v 2γ l ⋅ d l + C
ɶL C
ɶM v3
C
pα
pα
α
α
α
2
d
t
(8.363)
We can further simplify this expression by returning to Equation 8.346 and taking the dot
product of it with l. Through vector algebra and the use of the fact that (l · αR) = 0, we see that
ɶ Dv + (l ⋅ g) = − ρ SCDv + (l ⋅ g)
Vɺ = −C
2m
2
© 2014 by Taylor & Francis Group, LLC
(8.364)
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298
When this is substituted back into Equation 8.346, we get
ɶ D v + (l ⋅ g)l + V
−C
dl
ɶD v + C
ɶL v 2α R + C
ɶ N v(l × α R ) + g
= −C
α
pα
dt
(8.365)
which can be rewritten as
V
dl ɶ 2
ɶ N v(l × α R ) + g − (l ⋅ g)l
= CLα v α R + C
pα
dt
(8.366)
and by neglecting the Magnus force term as it is small and noting that l × (g × l) = g −
(l · g)l, we get
V
dl ɶ 2
= CLα v α R + [ l × (g × l)]
dt
(8.367)
Remember that we wish to find a useful form with which we can calculate the quasisteady state yaw of repose as shown in Equation 8.363. That equation encompasses different vector functions and time derivatives of l, but our ultimate goal is to find expressions
that only involve the measurable quantities of the aeroballistic coefficients, spin, gravity,
and velocity and not the unit vector, l. To do this, though it may seem a devious process,
we begin by taking the time derivative of Equation 8.367, getting
dl
d2l
d
+ V 2 = 0 + [ l × (g × l)]
Vɺ
dt
dt
dt
(8.368)
We use Equation 8.364 and substitute it in Equation 8.368, arriving at
2
ɶ Dv dl + (l ⋅ g) dl + V d l = dg − (l ⋅ g) dl − dl ⋅ g l − l ⋅ dg l
−C
dt
dt
dt 2 dt
dt dt dt
(8.369)
Noticing that dg/dt = 0 (if not, we really will have problems), we can rewrite this as
V
d 2l
dl d l ɶ d l
= −2(l ⋅ g) − ⋅ g l + C
Dv
dt 2
dt dt
dt
(8.370)
If we examine the RHS of Equation 8.370 term by term and realize from Equation 8.367 that
dl/dt can be solved for, then
−2(l ⋅ g)
dl
2
ɶL v 2α R + (l ⋅ g)g − (l ⋅ g)2 l
= − (l ⋅ g)C
α
V
dt
(8.371)
also
1 ɶ 2
dl
2
2
⋅ g l = CLα v (α R ⋅ g) + g − (l ⋅ g) l
V
dt
© 2014 by Taylor & Francis Group, LLC
(8.372)
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299
and
ɶ
ɶ Dv dl = CDv C
ɶL v 2α R + g − (l ⋅ g)l
C
α
dt V
(8.373)
Putting these expressions back into Equation 8.370 complicates things considerably, viz.,
V
2 ɶ 2
1 ɶ 2
d2l
2
= − C
Lα v ( l ⋅ g )α R + ( l ⋅ g )g − ( l ⋅ g ) l −
V CLα v (α R ⋅ g)l
V
dt 2
ɶ Dv dl
+ g 2l − (l ⋅ g)2 l + C
dt
(8.374)
ɶL , and drag, CD̃ , coefficients
If we assume that the terms containing our modified lift, C
α
are either small with respect to the other terms or cancel one another, the third term in
Equation 8.374 disappears and the others simplify to give eventually
V2
d2l
= 3(l ⋅ g)2 − g 2 l − 2(l ⋅ g)g
dt 2
(8.375)
Recalling that the triad (l, m, n) are unit vectors, with l in the direction of the velocity vector, v, we can write l = v/v′ and can transform Equation 8.367 into
1
dl 1 ɶ 2
= CLα v α R + 2 [ v × (g × v )]
dt V
v
(8.376)
Likewise our Equation 8.374, by dividing both sides by V, can be transformed to
d2l
2
=− 2
2
dt
V
−
1
V2
1
1
ɶ
2
CLα v( v ⋅ g )α R + v (v ⋅ g) g − v 3 (v ⋅ g ) v
ɶ v dl
ɶ
C
g2
1
v − 3 (v ⋅ g )2 v + D
CLα v(α R ⋅ g )v +
v
v
V dt
(8.377)
Now, if we examine each term of Equation 8.363, we will have to deal with such terms
d2l d2l
dl
d2l
as l × , l ⋅ 2 , and l ⋅ 2 l − 2 . We can perform all the substitutions of these terms
dt
dt
dt dt
with what we have shown in Equations 8.376 and 8.377, and then examine the resulting
Equation 8.363 for terms that can be neglected for comparative magnitudes. When all the
algebra is completed, the resulting equation, applicable to spinning projectiles [1], is
dV
−2I P p v ×
dt
αR =
ρ Sdv 4CMα
© 2014 by Taylor & Francis Group, LLC
(8.378)
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Ballistics: Theory and Design of Guns and Ammunition
For nonspinning projectiles, we can simplify Equation 8.363 by removing the spin terms
and neglecting terms of small magnitude. The resulting expression is
αR =
ɶ M γ 2 [ v × ( v × g )]
C
q
ɶ M Vv 3
C
(8.379)
α
The vector mechanics work out so that when there is a positive overturning moment (statically unstable projectile), the yaw of repose vector points to the right for a right-hand spin.
The yaw of repose for a statically stable nonspinning projectile is such that the nose points
slightly above the trajectory. Either Equation 8.378 or 8.379 can be inserted into Equation
8.361 and numerically integrated simultaneously with Equation 8.362 to yield the velocity
and position at any time. This forms the basis of the modified point mass method.
Problem 40
The Paris gun was built by Germany in the First World War to shell Paris from 75 miles
away. The weapon was a 210 mm diameter bore with the shells pre-engraved to account
for wear of the tube. During firing of this weapon, all things such as wind effects, Coriolis,
etc., had to be accounted for. When the United States entered the war, the doughboys (the
nickname for American troops) were to take the St. Mihiel salient where the gun was
located. We shall assume that the Germans have turned the gun to fire on the Americans.
The projectile is at some point in space defined later. To demonstrate your knowledge of
the modified point mass equations
1. Draw the situation.
2. Calculate the vector yaw of repose for this projectile using Equation 8.326.
Answer: αR = [0.008e2 + 0.002e3] [rad].
3. Write the acceleration vector for this projectile using Equation 8.339 at the instant
in its trajectory when the velocity (relative to the ground) is 2100 ft/s and the conditions given next apply.
Note: You do not need all of the information given next. It is provided to you so you can
compare the differences in formulations with the 6-DOF model.
Answer:
dV
ft
= [ −61e1 − 29e2 − 10e3 ] 2
dt
s
4. Why we do not need to obtain the angular acceleration vector dh/dt?
Positional information:
48° north latitude
Azimuth of velocity vector: 190° True
Angle of velocity vector to horizontal: +1°
Wind is blowing at 15 mph due south and horizontal
α = 0.5°, β = 0.25°
The projectile nose is rotating down at 1 rad/s
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301
Projectile information:
CMα = 3.50
(C
(C
CLα = 2.50
CMpα = 0.55
m = 220 [lbm]
lbm
ρ = 0.060 3
ft
rev
p = 130
s
CD = 0.28
)
) = 0.005
Mq
+ CMαɺ = −16.5
I P = 19.13 [lbm-ft 2 ]
Nq
+ CNαɺ
I T = 66.40 [lbm-ft 2 ]
CNpα = −0.02
C1P = −0.01
Supply all answers in an inertial coordinate system labeled 1, 2, and 3 with 1 being due
south and 3 being due west.
Problem 41
If we were to use a modified point mass assumption for both of the cases cited in
Problem 35
1. Calculate the vector yaw of repose for both cases.
Answer: αr = [0.0003e1 + 0.0060e2 − 0.0009e3] for the building
αr = [−0.0009e1 − 0.0007e2 − 0.0000e3] for the gunship
2. Draw and explain what this vector represents.
3. Comment on whether this model is applicable for each case and why.
8.7 Probability of First Round Hit
The theories describing the characteristics of exterior ballistic trajectories that we have just
completed, while increasingly complex, do not deal with the problems of real-life gunnery.
What an artillery commander or a tank commander wants to know is what the probability
will be that the first round fired will land where desired. This is less of a problem for the
indirect fire mode, where a spotter can call in corrective advice to the gunner, than it is in
the direct fire mode, where the urgency of a first round hit may be a life and death matter.
When you fire your weapon at a visible enemy, you may be subject to immediate return
fire. So a first round hit with high probability of target destruction is of primary interest to
gunner and commander.
Predicted fire is what we are talking about, particularly tank gun versus tank target and
the variables of the combat situation are what govern these probabilities. The advent of
smart, even autonomous munitions, has, of course, reduced the problems of predicted fire,
but at great monetary expense and the subjection of such munitions to countermeasures
that negate their usefulness. The study of predicted fire is still a worthwhile exercise.
Firing real projectiles from real weapon systems involves interactions among many variables. For example, the location (terrain) of the launching platform; the condition of the
gun, its age, wear history, mounting method; aiming procedures including fire control;
condition of the crew including fatigue, experience, fear of attack; weather conditions,
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Ballistics: Theory and Design of Guns and Ammunition
including temperature, humidity, air density and pressure; target location relative to the
launching platform, target size, and not least, the characteristics of the projectile itself.
Leaving these generalities, let us examine two of the areas where hit probabilities are
generated: the proving ground and troop training locations. In the proving ground, great
care can be taken to reduce, and even eliminate, some of the variables. For the ammunition
designer, this is the opportunity to assess how precisely the design functions round to round:
the gunner can take all the time he needs to aim the weapon—no one is shooting back at him.
The weapon can be new or in the early part of its wear life; the propellant temperature can be
closely controlled, reducing temperature effects on muzzle velocity variations. In some cases,
special mounts can be built so that even large caliber cannon can be fired from what amounts
to a bench rest similar to ones used with small arms. The object of such care is to find out
the round-to-round precision (sometimes called dispersion) of the ammunition. This is not
its accuracy, but is an important component of what is known as “first round hit probability.”
The group (usually five or more projectiles made as identical as possible) fired under
such controlled conditions clusters about an average point called the “center of impact”
or CI. It is the predictability and reproducibility of the CI that assures the gunner and
tank commander that when the first round of an engagement is fired it will go where it is
intended. Rarely are second and third rounds fired at the same target in combat.
The normal course of a development program sees multiple cycles of design, manufacture of test samples, test and evaluation until a stable design is achieved. These cycles are
usually followed by a series of confirmatory tests that evaluate the performance of the
stable design under varying conditions of temperature, gun wear, range performance, and
a host of other tests. By normalizing the gun super-elevations of each group after firing, it
is possible then to compare the CIs of the groups. At the proving ground, the azimuth or
left-right setting of the gun would be closely maintained as a constant. This comparison
speaks to the accuracy of the design, that is, the reproducibility of the CIs under varying
conditions of proving ground firings. The precision (dispersion) of the design would also
be found under these variations.
How would this scatter of the CIs relate to accuracy under real-world combat situations?
What could we then say about the probability of first round hit (PFRH) under combat conditions? To answer these questions, serious studies have been conducted in probabilistic
terms about the conditions to be found around the world in combat in what are termed
“Quasi-combat” conditions [6]. Extensive computer programs have been devised, as well,
more recently, to compute PFRHs for designs of direct fire munitions under “Quasicombat” conditions [7]. Much of these theories and computations are dependent on the
vehicle’s fire control system and data collections of conditions of terrain and atmospherics
around the world. The original work in this area was done by H. Brodkin of the PittmanDunn Laboratory at the now defunct Frankford Arsenal in 1958 and remains classified.
As shown by Christman [5], the PFRH is dependent on certain fixed biases, variable
biases, and random errors. Hit probability computed this way is heavily dependent, as it
should be, on both the type of fire control system employed and the flight properties of
the projectiles being judged. For the predicted fire hit probabilities Christman computed
in 1959, laser range finding with its very small ranging errors was not in use. Neither
were the short times of flight of the fin-stabilized, small diameter but highly dense, armor
defeating projectiles (APFSDS). They were not yet invented.
The super-elevation of the gun, that is, the amount of elevation above the line of sight to
the target that the gunner must use to lay the weapon, is greatly reduced when both the
flight velocity is increased and the drag of the projectile is reduced with its consequently
shortened time-of-flight. These factors have greatly aided the improved PRFH with the
© 2014 by Taylor & Francis Group, LLC
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303
latest tank cannon and fire control systems now in use. It is nevertheless instructive to
show the error sources considered in a stationary-to stationary engagement of a tank to a
vertical 2.3 × 2.3 m target. Such computations are now carried out on computer programs
similar to those shown in [6].
Some error sources are tabulated in Tables 8.1 and 8.2 [5].
As can be seen from Tables 8.1 and 8.2 there are many error sources that affect PFRH in
a quasi-combat situation. Some of these require explanation. For example:
Zeroing: The weapon is zeroed by firing a small group usually at 1200 m range, in a noncombat situation, recording the center of impact, CI, of the group and adjusting the optical
fire control system so that it has a basis from which range-affected variables can be set.
Once zeroed, the other fixed and variable biases and random errors remain to determine
the PFRH at the time.
Cant: This error, caused by the unevenness of terrain, causes a tilt of the weapon’s trunnion
in both the roll and pitch attitudes and thus a consequent error in super-elevation (gun
elevation above boresight) dictated by the fire control.
Drift: This error is usually the result of the Coriolis effect, that is, the effect of the earth’s
rotation under the trajectory causing a deviation in the CI. This is dependent on the latitude of the firing and target sites as well as the hemisphere (northern vs. southern hemispheres reverse the direction of the deviation).
Fire control and parallax: These errors are the result of the slop in the mechanical linkages of
the controls and the small offset in the alignment of the optical sight with the axis of the
gun tube.
TABLE 8.1
Horizontal Errors on First Rounds—
Quasi-Combat Conditions
Fixed biases
Drift
Fire control
Jump, mean
Parallax
Variable biases
Cant
Fire control
Jump variation
Wind, cross
Zeroing
Cant
Fire control
Group center of impact
Jump variation
Observation of CI
Wind, cross
Random errors
Round to round Dispersion
Laying error
© 2014 by Taylor & Francis Group, LLC
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304
TABLE 8.2
Vertical Errors on First Rounds—Quasi Battle Conditions
Fixed biases
Fire control
Jump
Parallax (at 0 range)
Range estimation
Variable biases
Jump variation
Cant
Fire control
Muzzle velocity (lot to lot)
Range estimation
Zeroing
Fire control
Cant
Group CI
Jump
Muzzle velocity (lot to lot)
Observation of CI
Range estimation
Random errors
Round to round dispersion
Laying error
Cross wind: A variable bias usually affecting the firing of different groups on different occasions
Laying error: This is the result of random performance of gunners sighting in on targets
and laying the gun from different directions occasion-to-occasion.
Range estimation: This is a bias which is fixed for laser ranging but range dependent for
optical ranging.
Round-to-round dispersion: This is a random error dependent on the type of ammunition
being fired. It also can vary from lot-to-lot in the same ammunition type due to muzzle
velocity variations shot-to-shot in the lot.
Jump: This is what is left over in the error budget when all other errors are accounted for.
The test procedure of [7] meticulously lays out the testing procedures used in qualifying
ammunition for service use. Appendix A of this reference provides a nomogram for computing PFRH. It also provides techniques for measuring some of the error sources mentioned.
We must emphasize that this discussion only points out the complexity of the problem
of computing PFRH. If the model by Bunn [6] is used to run such computations, it requires
considerable detailed data input, which for modern tank, fire control, and ammunition
designs is not available to general audiences. Furthermore, recall that the discussion was
only for stationary-to-stationary situations where the opposing tanks are firing head on.
Targets moving must provide lead data. Firing from moving vehicles with stabilized turrets as in the U.S. M1 Abrams tank, adds further error sources inherent in the stabilization
system, but whose advanced computers and fire control systems may eliminate some of
the other errors.
© 2014 by Taylor & Francis Group, LLC
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References
1. McCoy, R.L., Modern Exterior Ballistics, Schiffer Military History, Atglen, PA, 1999.
2. Murphy, C.H., Free flight motion of symmetric missiles, Ballistics Research Laboratory Report
No. 1216, Aberdeen Proving Ground, MD, 1963.
3. Jentz, T.L., Tank Combat in North Africa, the Opening Rounds, Schiffer Military History, Atglen,
PA, 1998.
4. Litz, B., Applied Ballistics for Long Range Shooting, 2nd edn., Applied Ballistics, Cedar Springs,
MI, 2011.
5. Christman, E.C., The effect of system design characteristics on first round hit probability of tank
fired projectiles, Ballistics Research Laboratory Report No. 1192, Aberdeen Proving Ground,
MD, 1959 (DTIC AD 316221).
6. Bunn, F.L., The tank accuracy model, U.S. Army Research Laboratory Report No. ARL-MR-A8,
Aberdeen Proving Ground, MD, 1993 (DTIC AD 262812).
7. U.S. Army Armor and Engineer Board, Common Service Test Procedure, Round-to-Round
Dispersion, U.S. Army Armor and Engineer Board, Fort Knox, KY, 1970 (DTIC AD 872085).
Further Readings
Capecelatro, A. and Salzarulo, L., Quantitative Physics for Scientists and Engineers—Mechanics, Auric
Associates, Newark, NJ, 1980.
Sears, F.W., Zemanski, M.W., and Young, H.D., University Physics, 6th edn., Addison-Wesley, Reading,
MA, 1982.
© 2014 by Taylor & Francis Group, LLC
© 2014 by Taylor & Francis Group, LLC
9
Linearized Aeroballistics
The aeroballistics topics discussed so far have built up to where the reader has an appreciation for the techniques required to analyze projectile motion to a great degree of
accuracy. The culmination of this study was the development of the equations for a six
degree-of-freedom (6-DOF) model, which accurately describes the motion of a rigid body
through air. With a 6-DOF model in hand, the aeroballistician can examine the effects of a
given configuration. The word given was italicized for emphasis because the aeroballistician must know the configuration properties before he or she analyzes the projectile. The
implications of this are that without other tools to determine what needs to be changed
in a design to alter the projectile behavior, one must simply guess at a new configuration,
determine the aerodynamic coefficients, and reanalyze. This process can be very inefficient. The solution to this problem is to develop a theory that can be used to quickly determine what must be changed in a projectile to alter its flight behavior, make the changes,
and reassess. This will be the topic for the remainder of this section.
Linearized theory was (at least in the opinion of the authors) refined to an exceptional
degree by Murphy [1] in 1963. Other authors before and since [2–4] have developed similar
theories, and a good description of these can be found in McCoy [5]. Linearized theory is
so named because the aerodynamic coefficients are assumed to be linear functions of the
angle of attack. In other words,
Fj ∝ C j0 (sin α t ) or
M j ∝ C j0 d(sin α t )
(9.1)
where the subscript j indicates any parameter of interest as introduced earlier. There are
good points and bad points (as always) with this technique. The good news is that the
mathematics become simple enough to determine quantities of interest extremely quickly
and find means of changing a projectile’s flight characteristics quickly. The bad news is
that the use of linear coefficients prevents us from duplicating some motions that occur
frequently enough in projectile flight to warrant the inclusion of their nonlinear brethren—
and the math becomes complicated to boot.
We will continue the practice of using the definitions of the appropriate vectors and
scalars based on Ref. [5]. The choice is somewhat arbitrary, but for several years now the
authors have used this lucid work as a supplementary textbook and it is a matter of convenience. Our coordinate system is defined as in Figure 9.1.
The aerodynamic coefficients introduced in the beginning of this chapter were written
for both forces and moments as
Fj =
1
ρV 2SC j
2
(9.2)
Mj =
1
ρV 2SdC j
2
(9.3)
307
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308
y, j ≈ n
CL
Y, J
α
CN = CY
x, i
Cn
I
CD
Z, K
X, I
V, I
Cm
α
CX
Projection
of the
velocity
vector
onto the
x–z plane
β
z, k ≈ m
CZ
FIGURE 9.1
Coordinate system for projectile aerodynamic coefficients.
We have also defined the angular rates of the projectile as
p = Roll (spin) rate
(9.4)
q = Pitch rate
(9.5)
r = Yaw rate
(9.6)
The projectile angular position with respect to the velocity vector was given by
α = Angle of attack
(9.7)
β = Angle of sideslip
(9.8)
The aerodynamic coefficients are functions of the rates expressed in Equations 9.4 through
9.6, as well as angular positions expressed in Equations 9.7 and 9.8. Additionally, these
coefficients are also functions of the time rate of change of α and β that do not normally
coincide with q and r. Thus, we can write
C j = C j (α , β , αɺ , βɺ , p, q, r )
(9.9)
With this nomenclature, any coefficient can normally be expressed as a series expansion
in the seven variables
βɺ d
qd
pd
rd
αɺ d
C j = C j0 + C jα α + C jβ β + C jα
+ C j p + C jq + C jr + ⋯
+ C jβ
V
V
V
V
V
(9.10)
In Equation 9.10, we have included the terms in parentheses to maintain the nondimensional characteristics of the coefficient. We can see that this expansion results in a large
number of terms that must be carried. Seldom in aeroballistics do we require terms in
this expression beyond second order, but they can be included if data are available. When
we discuss linear aeroballistics, we are limiting ourselves to the eight terms displayed in
Equation 9.10.
© 2014 by Taylor & Francis Group, LLC
Linearized Aeroballistics
309
The linearization implies that
C jk =
∂C j
∂k
(9.11)
k =0
Further simplifications will be made as we progress which will assist us in tackling the
mathematics. We shall make use in this section of starred coefficients. These coefficients
are defined in terms of their unstarred counterparts as
C j*k =
ρ Sd
C jk
2m
(9.12)
9.1 Linearized Pitching and Yawing Motions
In the beginning of this chapter, we discussed terminology that allowed us to describe the
pitching and yawing motion of a projectile. Because of the symmetry of typical projectiles,
we combined pitch and yaw into a total yaw without any explanation. In this section, we
will discuss the two motions separately and then formally make the assumptions that
allowed us to combine them. This approach was formulated by Murphy [1] and what follows is basically that development with the coordinate system altered to fit our needs.
If we have a projectile as depicted in Figure 9.1 and allow it only to move in a truly pitching motion, we can look down the z-axis and we would see what is depicted in Figure 9.2.
Some interesting observations can be made from this figure. First, we see that the velocity vector, V, and the associated unit vector, l, are pitched up at angle ϕ to the earth-fixed
coordinate system. The projectile is actually pointed above this angle by the pitch angle α.
The vector along which the projectile is pointed is the geometric axis unit vector, x, and the
spin (principal) axis unit vector, i. In an axially symmetric projectile, these are identical.
