schaub h junkins j l analytical mechanics of space systems

Analytical Mechanics of Space Systems

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Analytical Mechanics of Space Systems Second Edition Hanspeter Schaub University of Colorado Boulder, Colorado John L. Junkins Texas A&M University College Station, Texas EDUCATION SERIES Joseph A. Schetz Series Editor-in-Chief Virginia Polytechnic Institute and State University Blacksburg, Virginia Published by American Institute of Aeronautics and Astronautics, Inc. 1801 Alexander Bell Drive, Reston, VA 20191-4344

MATLAB1 is a registered trademark of The MathWorks, Inc. American Institute of Aeronautics and Astronautics, Inc., Reston, Virginia 1 2 3 4 5 Library of Congress Cataloging-in-Publication Data Schaub, Hanspeter. Analytical mechanics of space systems / Hanspeter Schaub, John L. Junkins. 2nd ed. p. cm. (AIAA education series) Includes bibliographical references and index. ISBN 978 1 60086 721 7 (alk. paper) 1. Celestial mechanics. 2. Differentiable dynamical systems. 3. Orbital mechanics. 4. Space vehicles Control systems. I. Title. QB350.5.S33 2009 521 dc22 2009032378 Copyright # 2009 by the American Institute of Aeronautics and Astronautics, Inc. All rights reserved. Printed in the United States of America. No part of this publication may be reproduced, distributed, or transmitted, in any form or by any means, or stored in a database or retrieval system, without the prior written permission of the publisher. Data and information appearing in this book are for informational purposes only. AIAA is not responsible for any injury or damage resulting from use or reliance, nor does AIAA warrant that use or reliance will be free from privately owned rights.D w U W M S m DO

Dedicated to Richard H. Battin A source of inspiration to all in this field.D w U W M S m DO

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AIAA Education Series Editor-in-Chief Joseph A. Schetz Virginia Polytechnic Institute and State University Editorial Board Takahira Aoki University of Tokyo Joa˜o Luiz F. Azevedo Instituto de Aerona´utica e Espac¸o Sa˜o Jose´ dos Campos, Brazil Karen D. Barker Robert H. Bishop University of Texas at Austin Aaron R. Byerley U.S. Air Force Academy Richard Colgren University of Kansas J.R. DeBonis NASA Glenn Research Center Kajal K. Gupta NASA Dryden Flight Research Center Rikard B. Heslehurst University of New South Wales Rakesh K. Kapania Virginia Polytechnic Institute and State University Brian Landrum University of Alabama in Huntsville Timothy C. Lieuwen Georgia Institute of Technology Michael Mohaghegh The Boeing Company Conrad F. Newberry Joseph N. Pelton George Washington University Mark A. Price Queen’s University Belfast David K. Schmidt University of Colorado, Colorado Springs David M. Van Wie Johns Hopkins UniversityD w U W M S m DO

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Foreword Analytical Mechanics of Space Systems by Hanspeter Schaub and John Junkins is a comprehensive treatment of a stimulating and challenging subject that is both old and new. The basics go back two centuries, and modern applications are constantly developing every day. We are delighted to welcome this volume to the AIAA Education Series for many reasons, including the fact that the outstanding authorship is a collaboration between a representative of industry and one from academia. This is a textbook in the classical fashion with examples and ample home- work problems. The material is arranged so that two or three different univer- sity courses at different academic levels can be based upon it. However, this text can also be used as a basis for continuing education short courses or inde- pendent self study. There are two main sections—Basic Mechanics and Celes- tial Mechanics—divided into 14 chapters and 7 appendices covering some 700 pages. The two sections conclude with chapters on nonlinear spacecraft control and spacecraft formation flying, respectively, two topics currently receiving considerable attention. The AIAA Education Series aims to cover a very broad range of topics in the general aerospace field, including basic theory, applications and design. The philosophy of the series is to develop textbooks that can be used in a college or university setting, instructional materials for intensive continuing education and professional development courses, and also books that can serve as the basis for independent self study for working professionals in the aero- space field. Suggestions for new topics and authors for the series are always welcome. Joseph A. Schetz Editor-in-Chief AIAA Education Series ixD w U W M S m DO

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Table of Contents Preface to the Second Edition . . . . . . . . . . . . . . . . . . . . . . . . . . xvii Preface to the First Edition . . . . . . . . . . . . . . . . . . . . . . . . . . . . xix Part 1 Basic Mechanics Chapter 1. Particle Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.2 Particle Position Description . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.3 Vector Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 Chapter 2. Newtonian Mechanics . . . . . . . . . . . . . . . . . . . . . . . . 31 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 2.2 Newton’s Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 2.3 Single Particle Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 2.4 Dynamics of a System of Particles. . . . . . . . . . . . . . . . . . . . . 48 2.5 Dynamics of a Continuous System . . . . . . . . . . . . . . . . . . . . 62 2.6 Rocket Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 Chapter 3. Rigid Body Kinematics. . . . . . . . . . . . . . . . . . . . . . . . 79 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 3.2 Direction Cosine Matrix. . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 3.3 Euler Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 3.4 Principal Rotation Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 3.5 Euler Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 3.6 Classical Rodrigues Parameters . . . . . . . . . . . . . . . . . . . . . . . 112 3.7 Modified Rodrigues Parameters . . . . . . . . . . . . . . . . . . . . . . . 117 3.8 Other Attitude Parameters. . . . . . . . . . . . . . . . . . . . . . . . . . . 126 3.9 Homogeneous Transformations . . . . . . . . . . . . . . . . . . . . . . . 132 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 xiD w U W M S m DO

Chapter 4. Eulerian Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . 143 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 4.2 Rigid Body Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 4.3 Torque-Free Rigid Body Rotation . . . . . . . . . . . . . . . . . . . . . 162 4.4 Dual-Spin Spacecraft . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172 4.5 Momentum Exchange Devices . . . . . . . . . . . . . . . . . . . . . . . 178 4.6 Gravity Gradient Satellite . . . . . . . . . . . . . . . . . . . . . . . . . . . 188 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 Chapter 5. Generalized Methods of Analytical Dynamics. . . . . . . . 207 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207 5.2 Generalized Coordinates. . . . . . . . . . . . . . . . . . . . . . . . . . . . 207 5.3 D’Alembert’s Principle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210 5.4 Lagrangian Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239 5.5 Quasi Coordinates. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 5.6 Cyclic Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272 5.7 Final Observations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 280 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281 Chapter 6. Variational Methods in Analytical Dynamics . . . . . . . . 289 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289 6.2 Fundamentals of Variational Calculus . . . . . . . . . . . . . . . . . . . 289 6.3 Hamilton’s Variational Principles . . . . . . . . . . . . . . . . . . . . . . 293 6.4 Hamilton’s Principal Function . . . . . . . . . . . . . . . . . . . . . . . . 298 6.5 Some Classical Applications of Hamilton’s Principle to Distributed Parameter Systems . . . . . . . . . . . . . . . . . . . . . . . 300 6.6 Explicit Generalizations of Lagrange’s Equations for Hybrid Coordinate Systems . . . . . . . . . . . . . . . . . . . . . . . . . 309 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317 Chapter 7. Hamilton’s Generalized Formulations of Analytical Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . 321 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321 7.2 Hamiltonian Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321 7.3 Relationship of Hamiltonian Function to Work=Energy Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 326 7.4 Hamilton’s Canonical Equations . . . . . . . . . . . . . . . . . . . . . . 331 7.5 Poisson’s Brackets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335 7.6 Canonical Coordinate Transformations . . . . . . . . . . . . . . . . . . 338 7.7 Perfect Differential Criterion for Canonical Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342 xiiD w U W M S m DO

7.8 Transformation Jacobian Perspective on Canonical Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347 Chapter 8. Nonlinear Spacecraft Stability and Control . . . . . . . . . 351 8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351 8.2 Nonlinear Stability Analysis . . . . . . . . . . . . . . . . . . . . . . . . . 351 8.3 Generating Lyapunov Functions. . . . . . . . . . . . . . . . . . . . . . . 366 8.4 Nonlinear Feedback Control Laws . . . . . . . . . . . . . . . . . . . . . 381 8.5 Lyapunov Optimal Control Laws . . . . . . . . . . . . . . . . . . . . . . 396 8.6 Linear Closed-Loop Dynamics . . . . . . . . . . . . . . . . . . . . . . . 402 8.7 Reaction Wheel Control Devices . . . . . . . . . . . . . . . . . . . . . . 408 8.8 Variable Speed Control Moment Gyroscopes . . . . . . . . . . . . . . 410 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 430 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432 Part 2 Celestial Mechanics Chapter 9. Classical Two-Body Problem. . . . . . . . . . . . . . . . . . . . 439 9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 439 9.2 Geometry of Conic Sections . . . . . . . . . . . . . . . . . . . . . . . . . 440 9.3 Coordinate Systems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 448 9.4 Relative Two-Body Equations of Motion. . . . . . . . . . . . . . . . . 455 9.5 Fundamental Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 459 9.6 Classical Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 470 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 487 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 487 Chapter 10. Restricted Three-Body Problem. . . . . . . . . . . . . . . . . 493 10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493 10.2 Lagrange’s Three-Body Solution . . . . . . . . . . . . . . . . . . . . . 493 10.3 Circular Restricted Three-Body Problem . . . . . . . . . . . . . . . . 508 10.4 Periodic Stationary Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . 528 10.5 Disturbing Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 529 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533 Chapter 11. Gravitational Potential Field Models . . . . . . . . . . . . . 537 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 537 11.2 Gravitational Potential of Finite Bodies . . . . . . . . . . . . . . . . . 538 11.3 MacCullagh’s Approximation . . . . . . . . . . . . . . . . . . . . . . . 541 11.4 Spherical Harmonic Gravity Potential . . . . . . . . . . . . . . . . . . 545 xiiiD w U W M S m DO

11.5 Multibody Gravitational Acceleration . . . . . . . . . . . . . . . . . . 555 11.6 Spheres of Gravitational Influence . . . . . . . . . . . . . . . . . . . . 557 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 560 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 560 Chapter 12. Perturbation Methods . . . . . . . . . . . . . . . . . . . . . . . 561 12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 561 12.2 Encke’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 562 12.3 Variation of Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . 564 12.4 State Transition and Sensitivity Matrix . . . . . . . . . . . . . . . . . 597 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 612 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 613 Chapter 13. Transfer Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . 617 13.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 617 13.2 Minimum Energy Orbit . . . . . . . . . . . . . . . . . . . . . . . . . . . 617 13.3 Hohmann Transfer Orbit. . . . . . . . . . . . . . . . . . . . . . . . . . . 621 13.4 Lambert’s Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 626 13.5 Rotating the Orbit Plane . . . . . . . . . . . . . . . . . . . . . . . . . . 639 13.6 Patched-Conic Orbit Solution . . . . . . . . . . . . . . . . . . . . . . . 643 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 667 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 667 Chapter 14. Spacecraft Formation Flying . . . . . . . . . . . . . . . . . . . 673 14.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 673 14.2 General Relative Orbit Description . . . . . . . . . . . . . . . . . . . . 674 14.3 Cartesian Coordinate Description . . . . . . . . . . . . . . . . . . . . . 676 14.4 Orbit Element Difference Description . . . . . . . . . . . . . . . . . . 684 14.5 Relative Motion State Transition Matrix . . . . . . . . . . . . . . . . 693 14.6 Linearized Relative Orbit Motion . . . . . . . . . . . . . . . . . . . . . 698 14.7 J2-Invariant Relative Orbits . . . . . . . . . . . . . . . . . . . . . . . . . 708 14.8 Relative Orbit Control Methods . . . . . . . . . . . . . . . . . . . . . . 729 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 749 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 751 Appendix A. Transport Theorem Derivation Using Linear Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 755 Appendix B. Various Euler Angle Transformations . . . . . . . . . . . . 759 Appendix C. MRP Identity Proof . . . . . . . . . . . . . . . . . . . . . . . . 763 Appendix D. Conic Section Transformations. . . . . . . . . . . . . . . . . 765 xivD w U W M S m DO

Appendix E. Numerical Subroutines Library . . . . . . . . . . . . . . . . 769 Appendix F. First-Order Mapping Between Mean and Osculating Orbit Elements . . . . . . . . . . . . . . . . . . . . 775 Appendix G. Direct Linear Mapping Between Cartesian Hill Frame Coordinates and Orbit Element Differences . . 779 Appendix H. Hamel Coefficients for the Rotational Motion of a Rigid Body. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 781 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 789 Supporting Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 794 Software download information can be found at the end of the book on the Supporting Materials page. xvD w U W M S m DO

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Preface to the Second Edition In the five years since publishing the first edition, we have had the opportunity to teach from this text and to hear feedback from other professors, students, and colleagues. We warmly acknowledge that the many communications and inputs received have contributed significantly to this new edition. Aside from eliminat- ing minor errors and unclear passages, we have made significant improvements and extensions of the original text. We have augmented the illustrative examples and added more homework problems, supported as appropriate with solutions available to instructors. There are numerous minor changes to the presentation and figures to minimize confusion or enhance ease of understanding. One significant set of additions concerns the Matlab1 codes for rigid body kinematics that we provided with the first edition on a CD. The second-edition software includes C-codes and Mathematica codes, in addition to Matlab codes, and also we have significantly expanded the scope of the software to include orbi- tal mechanics. In lieu of the CD, we will support the second edition through the AIAA Supporting Materials site from which you can download the most current versions of the software. Download information is provided on the last page of this book. Perhaps the most important changes are substantial text additions, the 10 major ones being the following: Expanded discussion in Chapter 3 on stereographic orientation parameters New section in Chapter 4 on dual-spin spacecraft Expanded discussion in Chapter 4 on the prolate/oblate spin characteristics of torque-free rigid body motion New discussion in Chapter 5 on cyclic coordinates and Routhian reductions New discussion in Chapter 6 on virtual work performed by internal forces New section in Chapter 9 on common astrodynamics coordinate frames used New discussion in Chapter 9 on computing satellite ground tracks Enhanced discussion in Chapter 10 on the stability of colinear libration points New discussion in Chapter 13 on the p-iteration method New discussion in Chapter 13 on the use of canonical units in astrodynamics Overall, the second edition has been true to our original concept with the length increased by less than 10 percent. If you are a fan of the first edition, xviiD w U W M S m DO

we trust that you will agree that this second edition is a substantial refinement. If you have not read the first edition, we encourage you to begin your study by reading the Preface to the first edition immediately following. Hanspeter Schaub John L. Junkins August 2009 xviiiD w U W M S m DO

Preface to the First Edition Writing a text in analytical dynamics is a task not to be taken lightly. From several perspectives, the subject matter may appear mature and already suffer- ing from an overabundance of available texts. However, even though the field has been the subject of intense effort over the past 200 years, the topic remains alive with modern developments necessarily having long roots. It is remarkable that the modern developments remain in close continuity with the classical material, and it is for this reason that modern research and development need always to be done in the context of historical treatments. It is evident that no single text can do justice to this field, because of the vast scope of classical material and especially the significant ongoing evolution of analytical dynamics, addressing many diverse applications. Thus the demand for new books in this field arises precisely because the field continues to evolve, and students (and practicing engineers) are simultaneously challenged by the volume of classical material and their need to efficiently develop a perspective and to integrate the recent developments with classical material—to address the challenges of particular applications. Another important issue is the trend toward unification of developments from analytical dynamics and related ideas from allied fields such as, for example, control of nonlinear systems. Classical mechanics had its roots in the work of Newton, Euler, Lagrange, Gauss, Gibbs, Hamilton, and Jacobi et al., all of whom sought to establish methodology and understand the behavior of natural dynamical systems under the influence of forces arising in nature. The modern evolution of analytical dynamics, since the early 1900s, has moved toward analysis of man-made systems under the influence of natural and especially active forces under our control. Thus the merger of many aspects of dynamical system analysis with control theory is natural and has shown signs of maturing rapidly during the past 20 years. Another important trend is adjusting analytical dynamics perspec- tives to accommodate the rapid advance of computational methods, high-level programming languages, and computer automation of algebra. All of these trends drive the demand for modern treatments of analytical dynamics that enable the 21st-century generation of students to efficiently own the classical developments and, importantly, facilitate efficient access of the classical material in a context that enables intelligent approaches to modern applications. The text evolved from our lecture notes and research over the course of our combined five decades in this field. We have written the text in such a fashion that either two or three courses can be taught using this book as the primary source. The first course is appropriate for junior- or senior-level undergradu- ates, and consists of the first four chapters. These chapters deal with basic kinematics and dynamics of particles and rigid bodies, formulated from the xixD w U W M S m DO

Newtonian and Eulerian perspectives. However, the systematic kinematic and vector=matrix treatments are carried out with sufficient rigor and depth to provide the foundation for the remainder of the book. These chapters are longer than typical and are well populated with illustrative examples and homework problems. We have attempted to balance rigor with an almost conversational style that we hope will be found readable and enjoyable by serious undergraduate students, yet not insult the intelligence of more advanced students seeking the notational and conceptual foundations needed for later chapters. A second course can be fashioned at the first-year graduate level by quickly reviewing the first four chapters to establish a notational framework, then spending the bulk of the course on Chapters 5–8. This is essentially a modern treatment of Lagrangian and Hamiltonian dynamics, as well as variational methods. The class of systems considered ranges from particles, to systems of particles and rigid bodies, and distributed parameter systems. Although this material has a classical core, the presentation represents a fresh style and provides substantial new insights. It is anticipated that this second course will make an efficient presentation of material that many first-year graduate students find intimidating and abstract. The final portion of the text, Chapters 9–14, is suitable for a third course: a first graduate course in celestial mechanics. The treatments of the two-body and three-body problems are classical, but written in an accessible style supple- mented by new material on the three-body problem. The treatment of perturba- tion methods is again classical, but the results presented on, for example, satellite constellations (or formation flying) give new life to this material in the context of an area of current research and soon-to-be-realized applications. Substantial new results and insights are presented on the formation flying problem, emphasizing the judicious use of nonlinear analysis and control stra- tegies to find families of near-invariant relative motions that cooperate with the perturbations in order to minimize fuel. This book has, of course, not been written in a vacuum. We stand on the shoulders of innumerable giants who built the foundation of analytical dynamics. During the modern era spanned by our careers to date, we have drawn insight and influence from many classical and recent texts. In particular, we were educated and influenced by the work of L. Pars, H. Goldstein, R. Battin, P. Likins, L. Meirovitch, and J. Papastavridis. We express our thanks to these authors for their significant contributions. Certainly our students have also served to temper and refine this material over the years, and we are thank- ful for their many contributions. Of special note, we mention the inspiration generated by our association with a living legend, Richard H. Battin. His immortal text Mathematics and Methods of Astrodynamics and his friendship have been vital stimuli for the evolution of our understanding and provided an opportunity for us to contract his especially virulent and most delightful terminal illness: love of analytical dynamics. While this condition is indeed terminal, our readers may find little comfort in the fact that it advances very slowly; evidently and thankfully, the time constant is about one century! Dick, saying thank you seems an xxD w U W M S m DO

inadequate tribute; we are very grateful to you for your Astrodynamics and much more. We have done our best to pass on your favor and your traditions to new generations of students and workers in this field. Hanspeter Schaub John L. Junkins February 2003 xxiD w U W M S m DO

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Part 1 Basic MechanicsD w U W M S m DO

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1 Particle Kinematics 1.1 Introduction Kinematics is a branch of dynamics that studies aspects of motion apart from considerations of masses and forces. Essentially, kinematics is a collection of vector=matrix methods to describe positions, velocities, and accelerations of particles and rigid bodies, as viewed from various reference frames. The subfield of particle kinematics considers only the motion of particles. This in itself can be quite challenging at times. As an example, consider a person driv- ing a car on the highway. The road itself is fixed to a constantly rotating Earth, which in turn is orbiting the sun. What is the velocity and acceleration of the person relative to a sun-fixed coordinate system? This chapter will help answer these and many related questions. 1.2 Particle Position Description 1.2.1 Basic Geometry When studying the kinematics of particle motion, one is not concerned about the physical dimensions or mass of a particle. Let P be a point in a three-dimensional space as illustrated in Fig. 1.1. To define the position of the point P, a coordinate system along with its origin must be chosen. Without this coordinate system, it is difficult to describe the position of point P. To visualize this problem, imagine one person A telling another person B that his or her location is ‘‘10 miles.’’ Without knowing from what reference point person A measured 10 miles and in what direction it was measured, it is impossible for person B to know the meaning of ‘‘10 miles.’’ Fig. 1.1 The Cartesian coordinate system. 3D w U W M S m DO

A coordinate system is defined by two items. First, a coordinate system origin O must be established to specify its position in space. Second, the orien- tation of the coordinate system must be chosen. By choosing the orientation of the coordinate system, a person will know what is considered ‘‘up’’ or ‘‘east’’ as measured within this coordinate system. Three perpendicular (or orthogonal) right-hand unit vectors are traditionally used to denote unit displacement direc- tions along the orthogonal axes. In Fig. 1.1a standard Cartesian coordinate system labeled as E is shown. The three unit vectors ^ee1, ^ee2, and ^ee3 are used to define the orientation of E, and the coordinate system origin is denoted by OE. All unit vectors will be labeled with a ( ^ ) symbol. When assigning the unit vectors to the coordinate system, the first two unit vectors typically span the local ‘‘horizontal plane,’’ while the third unit vector points in the upwards direction normal to the plane of the first two unit vectors. However, this sequence and interpretation is not required. A coordinate system, defined through the origin and the three unit direction vectors, is often referred to as a reference frame. Vectors with components taken in different coordinate systems are said to be written in different refer- ence frames. More generally, think of a reference frame as a rigid body. Although the Earth is a rigid body, there is an infinite set of coordinate systems that could be embedded in the Earth-fixed reference frame. For the present, we will usually associate only one coordinate system with a reference frame (rigid body). Let r ¼ OEP be the vector pointing from the coordinate origin OE to the point P. Note that there are an infinite number of ways to parameterize that vector in terms of orthogonal coordinate axis components. To write the posi- tion vector r in the Cartesian coordinate system E shown in Fig. 1.1, it is broken down (i.e., projected orthogonally) into the three components along the coordinate system unit axes. Let the ^ee1 component of r be called x, the ^ee2 component be called y, and the ^ee3 component be called z. Then the vector r is written in the E Cartesian coordinate system components as r ¼ x^ee1 þ y^ee2 þ z^ee3 ð1:1Þ Note that Eq. (1.1) is a vector equation where the vector equation elements (r and the unit direction vectors ^eei) are used without expressing them with respect to a particular coordinate frame. Instead, they are simply entities with a particu- lar magnitude and direction in three-dimensional space. Often it is convenient to express the vector components of r as a 3  1 column matrix. A left super- script label is used to denote the coordinate frame with respect to which the vector components have been taken Er ¼ x y z 0 @ 1 A E ð1:2Þ Considering the Cartesian coordinate system E, the ith entry of the matrix expression is the component of the r vector along the ith coordinate frame unit 4 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

direction vector ^eei. The short-hand notation Er is used to express the vector r as an equivalent matrix expression with E frame vector components. Thus Eq. (1.1) is to be treated as a vector equation, while Eq. (1.2) is to be treated as a matrix equation. The superscript coordinate system label is often omitted when it is clear in which system the vector components are taken. Care must be taken when performing vector operations if multiple coordi- nate systems are used. Writing a vector addition as q ¼ r þ p is correct because no coordinate systems have been assigned yet; this equation has an infinity of possible component descriptions. We mention that one of the subtle and powerful facts of vector algebra is the ability to derive vector equa- tions that hold for all possible component parameterizations of the vectors. However, if the vectors have specific coordinate systems components as shown in Eq. (1.2), then the following matrix vector addition would not be correct: q1 q2 q3 0 @ 1 A E ¼ r1 r2 r3 0 @ 1 A E þ p1 p2 p3 0 @ 1 A B The vector p is here written in B frame components while all other vectors are expressed in the E frame. To add the B frame components of the p vector to E frame vectors, these components would first have to be transformed (projected) from the B frame to the E frame. In Chapter 3 it will be shown how the direc- tion cosine matrix can be used to perform this transformation. 1.2.2 Cylindrical and Spherical Coordinate Systems Although the Cartesian coordinate system is the most common and the easiest one to visualize, many times it is not the easiest to use. This is particu- larly true if the motion of point P is of a rotational type or if the dominant forces are radial. In these cases it is usually easier to use either a cylindrical or spherical coordinate system. When we address particle kinetics in Chapter 2, we will provide some insight on coordinate system selection in the context of solving example problems. A cylindrical coordinate system C is illustrated in Fig. 1.2. Its orientation is defined through the triad of unit vectors f^ccd, ^ccy, ^cc3g. This system is particularly useful in describing particles rotating about an axis ^ee3 that are free to move parallel to the axis ^ee3. For a large number of problems having rotational symmetry of force fields or constraint surfaces, cylindrical coordinates would be an attractive choice. For example, consider a particle constrained to move on the surface of a cylinder. Contrary to the inertially fixed Cartesian coordi- nate system N , two unit orientation vectors of the cylindrical coordinate system are varying with y as seen from N . These are the unit vector ^ccd and ^ccy. They rotate in the horizontal plane perpendicular to the ^cc3 unit vector. The vector ^ccd tracks the heading of the projection of the r position vector in this PARTICLE KINEMATICS 5D w U W M S m DO

horizontal plane. The position vector r of point P is expressed in cylindrical coordinates as r ¼ d ^ccd þ z^cc3 () C d 0 z 0 @ 1 A ¼ Cr ð1:3Þ where the scalar d is the radial distance of point P from the ^cc3 axis. The second entry of the cylindrical coordinate system column vector in Eq. (1.3) will always be zero. Any particle position vector expressed in a cylindrical coordinate system will never have a component along the ^ccy direction. Note that in Eq. (1.3) the unit vector ^ccd has a variable direction as observed from N . The angle y describes how far ^ccd has rotated from the ^ee1 axis. Therefore, instead of using ðx, y, zÞ Cartesian coordinates to describe a position, cylindri- cal coordinates use d and z, and the angle y provides the azimuth angle of the unit vector ^ccd relative to ^ee1. Assuming ^cc3 is aligned with ^ee3, the unit vectors ^ccd and ^ccy can be related to ^ee1 and ^ee2 through ^ccd ¼ cos y^ee1 þ sin y^ee2 ð1:4aÞ ^ccy ¼  sin y^ee1 þ cos y^ee2 ð1:4bÞ A spherical coordinate system S is illustrated in Fig. 1.3 with its orientation defined through the triad of unit vectors f^ssr, ^ssy, ^ssfg. Note that all three unit orientation vectors are time varying for the spherical coordinate system as seen from N . The unit vector ^ssr now points from OE toward point P. Let the scalar r be the radial distance from the coordinate system center OE to the point P. Then the position vector r is expressed as components along the spherical coordinate triad f^ssr, ^ssy, ^ssfg as r ¼ r^ssr () S r 0 0 0 @ 1 A ¼ Sr ð1:5Þ A particle position vector written as a column vector with components taken in the S frame will have a non-zero entry only in the first position. As shown in Fig. 1.2 The cylindrical coordinate system. 6 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