We can see that through a rigid body rotation this forces the unit vectors of the transverse
geometric axis, y, and transverse principal axis, j, to be rotated from the earth-fixed Y-axis
through an angle of ϕ + α. If we assume that the projectile is constrained to pitch only, then
the time rate of change of this total angle is q and the yaw angle and yaw rate are equal to
zero as depicted in the figure.
d( + α)
=q
dt
β=0
y, j
r=0
Y, J
+α
x, i
α
V, l
X, I
Earth referenced
coordinate system
FIGURE 9.2
Projectile in a pure pitching motion.
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
310
d(θ + β)
=r
dt
α=0
q=0
X, I
θ
β
z, k
V, l
x, i
θ+β
Z, K
Earth referenced
coordinate system
FIGURE 9.3
Projectile in a pure yawing motion.
In a similar manner, we can constrain our projectile to motion in the yaw plane only,
which is depicted in Figure 9.3. In this case, the velocity and its associated unit vector are
yawed with respect to the earth-fixed coordinate system by angle θ. The projectile geometric
axis as well as the principal axis is yawed at angle β with respect to the velocity vector. This
results in a rotation of the transverse principle axis from the earth-fixed coordinate system
of θ + β as depicted in the illustration. The rate of change of this total angle is the yaw rate, r,
and because of our constraints there is no pitching motion as identified in the figure.
We shall now develop the equations of motion for each of these two specialized cases
with the purpose of combining them in the end. For the purpose of this development,
we shall define the force in the Y- and Z-directions using force coefficients CY and CZ,
respectively.
If we examine our projectile constrained to a pitching motion only, we can define the
force coefficient as
qd
αɺ d
CY = CY0 + CYα α + CYαɺ
+ CYq
V
V
(9.13)
where we have restricted ourselves to the linear coefficients. We can see that this pitching
motion causes a force in the Y-direction that is affected by angle of attack, rate of change of
angle of attack, and pitching rate. An item worthy of note is that for a perfectly symmetrical projectile, CY0 would be zero. It is included here for completeness and can be present if
an asymmetry exists.
The corresponding moment for pitching motion only is given by
qd
αɺ d
Cm = Cm0 + Cmα α + Cmαɺ
+ Cm q
V
V
(9.14)
where the same comments about the nondimensionalization and Cm0 apply as well.
Now we will examine the equations of motion. The force and moment equations are
given by
© 2014 by Taylor & Francis Group, LLC
F = ma
(9.15)
M = Iαɺ
(9.16)
Linearized Aeroballistics
311
If we define angles ϕ̂ and θ̂ as
φˆ = φ + α
(9.17)
θˆ = θ + β
(9.18)
then our scalar equations of a projectile in flight exhibiting pure pitching motion are
dVx
dV
≈m
= Fx
dt
dt
(9.19)
d 2Y
= FY cos φˆ + Fx sin φˆ − mg
dt 2
(9.20)
m
m
d 2φˆ MZ
=
dt 2
IZ
(9.21)
If we examine a small time of the projectile flight, we can assume constant velocity. If we
further limit the pitching motion to small angles, we can assume
Fx = − FD ≈ 0
(9.22)
cos φ̂ ≈ 1
(9.23)
sin φˆ ≈ φˆ
(9.24)
These assumptions can be used in Equations 9.19 and 9.20 to yield
dV
=0
dt
(9.25)
d 2Y
= FY − mg
dt 2
(9.26)
m
m
We know that
FY =
1
ρV 2SCY
2
(9.27)
If we then substitute Equations 9.13 and 9.27 into Equation 9.26, we obtain
m
d 2Y 1
= ρV 2S CY 0 + CYαα + CY αɺ
2
2
dt
© 2014 by Taylor & Francis Group, LLC
qd
αɺ d
+ CYq − mg
V
V
(9.28)
Ballistics: Theory and Design of Guns and Ammunition
312
or, using our definition of starred coefficients, we have
d 2Y V 2 *
qd
αɺ d
CY0 + CY*α α + CY*α
=
+ CY*q − g
2
d
dt
V
V
(9.29)
Equation 9.29 can be combined with Equation 9.21 to develop a single equation for projectile motion. With this, the dynamic equation for the pure pitching motion of a projectile
can then be described as
ˆ + Gˆ
ˆ 1α = A
αɺɺ + Hˆ 1dαɺ − M
1
d
(9.30)
This linear, second-order differential equation with constant coefficients was established
by Murphy [1] and modified here (the terms with the “d” subscript) to account for the
assumption of zero drag. In this expression, we identify the coefficients as follows:
)
(
1
* αɺ V
Ĥ1d = − CY*α + 2 Cm* q + Cm
d
k
Z
1 * V
M̂1 = 2 Cm
α
kZ
d
ˆ = 1 C* V
A
1
k 2 m0 d
Z
2
(9.32)
2
(9.33)
1 * g
Ĝd = − 2 Cm
q
kZ
d
k Z2 =
(9.31)
(9.34)
IZ
md 2
(9.35)
If we include drag (and thus ignore Equation 9.22) yet leave all of the other assumptions
in place, we obtain a result identical to Murphy [1]. This results in Equations 9.30, 9.31, and
9.34 being modified to
ˆ + Gˆ
ˆ 1α = A
αɺɺ + Hˆ 1αɺ − M
1
(
(9.36)
)
1
* q + Cm
* αɺ V
Ĥ1 = − CY*α + CD* + 2 Cm
d
kZ
(9.37)
1 *
* g
Ĝ = − 2 Cm
q − CD
d
kZ
(9.38)
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Linearized Aeroballistics
313
It is more convenient to examine the differential Equations 9.30 and 9.36 with dimensionless distance (defined as s/d) instead of time as the independent variable. The time derivatives of dimensionless distance can then be written as
ds V
=
dt d
(9.39)
d 2s Vɺ
=
dt 2 d
(9.40)
d
d ds V d
() = ()
= ()
dt
ds dt d ds
(9.41)
2
2
Vɺ d
d2
V d
()
=
() + ()
2
2
dt
d ds
d ds
(9.42)
and
With this, we can use the relations
to rewrite Equations 9.30 and 9.36, respectively, as
α ′′ + H1dα ′ − M1α = A1 + Gd
(9.43)
α ′′ + H1α ′ − M1α = A1 + G
(9.44)
The coefficients in these equations are given by
)
(
1
* q + Cm
* αɺ
H1d = − CY*α + 2 Cm
kY
(
(9.45)
)
1
* q + Cm
* αɺ
H1 = − CY*α + 2CD* + 2 Cm
k
Z
M1 =
1 *
Cmα
k Y2
(9.47)
A1 =
1 *
Cm 0
k Z2
(9.48)
1 * gd
Gd = − 2 Cm
q
2
kZ
V0
(9.49)
1 *
* gd2
G = − 2 Cm
q − CD
k
Z
V0
(9.50)
where V0 is the muzzle (or a reference) velocity of the projectile.
© 2014 by Taylor & Francis Group, LLC
(9.46)
Ballistics: Theory and Design of Guns and Ammunition
314
A similar procedure can be followed to define motion constrained to the yaw plane only.
This gives the result (details covered in Ref. [1]) of
β ′′ + H 2d β ′ − M2 β = A2
(9.51)
β ′′ + H 2 β ′ − M2 β = A2
(9.52)
The coefficients in these equations are given by
(
)
1
H 2d = − CZ*β + 2 Cn*r + Cn*βɺ
k
Z
(
)
1
H 2 = − CZ*β + 2CD* 2 Cn*r + Cn*βɺ
kY
(9.53)
(9.54)
M2 =
1 *
Cm β
k Y2
(9.55)
A2 =
1 *
Cn 0
k Y2
(9.56)
IY
md 2
(9.57)
k Y2 =
Assuming a projectile is axially symmetric implies that any plane orthogonal to the polar
axis is a principal axis. This forces the two transverse moments of inertia to be equal and,
with an assumption of small yaw, allows us to write
IT = Iy = Iz ≈ IY = IZ
(9.58)
This symmetry also allows us to equate the pitch and yaw coefficients. Thus, we define
CNα ≡ CYα = CZβ
(9.59)
C M q ≡ Cm q = C n r
(9.60)
CMα ≡ Cmα = Cn β
(9.61)
CMαɺ ≡ Cmαɺ = Cn βɺ
(9.62)
Complex numbers are commonly used to define pitch and yaw angles. This is extremely
convenient because it allows us to collapse two differential equations into one. We shall
define the complex yaw angle as ξ, which shall be defined thus
ξ ≡ α + iβ
© 2014 by Taylor & Francis Group, LLC
(9.63)
Linearized Aeroballistics
315
α
iβ
FIGURE 9.4
Complex yaw plane.
This definition allows one to look downrange as a projectile flies along a trajectory and
visualize the imaginary part of the equation affecting the yaw of the projectile and the real
part of the equation as affecting pitch. This is illustrated in Figure 9.4. In this figure, the
origin is the trajectory of the projectile looking downrange.
The two differential equations of motion Equations 9.44 and 9.52 can then be combined
by first multiplying Equation 9.52 by the imaginary number, i, and adding them together.
This results in
V
ξ ′′ + Hξ ′ − Mξ = A + G 0
V
2
(9.64)
The coefficients in this equation are given by
(
)
1
* q + CM
* αɺ
H = − CN* α + 2CD* + 2 CM
k
T
M=
A=
1
*α
CM
kT2
(
1
* 0 + iCn*0
Cm
kT2
(9.65)
(9.66)
)
gd
1 *
* 2
G = − 2 CM
q − CD
kZ
V0
(9.67)
(9.68)
The solution to Equation 9.64 can be found for a nonspinning projectile to be [1]
ξ = K1 exp(iψ 1 ) + K 2 exp(iψ 2 ) + K 3 exp(iψ 30 ) + ξ g
(9.69)
In this equation, each term Kj is known as an arm to be described subsequently.
Mathematically, we can express these terms as
K j = K j0 exp(λjs)
© 2014 by Taylor & Francis Group, LLC
(9.70)
316
Ballistics: Theory and Design of Guns and Ammunition
Here we see that each arm is a function of its initial value (that occurring at the muzzle
of the gun) and an exponential damping term. The exponential damping term decides
whether the amplitude of the motion will decay, grow, or remain constant. The damping
terms are given by [1,5]
1
λ1 = λ2 = − H
2
(9.71)
The exponential terms in Equation 9.69 contain phase angles, ψ j. These phase angles represent the instantaneous angle that each arm makes with the imaginary axis. These can be
written in terms of their initial value and a turning frequency as
ψ j = ψ j0 + ψ ′js
(9.72)
The turning frequencies are given for a nonspinning projectile by [1,5]
ψ 1′ = −ψ 2′ = − M
(9.73)
The third term on the RHS of Equation 9.69 is the so-called trim arm. This is a measure
of the amount that a fin-stabilized projectile will trim (i.e., fly with constant pitch or yaw)
during flight. It is given by
K 3 exp(iψ 30 ) = −
i(Cm0 + iCn0 )
CMα
(9.74)
The fourth term on the RHS of Equation 9.69 is the yaw caused by interaction of the projectile with the gravity vector, sometimes called the yaw of repose. It is defined as
gd
i CMq − kT2CD 2
V
ξg =
CMα
(
)
(9.75)
To visualize the physical meaning of Equation 9.69, we shall imagine we have a projectile and we are looking downrange along the trajectory such that the complex plane lies
perpendicular to the trajectory curve. Our projectile will be at some arbitrary yaw angle.
This is depicted in Figure 9.5. We need to note that the arms usually do not point to the
nose of the projectile, they point to the symmetry axis; however, it is easiest to visualize
the situation by scaling them to point to the nose. Imagine that we follow the projectile
depicted in Figure 9.5 as it traverses the trajectory. We would see the nose motion swirling
around. Throughout this time, we would also see the length of each of the arms changing
(growing, decaying, or remaining the same) as dictated by Equation 9.69. Additionally, we
would see the arms rotating around their respective origins at rates described by Equation
9.72. All through this time, our viewpoint would be changing because we have our gaze
fixed on the complex plane and it is rotating into the paper because of the curvature of the
trajectory.
In the development of Equation 9.64 and its solution Equation 9.69, the spin of the projectile was neglected. Because of this, these equations are specific to fin- or drag-stabilized
projectiles that have relatively small spin rates. References [1,2,5] develop the equation of
© 2014 by Taylor & Francis Group, LLC
Linearized Aeroballistics
317
α (pitch)
K2
g
Flight path
2
1
K3
K1
iβ (yaw)
FIGURE 9.5
Example of tricyclic arms.
motion for spinning projectiles in exactly the same manner. The results essentially incorporate the third angular component known as the roll or spin. The differential equation
for a spinning projectile is given by
ξ ′′ + ( H − iP)ξ ′ − ( M + iPT )ξ = −iPG
(9.76)
In this formulation, we can utilize axial symmetry and, thus, define our coefficients as follows:
H = CL*α − CD* −
M=
(
1
* q + CM
* αɺ
CM
kT2
)
1
*α
CM
kT2
T = CL*α +
G=
I
P= P
IT
1
* pα
CM
kP2
(9.77)
(9.78)
(9.79)
gd
V02
(9.80)
pd
V
(9.81)
The solution to Equation 9.76 is
ξ = K10 exp[λ 1s]exp[i(ψ 10 + ψ 1′s)] + K 20 exp[λ 2s] exp[i(ψ 20 + ψ 2′ s)] + ξ g
(9.82)
This equation is essentially the same form as Equation 9.69 except for the deletion of
the trim arm. It is also noteworthy that we have expanded the slow and fast arm terms
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
318
α (pitch)
΄
K1
2
΄1
K2
iβ (yaw)
g
FIGURE 9.6
Example of tricyclic arms for fin-stabilized projectile.
and exponents to display their exponential behavior. The expression is also commonly
written as
ξ = K1 exp(iψ 1 ) + K 2 exp(iψ 2 ) + ξ g
(9.83)
where the definitions of Equations 9.70 and 9.71 apply. The λj terms are known as the exponential damping coefficients and the ψ j terms are the precessional and nutation frequencies of the projectile. These are commonly defined as a complex pair where
λ1, 2 + iψ 1, 2 =
1
− H + iP ± 4 M + H 2 − P 2 + 2iP(2T − H )
2
(9.84)
As a parting note, we need to discuss the behavior of the fast and slow arms and the associated motion that they undergo. For a nonspinning projectile, we shall examine Equation
9.73. In this expression, the sign of M is important. For a nonspinning projectile M is negative. That tells us that the arms turn in opposite directions with K1 being positive (clockwise) and K 2 negative (counter-clockwise). This is depicted in Figure 9.6.
Likewise for a spinning projectile, we need to examine the derivative with respect to s of
Equation 9.84. In this case, we would find that
ψ 1′, 2 =
1(
P ± P2 − 4M )
2
(9.85)
Here we shall see in the following section that for stability this must result in a solution
that has no imaginary part. So both values of the root will have the same sign, thus the two
arms turn in the same direction as shown in Figure 9.7.
Initial conditions that are present when the projectile leaves the muzzle of the weapon
are important as our starting point for the values of the fast and slow arms. These can even
cause drastically different flight behavior when nonlinear coefficients are introduced later.
The initial sizes of the fast and slow arms can be expressed as functions of the precession
and nutation rates, the damping exponents, and the initial yaw and yaw rates [5] as
K10 exp(iψ 10 ) =
© 2014 by Taylor & Francis Group, LLC
ξ 0′ − ( λ2 + iψ 2′ ) ξ0
λ1 − λ2 + i (ψ 1′ −ψ 2′ )
(9.86)
Linearized Aeroballistics
319
α (pitch)
K1
K2
΄
1
΄
2
iβ (yaw)
g
FIGURE 9.7
Example of tricyclic arms for spin-stabilized projectiles.
K 20 exp(iψ 20 ) =
ξ 0′ − ( λ1 + iψ 1′ ) ξ 0
λ2 − λ1 + i (ψ 2′ −ψ 1′ )
(9.87)
Because the damping exponents are usually an order of magnitude or more smaller than
the precession and nutation rates, these equations can be simplified to
K10 exp(iψ 10 ) =
iξ 0′ + ψ 2′ξ 0
ψ 2′ −ψ 1′
(9.88)
K 20 exp(iψ 20 ) =
iξ 0′ + ψ 1′ξ 0
ψ 1′ −ψ 2′
(9.89)
These equations are important because they allow one to determine the initial amplitudes
of the arms given an assumed or measured initial yaw, yaw rate, and muzzle exit conditions for a known projectile geometry.
The expressions introduced in this section are the basis for stability criterion to be established next. In the next section, we shall discuss the behavior of these equations and use
them to define stability criteria for a projectile.
Problem 1
A 155-mm M549A1 projectile has the following properties and initial conditions:
CD = 0.3
CLα = 0.13
CMα = 4.28
Clp = −0.024
CMq + CMαɺ = −26
CMpα = 0.876
© 2014 by Taylor & Francis Group, LLC
lbm
ρ = 0.0751 3
ft
I P = 505.5 [lbm-in.2 ]
d = 155 [mm]
I T = 6610 [lbm-in.2 ]
ft
Vmuzzle = 3000
s
m = 96 [ibm]
320
Ballistics: Theory and Design of Guns and Ammunition
At an instant in time after launch when
p = 220 [Hz]
φ = δ = 4°
ft
V = 1764
s
Determine
1. The yaw of repose
Answer:
β R = 0.00172 [rad]
2. The precessional frequency in Hz
Answer:
dψ 2
= 1.9 [Hz]
dt
3. The nutational frequency in Hz
Answer:
dψ 1
= 14.9 [Hz]
dt
Problem 2
For the projectile given in Chapter 8, problem 37, determine the precessional and nutational frequencies in Hertz.
9.2 Gyroscopic and Dynamic Stabilities
In the previous section, we developed a pair of equations and their solutions using linear
aeroballistic coefficients that allow us to examine the motion of a projectile in pitch, yaw,
and roll. These equations will now be examined in detail so that we can establish criteria
for a stable projectile. In so doing, we will examine some interesting characteristics of
motion, which will be displayed as curves in the complex plane.
We shall repeat the equations and their solutions here for ease of reference but leave
the coefficient definitions in Section 9.1 to preserve space. The governing equations are as
follows:
For a nonspinning or slowly spinning projectile,
V
ξ ′′ + Hξ ′ − Mξ = A + G 0
V
© 2014 by Taylor & Francis Group, LLC
2
(9.90)
Linearized Aeroballistics
321
with the solution
ξ = K10 exp[λ1s]exp i (ψ 10 + ψ 1′s ) + K 20 exp[λ2s] exp i (ψ 20 + ψ 2′ s ) + K 3 exp[iψ 30 ] + ξ g
(9.91)
For a spinning projectile,
ξ ′′ + ( H − iP)ξ ′ − ( M + iPT )ξ = −iPG
(9.92)
ξ = K10 exp[λ1s] exp i (ψ 10 + ψ 1′s ) + K 20 exp[λ2s] exp i (ψ 20 + ψ 2′ s ) + ξ g
(9.93)
with the solution
For our general development of stability, we shall focus on Equation 9.92 and its solution
(Equation 9.93), since the trim term in Equation 9.91 can be easily dealt with separately.
If we examine Equation 9.93, we can readily see that nasty things can happen to us mathematically because of the exponential terms. Since K10 and K 20 are constants (they are the
initial magnitudes of the fast and slow arms, respectively), we can focus on the exponential terms that they are multiplied by as a means of determining whether they will grow,
shrink, or remain the same.
We shall consider the exponential functions of ψ and ψ′ first using the fast arm terms as
examples. The term ψ 10 is a constant and will be ignored. This leaves the term ψ 1s, which is
multiplied by i in the exponent. If ψ 1′ is purely real, then, when multiplied by i, it becomes
purely imaginary in the exponent (because s must be real), the solution is oscillatory and
this will cause the fast arm to increase and decrease in amplitude (i.e., oscillate), neither
increasing nor decreasing beyond the established limits of oscillation. This would be a
gyroscopically stable projectile. If it has an imaginary component, then, when multiplied
by i in the exponent, the solution has a real part. This real part will be multiplied by s and
continue to grow throughout the flight as s continually increases. This would result in a
gyroscopically unstable projectile.
The question to answer at this point is “What governs whether the exponents have real
or imaginary parts?” This can be answered by examination of a version of Equation 9.84,
whereby all aerodynamic forces and moments are ignored except for the largest (pitching)
moment. This has been shown [1,5] to result in a governing equation of
ξ ′′ − iPξ ′ − Mξ = −iPG
(9.94)
ξ = K1 exp i (ψ 10 + ψ 1′s ) + K 2 exp i (ψ 20 + ψ 2′ s ) + ξ g
(9.95)
with the solution
resulting in
ψ 1′, 2 =
(
1
P ± P2 − 4M
2
)
(9.96)
where the subscripts 1 and 2 represent the fast and slow arms, respectively.
Using Equation 9.96, we recall that for a gyroscopically stable projectile, ψ′ must be real,
therefore for gyroscopic stability, we require that
(P 2 − 4 M ) > 0
© 2014 by Taylor & Francis Group, LLC
(9.97)
322
Ballistics: Theory and Design of Guns and Ammunition
This expression has some interesting implications. If we look back at the definition of our
parameter, M in Equation 9.66, we see that it is dependent upon the pitching moment
coefficient. This happens to always be negative for a fin-stabilized projectile since the fins
impart a restoring moment. Unless there is some unique drag device, this moment is positive in a non-fin-stabilized projectile. Because of this, a fin-stabilized projectile is always
gyroscopically stable because P2 must be positive. However, a non-fin-stabilized projectile
must have a spin sufficient to make P2 > 4M. We therefore define a statically stable projectile as one in which M < 0. With this definition, a statically stable projectile is always
gyroscopically stable.
Gyroscopic stability is a necessary but not sufficient condition for a stable projectile.
The second condition required is that of dynamic stability. Let us once again examine
Equation 9.93, but this time we shall assume that we have a gyroscopically stable projectile. This means that the exponential terms containing ψ′ decay or remain constant,
leaving the terms containing λ as potentially destabilizing. We can readily see that, since
these are multiplied by the downrange distance, s, they must be negative to assure that
the fast and slow arms decay in magnitude. With this, we shall define a dynamically
stable projectile as one in which both λ’s are negative throughout the flight. Recall that
we calculate λ as the real part of Equation 9.84. For convenience, we shall express them
directly as
1
P(2T − H )
λ1, 2 = − H ∓
2
P2 − 4M
(9.98)
It should be noted here that, as is common in ballistics, there are always exceptions to any
rule. Some successful projectiles have been fielded where instability occurs for a very short
time in a flight or in a range where a certain projectile will never be fired. Of course, it is
always best to avoid these situations but sometimes lack of design space makes it unavoidable. In these instances, rational examination of the instability is necessary and should be
well documented.