Fig. 1.3, the two angles y and f completely describe the orientation of the unit vectors ^ssr, ^ssy, and ^ssf relative to the three ^eei (i ¼ 1; 2; 3). Therefore, the f^ssr, ^ssy, ^ssfg projections onto f^ee1, ^ee2, ^ee3g with components a function of ðr, y, fÞ are ^ssr ¼ cos f cos y^ee1 þ cos f sin y^ee2 þ sin f^ee3 ð1:6aÞ ^ssy ¼  sin y^ee1 þ cos y^ee2 ð1:6bÞ ^ssf ¼  sin f cos y^ee1  sin f sin y^ee2 þ cos f^ee3 ð1:6cÞ Spherical coordinates and the associated triad of unit vectors f^ssr, ^ssy, ^ssfg are very useful when describing a particle motion on the surface of a sphere or a particle orbiting a body. Example 1.1 Given a vector r written in the Cartesian coordinate system E as r ¼ 2^ee1  3^ee2 þ 5^ee3 express r in terms of the cylindrical coordinate system C, where ^cc3 ¼ ^ee3. From Eqs. (1.4), we can express ^ee1 and ^ee2 in terms of ^ccd and ^ccy as ^ee1 ¼ cos y^ccd  sin y^ccy ^ee2 ¼ sin y^ccd þ cos y^ccy Using this relationship the vector r is expressed in the C frame as r ¼ 2 cos y  3 sin yð Þ ^ccd  2 sin y þ 3 cos yð Þ ^ccy þ 5^cc3 The angle y is resolved noting that in the C frame the ^ccy component must be zero. Therefore y must be y ¼  tan 1 3 2   ¼ 56:31 deg which brings r to the desired result r ¼ 3:61^ccd þ 5^cc3 Fig. 1.3 The spherical coordinate system. PARTICLE KINEMATICS 7D w U W M S m DO

1.3 Vector Differentiation 1.3.1 Angular Velocity Vector In planar motion it is easy to define and visualize the concept of angular velocity as is shown in Fig. 1.4a. For this single axis ^ee3 rotation case, the rota- tion angles and rotation rates (angular velocities) are only scalar quantities. The instantaneous angular rate o of a particle is given by o ¼ _yy ð1:7Þ where a positive rotation or rotation rate is defined to be in the increasing y (counterclockwise) direction shown. Angular velocity of a particle in a plane simply describes at what rate the radius vector locating the particle is orbiting the origin. For the general three-dimensional case, we will prove in Chapter 3 that a general large angular displacement is not a vector quantity; however, paradoxi- cally, angular velocity is a vector quantity. For the present, we limit ourselves to an argument based on small angular displacements to introduce the angular velocity vector. As the rigid body shown in Fig. 1.4b rotates about the body- and space-fixed ^ee axis by the small angle Dy, the body-fixed point at position P0 rotates to position P00. This rotation is described through the rotation vector Dh defined as Dh ¼ Dy^ee ð1:8Þ The angular velocity vector is the instantaneous angular rate at which this rota- tion occurs. Let the angular velocity vector magnitude be o, then the vector x can be written as x ¼ o^ee ð1:9Þ The unit direction vector ^ee defines an axis about which the rigid body or coordinate system is instantaneously rotating. For the case of planar rotations Fig. 1.4 Angular velocity vector. 8 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

in Fig. 1.4a, the rotation axis is simply ^ee3. Note that any orientation of a rigid body can be defined by the orientation of any body-fixed coordinate system. Therefore, position descriptions for rotating rigid bodies and rotating coordi- nate systems are actually the same problem geometrically, and there is no need to formally distinguish between the two. For the case of constant ^ee, it is natural to define x ¼ lim Dt!0 Dyy Dt ð1:10Þ The angular velocity vector x of a rigid body or coordinate system B relative to another coordinate system N is typically expressed in B frame components: x ¼ o1 ^bb1 þ o2 ^bb2 þ o3 ^bb3 ð1:11Þ Each component oi expresses the instantaneous angular rate of the body B about the ith coordinate axis ^bbi as shown in Fig. 1.5. The oi components are obviously the orthogonal components of x. As will be evident in Chapter 3, it is often convenient to describe x with non-orthogonal components as well. Note that any time numerical values of oi are used that units of radians per second must be used. Otherwise any formulas that use x will not yield correct results. 1.3.2 Rotation About a Fixed Axis It is instructive to study in detail the rotation of a rigid body about a fixed axis. In particular, the velocity vector _rr of a body-fixed point P is examined. Let a body B have a rod attached to it that is fixed in space at points A and B as shown in Fig. 1.6, so that the rod is the axis of rotation. The rigid body B is rotating about this rod with an angular velocity x. The origin O of the coordinate system for B is located on the axis of rotation. Let P be a body- fixed point located relative to O by the vector r. The angle between the angular velocity vector x and the position vector r is y. Studying Fig. 1.6, it is quite clear that the body-fixed point P will have no velocity component parallel to the angular velocity vector x, i.e., P moves in a Fig. 1.5 Illustration of angular velocity body frame components. PARTICLE KINEMATICS 9D w U W M S m DO

plane perpendicular to the x axis. If one would look down the angular velocity vector, one would see P moving on a circle with radius r sin y while being ‘‘transported’’ with the rotating rigid body. Thus, the speed of P is given by j_rrj ¼ ðr sin yÞo ð1:12Þ Studying Fig. 1.6 further, it is apparent that the inertial velocity vector of P will always be normal to the plane of r and x. This provides the direction of _rr, which can then be written as _rr ¼ ðr sin yÞo x  r jx  rj   ð1:13Þ However, note that jx  rj ¼ or sin y, so that the transport velocity is _rr ¼ x  r ð1:14Þ The only restriction for Eq. (1.14) is that r must be a body-fixed vector within B. As mentioned earlier, the concepts of rigid bodies and reference frames can be used interchangeably. The above result would also hold if we are finding the velocity vector fixed to any reference frame that is rotating relative to another; as is evident in the following section, this easily generalizes for three- dimensional motion. Fig. 1.6 Rigid body rotation about a fixed axis. 10 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

1.3.3 Transport Theorem As mentioned, it is simpler to define a particle position in terms of cylindri- cal or spherical coordinate systems. However, when computing the velocity of the particle and taking the time derivative of the position vector, one must take into account that the base vector directions of the chosen coordinate system may be time varying also. The following transport theorem allows one to take the derivative of a vector with respect to one coordinate system, even though the vector itself has its components taken in another, possibly rotating, coordi- nate system. Let N be an inertially fixed reference frame with a corresponding triad of N -fixed orthogonal base vectors f ^nn1, ^nn2, ^nn3g. Let B be another reference frame with the B-fixed base vectors f^bb1, ^bb2, ^bb3g. For simplicity, let the origin of the two associated reference frames be coincident. Let r be a vector written in the B coordinate system: r ¼ r1 ^bb1 þ r2 ^bb2 þ r3 ^bb3 ð1:15Þ We introduce the following notation: the angular velocity vector xB=N defines the angular velocity of the B frame relative to the N frame. An angular velo- city vector is typically written in the B frame. Therefore, we write xB=N as xB=N ¼ o1 ^bb1 þ o2 ^bb2 þ o3 ^bb3 ð1:16Þ At this point we introduce the notion of taking the vector time derivative while accounting for the reference frame from which the vector’s time variations are being observed. Imagine that you are standing still on Earth’s surface. Let B be an Earth-fixed coordinate system with the origin in the center of the Earth. Your position vector would point from the Earth’s center to your feet on the surface. By calculating the derivative of your position vector within B, you are determining how quickly this vector changes direction and=or magnitude as seen from the B system. You would find the time variation of your position to be zero when viewed from the Earth-fixed frame. This should be no big surprise; after all, you are standing still and not walking around on Earth. Now, let’s introduce another coordinate system N with the same origin, but this one is nonrotating and therefore fixed in space. Calculating the derivative of your position vector in the N frame, you wish to know how fast this vector is changing with respect to the fixed coordinate system N . Because Earth itself is rotating, in this case your position derivative would be non-zero. This is because relative to N , you are moving at constant speed along a circle about the Earth’s spin axis. To indicate that a derivative is taken of a generic vector x as seen in the B frame, we write Bd dt ðxÞ PARTICLE KINEMATICS 11D w U W M S m DO

The derivative of r given in Eq. (1.15) with components taken in the B coordi- nate system is denoted by Bd dt rð Þ ¼ Bd dt r1 ^bb1 þ r2 ^bb2 þ r3 ^bb3   ¼ _rr1 ^bb1 þ _rr2 ^bb2 þ _rr3 ^bb3 ð1:17Þ because the unit vectors ^bbi are fixed (i.e., time invariant) within the B frame, and therefore the terms Bd=dtð^bbiÞ are zero. When taking the inertial derivative of r, however, these unit vectors must now be considered time varying as seen in N . Therefore, using the chain rule of differentiation, we get1 N d dt ðrÞ ¼ _rr1 ^bb1 þ _rr2 ^bb2 þ _rr3 ^bb3 þ r1 N d dt ^bb1   þ r2 N d dt ^bb2   þ r3 N d dt ^bb3   ð1:18Þ However, because ^bbi are body-fixed vectors within B, Eq. (1.14) can be used to find their derivative in N : N d dt ^bbi   ¼ xB=N  ^bbi ð1:19Þ Using Eqs. (1.17) and (1.19), Eq. (1.18) is rewritten as N d dt ðrÞ ¼ Bd dt ðrÞ þ xB=N  r ð1:20Þ However, note that it is not necessary for the vector r to be written in the B coordinate frame for Eq. (1.20) to hold. Rather, components can be written in any arbitrary coordinate frame. This result leads to the general form of the transport theorem. Theorem 1.1 (Transport Theorem): Let N and B be two frames with a relative angular velocity vector of xB=N , and let r be a generic vector; then the derivative of r in the N frame can be related to the derivative of r in the B frame as N d dt rð Þ ¼ Bd dt rð Þ þ xB=N  r ð1:21Þ This formula allows one to relate a vector derivative taken relative to frame B to the corresponding vector derivative taken in frame N , where B and N are arbitrarily moving reference frames. This permits one to relate the deriva- tive of r as it would be seen from the N frame to the analogous rate of change of r as seen in the B frame. It is a very fundamental and important result that is used almost every time kinematic equations are derived. In parti- cular, we will find that vectors are typically differentiated with respect to an inertial frame called N . However, the notation N d=dt xð Þ becomes cumbersome 12 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

at times. When we want to compact the equation, we will use the following shorthand notation: N d dt xð Þ  _xx ð1:22Þ Example 1.2 The inertial velocity and acceleration vectors are sought for a general planar motion described in terms of polar coordinates with components taken along f^eer, ^eey, ^ee3g. The origin and base vectors of the polar coordinate system E are denoted E ¼ O; ^eer; ^eey; ^ee3   as shown in Fig. 1.7. The inertial coordinate system having the same origin O is denoted N ¼ O; ^nn1; ^nn2; ^nn3   where ^nn3 ¼ ^ee3. The position vector Er written in the E coordinate system is r ¼ r^eer Let xE=N be the angular velocity vector of E with respect to N . As is evident in Fig. 1.7, this is simply xE=N ¼ _yy^ee3 ¼ _yy ^nn3 Fig. 1.7 Polar coordinates illustration. PARTICLE KINEMATICS 13D w U W M S m DO

Using the transport theorem in Eq. (1.21), the inertial velocity vector of r is found to be _rr ¼ Ed dt ðrÞ þ xE=N  r Using the definition of r ¼ r^eer, it is clear that Ed dt rð Þ ¼ Ed dt r^eerð Þ ¼ _rr^eer After carrying out the cross-product term, the inertial velocity vector _rr is reduced to _rr ¼ _rr^eer þ r _yy^eey ð1:23Þ where _rr and r _yy are the radial and the transverse velocity components, respec- tively. The inertial acceleration €rr is found by taking the inertial derivative of _rr using the transport theorem: €rr ¼ Ed dt _rrð Þ þ xE=N  _rr Using the result for _rr that was just found, we obtain Ed dt _rrð Þ ¼ €rr^eer þ _rr _yy þ r €yy   ^eey Again, after carrying out the cross-product and collecting terms, the inertial acceleration vector €rr is found to be €rr ¼ ð€rr  r _yy 2Þ^eer þ r €yy þ 2_rr _yy   ^eey ð1:24Þ where €rr is the radial component, r _yy2 is the centrifugal component, r €yy is the tangential component, and 2_rr _yy is the Coriolis acceleration component. It is instructive to obtain Eq. (1.24) by ‘‘brute force.’’ Notice we can write the N frame rectangular components of position, velocity, and acceleration as r ¼ x ^nn1 þ y ^nn2 _rr ¼ _xx ^nn1 þ _yy ^nn2 €rr ¼ €xx ^nn1 þ €yy ^nn2 14 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

Because ^nni are fixed in N , the transport theorem is not required. Upon substi- tuting the polar coordinate transformations x ¼ r cos y y ¼ r sin y and taking two time derivatives, you can obtain the lengthy trigonometric func- tions axðr, yÞ and ayðr, yÞ in €rr ¼ axðr; yÞ ^nn1 þ ayðr; yÞ ^nn2 Finally, substituting ^nn1 ¼ cos y^eer  sin y^eey ^nn2 ¼ sin y^eer þ cos y^eey and performing considerable algebra, you will find all trigonometric functions of y cancel, leaving the same result as in Eq. (1.24). Example 1.3 Let us consider the motion of satellite B relative to satellite A illustrated in Fig. 1.8. Both satellites are orbiting the planet in the same orbit plane, and all relative motion is planar. The orbit frame A: f^iirA ; ^iiyA ; ^iihg has the first axis ^iirA track the radius vector to satellite A, while B: f^iirB ; ^iiyB ; ^iihg tracks the motion of satellite B. Note that ^iih is the common third frame unit direction vector that is orthogonal to the common orbit plane. Further, note that ^iih is fixed relative to the inertial frame N . Of interest is the motion of satellite B as seen by an observer on satellite A. The inertial satellite positions are given by rA ¼ rA(t)^iirA rB ¼ rB(t)^iirB ˆırA ˆırB ˆıθB ˆıθA A B θB θA ρ rA rB Fig. 1.8 Relative motion of two satellites on coplanar elliptic orbits. PARTICLE KINEMATICS 15D w U W M S m DO

The relative position vector q is obtained through the simple relationship q ¼ rB  rA The angular velocities of frames A and B relative to an inertial frame N are oA=N ¼ _yyA(t)^iih xB=N ¼ _yyB(t)^iih while the angular velocity of A relative to B is computed as xB=A ¼ xB=N  xA=N ¼ (_yyB  _yyA)^iih Next, to evaluate the rate of change of q as seen by an observer fixed to the A frame, we need to compute Ad dt qð Þ ¼ Ad dt rBð Þ  Ad dt rAð Þ Taking the A frame time derivative of rA, we find Ad dt rAð Þ ¼ _rrA^iih þ rA Ad dt ^iirA   ¼ _rrA^iirA because of ^iirA being fixed as seen in the A frame. Using the transport theorem, the A frame derivative of rB is Ad dt rBð Þ ¼ Bd dt rBð Þ þ xB=A  rB ¼ _rrB^iirB þ rB Bd dt ^iirB   þ (_yyB  _yyA)^iih h i  rB^iirB   ¼ _rrB^iirB þ rB(_yyB  _yyA)^iiyB Thus, the relative velocity of satellite B relative to A, as seen in the A frame, is given by Ad dt qð Þ ¼ _rrB^iirB  _rrA^iirA þ rB(_yyB  _yyA)^iiyB Note that this relative velocity solution is computed using general vector expressions and without specifying how the A and B frame unit direction vectors are oriented relative to some inertial frame. Also, it is possible to express the final answer using several rotating coordinate frames. This often leads to a simpler expression. If required, the unit direction vectors could all be expressed with respect to a common frame at this point. 16 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

To obtain the relative acceleration as seen in the A frame, the preceding differentiation process is repeated to yield Ad2 dt2 qð Þ ¼ €rrB^iirB  €rrA^iirA þ 2_rrB(_yyB  _yyA) þ rB(€yyB  €yyA) h i ^iiyB  rB(_yyB  _yyA)2^iirB Let us consider the following special orbit cases. First, assume that spacecraft A and B are on perfect circular orbits where rA 6 ¼ rB as illustrated in Fig. 1.9a. Note that in this case the orbit radial rates _rri and accelerations €rri are zero, as well as the orbit angular accelerations €yyi. However, the orbit rates _yyB and _yyA are not equal. The A frame perceived relative velocity and acceleration expres- sions simplify here to Ad dt qð Þ ¼ rB(_yyB  _yyA)^iiyB Ad2 dt2 qð Þ ¼ rB(_yyB  _yyA)2^iirB Note that these expressions are time varying when expressed with respect to the A frame caused by ^iirB and ^iiyB rotating at a different rate relative to A. The case illustrated in Fig. 1.9b has both craft on identical circular orbits. Compared to the preceding case, the orbits rates are now equal with _yyA ¼ _yyB. This reduces the A frame perceived relative velocity and acceleration expres- sions to the expected result Ad dt qð Þ ¼ 0 Ad2 dt2 qð Þ ¼ 0 With both satellites following the same circular trajectory, the perceived rela- tive velocity and acceleration are zero, while the inertial satellite acceleration is, of course, nonzero. A B ˆırA ˆıθA ρ ˆırB ˆıθB rA rB a) Two circular orbits with differ- ent radii where rA  = rB A B ˆırA ˆıθA ρ ˆırB ˆıθB rB rA b) Two identical circular orbits with rA = rB Fig. 1.9 Special orbit considerations. PARTICLE KINEMATICS 17D w U W M S m DO

1.3.4 Particle Kinematics with Moving Frames So far all coordinate systems or reference frames discussed have been considered nontranslating. Their origins have been fixed inertially in space. Now a more general problem will be discussed in which the coordinate frame origins are free to translate, while the frame orientations (defined through the three respective unit direction vectors) might be rotating. Let P be a generic particle in a three-dimensional space. Assume two differ- ent frames A ¼ Of , ^aa1, ^aa2, ^aa3g and B ¼ O0 f , ^bb1, ^bb2, ^bb3g exist as shown in Fig. 1.10. The position of O0 relative to O is given by the vector R. Note that these two coordinate frames could be actually attached to some rigid bodies and define their position and orientation in space, or they could simply be some artificial coordinate sets placed there without any other physical significance. In this discussion, however, it is frequently useful to think of reference frames A and B as rigid bodies. Let the vectors r and q be the position vectors of particle P in the A and B frames, respectively. The angular velocity vector of B relative to A is given by xB=A. Observe that the position vector r of P in the A frame can be related to R and q through the vector addition r ¼ R þ q ð1:25Þ The shorthand notation vP  B is used in this section to express the velocity vector of particle P with the derivative taken relative to the B frame: vP  B  Bd dt qð Þ ð1:26Þ The velocity vector vvvP  A of P relative to the A frame is given by vP  A  Ad dt rð Þ ¼ Ad dt R þ qð Þ ¼ Ad dt Rð Þ þ Ad dt qð Þ ð1:27Þ Fig. 1.10 Two coordinate frames with moving origins. 18 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

The velocity vector of the origin O0 in the A frame is defined to be vvO0   A¼ Ad dt Rð Þ ð1:28Þ Using the transport theorem and the definition in Eq. (1.28), the velocity vector vP  A of Eq. (1.27) can be written as vP  A ¼ vO0   Aþ vP  B þ xB=A  q ð1:29Þ To find the acceleration aP  A of particle P in the A frame, the derivative of Eq. (1.29) is taken in the A frame: aP  A ¼ Ad dt vP  A   ¼ Ad dt vO0   A þ vvP  B þ xB=A  q   ð1:30Þ Allowing the differentiation operator to apply term by term in the last term, and using the transport theorem, aP  A becomes aP  A ¼ Ad dt vvO0   A   þ Bd dt vvP  B   þ xB=A  vvP  B þ Ad dt xB=A   q þ xB=A  Bd dt qð Þ þ xB=A  q   ð1:31Þ Looking at the first term, the acceleration of the origin O0 in the A frame is defined to be aO0   A ¼ Ad dt vO0   A   ð1:32Þ While looking at the second term, the acceleration of particle P in the B frame is aP  B ¼ Bd dt vP  B   ð1:33Þ The angular acceleration vector of the B frame relative to the A frame is defined to be aB=A ¼ Ad dt xB=A  ð1:34Þ Using the definitions in Eqs. (1.26) and (1.32–1.34), the particle P accelera- tion vector aP  A can be written as the useful result2 aP  A ¼ aO0   A þ aP  B þ aB=A  q þ 2xB=A  vvP  B þ xB=A  xB=A  q  ð1:35Þ PARTICLE KINEMATICS 19D w U W M S m DO

The term 2xB=A  vP  B defines the Coriolis acceleration, and the term xB=A  xB=A  q  is the centrifugal acceleration. The latter term can also be expressed as xB=A  xB=A  q  ¼ ðxB=A  qÞxB=A  jxB=Aj2q ð1:36Þ which immediately reveals the centripical acceleration vector components along xB=A and q. Note that Eq. (1.35) holds between any two reference frames. It is not necessary that A or B be inertially fixed. The vector components used in the various terms on the right-hand side of Eq. (1.35) can be taken along any choice of unit vectors. It is important that we recognize the complete freedom we have to use any basis vectors we wish to express components of any vector in Eq. (1.35). Example 1.4 Let us investigate the inertial velocity of an Earth-fixed observer. Let N : f ^nn1; ^nn2; ^nn3g be an inertial coordinate frame, while E: f^ee1; ^ee2; ^ee3g is an Earth- fixed frame as illustrated in Fig. 1.11a. The first two frame vectors of N and E span the Earth’s equatorial frame, while the polar axis is described by both ^ee3 and ^nn3. The latitude angle of the observer is given by f. The local topographic frame T : f^ee; ^nn; ^uug is centered at the observer. Note that ^ee is the local east direc- tion, ^nn the local north, while ^uu is the local up direction. Fig. 1.11 Coordinate frame illustrations. 20 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

The E and N frames only differ through a single-axis rotation about polar axis through the ‘‘sidereal time’’ angle gg(t). Because the Earth completes a revolution in about 24 hours, we find that _ggg  360 deg 24 h  0:2618 rad=s The angular velocity of the Earth-fixed frame E and the inertial frame N is xE=N ¼ _ggg ^ee3 ¼ _ggg ^nn3 Because the topographic frame T is also an Earth-fixed frame, the relative angular velocity to E is xT =E ¼ 0 Note that if the observer would be moving relative to the Earth’s surface, then xT =E would no longer be zero. Finally, the angular velocity of the Earth stationary observer relative to the inertial frame is xT =N ¼ _ggg ^nn3 The observer position vector relative to the inertial frame is r ¼ re ^uu where re is the Earth’s radius, assuming a spherical Earth. Using the transport theorem to take the inertial time derivative of r yields _rr ¼ T d dt rð Þ þ xT =N  r ¼ re _ggg( ^nn3  ^uu) because of _rre ¼ 0. To evaluate the cross product of the expression ^nn3  ^uu, we need to either map ^nn3 into the T frame or ^uu into the N frame. Note that the polar axis ^nn3 lies in the plane spanned by ( ^nn; ^uu). This allows us to draw this plane as illustrated in Fig. 1.11b to examine the geometric relationship to find the orthogonal components ^nn3 ¼ cos f ^nn þ sin f ^uu Substituting this projection of ^nn3 into the earlier _rr expression yields _rr ¼ re _ggg(cos f ^nn þ sin f ^uu)  ^uu ¼ _gggre cos f^ee The answer should make intuitive sense. All Earth-fixed objects are moving in the local eastward direction, and the velocity magnitude should go to zero when the point is located at the poles with f ¼ 90 deg. PARTICLE KINEMATICS 21D w U W M S m DO

Next, let us consider the case where the object on the Earth’s surface is traveling purely in the north–south direction. In this case the longitude angle f(t) becomes time varying. Using the ^nn3 expression in the T frame, the topographic frame angular velocity relative to the inertial frame is xT =N ¼ xT =E þ xE=N ¼  _ff2 ^ee þ _gg cos f ^nn þ _gg sin f ^uu Using the transport theorem, the inertial velocity is _rr ¼ T d dt rð Þ þ xT =N  r ¼ re _gg cos f^ee þ re _ff ^nn Taking another inertial derivative and simplifying the resulting algebra lead to the following inertial acceleration expression: €rr ¼ 2re sin f_gg _ff^ee þ re cos f sin f_gg2 ^nn  re cos2 f_gg2 þ _ff2   ^uu Example 1.5 A disk of radius r, attached to a rod of length L, is rolling on the inside of a circular tube of radius R as shown in Fig. 1.12. The rod is rotating at constant rate o ¼ _yy. Three different reference frames are defined. The inertially fixed frame is N ¼ fO, ^nn1, ^nn2, ^nn3g with the origin at the center of the tube. The second coordinate frame E ¼ fO, ^eeL, ^eey, ^ee3g has the same origin, but the direction axes track the center of disk O0. The third frame B ¼ fO0, ^bbr, ^bbf, ^bb3g has the origin in the center of the disk, and the direction unit vectors track a point P on the disk edge. Note that ^nn3 and ^ee3 point out of the paper and ^bb3 ¼  ^nn3 points into the paper. What is the inertial acceleration €rr of point P Fig. 1.12 Disk rolling inside circular tube. 22 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

expressed in E frame components? Note that because three frames are present, we cannot directly use Eq. (1.35). Instead, the result will be derived by differ- entiation of the position vector by applying the transport theorem. First, let’s determine an expression for relating the angular rates _ff and _yy ¼ o. Because there is no slippage between the disk and the tube, notice that the ‘‘contact arcs’’ must be equal on the tube and the cylinder, giving the constraint yR ¼ fr Taking the derivative of this expression and using _yy ¼ o, the term _ff is given as _ff ¼ R r o The angular velocity vectors of frame E relative to N and frame B relative to E are xE=N ¼ o ^nn3 xB=E ¼ _ff^bb3 ¼  R r o ^nn3 The angular velocity vector of frame B relative to frame N is xB=N ¼ xB=E þ xE=N ¼  R  r r o ^nn3 The position vector r of point P relative to the origin O is r ¼ L^eeL þ r^bbr Using the transport theorem in Eq. (1.21), the inertial velocity vector _rr of P is _rr ¼ Ed dt L^eeLð Þ þ xE=N  L^eeL þ Bd dt r^bbr   þ xB=N  r^bbr Note that because L and r are constants for this system, the derivatives within the E and B frames are zero because ^eeL is fixed in E and ^bbr is fixed in B, and so _rr ¼ oL^eey þ R  rð Þo^bbf The inertial acceleration vector €rr of P is found by taking the derivative of _rr in the N frame: €rr ¼ Ed dt oL^eeyð Þ þ xE=N  oL^eeyð Þ þ Bd dt R  rð Þo^bbf   þ xB=N  R  rð Þo^bbf   PARTICLE KINEMATICS 23D w U W M S m DO