We have shown mathematically how we define stability and the parameters that affect
stability. Sometimes, it is desirable to quantify how stable a projectile is. We do this
through use of a gyroscopic and dynamic stability factors. We define the gyroscopic
stability factor as
Sg =
P2
4M
(9.99)
Here, with our earlier discussion, Sg > 1 to assure gyroscopic stability. In a similar fashion,
we can define a dynamic stability factor as
Sd =
2T
H
(9.100)
where for a symmetric projectile to be deemed stable, whether spinning or nonspinning,
we require
1
< Sd (2 − Sd )
Sg
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(9.101)
Linearized Aeroballistics
323
For a statically stable projectile, we require that 0 < Sd < 2 for dynamic stability. This leads
to an interesting condition where one can spin a statically stable projectile too fast, resulting in instability. This condition translated to dimensionless spin rate is given by
P<
4M
Sd (2 − Sd )
(9.102)
for a statically stable projectile.
It is interesting to combine Equations 9.96 and 9.98 in various ways writing them in
terms of the dimensionless parameters P, M, H, and T. The details of this can be found in
Refs. [1,5] with the result
P = ψ 1′ + ψ 2′
(9.103)
M = ψ 1′ψ 2′ − λ1λ2
(9.104)
H = −(λ1 + λ 2)
(9.105)
PT = − (ψ 1′λ 1 + ψ 2′ λ 2 )
(9.106)
If we again examine Equation 9.91 or 9.93, we see that the magnitude of the precessional
and nutational arms is highly dependent upon initial conditions. Without going into
details (which are described quite well in Ref. [5]), we can express these initial conditions
in terms of the complex angle of attack and damping parameters as
K10 exp[iψ 10 ] =
ξ 0′ − ( λ2 + iψ 2′ ) ξ0
λ1 − λ2 + i (ψ 1′ −ψ 2′ )
(9.107)
K 20 exp[iψ 20 ] =
ξ 0′ − ( λ1 + iψ 1′ ) ξ 0
λ2 − λ1 + i (ψ 2′ −ψ 1′ )
(9.108)
In these equations, ξ0 and ξ 0′ are the initial complex yaw and yaw rates, respectively. These
parameters are determined by measurements as the projectile leaves the gun tube or are
assumed values.
We now have solid criteria by which we can determine whether a projectile will be
stable or not. These developments have been made assuming that the projectile aerodynamic coefficients behave in a linear fashion. As such, a projectile is either stable or it is
not. This stability, even with our linear model, will change during the flight based on
Mach number and angle of attack. We will discuss in a later section how a nonlinearity
can help or hurt matters. The true power of these equations is that they can tell us which
coefficients need to be altered to affect stability. This can be used in instances where we
want to change a physical configuration to make a projectile “drop out of the sky” or
design a round such that it damps more quickly and thus can fly with lower drag. Other
uses for these equations allow for tweaking the flight characteristics for better flight
behavior in general.
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324
Problem 3
Up until the late 1960s, many U.S. and foreign ships carried the Bofors 40 mm gun as a
general light support weapon. Originally designed as an antiaircraft weapon, this gun
served as an antitank weapon if the situation required it and its high rate of fire made it
quite successful as an antipersonnel weapon. Assume the properties of the system are
given next:
1. Calculate the gyroscopic stability factor at the beginning and at the end of the
flight assuming a terminal velocity of 2450 ft/s and the spin rate is 10% lower than
the initial value.
Answer: At the beginning of flight Sg = 5.814
2. Is the projectile stable throughout the flight?
Answer: Yes
3. Assuming that this is the longest time of flight for the projectile, at what spin rate
will the projectile become unstable?
Answer:
rad
punstable < 1887
s
4. Where will the instability occur?
Answer: At the muzzle of the weapon
Projectile and weapon information
lbm
ρ = 0.067 3
ft
CMα = 3.10
Clp = − 0.011
d = 40 [mm]
ft
Vmuzzle = 2850
s
I P = 1.231 [lbm-in.2 ]
I T = 6.263 [lbm-in.2 ]
m = 1.985 [ibm]
n=
1 rev
30 cal
Please note that this weapon actually has a progressive twist, but when faced with this
situation you only need the muzzle velocity and the twist at the muzzle to calculate initial
spin.
Problem 4
For the projectile described in Problem 35 of Chapter 8,
1. Determine the precessional damping exponent.
Answer: λ2 = −0.0000608
2. Determine the nutational damping exponent.
Answer: λ1 = −0.001135
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325
3. With (1) and (2) above which of these modes will damp out?
Answer: Both
4. Determine the dynamic stability factor, Sd.
Answer: Sd = 0.74
5. Consider a cargo projectile with identical properties to our projectile in Chapter 8,
Problem 35. The designer did not secure the cargo well enough so that the cargo
fails to spin up completely during gun launch in a worn tube. When this happens,
immediately after muzzle exit, the round spins down (and the cargo spins up a
little more) so that the projectile finally reaches a spin rate of 100 Hz. The velocity
is unaffected.
a. Determine the gyroscopic stability factor for each of the two situations.
Answer: Sg = 3.192 and Sg = 0.659.
b. Will both projectiles fly properly? Why or why not?
Answer: No, the second projectile will tumble.
Problem 5
A 155 mm HE projectile is fired from a cannon. The muzzle velocity of the projectile is 800
m/s and the twist of the rifling is 1:20. The projectile and filler properties are given next.
Assuming the aerodynamic forces and moments are negligible and that the projectile is
dynamically stable:
1. The initial spin rate of the complete projectile.
Answer:
rad
pmuzzle = 1621.5
s
2. The spin rate of the projectile in flight assuming the fill does not spin up in the
bore and both shell and fill come into dynamic equilibrium.
Answer:
rad
ptotal = 1259.2
s
3. Determine the gyroscopic stability factors for (1) and (2).
Answer:
Sg = 16.54 and Sg = 9.97
4. Is the projectile stable in (1) and (2)?
Answer: Yes to both.
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326
Projectile and weapon information
CMα = 1.07
Clp = − 0.012
I Pshell = 431 [lbm-in.2 ]
I Pfill = 124 [lbm-in.2 ]
lbm
ρ = 0.067 3
ft
I P total = 555 [lbm-in.2 ]
d = 155 [mm]
m = 106 [lbm]
m
Vmuzzle = 800
s
n=
I Ttotal = 3335 [lbm-in.2 ]
1 rev
20 cal
Problem 6
For the projectile given in Problem 5, determine the nutation and precession frequencies
in Hertz.
Answer:
dψ 1
= 42.05 [Hz] and
dt
dψ 2
= 0.67 [Hz]
dt
Problem 7
Assume the projectile in Problem 1 has slipped its rotating band and the spin at the same
instant in time is 130 Hz. Is the projectile stable?
Answer: No.
Problem 8
What is the minimum spin (Hz) required to stabilize the projectile in Problem 1?
Answer: pmin = 140 [Hz].
Problem 9
For the projectile given in Chapter 8, Problem 37, determine the minimum spin rate for
stability. (Hint: remember when a projectile is least stable.)
Problem 10
A right circular cylinder is to be fired horizontally for an impact test. If the cylinder is
made of steel (ρ = 0.283 lbm/in.3) and it is 0.5 in. in diameter and 0.75 in. long, determine
a. The spin rate required to stabilize the projectile (if it can be stabilized)
b. Comment on the answer above—what is dominant in the problem?
c. The precessional and nutational frequencies of the projectile at this spin rate in Hz
The projectile properties are provided next:
CD = 0.4
CLα = 0.18
lbm
ρair = 0.0751 3
ft
CMα = 8.0
d = 0.5 [in]
Clp = −0.02
l = 0.75 [in]
CMq + CMαɺ = −29
ft
Vmuzzle = 6000
s
CMpα = 0.92
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Linearized Aeroballistics
327
Problem 11
Modifications are made to a 155 mm M483A1 projectile so that it has the following properties and initial conditions:
CD = 0.2
CLα = 1.975
CMα = 4.573
Clp = −0.0285
CMq + CMαɺ = −15.2
lbm
ρ = 0.0751 3
ft
I P = 537.1 [lbm-in.2 ]
d = 155 [mm]
I T = 5753 [lbm-in.2 ]
ft
Vmuzzle = 2900
s
m = 103 [lbm]
CMpα = 1.20
At an instant in time after launch when
p = 100 [Hz]
φ = δ = 2°
ft
V = 1000
s
determine
a. If the projectile is stable
b. The precessional frequency in Hz
c. The nutational frequency in Hz
Problem 12
What is the minimum spin (Hz) required to stabilize the projectile in Problem 11?
Problem 13
For the projectile given in Chapter 8, Problem 25, determine the precessional and nutational
frequencies in Hertz. Determine the minimum spin rate for the projectile to be stable.
Problem 14
For the projectile given in Chapter 8, Problem 27, determine the precessional and nutational
frequencies in Hertz. Determine the minimum spin rate for the projectile to be stable.
Problem 15
For the projectile given in Chapter 8, Problem 28, determine the precessional and nutational frequencies in Hertz. Determine the minimum spin rate for the projectile to be
stable. Does it differ depending on which side of the aircraft it is fired from? Calculate it
for both cases.
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328
9.3 Yaw of Repose
In Section 9.1, we introduced the yaw of repose for a projectile and defined it in Equation
9.75 using the symbol ξg. The subscript “g” was used to denote that this quantity comes
about through the action of gravity on the projectile. In terms of our dimensionless parameters, we can rewrite Equation 9.75 as
ξg =
PG
M + iPT
(9.109)
A qualitative look at this expression leads to some extremely interesting results. First and
foremost is that the spin rate directly affects the yaw. The greater the spin (and therefore
the larger the value of P), the greater the yaw of repose is.
The second useful item to note is that the more abrupt the trajectory curve is, the greater the
yaw of repose is. In fact, if we look at the term G, it is linear in the cosine of angle of attack, ϕ.
For details of this form that G takes, the reader is referred to Ref. [5]. Thus, when the projectile
approaches maximum ordinate, the yaw of repose should be a maximum given all of the other
parameters remain constant. Because of the decay of the other terms, the result is that the yaw
of repose is usually a maximum shortly before or after reaching maximum ordinate.
The sign of the yaw of repose is important. In our convention, the term P is positive for a
right-hand twist. Thus, a positive value of ξg causes the projectile to nose over to the right.
Note that there can also be a significant pitch component to this quantity; this is easily seen
as the real part of Equation 9.109.
If we examine a plot of pitch (α) versus yaw (β) for a British 14-in. projectile in Figure 9.8,
we can imagine the yaw of repose as the vector pointing to the right (viewed from the rear)
to the center of the precessional path similar to Figure 9.7. We can see that the magnitude
as well as the direction of this vector change as the projectile moves downrange. In Figure 9.8,
the projectile was analyzed using the PRODAS software and was fired with a muzzle
velocity of 2483 ft/s, spin rate of 71 Hz corresponding to a 1:30 twist with an initial pitch
α versus β
0.18
0.15
0.12
0.09
α (deg)
0.06
0.03
0.00
–0.03
–0.06
–0.09
–0.12
0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 0.22 0.24 0.26 0.28 0.30
β (deg)
FIGURE 9.8
Pitching and yawing motion for a British 14-in. Mk.I projectile fired at 2483 ft/s with a 0.1° initial pitch angle.
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Linearized Aeroballistics
329
angle of 0.1°. There was no initial yaw or pitch/yaw rate. This projectile has progressed
through only one and one-half yaw cycles (about 1.7 s) when the analysis was stopped to
yield a nice clear illustration.
9.4 Roll Resonance
Until this point, we have assumed that the projectiles under study have been axially symmetric. This rarely happens in practice because of manufacturing tolerances in a given
projectile design. In Chapter 10, we shall discuss the means of handling a slight mass
asymmetry. In this section, we shall discuss the implications of a geometric (including
slight mass) asymmetry as applied to a fin-stabilized projectile.
Fin asymmetries commonly occur when a finned projectile is manufactured or can be
the result of damage owing to rough handling. In the field of explosively formed penetrators which are normally drag- or fin-stabilized, inconsistencies can (and usually do) arise
due to the explosive formation process. In either case, this effect may be coupled with some
mass asymmetry as well.
In Equation 9.74, the trim arm was introduced, which would force a statically stable
projectile to fly with an angle of attack. It is for this reason that all fin- and drag-stabilized
projectiles are designed to roll slightly to increase accuracy. One can see from the way that
this equation was written there is no change in the orientation of K30. It was fixed, oriented
at the initial angle ψ 30.
To begin our assessment of this specific type of asymmetry, we shall start with the governing equation for a spin-stabilized projectile (Equation 9.76) because the roll is going to
play a part. We shall alter the RHS to incorporate a forcing term representing the lifting
force and moment that is caused by the asymmetry (say, e.g., a bent fin). We shall write this
in such a way that the direction of the applied force and moment rotates with the projectile:
ξ ′′ + ( H − iP)ξ ′ − ( M + iPT )ξ = −iA3 exp(iψ )
(9.110)
ρ Sd 1
A3 =
2 ( Cm0 + iCn0 ) + (ψ ′ − P) ( CZ0 + iCY0 )
2m k T
(9.111)
where
ψ′ =
pd
, dimensionless turning rate
V
(9.112)
s
∫
ψ = ψ ′ds, dimensionless distance
(9.113)
0
This development was put forth in Refs. [1,4,5]. If we look closely at these equations, we see
that the forcing function, A3, rotates with the projectile.
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330
If we solve Equation 9.110, assuming a solution for the particular part of
ξ p = K 3 exp [ i(ψ + ψ 0 )]
(9.114)
where ψ 0 is some arbitrary angle that contains the plane of the asymmetry, we obtain a
general solution for a constant roll rate of
ξ = K1 exp[iψ 1 ] + K 2 exp[iψ 2 ] + K 3 exp[i(ψ + ψ 0 )]
(9.115)
and, after inserting the initial conditions, say, of ψ 0 = 0, we obtain
K3 =
−iA3
ψ ′2 − Pψ ′ + M − i(ψ ′H − PT )
(9.116)
This is the expression for the yaw component caused by a lift force and corresponding
moment constrained to rotate at the projectile spin rate. If the spin rate is zero, the orientation of this lift force will be fixed and the projectile will drift more and more in that direction. This is not desirable from an accuracy standpoint so we must have some spin.
The denominator in Equation 9.116 is normally dominated by its real part because H and
the product PT are small by comparison. However, much like a resonance in a spring-mass
system, if the roll frequency ever approaches either one of the precession or nutation frequencies (and remains there for some time), the denominator in Equation 9.114 approaches
zero and the yaw becomes very large [1,5]. This usually occurs when the nutational frequency is approached and is called roll resonance or spin-pitch resonance [1]. Since projectiles are usually changing spin rate throughout their flight, this is only a problem if there
is a slow change of spin rate when the frequencies are close.
Another way of looking at this is to imagine a projectile where this asymmetry is present. Since the asymmetry is at the same frequency as the nutation rate, every time the
projectile is at the outer limit of its motion it gets kicked a little further, similar to pushing a child on a swing. This disturbance grows as long as the two motions stay coupled
(i.e., at the same frequency); however, if they became out of phase, the problem would
correct itself.
1000
600
400
200
0
–200
0
100
200
300
400
500
600
Attitude (deg)
Spin rate (rad/s)
800
–400
90
80
70
60
50
40
30
20
10
0
Tumbled
0
(a)
Range (ft)
(b)
100
200
300
400
500
600
Range (ft)
FIGURE 9.9
Explosively formed penetrator experiencing roll resonance. (Courtesy of Eric Volkmann, Alliant Techsystems,
Hopkins, MN.)
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Linearized Aeroballistics
331
An example of roll resonance is depicted in Figure 9.9. In this case, an explosively formed
penetrator (EFP) was the device under test. Keep in mind that only total angle of attack is
measured here so the yawing motion is not constrained to a single plane. We see that as
the EFP approached a spin rate of ∼300 rad/s it locked in and flew very far off of the target.
Problem 16
Roll resonance of a projectile occurs when the spin rate approaches a “forcing” frequency
of the projectile. This occurs more frequently in fin-stabilized projectiles than spin-stabilized
projectiles because, in the latter, the spin rate is usually quite high in order to maintain
stability, the overturning moment is positive and these forcing functions unless they are
intentional—like thrusters are usually due to asymmetries (like bent fins) usually are
small. If we examine the projectile of Chapter 8, Problem 37, instead as a fin-stabilized
projectile, we can write the equation for the pointing direction as
ξ = K1 exp[iψ 1 ] + K 2 exp[iψ 2 ] + K 3 exp[i(ψ + ψ 0 )]
where the subscripts 1 and 2 represent the fast and slow modes, respectively. The third
term is the forcing function where we can define
ψ′ =
pd
V
(1)
and, after inserting the initial conditions, say, of ψ 0 = 0, we obtain
K3 =
−iA3
ψ ′2 − Pψ ′ + M − i(ψ ′H − PT )
(2)
Based on what you know about the behavior of imaginary numbers and expressions (1)
and (2), determine the spin rate at which catastrophic yaw will occur. Use the aerodynamic
properties of the projectile from Chapter 8, Problem 37, but assume that the overturning
moment is the negative of what was provided (we are essentially faking a fin-stabilized
version). You may assume the velocity stays constant at 750 ft/s.
References
1. Murphy, C.H., Free Flight Motion of Symmetric Missiles, Ballistics Research Laboratory Report
No. 1216, Aberdeen Proving Ground, MD, 1963.
2. Vaughn, H., A detailed development of the tricyclic theory, Sandia National Laboratories
Report No. SC-M-67–2933, Albuquerque, NM, February 1968.
3. McShane, E.J., Kelley, J.L., and Reno, F.V., Exterior Ballistics, University of Denver Press, Denver,
CO, 1953.
4. Nicolaides, J.D., On the free flight motion of missiles having slight configurational asymmetries, Ballistics Research Laboratory Report No. 858, Aberdeen Proving Ground, MD, 1953.
5. McCoy, R.L., Modern Exterior Ballistics, Schiffer Military History, Atglen, PA, 1999.
© 2014 by Taylor & Francis Group, LLC
© 2014 by Taylor & Francis Group, LLC
10
Mass Asymmetries
Until this point we have assumed that the projectile has been an axially symmetric body.
This allowed us to simplify the equations of motion considerably. Projectiles are rarely axially symmetric. The asymmetry usually comes about through manufacturing tolerances,
damage due to rough handling, cargo slippage or, more recently, they are simply designed
that way. The purpose of this section is simply to introduce the geometry of mass asymmetries, which will be introduced into the equations of motion for the projectile in later
sections.
Mass asymmetries come in two categories: static imbalance and dynamic imbalance.
In a static imbalance, the center of gravity (CG) of the projectile is not located on the geometric axis of symmetry. The geometric axis of symmetry can be defined by imagining a
projectile with the same exterior dimensions as the unbalanced projectile but of uniform
density. The symmetry axis would then be centrally located in the body of revolution (i.e.,
a perfectly axially symmetric body). In a statically imbalanced projectile, this axis would
be shifted to pass through the CG but remain parallel to the geometric axis. This is illustrated in Figure 10.1.
A dynamically imbalanced projectile also has a CG that is offset from the geometric
axis of symmetry. In this case, however, the mass distribution is such that the principal
axis of inertia resides as some angle to the geometric axis as well. This is illustrated in
Figure 10.2.
Whether a projectile is statically or dynamically imbalanced, we shall define the
plane in which the CG offset is located relative to some reference plane (we shall arbitrarily use the x–y plane as the reference, which we have defined in earlier sections)
using the symbol Φ. This is illustrated in Figure 10.3 as viewed from the rear of the
projectile.
The effect of these mass asymmetries on projectile flight can dramatically affect accuracy, especially in direct fire systems. Consider a projectile with an imbalance in the gun
tube. While in the tube, the projectile is constrained to rotate about the tube geometric
axis. If we idealize this situation to say that the tube is perfectly straight, inflexible, and fits
the projectile snugly, we can further state that the projectile is constrained to rotate about
its own geometric axis. Note that there is a wealth of literature dedicated to the real situation (e.g., [1–10]).
333
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Ballistics: Theory and Design of Guns and Ammunition
334
CG
Principal axis of inertia
ε
Geometric axis
FIGURE 10.1
Statically imbalanced projectile.
Principal axis of inertia
CG
Φ
ε
Geometric axis
FIGURE 10.2
Dynamically imbalanced projectile.
Φ
CG
ε
Geometric axis
FIGURE 10.3
Center of gravity (CG) offset viewed from rear of projectile.
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Mass Asymmetries
335
References
1. Berger, M.P., Position and form of bands for projectiles, Memoires Militaires et Scientifiques, Publies
par le Department de la Marine, Translated by LT C.C. Morrison, Notes on the Construction of
Ordnance, Washington, DC, June 10, 1884.
2. Kent, R.H. and Hitchcock, H.P., Comparison of Predicted and Observed Yaw in Front of the Muzzle
of a 12′ Gun, Report No. 990 AD-116-140, USA BRL, Aberdeen Proving Ground, MD, July 1956.
3. Kent, R.H. and McShane, E.J., An Elementary Treatment of the Motion of a Spinning Projectile about
Its Center of Gravity, Report No. 459 AD-491-943, USA BRL, Aberdeen Proving Ground, MD,
April 1944.
4. Heppner, L.D., Setback and Spin for Artillery, Mortar, Recoilless Rifle and Tank Ammunition, Report
No. DPS-2611, USA BRL, Aberdeen Proving Ground, MD, January 1968.
5. Gay, H.P. and Elder, A.S., The Lateral Motion of a Tank Gun and Its Effect on the Accuracy of Fire,
Report No. 1070 AD-217-657, USA BRL, Aberdeen Proving Ground, MD, March 1959.
6. Kirkendall, R.D., The Yawing Motions of Projectiles in the Bore, Technical Note No. 1739 AD-878327-L, USA BRL, Aberdeen Proving Ground, MD, September 1970.
7. Zaroodny, S.J., On Jump due to Muzzle Disturbances, Report No. 703 AD-805-876, USA BRL,
Aberdeen Proving Ground, MD, June 1949.
8. Gay, H.P., On the Motion of a Projectile as It Leaves the Muzzle, Technical Note No. 1425 AD-801974, USA BRL, Aberdeen Proving Ground, MD, August 1961.
9. Sterne, T.E., On Jump due to Bore Clearance, Report No. 491 AD-491-938, USA BRL, Aberdeen
Proving Ground, MD, September 1944.
10. Line, L.E., The Erosion of Guns at the Muzzle, NRDC Report No. A-357, OSRD Report No.
6322, National Defense Research Committee, Office of Scientific Research and Development,
Washington, DC, November 1945.
© 2014 by Taylor & Francis Group, LLC
© 2014 by Taylor & Francis Group, LLC
11
Lateral Throwoff
Earlier in the text, we stated that projectiles rarely leave the tube with their velocity vectors aligned with the geometric axis of the gun tube. This chapter and Chapter 12 describe
this behavior. The result of this behavior is weapon inaccuracy and it must be well understood by the practicing ballistician because, although it is not practical to completely eliminate the behavior, we would like to reduce it to acceptable levels. The first component of
this behavior is known as lateral throwoff. It is a dynamic response of the projectile to
either a static or a dynamic imbalance and will now be described in detail.