Because o is constant, the inertial acceleration is then written as the simple expression €rr ¼ o2L^eeL  R  rð Þ2 r o2 ^bbr To express the inertial acceleration only in unit direction vectors of, for exam- ple, the E frame, we eliminate ^bbr by making use of the identity ^bbr ¼  cos f^eeL þ sin f^eey to obtain the final result €rr ¼  o2L  R  rð Þ2 r o2 cos f " # ^eeL  R  rð Þ2 r o2 sin f^eey Although the result in Eq. (1.35) can be quite useful at times, when more than two frames are present, it is typically easier to derive the acceleration terms by differentiating the position vector twice, as in this example. References 1Likins, P. W., Elements of Engineering Mechanics, McGraw Hill, New York, 1973. 2Greenwood, D. T., Principles of Dynamics, 2nd ed., Prentice Hall, Englewood Cliffs, NJ, 1988. Problems 1.1 The particle P moves along a space curve described by the Cartesian coordinates xðtÞ ¼ cosðtÞ yðtÞ ¼ sinðtÞ zðtÞ ¼ sinðtÞ Describe the given motion in terms of cylindrical and spherical coordi- nates by finding explicit equations for the coordinates. 1.2 Assume a particle has a mass of m ¼ 10 kg, and the inertial position vector r and velocity vector vvv, and mass m ¼ 10 r ¼ 2^ii þ 2^||| þ 1^kk   m (P1:2:1) v ¼ 3^ii þ 1^|| þ 2^kk   m=s (P1:2:2) Using these states, determine the following: (a) r  v (b) The angle between r and vvv 24 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

(c) r  v (d) Normalized r and v vectors (e) jrj  jvj (f) r  v (g) Angular momentum vector (h) Linear momentum vector (i) Kinetic energy 1.3 The planar point acceleration vector is given in the Cartesian coordinates as €rr ¼ €xx^ee1 þ €yy^ee2 Directly transform this vector into polar coordinates r, y, ^eer, and ^eey by substituting x ¼ r cos y, y ¼ r sin y. Verify the result in Eq. (1.24) obtained through the transport theorem. 1.4 Let a particle P be free to slide radially in a rotating tube as shown in Fig. P1.4. Assume the tube is rotating at a constant angular velocity o. What is the inertial velocity and acceleration of the particle P? Express your answer as functions of r, y, ^eer, and ^eey. 1.5 Let N ¼ Of , ^nn1, ^nn2, ^nn3g be an inertial, nonrotating reference frame with its origin in the center of Earth. The Earth-fixed, equatorial coordinate frame E ¼ Of , ^ee1, ^ee2, ^ee3g has the same origin, but the unit direction vectors are fixed in the Earth. The Earth-fixed, topocentric coordinate frame T ¼ O0  , ^uu, ^ee, ^nng tracks a stationary point on Earth as shown in Fig. P1.5. Notice the local ‘‘geometric’’ interpretation: ^uu ¼ up, ^ee ¼ east, and ^nn ¼ north. Assuming that a stationary person is at a latitude of f ¼ 40 deg and a longitude of l ¼ 35 deg, what is the inertial velocity and acceleration of the point O0? Express your answer in both f ^nng and f^eeg components as functions of r, y, l, f, and derivatives thereof. 1.6 When launching a vehicle into orbit, one typically tries to make use of Earth’s rotation when choosing a launch site. From what place on Earth Fig. P1.4 Particle in rotating tube. PARTICLE KINEMATICS 25D w U W M S m DO

would it be the simplest (i.e., require least additional energy to be added) to launch vehicles into space, and how much initial eastward velocity (as seen in an Earth-fixed frame) would a vehicle have there because of Earth’s rotation? 1.7 The person in problem 1.4 has boarded a high-speed train and is traveling due south at a constant 450 km=h as seen in an Earth-fixed reference frame. What is the inertial velocity and acceleration now? 1.8 A constantly rotating disk is mounted on a moving train as shown in Fig. P1.8. The train itself is moving with a time varying velocity of vðtÞ. Assume the particle P is fixed on the disk, what are its inertial velocity and acceleration? Express your answer with f ^ddg components as functions of r, o, and vðtÞ. Fig. P1.5 Coordinate frames of a person on Earth. Fig. P1.8 Rotating disk on train. 26 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

1.9 Repeat problem 1.8, but this time assume that the particle P is free to move radially on the disk. Again find the corresponding inertial velocity and acceleration. 1.10 Two rotating disks are arranged as shown in Fig. P1.10. Relative to an inertial reference frame N , disk A has a relative angular velocity xA=N and disk B has a relative angular velocity xB=N. Each disk has a particle A or B, respectively, fixed to its rim. The orientation of the A frame is given by f^aar , ^aat, ^aa3g and the orientation of the B frame is given by f^bbr , ^bbt, ^bb3g. (a) What is the relative inertial velocity _rr and acceleration €rr of particle B vs A? (b) As seen from particle A, what is the relative velocity and accelera- tion of particle B? It is recommended that part (b) be solved in two ways: 1) by using Eqs. (1.29) and (1.35), and 2) by differentiation of the position and velo- city vector using the transport theorem. 1.11 Consider the overly simplified planetary system shown in Fig. P1.11. The Earth is assumed to have a circular orbit of radius R about the sun and is orbiting at a constant rate _ff. The moon is orbiting Earth also in a circular orbit at a constant radius r at a constant rate _yy. Assume the sun is iner- tially fixed in space by the frame f ^nn1, ^nn2, ^nn3g. Further, a UFO is orbiting the sun at a radius R2 at fixed rate _gg. Let the Earth frame E be given by Fig. P1.10 Two rotating disks. PARTICLE KINEMATICS 27D w U W M S m DO

the direction vectors f^eer, ^eef, ^ee3g, the moon frame M by f ^mmr , ^mmy, ^mm3g, and the UFO frame U by f ^uur , ^uug, ^uu3g. (a) Find the inertial velocity and acceleration of the moon relative to the sun. (b) Find the position vector of the moon relative to the UFO. (c) Find the angular velocity vectors xE=U and xM=U . (d) What are the velocity and acceleration vectors of the moon as seen by the UFO frame? 1.12 A disk of constant radius r is attached to a telescoping rod that is extend- ing at a constant rate as shown in Fig. P1.12. Both the disk and the rod are rotating at a constant rate. Find the inertial velocity and acceleration of point P at the rim of the disk. 1.13 A disk is rolling at a constant rate _yy on a moving conveyor belt as shown in Fig. P1.13. The conveyor belt speed v is constant. Find the inertial velocity and acceleration of point P relative to the N frame. Fig. P1.11 Planar planetary system. Fig. P1.12 Rotating disk attached to telescoping rod. 28 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

1.14 A vertical disk of radius r is attached to a horizontal shaft of length R as shown in Fig. P1.14. The shaft is rotating at a time varying rate _ff, while the disk is rotating at a time varying rate _yy. A fixed point P is on the rim of the disk, while a missile is flying overhead at a fixed height h with the trajectory rm ¼ h ^nn3  t ^nn2. (a) Find the inertial velocity and acceleration of point P. (b) What is the velocity and acceleration of point P as seen by the missile. 1.15 Two disks are rotating at constant rates _yy and _ff a fixed distance L apart as shown in Fig. P1.15. The origins of both disks are inertially fixed. The radius of the left disk is r and the radius of the right disk is R. (a) What is the inertial velocity and acceleration of point B on the right disk? (b) As seen from point A on the left disk, what is the relative velocity and acceleration of point B? Fig. P1.13 Disk rolling on a conveyor belt. Fig. P1.14 Grinding disk. PARTICLE KINEMATICS 29D w U W M S m DO

1.16 A person A is descending in an elevator at a constant velocity v. A second person B is riding a big wheel of radius R whose center is a distance L away from the elevator as shown in Fig. P1.16. What is the relative velocity and acceleration of person B as seen from person A? Fig. P1.15 Two rotating disks. Fig. P1.16 Person riding large wheel. 30 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

2 Newtonian Mechanics 2.1 Introduction The previous chapter on particle kinematics dealt with vector methods for describing a motion. Now we want to establish complete motion models that permit us to solve for the motion once the system forces and torques are given. Mass distribution and point of application of forces of a dynamical system clearly affect the resulting motion and must be taken into account. The motions are found by solving the system equations of motion that form the cause=effect model between the forces acting on the system and the resulting translational, rotational, and deformational accelerations. In this chapter, we first consider the dynamics of a single particle and then that of a system of particles. An example of a system of particles would be the solar system with the various planets within it idealized as particles. The parti- cle mechanics results will then be generalized to derive formulations for the dynamics of a continuous system. Examples of such systems include a vibrat- ing beam or a generally deformable collection of matter (such as a bowl of gelatin) where the system shape may be time varying. 2.2 Newton’s Laws The following laws of nature were discovered by Sir Isaac Newton over 200 years ago in England. Later in the early 20th century, Albert Einstein theorized in his papers about special relativity that these basic laws were only a low- speed approximation. However, relativistic effects become significant only when the velocity of a particle or body approaches that of the speed of light. In this discussion we will assume that all systems studied are moving much slower than the speed of light, and we will therefore neglect relativistic effects. The following three laws are commonly known as Newton’s laws of motion.1 3 Newton’s First Law: Unless acted upon by a force, a particle will maintain a straight line motion with constant inertial velocity. Newton’s first law is the most easily overlooked law because it is a special case of the second law. It simply states that unless something pushes against the particle, it will keep on moving in the same direction with constant velo- city. 31D w U W M S m DO

Newton’s Second Law: Let the vector F be the sum of all forces acting on a particle having a mass m with the inertial position vector r. Assume that N is an inertial reference frame, then F ¼ N d dt m_rrð Þ ð2:1Þ Or in words, the force acting on m is equal to the inertial time rate of change of the particle linear momentum p ¼ m_rr. If the mass m is constant, then this result simplifies to the well known result F ¼ m€rr ð2:2Þ We observe that if units are not chosen consistent with Eqs. (2.1) and (2.2), Newton’s second law requires an additional proportionality factor. Note that all derivatives taken in Newton’s second law must be inertial derivatives. Because it is typically necessary to also describe a position vector in a non-inertial co- ordinate frame, the importance of proper kinematics skills becomes apparent. Without correctly formulated kinematics, the dynamical system description will be incorrect from the start. We mention that a large fraction of errors made in practice have their origin in kinematics errors formulating €rr and similar vector derivatives. Newton’s Third Law: If mass m1 is exerting a force F21 on mass m2; then the force F12 experienced by m1 due to interaction with m2 will be F12 ¼ F21 ð2:3Þ This conforms to our intuitive experience. Anytime one pushes against an object, the reaction force from the object to one’s hand is of equal magnitude. Be sure to keep that in mind when contemplating punching a solid wall, or jumping from a canoe. To write down Newton’s laws, it is important to make use of force and moment sketches known as free body diagrams (FBDs). In essence, FBDs are used to specify and determine the force vector F in Eq. (2.2). Figure 2.1 is an example of a FBD. There are several conventions for free body diagrams; we adopt the following rule. The FBD should show all forces and moments acting on the system. We exclude from our FBDs acceleration vectors and so-called inertia forces that are subsets of the m€rr terms in Eq. (2.2) that may arise in rotating coordinate systems. Fig. 2.1 Newton’s law of universal gravitation. 32 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

Sir Isaac Newton is probably best known for the development of calculus and the laws of gravity that by popular account were initiated when an apple fell on his head while he was sitting under a tree. However, his laws of motion form the foundation of all modern sciences and engineering. Newton’s Law of Universal Gravitation: Let the vector r12 ¼ r2  r1 describe the position of mass m2 relative to mass m1 as shown in Fig. 2.2. Then the mutually attractive gravitational force between the objects will be F12 ¼ F21 ¼ Gm1m2 jr12j2 r12 jr12j ð2:4Þ where G ffi 6:6732  10 11 m3 s2kg is the universal gravity constant. For example, this law of universal gravitation allows one to model accu- rately the attractive forces between spacecraft and planets. Note, however, that because the universal gravity constant G is relatively small, the gravitational attraction between two everyday objects such as a house and a car is very small and typically ignored. Even Mount Everest makes a barely measurable perturbation in the Earth’s total gravitational attraction on objects in the immediate vicinity of Mount Everest. One important aspect of the law of universal gravitation is that the gravity force is conservative and can be calculated from a gravity field potential energy function. A general potential energy function V ðrÞ is a scalar function that depends on the system position vector r. The potential function measures how much work has to be done to the system to move an object from rest at reference position r0 to rest at position r. A conservative force is defined as a force derivable by taking the gradient of a corresponding potential energy func- tion V ðrÞ as FðrÞ ¼ HV ðrÞ ð2:5Þ Fig. 2.2 Illustration of a simple spacecraft FBD. NEWTONIAN MECHANICS 33D w U W M S m DO

Given V, we can derive F from the gradient operator as in Eq. (2.5). Given F, we can derive V by integration. Note that conservative forces depend only on the position vector r and not on the velocity vector _rr or time t. For example, the classical viscous drag force F ¼ c_rr would not be a conservative force. The gravity potential energy function VG experienced by the masses m1 and m2 is1;3 VGðr12Þ ¼  Gm1m2 jr12j ¼  Gm1m2 r12 ð2:6Þ VGðr12Þ is energy required to separate the two masses from the current distance of jr12j to an infinite separation. We will subsequently consider (in Section 2.3.3) the relationship of potential energy and work in more detail. Let’s describe the r12 vector through Cartesian coordinates as r12 ¼ x1 x2 x3 0 @ 1 A ð2:7Þ The magnitude of r12 is defined as jr12j ¼ x2 1 þ x2 2 þ x2 3 q ð2:8Þ and the partial derivatives of jr12j with respect to the Cartesian coordinates xi are given by @jr12j @xi ¼ xi jr12j ð2:9Þ The gradient of the potential field VG is given by @VG @xi ¼ Gm1m2 jr12j2 @jr12j @xi ¼ Gm1m2 jr12j2 xi jr12j ð2:10Þ The gravitational force F21 that the mass m2 experiences due to the mass m1 at the relative position r12 is given by F21 ¼ HVG ¼  Gm1m2 jr12j2 1 jr12j x1 x2 x3 0 @ 1 A ¼  Gm1m2 jr12j3 r12 ð2:11Þ Another example of a conservative force is the force exerted by a spring. Let the spring have a spring constant k and a linear deflection x. Then its potential function VS is given by VSðxÞ ¼ 1 2 kx2 ð2:12Þ 34 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

The spring potential energy indicates how much work was performed to stretch the spring from a zero reference deflection state to the deflection x. The force exerted by the spring on a mass m is given by the famous Hook’s law: F ¼ HVG ¼ kx ð2:13Þ Example 2.1 Let us find a first-order approximation of the gravity potential function in Eq. (2.6) that a body with m would experience near the Earth’s surface (Fig. 2.3). Assume a spherical Earth with radius Re and mass me. The radial distance r of the body to the center of Earth is written as r ¼ Re þ h where h is the height above the Earth’s surface. The gravity potential experi- enced by the body m due to Earth is V ðrÞ ¼  Gmem r The function V ðrÞ can be approximated about the distance Re through the Taylor series expansion V ðrÞ ¼ V ðReÞ þ 1 1! @V @r Re h þ 1 2! @2V @r2 Re h2 þ    Fig. 2.3 Mass m moving near the Earth. NEWTONIAN MECHANICS 35D w U W M S m DO

The local gravity potential Vlocal uses Re as its reference potential and can be approximated by VlocalðhÞ ¼ V ðrÞ  V ðReÞ ’ @V @r Re h þ Oðh2Þ After carrying out the partial derivative, the local gravity potential function for the special case of a constant gravity field is found to be VlocalðhÞ ¼ Gme R2 e mh ¼ mgh where g ¼ Gme=R2 e is the local gravitational acceleration. 2.3 Single Particle Dynamics The equation of motion for a single particle is given by Newton’s second law in Eq. (2.2), where it is assumed that the particle mass m is constant and €rr is the second inertial derivative of the position vector r. The following two sections treat two cases of this simple dynamical system. In the first case the force being applied to the mass is assumed to be constant and in the second case it is assumed to be time varying. 2.3.1 Constant Force If the force F being applied to the mass m is a constant vector, then the equations of motion m€rr ¼ F ¼ const ð2:14Þ can be solved for the time varying position vector rðtÞ. Eq. (2.14) can be solved for the inertial acceleration vector €rr as €rrðtÞ ¼ F m ð2:15Þ After integrating this equation once from an initial time t0 to an arbitrary time t, we obtain the following velocity formulation for mass m: _rrðtÞ ¼ _rrðt0Þ þ F m t  t0   ð2:16Þ After integrating the velocity formulation, an expression for the time varying position vector rðtÞ of the mass m is found: rðtÞ ¼ rðt0Þ þ _rrðt0Þ t  t0   þ F 2m t  t0  2 ð2:17Þ 36 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

Note that Eqs. (2.15–2.17) are actually each three sets of equations since r ¼ ðx1; x2; x3ÞT and F ¼ ðF1; F2; F3ÞT are each three-dimensional vectors. Given an initial velocity vector _rrðt0Þ, the time required to reach a final velocity under a constant driving force F can be solved from Eq. (2.16): t  t0   ¼ _xxiðtÞ  _xxiðt0Þ   m Fi ð2:18Þ Given an initial position vector rðt0Þ, the time required to reach a final position vector under constant driving force is found by solving the quadratic equation in Eq. (2.17) for the time t: t  t0 ¼ m Fi  _xx2 i ðt0Þ þ 2Fi m xiðtÞ  xiðt0Þ   r  _xxiðt0Þ ! ð2:19Þ Given an initial position and velocity vector and a final position vector, the corresponding final velocity vector is found by substituting Eq. (2.18) into Eq. (2.17) and solving for _rrðtÞ: _xx2 i ðtÞ ¼ _xx2 i ðt0Þ þ 2 Fi m xiðtÞ  xiðt0Þ   ð2:20Þ Example 2.2 The trajectory of a mass m is studied as it travels in a vertical plane under the influence of a constant gravitational force F. Determine an equation that relates an arbitrary target location ðx1; x2Þ to the corresponding launch velocity v0 and flight path angle g0. As shown in Fig. 2.4, the mass is at the coordinate center at time zero with a speed of v0 and an elevation angle of g0. The Carte- sian components of the initial position and velocity vectors are therefore given by rðt0Þ ¼ 0 0   _rrðt0Þ ¼ v0 cos g0 sin g0   Because the gravitational force F acts only along the vertical direction, the equations of motion are given as €rrðtÞ ¼ 1 m 0 F   ¼ 0 g   NEWTONIAN MECHANICS 37D w U W M S m DO

where g ¼ F=m is the local constant gravitational acceleration. Using Eq. (2.16), the velocity vector _rrðtÞ is _rrðtÞ ¼ v0 cos g0 sin g0    0 gt   The position vector rðtÞ is found through Eq. (2.17): rðtÞ ¼ x1ðtÞ x2ðtÞ   ¼ v0t cos g0 sin g0    0 gt2=2   By solving the x1ðtÞ equation for the time t and substituting it into the x2ðtÞ equation, one obtains the parabola expression relating x2 to x1 (the equation of the path or trajectory): x2 ¼ x1 tan g0  g sec2 g0 2v2 0 x2 1 An interesting question now arises. Given an initial speed v0, what would the initial elevation angle g0 have to be to make the mass m hit a target at coordinates ð~xx1; ~xx2Þ? To answer this, we rewrite the preceding expression relat- ing x1 and x2, making use of the trigonometric identity sec2 g0 ¼ 1 þ tan2 g0: tan2 g0  2v2 0 g~xx1 tan g0 þ 2v2 0 ~xx2 g~xx2 1 þ 1 ¼ 0 This quadratic equation can be solved explicitly for tan g0: tan g0   1=2¼ v2 0 g~xx1  v0 g~xx1 v2 0  2g~xx2  g2 ~xx2 1 v2 0 s Fig. 2.4 Ballistic trajectories under constant gravity force. 38 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

If the point ð~xx1; ~xx2Þ is within the range limit, then this formula will return two real answers. One corresponds to a lower trajectory and the other to a higher trajectory as illustrated in Fig. 2.4. If the point ð~xx1; ~xx2Þ is on the range limit, then the formula will return a double root. If the real point ð~xx1; ~xx2Þ is outside the range limit, then two complex variables will be returned, indicating the reasonable truth that no real solutions exist. To find the envelope of all possible trajectories, the case in which only double roots exist is examined. Setting the square root term to zero, the follow- ing parabola is found: x2 ¼ v2 0 2g  g 2v2 0 x2 1 Any targets that are accessible with the given v0 must lie within this parabola. The trajectory envelope parabola is shown as a dashed line in Fig. 2.4. As can be verified, the special case in which ~xx1 ¼ 0 gives ~xx2 ¼ v2 0=2g. One can readily show that this is the apogee of a vertically launched projectile with launch velocity v0. Another special case is the one in which ~xx2 ¼ 0, which provides the maximum impact range x1 ¼ v2 0=g if the surface is flat. Figure 2.5 compares the various launch angles required to hit a target at a distance x1 away with a given initial velocity v2 0 . For this constant gravity field case, the maximum range launch angle is always 45 deg. Later this problem is revisited in celestial mechanics, where the inverse square gravity field case is considered. 2.3.2 Time-Varying Force When the force F acting on the mass m is time varying, then there are typi- cally no closed form solutions for the velocity and position vectors. The equa- tions of motion are given as €rr ¼ 1 m FðtÞ ð2:21Þ Fig. 2.5 Ballistic trajectories under constant gravity force. NEWTONIAN MECHANICS 39D w U W M S m DO

Upon integrating Eq. (2.21) from t0 to t, the velocity vector _rrðtÞ at time t is given as _rrðtÞ ¼ _rrðt0Þ þ 1 m ðt t0 FðtÞ dt ð2:22Þ The position vector rðtÞ is obtained by integrating the velocity vector: rðtÞ ¼ rðt0Þ þ _rrðt0Þ t  t0   þ 1 m ðt t0 ðt2 t0 Fðt1Þ dt1 dt2 ð2:23Þ Finding the time required to accelerate from one velocity to another or to travel from one position to another under the influence of FðtÞ cannot be found generically as for the case of constant F. These results would have to be found explicitly for a given problem statement or through a numerical method if no closed-form solution exists. Example 2.3 Let the mass m be restricted to travel only in one dimension. It is attached to the coordinate frame origin through a linear spring with spring constant k. The force acting on mass m is then given through Hook’s law as F ¼ kx and the equations of motion are then given through Newton’s second law in Eq. (2.21) as €xx ¼ 1 m kxð Þ This can be rewritten in the form of the standard unforced oscillator differential equation: m€xx þ kx ¼ 0 The oscillator problem is known to have a solution of the type xðtÞ ¼ A cos ot þ B sin ot where the constants A, B, and o are yet to be determined. The velocity and acceleration expressions are then given as _xxðtÞ ¼ Ao sin ot þ Bo cos ot €xxðtÞ ¼ Ao2 cos ot  Bo2 sin ot ¼ o2xðtÞ 40 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

Substituting the expression for €xxðtÞ into the equation of motion the following expression is obtained: mo2 þ k  x ¼ 0 which must hold for any position x. Therefore, the natural frequency o is given by4 o ¼ k m r The constants A and B would be found through enforcing the solution to satisfy the initial conditions xðt0Þ ¼ A and _xxðt0Þ ¼ oB. Example 2.4 Consider the motion of a planar pendulum of length L as illustrated in Fig. 2.6. The position vector r of mass m is given by r ¼ L^eer The angular position of the mass is given by y. Because L is constant, the iner- tial acceleration of m is given by €rr ¼ L_yy2 ^eer þ L€yy^eey The constant gravitational acceleration vector Fg is expressed in the rotating E frame as Fg ¼ mg cos y^eer  mg sin y^eey Fig. 2.6 Planar pendulum illustration. NEWTONIAN MECHANICS 41D w U W M S m DO

The cable tension is given by FL ¼ N ^eer with N > 0. Let us write Newton’s second law m€rr ¼ P Fi with components in the ^eer and ^eey directions: ^eer : mL_yy2 ¼ N þ mg cos y ^eey : mL€yy ¼ mg sin y Using the ^eer components, the tension force magnitude is expressed as N ¼ mL_yy2 þ mg cos y Using the ^eey components, we find the nonlinear differential equations of motion of the spherical pendulum mass m: L€yy þ g sin y ¼ 0 If the pendulation amplitude y is small, then we can approximate sin y  y and find the linear differential equations of motion: L€yy þ gy ¼ 0 Note that this differential equation is equivalent to that of the spring-mass system in Example 2.3. The solution of yðtÞ is then of the form yðtÞ ¼ A cos ot þ B sin ot with o being the natural pendulation frequency. Substituting yðtÞ into the differ- ential equations of motion, we can solve for the frequency o: o ¼ g L r 2.3.3 Kinetic Energy The kinetic energy T of a particle of mass m is given by T ¼ 1 2 m_rr  _rr ð2:24Þ To find the work done on the particle, we investigate the time derivative of the kinetic energy T: dT dt ¼ m€rr  _rr ð2:25Þ 42 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

After using Eq. (2.14), the kinetic energy rate, or power, is given as dT dt ¼ F  _rr ð2:26Þ If the force F is conservative, it can be expressed as the negative gradient of a potential function V: dT dt ¼  @V @r  _rr ð2:27Þ Noting that @V @r _rr ¼ dV dt Eq. (2.27) can be written as dT dt þ dV dt ¼ 0 ð2:28Þ Therefore, the total system energy E ¼ T þ V is conserved. For conservative systems it is often convenient to obtain an expression relating coordinates and their time derivatives using the system energy. This avoids having to perform difficult integrations of the acceleration expressions to obtain the same relation- ship. Let W be the work performed between times t1 and t2. Upon integrating Eq. (2.26) from time t1 to t2, the following work=energy equation is obtained: T ðt2Þ  Tðt1Þ ¼ ðt2 t1 F  _rr dt ¼ ðrðt2Þ rðt1Þ F  dr  W ð2:29Þ Example 2.5 A mass m of 10 kg has an initial kinetic energy of 40 J (1 J ¼ 1 kg  m2= s2 ¼ 1 N  mÞ. A constant force F ¼ 4 N is acting on this mass from the initial position rðt0Þ ¼ 0 m to the final position at rðtf Þ ¼ 10 m. What is the work done on the mass and what is the final velocity at tf ? Using Eq. (2.29), the work W done to the mass m is W ¼ ðrðtf Þ rðt1Þ F  dr ¼ ð10 m 0 m 4 N  dr ¼ 40 N  m ¼ 40 J The energy at tf is given by Tðtf Þ ¼ Tðt0Þ þ W ¼ 40 J þ 40 J ¼ 80 J NEWTONIAN MECHANICS 43D w U W M S m DO