If we imagine a projectile with a mass asymmetry as depicted in Figure 10.3, we can
imagine the spinning motion as viewed from the rear. If we ignore the axial velocity by
simply spinning the projectile at a high rate, say, between two flexible supports on a test
stand, we would see a wobble develop as a result of the centrifugal action on the center of
mass. All the time the projectile is being spun up in the gun, the tube walls and stiffness
of the supporting members prevent this wobble (to the extent the clearances allow) from
developing. At the instant, the projectile is free from the constraints of the tube we expect
it to become affected by this centrifugal loading. This is lateral throwoff because the effect
is to fling the projectile in a direction off the tube centerline.
We can use the analogy of a vacuum trajectory to examine the lateral throwoff effect
generated by either a static or a dynamic imbalance. Consider the projectile asymmetry
from Figure 10.3. If we examine the projectile over a short period of flight, ignoring gravity as well as assuming no drag because of the vacuum assumption, we would see the
dynamic forces acting on the projectile as depicted in Figure 11.1. In this figure, the only
force acting is the centrifugal force due to spin. This dynamic action will result in the force
vector changing direction, though since there is no angular acceleration or deceleration it
maintains a constant magnitude. It is worth noting that we have resorted to our complex
plane in this example as it is convenient to use in our development. At the instant, in time
depicted here, we can break the force into a component in the y-direction and one in the
iz-direction.
We are not necessarily concerned with the force acting on the CG per se. We want to
see where the projectile moves because of this force. To accomplish this, we need to use
Newton’s second law. We know that
Fr = mar
(11.1)
This is the centripetal force. The centrifugal force would be equal but opposite in sign.
From dynamics [1], we recall that
ar = −rp 2
(11.2)
In the case we are considering here, we see that
r =ε
and Φ = pt
(11.3)
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Ballistics: Theory and Design of Guns and Ammunition
338
y
Fr [i sin(Φ)]
Φ
Fr [cos(Φ)]
Fr
iz
Center of gravity
ε
Geometric axis
FIGURE 11.1
Dynamic force acting on a statically or dynamically imbalanced projectile.
With this, we can write the magnitude of the force as
Fr = mε p 2
(11.4)
and the centripetal acceleration in the complex plane as
a=−
Fr
[cos( pt) + i sin( pt)] = −ε p 2 [cos( pt) + i sin( pt)]
m
(11.5)
The complex velocity can therefore be expressed as
t
∫
V = −ε p 2 [cos( pt) + i sin( pt)]dt
(11.6)
0
Evaluating the integral and assuming that as the projectile leaves the muzzle we have an
initial orientation of the mass asymmetry of Φ = Φ 0 yields
V = −ε p[sin( pt + Φ0 ) − i cos( pt + Φ0 )] = ε p[− sin( pt + Φ0 ) + i cos( pt + Φ0 )]
(11.7)
To see how much lateral movement has developed, we can integrate again
t
∫
r = ε p [− sin( pt + Φ0 ) + i cos( pt + Φ0 )] dt
0
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(11.8)
Lateral Throwoff
339
The evaluation of which yields
r = ε [cos( pt + Φ 0) + i sin( pt + Φ 0)]
(11.9)
As an example, if we were only concerned with motion in the crossrange direction, we
could state
z = Im{ε [cos( pt + Φ 0) + i sin( pt + Φ 0)]} = ε sin( pt + Φ 0)
(11.10)
To apply numbers to this example, let us consider a projectile that weighs 100 lbm and is
spinning at a rate of 270 Hz. We shall assume the projectile has a CG offset of 0.25 in. If this
were the case, the velocity in the z-direction as well as the motion for the first 4 s of flight
can be seen in Figures 11.2 and 11.3. Here we have assumed that the CG offset has emerged
from the weapon at the 12 o’clock position.
0.2
0.196
Im (Vr(t))
0.15
0.1
0.05
0
0
1
0
0
2
t
3
4
4
FIGURE 11.2
Velocity in the z-direction of a 100-lbm projectile spinning at 270 Hz with a 0.25 in. CG offset.
0.4
z (ft)
0.3
0.2
0.1
0
0
1
2
3
4
t (s)
FIGURE 11.3
Displacement in the z-direction of a 100 lbm projectile spinning at 270 Hz with a 0.25 in. CG offset.
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
340
The most interesting observation between the figures is that for this arbitrary emergence of the CG offset, we see that the projectile would like to move laterally to the
right for a right-hand spin. This is commonly known as drift. Just to put things into
perspective, the muzzle velocity consistent with the 270 Hz spin rate is about 2,750 ft/s
so the projectile would only have gone about 0.4 ft to the right after it traversed 11,000 ft
downrange.
We must always bear in mind that this example was an idealized situation. In the case
of a real projectile, there are other forces acting which complicate the motion; however, it is
instructive to look at simplifications such as this to see the phenomenon at work. We will
now move on to examine the dynamic behavior in terms of the equations of motion of a
projectile from statically imbalanced and dynamically imbalanced projectiles. We shall
see how this affects lateral throwoff.
11.1 Static Imbalance
In Figure 10.1, we saw the effect on the principal axis of a static imbalance. Although this
rarely happens in production (imbalances are usually of the dynamic type) it can happen
and presents an interesting case. We shall follow the analysis procedure documented by
McCoy [2] in the development, correcting terms to fit our coordinate system.
If we examine the velocity of the center of mass of the projectile as it leaves the gun
tube, we see a scene as depicted in Figure 11.4. If the projectile is constrained as it continues down the gun tube, the motion of the CG would resemble a spiral or helix similar
to a thread on a bolt, except that the pitch of the helix would continue to increase as the
axial velocity increases. We could express this mathematically using a cylindrical set of
coordinates with x indicating the axial distance, r indicating the radius of the CG from the
Φ
Center of gravity
VΦ
ε
Geometric axis
FIGURE 11.4
Velocity of a statically imbalanced projectile’s CG.
© 2014 by Taylor & Francis Group, LLC
Lateral Throwoff
341
centerline, and Φ indicating the angular position from the vertical plane. If we assume
the tube is straight, then the axial component of the velocity vector will be constrained
along the tube and our unit vector, l, will describe the direction adequately. We shall use
the unit vectors er and eΦ to represent the radial and angular positions, respectively. If we
use our instantaneous spin rate, p, as defined in Equation 11.3, we can write the tangential
component of velocity as
VΦ = rpeΦ = ε peΦ
(11.11)
Vx = V l
(11.12)
The axial velocity is simply
Then the velocity vector could be written in cylindrical coordinates as
V = V l + ε pe Φ
(11.13)
Or, if we like to remain in Cartesian coordinates, we can combine Equation 11.13 with
Equation 11.7 to yield
V = V l + ε p(− sin( pt + Φ 0)n + i cos( pt + Φ 0)m)
(11.14)
These Cartesian coordinates are useful when we want to write the velocity vector at the
muzzle of the weapon. The lateral throwoff caused by a static imbalance can be described
as the tangent of the angle of the projectile CG as it exits. For small angles (usually the
case), this is approximately the angle itself in radians. With this, we can define the lateral
throwoff at the muzzle owing to a static imbalance as
TL =
ε p0
[− sin(Φ 0) + i cos(Φ 0)]
V0
(11.15)
where we have used t = 0 at the muzzle and specified the spin rate and muzzle velocity.
We must keep in mind that this is an angular measure for small angles or, more precisely,
a tangent of an angle. We can use the relationship
i exp(iθ ) = − sin θ + i cos θ
(11.16)
to write
TL = i
ε p0
exp(iΦ0 )
V0
(11.17)
If the projectile has a rotating band that forces it to spin based on the rifling twist, this
expression can be written in terms of the projectile diameter and twist rate as well. This is
extremely straightforward and left as an exercise for the reader.
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
342
11.2 Dynamic Imbalance
The diagram of Figure 10.2 represents the most common case of a projectile asymmetry, a
dynamic imbalance. In this case, a lateral throwoff effect as described in Section 11.1 will
result as the projectile leaves the muzzle of the gun and there will also be significant flight
dynamic effects as the projectile moves downrange. Usually, this mass asymmetry is small
and can be treated as a small amount of mass removed from or added to a projectile at a
point defined by a radial set of coordinates from the CG. We shall use the former approach
following the development of Ref. [2]. This is depicted in Figure 11.5. Figure 11.6 depicts
how this removed mass is oriented relative to the CG offset in the radial direction.
The development put forward in Ref. [2] assumes that the removed mass is much smaller
than the overall mass of the projectile. As established earlier, we will use two orthogonal
coordinate systems. The first is our i-j-k triad which is oriented along the projectile axis as
depicted in Figure 9.1. This coordinate system does not roll with the projectile. We shall
also make use of a second non-rolling coordinate system using the l-n-m system depicted
in the same figure. In this case, the coordinate system is oriented along the velocity vector.
Principal axis of inertia
Center of gravity
rE
ε
lE
rE
Mass removed, mE
FIGURE 11.5
Dynamically imbalanced projectile with mass removed.
Φ
Center of gravity
VΦ
Geometric axis
ε
mE
rE
FIGURE 11.6
Velocity of a dynamically imbalanced projectile’s CG.
© 2014 by Taylor & Francis Group, LLC
Lateral Throwoff
343
The coordinate systems are related to one another, assuming small yaw angles, through
the relationships
i = γ l +αn + βm
(11.18)
j = −α l + n
(11.19)
k = −β l + m
(11.20)
γ = cos α cos β ≈ 1
(11.21)
where
α is the pitch angle
β is the yaw angle
γ is defined as
The angular momentum of the projectile is the vector sum of all of the angular momenta
and is closely approximated by
di
H = I P pi + I T i × − mE (rE × v E )
dt
(11.22)
Here
H is the total angular momentum
v E is the velocity of the removed mass
This velocity can be broken into two components, one owing to the rotation about the spin
axis and the other owing to the yawing motion of the projectile as follows:
di
v E = p(i × rE )+ i × × rE
d
t
(11.23)
Then inserting this relationship into Equation 11.22 and combining terms gives us, after
utilization of the vector triple product
di
H = (I p − mErE2 )pi + IT i × − mE
dt
di
di
× − p(rE ⋅ i)rE − rE × (rE × i) + (rE ⋅ i) rE ×
dt
dt
(11.24)
If we examine Figure 11.5, we see that
(rE ⋅ i) = lE
© 2014 by Taylor & Francis Group, LLC
(11.25)
Ballistics: Theory and Design of Guns and Ammunition
344
And we note that for a spin-stabilized projectile, the yaw rate, di/dt, is much smaller than
the spin rate, p, we can eliminate terms in Equation 11.24 to yield
di
H ≈ I P pi + I T i × + mE plErE
dt
(11.26)
We can express the mass asymmetry vector r E in terms of the projectile geometric axes as
rE = lEi + rE cosΦ j + rE sin Φ k
(11.27)
This can be expressed in our coordinate system attached to the velocity vector through the
relationships in Equations 11.18 through 11.20 as
rE = (lEγ − rEα cosΦ − rE β sin Φ )l + (lEα + rE cos Φ )n + (lE β + rE sin Φ )m
(11.28)
We can simplify this expression somewhat if we use the fact that both α and β are much
smaller than γ. In this case, the expression would simplify to
rE = (lEγ )l + (lEα + rE cos Φ )n + (lE β + rE sin Φ )m
(11.29)
We can take the derivative of Equation 11.29 using the fact that the coordinate system is
effectively not rotating to write
drE
= (lEγɺ )l + (lEαɺ − rE p sin Φ )n + (lE βɺ + rE p cos Φ )m
dt
(11.30)
Here we have used the fact that
dΦ
=p
dt
(11.31)
As in our previous analyses, we shall consider a short period of flight. By doing this,
we can neglect all forces and moments except the pitching (overturning) moment. This
allows us to equate the rate of change of angular momentum to the applied pitching
moment
d 2i
dH
di
dr
* α V 2 ( l × i)
≈ I P p + I T i × 2 + mE plE E = mCM
dt
dt
dt
dt
(11.32)
This expression can be written as a set of three equations in terms of each component as
follows:
d2β
d 2α
I P pγɺ + I T α 2 − β 2 + mE plE2γɺ = 0
dt
dt
© 2014 by Taylor & Francis Group, LLC
(11.33)
Lateral Throwoff
345
d 2γ
d2β
* αV 2 β
I P pαɺ + I T β 2 − γ
+ mE plE (lEαɺ − rE p sin Φ ) = −mCM
2
d
d
t
t
(11.34)
d 2α
d 2γ
* αV 2α
I P pβɺ + I T γ
− α 2 + mE plE (lE βɺ + rE p cosΦ ) = mCM
2
d
d
t
t
(11.35)
The details of this are provided in Ref. [2]. If we change the temporal derivatives into spatial
derivatives along a dimensionless downrange distance, s, and define the following:
I
P= P
IT
M=
pd
V
(11.36)
md 2
*α
CM
IT
(11.37)
I E = mErElE
(11.38)
We can rewrite Equations 11.33 through 11.35 as
m l2
P 1 + E E γ ′ + αβ ′′ − βα ′′ = 0
IT
m l2
P 1 + E E α ′ + βγ ′′ − γβ ′′ + Mβ
IT
I I
− E 2T
IP
(11.39)
2
P sin Φ = 0
(11.40)
m l2
I I
P 1 + E E β ′ + γα ′′ − αγ ′′ − Mα + E 2T P 2 cosΦ = 0
I
T
IP
(11.41)
Here the primed quantities are differentiated with respect to s. With small yaw as well as
classical size assumptions (see Refs. [2,3]), we can neglect several of these terms because
they are either products of small numbers or summed with a much larger number. This
results in Equation 11.39 vanishing altogether and the other two transforming into
I I
Pα ′ − β″ + Mβ = E 2T
IP
2
P sinΦ
(11.42)
I I
Pβ ′ + α ′′ − Mα = − E 2T
IP
2
P cosΦ
(11.43)
If we now multiply Equation 11.42 by −i and add it to Equation 11.43, we obtain
I I
(α ′′ + iβ ′′) + P( β ′ − iα ′) − M(α + iβ ) = − E 2T
IP
© 2014 by Taylor & Francis Group, LLC
2
P (cosΦ + i sinΦ )
(11.44)
Ballistics: Theory and Design of Guns and Ammunition
346
If we invoke our definition of complex yaw angle, we can write this as
I I
ξ ′′ − iPξ − Mξ = E 2T
IP
2
P exp(i Φ )
(11.45)
The solution to this differential equation was discussed in Section 9.1. The difference here
is that the forcing term on the RHS is somewhat different. If we use a solution written as
ξ = K1 exp(iψ 1 ) + K 2 exp(iψ 2 ) + K 4 exp(iΦ )
(11.46)
where K1 and K2 are the solutions to the homogeneous part of the equation and K4 is our new
term which depends on the spin rate and the mass asymmetry we can solve for the magnitude of the trim arm caused by the asymmetry. If we solve Equation 11.46 for the particular
solution, we find that this new trim arm caused by the mass asymmetry is given by
K4 =
IE
I p2 M
IT − IP +
ITP
(11.47)
The third term in the denominator is usually very small so this term has been approximated (see Refs [2,4].)
K4 ≈
IE
IT − IP
(11.48)
This trim arm due to a mass asymmetry is usually small.
Throughout this development, IP and IT have been used as the moments of inertia even
though, in the purest sense, the mass asymmetry removes the axially symmetric properties of the projectile. For most cases, it is sufficient to use these quantities based on an axially symmetric projectile.
References
1. Greenwood, D.T., Principles of Dynamics, 2nd edn., Prentice Hall, Englewood Cliffs, NJ, 1988.
2. McCoy, R.L., Modern Exterior Ballistics, Schiffer Military History, Atglen, PA, 1999.
3. Murphy, C.H., Free Flight Motion of Symmetric Missiles, Ballistics Research Laboratory Report
No. 1216, Aberdeen Proving Ground, MD, 1963.
4. Murphy, C.H., Yaw Induction by Means of Asymmetric Mass Distributions, Ballistics Research
Laboratory Memorandum Report No. 2669, Aberdeen Proving Ground, MD, 1976.
© 2014 by Taylor & Francis Group, LLC
12
Swerve Motion
Following our procedure of slowly introducing complexity into the description of projectile behavior we shall now develop equations to characterize the remainder of what is
known in general as swerve motion. We saw in Chapter 11 that a mass asymmetry can
cause projectile motion transverse to the original line of fire even in a vacuum. We stated
in that section that a dynamic projectile imbalance was more common than a static imbalance but either can actually occur.
Chapter 6 explained many aspects of projectile behavior that arise due to the presence
of the air stream. All of the coefficients were functions of the angle of the attack observed
by the projectile relative to that air stream. If we examine how a statically or dynamically
imbalanced projectile would behave as viewed from above the trajectory curve based on
its spin, we would see motion as depicted in Figures 12.1 and 12.2. We must keep in mind
that the motion in these figures is greatly exaggerated for ease of viewing.
We can imagine, by looking at these figures that the aerodynamic forces would be considerable because even in the case of the statically imbalanced projectile, motion laterally
across the trajectory will manifest itself in an angle of attack and therefore affect the flight
characteristics.
In this section, we shall describe and evaluate the aerodynamic forces that arise from
this behavior and include them in our equations of motion for projectile flight. We shall
also include the effect of configurational asymmetries such as bent fins or damaged form
because these will result in similar behavior even without the mass asymmetry present. In
fact, to a varying degree, every projectile has a combination of both form and mass asymmetries present.
12.1 Aerodynamic Jump
McCoy [1] has shown that the equation of motion for the point mass solution plus swerving
motion is given by
(
gd
d2 y
d2z
i
+
= CLα *ξ − 2 exp 2CD* s
2
2
V0
ds
ds
)
(12.1)
with a solution of
dy
y + iz = ( y0 + iz0 ) +
ds
(
)
s 2 gd exp 2CD* s − 2CD* s − 1
dz
*
+i
s + CLα I L −
2
ds 0
2V02
0
2 CD* s
(
)
(12.2)
347
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
348
Trajectory
FIGURE 12.1
Motion of a statically imbalanced projectile.
Trajectory
FIGURE 12.2
Motion of a dynamically imbalanced projectile.
where
s s1
IL =
∫ ∫ ξ ds ds
1
2
(12.3)
0 0
Here we have used s1 and s2 as dummy variables representing integrations with respect
to s. Equation 12.2 describes the position of the projectile in a direction perpendicular to
the trajectory based on flat fire point mass assumptions.
© 2014 by Taylor & Francis Group, LLC
Swerve Motion
349
Equation 9.72 was developed as a solution for ξ. Reference [1] has shown that the solution
to the double integral of Equation 12.3 can be obtained by substitution of Equation 9.72 into
Equation 12.3 resulting in
λ − iψ 1′
λ − iψ 1′
K exp(iψ 20 ) s
I L = − 12
K exp(iψ 10 ) + 22
2 20
2 10
λ1 + ψ 1′
λ2 + ψ 2′
+ (R11 − iR12 )K10 exp(iψ 10 ) {exp[(λ1 − iψ 1′ )s] − 1}
+ (R21 − iR22 )K 20 exp(iψ 20 ) {exp[(λ2 − iψ 2′ )s] − 1} + i
(
)
PG0 2 exp 2CD* s − 2CD* s − 1
(12.4)
s
2
M
*
C
s
2
D
(
)
Here we have used
G0 =
R11 =
R12 =
R21 =
R22 =
gd
V02
(12.5)
λ12 − ψ 1′2
(λ
2
1
+ ψ 1′2
)
(12.6)
2
2λ1ψ 1′
(λ
2
1
+ ψ 1′2
)
2
(12.7)
2
(12.8)
λ22 − ψ 2′2
(λ
2
2
(λ
+ ψ 2′2
2λ2ψ 2′
2
2
+ ψ 2′2
)
)
(12.9)
2
If we make the assumption that λ12, 2 ≪ ψ 1′2, 2 , the aforementioned parameters become
R11 ≈ −
1
1
, R12 ≈ 0, R21 ≈ − 2 , and R22 ≈ 0
2
ψ 1′
ψ 2′
(12.10)
Inserting these assumptions into Equation 12.4 yields the following result:
1
1
1
I L = i K10 exp(iψ 10 ) +
K 20 exp(iψ 20 ) s − 2 K10 exp(iψ 10 ) {exp[(λ1 + iψ 1′ )s] − 1}
′
′
ψ2
ψ 1′
ψ1
−
(
)
1
PG0 2 exp 2CD* s − 2CD* s − 1
′
i
s
K
i
+
i
s
−
1
+
exp(
)
exp[(
)
]
ψ
λ
ψ
{
}
20
2
2
20
2
M
ψ 2′2
2CD* s
© 2014 by Taylor & Francis Group, LLC
(
)
(12.11)
Ballistics: Theory and Design of Guns and Ammunition
350
This result is important because it depicts the three components of swerve motion. The
first term on the RHS is called the aerodynamic jump, JA, and it is what we will examine
for the remainder of this section. The second two terms are the epicyclic swerve, SE, and
will be discussed in Section 12.2. The third term is called drift, DR, and will be discussed
in Section 12.3. To keep things simple, we will restate the aerodynamic jump as
1
1
J A = iCL*α K10 exp(iψ 10 ) +
K 20 exp(iψ 20 )
ψ 2′
ψ 1′
(12.12)
We should note a few things about Equation 12.12. First, we must keep in mind that in
Equation 12.11 this aerodynamic jump term is multiplied by a downrange distance, s,
implying that it is actually an angular measure (for small angles). A second observation
is that the aerodynamic jump is completely dependent upon the initial conditions of the
projectile and how these couple in with the fast and slow arm turning rates.
If we insert our approximated initial fast and slow arm amplitudes from Equations 9.78
and 9.79 into Equation 12.12, we obtain
−iξ 0′ − ψ 2′ξ 0′
iξ ′ + ψ 1′ξ 0′
+ 0
J A = iCL*α
ψ 1′ (ψ 1′ − ψ 2′ ) ψ 2′ (ψ 1′ − ψ 2′ )
(12.13)
This can be rewritten [1] as
C
J A = kT2 Lα
CMα
(iPξ 0 − ξ 0′ )
(12.14)
Downrange
This result shows that by knowing the projectile mass properties and launch conditions,
we can determine to what angle a projectile will “jump.” We can envision this jump effect
as shown in Figure 12.3.
While we have said a great deal mathematically about aerodynamic jump, we have not
really described the physics behind it. Because of the presence of aerodynamic lift on the
projectile, there is a strong influence of angle of attack on the resultant motion. We saw
Line of departure
of projectile
tan–1( JA)
Mean motion due
to aerodynamic
jump
Crossrange
FIGURE 12.3
Graphical representation of aerodynamic jump.