Using Eq. (2.24), the final velocity _rrðtf Þ is found to be _rrðtf Þ ¼ 2Tðtf Þ m r ¼ 4 m=s 2.3.4 Linear Momentum The linear momentum vector p of a particle is defined as p ¼ m_rr ð2:30Þ The momentum measure provides a sense of how difficult it will be to change a motion of a particle. Assume a locomotive has a large mass m and a very small inertial velocity _rr. Despite the slow motion, it makes intuitive sense that it would be very difficult to stop the motion of this large object. The linear momentum p of the locomotive is large because of the large mass. Similarly, consider a bullet with a small mass and a very high inertial velocity. Again, it makes intuitive sense that it would be difficult to deflect the motion of the bullet once it has been fired. In this case the linear momentum of the bullet is large not because of its mass, but because of its very large inertial velocity. Using the linear momentum definition, we are able to rewrite Newton’s second law in Eq. (2.1) in terms of p as F ¼ N d dt m_rrð Þ ¼ N d dt pð Þ ð2:31Þ Thus, the force acting on a particle can be defined as the inertial time rate of change of the linear momentum of the particle. Note that the mass m in Eq. (2.31) cannot be time varying in this equation because of the assumption that the system consists of a single particle. If m varies with time, then mass parti- cles are ejected, and the system must be considered a multiparticle system. If no force is acting on the particle, then _pp is zero and the linear momentum is constant. For the single particle system, this is a rather trivial result. However, using the analogous arguments on a multiparticle system will yield some very powerful conclusions. 2.3.5 Angular Momentum Let P be an arbitrary point in space with the inertial position vector rP and let the mass m have an inertial position vector r. The relative position of m to point P is given through s ¼ r  rP ð2:32Þ 44 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

The angular momentum vector H P of the particle m about point P is defined as H P ¼ s  m _ ss ð2:33Þ Taking the time derivative of H P, we find _HH P ¼ _ ss  m _ ss þ s  m € ss ð2:34Þ After noting that € ss ¼ €rr  €rrP and that a vector cross product with itself is zero, the vector _HH P is _HH P ¼ s  m€rr  s  m€rrP ð2:35Þ Using Eq. (2.14), this is rewritten as _HH P ¼ s  F þ m€rrP  s ð2:36Þ Note that the term s  F is the moment (or torque) vector LP due to force F about point P. The angular momentum time derivative can then be written in its most general form _HH P ¼ LP þ m€rrP  s ð2:37Þ Note that if the reference point P is inertial (nonaccelerating), then Eq. (2.37) is reduced to the famous Euler’s equation1;2: _HH P ¼ LP ð2:38Þ Example 2.6 A weightless cylinder of radius R with a mass m embedded in it is rolling down a slope of angle a without slip under the influence of a constant gravity field as shown in Fig. 2.7. The mass is offset from the cylinder center by a distance l. Let N : fO; ^nn1; ^nn2; ^nn3g be an inertial frame and E : fO 0; ^eer; ^eey; ^ee3g be a rotating frame tracking the point mass within the cylinder. Note that ^ee3 ¼  ^nn3. The angular velocity vector between the E and the N frame is vE=N ¼ _yy^ee3 ¼ _yy ^nn3 Because of the no-slip condition, the distance d that the center of the cylinder travels downhill is related to rotation angle y through d ¼ Ry NEWTONIAN MECHANICS 45D w U W M S m DO

The position vector r of the point mass relative to O is written as r ¼ d ^nn1 þ l^eer ¼ Ry ^nn1 þ l^eer Using the transport theorem, the inertial velocity and acceleration vectors are found to be _rr ¼ R_yy ^nn1 þ l _yy^eey €rr ¼ R€yy ^nn1 þ l €yy^eey  l _yy2 ^eer The E frame unit vectors are expressed in terms of N frame components as ^eer ¼ sin y ^nn1 þ cos y ^nn2 ^eey ¼ cos y ^nn1  sin y ^nn2 The acceleration vector of the point mass m is then expressed in the N frame as N €rr ¼ R€yy þ l €yy cos y  l _yy2 sin y   ^nn1  l €yy sin y þ l _yy2 cos y   ^nn2 The forces acting on the rolling cylinder are the gravitational force Fg, Fg ¼ mg sin a ^nn1  cos a ^nn2ð Þ the normal force N pushing perpendicular from the surface, N ¼ N ^nn2 and the frictional force Ff , which is keeping the cylinder from slipping, Ff ¼ Ff ^nn1 Fig. 2.7 Cylinder with offset mass rolling down a slope. 46 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

Newton’s second law states that m€rr ¼ Fg þ N þ Ff After substituting N €rr and the expressions for the forces into the preceding equation and equating the N frame components, the following two relation- ships are found: mðR€yy þ l cos y€yy  l _yy2 sin yÞ ¼ mg sin a  Ff  mðl sin y€yy þ l _yy2 cos yÞ ¼ mg cos a þ N Once an expression for €yy is found, the second equation could be used to solve for the time varying normal force component N. To solve the first equation for the angular acceleration, an expression for the frictional force component Ff must be found. To do so, we examine the angular momentum vector of the point mass about the E frame origin O 0. The relative position vector s of the point mass to O 0 and its inertial derivative are given bys ¼ l^eer _ ss ¼ l _yy^eey The angular momentum vector H O0 can then be written as H O0 ¼ s  m _ ss ¼ ml2 _yy ^nn3 and its inertial derivative is given by _HH O 0 ¼ ml2 €yy ^nn3 The torque LO0 about point O 0 is written as LO0 ¼ s  Fg  R ^nn2  Ff þ N   ¼ mgl sin y þ að Þ ^nn3  RFf ^nn3 The inertial position vector rO 0 of point O 0 and its second inertial derivative are given by rO0 ¼ d ^nn1 ¼ Ry ^nn1 €rrO0 ¼ R€yy ^nn1 Euler’s equation with moments about a general point in Eq. (2.37) is for this case _HH O0 ¼ LO0 þ m€rrO 0  s NEWTONIAN MECHANICS 47D w U W M S m DO

which leads to the desired expression for Ff in terms of €yy: RFf ¼ ml2 €yy  mgl sin y þ að Þ þ mRl €yy cos y Substituting this expression back into the previous equation relating €yy and Ff results in the equations of motion in terms of the rotation angle y: R2 þ l2 þ 2Rl cos y  €yy  Rl _yy2 sin y  gR sin a  gl sin y þ að Þ ¼ 0 This equation could be solved for the angular acceleration €yy, which could then be used to find the normal force component N purely in terms of y and _yy. 2.4 Dynamics of a System of Particles 2.4.1 Equations of Motion Until now we have considered only dynamical systems with a single parti- cle. In this section we will discuss systems of N particles, each with a constant mass mi. An example of such a dynamical system would be our solar system. To study the translational (orbital) motion of the planets and moons, due to the large distances involved, they can usually be considered to be point masses with each having different masses mi. Because we are now dealing with a finite number of masses, we write Newton’s second law in index form as Fi ¼ mi €RRi ð2:39Þ where €RRi is the inertial acceleration vector of mi as shown in Fig. 2.8. The force acting on mi can be broken down into two subsets of forces as Fi ¼ FiE þ PN j¼1 Fij ð2:40Þ where FiE is the vector sum of all external forces acting on mass mi, and Fij is an internal force vector due to the influence of the jth masses on the ith mass. The total force vector F acting on the system of N particles is defined to be F ¼ PN i¼1 Fi ¼ PN i¼1 FiE ð2:41Þ The internal forces Fij do not appear in F because of Newton’s third law, which states that Fij ¼ Fji, i.e., internal forces cancel in pairs. The total mass M of the N particles is defined as M ¼ PN i¼1 mi ð2:42Þ 48 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

The system center of mass position vector Rc is defined such that PN i¼1 miri ¼ 0 ð2:43Þ where ri ¼ Ri  Rc   is the position vector of mi relative to Rc. Thus Eq. (2.43) can be rewritten as PN i¼1 miRc ¼ PN i¼1 miRi ð2:44Þ which is further simplified using the system mass definition in Eq. (2.42) to M Rc ¼ PN i¼1 miRi ð2:45Þ The center of mass position vector Rc is expressed in terms of the individual inertial mass position vectors Ri as Rc ¼ 1 M PN i¼1 miRi ð2:46Þ After taking two inertial derivatives of Eq. (2.45), we obtain M €RRc ¼ PN i¼1 mi €RRi ¼ PN i¼1 Fi ð2:47Þ After substituting Eq. (2.41), we obtain the final result M €RRc ¼ F ð2:48Þ Fig. 2.8 System of N particles. NEWTONIAN MECHANICS 49D w U W M S m DO

also known as the superparticle theorem. The dynamics of the mass center of the system of N particles under the influence of the total external force vector F is the same as the dynamics of the superparticle M. Note that the superpar- ticle theorem tracks only the center of mass motion of the system. No infor- mation is obtained about the size, shape, or orientation of the cloud of N particles. Example 2.7 Let three masses be connected through springs with a spring stiffness constant k as shown in Fig. 2.9. The second and third mass each are subjected to a constant force where F2 ¼ f and F3 ¼ 2f . The total system mass M is given through M ¼ 2m þ m þ m ¼ 4m and the total external force F being applied to the system is F ¼ f þ 2f ¼ 3f The center of mass of the three-mass system is found through Eq. (2.45) to be rc ¼ 2mr1 þ mr2 þ mr3 M ¼ 2r1 þ r2 þ r3 4 Using the superparticle theorem in Eq. (2.48), the equation of motion for the center of mass of the three-mass system is 4m€rrc ¼ 3f Assuming that the rc is originally at rest at the origin, the system center of mass location is then integrated to obtain rcðtÞ ¼ 3f 8m t2 Fig. 2.9 Three-mass system. 50 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

To find the equations of motion of the individual masses, we need to write Eq. (2.39) for each mass: 2m€rr1 ¼ kðr2  r1Þ m€rr2 ¼ kðr2  r1Þ þ kðr3  r2Þ þ f m€rr3 ¼ kðr3  r2Þ þ 2f This can be written in a standard ordinary differential equation (ODE) matrix form for a vibrating system 2m 0 0 0 m 0 0 0 m 2 4 3 5 €rr1 €rr2 €rr3 0 @ 1 A þ k k 0 k 2k k 0 k k 2 4 3 5 r1 r2 r3 0 @ 1 A ¼ 0 f 2f 0 @ 1 A which can be solved given a set of initial conditions for riðt0Þ and _rriðt0Þ. 2.4.2 Kinetic Energy The total kinetic energy T of a cloud of N particles can be written as the sum of the kinetic energies of each particle: T ¼ 1 2 PN i¼1 mi _RRi  _RRi ð2:49Þ After making use of the expression _RRi ¼ _RRc þ _rri, the total kinetic energy is rewritten as T ¼ 1 2 PN i¼1 mi   _RRc  _RRc þ _RRc  PN i¼1 mi_rri   þ 1 2 PN i¼1 mi _riri  _rri ð2:50Þ where the middle term PN i¼1 mi_rri is zero due to the definition of the center of mass in Eq. (2.43). The total kinetic energy of a system of N constant mass particles mi can therefore be written as T ¼ 1 2 M _RRc  _RRc þ 1 2 PN i¼1 mi_rri  _rri ð2:51Þ where the first term contains the system translational kinetic energy and the second contains the system rotation and deformation kinetic energy. To find the work done on the system, we examine the energy rate dT =dt: dT dt ¼ M €RRc  _RRc þ PN i¼1 mi€rri  _rri ð2:52Þ NEWTONIAN MECHANICS 51D w U W M S m DO

After making use of the facts that M €RRc ¼ F and that €rri ¼ €RRi  €RRc, the energy rate is written as dT dt ¼ F  _RRc þ PN i¼1 mi €RRi  _rri  €RRc  PN i¼1 mi_rri   ð2:53Þ After using Eqs. (2.39) and (2.43), the energy rate or power is written in the final form as dT dt ¼ F  _RRc þ PN i¼1 Fi  _rri ð2:54Þ If only conservative forces are acting on mi, then the forces Fi can be written as the gradient of a potential function ViðriÞ: Fi ¼  @Vi @ri ð2:55Þ Noting that @Vi @ri _riri ¼ _VVi and defining the total conservative potential function to be d dt V ¼ PN i¼1 _VVi ð2:56Þ Eq. (2.54) can be written as dT dt þ dV dt ¼ F  _RRc ð2:57Þ Studying Eq. (2.57), it is clear that for systems in which the total applied force vector F is zero, the total system energy E ¼ T þ V is conserved. If the total resultant force F is itself a conservative force due to a potential function VcðRcÞ, then Eq. (2.57) can be written as dT dt þ dV dt þ dVc dt ¼ 0 ð2:58Þ and the total system energy E ¼ T þ V þ Vc is also conserved. 52 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

After integrating the kinetic energy rate equation in Eq. (2.54) with respect to time, the change in kinetic energy between two times is given by the work= energy equation Tðt2Þ  Tðt1Þ ¼ ðt2 t1 F  _RRc dt þ PN i¼1 ðt2 t1 Fi  _rri dt ð2:59Þ which can also be written as the spatial integral Tðt2Þ  Tðt1Þ ¼ ðRcðt2Þ Rcðt1Þ F  dRc þ PN i¼1 ðrðt2Þ rðt1Þ Fi  dri ð2:60Þ The first term on the right-hand side of Eq. (2.60) is the translational work done, and the second term is the rotation and deformation work done on the system. Example 2.8 Let us first examine the simple two-particle system illustrated in Fig. 2.10a. Let ^iir ¼ r2=r2, where the notation ri ¼ jrij is used. The forces Fi are collinear with the center-of-mass relative position vectors ri. Using the center-of-mass definition, we find r1 ¼  m2 m1 r2 or _rr1 ¼  m2 m1 _rr2 If the forces are chosen such that F1 ¼ m1 m2 F2 Fig. 2.10 Illustration of forces doing work on a two-particle system. NEWTONIAN MECHANICS 53D w U W M S m DO

then the relative acceleration of particle 2 with respect to particle 1 is €RR2  €RR1 ¼ F2 m2  F1 m1 ¼ F2 m2  F2 m1 m1 m2 ¼ 0 For the center-of-mass relative position coordinates this implies €RR2  €RR1 ¼ €rr2  €rr1 ¼ €rr2 1 þ m2 m1   ¼ 0 ) €rr2 ¼ 0 ¼ €rr1 The inertial time derivatives of ri thus are the constant expressions _rriðtÞ ¼ _rriðt0Þ With these particular collinear forces Fi the two particles will act as a virtual rigid body and not change their relative shape. The second term of the power equation in Eq. (2.54) is expressed as P2 i¼1 Fi  _rriðt0Þ ¼ F2 m1 m2   m2 m1 _rr2ðt0Þ   þ F2  _rr2ðt0Þ ¼ 0 Thus, these forces will not perform any work associated with the relative motion of the masses of the two-particle system. The system power equation is then given by dT dt ¼ F  _RRc ¼ 1 þ m1 m2   F2  _RRc The forces only perform work on the two-particle center-of-mass translation (i.e., the same work one would obtain if all mass were concentrated to a parti- cle at the mass center). Next, consider the setup illustrated in Fig. 2.10b, where the force directions are aligned such that they are always orthogonal to ri. Here the total external forces is F ¼ 0, and the ‘‘translational work’’ done on the center-of-mass motion is F  _RRc ¼ 0 In contrast, however, the work done on the cluster shape is P2 i¼1 Fi  _rri ¼ F1  _rr1  F1  _rr2 ¼ F1  _rr2 1 þ m2 m1   6 ¼ 0 Because the cluster center of mass is an inertial point, the cluster will rotate and deform about the center of mass in this second case, and all work done on the system will be associated with deformation and rotation. 54 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

2.4.3 Linear Momentum In Eq. (2.30) the linear momentum pi of a single particle is defined. For a system of particles, the total linear momentum of the system is defined as the sum p ¼ PN i¼1 pi ¼ PN i¼1 mi _RRi   ð2:61Þ Let ri be the ith particle position vector relative to the system center of mass as defined in Eq. (2.43). Taking the derivative of Eq. (2.45), we can write the total linear momentum expression in Eq. (2.61) in terms of the total system mass M and the center of mass inertial velocity vector _RRc: p ¼ M _RRc ð2:62Þ Note that the superparticle theorem introduced in Eq. (2.48) also holds for the linear momentum of a system of particles. The linear momentum of the mass center of the system of N particles is the same as the linear momentum of the superparticle M. Let Fi be the force acting on the ith particle. Note that Fi is composed both of a net external force component FiE and the internal force component Fij due to interaction with other particles [see Eq. (2.40)]. Using the particle equa- tions of motion in Eq. (2.39), the inertial time rate of change of the total linear momentum of the particle system is expressed as _pp ¼ PN i¼1 mi €RRi   ¼ PN i¼1 Fi   ð2:63Þ Because the inertial forces Fij will cancel each other in this summation due to Newton’s third law, the time rate of change of the linear momentum of a parti- cle system is equal to the total external force acting on the system: F ¼ N d dt pð Þ ð2:64Þ If no external force F is present, then the total system linear momentum vector p will be constant. This leads to the important law of conservation of linear momentum. Unless an external force is acting on a system of N parti- cles, the total linear momentum of the system is conserved. This property is used extensively in collision problems or in the rocket propulsion problem. If two bodies collide, then energy is used to deform the bodies. The total system energy is not conserved during the collision. However, momentum is conserved and can be used to compute the velocities of the bodies after the collision. NEWTONIAN MECHANICS 55D w U W M S m DO

Example 2.9 Let us verify the linear momentum properties in Eqs. (2.62) and (2.64) for the simple two-particle system illustrated in Fig. 2.11. Let Ri be the inertial position vectors of the particles of mass mi, while Rc is the cluster inertial center-of-mass vector. The particle positions ri are taken relative to Rc and satisfy the center-of-mass property m1r1 þ m2r2 ¼ 0 while the inertial position vectors are expressed as R1 ¼ Rc þ r1 R2 ¼ Rc þ r2 The linear momentum p is found through p ¼ m1 _RR1 þ m2 _RR2 ¼ m1 _RRc þ m1_rr1 þ m2 _RRc þ m2_rr2 Using the derivative of the center-of-mass property, we find m1_rr1 þ m2_rr2 ¼ 0 which simplifies the linear momentum expression p to p ¼ ðm1 þ m2Þ _RRc ¼ M _RRc where M ¼ m1 þ m2 is the total cluster mass. This p expression matches the result found in Eq. (2.62). Next, let us compute the inertial time derivative of the linear momentum expression. Differentiating directly Eq. (2.61) leads to _pp ¼ m1 €RR1 þ m2 €RR2 ¼ F1 þ F2 ¼ F where F is the total external force acting on this system. Fig. 2.11 Two-particle system illustration. 56 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

Example 2.10 Assume the dynamical system of interest consists of only two particles m1 and m2 moving along a one-dimensional, frictionless track at different rates. Before a collision at time t0, they each have a constant speed of v1ðt0 Þ and v2ðt0 Þ, respectively, with v1ðt0 Þ > v2ðt0 Þ. The total energy before the impact is given by T ðt0 Þ ¼ 1 2 m1v1ðt0 Þ2 þ m2v2ðt0 Þ2  The total linear momentum is pðt0 Þ ¼ m1v1ðt0 Þ þ m2v2ðt0 Þ First, let us assume that the collision is perfectly elastic. In this case any energy used to deform the bodies during the collision is regained when the body shapes are restored (e.g., think of two rubber balls colliding). Both the total energy T ðtþ 0 Þ T ðtþ 0 Þ ¼ 1 2 m1v1ðtþ 0 Þ2 þ m2v2ðtþ 0 Þ2  and the linear momentum pðtþ 0 Þ pðtþ 0 Þ ¼ m1v1ðtþ 0 Þ þ m2v2ðtþ 0 Þ are conserved during the collision. Setting Tðt0 Þ ¼ Tðtþ 0 Þ and pðt0 Þ ¼ pðtþ 0 Þ, we can express the particles speeds after the collision as v1ðtþ 0 Þ ¼ 1 M v1ðt0 Þðm1  m2Þ þ 2v2ðt0 Þm2   v2ðtþ 0 Þ ¼ 1 M v2ðt0 Þðm2  m1Þ þ 2v1ðt0 Þm1   with M ¼ m1 þ m2 being the total system mass. Note that if m1 ¼ m2, then v1ðtþ 0 Þ ¼ v2ðt0 Þ and v2ðtþ 0 Þ ¼ v1ðt0 Þ, and the particles swap velocities during the collision. Second, we assume that the collision is such that the two particles join and become one (e.g., think of two chunks of clay colliding). In this case the total energy T ðtþ 0 Þ after the collision is given by Tðtþ 0 Þ ¼ 1 2 M v2 where v is the speed of the joined particles after the collision. The linear momentum of the joined particle system is pðtþ 0 Þ ¼ Mv NEWTONIAN MECHANICS 57D w U W M S m DO

Note that this collision is not perfectly elastic and that energy is not conserved. However, linear momentum is conserved, and we can set pðt0 Þ ¼ pðtþ 0 Þ to solve for the velocity v of the joined particle after the collision: v ¼ 1 M m1v1ðt0 Þ þ m2v2ðtþ 0 Þ   The total energy after the collision is given by Tðtþ 0 Þ ¼ 1 2 M v2 ¼ 1 2M m1v1ðt0 Þ þ m2v2ðtþ 0 Þ  2¼ p2 2M The change in energy DT ¼ Tðtþ 0 Þ  Tðt0 Þ is given by DT ¼  m1m2 2M v1ðt0 Þ  v2ðt0 Þ  2 The energy lost during this plastic collision is used to permanently deform the two bodies, as well as to radiate heat and produce sound waves. These two examples are idealized situations. In reality the collisions are never perfectly elastic or plastic. In this case more knowledge is required about how the bodies will deform to predict the motion after the collision. 2.4.4 Angular Momentum As was done for the case of a single particle, let’s find the angular momen- tum of the N particle system about an arbitrary point P given by the inertial position vector RP. The relative position of each mass mi is given through the vector si ¼ Ri  RP ð2:65Þ The total system angular momentum vector H P about the point P is given as the sum of all the single particle angular momentum vectors about this point: H P ¼ PN i¼1 si  mi _ ss i ð2:66Þ Taking the time derivative of H P, we get _HH P ¼ PN i¼1 _ ssi  mi _ ssi þ PN i¼1 si  mi € ssi ð2:67Þ 58 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

After performing similar arguments as in the single particle case, this expres- sion is rewritten as _HH P ¼ PN i¼1 si  mi €RRi  PN i¼1 simi    €RRP ð2:68Þ Using Eqs. (2.45) and (2.65), the following mass center identity is found: PN i¼1 s imi ¼ PN i¼1 Rimi  PN i¼1 mi   RP ¼ M Rc  RP   ð2:69Þ The total external moment LP applied to the system is defined to be LP ¼ PN i¼1 si  mi €RRi ¼ PN i¼1 s i  Fi ð2:70Þ Using Eqs. (2.69) and (2.70), the system angular momentum derivative _HH P about a point P is3 _HH P ¼ LP þ M €RRP  Rc  RP   ð2:71Þ Note that if either Rc ¼ RP or RP is nonaccelerating inertially, then Eq. (2.71) reduces to the most familiar Euler equation _HH P ¼ LP ð2:72Þ Analogously to the linear momentum development, if no external torque LP is acting on the system of particles, then the total angular momentum vector H P is constant. Example 2.11 Two particles are attached on strings and are moving in a planar, circular manner as shown in Fig. 2.12. The plane on which the particles are moving is level compared to the gravity field. Thus, given an initial velocity and ignoring frictional effects, the particles will continue to move at a constant rate. Assume that the two circular paths meet tangentially at one point. We would like to investigate how the velocities will change if the particles meet at time t0. This condition is shown in gray in the figure. The total kinetic energy before the collision is Tðt0 Þ ¼ m1 2 v1ðt0 Þ2 þ m2 2 v2ðt0 Þ2 while the momentum H about O along the plane normal direction ^iin is Hðt0 Þ ¼ R1m1v1ðt0 Þ  R1m2v2ðt0 Þ ¼ R1½m1v1ðt0 Þ  m2v2ðt0 ފ NEWTONIAN MECHANICS 59D w U W M S m DO

Assuming the collision is perfectly elastic, then both the total energy and angu- lar momentum are conserved. After the collision, we express them as T ðtþ 0 Þ ¼ m1 2 v1ðtþ 0 Þ2 þ m2 2 v2ðtþ 0 Þ2 Hðtþ 0 Þ ¼ R1½m1v1ðtþ 0 Þ  m2v2ðtþ 0 ފ Setting T ðt0 Þ ¼ T ðtþ 0 Þ and Hðt0 Þ ¼ Hðtþ 0 Þ, we are able to solve for the particle velocities after the collision: v1ðtþ 0 Þ ¼ m1v1ðt0 Þ  m2½v1ðt0 Þ þ 2v2ðt0 ފ m1 þ m2 v2ðtþ 0 Þ ¼ m2v2ðt0 Þ  m1½v2ðt0 Þ þ 2v1ðt0 ފ m1 þ m2 Next, let us examine the angular momentum H P expressed in Eq. (2.66) about the arbitrary point P by rewriting the inertial position vectors again as Ri ¼ Rc þ ri as illustrated in Fig. 2.8. H P ¼ PN i¼1 ðRc þ ri  RpÞ  mið _RRc þ _rri  _RRpÞ ð2:73Þ Expanding the vector algebra leads to H P ¼ ðRc  RpÞ  PN i¼1 mi   ð _RRc  _RRpÞ þ ðRc  RpÞ  PN i¼1 mi_rri   þ PN i¼1 miri    ð _RRi  _RRPÞ þ PN i¼1 ri  mi_rri ð2:74Þ R1 R2 m1 m2 O ˆın Fig. 2.12 Illustration of two particles moving in a circular manner on a level plane. 60 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

Using the center-of-mass definition PN i¼1 miri ¼ 0 and the total mass expres- sion PN i¼1 mi ¼ M , the angular momentum about point P is H P ¼ ðRc  RpÞ  M ð _RRc  _RRpÞ þ PN i¼1 ri  mi_rri ð2:75Þ Note that this algebraic expression separates conveniently the angular momen- tum of the cluster of particles into the angular momentum caused by the center- of-mass motion and that of the particle motion about the center of mass. If the point P is the origin O of an inertial frame, then RP ¼ _RRP ¼ 0, and H O ¼ Rc  M _RRc þ PN i¼1 ri  mi_rri ð2:76Þ Example 2.12 Let us consider the angular momentum for the simple two-particle systems illustrated in Fig. 2.10 about the inertial point O. In case a) shown in Fig. 2.10a the forces Fi are collinear and satisfy F1 ¼ m1 m2 F2 Here the super-particle theorem yields M €RRc ¼ F ¼ F1 1 þ m2 m1   Thus the center-of-mass motion is noninertial with both Rc and _RRc varying with time. Using the center-of-mass definition m1r1 þ m2r2 ¼ 0, Example 2.8 shows that €rri ¼ 0, and thus r2 can vary with time, while _rr2 is constant. The angular momentum H O is expressed as H O ¼ RcðtÞ  M _RRcðtÞ þ 1 þ m2 m1   r2ðtÞ  m2_rr2ðt0Þ Taking the inertial derivative of H O yields _HH O ¼ Rc  M €RRc ¼ Rc  F ¼ L Next, let us consider the second two-particle system illustrated in Fig. 2.10b, where the forces Fi are orthogonal to ri and F2 ¼ F1. In this case the NEWTONIAN MECHANICS 61D w U W M S m DO

total force acting on the system is F ¼ 0, and the super-particle theorem yields M €RRc ¼ F ¼ 0 Thus the inertial center-of-mass velocity will be constant in this case. However, the relative position and velocity vectors ri and _rri are time varying. The angu- lar momentum about O is H O ¼ RcðtÞ  M _RRcðt0Þ þ 1 þ m2 m1   r2ðtÞ  m2_rr2ðtÞ Because €RRc ¼ 0 in this case, note that €rr2 ¼ €RR2. Taking the inertial derivative of this momentum expression yields _HH O ¼ 1 þ m2 m1   r2  m2€rr2 ¼ 1 þ m2 m1   r2  m2 €RR2 ¼ 1 þ m2 m1   r2  F2 ¼ L 2.5 Dynamics of a Continuous System 2.5.1 Equations of Motion The development of the dynamical equations of motion of a continuous system parallels that of the system of N particles. Any finite sums over all particles are generally replaced with volume integrals over the body B. This allows us to describe any constant mass body, even if it is flexible or does not have a constant shape, as in a chunk of gelatin. However, care must be taken to define a control volume that contains the instantaneous mass of the system when actually carrying out volume integrations. Let dm be an infinitesimal body element with the corresponding inertial position vector R as shown in Fig. 2.13. Then, as such, it can be considered to be a particle that abides by Newton’s second law. The equations of motion for this infinitesimal element are dF ¼ €RR dm ð2:77Þ where dF is the total force acting on dm. The force vector dF is broken up into external and internal components as dF ¼ dFE þ dFI ð2:78Þ To express the volume integral over the body B, let us use the shorthand nota- tion Ð B ¼ Ð Ð Ð B. The total force F acting on this continuous body is given by F ¼ ð B dF ¼ ð B dFE ð2:79Þ 62 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

where the internal forces again cancel because of Newton’s third law. The total body mass is given by M ¼ ð B dm ð2:80Þ The system center of mass is defined such that ð B r dm ¼ 0 ð2:81Þ where r ¼ R  Rc is again the internal position vector of dm relative to Rc. Therefore, Eq. (2.81) can be rewritten as M Rc ¼ ð B R dm ð2:82Þ The center of mass vector Rc is then expressed as Rc ¼ 1 M ð B R dm ð2:83Þ After twice differentiating Eq. (2.82), we obtain M €RRc ¼ ð B €RR dm ¼ ð B dF ð2:84Þ After substituting Eq. (2.79), we obtain the equivalent superparticle theorem for a continuous body: M €RRc ¼ F ð2:85Þ Fig. 2.13 Mass element of a continuous system. NEWTONIAN MECHANICS 63D w U W M S m DO

2.5.2 Kinetic Energy Let the inertial vector R define the position of the infinitesimal mass ele- ment dm. The kinetic energy of the entire continuous body B is then given as T ¼ 1 2 ð B _RR  _RR dm ð2:86Þ After substituting _RR ¼ _RRc þ _rr, the kinetic energy is expressed as T ¼ 1 2 ð B dm   _RRc  _RRc þ _RRc  ð B _rr dm þ 1 2 ð B _rr  _rr dm ð2:87Þ Making use of Eqs. (2.80) and (2.81), the kinetic energy for a continuous body B is written as T ¼ 1 2 M _RRc  _RRc þ 1 2 ð B _rr  _rr dm ð2:88Þ The first term in Eq. (2.88) represents the translational kinetic energy, and the second term represents the rotational and deformational energy. To find the work done on the continuous body B, the kinetic energy rate is found: dT dt ¼ M €RRc  _RRc þ ð B _rr  €rr dm ð2:89Þ After using Eq. (2.85) and the fact that €rr ¼ €RR  €RRc, the kinetic energy rate is given as dT dt ¼ F  _RRc þ ð B €RR dm    _rr  €RRc  ð B _rr dm ð2:90Þ Using Eqs. (2.77) and (2.81), the kinetic energy rate or power for a continuous, constant mass body B is given by dT dt ¼ F  _RRc þ ð B dF  _rr ð2:91Þ The change in kinetic energy or the work performed between two times is found by integrating the kinetic energy rate expression with respect to time: T ðt2Þ  Tðt1Þ ¼ ðt2 t1 F  _RRc dt þ ðt2 t1 ð B dF  _rr dt ð2:92Þ 64 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

This can also be written alternatively as a spatial integration: Tðt2Þ  Tðt1Þ ¼ ðRðt2Þ Rðt1Þ F  dRc þ ðrðt2Þ rðt1Þ ð B dF  dr ð2:93Þ where the first term expresses the translational work and the second term is the rotational and deformational work done on the system. 2.5.3 Linear Momentum To determine the total linear momentum of a continuous body B, we express the linear momentum of an infinitesimal body element dm as dp ¼ _RR dm ð2:94Þ Integrating the infinitesimal linear momentum contributions over the entire body, the total linear momentum is given by p ¼ ð B dp ¼ ð B _RR dm ð2:95Þ Using the center of mass property in Eq. (2.82), the total linear momentum of the body is written directly in terms of the body mass M and the center of mass motion _RRc: p ¼ M _RRc ð2:96Þ Again the superparticle theorem applies to the continuous body. The sum of the individual infinitesimal linear momenta of the body is the same as the linear momenta of a particle of mass M with the same velocity vector as the body center of mass motion. Note that the body B is not restricted to be a rigid body in this section. If the body center of mass is inertially stationary (i.e., the body has zero linear momentum), it is still possible for various body components to be moving inertially. For example, consider a heap of gelatin floating in space. It is possible for the gelatin to be deforming without moving. Although the individual components of gelatin might have some linear momen- tum, the total sum of these components cancel each other out to result in a zero net motion of the body center of mass. Taking the inertial derivative of Eq. (2.96) and making use of the inter- nal=external force properties in Eqs. (2.78) and (2.79), we express the total linear momentum rate as _pp ¼ ð B €RR dm ¼ ð B dF ¼ F ð2:97Þ Thus, the time rate of change of the total linear momentum of a continuous body B is equal to the total external force vector being applied to this body. If NEWTONIAN MECHANICS 65D w U W M S m DO

no external force vector is applied, then the total linear momentum is con- served and its rate is zero. 2.5.4 Angular Momentum To find the angular momentum vector of the continuous body B about an arbitrary point P, we write the relative position vector s of dm to P as s ¼ R  RP ð2:98Þ The total system angular momentum vector H P about P is then given by H P ¼ ð B s  _ ss dm ð2:99Þ Taking the derivative of H P, we get _HH P ¼ ð B _ ss  _ss dm þ ð B s  € ss dm ð2:100Þ which can be rewritten as _HH P ¼ ð B s  €RR dm  ð B s dm    €RRP ð2:101Þ The term in the parentheses can be expanded to ð B s dm ¼ ð B R dm  ð B dm   RP ¼ M ðRc  RPÞ ð2:102Þ The total external moment LP applied to the system is defined to be LP ¼ ð B s  €RR dm ¼ ð B s  dF ð2:103Þ Using these two identities in Eqs. (2.102) and (2.103), the system angular momentum derivative vector _HH P about P is _HH P ¼ LP þ M €RRP  Rc  RP   ð2:104Þ As was the case with the system of N particles, if either Rc ¼ RP or the vector RP is nonaccelerating inertially, then Eq. (2.104) reduces to the Euler equa- tion2 _HH P ¼ LP ð2:105Þ 66 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

As was the case with the dynamical system of finite particles, the angular momentum of a continuous body is constant if no external torque vector LP is applied. 2.6 Rocket Problem In this section we investigate the thrust that a rocket motor produces by expelling propellant at a high velocity from the spacecraft. Consider the one- stage rocket shown in Fig. 2.14. Let m be the mass of the rocket including any propellant that is currently on board. The propellant fuel is being burned and ejected at a mass flow rate of _mm. The current velocity vector of the rocket is v, while the exhaust velocity of the ejected propellant particles dm relative to the rocket is ve. Note that the orientation of the exhaust velocity vector ve does not have to point backwards. If the nozzle would be pointing forward, then the engine would be used to perform a breaking maneuver. The rocket is assumed to be flying through an atmosphere with an ambient pressure Pa. At the point where the exhaust gases escape the engine nozzle, the exhaust pressure is given by Pe. We would like to develop the thrust vector that the rocket engine is exerting onto the spacecraft. To do so, we utilize Eq. (2.64) or (2.97), which state that the external force F exerted onto a system of particles or a continuous body is equal to the time rate of change in linear momentum. Let us treat the rocket mass m and the expelled propellant particle Dm as a two-particle system and track their linear momentum change over a small time interval Dt. Using Eq. (2.72), we can write the linear momentum equation as FDt ¼ pðt þ DtÞ  pðtÞ ð2:106Þ The quantity FDt is the impulse being applied to the system over the time interval dt. At time t the rocket and propellant mass are still combined as m. At time t þ Dt, the rocket mass has been reduced to m  dm, and the propel- lant particle Dm is about to leave the engine nozzle. Assume that the only external force acting on this two-particle system is due to pressure differential Fig. 2.14 One-stage rocket expelling a propellant particle Dm with an ambient atmosphere Pa. NEWTONIAN MECHANICS 67D w U W M S m DO

at the engine nozzle. Let A be the nozzle cross-sectional area, then the external force F is expressed as F ¼  ve ve A Pe  Pa   ð2:107Þ More generally, however, we write the external force vector F as F ¼  ve ve A Pe  Pa   þ Fe ð2:108Þ where Fe is the net sum of non-pressure-related external forces such as gravita- tional forces acting on the system. The pressure-induced force is assumed to be collinear with the exhaust velocity vector ve. Note that if Pa ¼ Pe (exhaust expands to ambient pressure) or Pa ¼ Pe ¼ 0 (operating in a vacuum and exhaust expanding to zero pressure), then the net external force on the system is zero. Further, if the direction of the exhaust velocity vector ve is in the opposite direc- tion to the rocket velocity vector v, then a positive pressure differential Pe  Pa > 0 results in an acceleration in the rocket velocity direction. The linear momentum p of the system at time t is pðtÞ ¼ mv ð2:109Þ because the propellant particle dm is still joined with the rocket. At time t þ Dt, the small propellant mass Dm is being ejected from the rocket with a relative velocity vector ve . Because the rocket is losing mass, the mass difference Dm over time dt is a negative quantity. The linear momentum at time t þ Dt is pðt þ DtÞ ¼ ðm þ DmÞðv þ DvÞ  Dmðv þ veÞ ð2:110Þ where ðm þ DmÞ is the rocket mass without the escaping fuel particle, and Dv is the change in rocket velocity vector over the time interval Dt. Dropping higher order differential terms in Eq. (2.110) and substituting the F; pðtÞ and pðt þ DtÞ expressions into Eq. (2.106) leads to  ve ve A Pe  Pa  Dt þ FeDt ¼ mDv  Dmve ð2:111Þ Dividing both sides by Dt and solving for the acceleration term, we find m Dv Dt ¼  ve ve A Pe  Pa   þ Dm Dt ve þ Fe ð2:112Þ Allowing the time step Dt to become infinitesimally small, we arrive at the rocket equations of motion: m dv dt ¼ ve A ve Pe  Pa    dm dt   |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Fs þ Fe ¼ Fs þ Fe ð2:113Þ 68 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

The Fs force component is called the static thrust of the rocket engine. If the rocket were attached to a test stand, then it would require a force Fs to keep the rocket immobile during the engine test firing. If the exhaust velocity vector is in the opposite direction to the rocket velo- city vector v as shown in Fig. 2.14, and the rocket is operating in a weightless environment with Fe ¼ 0, then the rocket equations of motion simplify to the famous one-dimensional form m dv dt ¼ A Pe  Pa    dm dt ve ¼ Fs ð2:114Þ with the parameter Fs being the scalar static rocket thrust. Let us assume that over a time interval from t0 to tf that the negative mass flow rate _mm is constant. Equation (2.114) can be rewritten as m dv dm _mm ¼ Fs ð2:115Þ Rearranging this equation by separating the dv and dm terms, and integrating from t0 to tf , we find ðvf v0 dv ¼ vf  v0 ¼ Fs _mm ðmf m0 dm m ¼  Fs _mm ln m0 mf   ð2:116Þ Note that Fs and _mm can be taken outside the integral sign since they are constants in this investigation. The scalar velocity v0 is the velocity that the rocket possessed at t0, while vf is the rocket velocity at the thruster burnout at tf . The initial rocket mass is m0 and the smaller, final rocket mass is mf . The burnout velocity vf can be solved for in terms of the initial rocket velocity and mass, as well as the final burnout mass mf : vf ¼ v0  Fs _mm ln m0 mf   ð2:117Þ The second term in Eq. (2.117) is a positive quantity since m0 > mf and the mass flow rate _mm is a negative quantity. Let Dm < 0 be the amount of fuel mass lost over the given time interval. Then mf ¼ m0 þ Dm. The change in velocity Dv ¼ vf  v0 that results from ejecting Dm of fuel is given by Dv ¼  Fs _mm ln 1 1 þ E   ð2:118Þ where E ¼ Dm=m0 is the ratio of fuel spent and the initial rocket mass over the time interval Dt. Note that this change in velocity depends only on the amount of fuel spent and Fs , not on the length of the burning time. Thus, if a thruster produces half the mass flow rate _mm as another thruster, but burns for twice as long, then both thrusters will produce the same velocity change Dv. However, this result is true only if no other external forces are acting on the body. If gravity NEWTONIAN MECHANICS 69D w U W M S m DO

is pulling on the rocket, then the amount of time spent trying to accelerate the rocket will have a drastic effect on the rocket velocity at burnout time. A common measure of rocket thruster efficiency is the specific impulse Isp, defined as3;5 Isp ¼ Fs ð _mmÞg ð2:119Þ and having units of seconds. The gravitational acceleration g used here is that experienced on the Earth’s surface. The higher this Isp value is, the more force the rocket thruster is able to produce for a given mass flow rate. If the exhaust pressure Pe is close to the ambient pressure Pa, the pressure contribution to the static thrust Fs in Eq. (2.114) is negligible. In this case Fs   _mmve and the specific impulse simplifies to Isp  ve g ð2:120Þ From this simplification it is evident that to achieve higher thruster efficiencies, the exhaust velocity ve should be as high as possible. The faster a given fuel particle is ejected from the rocket, the larger a momentum change (i.e., rocket speed up) it will cause. Using the specific impulse definition, the rocket velo- city change Dv for a given fuel ratio E burned is given by Dv ¼ Ispg ln 1 1 þ E   ð2:121Þ The specific impulse ranges for different rocket thruster systems are shown in Table 2.1.5 Note that the higher specific impulse propulsion methods, such as Table 2.1 Specific impulse and thrust ranges for different rocket thruster designs Thruster type Vacuum Isp, s Thrust range, N Comments Solid motor 280 300 50 5  106 Simple, reliable low cost design with a low performance, but a high thrust Cold gas 50 75 0.05 200 Extremely simple and reliable design with a very low performance and heavy weight for the small thrust produced Liquid motor 150 450 5 5  106 Higher performance thrusters at the cost of a more complicated mechanical and cryogenic design Electrothermal arcjet 450 1500 0.05 5 Higher performance, low thrust system with a complicated thermal interface Ion 2000 6000 5  106 0.5 Very high performance system with typically a very low thrust 70 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

the ion or arcjet thrusters, typically produce only a very small thrust. Such modes of propulsion are able to achieve a desired Dv with a much smaller amount of fuel mass Dm than a propulsion method with a lower Isp. However, because of the small amount of thrust produced, these efficient pro- pulsion methods will take a much longer time to produce this desired velocity change. Example 2.13 Assume we are trying to launch an initially at rest sounding rocket vertically from the Earth’s surface and it is to fly only several miles high. For these small altitudes, we are still able to assume that the gravitational attraction g is constant during the flight. The solid rocket motor produces a constant Isp for the duration of its burn. Because the only external force acting on the rocket is the constant gravitational acceleration, the rocket equations of motion in the vertical direction are given by Eq. (2.113): m_vv ¼ Fs  mg ¼ g m þ Isp _mm   ð2:122Þ Note that _mm < 0. This equation illustrates the challenge that a highly efficient ion propulsion system would have in attempting to launch this sounding rocket. The change in velocity expression given in Eq. (2.118) assumes that no external forces are acting on the rocket except for the ambient and exhaust pressure. With the gravity force acting on our sounding rocket, the thruster is constantly battling the gravitational acceleration. In fact, if the rocket thrust is less than the weight mg of the rocket, then the propulsion system will not be able to lift the rocket off the launch pad. Thus, although an ion high perfor- mance propulsion system is very effective in accelerating a spacecraft in a weightless or free-falling environment, it would be an inappropriate propulsion choice to launch a rocket off a planet’s surface. The rocket velocity at burnout time tf is then given by vf ¼ g tf þ Isp ln m0 mf    ð2:123Þ The longer the thruster takes to accelerate the rocket to the desired velocity, the longer the thruster must combat the gravitational acceleration. Because the efficient ion propulsion system requires a large time tf to achieve a desired Dv, the gravity is also given a large amount of time to counter the achievements of the ion thruster. This is why it is common to use solid or liquid chemical propulsion systems to launch a rocket from the planet’s surface to a low Earth orbit. Although these propulsion choices are less efficient, they provide a thrust that is much larger than the rocket weight. With this large static thrust, the rocket is propelled to the desired velocity quickly and the gravity field has less time to decelerate the craft. NEWTONIAN MECHANICS 71D w U W M S m DO

References 1Wiesel, W. E., Spaceflight Dynamics, McGraw Hill, New York, 1989. 2Junkins, J. L., and Turner, J. D., Optimal Spacecraft Rotational Maneuvers, Elsevier, Amsterdam, 1986. 3Greenwood, D. T., Principles of Dynamics, 2nd ed., Prentice Hall, Englewood Cliffs, NJ, 1988. 4Craig, R. R., Structural Dynamics, Wiley, New York, 1981. 5Wertz, J. R., and Larson, W. J., Space Mission Analysis and Design, Kluwer, Dordrecht, The Netherlands, 1991. 6Nelson, R. C., Flight Stability and Automatic Control, McGraw Hill, New York, 1989. Problems 2.1 Plot the magnitude of the gravitational acceleration as it varies from Earth’s surface to a height of 300 km. 2.2 Given a spring with a spring stiffness constant of k ¼ 5 kg=s2 and a stored potential energy of 100 N  m, find the spring deflection x and the force F required to keep the spring at this deflection. 2.3 A mass m is sliding down a constant, frictionless slope of 10 deg with an initial velocity of vðt0Þ ¼ 1 m=s. How long will it take this mass to accel- erate to a velocity of vðtf Þ ¼ 10 m=s and what distance will it have traveled? 2.4 A skydiver exits an aircraft at an altitude of 3000 m. The aircraft is flying horizontally at 36 m=s. The skydiver has a mass m of 80 kg, a forward projected surface area A of 0.75 m2, and a coefficient of drag cd of 0.555. Assume a uniform gravitation field with a gravitational acceleration of 9.81 m=s2. The air density r is 1.293 kg=m3. Recall the relationship Drag¼ 1 2 rv2cdA, and this force is opposite to the velocity vector.6 (a) What is the theoretical terminal velocity of this skydiver? (b) Find the skydiver equations of motion and solve them numerically for a 45-s freefall. Plot the altitude vs horizontal position, the sky- diver speed vs time, and the horizontal=vertical velocity vs time. (c) Taking into account that the more air speed a skydiver has, the better and faster the parachute will open, what is the ‘‘worst’’ time for a skydiver to try to open the parachute? (d) How long does it take for the skydiver to reach 95% of the terminal velocity? (e) What acceleration does the skydiver experience at terminal velocity? (f) How far forward does the skydiver get thrown on exit before he or she essentially descends vertically? 72 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

2.5 A ball of mass m is sliding in a frictionless tube as shown in Fig. P2.5. The tube is rotating at a constant angular velocity v. Initially the ball is at rest relative to the tube at point A at r ¼ L^eer. (a) What is the velocity vector when the ball exits the tube? (b) Up to the point where the ball exists the tube, how much work has been performed onto the ball? (c) Find an expression for the angular momentum vector H A of the mass m about point A. 2.6 A cannon tries to hit a target that is a distance R away with a projectile of mass m as shown in Fig. P2.6. However, at a distance R=4 there is an obstacle of height H present. What is the smallest elevation angle g0 and corresponding initial speed v0 the projectile m must possess initially to hit the target and miss the obstacle? Assume a constant gravity field is present. Fig. P2.5 Ball in rotating tube. Fig. P2.6 Ballistic trajectory problems: clearing an obstacle. NEWTONIAN MECHANICS 73D w U W M S m DO

74 ANALYTICAL MECHANICS OF SPACE SYSTEMS 2.7 A cannon tries to hit a target that is a distance R away and elevated off the ground by a height H with a projectile m as shown in Fig. P2.7. What is the smallest initial velocity v and corresponding heading angle g the particle may have to hit this target? Assume a constant gravity field is present. 2.8 As shown in Fig. P2.8, a ball with mass m is propelled by a spring with a spring stiffness k to roll without friction on a surface until it is launched into the air by a ramp of height h. The departure angle g is fixed by the ramp and is not a variable. The goal is to hit a target on ground level a distance d away from the ramp. (a) Find the initial velocity v0 the ball must have when leaving the ramp to hit the target. (b) What is the initial compression z the spring must have such that the mass will have the necessary velocity v0 when leaving the ramp? Fig. P2.7 Ballistic trajectory problem: hitting elevated target. Fig. P2.8 Spring propelled mass.D w U W M S m DO

2.9 Consider the two-particle system studied in Example 2.8. Verify the results shown for the elastic collision by providing all of the algebra required to complete the steps outlined. 2.10 A massless cylinder is rolling down a slope with an inclination angle a under the influence of a constant gravity field. A mass m is attached to the cylinder and is offset from the cylinder center by R=2 as shown in Fig. P2.10. (a) Find the equations of motion of the mass m in terms of the angle y. (b) What is the normal force N ¼ N ^nn2 that the ground is exerting against the cylinder? 2.11 A ball m is freely rolling in the lower half of a sphere under the influence of a constant gravity field as shown in Fig. P2.11. The sphere has a constant radius r. Assume that _ffðt0Þ is zero and that yðt0Þ, _yyðt0Þ, and fðt0Þ are given. NEWTONIAN MECHANICS 75 Fig. P2.10 Rolling cylinder with offset mass. Fig. P2.11 Ball rolling inside a sphere.D w U W M S m DO

(a) Find the equation of motion of the ball rolling without slip inside the sphere in terms of the spherical angle f: Hint: The angular momentum about the ^nn3 axis is conserved. (b) What is the normal force that the wall of the sphere exerts onto the ball at any point in time? (c) Since _ffðt0Þ ¼ 0, the ball is starting out on an extrema. Find an expression in terms of y0; _yy0, and _ff0 that determines the other motion extrema where _ff ¼ 0. Hint: Use conservation of energy. 2.12 A cloud contains four particles with masses m1 ¼ m2 ¼ 1 and m3 ¼ m4 ¼ 2. The position vector of each particle is R1 ¼ 1 1 2 0 @ 1 A R2 ¼ 1 3 2 0 @ 1 A R3 ¼ 2 1 1 0 @ 1 A R4 ¼ 3 1 2 0 @ 1 A and their respective velocity vectors are _RR1 ¼ 2 1 1 0 @ 1 A _RR2 ¼ 0 1 1 0 @ 1 A _RR3 ¼ 3 2 1 0 @ 1 A _RR4 ¼ 0 0 1 0 @ 1 A (a) How much of the total cloud kinetic energy is translational kinetic energy and how much is rotation and deformation energy? (b) What is the cloud angular momentum vector about the origin and about the center of mass? 2.13 Two particles with mass m=2 are attached by a linear spring with a spring constant k as shown in Fig. P2.13. Consider arbitrary initial position and velocity of each mass on the plane. For simplicity, however, assume that the initial separation 2r0 is the unstretched length of the spring, and that the mass center has zero inertial velocity initially. (a) Determine the differential equations of motion whose solution would give rðtÞ and yðtÞ as functions of time and initial conditions; it is not necessary to solve these differential equations. 76 ANALYTICAL MECHANICS OF SPACE SYSTEMS Fig. P2.13 Two masses moving in a plane.D w U W M S m DO

(b) Determine an expression that relates the radial velocity _rr and the angular velocity _yy as functions of r, y, and initial conditions. 2.14 A particle of mass m is free to sling along a vertical ring as shown in Fig. P2.14. The ring itself is rotating at a constant rate _ff. (a) Determine the equations of motion of the particle in terms of y. (b) What are the normal forces produced by the ring onto the particle? 2.15 Newton’s second law for a particle of mass m states that F ¼ d=dtðmvÞ. If m is time varying, then one might expect F ¼ _mmv þ m_vv to be true. Explain why this logic is incorrect and does not lead to the correct rocket thrust equation. 2.16 The static thrust Fs of a rocket is given in Eq. (2.109). Draw a freebody diagram of a rocket engine test stand and verify that this is indeed that static force required to keep the rocket in place. NEWTONIAN MECHANICS 77 Fig. P2.14 Particle sliding along a rotating ring.D w U W M S m DO