© 2014 by Taylor & Francis Group, LLC
Swerve Motion
351
earlier that a projectile, through purely dynamic means, can yaw because of either spin or
some geometric asymmetry. When this happens, the aerodynamic forces change, either
improving or worsening the situation. This interaction of the aerodynamic forces with the
projectile manifests itself in the jump angle as depicted in Figure 12.3.
12.2 Epicyclic Swerve
The second two terms in Equation 12.11 describe the epicyclic swerve of a projectile. We
can define this parameter specifically [1] as
1
1
SE = −CL* α 2 K 20 exp(iψ 20 ) {exp[(λ1 + iψ 1′ )s] − 1} + 2 K 20 exp(iψ 20 ) {exp[(λ2 + iψ 2′ )s] − 1}
′
′
ψ
ψ
2
1
(12.15)
McCoy [1] has shown that this equation can be put into a more useful form through use of
the relation
C
CL*α = kT2 Lα
CMα
ψ 1′ψ 2′
(12.16)
If we insert Equation 12.16 into Equation 12.15, we obtain
C ψ ′
ψ′
SE = −kT2 L α 2 K10 exp(iψ10 ) {exp[(λ1 + iψ 1′ )s] − 1} + 1 K 20 exp(iψ 20 ) {exp[(λ2 + iψ 2′ )s] − 1}
′
′
ψ
ψ
C
M
2
1
α
(12.17)
Following McCoy, we shall examine two special cases of this equation. The first is where
we have a projectile that is non-spinning (statically stable) and the second is a spin-stabilized
projectile with a good gyroscopic stability (measured at muzzle exit) of at least 1.5.
For the non-spinning projectile, the following conditions apply:
M < 0, P = 0, and ψ 2′ = − ψ 1′
(12.18)
If we insert these conditions into Equation 12.17, we get
C
SEnon-spin = kT2 Lα ( K10 exp(iψ 10 ) {exp[(λ1 + iψ 1′ )s] − 1} + K 20 exp(iψ 20 ) {exp[(λ2 + iψ 2′ )s] − 1})
CMα
(12.19)
Now we can invoke the fact that the spin is equal to zero and insert Equation 9.59, in which
we shall neglect the trim and yaw of repose, into Equation 12.19 to yield
C
S Enon-spin = kT2 Lα (ξ − ξ 0 )
CMα
© 2014 by Taylor & Francis Group, LLC
(12.20)
Ballistics: Theory and Design of Guns and Ammunition
352
This relationship will produce a motion in exactly the same manner as the aerodynamic
jump developed in Section 12.1. It essentially couples the yawing motion of the projectile
to the swerving motion. Both will thus damp together and the more yaw, the greater the
epicyclic swerve.
If we examine the spin-stabilized projectile, we can write
M > 0 and ψ 1′2 ≫ ψ 2′2
(12.21)
With the aforementioned mathematical statements, McCoy [1] has stated that an excellent
approximation of Equation 12.15 for a spinning projectile is
C ψ ′
SEspin = −kT2 Lα 1 ( K 20 exp(iψ20 ) {exp[(λ2 + iψ 2′ )s] − 1})
CMα ψ 2′
(12.22)
An interesting comparison may be drawn between the epicyclic swerving behavior of
a spinning projectile and a non-spinning projectile. If we compare Equations 12.22 and
12.20, we see that in the latter the yawing motion and swerve are locked together and
operate in the same fixed plane. This is because the lift generated by the motion never
rotates. In a spin-stabilized projectile, the lift vector is always rotating, thus the center of
mass of the projectile will move in a helical manner around the flight path. Furthermore,
the motion will be locked to the rate of turning of the slow arm and will damp or increase
as the slow arm does.
12.3 Drift
The last term in Equation 12.11 describes the drift of a projectile. We can define this parameter specifically [1] as
DR = i
(
)
PG0 2 exp 2CD* s − 2CD* s − 1
s
2
M
2CD* s
(
)
(12.23)
If we expand the term in brackets in a power series, we can rewrite this equation as
PG0
DR = i
M
2
1
2 2 *
s 1 + CD s + CD* s + ⋯
3
3
(
)
(
)
(12.24)
Examination of the drift equation in this form has some advantages. First, we can see that
if a projectile has no spin, P = 0 and there is no drift. If we look at a fin- or drag-stabilized
projectile where M < 0, we see that the projectile will drift in the direction opposite to
© 2014 by Taylor & Francis Group, LLC
Swerve Motion
353
the spin. That is, a left-hand spin will produce a right-hand drift and vice versa. In a statically unstable (spin-stabilized) projectile where M > 0, we see that the projectile will drift
in the same direction as the spin. It must be noted that this drift is very small compared
to the other swerve components as well as Coriolis drift. In fact, to even measure it, some
researchers [1] have fired two projectiles simultaneously out of side-by-side gun barrels with
both left- and right-hand twist to remove Coriolis and wind drift components which would
affect each equally.
The interested reader should consult Ref. [1] for further information on this topic.
Reference
1. McCoy, R.L., Modern Exterior Ballistics, Schiffer Military History, Atglen, PA, 1999.
© 2014 by Taylor & Francis Group, LLC
© 2014 by Taylor & Francis Group, LLC
13
Nonlinear Aeroballistics
Until this point, we have concerned ourselves with linear behavior of the aerodynamic
coefficients only. This is very convenient for direct fire projectiles and projectiles that fly
with very little yaw. It had the benefit of allowing us to make a black or white decision with
regard to projectile stability as well—the projectile was either stable or not. In real systems,
several of the coefficients are only linear over a small range of angles of attack. This can be
either helpful or hurtful to a particular design.
Limit-cycle motion is motion that develops over time in a projectile, whereby the projectiles angle of attack grows until a certain (sometimes rather large) angle is achieved. As
the angle of attack increases (or some other parameter such as the air density changes), the
coefficients change so that the projectile will actually become stable at some large angle of
attack. At first, this may seem like it is a desirable quality in a projectile; however, range
is sacrificed due to the larger drag generally associated with this large yaw. Some systems
have been fielded unwittingly in this condition and it was only after a large number of firings in the field that this was determined to be an issue.
This nonlinear behavior arises out of the interaction between the air and the surfaces of
the projectile. It is a rather complicated mechanism that can arise (many times in a discontinuous manner) from boundary layer separation, fin masking, vortex shedding, etc. All
of which are fluid dynamic phenomenon. This is and continues to be a challenging area of
aeroballistic research, where experimental, theoretical, and computational techniques are
pushed to the limit of their usefulness.
The next two sections will look at this behavior to some degree of detail; however,
because of space constraints, the reader is encouraged to consult the literature for more
detailed mathematical and theoretical treatment.
13.1 Nonlinear Forces and Moments
In general, we can divide nonlinear forces and moments into two categories: geometric
and aerodynamic nonlinearities. The geometric nonlinearities arise from the cosine terms
in the equations of motion that were eliminated when we assumed a small yaw angle.
This small angle assumption is generally valid for most projectiles in flight. If a projectile
is flying with large yaw, the cosine terms must be retained and the resulting equations are
more difficult to solve. Since this behavior is usually designed out of projectiles, we shall
focus on the second type of nonlinearity, the aerodynamic nonlinearity.
The aerodynamic nonlinearity can exist even at angles of attack that are consistent with
the small yaw assumption. They arise due to the fluid–mechanic interaction of the air with
the solid projectile body. This interaction can consist of phenomena such as vortex shedding, separation, shock interactions, etc.
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The most dominant force acting on the body is the drag force. In all of our previous discussions, we have stated that the forces that arise due to other sources are small, and that is
still true for the case of nonlinearities; however, the moments caused by these other forces
cannot be neglected. We can define a nonlinear drag coefficient as
CD = CD0 + CDδ 2δ 2 + ⋯
(13.1)
In this equation, the first term on the RHS is the zero-yaw-drag coefficient and the second
term is the yaw-drag coefficient. More coefficients can be added but typically the expression is truncated at the yaw-drag term.
There are essentially two common ways of determining the yaw-drag coefficient: experimentally or computationally. Experimental evaluation is more common although recent
advances in computational fluid dynamics (CFD) [1] have shown that it is possible to
extract coefficients directly from analyses. In either case, the overall drag coefficient at
multiple angles of attack is determined from either a direct force measurement (in the case
of a wind tunnel or CFD model) or the velocity decay (in a free flight test), and the results
are plotted as CD versus angle of attack. The slope of the resulting line (hopefully, it is a
line) is then the yaw-drag coefficient and the y-intercept is the zero-yaw-drag coefficient. In
the case of a free flight firing where the projectile is dragging down continuously, Murphy
[2] and McCoy [3] suggest an averaging scheme that has been successfully demonstrated
based on a great deal of experience.
The aforementioned technique is known as a quasi-linear approach because it defines
a linear function that is a solution to a nonlinear equation. The same approach is used to
determine the nonlinear moments, which are generally assumed to have the same form
as Equation 13.1.
In general, both the zero-yaw-drag coefficient and the yaw-drag coefficient are positive
values. In the case of the pitching or overturning moment of a spin-stabilized projectile, the
zero-yaw overturning moment coefficient is positive while the cubic overturning moment
coefficient is negative [3]. This condition can have some interesting effects on stability as
summarized by McCoy [3].
The overall equation of motion that includes all of the nonlinear terms that is equivalent
to our linear equation (Equation 9.76) with the gravitational term neglected is
ξ ′′ + ( H 0 + H 2δ 2 − iP)ξ ′ − [ M0 + M2δ 2 + iP(T0 + T2δ 2 )]ξ = 0
(13.2)
We can define our coefficients as follows:
H0 =
ρ Sd
1
CLα 0 − CD0 − 2 (CMq + CMαɺ )0
2m
kT
(13.3)
H2 =
ρ Sd
1
CLα 2 − CD2 − 2 (CMq + CMαɺ )2
2m
kT
(13.4)
M0 =
ρ Sd 1
CMα0
2m kT2
(13.5)
M2 =
ρ Sd 1
CMα2
2m kT2
(13.6)
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Nonlinear Aeroballistics
357
T0 =
ρ Sd
1
CLα0 + 2 CMpα0
2m
kP
(13.7)
T2 =
ρ Sd
1
CLα 2 + 2 CMpα2
2m
kP
(13.8)
I
P= P
IT
pd
V
(13.9)
The solution to Equation 13.2 is
ξ = K10 exp[λ1s] exp[i(ψ 10 + ψ 1′s)] + K 20 exp[λ2s] exp[i(ψ 20 + ψ 2′ s)]
(13.10)
where again, we are reminded that the gravitational term has been neglected. McCoy [3]
has written expressions for the coefficients in terms of the damping exponents and turning rates as follows:
δ e12 = K12 + 2K 22
(13.11)
2
δ e2
= K 22 + 2K12
(13.12)
K 2 + K 22
ψ 1′ + ψ 2′ = P + M2 1
≈P
ψ 1′ −ψ 2′
(13.13)
ψ 1′δ e22 −ψ 2′δ e21
ψ 1′ −ψ 2′
(13.14)
ψ 1′ψ 2′ = M0 + M2δ e2
(13.15)
δ e2 =
λ1 =
)
(
)
(
)
− H 0ψ 1′ + P T0 − T2δ e21 − H 2 ψ 1′ K12 + K 22 + ψ 1′K 22
ψ 1′ −ψ 2′
(
)
− H 0ψ 2′ + P T0 − T2δ e22 − H 2 ψ 2′ K12 + K 22 + ψ 1′K12
λ2 =
ψ 1′ −ψ 2′
(
(13.16)
(13.17)
In terms of some of these parameters, McCoy [3] has derived a form for the nonlinear lift
coefficient as
ψ ′2δ 2 K eiψ 1 + ψ 1′2δ e22K 2eiψ 2
CLα = CLα 0 + CLα 2 2 e12 1 iψ 1
iψ 2
2
ψ 2′ K1e + ψ 1′ K 2e
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(13.18)
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358
As is readily apparent, these expressions are significantly more complex than their linear
cousins. Because of this, they are generally solved using numerical schemes. The interested reader is referred to Refs. [4–12] for a more detailed treatment as well as examples of
this behavior.
Problem 1
If the projectile in Problem 1 of Chapter 9 happens to be flying at a limit-cycle yaw of
4° with a spin rate of 130 Hz and velocity 1764 ft/s. What would the nonlinear pitching
moment have to be for the projectile to be marginally stable?
Hints:
1. Assume all of the other coefficients are linear.
2. Recall the definition of the nonlinear pitching moment (you have the linear part in
Problem 1 of Chapter 9).
Answer: CMα 2 = −119.079
13.2 Bilinear and Trilinear Moments
We have discussed nonlinear forces and moments and their implications in the previous
section. At this point, we shall turn our attention to nonlinear moments in which the cubic
behavior itself can be described by a bilinear or trilinear curve. This is evident when the
cubic coefficient is plotted versus yaw angle. A bilinear coefficient would have two different linear slopes, while a trilinear moment would have three. This is quite useful since
many experimental data can be fitted using these curves. In particular, we shall examine
the Magnus moment and its implications because this is the dominant moment in spinstabilized projectile flight behavior [3].
If we are examining projectile flight data, it is often tempting to fit a higher order polynomial curve to deal with the nonlinearity. This is usually not advisable since the abrupt
changes in behavior at certain angles of attack are caused by fluid–solid interactions such
as boundary layer separation, vortex shedding, etc.
To describe the behavior of projectiles with nonlinear Magnus moment coefficients, we
shall use two examples: one with a linear cubic Magnus moment and one with a bilinear
Magnus moment. We are interested in two things: first, the effect of initial conditions on
projectile stability and second, limit-cycle motion.
In the excellent treatment by McCoy [3], for illustrative purposes, the author suggested
assuming a linear pitch damping moment with a cubic Magnus moment coefficient. This
will force H2 to be zero and allow Equations 13.16 and 13.17 to be written as
λ1 =
λ2 =
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(
− H 0ψ 1′ + P T0 + T2δ e21
)
ψ 1′ −ψ 2′
(
− H 0ψ 2′ + P T0 + T2δ e22
ψ 1′ −ψ 2′
)
(13.19)
(13.20)
Nonlinear Aeroballistics
359
We can put these equations into the form
λ1 = λ10 + λ12 δ e21
(13.21)
λ2 = λ20 + λ22 δ e22
(13.22)
where, we can define
λ10 =
− H 0ψ 1′ + PT0
ψ 1′ −ψ 2′
(13.23)
λ20 =
− H 0ψ 2′ + PT0
ψ 1′ −ψ 2′
(13.24)
PT2
ψ 1′ −ψ 2′
(13.25)
λ12 = −λ22 =
With these expressions, we can draw plots of damping coefficients versus yaw angle in a
manner similar to the coefficients.
At this juncture, we need to recall that these damping exponents will decrease the yaw
of their particular mode if they are negative, and increase the yaw if they are positive.
Thus, negative values are stabilizing and positive values are destabilizing. As a simple
example, let us look at a projectile that has a linear cubic Magnus moment. In analyzing
this projectile, we create two plots of damping coefficient versus yaw. These are depicted
as in Figures 13.1 and 13.2.
In Figure 13.1, we can see the fast mode damping coefficient is negative for all yaw angles
of interest (if the projectile is flying at an angle above 11°, we probably have a problem).
0.0006
0.0004
λ1
0.0002
0.01
0.02
0.03
0.04
0.05
δ 2e1
–0.0002
–0.0004
λ10
–0.0006
FIGURE 13.1
Plot of fast mode damping coefficient versus yaw for linear cubic fast mode.
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Ballistics: Theory and Design of Guns and Ammunition
360
0.0006
λ2 0.0004
0.0002
0.01
0.02
0.03
0.04
0.05
δ 2e2
–0.0002
–0.0004
λ20
–0.0006
FIGURE 13.2
Plot of slow mode damping coefficient versus yaw for linear cubic slow mode.
Thus, the fast mode will always damp for this projectile. Examination of Figure 13.1 reveals
that as long as the projectiles yaw angle is below 5.73°, the slow arm will damp to zero
(recall that the yaw angle is equal to sin(δ)); above this angle, it will grow without bound.
Although this angle is fairly large for a projectile, there have been instances documented
where a slowly launched missile was stable when fired from one side of a fast warship,
but unstable when launched from the other [2,8]. The instability was caused by the vector
addition of the ships own speed with the launch velocity.
Figures 13.3 and 13.4 show the fast and slow damping exponents for a projectile with
bilinear cubic Magnus moment behavior. This is an interesting example because it
0.0006
λ1 0.0004
0.0002
0.01
0.02
0.03
0.04
0.05
δ 2e1
–0.0002
–0.0004
λ10
–0.0006
FIGURE 13.3
Plot of fast mode damping coefficient versus yaw for bilinear cubic fast mode.
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Nonlinear Aeroballistics
361
0.0006
0.0004
λ20
λ2
0.0002
0.01
0.02
0.03
0.04
0.05
δ 2e2
–0.0002
–0.0004
–0.0006
FIGURE 13.4
Plot of slow mode damping coefficient versus yaw for bilinear cubic slow mode.
illustrates how a projectile can enter into limit-cycle motion. Limit-cycle motion is motion
in which the projectile cones in a predictable manner about the velocity vector.
If we examine Figure 13.3, we see that, similar to our earlier case, the fast arm damping
coefficient is everywhere negative. Because of this, the fast mode will always damp to zero.
The interesting part of the story is shown in Figure 13.4. Here, we see that for small angles,
the projectiles slow arm will continue to grow because the damping exponent is positive.
Once the amplitude of the motion grows beyond 5.74°, the sign of the coefficient changes
driving the motion back to zero. However, the motion cannot be driven all the way back
to zero because as soon as the angle decreases below 5.74°, the now positive damping coefficient will again cause it to increase. The end result will be a projectile that cones about
the velocity vector at a 5.74° angle.
These examples assumed that the velocity of the projectile has had no affect on the exponents. We must always keep in mind that there are many interrelated phenomena that
affect these coefficients—the real world is a complicated place. This discussion should
provide you with a feel for the physics of the projectile behavior.
References
1. DeSpirito, J. and Heavey, K.R., CFD computation of magnus moment and roll damping
moment of a spinning projectile, AIAA Paper No. 2004-4713, American Institute of Aeronautics
and Astronautics, New York, August 2004.
2. Murphy, C.H., Free flight motion of symmetric missiles, Ballistics Research Laboratory Report
No. 1216, Aberdeen Proving Ground, MD, 1963.
3. McCoy, R.L., Modern Exterior Ballistics, Schiffer Military History, Atglen, PA, 1999.
4. Murphy, C.H., Data reduction for the free flight spark ranges, Report No. 900, USA BRL,
Aberdeen Proving Ground, MD, February 1954.
5. Murphy, C.H., Limit cycles for non-spinning statically stable symmetric missiles, Report No.
1071, USA BRL, Aberdeen Proving Ground, MD, March 1959.
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362
Ballistics: Theory and Design of Guns and Ammunition
6. Murphy, C.H., The measurement of non-linear forces and moments by means of free flight
tests, Report No. 974, USA BRL, Aberdeen Proving Ground, MD, February 1956.
7. Murphy, C.H., Advances in the dynamic analysis of range data, Memorandum Report No.
1270, USA BRL, Aberdeen Proving Ground, MD, May 1960.
8. Platus, D.H., Dynamic instability of finned missiles caused by unequal effectiveness of windward and leeward fins, AIAA Paper No. 70–206, American Institute of Aeronautics and
Astronautics, New York, January 1970.
9. Tobak, M., Schiff, L.B., and Peterson, V.L., Aerodynamics of bodies of revolution in coning
motion, AIAA Journal, 7(1), January 1969, 95–99.
10. Seginer, A. and Rosenwasser, I., Magnus effect on spinning transonic finned missiles, AIAA
Paper No. 83–2146, American Institute of Aeronautics and Astronautics, New York, August 1983.
11. Platou, A.S., Magnus characteristics of finned and nonfinned projectiles, AIAA Journal, 3(1),
American Institute of Aeronautics and Astronautics, New York, January 1965.
12. Cohen, C.J., Clare, T.A., and Stevens, F.L., Analysis of the non-linear rolling motion of finned
missiles, AIAA Paper No. 72-980, American Institute of Aeronautics and Astronautics,
New York, September 1972.
© 2014 by Taylor & Francis Group, LLC
Part III
Terminal Ballistics
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© 2014 by Taylor & Francis Group, LLC
14
Introductory Concepts
Terminal ballistics is the regime that the projectile enters at the conclusion of its flight. It
has been delivered into its flight by the interior ballistician, pursued and guided through
its flight by the exterior ballistician, and now at its target becomes the responsibility of the
terminal ballistician. The basic objective of firing the projectile is to defeat some type of
target and we will study the widely varying phenomena of terminal effects that are the
tools of the terminal ballistician. These end effects are dependent on the design and mission of the projectile. The most common of the missions are as follows: fragmentation of
the projectile body by its cargo of high explosives; penetration or perforation of the target
by the application of kinetic or chemical energy; blast at the target area delivered by the
chemical energy of the explosive cargo; and the dispersal of the cargo for lethal or other
missions, e.g., smoke, illumination, propaganda dispersal, etc.
Since most terminal ballistic phenomena involve the generation and effects of stress
waves in solids, we will spend some time examining the details of this field. We must
gain some knowledge of terminal ballistic terminology to be able to study the theories of
kinetic energy penetration of solid targets; detonation, deflagration, and burning of energetic materials; the fundamentals of shaped charges; fragmentation theories; blast effects;
and lethality with the study of wound ballistics.
We shall begin by introducing some concepts that we shall use throughout our study of
this field.
In examination of penetration theories, we need to consider the following items: What
constitutes defeat of the target? What is the source of the data for which we have to create a
theory? Does the theory track with respect to momentum balance or energy balance? How
many empirically derived constants are there in the model (this tells us how universal the
theory will be)? What simplifications and assumptions were made?
Penetration is defined as an event during which a projectile creates a discontinuity in
the original surface of the target. Perforation requires that, after projectile or its remnants
are removed, light may be seen through the target. Since penetration is a somewhat stochastic event, we need to define some statistical parameters. V10 is the velocity at which
a given projectile will defeat a given target 10% of the time. V50 is the velocity at which a
given projectile will defeat a given target 50% of the time, and V90 is the velocity at which
a given projectile will defeat a given target 90% of the time. These quantities are depicted
in Figure 14.1.
The 50% penetration velocity is commonly used as both experimental measurement as
well as a production check. The following procedure illustrates its usage in an experiment.
The reader should refer to Figure 14.2 to illustrate the meaning. First, we should estimate
V50 through a calculation. Once this is accomplished, we fire a projectile with a Vs as close
to V50 as we can achieve. Let us say, the velocity of this experimental firing is a bit over our
estimate (at 1 in Figure 14.2). Assuming shot 1 only partially penetrated, we increase the
velocity considerably, and let us say that we achieve complete penetration at 2 in the figure.
We now assume that V50 is midway between 1 and 2. We now would attempt to fire at the
velocity halfway between 1 and 2 (at 3) and, say, we get complete penetration. We would
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366
Probability of success
0.90
0.50
0.10
V
V10
V50
V90
FIGURE 14.1
Statistical velocities defined.