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3 Rigid Body Kinematics 3.1 Introduction Attitude coordinates (sometimes also referred to as attitude parameters) are sets of coordinates fx1; x2; . . . ; xng that completely describe the orientation of a rigid body relative to some reference frame. There is an infinite number of atti- tude coordinates to choose from. Each set has strengths and weaknesses compared to the other sets. This is analogous to choosing among the infinite sets of translational coordinates such as Cartesian, polar, or spherical coordi- nates to describe a spatial position of a point. However, describing the attitude of an object relative to some reference frame does differ in a fundamental way from describing the corresponding relative spatial position of a point. In Carte- sian space, the linear displacement between two spatial positions can grow arbitrarily large. On the other hand, two rigid body (or coordinate frame) orien- tations can differ at most by a 180 deg rotation, a finite rotational displacement. If an object rotates past 180 deg, then its orientation actually starts to approach the starting angular position again. This concept of two orientations differing only by finite rotations is important when designing control laws. A smart choice in attitude coordinates can exploit this fact and produce a control law that can intelligently handle very large orientation errors. The quest for ‘‘the best rigid body orientation description’’ is a very funda- mental and important one. It has been studied by such great scholars as Euler, Jacobi, Hamilton, Cayley, Klein, Rodrigues, and Gibbs and has led to a rich collection of elegant results. A good choice for attitude coordinates can greatly simplify the mathematics and avoid such pitfalls as mathematical and geometri- cal singularities or highly nonlinear kinematic differential equations. Among other things, a bad choice of attitude coordinates can artificially limit the operational range of a controlled system by requiring it to operate within the nonsingular range of the chosen attitude parameters. The following list contains four truths about rigid body attitude coordinates that are listed without proof 1: 1) A minimum of three coordinates is required to describe the relative angular displacement between two reference frames F 1 and F 2. 2) Any minimal set of three attitude coordinates will contain at least one geometrical orientation where the coordinates are singular, namely at least two coordinates are undefined or not unique. 3) At or near such a geometric singularity, the corresponding kinematic differential equations are also singular. 4) The geometric singularities and associated numerical difficulties can be avoided altogether through a regularization. Redundant sets of four or more 79D w U W M S m DO

coordinates exist that are universally determined and contain no geometric singularities. 3.2 Direction Cosine Matrix Rigid body orientations are described using displacements of body-fixed referenced frames. The reference frame itself is usually defined using a set of three orthogonal, right-hand unit vectors. For notational purposes, a reference frame (or rigid body) is labeled through a script capital letter such as F , and its associated unit base vectors are labeled with lowercase letters such as ^ffi. There is always an infinity of ways to attach a reference frame to a rigid body. However, typically the reference frame base vectors are chosen such that they are aligned with the principal body axes. Let the two reference frames N and B each be defined through sets of orthonormal right-hand sets of vectors f ^nng and f^bbg where we use the shorthand vectrix notation f ^nng  ^nn1 ^nn2 ^nn2 8 < : 9 = ; f^bbg  ^bb1 ^bb2 ^bb2 8 < : 9 = ; ð3:1Þ The sets of unit vectors are illustrated in Fig. 3.1. The reference frame B can be thought of as being a generic rigid body, and the reference frame N could be associated with some particular inertial coordinate system. Let the three angles a1i be the angles formed between the first body vector ^bb1 and the three inertial axes ^nn1, ^nn2, and ^nn3. The cosines of these angles are called the direction Fig. 3.1 Direction cosines. 80 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

cosines of ^bb1 relative to the N frame. The unit vector ^bb1 can be projected onto f ^nng as ^bb1 ¼ cos a11 ^nn1 þ cos a12 ^nn2 þ cos a13 ^nn3 ð3:2Þ Clearly the direction cosines cos a1j are the three orthogonal components of ^bbj. Analogously, the direction angles a2i and a3i between the unit vectors ^bb2 and ^bb3 and the reference frame N base vectors can be found. These vectors are then expressed as ^bb2 ¼ cos a21 ^nn1 þ cos a22 ^nn2 þ cos a23 ^nn3 ð3:3Þ ^bb3 ¼ cos a31 ^nn1 þ cos a32 ^nn2 þ cos a33 ^nn3 ð3:4Þ The set of orthonormal base vectors f^bbg can be compactly expressed in terms of the base vectors f ^nng as f^bbg ¼ cos a11 cos a12 cos a13 cos a21 cos a22 cos a23 cos a31 cos a32 cos a33 2 4 3 5f ^nng ¼ ½CŠf ^nng ð3:5Þ where the matrix ½CŠ is called the direction cosine matrix. Note that each entry of ½CŠ can be computed through Cij ¼ cosðff ^bbi; ^nnjÞ ¼ ^bbi  ^nnj ð3:6Þ Analogously to Eq. (3.5), the set of f ^nng vectors can be projected onto f^bbg vectors as f ^nng ¼ cos a11 cos a21 cos a31 cos a12 cos a22 cos a32 cos a13 cos a23 cos a33 2 4 3 5f^bbg ¼ ½CŠT f^bbg ð3:7Þ Substituting Eq. (3.7) into (3.5) yields f^bbg ¼ ½CŠ½CŠT f^bbg ð3:8Þ which requires that ½CŠ½CŠT ¼ ½I33Š ð3:9Þ Similarly, substituting Eq. (3.5) into (3.7) yields ½CŠT ½CŠ ¼ ½I33Š ð3:10Þ RIGID BODY KINEMATICS 81D w U W M S m DO

Eqs. (3.9) and (3.10) show that the direction cosine matrix ½CŠ is ortho- gonal.1–4 Therefore, the inverse of ½CŠ is the transpose of ½CŠ: ½CŠ 1 ¼ ½CŠT ð3:11Þ Thanks to the orthogonality of the direction cosine matrix ½CŠ, we will see in the following that the forward and inverse transformation (projection) of vectors between rotationally displaced reference frames can be accomplished without arithmetic. Another important property of the direction cosine matrix is that its determi- nant is 1. This can be shown as follows.5 From Eq. (3.9) it is evident that detðCCT Þ ¼ detð½I33ŠÞ ¼ 1 ð3:12Þ Because ½CŠ is a square matrix, this can be written as6 detðCÞ detðCT Þ ¼ 1 ð3:13Þ Because detðCÞ is the same as detðCT Þ, this is further reduced to6 detðCÞð Þ2¼ 1() detðCÞ ¼ 1 ð3:14Þ As is shown by Goldstein in Ref. 7, if the reference frame base vectors f^bbg and f ^nng are right-handed, then detðCÞ ¼ þ1. Goldstein also shows that the 3  3 direction cosine matrix ½CŠ will have only one real eigenvalue of 1. Again, it will be þ1 if the reference frame base vectors are right-handed. In a standard coordinate transformation setting, the ½CŠ matrix is typically not restricted to projecting one set of base vectors from one reference frame onto another. Rather, the most powerful feature of the direction cosine is the ability to directly project (or transform) an arbitrary vector, with components written in one reference frame, into a vector with components written in another reference frame. To show this, let v be an arbitrary vector and let the reference frames B and N be defined as earlier. Let the scalars vbi be the vector components of v in the B reference frame: v ¼ vb1 ^bb1 þ vb2 ^bb2 þ vb3 ^bb3 ¼ fvbgT f^bbg ð3:15Þ Similarly v can be written in terms of N frame components vni as v ¼ vn1 ^nn1 þ vn2 ^nn2 þ vn3 ^nn3 ¼ fvngT f ^nng ð3:16Þ Substituting Eq. (3.7) into Eq. (3.16), the v vector components in the N frame can be directly projected into the B frame: vb ¼ ½CŠvn ð3:17Þ 82 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

Because the inverse of ½CŠ is simply ½CŠT, the inverse transformation is vn ¼ ½CŠT vb ð3:18Þ The fact that Eqs. (3.17) and (3.18) are exactly analogous to Eqs. (3.5) and (3.7) is a fundamental property of Gibbsian vectors and, more generally, Carte- sian tensors. Another common problem is that several cascading reference frames are present where each reference frame orientation is defined relative to the previous one, and it is desired to replace the sequence of projections by a single projection. Let f^rrg contain the base vectors of the reference frame R whose relative orientation to the B frame is given through ½C0Š: f^rrg ¼ ½C0Šf^bbg ð3:19Þ The basis vectors f ^nng in the N frame can be projected directly into the R frame through f^rrg ¼ ½C0Š½CŠf ^nng ¼ ½C00Šf ^nng ð3:20Þ where the direction cosine matrix ½C00Š ¼ ½C0Š½CŠ projects vectors in the N frame to vectors in the R frame. The direct transformation matrix from the first to the last cascading reference frame is clearly found by successive matrix-multiplications of each relative transformation matrix in reverse order as just shown. This property ½C00Š ¼ ½C0Š½CŠ for composition of successive rota- tions is very important. When rotational coordinates are introduced to parame- terize the ½CŠ matrix, the corresponding ‘‘composition’’ relationship among the three sets of coordinates is also of fundamental importance. The direction cosine matrix is the most fundamental, but highly redundant, method of describing a relative orientation. As was mentioned earlier, the mini- mum number of parameters required to describe a reference frame orientation is three. The direction cosine matrix has nine entries. The six extra parameters in the matrix are made redundant through the orthogonality condition ½CŠ½CŠT ¼ ½I33Š. This is why in practice the elements of the direction cosine matrix are rarely used as coordinates to keep track of an orientation; instead less redundant attitude parameters are used. The biggest asset of the direction cosine matrix is the ability to easily transform vectors from one reference frame to another. Example 3.1 Let the two reference frames B and F be defined relative to the inertial reference frame N by the orthonormal unit base vectors ^bb1 ¼ 0; 1; 0  T ^bb2 ¼ 1; 0; 0  T ^bb3 ¼ 0; 0; 1  T ^ff1 ¼ 1 2 ; 3 p 2 ; 0  T ^ff2 ¼ 0; 0; 1  T ^ff3 ¼ 3 p 2 ;  1 2 ; 0  T RIGID BODY KINEMATICS 83D w U W M S m DO

where the ^bbi and ^ffi vector components are written in the inertial N frame. Let us use the following notation to label the various direction cosine matrices. The matrix ½BNŠ maps vectors written in the N frame into vectors written in the B frame. Analogously, the matrix ½FBŠ maps vectors in the B frame into F frame vectors and so on. To find the entries of the various relative rotation matrices, note the following useful identity: ½FBŠij ¼ cos aij ¼ ^ffi  ^bbj Given the base vectors of each frame, it is not necessary to find the angles between each set of vectors to find the appropriate direction cosine matrix. Because all base vectors have unit length, the inner product of the correspond- ing vectors will provide the needed direction cosines. The rotation matrices ½BNŠij ¼ ^bbi  ^nnj, ½FNŠij ¼ ^ffi  ^nnj, and ½FBŠij ¼ ^ffi  ^bbj are ½BN Š ¼ 0 1 0 1 0 0 0 0 1 2 6 4 3 7 5 ½FN Š ¼ 1 2 3 p 2 0 0 0 1 3 p 2  1 2 0 2 6 6 6 6 6 4 3 7 7 7 7 7 5 ½FBŠ ¼ 3 p 2 1 2 0 0 0 1  1 2 3 p 2 0 2 6 6 6 6 6 4 3 7 7 7 7 7 5 Instead of calculating the rotation matrix ½FBŠ from dot products of the respec- tive base vectors, it could also be calculated using Eq. (3.20): ½FBŠ ¼ ½FNŠ½BN ŠT ¼ 3 p 2 1 2 0 0 0 1  1 2 3 p 2 0 2 6 6 6 6 4 3 7 7 7 7 53 To find the kinematic differential equation in terms of the direction cosine matrix ½CŠ, let us write the instantaneous angular velocity vector v of the B frame relative to the N frame in B frame orthogonal components asv ¼ o1 ^bb1 þ o2 ^bb2 þ o3 ^bb3 ð3:21Þ Let N d=dtf^bbg be the derivative of the B frame base vectors taken in the N frame. Using the transport theorem we find 8 N d dt f^bbig ¼ Bd dt f^bbig þ v  f^bbig ð3:22Þ 84 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

Because the B frame base vectors are fixed within their frame, the expression Bd=dtf^bbg is zero. After introducing the skew-symmetric tilde matrix operator ½~xxŠ ¼ 0 x3 x2 x3 0 x1 x2 x1 0 2 4 3 5 ð3:23Þ Eq. (3.22) leads to the vectrix equation N d dt f^bbg ¼ ½ ~ vvŠf^bbg ð3:24Þ Taking the time derivative of the right-hand side of Eq. (3.5), we find N d dt ½CŠf ^nngð Þ ¼ d dt ½CŠð Þf ^nng þ ½CŠ N d dt f ^nnð gÞ ¼ ½ _CCŠf ^nng ð3:25Þ where the shorthand notation d=dtð½CŠÞ ¼ ½ _CCŠ is used. Using Eq. (3.5), Eqs. (3.24) and (3.25) are combined to ½ _CCŠ þ ½ ~ vvŠ½CŠ  f ^nng ¼ 0 ð3:26Þ Because Eq. (3.26) must hold for any set of f ^nng, the kinematic differential equation satisfied by the direction cosine matrix ½CŠ is found to be1;9 ½ _CCŠ ¼ ½ ~ vvŠ½CŠ ð3:27Þ Using explicit frame labeling, Eq. (3.27) is written as ½ _BNBN Š ¼ ½ ~xxB=N Š½BN Š ð3:28Þ Note that the 3  3 matrix ½ ~xxB=N Š in Eq. (3.28) is obtained by expressing the xB=N vector in B frame components. It can easily be verified that Eq. (3.9) is indeed an exact solution of the preceding differential equation. Take the derivative of ½CŠ½CŠT d dt ½CŠ½CŠT   ¼ ½ _CCŠ½CŠT þ ½CŠ½ _CCŠT ð3:29Þ and then substitute Eq. (3.27) to obtain d dt ½CŠ½CŠT   ¼ ½ ~ vvŠ½CŠ½CŠT  ½CŠ½CŠT ½ ~ vvŠT ð3:30Þ RIGID BODY KINEMATICS 85D w U W M S m DO

Making use of the orthogonality of ½CŠ and since ½ ~ vvŠ ¼ ½ ~ vvŠT is skew- symmetric, this simplifies to d dt ½CŠ½CŠT   ¼ ½ ~ vvŠ þ ½ ~ vvŠ ¼ 0 ð3:31Þ Because ½CŠ½CŠT is a constant solution of the differential equation in Eq. (3.27), and Eq. (3.9) is satisfied initially, the solution of Eq. (3.27) will theore- tically satisfy the orthogonality condition for all time. In practice, numerical solutions of Eq. (3.27) will slowly accumulate arithmetic errors so that the orthogonality condition ½CŠ½CŠT  ½I33Š ¼ 0 is slightly in error. There are several ways to resolve this minor difficulty. Given an arbitrary time history of vðtÞ, Eq. (3.27) represents a rigorously linear differential equation that can be integrated to yield the instantaneous direction cosine matrix ½CŠ. A major advantage of the kinematic differential equation for ½CŠ is that it is linear and universally applicable. There are no geometric singularities present in the attitude description or its kinematic differ- ential equations. However, this advantage comes at the cost of having a highly redundant formulation. Several other attitude parameters will be presented in the following sections that include a minimal number (3) of attitude para- meters. However, all minimal sets of attitude coordinates have kinematic differ- ential equations that contain some degree of nonlinearity and also embody geometric and=or mathematical singularities. Only the once-redundant Euler parameters (quaternions) will be found to retain a singularity-free description and possess linear kinematic differential equations analogous to the direction cosine matrix. 3.3 Euler Angles The most commonly used sets of attitude parameters are the Euler angles. They describe the attitude of a reference frame B relative to the frame N through three successive rotation angles ðy1; y2; y3Þ about the sequentially displaced body-fixed axes f^bbg. Note that the order of the axes about which the reference frame is rotated is important here. Performing three successive rota- tions about the third, second, and first body axis, labeled (3-2-1) for short, does not yield the same orientation as if instead the rotation order is (1-2-3). Note that these sequential rotations provide an instantaneous geometrical recipe for N . Clearly, for B undergoing a general motion, the yiðtÞ are time varying in a general manner. Aircraft and spacecraft orientations are commonly described through the Euler angles yaw, pitch, and roll (c; y; f) as shown in Fig. 3.2. They are usually measured relative to axes associated with a nominal flight path. The position of f^bbg relative to f ^nng is described by a sequence of three rigid rota- tions about prescribed body-fixed axes. While the conceptual description is a sequence of rotations, we can consider the instantaneous values of these three angles and thereby establish a means for describing general, nonsequential rotations. The popularity of Euler angles stems from the fact that the relative attitude is easy to visualize for small angles. To transform components of a 86 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

vector in the N frame into the B frame through a sequence of Euler angle rotations, the reference axes are first rotated about the ^bb3 axis by the yaw angle c, then about the ^bb2 axis by the pitch angle y, and finally about the ^bb1 axis by the roll angle f, as is shown in Fig. 3.3. Thus the standard yaw-pitch- roll (c; y; f) angles are the (3-2-1) set of Euler angles.11 Another very popular set of Euler angles is the (3-1-3) set of Euler angles. These angles are commonly used by astronomers to define the orientation of orbit planes of the planets relative to the Earth’s orbit plane.1 While the (3-2-1) Fig. 3.2 Yaw, pitch, and roll Euler angles. Fig. 3.3 Successive yaw, pitch, and roll rotations. RIGID BODY KINEMATICS 87D w U W M S m DO

Euler angles are considered an asymmetric set, the (3-1-3) Euler angles are a symmetric set because two rotations about the third body axis are performed. Instead of being called yaw, pitch, and roll angles, the (3-1-3) Euler angles are called longitude of the ascending node O, inclination i, and argument of the perihelion o and are illustrated in Fig. 3.4.1;11 The direction cosine matrix introduced in Section 3.2 can be parameterized in terms of the Euler angles. Since each Euler angle defines a successive rota- tion about the ith body axis, let the three single-axis rotation matrices ½Miðyފ be defined as ½M1ðyފ ¼ 1 0 0 0 cos y sin y 0  sin y cos y 2 6 4 3 7 5 ð3:32aÞ ½M2ðyފ ¼ cos y 0  sin y 0 1 0 sin y 0 cos y 2 6 4 3 7 5 ð3:32bÞ ½M3ðyފ ¼ cos y sin y 0  sin y cos y 0 0 0 1 2 6 4 3 7 5 ð3:32cÞ Let the ða; b; gÞ Euler angle sequence be ðy1; y2; y3Þ. Using Eq. (3.20) to combine successive rotations, the direction cosine matrix in terms of the ða; b; gÞ Euler angles is written as1 ½Cðy1; y2; y3ފ ¼ ½Mgðy3ފ½Mbðy2ފ½Maðy1ފ ð3:33Þ Fig. 3.4 (3-1-3) Euler angle illustration. 88 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

In particular, the direction cosine matrix in terms of the (3-2-1) Euler angles ðy1; y2; y3Þ ¼ ðc; y; fÞ is10 ½CŠ ¼ cy2cy1 cy2sy1 sy2 sy3sy2cy1  cy3sy1 sy3sy2sy1 þ cy3cy1 sy3cy2 cy3sy2cy1 þ sy3sy1 cy3sy2sy1  sy3cy1 cy3cy2 2 6 6 4 3 7 7 5 ð3:34Þ where the shorthand notation cx ¼ cos x and sx ¼ sin x is used. The inverse transformations from the direction cosine matrix ½CŠ to the ðc; y; fÞ angles are c ¼ y1 ¼ tan 1 C12 C11   ð3:35aÞ y ¼ y2 ¼  sin 1ðC13Þ ð3:35bÞ f ¼ y3 ¼ tan 1 C23 C33   ð3:35cÞ In terms of the (3-1-3) Euler angles ðy1; y2; y3Þ ¼ ðO; i; oÞ, the direction cosine matrix ½CŠ is written as1 ½CŠ ¼ cy3cy1  sy3cy2sy1 cy3sy1 þ sy3cy2cy1 sy3sy2 sy3cy1  cy3cy2sy1 sy3sy1 þ cy3cy2cy1 cy3sy2 sy2sy1  sy2cy1 cy2 2 6 6 4 3 7 7 5 ð3:36Þ The inverse transformations from the direction cosine matrix ½CŠ to the (3-1-3) Euler angles ðO; i; oÞ are O ¼ y1 ¼ tan 1 C31 C32   ð3:37aÞ i ¼ y2 ¼ cos 1ðC33Þ ð3:37bÞ o ¼ y3 ¼ tan 1 C13 C23   ð3:37cÞ The complete set of 12 transformations between the various Euler angle sets and the direction cosine matrix can be found in the Appendix B. We empha- size that while Eqs. (3.33–3.37) are easily established by sequential angular displacements, we consider the inverse situation; given a generally varying ½CŠ matrix, we can consider equations such as Eqs. (3.33–3.37) to hold at any instant in the motion, and thus fcðtÞ; yðtÞ; fðtÞg or fOðtÞ; iðtÞ; oðtÞg can be considered as candidate coordinates for general rotational motion. Note that each of the 12 possible sets of Euler angles has a geometric singularity where two angles are not uniquely defined. For the (3-2-1) Euler angles, pitching up or down 90 deg results in a geometric singularity. If the RIGID BODY KINEMATICS 89D w U W M S m DO

pitch angle is 90 deg, then it does not matter if c ¼ 0 and f ¼ 10 deg or c ¼ 10 and f ¼ 0 deg. Only the sum c þ f is unique in this case. For the (3- 1-3) Euler angles, the geometric singularity occurs for an inclination angle of 0 or 180 deg. This geometric singularity also manifests itself in a mathematical singularity of the corresponding Euler angle kinematic differential equation. Let u ¼ fy1; y2; y3g and f ¼ ff1; f2; f3g be two Euler angle vectors with identical rotation sequences. Often it is necessary to find the attitude that corre- sponds to performing two successive rotations, i.e., ‘‘adding’’ the two rotations. If a rigid body first performs the rotation u and then the rotation f , then the final attitude is expressed relative to the original attitude through the vectorw ¼ fj1; j2; j3g defined through ½FNð wފ ¼ ½FBð f ފ½BN ð u ފ ð3:38Þ Equation (3.38) could be used to solve for w in terms of the vector compo- nents of f and u . This process is very tedious and typically does not provide any simple, compact final expressions. However, for the case in which u andf are vectors of symmetric Euler angles, then it is possible to obtain relatively compact transformations from the first two vectors into the overall vector using spherical geometry relationships.4;12 A sample spherical triangle is shown in Fig. 3.5. The following two spheri- cal triangle laws are the only two required in deriving the symmetrical Euler angle successive rotation property. The spherical law of sines states that sin A sin a ¼ sin B sin b ¼ sin C sin c ð3:39Þ and the spherical law of cosines states that cos A ¼  cos B cos C þ sin B sin C cos a ð3:40aÞ cos B ¼  cos A cos C þ sin A sin C cos b ð3:40bÞ cos C ¼  cos A cos B þ sin A sin B cos c ð3:40cÞ Figure 3.6a illustrates the orientation of the first body axis as it is first rotated from N to B with the (3-1-3) Euler angle vector u and then from B to F with the (3-1-3) vector f. The (3-1-3) Euler angle description of the direct rotation Fig. 3.5 Spherical triangle labels. 90 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

from N to F is clearly given by the angles j1, j2, and j3. To obtain direct transformations from u and f to w, the bold spherical triangle in Fig. 3.6a is used. The spherical arc lengths and angles of this triangle are labeled in Fig. 3.6b. Using the spherical law of cosines, we find that cosðp  j2Þ ¼  cos y2 cos f2 þ sin y2 sin f2 cosðy3 þ f1Þ ð3:41Þ This is trivially solved for the angle j2 as j2 ¼ cos 1 cos y2 cos f2  sin y2 sin f2 cosðy3 þ f1Þ   ð3:42Þ Using the spherical laws of sines, we are able to find the following expressions for j1 and j3: sinðj1  y1Þ ¼ sin f2 sin j2 sinðy3 þ f1Þ ð3:43Þ sinðj3  f3Þ ¼ sin y2 sin j2 sinðy3 þ f1Þ ð3:44Þ Fig. 3.6 Illustration of successive (3-1-3) Euler angle rotations: a) successive (3-1-3) Euler angles and b) spherical triangle. RIGID BODY KINEMATICS 91D w U W M S m DO

To avoid quadrant problems, we prefer to find expressions of j1 and j3 that involve the tanðÞ function instead of the sinðÞ function. To accomplish this, using the spherical law of cosines, we find the following two relationships: cosðj1  y1Þ ¼ cos f2  cos y2 cos j2 sin y2 sin j2 ð3:45Þ cosðj3  f3Þ ¼ cos y2  cos f2 cos j2 sin f2 sin j2 ð3:46Þ Combining Eqs. (3.43–3.46), we are able to solve for j1 and j3 using the inverse tan function: j1 ¼ y1 þ tan 1 sin y2 sin f2 sinðy3 þ f1Þ cos f2  cos y2 cos j2   ð3:47Þ j3 ¼ f3 þ tan 1 sin y2 sin f2 sinðy3 þ f1Þ cos y2  cos f2 cos j2   ð3:48Þ Using Eqs. (3.42), (3.47), and (3.48) to solve for f instead of back-solving w out of the direction cosine matrix in Eq. (3.38) is numerically more efficient. Although the Euler angle successive or composite rotation was developed for the (3-1-3) special case, the transformations in Eqs. (3.42), (3.47), and (3.48) actually hold for any symmetric rotation sequence.4;12 Asymmetric sets, however, will have to be composited using the corresponding direction cosine matrices. On occasion it is required to find the relative attitude vector between two reference frames. For example, given the symmetric Euler angle vectors u andw, find the corresponding vector f that relates B to F . Using the same spheri- cal triangle in Fig. 3.6b, we find the following closed form expressions for f : f1 ¼ y3 þ tan 1 sin y2 sin j2 sinðj1  y1Þ cos y2 cos f2  cos j2   ð3:49Þ f2 ¼ cos 1 cos y2 cos j2 þ sin y2 sin j2 cosðj1  y1Þ   ð3:50Þ f3 ¼ j3  tan 1 sin y2 sin j2 sinðj1  y1Þ cos y2  cos f2 cos j2   ð3:51Þ Similar expressions can be found to express u in terms of f and w. Example 3.2 Let the orientations of two spacecraft B and F relative to an inertial frame N be given through the asymmetric (3-2-1) Euler angles u B ¼ ð30; 45; 60ÞT andu F ¼ ð10; 25; 15ÞT degrees. What is the relative orientation of spacecraft B relative to F in terms of (3-2-1) Euler angles? 92 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

The orientation matrices ½BN Š and ½FN Š are found using Eq. (3.34): ½BN Š ¼ 0:612372 0:353553 0:707107 0:78033 0:126826 0:612372 0:126826 0:926777 0:353553 2 6 4 3 7 5 ½FN Š ¼ 0:892539 0:157379 0:422618 0:275451 0:932257 0:234570 0:357073 0:325773 0:875426 2 6 4 3 7 5 The direction cosine matrix ½BFŠ that describes the attitude of B relative to F is computed by using Eq. (3.20): ½BFŠ ¼ ½BNŠ½FN ŠT ¼ 0:303372 0:0049418 0:952859 0:935315 0:1895340 0:298769 0:182075 0:9818620 0:052877 2 4 3 5 Using the transformations in Eq. (3.34), the relative (3-2-1) Euler angles are c ¼ tan 1 0:0049418 0:303372   ¼ 0:933242 deg y ¼  sin 1 0:952859ð Þ ¼ 72:3373 deg f ¼ tan 1 0:298769 0:052877   ¼ 79:9635 deg Because f is much larger than c and y, the attitude of B could be described qualitatively to differ from F by a 57:6-deg roll. This result was not immedi- ately obvious studying the original Euler angle vectors u B and u F . Let the vector v define the instantaneous rotational velocity of the B frame relative to the N frame. To avoid having to integrate the direction cosine matrix directly given an v time history, the Euler angle kinematic differential equations are needed. The (3-2-1) Euler kinematic differential equation is derived in the following. The methodology can be used for any set of Euler angles. The vector v is written in body frame components asv ¼ o1 ^bb1 þ o2 ^bb2 þ o3 ^bb3 ð3:52Þ From Fig. 3.3 it is evident that the B frame rotation can also be written in terms of the Euler angle rates ð _cc; _yy; _ffÞ asv ¼ _cc ^nn3 þ _yy^bb0 2 þ _ff^bb1 ð3:53Þ RIGID BODY KINEMATICS 93D w U W M S m DO