Vs
Complete penetration
Complete penetration
Partial penetration
2
6
Complete penetration
3
1
5
Partial penetration
Initial estimate
Partial penetration
4
FIGURE 14.2
Illustration of the V50 experimental procedure.
next lower the velocity to get a partial penetration, say at 4, then we would increase it to
get a complete penetration (but let us say, we get only a partial penetration at 5). We would
then have to increase the next shot velocity to 6.
We would continue the aforementioned procedure, commonly known as an up and down
test, until we obtained three complete penetrations and three partial penetrations with the
difference between the highest and the lowest velocities in the set less than 200 ft/s. At that
point, we would calculate the experimental V50 from
V50
∑
=
6
i=1
6
Vi
(14.1)
The limit velocity, Vi (sometimes called the ballistic limit when referring to the armor), is
the velocity below which a given projectile will not defeat a given target. The technique
for determining it was invented by the U.S. Army Ballistics Research Laboratory (BRL),
Aberdeen, Maryland. The object is to fire a few projectiles that achieve complete penetration, measuring the residual velocity through the use of flash x-rays, and then generate a
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Introductory Concepts
367
Vr
Vl
Vs
FIGURE 14.3
Limit velocity illustrated.
curve as shown in Figure 14.3. We then plot the residual velocity after penetration versus
the striking velocity. Usually, there will be a lower limit that develops below which the
armor is not penetrated or the projectile gets stuck in the armor.
From experimental evidence, we know that the following factors affect the limit velocity:
material hardness, yaw at impact, projectile density, projectile nose shape, and length to
diameter ratio of the projectile. For the material hardness, in general, the harder the target,
the higher V50 becomes; while the harder the penetrator, the lower V50 becomes and there is
more residual penetrator. With respect to yaw at impact, the more yaw, the greater chance
for breakup or ricochet and the higher V50 becomes. With projectile density, we find that
the more dense the projectile is, the lower V50 becomes. A blunter nose translates, in general, to a higher V50. If the target is overmatched significantly, however, the nose shape has
negligible effect. The length to diameter ratio can go either way and a great deal depends
on the obliquity of impact.
We will now introduce some concepts, which we shall use in our examination of penetration events.
© 2014 by Taylor & Francis Group, LLC
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15
Penetration Theories
Now that we have a firm grounding in some penetration concepts such as limit velocity, we can proceed to discuss various penetration theories. We shall discuss, in some
detail, penetration mechanisms in a variety of materials, all of which, to different degrees,
serve to protect some vital target. Because these materials behave very differently from
one another, they must be treated separately. It is this large difference in behavior, as well
as mechanical properties, which makes the selection of a material for ballistic protection
an important one.
We shall move successively through metals, concrete, soil, ceramic, and composite
armors so that the reader gets a feel for how they behave. In all instances, the day-to-day
analysis techniques of these materials are progressing, especially in the areas of numerical
methods.
15.1 Penetration and Perforation of Metals
Metals are by and large the most common target of medium to large caliber projectiles.
Although small caliber ammunition is generally used against soft targets, there are times
when even they are called upon to penetrate metal objects. This section will discuss several models of penetration into two of the most common metals: steel and aluminum.
While these formulas are not exactly perfect for other metals, usually a metal will behave
like one or the other.
Projectiles may impact metallic targets under a wide range of velocities. The nature of
the target material is such that different velocities must be handled using somewhat different techniques. At very low velocities (<250 m/s), the penetration is usually coupled to
the overall structural dynamics of the target. Responses are on the order of 1 ms. As the
impact velocity increases (500–2000 m/s), the local behavior of the target (and sometimes
penetrator) material dominates the problem. This local zone is approximately 2–3 projectile diameters from the center of impact. With further increases in velocity (2000–3000 m/s),
the high pressures involved allow the materials to be modeled as fluids in the early stages
of impact. At impact speeds greater than 12,000 m/s, energy exchange occurs at such a
high rate that some of the colliding material will vaporize. This energy exchange must be
accounted for. We will not treat this last case as it is beyond the normal scope of military
applications.
A typical sequence of events that occur during a projectile impact is developed here [1].
Given that a projectile strikes a target, compressive waves propagate into both the projectile and the target. Relief waves propagate inward from the lateral free surfaces of the
penetrator, cross at the centerline, and generate a high tensile stress. If the impact were
normal, we would have a two-dimensional stress state. If the impact were oblique, bending stresses will be generated in the penetrator. When the compressive wave reached the
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370
free surface of the target, it would rebound as a tensile wave. The target may fracture at
this point as will be seen in Section 16.3. The projectile may change direction if it perforates (usually toward the normal of the target surface).
Because of the differences in target behavior based on the proximity of the distal surface,
we must categorize targets into four broad groups. A semi-infinite target is one where
there is no influence of distal boundary on penetration. A thick target is one in which the
boundary influences penetration after the projectile is some distance into the target. An
intermediate thickness target is a target where the boundaries exert influence throughout
the impact. Finally, a thin target is one in which stress or deformation gradients are negligible throughout the thickness.
There are several methods by which a target will fail when subjected to an impact. The
major variables are the target and penetrator material properties, the impact velocity, the
projectile shape (especially the ogive), the geometry of the target supporting structure, and
the dimensions of the projectile and target.
The failure modes of the target are depicted in Figure 15.1. They will now be described.
Spalling is very common and is the result of wave reflection from the rear face of the plate.
It is common for materials stronger in compression than in tension. Scabbing is similar
to spalling, but the fracture results predominantly from large plate deformation, which
begins a crack at a local inhomogeneity. These failure mechanisms will be expounded
upon in Section 16.3. Brittle fracture occurs usually in weak and lower density targets.
Radial cracking is common in ceramic type materials where the tensile strength is lower
than the compressive strength, but it does occur in some steel armor. Plugging occurs in
materials that are fairly ductile and usually when the projectile impact velocity is very
close to the ballistic limit. Petaling occurs when the radial and circumferential stresses are
high and the projectile impact velocity is close to the ballistic limit.
Perforation
Non-perforation
Plugging
Piercing
(ductile)
Petaling
Scabbing
Scab
Spalling
(brittle)
Spall ring
FIGURE 15.1
Target failure modes.
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Star crack
Penetration Theories
371
Because of the very high loading rates and correspondingly high temperatures, we need
to describe some phenomena that occur during penetration events. Terms such as these
occur throughout the literature, so it is good to understand what they mean.
The concept of adiabatic shearing is encountered in impacts where a plug has been
formed. On initial impact, a local ring of intense shear is generated. Since this occurs
very quickly (∼μs), the target does not have sufficient time to build up any motion.
Locally intense heat is generated. Because of the time scale and a large deformation
rate, the heat cannot be conducted away. Since the material properties are weaker at
this high temperature, the material tends to yield readily and flow plastically. The
process then feeds on itself. Finally, a plug is formed and breaks free. If the minimum
perforation velocity is exceeded by more than about 5%–10%, the plug will usually
break up. Blunt noses on projectiles tend to increase the propensity to fail a target by
adiabatic shear.
Hydrodynamic erosion is an important concept in terminal ballistics. Metal cutting tools
such as water jets or soft metal penetrators and shaped charge jets can defeat a target by
hydrodynamic erosion. During hydrodynamic erosion, the penetrator material forces the
target material aside in a manner similar to a punch being pushed into the target material
except that the hole will be larger. This phenomenon usually occurs at impact velocities
over 1000 m/s. Deposition of the penetrator material on the walls of the hole is an indication that this failure mechanism played a part in the penetration.
The hydrodynamic transition velocity is the velocity below which the projectile and
target act as essentially elastic bodies and above which both target and projectile can be
treated as fluids. This concept is illustrated by the penetration sequence of Brooks [1]. For
all penetration velocities, the target material is accelerated radially away from the axis of
penetration. At low velocities, elastic strain keeps the target material in contact with the
penetrator. At high velocities, the material is thrown away from the projectile, so that the
hole becomes bigger than the projectile diameter. The radial acceleration of the material
is greatest at the tip of the projectile. At the hydrodynamic transition velocity, the tip of
the penetrator deforms laterally. The projectile tip becomes spherically blunted and forms
a stable shape that penetrates the target for the remainder of the event. The transition
velocity varies inversely with the tip radius. Hydrodynamic transition velocity is possibly
related to the rate of rod erosion and plastic wave propagation.
Shear banding is a form of adiabatic shearing in which layers of material in a like state
of shear tend to form. There are discontinuities in stress and strain instead of a gradual
increase in shear strain near the disturbed region. Uranium and tungsten tend to display
this phenomenon. Normal material models used in finite element codes do not show this
effect. A model that includes thermal softening is required.
The analytical models in use today to solve these types of problems can be organized
into three broad categories: empirical or quasi-analytical, approximate analytical, and
numerical. In empirical or quasi-analytical models, algebraic equations are developed
from large amounts of experimental data. These models are generally curve fits (results
based). They usually do not incorporate physics and tend to be configuration dependent.
An approximate analytical model attempts to examine the physics of a particular aspect
of the penetration process or failure mechanism such as petaling, plugging, etc. The mathematics becomes tractable because we must make simplifying assumptions. They are usually limited to particular situations. Numerical models usually attempt to solve the full
equations of continuum mechanics using finite difference or finite element techniques.
This is the most general method. The problem with numerical models is that good material
models are required and this can be expensive.
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Ballistics: Theory and Design of Guns and Ammunition
Most analytical models can only consider one damage mechanism (like plugging or
fracture) or conservation law before they become mathematically intractable. Some allow
as many as two mechanisms. The approach is to make simplifying assumptions. Typical
assumptions are to assume localized influence where the projectile is only influenced by
a small region of the target, to ignore rigid body motions, and to ignore thermal, friction,
shock heating, and any material behavioral changes due to these mechanisms, that the
target is initially stress free, etc. One important thing to recognize is that a complicated
model does not necessarily yield a more accurate answer.
Perforation of finite thickness plates in which plugging is the predominant penetration
mode is divided into three stages. In the first stage, locally, the material ahead of the projectile is compressed and the mass is added to the projectile (i.e., the projectile decelerates
somewhat and the added mass accelerates). In the second stage, more material is accelerated but shearing is occurring on the surface area of the plug. In the third stage, the plug
has completely sheared out and both the plug as well as the projectile move with the same
velocity. If this model is used for an oblique impact, one must use the line-of-sight thickness. At velocities from 1200 to 5000 m/s, the model used usually involves hydrodynamic
erosion of the projectile tip as the first stage as well. This can be followed by both plugging
and further tip erosion. In the third stage, we usually consider the projectile to be completely eroded and the plug is ejected from the armor [1].
Some models account for the flexibility of the target. This is usually required as the
impact velocities approach the limit velocity. In this case, a significant amount of energy is
consumed in both elastically and plastically bending the target plate.
We shall examine the underlying assumptions in a few penetration theories before
moving on to detailed examination of the theories themselves. Theories that are derived
from a momentum balance are typically used for thin plates. These theories can be used
with minor modifications when the target petals. They usually require that the projectile
remains intact.
Theories that are derived from an energy balance are typically used for thick and moderately thick plates. With moderately thick targets, plugging can occur. Thick plates are
usually defeated by a piercing phenomenon that also has distinct phases. The first phase
is a radial displacement of the target material. Sometimes, there is plugging at this stage.
This stage is followed by plastic flow and yielding of the target. The target material may
well be able to be treated like a fluid during this phase.
Many empirically based predictive relationships are based on energy approaches. A particularly popular model takes the form of
E = kd mt n
(15.1)
m+n ≈ 3
(15.2)
where we have
In these equations
E is the perforation energy
d is the projectile diameter
t is the plate thickness
k is an empirically derived constant (see Figure 15.2)
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Penetration Theories
373
θ
t
d
FIGURE 15.2
Projectile impact problem illustrated.
If we let m = 1.5 and n = 1.4, we get the famous DeMarre formula for normal impact. If we
would like to include an angle of obliquity in the aforementioned formula, it is common
practice to use
E = kd mt nsec pθ
(15.3)
Here p is an experimental parameter based on the projectile–armor combination and
θ is the angle of obliquity measured from the normal to the plate. Sometimes, the armor
fabrication process will affect the penetration. In this case, there is a function called the
figure of merit (FOM) where the perforation velocity of the armor is compared to that of
mild steel.
FOM =
Vl
Vlmild steel
(15.4)
Note that in Equation 15.4, the velocity used does not necessarily have to be the limit velocity. Another useful relationship commonly employed by the projectile designer is
Eperf =
1
2
mVperf
2
(15.5)
Inserting Equation 15.3 into Equation 15.5 yields
2
Vperf
= 2k
d mt n
sec p θ
m
(15.6)
Now taking the square root and assimilating terms, we get
Vperf = k
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d mt n
sec j(θ ) θ
m
(15.7)
374
Ballistics: Theory and Design of Guns and Ammunition
In 1886, DeMarre developed a famous formula for the penetration of a plate given a normal
impact:
mV 2
t1.4
α
=
d3
d1.5
(15.8)
where
m is the penetrator mass
V is the impact velocity
d is the diameter of the projectile
t is the plate thickness with α being an empirically derived constant
As a word of caution, many of these formulas are dangerous because of the units in the
empirically derived constant, it is commonplace to see CGS units in these formulas as well.
Over time, many have modified the DeMarre formula and used it in this form
mV 2
t
=α
d3
d
β
(15.9)
Here β is an empirically derived constant as well. In the aforementioned form, the DeMarre
formula is used when considering a normal impact. Some researchers have extended its
use to include an oblique impact and it would then take the following form:
mV 2
tg(θ )
=α
d3
d
β
(15.10)
where g(θ) is a function of the angle of obliquity and is most often taken as sec θ.
We sometimes define the specific limit energy (SLE) as
mVl2
≡ SLE
d3
(15.11)
Hans Bethe, a physicist at Cornell University in 1941, determined that for piercing type
problems (i.e., thick plate perforation where a hole is laterally or radially widened by the
penetrator), the constant, β, should be equal to 1, thus yielding
mVl2 ∼ td 2
(15.12)
Around the same time (1942), Zener and Holloman from Watertown Arsenal came up with
a formula for use when plugging or petaling is the predominant penetration mode. They
stated that in this case, β should equal to 2, thus yielding
mVl2 ∼ t 2d
(15.13)
In 1943, Curtis and Taub attempted to modify the DeMarre formula to account for a mode
change during the penetration event. In a thick plate, the mode changes at some point
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Penetration Theories
375
t
t΄
d
FIGURE 15.3
Section of a target plate that defines t and t′.
from a piercing to a plugging at the rear surface. This results in a decrease in energy consumed per unit path length, so the DeMarre formula had to be further modified to
mVl2
t
=α +γ
d3
d
(15.14)
Here α and γ are constants and γ < 0. If we define t′ as depicted in Figure 15.3, then γ is a
quadratic function of t′. Also t′ ∼ d and is the distance after the mode changes.
S. Jacobson, working at the Picatinny Arsenal in New Jersey further refined the concept
that there is a different energy relationship for each of the two modes. For plugging, this is
Eplug = force ⋅ distance ≈ π dt Ys ⋅ t
(15.15)
where Ys is the shear yield strength of the material. For the piercing mode, we have
E piercing = Yflow ⋅ V ≈
π d2
tYflow
4
(15.16)
where
V is the volume of the plug
Yflow is the flow or plastic yield stress of the target material
We can rewrite Equations 15.15 and 15.16 as
2
t
Eplug = k plug d 3 Ys
d
(15.17)
t
Epiercing = k piercing d 3 Yflow
d
(15.18)
If we graph both expressions, we obtain a plot as illustrated in Figure 15.4. To obtain t/dcrit,
we solve Equations 15.17 and 15.18 where
E plug = E piercing
(15.19)
Ys ≈ 0.6Yflow
(15.20)
using the relations that
k piercing =
© 2014 by Taylor & Francis Group, LLC
π
4
and k plug = π
(15.21)
Ballistics: Theory and Design of Guns and Ammunition
376
E
Plugging mode
Piercing mode
t/d
t/dcrit
FIGURE 15.4
Energy in penetration modes based on the model of Jacobson.
Then combining Equations 15.17 and 15.18, we get
2
π 3 t
t
t
d Yflow = π d 3 0.6Yflow → = 0.42
4 d
d crit
d
(15.22)
This value of t/d is the point where the mode of penetration changes from plugging to
piercing. Thus, against targets whose thickness is such that an attack by a penetrator
whose t/d ratio is greater than 0.42, we can expect that the penetration mode will be piercing, otherwise plugging is to be expected.
Lambert and Zukas proposed a model in 1982 while working at the BRL to cover more
general cases of penetration. If we examine Equation 15.14, we can see that as the plate
thickness goes to zero, the residual velocity should approach the striking velocity and the
limit velocity should approach zero. Expressed mathematically, we require that
lim Vl → 0
t →0
(15.23)
However, if we look at Equation 15.14, we note that if Vl = 0 and t = 0 it requires the product
γα to equal zero, which is not physically possible. Therefore, the Lambert model replaces γ
by [exp (−t/d) − 1] as
t
mVl2
t
= α + exp − − 1
3
d
d
d
(15.24)
Vl = 0 at t = 0
(15.25)
Vl = ∞ at t = ∞
(15.26)
This forces
and
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Penetration Theories
377
Since the penetrator volume is proportional to d2l and since there should be a dependence
on this volume in the specific limit energy, we want to keep the dimension of diameter
cubed in Equation 15.24, thus we shall write
2 l
d 3 → d 3 − cl c = d l
d
c −1
l
= d3
d
c
(15.27)
where c is a constant. We can then incorporate this into Equation 15.24 as
c
mVl2 l t
t
= α + exp − − 1
3
d
d
d d
(15.28)
Next we will include obliquity effects by adding in the angle of obliquity, θ through
replacement of t by t seck θ. In this case, if k = 1, we have the true path length through the
armor plate (line-of-sight thickness). We shall define
z=
t
sec k θ
d
(15.29)
We can now rewrite Equation 15.29 as
c
mVl2
t
l t
= α sec k θ + exp − sec k θ − 1
3
d
d
d d
(15.30)
If we solve Equation 15.30 for the limit velocity, we obtain
c
3
t
d
l t
Vl = α sec k θ + exp − sec k θ − 1
d
m
d d
(15.31)
The Lambert model was used to examine the firing of 200 long-rods into rolled homogeneous armor (RHA). The test conditions were as follows:
0.5 ≤ m [g] ≤ 3630
0.6 ≤ t [cm] ≤ 15
0.2 ≤ d [cm] ≤ 0.5
0° ≤ θ ≤ 60°
4≤
l
≤ 30
d
g
7.8 ≤ ρ 3 ≤ 19.0
cm
A least-squares fit of the results yielded the following: α = (4000)2, c = 0.3, and k = 0.75. If we
insert these into Equation 15.31, we get
l
Vl =
d
0.15
( 4000)
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d3 t
t
m
sec 0.75 θ + exp − sec 0.75 θ − 1
m d
d
s
(15.32)
Ballistics: Theory and Design of Guns and Ammunition
378
Please note the CGS units. The authors suggest that the model is applicable where t/d > 1.5.
Also we must note that nose geometry has a significant influence for t/d < 1.0. RHA or good
quality steel is the target (the specific properties are unimportant).
One measure of lethal effects once a projectile has perforated the target material is the
residual velocity. Vr is the symbol for the residual velocity of the penetrator. That is the
velocity that the penetrator moves with once it perforates the target. Mathematically, it is
defined in the Lambert model as
0,
0 ≤ Vs ≤ Vl
Vr =
1
p
a (Vsp − Vlp ) , Vs > Vl
(15.33)
If we assume that Vs is large so that the absorption of momentum by the target is negligible, then the momentum balance can be written in terms of identifiable penetrator mass
and velocity (mr and Vr), and the large quantity of unidentifiable target and penetrator
ejecta with each particle mi having a particular velocity, Vi. Thus, the momentum balance is
n
mrVr +
∑ m V → m V as V → ∞
i
i
s s
(15.34)
s
i =1
Even though Equation 15.34 is mathematically satisfying, in practice, it is usually difficult
to measure the mass and velocity of all of the fragments, so most of the miVi will remain
unknown.
We shall now consider a general case of impact as illustrated in Figure 15.5. Here we
shall let m′ be the mass of the ejecta. We can then write
π 3
d z
4
(15.35)
t
sec 0.75 θ
d
(15.36)
m′ = ρ
where
z=
t
m
θ
Vs
m΄
FIGURE 15.5
General case of projectile impact.
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d
Penetration Theories
379
therefore
m′ = ρ
π 3 t
π
d sec 0.75 θ = ρ d 2t sec 0.75 θ
4 d
4
(15.37)
If we now assume that
n
∑ m V = hm′V
i i
(15.38)
r
i =1
This is equivalent to stating that m′ is the mass of material pushed ahead of the penetrator,
m′ is ejected with speed Vr (plugging theory), and the total momentum of the ejecta jumble
is proportional to m′Vr. We can also write, in the limiting case, that the residual momentum
approaches the initial momentum or, mathematically
Mr
→1
M
(15.39)
If we substitute Equation 15.38 into Equation 15.34, we get
mVr + hm′Vr → msVs as Vs → ∞
(15.40)
which can be rearranged to yield
Vr
ms
→
as Vs → ∞
Vs
mr + hm′
(15.41)
We know that if penetration occurred, Equation 15.33 applies, so we have
(
Vr = a Vsp − Vlp
)
1
p
(15.42)
We can divide Equation 15.42 by Vs to get
1
V p p
Vr
= a 1 − l
Vs
Vs
(15.43)
which means that as Vs approaches infinity, the second term in the parentheses approaches
zero or
Vr
→ a as Vs → ∞
Vs
This is illustrated in Figure 15.6.
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(15.44)
Ballistics: Theory and Design of Guns and Ammunition
380
ms
Vr
mr + hm΄
Vr
Vs
Vs
Vl
FIGURE 15.6
Asymptote on limit velocity.