The unit vector ^bb0 2 is the direction of the body-fixed axis ^bb2 before performing a roll f about ^bb1, as is shown in Fig. 3.3. It can be written in terms of f^bbg as ^bb0 2 ¼ cos f^bb2  sin f^bb3 ð3:54Þ The direction cosine matrix in terms of the (3-2-1) Euler angles in Eq. (3.34) is used to express ^nn3 in terms of f^bbg: ^nn3 ¼  sin y^bb1 þ sin f cos y^bb2 þ cos f cos y^bb3 ð3:55Þ After substituting Eqs. (3.54) and (3.55) into Eq. (3.53) and then comparing terms with Eq. (3.52), the following kinematic equation is found: B o1 o2 o3 0 B B @ 1 C C A ¼  sin y 0 1 sin f cos y cos f 0 cos f cos y  sin f 0 2 6 6 4 3 7 7 5 _cc _yy _ff 0 B B @ 1 C C A ð3:56Þ The kinematic differential equation of the (3-2-1) Euler angles is the inverse of Eq. (3.56): _cc _yy _ff 0 B B @ 1 C C A ¼ 1 cos y 0 sin f cos f 0 cos f cos y  sin f cos y cos y sin f sin y cos f sin y 2 6 6 4 3 7 7 5 B o1 o2 o3 0 B B @ 1 C C A ¼ ½Bðc; y; fފ B v ð3:57Þ Similarly, the kinematic differential equations for the (3-1-3) Euler angles are found be Bx ¼ sin y3 sin y2 cos y3 0 cos y3 sin y2  sin y3 0 cos y2 0 1 2 6 6 4 3 7 7 5 _yy1 _yy2 _yy3 0 B B @ 1 C C A ð3:58Þ with the inverse relationship _yy1 _yy2 _yy3 0 B B @ 1 C C A ¼ 1 sin y2 sin y3 cos y3 0 cos y3 sin y2  sin y3 sin y2 0  sin y3 cos y2  cos y3 cos y2 sin y2 2 6 6 4 3 7 7 5 Bx ¼ ½Bð u ފ Bx ð3:59Þ The complete set of 12 transformations between the various Euler angle rates and the body angular velocity vector can be found in Appendix B. Note 94 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

that the Euler angle kinematic differential equations encounter a singularity either at y2 ¼ 90 deg for the (3-2-1) set or at y2 ¼ 0 or 180 deg for the (3-1- 3) set. It turns out that all Euler angles sets encounter a singularity at specific values of the second rotation angle y2 only. The first and third rotation angles y1 and y3 never lead to a singularity. In all cases, it can be verified that the singularity occurs for those y2 values that result in y1 and y3 being measured in the same plane. If the Euler angle set is symmetric, then the singular orien- tation is at y2 ¼ 0 or 180 deg. If the Euler angle set is asymmetric, then the singular orientation is y2 ¼ 90 deg. Symmetric sets as the (3-1-3) Euler angles would not be convenient to describe small departure rotations of f^bbg from the f ^nng axes since for small angles one would always operate very close to the singular attitude at y2 ¼ 0. The Euler angles provide a compact, three-parameter attitude description whose coordinates are easy to visualize. One major drawback of these angles is that a rigid body or reference frame is never further than a 90-deg rotation away from a singular orientation. Therefore, their use in describing large, and in particular arbitrary, rotations is limited. Also, their kinematic differential equations are fairly nonlinear, containing computationally intensive trigono- metric functions. The linearized Euler angle kinematic differential equations are valid only for a relatively small domain of rotations. 3.4 Principal Rotation Vector The following theorem has been very fundamental in the development of several types of attitude coordinates and is generally referenced to Euler.13;14 Theorem 3.1 (Euler’s Principal Rotation): A rigid body or coordinate reference frame can be brought from an arbitrary initial orientation to an arbi- trary final orientation by a single rigid rotation through a principal angle F about the principal axis ^ee; the principal axis is a judicious axis fixed in both the initial and final orientation. This theorem can be visualized using Fig. 3.7. Let the principal axis unit vector ^ee be written in B and N frame components as ^ee ¼ eb1 ^bb1 þ eb2 ^bb2 þ eb3 ^bb3 ð3:60aÞ ^ee ¼ en1 ^nn1 þ en2 ^nn2 þ en3 ^nn3 ð3:60bÞ Implicit in the theorem we see that ^ee will have the same vector components in the B as in the N reference frame; i.e., ebi ¼ eni ¼ ei. Equation (3.5) shows that e1 e2 e3 0 @ 1 A ¼ ½CŠ e1 e2 e3 0 @ 1 A ð3:61Þ must be true. Therefore, the principal axis unit vector ^ee is the unit eigenvector of ½CŠ corresponding to the eigenvalue þ1. Thus the proof of the principal rotation theorem reduces to proving the ½CŠ has an eigenvalue of þ1. This RIGID BODY KINEMATICS 95D w U W M S m DO

proof is given in Goldstein in Ref. 7. The eigenvalue þ1 is unique and the corresponding eigenvector is unique to within a sign of F and ^ee, except for the case of a zero rotation. In this case ½CŠ ¼ ½I33Š and F would be zero, but there would be an infinity of unit axes ^ee such that ^ee ¼ ½I33Š^ee. For the general case, the lack of sign uniqueness of F and ^ee will not cause any practical problems. The sets ð^ee; FÞ and ð^ee; FÞ both describe the same orientation. The principal rotation angle F is also not unique. Figure 3.7 shows the direction of the angle F labeled such that the shortest rotation about ^ee will be performed to move from N to B. However, this is not necessary. If so desired, one can also rotate in the opposite direction by the angle F0 and achieve the exact same orientation as shown in Fig. 3.8. The difference between F and F0 will always be 360 deg. In most cases the magnitude of F is simply chosen to be less than or equal to 180 deg. Fig. 3.7 Illustration of Euler’s principal rotation theorem. Fig. 3.8 Illustration of both principal rotation angles. 96 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

To find the direction cosine matrix ½CŠ in terms of the principal rotation components ^ee and F, the fact is used that each reference frame base vector ^nni is related to ^bbi through a single axis rotation about ^ee. Let the unit principal axis vector be written as ^ee ¼ e1 ^nn1 þ e2 ^nn2 þ e3 ^nn3 ð3:62Þ and let xi be the angle between ^nni and ^ee as shown in Fig. 3.9. Let’s note the following useful identity ^ee  ^nni ¼ cos xi ¼ ei ð3:63Þ Studying Fig. 3.9, the base vector ^bbi can be written as ^bbi ¼ cos xi ^ee þ sin xi ^uu0 ¼ ei ^ee þ sin xi ^uu0 ð3:64Þ The unit vector ^uu0 is given by ^uu0 ¼ cos F ^uu þ sin F^vv ð3:65Þ It follows from the geometry of the single axis rotation that ^vv ¼ ^ee  ^nni j^ee  ^nnij ¼ 1 sin xi ^ee  ^nni   ð3:66Þ ^uu ¼ ^vv  ^ee ¼ 1 sin xi ^ee  ^nni    ^ee ð3:67Þ The expression for ^uu can be further reduced by making use of the triple cross product identity a  ðb  cÞ ¼ ða  cÞb  ða  bÞc ð3:68Þ Fig. 3.9 Mapping ^nni into ^bbi base vectors. RIGID BODY KINEMATICS 97D w U W M S m DO

to the simpler form ^uu ¼ 1 sin xi ^nni  ei ^ee   ð3:69Þ After substituting Eqs. (3.65), (3.66), and (3.69) into Eq. (3.64), each base vector ^bbi is expressed in terms of reference frame N base vectors: ^bbi ¼ cos F ^nni þ 1  cos Fð Þ^ee^eeT ^nni þ sin F ^ee  ^nni   ð3:70Þ where ^ee^eeT is the outer vector dot product of the vector ^ee. Making use of the definition of ½~eeŠ in Eq. (3.23), the set of base vectors f^bbg can be expressed as f^bbg ¼ cos F½I33Š þ 1  cos Fð Þ^ee^eeT  sin F½~eeŠ  f ^nng ð3:71Þ Using the relationship f^bbg ¼ ½CŠf ^nng, the direction cosine matrix can be directly extracted from Eq. (3.70) to be ½CŠ ¼ e2 1S þ cF e1e2S þ e3sF e1e3S  e2sF e2e1S  e3sF e2 2S þ cF e2e3S þ e1sF e3e1S þ e2sF e3e2S  e1sF e2 3S þ cF 2 6 6 4 3 7 7 5 ð3:72Þ where S ¼ 1  cF. Again the shorthand notation cF ¼ cos F and sF ¼ sin F was used here. The direction cosine matrix ½CŠ depends on four scalar quanti- ties e1; e2; e3, and F. However, only three degrees of freedom are present since the vector components ei must abide by the unit constraint P3 i e2 i ¼ 1. By inspection of Eq. (3.72), the inverse transformation from the direction cosine matrix ½CŠ to the principal rotation elements is found to be cos F ¼ 1 2 C11 þ C22 þ C33  1   ð3:73Þ ^ee ¼ e1 e2 e3 0 B @ 1 C A ¼ 1 2 sin F C23  C32 C31  C13 C12  C21 0 B @ 1 C A ð3:74Þ Note that Eq. (3.73) will yield a principal rotation angle within the range 0  F  180 deg. The direction of ^ee in Eq. (3.74) will be such that the princi- pal rotation parameterizing ½CŠ will be through a positive angle F about ^ee. To find the second possible principal rotation angle F0, one subtracts 360 deg from F: F0 ¼ F  2p ð3:75Þ The angle F0 is equally valid as F and yields the same principal rotation axis ^ee. The only difference is that a longer rotation (for jFj  p) is being performed 98 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

in the opposite direction. As with the sequential Euler angle rotations, the instantaneous principal rotation parameters fe1ðtÞ; e2ðtÞ; e3ðtÞ; FðtÞg can be con- sidered coordinates associated with the instantaneous direction cosine matrix ½Cðtފ, and obviously does not restrict the body to actually execute the principal rotation. Example 3.3 Let the B frame attitude relative to the N frame be given by the (3-2-1) Euler angles ð10; 25; 15Þ deg. Find the corresponding principal rotation axis and angles. Using Eq. (3.33) the direction cosine matrix ½BN Š is ½BN Š ¼ 0:892539 0:157379 0:422618 0:275451 0:932257 0:234570 0:357073 0:325773 0:875426 2 4 3 5 The first principal rotation angle F is found through Eq. (3.73): F ¼ cos 1 1 2 ð0:892539 þ 0:932257 þ 0:875426  1Þ   ¼ 31:7762 deg The corresponding principal rotation axis is given though Eq. (3.74): ^ee ¼ 1 2 sinð31:7762 degÞ 0:23457  0:325773 0:357073  ð0:422618Þ 0:157379  ð0:275451Þ 0 @ 1 A ¼ 0:532035 0:740302 0:410964 0 @ 1 A The second principal rotation angle F0 is calculated using Eq. (3.75): F0 ¼ 31:7762 deg  360 deg ¼ 328:2238 deg Either principal rotation element sets ð^ee; FÞ or ð^ee; F0Þ describes the identical attitude as the original (3-2-1) Euler angles. Many important attitude parameters that are derived from Euler’s principal rotation axis ^ee and angle F can be written in the general form p ¼ f ðFÞ^ee ð3:76Þ where f ðFÞ could be any scalar function of F. All these attitude coordinate vectors have the same direction and differ only by their magnitude j pj ¼ f ðFÞ. The principal rotation vector g is simply defined as g ¼ F^ee ð3:77Þ Therefore, the magnitude of g is f ðFÞ ¼ F. This attitude vector has a very interesting relationship to the direction cosine matrix that can be verified to also hold for higher dimensional orthogonal projections, as shown in Ref. 15. RIGID BODY KINEMATICS 99D w U W M S m DO

To gain more insight, consider the special case of a pure single axis rotation about a fixed ^ee with the rotation angle being F. The angular velocity vector for this case is v ¼ _FF^ee ð3:78Þ or in matrix form: ½ ~ vvŠ ¼ _FF½~eeŠ ð3:79Þ Substituting Eq. (3.78) into Eq. (3.27) leads to the following development: d½CŠ dt ¼  dF dt ½~eeŠ½CŠ d½CŠ dF ¼ ½~eeŠ½CŠ ð3:80Þ ½CŠ ¼ e F½~eeŠ The last step holds true for ½~eeŠ being a constant matrix for a rotation about a fixed axis. Because of Euler’s principal rotation theorem, however, any arbitrary rotation can be instantaneously described by the equivalent single axis rotation. Euler’s theorem means that Eq. (3.80) holds at any instant for an arbitrary time varying direction cosine matrix ½CŠ. Note for time varying ½CŠ, however, that ^ee and F must be considered time varying. Using Eq. (3.77), the rotation matrix ½CŠ is related to g through ½CŠ ¼ e ½ ~ ggŠ ¼ P1 n¼0 1 n! ½ ~ ggŠð Þn ð3:81Þ It turns out that this mapping also holds for higher-dimensional proper ortho- gonal matrices ½CŠ. For the case of three-dimensional rotations, the infinite power series in Eq. (3.81) can more conveniently be written as a finite, closed- form solution4;15: ½CŠ ¼ e F½~eeŠ ¼ ½I33Š cos F  sin F½~eeŠ þ ð1  cos FÞ^ee^eeT ð3:82Þ To find the inverse transformation from ½CŠ to g, the inverse matrix logarithm is taken: ½ ~ ggŠ ¼  ln½CŠ ¼ P1 n¼0 1 n ð1  ½CŠÞn ð3:83Þ This inverse mapping is defined everywhere except for F ¼ 0 and F ¼ 180 deg rotations. For these rotations, the non-uniqueness of the g vector leads to mathematical difficulties. Otherwise, a vector g is reliably returned corresponding to a principal rotation of less than or equal to 180 deg. 100 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

Example 3.4 In Example 3.3 it was shown that the direction cosine matrix ½BN Š ¼ 0:892539 0:157379 0:422618 0:275451 0:932257 0:234570 0:357073 0:325773 0:875426 2 4 3 5 represents the equivalent orientation as the principal rotation vector g ¼ 0:55460 rad 0:532035 0:740302 0:410964 0 @ 1 A ¼ 0:295067 0:410571 0:227921 0 @ 1 A To verify the mapping in Eq. (3.81), let’s write ½ ~ ggŠ using the definition of tilde matrix operator in Eq. (3.23): ½ ~ ggŠ ¼ 0 0:227921 0:410571 0:227921 0 0:295067 0:410571 0:295067 0 2 4 3 5 Using software packages such as Mathematica1 or MATLAB1, the matrix exponential mapping in Eq. (3.81) can be solved numerically for the corre- sponding direction cosine matrix ½BN Š: ½BNŠ ¼ e ½ ~ ggŠ ¼ 0:892539 0:157379 0:422618 0:275451 0:932257 0:234570 0:357073 0:325773 0:875426 2 4 3 53 Let ðF1; ^ee1Þ be the principal rotation elements that relate the B frame rela- tive to the N frame, while ðF2; ^ee2Þ orients the F frame relative to the B frame. The F frame is related directly to the N frame by the elements ðF; ^eeÞ through the relationship ½FN ðF; ^eeފ ¼ ½FBðF2; ^ee2ފ½BN ðF1; ^ee1ފ ð3:84Þ Instead of solving for the overall principal rotation elements through the corre- sponding direction cosine matrix, it is possible to express ðF; ^eeÞ directly in terms of ðF1; ^ee1Þ and ðF2; ^ee2Þ through4 F ¼ 2 cos 1 cos F1 2 cos F2 2  sin F1 2 sin F2 2 ^ee1  ^ee2   ð3:85Þ ^ee ¼ cos F2 2 sin F1 2 ^ee1 þ cos F1 2 sin F2 2 ^ee2 þ sin F1 2 sin F2 2 ^ee1  ^ee2 sin F 2 ð3:86Þ RIGID BODY KINEMATICS 101D w U W M S m DO

This composite rotation property is easily derived from the Euler parameter composite rotation property shown in the next section. Given the two principal rotation element sets ðF1; ^ee1Þ and ðF; ^eeÞ, the relative orientation set ðF2; ^ee2Þ is expressed similarly through F2 ¼ 2 cos 1 cos F 2 cos F1 2 þ sin F 2 sin F1 2 ^ee  ^ee1   ð3:87Þ ^ee2 ¼ cos F1 2 sin F 2 ^ee  cos F 2 sin F1 2 ^ee1 þ sin F 2 sin F1 2 ^ee  ^ee1 sin F2 2 ð3:88Þ The kinematic differential equation of the principal rotation vector g is given by4;16 18 _ gg ¼ ½I33Š þ 1 2 ½ ~ ggŠ þ 1 F2 1  F 2 cot F 2    ½ ~ ggŠ2   Bx ð3:89Þ where F ¼ k gk. The inverse transformation of Eq. (3.89) is B x ¼ ½I33Š  1  cos F F2   ½ ~ ggŠ þ F  sin F F3   ½ ~ ggŠ2   _ gg ð3:90Þ As expected, the kinematic differential equation in Eq. (3.89) contains a 0=0 type mathematical singularity for zero rotations where F ¼ 0 deg. Therefore, the principal rotation vector is not well suited for use in small motion feedback control type applications where the reference state is the zero rotation. Further, the mathematical expression in Eq. (3.89) is rather complex, containing poly- nomial fractions of degrees up to three in addition to trigonometric functions. This makes g less attractive to describe large arbitrary rotations as compared to some other, closely related, attitude parameters that will be presented in the next few sections. Example 3.5 Given the prescribed body angular velocity vector v ¼ oðtÞ^ee for a single axis rotation, Eq. (3.89) yields the following kinematic differential equation for the principal rotation vector g ¼ F^ee: _ gg ¼ ½I33Š  F 2 ½ ~ ggŠ þ 1 F2 1  F 2 cot F 2    F2½ ~ ggŠ2   oðtÞ^ee 102 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

Noting that ½ ~ ggŠ^ee ¼ F½~eeŠ^ee ¼ 0, this is simplified to _ gg ¼ oðtÞ^ee Therefore, the general expression in Eq. (3.89) simplifies to the single axis result in Eq. (3.78). The principal rotation elements ^ee and F have had a fundamental influence on the derivation of many sets of attitude coordinates. All of the following atti- tude parameters will be directly derived from these principal rotation elements. 3.5 Euler Parameters Another popular set of attitude coordinates are the four Euler parameters (quaternions). They provide a redundant, nonsingular attitude description and are well suited to describe arbitrary, large rotations. The Euler parameter vectorb is defined in terms of the principal rotation elements as b0 ¼ cos F=2ð Þ ð3:91aÞ b1 ¼ e1 sin F=2ð Þ ð3:91bÞ b2 ¼ e2 sin F=2ð Þ ð3:91cÞ b3 ¼ e3 sin F=2ð Þ ð3:91dÞ It is evident since e2 1 þ e2 2 þ e2 3 ¼ 1, that the bi satisfy the holonomic constraint b2 0 þ b2 1 þ b2 2 þ b2 3 ¼ 1 ð3:92Þ Note that this constraint geometrically describes a four-dimensional unit sphere. Any rotation described through the Euler parameters has a trajectory on the surface of this constraint sphere. Given a certain attitude, there are actu- ally two sets of Euler parameters that will describe the same orientation. This is due to the non-uniqueness of the principal rotation elements themselves. Switching between the sets ð^ee; FÞ and ð^ee; FÞ will yield the same Euler parameter vector b. However, if the second principal rotation angle F0 is used, another Euler parameter vector b 0 is found. Using Eq. (3.75), one can show that b0 0 ¼ cos F0 2   ¼ cos F 2  p   ¼  cos F 2   ¼ b0 b0 i ¼ ei sin F0 2   ¼ ei sin F0 2  p   ¼ ei sin F 2   ¼ bi Therefore, the vector b 0 ¼  b describes the same orientation as the vector b. This results in the following interesting observation. Because any point on the unit constraint sphere surface represents a specific orientation, the anti-pole to that point represents the exact same orientation. The difference between the two attitude descriptions is that one specifies the orientation through the RIGID BODY KINEMATICS 103D w U W M S m DO

shortest single axis rotation, the other through the longest. From Eq. (3.91a) it is clear that in order to choose the Euler parameter vector corresponding to the shortest rotation (i.e., jFj  180 deg), the coordinate b0 must be chosen to be nonnegative. Using the trigonometric identities sin F ¼ 2 sinðF=2Þ cosðF=2Þ cos F ¼ 2 cos2ðF=2Þ  1 in Eq. (3.72), the direction cosine matrix can be written in terms of the Euler parameters as ½CŠ ¼ b2 0 þ b2 1  b2 2  b2 3 2ðb1b2 þ b0b3Þ 2ðb1b3  b0b2Þ 2ðb1b2  b0b3Þ b2 0  b2 1 þ b2 2  b2 3 2ðb2b3 þ b0b1Þ 2ðb1b3 þ b0b2Þ 2ðb2b3  b0b1Þ b2 0  b2 1  b2 2 þ b2 3 2 6 6 4 3 7 7 5 ð3:93Þ The fact that b and  b produce the same direction cosine matrix ½CŠ can be easily verified in Eq. (3.93). All Euler parameters appear in quadratic product pairs, thus changing the signs of all bi components has no effect on the result- ing ½CŠ matrix. It is evident that the most general angular motion of a reference frame generates two arcs on the four-dimensional unit sphere [the geodesic arcs generated by b ðtÞ and  b ðtÞ]. This elegant description is universally nonsingular and is unique to within the sign  b ðtÞ. The inverse transforma- tions from ½CŠ to the Euler parameters can be found through inspection of Eq. (3.93) to be b0 ¼  1 2 C11 þ C22 þ C33 þ 1 p ð3:94aÞ b1 ¼ C23  C32 4b0 ð3:94bÞ b2 ¼ C31  C13 4b0 ð3:94cÞ b3 ¼ C12  C21 4b0 ð3:94dÞ Note that the non-uniqueness of the Euler parameters is evident again in this inverse transformation. By keeping the þ sign in Eq. (3.94a), one restricts the corresponding principal rotation angle F to be less than or equal to 180 deg. From a practical point of view this non-uniqueness does not pose any difficul- ties. Initially one simply picks an initial condition on one Euler parameter trajectory and then remains with it either through solving an associated kine- matic differential developed next, or using elementary continuity logic. 104 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

Clearly Eq. (3.94) has a 0=0 type mathematical singularity whenever b0 ! 0. This corresponds to the b vector describing any 180 deg principal rotation. A computationally superior algorithm has been developed by Sheppard in Ref. 19. First the four b2 i terms are computed: b2 0 ¼ 1 4 ð1 þ trace ½CŠÞ ð3:95aÞ b2 1 ¼ 1 4 ð1 þ 2C11  trace ½CŠÞ ð3:95bÞ b2 2 ¼ 1 4 ð1 þ 2C22  trace ½CŠÞ ð3:95cÞ b2 3 ¼ 1 4 ð1 þ 2C33  trace ½CŠÞ ð3:95dÞ Then Sheppard takes the square root of the largest b2 i found in Eq. (3.95) where the sign of bi is arbitrarily chosen to be positive. The other bj are found by dividing the appropriate three of the following six in Eq. (3.96) by the chosen largest bi coordinate: b0b1 ¼ ðC23  C32Þ=4 ð3:96aÞ b0b2 ¼ ðC31  C13Þ=4 ð3:96bÞ b0b3 ¼ ðC12  C21Þ=4 ð3:96cÞ b2b3 ¼ ðC23 þ C32Þ=4 ð3:96dÞ b3b1 ¼ ðC31 þ C13Þ=4 ð3:96eÞ b1b2 ¼ ðC12 þ C21Þ=4 ð3:96f Þ To find the alternate set of Euler parameter, the sign of the chosen bi would simply be set negative. Example 3.6 Let’s use Sheppard’s method to find the Euler parameters of the direction cosine matrix ½CŠ. ½CŠ ¼ 0:892539 0:157379 0:422618 0:275451 0:932257 0:234570 0:357073 0:325773 0:875426 2 4 3 5 Using the expressions in Eq. (3.95), the absolute values of the four Euler para- meter are found: b2 0 ¼ 0:925055 b2 1 ¼ 0:021214 b2 2 ¼ 0:041073 b2 3 ¼ 0:012657 RIGID BODY KINEMATICS 105D w U W M S m DO

The b0 term is selected as the largest element and used in Eqs. (3.96a–3.96c) to find the Euler parameter vector: b ¼ ð0:961798; 0:14565; 0:202665; 0:112505ÞT The alternate Euler parameter vector would be found by simply reversing the sign of each element in b . Example 3.7 To transform between different attitude coordinate sets, it is always possible to map the attitude coordinates first to the rotation matrix and then extract the desired new attitude coordinate set from the direction cosine matrix. However, for particular attitude-to-attitude coordinate transformations, there exist more convenient and elegant direct formulas. One such set of transformations are the mappings Euler-angle-to-Euler-parameters sets. Let us first consider how to translate the asymmetric ð3-1-3Þ set ðO; i; oÞ into equivalent Euler parameters or elements of the quaternion ðb0; b1; b2; b3Þ. From Eq. (3.94a) we find that b0 ¼  1 2 traceð½CŠÞ þ 1 p Using the direction cosine matrix definition in terms of ð3; 2; 1Þ Euler angles in Eq. (3.36), note that traceð½CŠÞ þ 1 ¼ C11 þ C22 þ C33 þ 1 ¼ cocO  socisO  sosO þ cocicO þ ci þ 1 ¼ ð1 þ cocO  sosOÞð1 þ ciÞ ¼ ½1 þ cosðo þ Oފð1 þ cos iÞ ¼ 4 cos o þ O 2  2 cos i 2  2 Substituting this result into the preceding b0 expression, and choosing the ‘‘þ’’ sign, leads to direct 3-1-3 to Euler parameter b0 mapping: b0 ¼ cos i 2   cos o þ O 2   Using similar algebra, the remaining Euler parameters can be expressed as b1 ¼ sin i 2   cos o  O 2   b2 ¼ sin i 2   sin o  O 2   b3 ¼ cos i 2   sin o þ O 2   106 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