If we look at Equations 15.44 and 15.41, we see that
ms
a=
mr + hm′
(15.45)
Furthermore, we can assume in the plugging mode that the penetrators’ mass does not change
significantly during penetration, so we get ms = mr = m. We can then write Equation 15.45 as
m
a=
m + hm′
(15.46)
There is empirical evidence that suggests that h ≈ 1/3, so we can write
m
a =
1
m + 3 m′
(15.47)
If we assume that the penetrator remains intact throughout the perforation event, we can
write
(15.48)
KEimpact = KElimit + KEresidual
This can also be expressed as
(
Vr2 ∼ Vs2 − Vl2 → Vr ∼ Vs2 − Vl2
)
1
2
(15.49)
which, if written as
(
Vr = a Vs2 − Vl2
)
1
2
(15.50)
would say that p = 2. If we looked at momentum, we would get
Vs ~ Vl + Vr
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(15.51)
Penetration Theories
381
which could be written as
Vs = a(Vl + Vr )
(15.52)
Equation 15.52 implies that for a momentum balance, p = 1. Thus, it is clear that the value
for p should fall between 1 and 2. Lambert accounted for this by choosing
t
sec 0.75 θ
d
(15.53)
t
π
→ ∞ and/or θ →
d
2
(15.54)
p → 2 as t → 0
(15.55)
p = 2+z = 2+
where both p and z grow monotonically as
also
Lambert also found that a better empirical fit was obtained if he let
p = 2+
z
t
= 2+
sec 0.75 θ
3
3d
(15.56)
A numerical model for penetration was proposed by A. Tate to determine penetration of
metals [2]. The base equation for this model is
1
1
ρ p ( Vi − u)2 + Yp = ρ t u2 + Rt
2
2
(15.57)
where
ρp is the density of the projectile material
ρt is the density of the target material
Vi is the impact velocity
u is the instantaneous projectile velocity
Yp and Rt are the ballistic resistances of the projectile and target, respectively, defined as
Yp = 1.7σ p
(15.58)
2
E
Rt = σ t + ln 0.57 t
3
σ
t
(15.59)
where
σ P is the yield strength of the projectile material
σt is the yield strength of the target material
Et is the modulus of elasticity of the target material
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Ballistics: Theory and Design of Guns and Ammunition
382
The way the Tate model is used is to integrate Equation 15.57 numerically until the velocity
goes to zero or perforation occurs. When the projectile stops, a second integration determined
the depth of penetration. When perforation occurs, the value of u is the residual velocity. Tate
states that the accuracy of this method is within 20%. One of the model’s downsides is that it
does not handle oblique impacts but it can at least be altered by the line-of-sight thickness.
If a penetrator hits a target at a great enough angle, it may ricochet. The ricochet process
can be described as follows. During impact, both the projectile and the target are compressed elastically. When this energy is released, it will change the projectile’s motion.
Deformations because of resisting force of the target will change the direction of the penetrator. Rotating moments are generated by internal forces acting within the projectile.
In general, thin plates do not allow ricochet except at extreme angles of attack. Tate
has produced a ricochet formula for the critical ricochet angle (oblique impacts at angles
greater than this will ricochet):
1
2 ρ pV 2 L2 + d 2 ρ p 2
tan β >
1+
3 YP Ld ρ t
3
(15.60)
Here Yp is a characteristic strength usually taken as the Hugoniot elastic limit (described
in Section 16.3), the subscripts “p” and “t” are projectile and target, respectively, and L, d,
and V are the length, diameter, and velocity of the penetrator, respectively.
As vehicles become lighter weight, aluminum is being used more and more as armor. It is
therefore necessary to determine the penetration capabilities of projectiles into aluminum.
Aluminum behaves a little differently than steel during penetration by ogival projectiles in its tendency to be pierced rather than to develop plugs. A penetrator is usually of
significantly greater density than the target in most cases. One significant difference is the
evidence of a layer of aluminum with an altered microstructure on the penetrated surface.
This indicates a melt layer that is believed to assist in penetration.
A simple model of projectile penetration into aluminum was put forward by Forrestal et al.
in 1992 [3]. A distinct advantage of this model is its simplicity. A possible disadvantage is
that the empirical nature is not universal. Even though the study was performed specifically with 7075-T651 targets, it yields a fairly good representation of aluminum penetration. The model assumes normal impact of the projectile and that the projectile is rigid.
This may, at first, seem to be a restrictive assumption, but the method provides reasonable
estimates for slightly yawed projectiles if the angle is below about 5° and possibly further.
We first define the caliber-radius-head as
s
d
(15.61)
d
4ψ − 1
2
(15.62)
ψ =
where
d is the diameter of the projectile
ψ is the caliber-radius-head
s is the ogive radius
We can also define a nose length as
l=
This geometry is illustrated in Figure 15.7.
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Penetration Theories
383
L
l
d
s
θ
FIGURE 15.7
Ogival penetrator for the model of Forrestal et al.
We shall say that the resistance force of the aluminum target on the penetrator in this
case will have two components: one normal to the surface (normal stresses) and one tangential to the surface (shear stresses and friction). If we lump the shear stress in with the
stress owing to friction and furthermore assume that the tangential stress is proportional
to normal stress, we can write
σ t = µσ n
(15.63)
where
σt is the tangential stress
σn is the normal stress
μ is the proportionality constant (coefficient of sliding friction)
Forrestal et al. [4] developed a formula for the axial force on an ogival nose:
π 2
Fz = 2πs
∫
θ0
d
s−
sin θ −
2
s
σ (V , θ ) dθ
+
θ
µ
θ
(cos
sin
n z
(15.64)
where
d
s− 2
θ 0 = sin
s
−1
(15.65)
Here Vz is the instantaneous velocity during penetration. The stress function σn(Vz, θ) is
assumed to be similar to that of a spherically symmetric expanding cavity (defined later).
If we let V be the constant velocity at which the tip of the projectile radially expands the
hole, then we can write the radial stress at the cavity surface as
ρt
σr
= A + B
V
Y
Y
where
σr is the radial stress
Y is the material yield stress
ρt is the target density
A and B are constants defined as
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2
(15.66)
Ballistics: Theory and Design of Guns and Ammunition
384
A=
n
2 2E
1 +
I
3 3Y
(15.67)
3
2
(15.68)
B=
where
( )
1− 32YE
I=
∫
0
(− ln x)n
dx
1− x
(15.69)
In these expressions, E is Young’s modulus and n is the strain hardening exponent
(assumes power-law strain hardening). For an assumed incompressible 7075-T651 aluminum, Forrestal et al. [3] provide I = 3.896 and A = 4.609.
Empirically, curve-fitting the stress–strain curves (thus including compressibility) for
7075-T651 yielded slightly different results with A = 4.418 and B = 1.068.
To approximate the normal stress on the ogive, we can replace the spherically symmetric
velocity, V in Equation 15.66 with Vz cos θ, then we have
ρt
σ n (Vz , θ )
= A + B
Vz , cos θ
Y
Y
2
(15.70)
If we insert Equation 15.70 into Equation 15.64, we obtain
π 2
Fz = 2π sY
∫
θ0
s − d2
ρt 2
2
sin θ −
(cos θ + µ sin θ A + B Vz cos θ dθ
s
Y
(15.71)
Now we integrate to obtain
Fz =
π d2
ρ V2
Y α + β t z
4
Y
(15.72)
where
π
α = A 1 + 4 µψ 2 − θ 0 − µ (2ψ − 1) 4ψ − 1
2
2
8ψ − 1
µ (2ψ − 1)(6ψ + 4ψ − 1) 4ψ − 1
2π
−
µψ
θ
β = B
+
−
0
2
24ψ 2
2
24ψ
(15.73)
(15.74)
Now that we have an expression for force as a function of velocity, we need to come up
with how this varies during penetration.
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Penetration Theories
385
We can write Newton’s second law as
dVz
dt
− Fz = m
(15.75)
We can convert this time integral to a distance integral and rewrite it as follows:
dVz
dz
− Fz = mVz
(15.76)
One can write the mass of our projectile in terms of the parameters we have already
described. The mass of the cylindrical section of the projectile is
mcylinder = ρ p
π d2
L
4
(15.77)
We can write the mass of the ogive as
mogive = ρ p
π d3
k
8
(15.78)
where
4ψ − 1
4ψ 1
k = 4ψ 2 −
+ 4ψ − 1 − 4ψ 2 (2ψ − 1)sin −1
3 3
2ψ
(15.79)
Now the total mass of the projectile is
m = mcylinder + mogive = ρ p
π d2
kd
L+
4
2
(15.80)
If we insert Equations 15.80 and 15.72 into Equation 15.76, we get, after some rearrangement
kd
Vz
dVz
−dz = ρ p L +
+
2
Y
α
βρ tVz2
(15.81)
This can be integrated as
P
0
kd
Vz
dVz
− dz = ρ p L +
+
2
Y
α
βρ tVz2
V
0
∫
∫
(15.82)
0
The result of this integration is
P=
1 ρp
kd β ρ tV02
L + ln 1 +
2β ρt
2 α Y
where
P is the final penetration depth
V0 is the impact velocity
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(15.83)
Ballistics: Theory and Design of Guns and Ammunition
386
If the penetration depth, P, is greater than the target thickness, perforation will occur.
When this is the case, it is useful to be able to calculate the residual velocity of the penetrator, which we do by integrating Equation 15.82 with different limits of integration:
Vr
T
kd
Vz
dVz
− dz = ρ p L +
2 α Y + βρ tVz2
V
0
∫
∫
(15.84)
0
where T is the target thickness. Performing the integration yields
αY
2βρ tT α Y
Vr =
+ V02 exp −
−
kd
βρ t
ρ p ( L + 2 ) βρ t
(15.85)
This model has proven to be fairly accurate (within 15%) once the coefficients have been
tuned. It is fairly sensitive to the friction coefficient, μ, incorporated in both α and β, which
Forrestal et al. [4] suggest should be between 0 and 0.06.
Problem 1
A German 280 mm armor-piercing projectile weighs 666 lbm and is about 34 in. in length.
It strikes a British warship in the 1/2 in. thick vertical side plating at an angle of 12° from
horizontal along the path depicted below. The initial impact velocity is 2000 ft/s. Determine
the residual velocity of the shell after passing through each compartment and how far
through the ship it will go (i.e., in which compartment will it stop).
Assume the density of the armor plate to be ρ = 0.283 lbm/in.3
7.00-in. thick
0.25-in. thick
0.50-in. thick
Path of shell
1.25-in. thick
12°
4.00-in. thick
(Assume normal to shell path)
Answer: The projectile is arrested by the 1.25 in. deck.
Problem 2
An explosively formed penetrator impacts a 4 in. thick RHA plate at a velocity of 1500 m/s.
The penetrator parameters are given later. Determine if the penetrator will perforate the
target using the Lambert/Zukas model given
1. A normal impact
m
Answer: Vl = 1299 yes
s
2. An impact at 30° obliquity
m
Answer: Vl = 1389 yes
s
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Penetration Theories
387
Penetrator information:
l = 95 [mm]
m = 1.25 [lbm]
m
d = 22 [mm] Vs = 1500
s
Problem 3
A German 7.5 cm Gr 34A1 projectile is fired at a 2 in. thick armor plate at a 30° obliquity.
The impact velocity is 400 m/s. The penetrator parameters are given later.
1. Determine whether the penetration mode will be plugging or piercing through
use of the Jacobson model for a normal impact.
Answer: Piercing
2. Determine if the penetrator will perforate the armor though use of the Lambert model.
Answer: No
3. Comment on the validity of the model.
Penetrator information:
l = 39 [cm]
m = 1.25 [kg]
m
d = 7.5 [cm] Vs = 400
s
Problem 4
A Japanese 20 mm projectile with the properties given later impacts the 1/2 in. thick aluminum armor plate on a U.S. plane’s rear gun mount at 30° obliquity. If the projectile and
the armor have the following properties:
Determine how deep the projectile will penetrate into the armor (assume μ = 0.03).
Answer: P = 53.1 [mm] = 2.09 [in.]
1. If the projectile perforates the armor, determine its residual velocity.
m
Answer: Vr = 423
s
Estimated penetrator information:
s = 40 [mm] m = 128 [g]
m
d = 20 [mm] Vs = 500
s
lbm
ρ p = 0.283 3
in.
L = 60 [mm]
Estimated armor information:
lbm
A = 4.418Y = 39, 000 [psi] ρ t = 0.098 3
in.
B = 1.068
© 2014 by Taylor & Francis Group, LLC
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388
Problem 5
You are asked to design a “slug butt” for a test gun. The requirement for thus device is
to stop a 106 lbm steel projectile moving at 1000 m/s. You have available a pile of 2″ thick
Rolled Homogeneous Armor (RHA) plates that you can stack together. Note that stacking plates together results in slightly worse performance than a solid plate—but we can
account for this by adding some extra material to provide “margin.” Normally slug butts
are constructed at an angle to deflect ricochets downward and improve ballistic performance but this decreases the “target” area. Given the information provided earlier and
later, determine the number of plates you need to prevent a penetration and determine the
angle of the slug butt—do not use an angle greater than 40° as the “target area” will be too
small. List all assumptions and comment on your design.
Projectile information:
l = 40 [in]
m = 106 [lbm]
d = 155 [mm]
m
Vs = 1000
s
lbm
ρ p = 0.283 3
in.
Armor information:
lbm
ρs = 0.283 3 available in 2 in. thick plates
in.
Problem 6
A German 305 mm armor piercing projectile weighs 894 lbm and is about 35.2 in. in length.
It strikes a British warship in the 3-1/4 in. thick turret crown at an angle of 20° from horizontal along the path depicted later. The initial impact velocity is 1800 ft/s. Determine the
residual velocity of the shell after passing through each compartment and how far through
the ship it will go (i.e., in which compartment will it stop).
Assume the density of the armor plate to be ρ = 0.283 lbm/in.3.
3.25˝ thick
15°
9.00˝ thick
1.50˝ thick
20°
8.00˝ thick
Path of shell
1.00˝ thick
6.00˝ thick
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Penetration Theories
389
Problem 7
A German 280 mm armor piercing projectile weighs 666 lbm and is about 34 in. in length.
It strikes a British warship in the 0.43″ thick vertical side plating at an angle of 10° from
horizontal along the path depicted later. The initial impact velocity is 1900 ft/s. Determine
the residual velocity of the shell after passing through each compartment and how far
through the ship it will go (i.e., in which compartment will it stop)
Assume the density of the armor plate to be ρ = 0.283 lbm/in3.
3.25˝ thick
15°
9.00˝ thick
0.43˝ thick
1.50˝ thick
8.00˝ thick
10°
Path of shell
1.00˝ thick
6.00˝ thick
Problem 8
A British Short Magazine Lee Enfield (SMLE) is fired at a sniper plate across no-man’s land in
WWI (because of the static nature of the fighting, snipers in the opposing lines fired through
small holes in thick metal plates to minimize exposure). The projectile has a mass of 175 grains
and the projectile is 1.1 in. long and made of lead (ignoring the copper jacket). The diameter is
0.310 in. The range to the target is 200 yards so we can assume an impact velocity of 1512 ft/s.
The angle of impact is 12.85 min from the normal. Using the Lambert–Zukas model, determine
the thickness of armor plate that the projectile will penetrate (i.e., obtain V50).
Assume the density of the armor plate to be ρt = 0.283 lbm/in3. Assume the density of
lead to be ρp = 0.407 lbm/in.3.
Problem 9
Using the same information in Problem 8, determine how deep a projectile will penetrate into
a 1 in. thick steel sniper plate assuming the impact is normal this time and the impact velocity
is 1800 ft/s. Use the Tate formula. The additional target and bullet properties are as follows:
lbf
σ p = 10, 000 2
in.
lbf
lbf
σ t = 36,000 2 Et = 29,000,000 2
in.
in.
The Tate formula shows the energy balance between the projectile and the target. When
solved for the penetration depth, the following equation results
ρp
P=
Yp
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Vi
∫ u(v)l(v)dv
vc
Ballistics: Theory and Design of Guns and Ammunition
390
In this formula, u(v) is the instantaneous velocity of the base of the hole, v is the instantaneous velocity of the projectile and l(v) is the instantaneous projectile length. The limits of
integration are between the impact velocity, Vi, and a cutoff velocity, vc, that depends on
whether the projectile is stronger than the target or not. For this case, we can use
vc = 2
Rt − Yp
ρp
for Rt > Yp
The other terms are as follows:
)
(
1
u(v) =
v − γ v2 + A γ =
2
−
γ
1
A=
ρt
ρp
2(Rt − Tp )(1 − γ 2 )
ρt
l(v) v + v 2 + A
=
Vi + Vi + A
L
Rt − Yp
γ Yp
)(
(
γρ p
v v2 + A − γ v2 − V V 2 + A − γ V 2
exp
i
i
i
2
2(1 − γ )Yp
)
where L is the initial length of the projectile. Comment on the results.
Problem 10
At the end of World War I, the German navy surrendered to the British at Scapa Flow
(a large anchorage in northern Scotland). When it appeared that surrender negotiations
were breaking down, the German sailors opened the sea cocks (valves in the bottom of
the ships) and sank most of their ships within sight of astounded British onlookers. In
the 1920s, the British raised what ships they could and used them as targets to assess the
penetration and bursting characteristic of their heavy shell [5]. One such test was against
the side armor of S.M.S. Baden, the largest warship built by Germany during the war. We
would like to examine two impacts against the armor of this vessel. In each case, the projectile data are given next:
Projectile information [6]:
l = 66 [in.]
d = 15 [in.]
s = 90 [in.]
m = 1938 [lbm]
lbm
ρ p = 0.283 3
in.
ft
Vs = 1550
s
Armor information:
lbm
ρs = 0.283 3
in.
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Penetration Theories
391
For each of the following situations, determine the residual velocity of the projectile
(if any):
(a) Gun deck
( b) B-Turret roof
θ = 16.5 [degrees]
θ = 75.25 [degrees]
t = 1.187 [in.]
t = 4 [in.]
lbm
ρ t = 0.283 3
in.
lbm
ρ t = 0.283 3
in.
15.2 Penetration and Perforation of Concrete
Concrete penetrating munitions have always been important in the military arsenal.
Bunkers, buildings, and walls are used as cover by an enemy and it is required to
perforate the structure and deliver some type of lethal or nonlethal effect behind the
obstruction.
Concrete comes in a variety of forms that have variable strengths, reinforcement geometry, and material properties owing to curing. Each of these forms behaves somewhat
differently when impacted by a projectile. There is some evidence that once the impact
velocity of a projectile is great enough, one can ignore reinforcement and only the concrete
strength becomes important. As a consequence of the high compressive strength of concrete relative to its tensile strength, it tends to spall readily.
A relatively simple model of projectile penetration into concrete was put forward by
Forrestal et al. in 1994 [7]. This model has an advantage in its simplicity. But a slight disadvantage is that its empirical nature makes its global applicability somewhat limited. We
shall use this model as a fairly good representation of concrete penetration physics. The
model assumes normal impact of the projectile and that the projectile is rigid. This may
seem to be restrictive assumptions; however, the method provides reasonable estimates for
slightly yawed projectiles if the angle is below about 5° based on this author’s own work.
The point of departure is the determination of the force on the nose of the projectile that
is defined in a manner similar to a fluid mechanics analysis as
F=
π d2
(τ 0 A + NBρV 2 )
4
(15.86)
8ψ − 1
24ψ 2
(15.87)
With N defined as
N=
In these equations, the projectile properties are as follows (see Figure 15.7): d is the diameter of the projectile; ψ is the caliber-radius-head, defined in Equation 15.88; V is the projectile velocity (assuming rigid body motion); and s (used in Equation 15.88) is the ogive
radius.
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Ballistics: Theory and Design of Guns and Ammunition
392
The caliber-radius-head is defined as
ψ =
s
d
(15.88)
The target properties used in Equation 15.86 are as follows: ρ is the density of the target,
the product τ 0A is a shear strength parameter obtained from a triaxial strength test, and B
is a compressive strength parameter. In this model, the parameters are set as
B=1
(15.89)
τ 0 A = Sf c′
(15.90)
Here S is a dimensionless empirical constant that depends upon the unconfined compressive strength f c′.
If we define the instantaneous depth of penetration as z, we find that for z > 2d, we can
write
F=
π d2
( Sfc′ + NρV 2 ) , z > 2d
4
(15.91)
This equation is valid for deep penetration depths. For depths less than two projectile diameters, the penetration process is affected by surface cratering. Beyond two projectile diameters,
the hole caused by the projectile will be approximately equal to the projectile diameter. This is
known as the tunnel region. We shall define the penetration depth as P.
Below two projectile diameters, the damage to the concrete will, in general, be a conical
taper called the crater. This is illustrated in Figure 15.8.
In the surface crater region, the force on the projectile nose is proportional to the penetration depth or, mathematically
F = cz , 0 < z < 2d
(15.92)
Here c is a constant, which we will soon define.
Crater region
Tunnel region
Direction of
penetration
d
2d
z
FIGURE 15.8
Illustration of a concrete penetration.
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Penetration Theories
393
If we begin with Newton’s second law, we see that
F = ma = m
d2z
dt 2
(15.93)
Here m is the mass of the projectile. Since we know the force acting on the projectile will
tend to slow it down, we can equate Equations 15.92 and 15.93:
d2z
= −cz
dt 2
(15.94)
d2z
= −ω 2 z
dt 2
(15.95)
m
We can rewrite Equation 15.94 as
where we have defined
ω2 =
c
m
(15.96)
If we assume a solution of the form
z = A1sin ωt
(15.97)
dz
= A1ω cos ωt
dt
(15.98)
d2z
= − A1ω 2 sin ωt
dt 2
(15.99)
we can write
Our initial conditions are such that at t = 0, dz/dt = Vs, where Vs is our striking velocity, so
Vs = A1ω → A1 =
Vs
ω
(15.100)
Then, we have for z < 2d
z=
© 2014 by Taylor & Francis Group, LLC
Vs
sin ωt
ω
(15.101)
dz
= Vs cos ωt
dt
(15.102)
d2z
= −ωVs sin ωt
dt 2
(15.103)
Ballistics: Theory and Design of Guns and Ammunition
394
We now use a compatibility condition that at z = 2d, both Equations 15.103 and 15.91 must
yield the same answer. We shall call the time it takes the projectile to reach 2d, t1 and the
velocity at that point will be V1, thus at z = 2d we have
F t=t1 =
πd 2
Sf c′ + NρV12 , z = 2d
4
(
)
d2z
= −ωVs sin ωt1 , z = 2d
dt 2
(15.104)
(15.105)
Since F = ma, we can combine the aforementioned equations to write
mωVs sin ωt1 =
πd 2
Sf c′ + NρV12 , z = 2d
4
(
)
(15.106)
Also at t = t1, Equations 15.101 and 15.102 can be written as
Vs
sin ωt1
ω
(15.107)
V1 = Vscos ωt1
(15.108)
2d =
We now rearrange Equation 15.108 to
Vs =
ω 2d
cos ωt1
(15.109)
Now insert Equation 15.109 into Equation 15.106 giving us
mω 2 2 d =
πd 2
Sf c′ + NρV12
4
(
)
(15.110)
And if we make use of Equation 15.96, we can obtain c as
c=
πd
Sfc′ + N ρV12
8
(
)
(15.111)
We now need to find V1, which we do by squaring Equations 15.107 and 15.108 and adding
them, resulting in
V12 +
4cd 2
= Vs2sin 2 ωt1 + Vs2cos 2 ωt1
m
(15.112)
Making use of a trigonometric identity and rearranging brings us to
c=
© 2014 by Taylor & Francis Group, LLC
(
m
Vs2 − V12
4d 2
)
(15.113)
Penetration Theories
395
If we now equate Equations 15.111 and 15.113, we get
2 mVs2 − πd 3 Sfc′
2 m + πd 3 N ρ
V12 =
(15.114)
Once we have V1 and c, the determination of the time t1 is found simply through use of
Equation 15.108:
t1 =
1
V
cos −1 1 =
ω
Vs
m
V
cos −1 1
c
Vs
(15.115)
To summarize the analysis procedure for the crater region, we must first find V1 through
use of Equation 15.114, then we find c through use of Equation 15.113, and finally, we find
t1 through use of Equation 15.115.
If V goes to zero before time, t1 is reached, the projectile never penetrates deeper than
the crater region and our analysis would be complete. The depth of penetration in this case
would be found from Equation 15.102:
V = 0 = Vs cos ωt
(15.116)
This would occur when
ωt =
π
π m
t=
sin ωt = 1
2
2 c
(15.117)
If we insert this result into Equation 15.101, we obtain the achieved depth of penetration, P:
P = Vs
m
c
(15.118)
The striking velocity that would make this true would be determined from Equation
15.114 with V1 set equal to zero. So for a projectile to stop before creating a tunnel, the
velocity is given by
VsNotunnel ≤
2πd 3Sf c′
2m
(15.119)
If the projectile penetrates beyond 2 diameters into the concrete, it will enter the so-called
tunnel region. When the projectile continues into the tunnel region, there is a change in
the governing equation as discussed earlier. To determine the depth of penetration, we
begin by combining Equations 15.91 and 15.93 to obtain
m
d 2 z πd 2
Sf c′ + NρV 2 , 2d < z < P
=
dt 2
4
(
)
(15.120)
We can transform our independent variable from time to distance and we can write
mV
© 2014 by Taylor & Francis Group, LLC
dV πd 2
Sf c′ + NρV 2 , 2d < z < P
=
dz
4
(
)
(15.121)
Ballistics: Theory and Design of Guns and Ammunition
396
If we rewrite Equation 15.121 as follows:
dV πd 2 Sf c′
=
+ Nρ V
dz
4m V
(15.122)
Now we integrate it from V1 to zero and 2d to P, so we can write
0
0
P dV
dz dV =
dz
V1
2d
∫ ∫
V1
P πd 2 Sf ′
c
+ NρV dz dV
4m V
2d
∫ ∫
(15.123)
which results in
P=
NρV12
2m
ln 1+
− 2d , 2d < P
2
πd Nρ
Sf c′
(15.124)
If we have determined through use of Equations 15.113 through 15.115 that a projectile will
penetrate beyond the tunnel region, we can write a procedure to determine the depth of
penetration as follows. First, calculate V1, c, and t1 as described earlier for the crater region.
Then calculate P from Equation 15.124. If this is greater than the concrete thickness, the
projectile will perforate. If not, the projectile will penetrate to depth P. It would be good to
see if a spall thickness is created (as will be described in Section 16.3) by the impact and
if this is the case, we could add the spall thickness to P and perforation may still result
(though with low residual velocity).
Since the dimensionless parameter S is obtained or verified experimentally, it would be
nice to know how close we came to our estimate by direct calculation. If one had an experiment where a given projectile penetrated to depth P, we can back calculate S as follows. We
start with Equation 15.124 and rearrange thusly
S=
1
NρV12
f c′
πd 2 Nρ
exp (P − 2d)
− 1
2m
(15.125)
In an experiment, we usually are given the striking velocity, so we want to replace V1 in
this equation with Vs, so we use Equation 15.114:
S=
(
)
N ρ 2 mVs2 − πd 3 Sfc′
fc′ (2 m + πd N ρ )
3
1
πd 2 N ρ
− 1
exp ( P − 2 d)
2 m
(15.126)
which can be simplified to
π d 3 Sfc′
N ρVs2 1 −
2 mVs2
1
S=
3
πd N ρ
πd 2 N ρ
fc′ 1 +
− 1
exp ( P − 2 d)
2 m
2 m
© 2014 by Taylor & Francis Group, LLC
(15.127)
Penetration Theories
397
Dimensionless empirical constant versus unconfined compressive strength
25
S (dimensionless)
20
15
y = 86.431x –0.5158
10
y = 93.48x –0.5603
y = 103.71x –0.6142
5
0
0
20
40
60
80
100
120
f ΄c(MPa)
FIGURE 15.9
Determination of dimensionless parameter S for Forrestal et al. [8] concrete penetration model.
With this equation, one can find S if you know the striking velocity and the concrete
strength. Forrestal et al. [7] have calibrated this equation with several experiments.
A reproduction of their chart is shown in Figure 15.9 with the addition of upper and lower
bounds based on their data. The equation used to determine S given the unconfined compressive strength f c′ is
S = 93.48 f c′−0.5603
(15.128)
Here recall that f c′ is in MPa and S is dimensionless. Bounding equations are shown in
Figure 15.9. These equations were obtained through use of a curve fit routine.
Problem 11
A 0.50 caliber projectile is fired at an extremely thick concrete wall of 2100 psi unconfined
compressive strength and density of 0.084 lbm/in.3 It strikes with no obliquity and a 2000
ft/s velocity. How far does it penetrate?
Answer: P = 20.4 [cm]
Projectile information:
s = 63.50 [mm] m = 662 [grains]
ft
d = 12.70 [mm] Vs = 2000
s
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398
Ballistics: Theory and Design of Guns and Ammunition
Problem 12
Areal density is a measure of the mass of one square inch (or sometimes a square foot
or meter) of a material used as armor. Given a threat projectile with the properties provided later, find the armor solution with the lowest areal density that prevents perforation if your choices are steel with density 0.283 lbm/in.3, concrete of 2500 psi unconfined
compressive strength and density of 0.084 lbm/in.3, and aluminum with the following
properties:
A = 4.418
B = 1.068
Y = 39, 000 [psi]
lbm
ρ t = 0.098 3
in.
Assume the projectile strikes with 15° of obliquity.
Penetrator information:
s = 35 [mm]
d = 12 [mm]
L = 15 [mm]
m = 0.03 [kg]
m
Vs = 600
s
lbm
ρ p = 0.283 3
in.
Problem 13
We would like to compare defenses against the projectile in Problem 10. Assuming this
projectile impacts a steel plate at zero degrees obliquity
a. The thickness of the armor required to prevent penetration.
b. The thickness of 1500 psi unconfined compressive strength concrete required to
do the same assuming that the concrete does not spall—assume a density of 0.084
lbm/in.3 for the concrete—note that this particular shell has a secant ogive so as
an estimate for the purposes of this problem—divide the resultant ogive length
by 2—please note that this is simply a guess and not based on physics.
c. Comment on the validity of the answer to part b.
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Penetration Theories
399
15.3 Penetration and Perforation of Soils
In recent times, the penetration of soils has gained importance in the terminal ballistic
field. Enemy strong points have been encountered below a soil layer. Land mines need to
be defeated below various types of soils as well. It is therefore necessary to determine the
penetration capabilities of projectiles into soils with the intention of defeating a buried
target.
As a reasonable approach to determine soil penetration, we shall use the method of
Forrestal and Luk [8]. While other approaches exist, this rather simple procedure is excellent for introducing the physics of the problem.
Soils vary widely in their behavior under penetration loadings. Because the behavior
is somewhat complicated, more parameters are needed to describe a soil than a material
such as a metal. The first thing we have to realize is that soil can be in a state where the
density is less than its locked density. The locked density is where the soil behaves like a
solid or fluid in compression (i.e., its states are defined by a hydrostat). We therefore need
to introduce two densities: p0, its initial density and ρ*, its locked density. We also need to
define η*, its locked volumetric strain. Here we define η* as
η* = 1 −
ρ0
ρ*
(15.129)
Two typical models used for soils come directly from our failure theories of structures.
They are the Tresca (maximum shear stress) theory and the Mohr–Coulomb theory of failure. Both of these were introduced in Section 4.2. Here we shall use a combination of the
two. A Mohr–Coulomb yield criteria with a Tresca flow rule. For the Tresca criterion, once
a shear stress failure level is achieved, the material strength is not increased with increasing load. With the Mohr–Coulomb criterion, the yield stress in the material increases with
compressive load. The combination of the two allows the material to resist more load as
compression is applied up to a point, then further increase in the compressive loading will
not affect the material strength.
Similar to the aluminum penetration model, we again define the caliber-radius-head as
s
d
(15.130)
d
4ψ − 1
2
(15.131)
ψ =
where
d is the diameter of the projectile
ψ is the caliber-radius-head
s is the ogive radius
We again define nose length as
l=
The method considers the resistance force of the soil on the penetrator to have two components: a normal force (normal stresses) and a tangential force (shear stresses and friction).
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
400
If we lump the shear stress in with the stress owing to friction and furthermore assume
that the tangential stress is proportional to normal stress, we can again write
σ t = µσ n
(15.132)
where
σt is the tangential stress
σn is the normal stress
μ is the proportionality constant (a coefficient of sliding friction)
Forrestal et al. [4] developed a formula for the axial force on an ogival nose that we introduced in Section 15.1 and we again use here
π /2
Fz = 2π s
s − d2
(cos θ + µ sin θ ) σ n (Vz ,θ )dθ
s
∫ sin θ −
θ0
(15.133)
where
s − d2
θ 0 = sin −1
s
−1 2ψ − 1
= sin
2ψ
(15.134)
Here Vz is the instantaneous velocity during penetration. The stress function σn (Vz, θ) is
assumed to be similar to that of a spherically symmetric expanding cavity.
At this point, we are going to depart from the mathematics to look at the penetration
event in a qualitative manner. Let us assume that we are at some axial location in the
ogive of the projectile and we are looking in the direction of penetration at time, t. What
we would see is illustrated in Figure 15.10. The projectile would be opening a cavity at a
rate that we shall call Vt. The plastic zone would be expanding at some rate ct. Here c is the
speed of the plastic wave (dependent upon the Hugoniot jump conditions to be discussed
Elastic region
c1
c
Plastic region
Undisturbed region
Vt
Projectile
FIGURE 15.10
Elastic and plastic compression zones at a section of an ogive penetrating into soil looking in the direction of
penetration.
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Penetration Theories
401
in Section 16.1). The elastic zone would be expanding at a rate c1t. Here c1 is the speed of the
dilatational wave in the material. It can be shown that V and c are related (as per the shock
theory that will follow in Chapter 16) and we can define a parameter, γ as
1
3
3
τ
V
γ = = 1 + c − (1 − η *)
c
2E
(15.135)
where E is Young’s modulus. With the aforementioned physics, Forrestal and Luk [6]
derived material response models for each of the three failure models we have discussed
earlier. For the detailed derivation, the interested reader is referred to that paper. The basic
idea was to have a general function for the force acting on the projectile nose that we can
integrate using Newton’s second law to obtain the velocity and penetration distance as a
function of time.
If we insert expressions that relate the radial expansion velocity of the cavity, V, to the
projectile penetration velocity, Vz, we can put the expression for the retarding force in this
form
Fz = α s + β s Vz2
(15.136)
where
πd 2
π
τ c A 1+ 4 µψ 2 − θ 0 − µ (2ψ − 1) (4ψ − 1)
4
2
(15.137)
2
8ψ − 1
πd 2
µ (2ψ −1)(6ψ + 4ψ − 1) 4ψ − 1
2π
−
+
−
µψ
θ
ρ 0B
0
2
4
24ψ 2
2
24ψ
(15.138)
αs =
βs =
Here the coefficients A and B are dependent upon the material model used for the soil.
Recall that the definition for the Tresca criterion implies that once a material reaches
its state of maximum shear stress, it begins to deform plastically and cannot support any
more load. For a soil that behaves in a Tresca type manner, we have
3
τc
− (1 − η *)
1+
2
2E
A = 1 − ln
3
3
τc
1+
2E
(15.139)
1
3
3
τc
3τ c
3τ c
η
(
*)
−
−
+
1
1
+η * 1 −
2E
3
2E
E
+
B=
2 −
4
2(1 − η *)
τ
3
3
21+ c
τc
1
1
η
(
*)
−
−
+
2E
2E
2
© 2014 by Taylor & Francis Group, LLC
3
τc
31+
2E
1 +
(1 − η *)
(15.140)
Ballistics: Theory and Design of Guns and Ammunition
402
Recall that the definition for the Mohr–Coulomb criterion implies that as the compressive
forces increase, it becomes harder to have the material fail in shear. For a soil that behaves
in a Mohr–Coulomb type manner, we have
τc
1 1 + 2E
A=
α γ
B=
2α
−
1
λ
(15.141)
3
(1 − η *)(1 − 2α )(2 − α )
τc
1 1 + 2E
+ 2
γ γ
2α
2
3
3
3
τ
τ
γ
[
(
η
*)(
α
)
+
3
γ
]
3
2
1
−
2
−
c
+η * 1 − c −
4
E
E
2
τ
(1 − η *)(1 − 2α )(2 − α ) 1 + c
2E
(15.142)
Also note that the Tresca criterion behaves the same as the Mohr–Coulomb criteria with
λ = 0. We define
α=
3λ
3 + 2λ
(15.143)
Because of a singularity in the governing equations, there is a special set of equations for
the Mohr–Coulomb criterion when we have λ = 3/4. In this case,
τc
1+ 2E 4
A = 2
−
γ 3
2
τ c 3τ c
3τ c
η
*
1
+
−
+
1
2E E
2E 2
−2 ln γ
+
−
B=
(1 − η *)
γ3
3
τ
3 ln 1 + c
1
2E
3 −
(
*)
η
1
−
τc
1 +
2E
(15.144)
(15.145)
When a material behaves according to the model that combines both Mohr–Coulomb and
Tresca behaviors, things become slightly more complicated. The parameters A and B will
be dependent upon the rate of loading. One must keep in mind that this failure criterion
implies that up to some stress level, the material will have improved resistance to compressive loading because of the internal friction of the grains and after a limit load is reached
(τm), the material simply yields regardless of load. Thus, we can consider three velocity
regimes: V < Vmin, where the yielding is completely Mohr–Coulomb behavior; Vmin < V <
Vmax, where the yielding closest to the projectile is by Tresca criterion and the yielding near
the elastic–plastic interface is according to the Mohr–Coulomb criterion; and V > Vmax,
© 2014 by Taylor & Francis Group, LLC
Penetration Theories
403
where the entire yield region is according to the Tresca model. We shall consider each of
these cases.
If V < Vmin, we stated that the yielding is completely according to the Mohr–Coulomb
model. Thus, Equations 15.141 through 15.145 apply. The equation required to determine
Vmin is
Vmin
2α
τc
1+
τ c τ m
2E
−
=
αρ0B τ c γ
(15.146)
Recall that τm is the stress level at which the material behaves according to the Tresca
model. We shall discuss how we determine V shortly.
If Vmin < V < Vmax, the zone of yielding material has two subzones: a zone next to the projectile that behaves according to the Tresca model and a zone next to the elastic region that
behaves according to the Mohr–Coulomb model. We shall first write the equation for Vmax.
Vmax =
1 τm
3τ c
3τ c α τ c
ρ0
+η * 1 −
2E
E
τ cγ 2
2
1 2
− λ − 3
(15.147)
If we define a coordinate, ξ, that varies from 0 at the projectile surface to 1 at the elastic–
plastic interface, we can determine a coordinate, ξm, where the yield behavior changes from
Tresca to Mohr–Coulomb. Unfortunately, this crossover point has to be solved numerically
with the equation that follows:
τc
1+
E
2
2α
(1 − η *)ξ m3 + γ 3
+
2α
3
γ 3 2(1 − η *)(2 − α )ξ m3 + 3γ 3
αρ 0V 2
+
4
2
τ cγ
(1 − η *)(1 − 2α )(2 − α ) (1 − η *)ξ m3 + γ 3 3
τ
α ρ 0V 2 1 + c
2
E
2α
τ cγ 2 (1 − η *)ξ m3 + γ 3
2α
3
3
3
2
γ 2(1 − η *)(2 − α ) + 3γ τ m
3τ
3τ c
+η * 1 − c −
=0
4 −
2E
τc τc
E
(1 − η *)(1 − 2α )(2 − α ) 1 +
2E
(15.148)
Keep in mind here that we know all of the information (including V) and we are solving
for ξm. A good math code will generally solve this equation quickly.
Once we have ξm, then A and B are given at the projectile surface (Tresca) by
A=
1 τm
α τ c
© 2014 by Taylor & Francis Group, LLC
3
ξm
1 2 τm
η
ln
1
+
(
*)
−
+
−
1
λ 3τ
c
γ
(15.149)
Ballistics: Theory and Design of Guns and Ammunition
404
3
ξm
3 + 4(1 − η *)
γ
1
B=
3 −
4
2(1 − η *)
3 3
ξ
1 + (1 − η *) m
γ
(15.150)
These equations account for the fact that the yielding is Mohr–Coulomb outside of ξ = ξm.
If V > Vmax, the yielding is completely according to the Tresca model. Thus, A and B are
given by
A=
2
τ
− 2 m
3
τc
γ
ln
τ
1+ c
2E
2
3τ c
3τ
+η * 1 − c
γ
3
2E
E
B=
−
+
4
2
γ
2(1 − η *)
τ
21+ c
2E
(15.151)
3
τc
+
3
1
2E
1 +
(1 − η *)
(15.152)
To approximate the normal stress on the ogive, we can replace the spherically symmetric
velocity, V, in our previous equations with Vz cos θ. We can write an equation for the normal stress function on the ogive [6] as
σ n (Vz , θ ) = τ c A + ρ0B[Vzcos θ ] 2
(15.153)
We can write Newton’s second law as
− Fz = m
dVz
dt
(15.154)
We can then convert this time integral to a distance integral as before to yield
− Fz = mVz
dVz
dz
(15.155)
If we substitute Equation 15.136 into the aforementioned equation and integrate, we get an
equation for the acceleration, velocity, and depth of penetration, respectively, as a function
of time:
a=−
cos tan −1
2
© 2014 by Taylor & Francis Group, LLC
αs
m
βs t α s βs
V0 −
α s
m
(15.156)
Penetration Theories
405
Vz =
βs t α s βs
αs
tan tan −1
V0 −
βs
m
αs
βs
t α s βs
cos tan −1
V0 −
m
m
αs
ln
z=
βs
βs
V0
cos tan −1
α s
(15.157)
(15.158)
If we determine the distance where the velocity of the projectile slows to zero, we obtain
the depth of penetration as
P=
β V2
m
ln 1 + s 0
2β s
αs
(15.159)
where
P is the final penetration depth
V0 is the impact velocity
So now that we have developed penetration formulas for soils, what do we do with
them? The use of models such as this one, as nice as it is, usually carries with it some
practical issues. A detailed model like this requires detailed material properties that, in
practice, one rarely has. It usually will require a test or two to calibrate it. Forrestal and Luk
[8] suggest using a value of 0.13 for η*. The authors claim the model is relatively insensitive
to it. The model was derived for normal penetration, but the authors claim good results up
to impact yaw angles of 30°. In this case, they used the line-of-sight penetration depth. As
one might expect, the accuracy of this particular model varies significantly with the properties of the soil. Rocks, roots, and soil layers further complicate everything. Nevertheless,
the model is an excellent tool and describes the physics of soil penetrations well. This is a
highly active area of current research.
Problem 14
A 0.50 caliber projectile is fired at a soil berm with properties established later. How far
does it penetrate?
Answer: P = 84.9 [cm]
Projectile information:
s = 63.50 [mm]
m = 662 [grains]
d =12.70 [mm]
ft
Vs = 2000
s
© 2014 by Taylor & Francis Group, LLC
Ballistics: Theory and Design of Guns and Ammunition
406
Soil information (assume Mohr–Coulomb behavior):
Initial density
kg
ρ0 = 1860 3
m
µ = 0.1
Locked density
kg
ρ * = 2125 3
m
λ = 0.33
lbf
τ c = 1500 2
in.
lbf
E = 2×107 2
in.
lbf
τ m = 2500 2
in.
15.4 Penetration and Perforation of Ceramics
The desire to decrease the weight of vehicles coupled with constant improvements in manufacture has increased interest in the use of ceramics as armor. The design of an armored
vehicle using ceramics requires an understanding of their behavior under impact loads.
The advantages of ceramic armor are its relatively low density, high hardness, and high
compressive strength. The disadvantages are that ceramics are usually brittle, have low
tensile strength that when coupled with high compressive strength can be a problem from
a spallation standpoint, they allow the protection to be degraded in a multihit situation,
and they are somewhat expensive. Their complex structural behavior makes them difficult
to model although this is only a disadvantage to the designers.
The response of a ceramic to penetration is unique amongst all of the other materials discussed in this text. The material behaves differently depending on the radial confinement
and whether it is backed or not. For reasons such as spallation, they are usually backed by
a fiber reinforced composite, plastic, elastomer, or metal plate. If a ceramic is not backed, it
will most likely spall when subjected to a high-shock load. This spallation can be analyzed
by the techniques we will discuss in Chapter 16 on shock theory.
Although not exhaustive, this is a list of common ceramics currently either in use or
being studied for armor applications:
Boron carbide (B4C)
Silicon carbide (SiC)
Titanium di-boride (TiB2)
Aluminum nitride (AlN)
Alumina (Al2O3)
Historically, terra cotta (ceramic) armor has been found in Chinese tombs dating from 400
BC. Before First World War, the practice of placing coal bunkers around magazines to take
advantage of comminution, a phenomena that we shall discuss shortly.
If a ceramic is backed, one can take advantage of its high compressive strength to resist
penetration. This will cause the tip of the penetrator to deform. Large stresses then build
© 2014 by Taylor & Francis Group, LLC
Penetration Theories
407
up in the penetrator. If the striking velocity is low enough, the penetrator will break up or
ricochet. This process is called interface defeat or infinite dwell. If the penetrator survives
the initial impact, the ceramic begins to fail. This process is complicated, which is why it
is difficult to model, but it is key to understanding the behavior and utilizing the ceramic
to the maximum extent possible.
The ceramic penetration process has been documented by Cheeseman [9]. After an initial dwell and several reflections of the shocks and rarefactions, the following events occur
and will either continue to perforation or stop when the penetration is arrested. Initially,
tensile cracks appear near the penetrator forming circular rings. These cracks propagate
along the principal stress planes that are usually 25°–75° from the surface normal.
Once the cracks reach the distal boundary, they coalesce into conical form. At this point,
if the ceramic was not backed, a plug would be ejected and the material would be perforated. If the plate is backed, then at the time when the conoid is formed, the stress is redistributed circumferentially and radial cracks appear. After the appearance of radial cracks,
lateral cracking in the plane of the impact surface forms. This process is illustrated in
Figure 15.11. With backing material present that holds the ceramic plug in place, the material has nowhere to go so micro-cracking begins. This pulverizes the ceramic material.
This is known as the comminuted zone. The process of comminution and the sand-like
character of the comminuted material erode the penetrator at a rapid rate. The powdered
material continually gets in the way of the penetrator. This material flows radially outward
and rearward. A similar effect occurs during shaped charge jet penetration into sand bags.
The penetration of a ceramic armor is highly dependent upon the boundary conditions. It
is known that confinement increases the penetrati
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