The other symmetric Euler angle mapping to Euler parameters can be obtained using a similar procedure. However, this direct trigonometric method does not lead to as elegant results for the asymmetric Euler angles. A complete list of all direct Euler-angle-to-Euler-parameter transformations is found in Ref. 1. A very important composite rotation property of the Euler parameters is the manner in which they allow two sequential rotations to be combined into one overall composite rotation. Let the Euler parameter vector b 0 describe the first,b 00 the second, and b the composite rotation. From Eq. (3.20) it is clear that ½FN ð b ފ ¼ ½FBð b 00ފ½BNð b 0ފ ð3:97Þ Using Eq. (3.93) in Eq. (3.97) and equating corresponding elements leads to the following elegant transformation that bilinearly combines b 0 and b 00 intob : b0 b1 b2 b3 0 B B B B @ 1 C C C C A ¼ b00 0 b00 1 b00 2 b00 3 b00 1 b00 0 b00 3 b00 2 b00 2 b00 3 b00 0 b00 1 b00 3 b00 2 b00 1 b00 0 2 6 6 6 6 4 3 7 7 7 7 5 b0 0 b0 1 b0 2 b0 3 0 B B B B @ 1 C C C C A ð3:98Þ By transmutation of Eq. (3.98), an alternate expression b ¼ ½Gð b 0ފ b 00 is found b0 b1 b2 b3 0 B B B B @ 1 C C C C A ¼ b0 0 b0 1 b0 2 b0 3 b0 1 b0 0 b0 3 b0 2 b0 2 b0 3 b0 0 b0 1 b0 3 b0 2 b0 1 b0 0 2 6 6 6 6 4 3 7 7 7 7 5 b00 0 b00 1 b00 2 b00 3 0 B B B B @ 1 C C C C A ð3:99Þ where the components of the matrix ½Gð b 0ފ are given in Eq. (3.99). Note the useful identity ½Gð b ފT b ¼ 1 0 0 0 0 B B @ 1 C C A ð3:100Þ By inspection, it is evident that the 4  4 matrices in Eqs. (3.98) and (3.99) are orthogonal. These transformations provide a simple, nonsingular, and bilinear method to combine two successive rotations described through Euler parameters. For other attitude parameters such as the Euler angles, this same composite transformation would yield a very complicated, transcendental expression. RIGID BODY KINEMATICS 107D w U W M S m DO

Example 3.8 Using Stanley’s method, the direction cosine matrices ½BN Š and ½FBŠ defined in Example 3.1 can be parameterized through the Euler parameter vectors b 0 and b 00 respectively as ½BN Š ) b 0 ¼ 0; 1 2 p ; 1 2 p ; 0  T ½FBŠ ) b 00 ¼ 1 2 3 p 2 þ 1 s ;  1 2 3 p 2 þ 1 s ;  2 p 4 2 þ 3 pp ; 2 p 4 2 þ 3 pp 0 @ 1 A T Note that the vector b 0 describes the attitude of the B frame relative to the N frame, while the vector b 00 describes the F frame attitude relative to the B frame. Eq. (3.98) can be used to combine the two successive attitude vectors into one vector b that directly describes the F frame orientation relative to the N frame: b ¼ 1 2 2 p ð 3 p ; 3 p ; 1; 1ÞT To verify that b does indeed parameterize the direction cosine matrix ½FN Š given in Example 3.1, it can be substituted back into Eq. (3.93) to yield ½FN Š ¼ 1 2 3 p 2 0 0 0 1 3 p 2  1 2 0 2 6 6 6 6 6 4 3 7 7 7 7 7 5 Example 3.9 The direct derivation of the kinematic transformation of symmetric Euler angles to Euler parameters used in Example 3.7 does not work very well with asymmetric Euler angles. Reference 1 presents an elegant general methodology to derive the Euler-angle-to-Euler-parameter transformation using the Euler parameter composite rotation property. Let us investigate the ð3-2-1Þ Euler angle ðy1; y2; y3Þ ¼ ðc; y; jÞ mapping to the equivalent Euler parameters ðb0; b1; b2; b4Þ. Note that the ð3-2-1Þ Euler angles define a sequential rotation sequence about the current 3, 2, and 1 body axes. Let N be the inertial frame, while B is the actual body fixed frame. Performing the ð3-2-1Þ rotation sequence, let B0 be the frame after the yaw rotation c about the 3 axis is performed, and B00 be the frame after the pitch rotation y about the current 2 axis is done. 108 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

Note that for these three single-axis rotations, the equivalent Euler parameters are given by b0B0=N ¼ cos c 2   b1B0 =N ¼ 0 b2B0=N ¼ 0 b3B0=N ¼ sin c 2   b0B00=B0 ¼ cos y 2   b1B00 =B0 ¼ sin y 2   b2B00=B0 ¼ 0 b3B00=B0 ¼ 0 b0B00=B ¼ cos f 2   b1B00 =B ¼ 0 b2B00=B ¼ 0 b3B00=B ¼ sin j 2   Using the convenient Euler parameter composite rotation property in Eq. (3.98), as well as the compact notation ci ¼ cosðyi=2Þ and si ¼ sinðyi=2Þ, the Euler parameter set ðb0; b1; b2; b4Þ of the body orientation relative to the inertial frame is expressed as b0 b1 b2 b3 0 B B @ 1 C C A ¼ c3 0 0 s3 0 c3 s3 0 0 s3 c3 0 s3 0 0 c3 2 6 6 4 3 7 7 5 c2 s2 0 0 s2 c2 0 0 0 0 c2 s2 0 0 s2 c2 2 6 6 4 3 7 7 5 c1 0 0 s1 0 B B @ 1 C C A Carrying out the matrix algebra, this method directly leads to the desired (3-2-1) Euler angle transformation into Euler parameters: b0 ¼ cos f 2   cos y 2   cos c 2    sin f 2   cos y 2   sin c 2   b1 ¼ cos f 2   sin y 2   cos c 2   þ sin f 2   sin y 2   sin c 2   b2 ¼ cos f 2   sin y 2   sin c 2    sin f 2   sin y 2   cos c 2   b3 ¼ sin f 2   cos y 2   cos c 2   þ cos f 2   cos y 2   sin c 2   The remaining Euler-angle-(both symmetric and asymmetric)-to-Euler-parameter transformations can be readily derived in a similar manner without resorting to extensive trigonometric identities to obtain a simple algebraic expression. The kinematic differential equation for the Euler parameters can be derived by differentiating the bi in Eq. (3.94). The following development will estab- lish the kinematic equation for _bb0 only; the remaining _bbi equations can be developed in an analogous manner. After taking the derivative of Eq. (3.94a), _bb0 is expressed as _bb0 ¼ _CC11 þ _CC22 þ _CC33 8b0 ð3:101Þ RIGID BODY KINEMATICS 109D w U W M S m DO

After using the expressions for _CCii given in Eq. (3.27), the term _bb0 is rewritten as _bb0 ¼ 1 2  C23  C32 4b0 o1  C31  C13 4b0 o2  C12  C21 4b0 o3   ð3:102Þ Using Eqs. (3.94b–3.94d), the _bb0 differential equation is simplified to _bb0 ¼ 1 2 ðb1o1  b2o2  b3o3Þ ð3:103Þ After performing a similar derivation for the _bb1, _bb2, and _bb3 terms, the four coupled kinematic differential equations for the Euler parameters are found to be the exceptionally elegant matrix form _bb0 _bb1 _bb2 _bb3 0 B B B B B @ 1 C C C C C A ¼ 1 2 0 o1 o2 o3 o1 0 o3 o2 o2 o3 0 o1 o3 o2 o1 0 2 6 6 6 6 6 4 3 7 7 7 7 7 5 b0 b1 b2 b3 0 B B B B B @ 1 C C C C C A ð3:104Þ or by transmutation of Eq. (3.104), the kinematic differential equation has the elegant form _bb0 _bb1 _bb2 _bb3 0 B B B B B @ 1 C C C C C A ¼ 1 2 b0 b1 b2 b3 b1 b0 b3 b2 b2 b3 b0 b1 b3 b2 b1 b0 2 6 6 6 6 6 4 3 7 7 7 7 7 5 0 o1 o2 o3 0 B B B B B @ 1 C C C C C A ð3:105Þ Note that the transformation matrix relating _ bb and v is orthogonal and singu- larity free. The inverse transformation from v to dð b Þ=dt is always defined. Further, the Euler parameter kinematic differential equation of Eq. (3.104) is rigorously linear if oiðtÞ are known functions of time only. If oiðtÞ are them- selves coordinates, then Eqs. (3.104) and (3.105) are more generally considered bilinear. This makes the Euler parameters very attractive attitude coordinates for attitude estimation problems where the kinematic differential equation is linearized. All three parameter sets of attitude coordinates always have kine- matic differential equations that are nonlinear and contain 0=0 type mathemati- cal singularities. In attitude estimation problems their linearization is only locally valid, whereas the linear (or bilinear) property of the Euler parameter 110 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

kinematic differential equation is globally valid. The Euler parameter kinematic differential equation in Eq. (3.105) can be written compactly as _ bb ¼ 1 2 ½Bð b ފ Bx ð3:106Þ where the 4  3 matrix ½Bð b ފ is defined as ½Bð b ފ ¼ b1 b2 b3 b0 b3 b2 b3 b0 b1 b2 b1 b0 2 6 6 4 3 7 7 5 ð3:107Þ By carrying out the matrix algebra, the following useful identities can easily be verified: ½Bð b ފT b ¼ 0 ð3:108Þ ½Bð b ފT b 0 ¼ ½Bð b 0ފT b ð3:109Þ It is easily verified that the normalization condition b T b ¼ 1 is a rigorous analytical integral of Eqs. (3.104) and (3.105). However, in practice the norm of b may slightly differ from 1 when numerically integrating Eq. (3.104). It is therefore necessary to take care to reimpose this condition differentially after each numerical integration step, if the solution is to remain valid over long time intervals. However, in contrast to the renormalization of ½Cðtފ to satisfy ½CŠT ½CŠ ¼ ½I33Š when solving Eq. (3.27), only one scalar condition needs to be considered when integrating b ðtÞ. In control applications, often the four Euler parameters are broken up into two groups. The parameter b0 is singled out because it contains no information regarding the corresponding principal rotation axis of the orientation being represented. In effect, it is a scalar measure of the three-dimensional rigid body attitude measure whose value is þ1 or 1 if the attitude is zero. The remaining three Euler parameters are grouped together into a three-dimensional vector as e  ðb1; b2; b3ÞT ð3:110Þ If the attitude goes to zero, then so will this vector. From the Euler parameter differential equation in Eq. (3.105), it is evident that the differential equations for _bb0 and _ ee are of the form _bb0 ¼  1 2 eT v ¼  1 2 vT e ð3:111Þ _ ee ¼ 1 2 ½TŠ v ð3:112Þ The 3  3 matrix ½TŠ is defined as ½Tðb0; eފ ¼ b0½I33Š þ ½ ~ eeŠ ð3:113Þ RIGID BODY KINEMATICS 111D w U W M S m DO

3.6 Classical Rodrigues Parameters The origin of the classical Rodrigues parameter vector q (or Gibbs vector) dates back over a hundred years to the French mathematician O. M. Rodrigues. This rigid body attitude coordinate set reduces the redundant Euler parameters to a minimal three-parameter set through the transformation qi ¼ bi b0 i ¼ 1; 2; 3 ð3:114Þ The inverse transformation from classical Rodrigues parameters to Euler para- meters is given by b0 ¼ 1 1 þ qT q p ð3:115aÞ bi ¼ qi 1 þ qT q p i ¼ 1; 2; 3 ð3:115bÞ Using the definitions in Eq. (3.91), the vector q is expressed directly in terms of the principal rotation elements as the elegant transformation q ¼ tan F 2 ^ee ð3:116Þ From Eqs. (3.114) and (3.116), it is evident that the classical Rodrigues para- meters go singular whenever F ! 180 deg. Very large rotations can be described with these parameters without ever approaching a geometric singular- ity. For rotations with jFj  90 deg, it is evident that qðtÞ locates points near the origin bounded by the unit sphere. Compare this 180 deg nonsingular range to the Euler angles where any orientation is never more than 90 deg away from a singularity. The small angle behavior of the classical Rodrigues parameters is also more linear than compared to the small angle behavior of any Euler angle set. Lin- earizing Eq. (3.116), it is evident that q  F 2 ^ee ð3:117Þ This means that classical Rodrigues parameters will linearize roughly to an ‘‘angle over 2’’ type quantity, whereas the Euler angles linearize as an angle type quantity well removed from singular points. As discussed in Ref. 20, the classical Rodrigues parameters can be viewed as a special set of stereographic orientation parameters. Stereographic projections are used to map a higher-dimensioned spherical surface onto a 112 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

lower-dimensioned hyperplane. In this case, the surface of the four-dimensional Euler parameter unit constraint sphere in Eq. (3.92) is mapped (projected) onto a three-dimensional hyperplane through Eq. (3.114). Figure 3.10 illustrates how such a projection would yield the classical Rodrigues parameters. The projection point is chosen to be the origin b ¼ 0, and the hyperplane upon which all Euler parameter coordinates are projected is the tangent surface at b0 ¼ 1. Note that on the constraint sphere surface b0 ¼ 1 corresponds to F ¼ 0 deg, b0 ¼ 0 corresponds to F ¼ 180 deg, and b0 ¼ 1 represents F ¼ 360 deg. The transformation in Eq. (3.114) maps any Euler parameter set on the unit constraint sphere surface onto a corresponding point located on the classical Rodrigues parameter hyperplane. All stereographic orientation parameters can be viewed as a projection of the constraint sphere onto some hyperplane. Because the Euler parameters themselves are not unique, the corresponding stereographic orientation para- meters are also generally not unique. The set corresponding to the projection of the Euler parameter set  b is referred to as the shadow set and is differen- tiated from the original set by a superscript S.20 However, it turns out that the shadow set of the classical Rodrigues parameters are indeed identical to the original classical Rodrigues parameters, as is easily verified by reversing the bi signs in Eq. (3.114) or by inspection of Fig. 3.10: qS i ¼ bi b0 ¼ qi ð3:118Þ Fig. 3.10 Stereographic projection of Euler parameters to classical Rodrigues parameters. RIGID BODY KINEMATICS 113D w U W M S m DO

The direction cosine matrix in terms of the classical Rodrigues parameters can be found by using their definition in Eq. (3.114) in the direction cosine matrix formulation in Eq. (3.93). The resulting parameterization is in matrix form3;20 ½CŠ ¼ 1 1 þ qT q 1 þ q2 1  q2 2  q2 3 2ðq1q2 þ q3Þ 2ðq1q3  q2Þ 2ðq2q1  q3Þ 1  q2 1 þ q2 2  q2 3 2ðq2q3 þ q1Þ 2ðq3q1 þ q2Þ 2ðq3q2  q1Þ 1  q2 1  q2 2 þ q2 3 2 6 6 4 3 7 7 5 ð3:119Þ and in vector form4;20 ½CŠ ¼ 1 1 þ qT q ðð1  qT qÞ½I33Š þ 2qqT  2½~qqŠÞ ð3:120Þ A direct mapping from the direction cosine matrix ½CŠ to the q parameters is through the matrix equation32 ½~qqŠ ¼ ½CŠT  ½CŠ z2 ð3:121Þ where z ¼ traceð½CŠÞ þ 1 p ¼ b0=2. As expected, this transformation is singular for principal rotations of 180 deg. Using the tilde matrix definition, the direct mapping from the direction cosine matrix ½CŠ to the Gibbs vector q is32 q ¼ q1 q2 q3 0 @ 1 A ¼ 1 z2 C23  C32 C31  C13 C12  C21 0 @ 1 A ð3:122Þ Note the following useful identity: ½CðqފT ¼ ½Cðqފ ð3:123Þ Because q defines the relative orientation of a second frame to a first frame, the relative orientation of the first frame relative to the second corresponds simply to reversing the sign of q, as in f ^nng ¼ ½CðqފT f^bbg ¼ ½Cðqފf^bbg ð3:124Þ This elegant property does not exist with Euler angles. Similar to the direction cosine matrices and Euler parameters, the classical Rodrigues parameter vectors have a composite rotation property. Given two attitude vectors q0 and q00, let the overall composite attitude vector q be defined through the quadratically nonlinear condition ½FN ðqފ ¼ ½FBðq00ފ½BN ðq0ފ ð3:125Þ 114 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

However, solving for an overall transformation from q0 and q00 to q using Eq. (3.125) is very cumbersome. Using the successive rotation property of the Euler parameters and the definition of the classical Rodrigues parameters in Eq. (3.114), the composite attitude vector q is expressed directly in terms of q0 and q00 through4;21 q ¼ q00 þ q0  q00  q0 1  q00  q0 ð3:126Þ Assume that the attitude vectors q and q0 are given and the relative attitude vector q00 is to be found. With direction cosine matrices and Euler parameters, the two attitude descriptions were related through an orthogonal matrix that made finding the relative attitude description trivial. This is no longer the case with the classical Rodrigues parameter composite rotation property. However, we can use Eq. (3.125) to solve for ½FBðq00ފ, first using the orthogonality of the direction cosine matrices: ½FBðq00ފ ¼ ½FN ðqފ½BNðq0ފT ð3:127Þ Using the identity in Eq. (3.123), this is rewritten as ½FBðq00ފ ¼ ½FN ðqފ½BN ðq0ފ ð3:128Þ which then leads to the desired direct transformation from q and q0 to the rela- tive orientation vector q00: q00 ¼ q  q0 þ q  q0 1 þ q  q0 ð3:129Þ A similar transformation could be found to express q0 in terms of q and q00. The kinematic differential equation of the classical Rodrigues parameters is found by taking the derivative of Eq. (3.114) and then substituting the corre- sponding expressions for _bbi given in Eq. (3.105). The resulting matrix formula- tion is3 _qq ¼ 1 2 1 þ q2 1 q1q2  q3 q1q3 þ q2 q2q1 þ q3 1 þ q2 2 q2q3  q1 q3q1  q2 q3q2 þ q1 1 þ q2 3 2 6 6 4 3 7 7 5 B o1 o2 o3 0 B B @ 1 C C A ð3:130Þ and the compact vector matrix form is _qq ¼ 1 2 ½I33Š þ ½~qqŠ þ qqT Bx ð3:131Þ Note that the preceding kinematic differential equation contains no trigono- metric functions and has only a quadratic nonlinearity. It is defined for any rotation except for F ¼ 180 deg. As qðtÞ approaches F ¼ 180 deg, both RIGID BODY KINEMATICS 115D w U W M S m DO

qðtÞ and _qqðtÞ diverge to infinity. The inverse transformation of Eq. (3.131) is given by4 B v ¼ 2 1 þ qT q ð½I33Š  ½~qqŠÞ_qq ð3:132Þ As is evident, for ðq; _qqÞ ! 1, the transformation of Eq. (3.132) exhibits an 1=1 type singular behavior near jFj ! 180 deg. There exists a very elegant, analytically exact transformation between the orthogonal direction cosine matrix ½CŠ and the classical Rodrigues parameter vector q called the Cayley Transform.3;4;9;15;22 What is remarkable is that this transformation holds for proper orthogonal matrices of dimensions higher than three. A proper orthogonal matrix is an orthogonal matrix with a determinant of þ1. Thus it is possible to parameterize any proper orthogonal ½CŠ matrix by a minimal set of higher-dimensional classical Rodrigues parameters. The Cayley transform parameterizes a proper orthogonal matrix ½CŠ as a function of a skew-symmetric matrix ½QŠ: ½CŠ ¼ ð½I Š  ½QŠÞð½I Š þ ½QŠÞ 1 ¼ ð½I Š þ ½QŠÞ 1ð½I Š  ½QŠÞ ð3:133Þ The matrix product order is irrelevant in this transformation. Another surpris- ing property of this transformation is that the inverse transformation from the skew-symmetric matrix Q back to the ½CŠ matrix has exactly the same form as the forward transformation in Eq. (3.133): ½QŠ ¼ ð½I Š  ½CŠÞð½I Š þ ½CŠÞ 1 ¼ ð½I Š þ ½CŠÞ 1ð½I Š  ½CŠÞ ð3:134Þ For the case in which ½CŠ is a 3  3 rotation matrix, the transformation in Eq. (3.133) yields the standard three-dimensional Rodrigues parameters. This can be verified by setting ½QŠ ¼ ½~qqŠ in Eq. (3.133), carrying out the 3  3 special case algebra implicit in Eq. (3.133), and comparing the result to Eq. (3.119). The kinematic differential equation of the ½CŠ is given in Eq. (3.27). This expression also holds for matrix dimensions higher than three.3;9 Because the Cayley transform parameterizes a proper orthogonal matrix in terms of an ‘‘orientation coordinate’’ type quantity ½QŠ, the matrix ½ ~ vvŠ represents an analo- gous ‘‘angular velocity’’ cross product matrix that can be defined as3;9;15 ½ ~ vvŠ ¼ 2ð½I Š þ ½QŠÞ 1½ _QQŠð½I Š  ½QŠÞ 1 ð3:135Þ The kinematic differential equation of the higher-dimensional Rodrigues para- meters is obtained by differentiation of Eq. (3.134), and substituting Eqs. (3.27) and (3.133) as ½ _QQŠ ¼ 1 2 ð½I Š þ ½QŠÞ½ ~ vvŠð½I Š  ½QŠÞ ð3:136Þ 116 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

It can readily be verified that the 3  3 special case of Eq. (3.136) is equivalent to Eq. (3.130), and so again the general n  n case contains the classical 3  3 results. Example 3.10 Given the orthogonal 4  4 matrix ½CŠ, ½CŠ ¼ 0:505111 0:503201 0:215658 0:667191 0:563106 0:034033 0:538395 0:626006 0:560111 0:748062 0:272979 0:228387 0:337714 0:431315 0:767532 0:332884 2 6 6 4 3 7 7 5 it is easy to verify that ½CŠ can be parameterized in terms of higher-dimensional classical Rodrigues parameters. Using MATLAB1 to solve Eq. (3.134), the skew-symmetric 4  4 matrix ½QŠ is found to be ½QŠ ¼ 0 0:5 0:2 0:3 0:5 0 0:7 0:6 0:2 0:7 0 0:4 0:3 0:6 0:4 0 2 6 6 4 3 7 7 5 where the six upper diagonal elements of ½QŠ are the higher-dimensional classi- cal Rodrigues elements. 3.7 Modified Rodrigues Parameters The modified Rodrigues parameters (MRPs) are an elegant recent addition to the family of attitude parameters.4;20;23 25 The MRP vector s is defined in terms of the Euler parameters as the transformation si ¼ bi 1 þ b0 i ¼ 1; 2; 3 ð3:137Þ The inverse transformation is given by b0 ¼ 1  s2 1 þ s2 bi ¼ 2si 1 þ s2 i ¼ 1; 2; 3 ð3:138Þ where the notation s2n ¼ sT s  n is introduced. Substituting Eq. (3.90) into Eq. (3.137), the MRP vector can be expressed in terms of the principal rotation elements as s ¼ tan F 4 ^ee ð3:139Þ RIGID BODY KINEMATICS 117D w U W M S m DO

Studying Eq. (3.139), it is evident that the MRP has a geometric singularity at F ¼ 360 deg. Any rotation can be described except a complete revolution back to the original orientation. This gives s twice the rotational range of the classical Rodrigues parameters. Also note that for small rotations the MRPs linearize as s  ðF=4Þ^ee. Observing Eq. (3.137), it is evident that these equations are well behaved except near the singularity at b0 ¼ 1, where F ! 360 deg. Also, the inverse transformation of Eq. (3.138) is well behaved everywhere except at j sj ! 1; we see from Eq. (3.139) that this again occurs at F ! 360 deg. The MRP vector s can be transformed directly into the classical Rodrigues parameter vector q through q ¼ 2 s 1  s2 ð3:140Þ with the inverse transformation beings ¼ q 1 þ 1 þ qT q p ð3:141Þ Naturally, these transformations are singular at F ¼ 180 deg because the clas- sical Rodrigues parameters are singular at this orientation. As are the classical Rodrigues parameters, the MRPs are also a particular set of stereographic orientation parameters. Equation (3.137) describes a stereo- graphic projection of the Euler parameter unit sphere onto the MRP hyperplane normal to the b0 axis at b0 ¼ 0, where the projection point is at b ¼ ð1; 0; 0; 0Þ. This is illustrated in Fig. 3.11. As a 360 deg principal rotation is approached (i.e., b0 ! 1), the projection of the corresponding point on the constraint sphere goes to infinity. This illustrates the singular behavior of the MRPs as they describe a complete revolution. However, contrary to the classical Rodrigues parameters, the projection of the alternate Euler parameter vector  b results in a distinct set of shadow (or image) MRPs as can be seen in Fig. 3.11. Each MRP vector is an equally valid attitude description satisfying the same kinematic differential equation. Therefore, one can arbitrarily switch between the two vectors through the mapping20;24 sS i ¼ bi 1  b0 ¼ si s2 i ¼ 1; 2; 3 ð3:142Þ where the choice as to which vector is the original and which is the shadow vector is arbitrary. We usually let s denote the mapping point interior to the unit sphere and sS the point exterior to the unit sphere. As with the non- uniqueness of the principal rotation vector g and the Euler parameter vector 118 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

b , one set of MRPs always corresponds to a principal rotation F  180 deg and the other to F  180 deg. From Eq. (3.139) it is clear that j sj  1 if F  180 deg j sj  1 if F  180 deg j sj ¼ 1 if F ¼ 180 deg ð3:143Þ The behavior is seen in Fig. 3.11. The unit sphere j sj ¼ 1, corresponding to all principal rotations of 180 deg from the origin, is of particular importance. As one set of MRPs exits the unit sphere, the other (shadow) set enters. The mapping in Eq. (3.142) can be written in terms of the principal rotation elements using the definitions of bi in Eq. (3.90) assS ¼ tan F  2p 4   ^ee ð3:144Þ Using Eq. (3.74), this can be written directly in terms of the alternate principal rotation angle F0: sS ¼ tan F0 4   ^ee ð3:145Þ Fig. 3.11 Stereographic projection of Euler parameters to modified Rodrigues parameters. RIGID BODY KINEMATICS 119D w U W M S m DO

Eq. (3.145) clearly shows that the shadow MRP vector is a direct result of the alternate principal rotation vector. The shadow MRPs have a singular orientation at F ¼ 0 deg as compared to the original MRPs, which are singular at F ¼ 360 deg. This allows one to avoid MRP singularities altogether by switching between original and shadow MRP sets as one MRP vector approaches a singular orientation. On which surface s T s ¼ c one switches is arbitrary. However, switching between the two MRPs whenever the vector s penetrates the surface sT s ¼ 1 has many positive aspects. For one, the mapping between the two MRP vectors simplifies on this surface to sS ¼  s. Further, the magnitude of s will remain bounded above by 1. Having a bounded norm of an attitude description is useful because it reflects the fundamental fact that two orientations can differ only by a finite rotation. Also, the current MRP attitude description will always describe the shortest principal rotation because of Eq. (3.143). Therefore, the combined set of original and shadow MRPs with the switching surfaces T s ¼ 1 provides for a nonsingular, bounded, minimal attitude description. It is ideally suited to describe large, arbitrary motions. The combined set is also useful in a feedback control type setting. For example, it linearizes well for small angles and has a bounded maximum norm of 1, which makes the selec- tion of feedback gains easier. The direction cosine matrix in terms of the MRP is found by substituting Eq. (3.138) into Eq. (3.92) and is given as4;20;24;25 ½CŠ ¼ 1 ð1 þ s2Þ2  4ðs2 1  s2 2  s2 3Þ þ ð1  s2Þ2 8s1s2 þ 4s3ð1  s2Þ 8s2s1  4s3ð1  s2Þ 4ðs2 1 þ s2 2  s2 3Þ þ ð1  s2Þ2    8s3s1 þ 4s2ð1  s2Þ 8s3s2  4s1ð1  s2Þ 2 6 4 8s1s3  4s2ð1  s2Þ    8s2s3 þ 4s1ð1  s2Þ 4ðs2 1  s2 2 þ s2 3Þ þ ð1  s2Þ2 3 7 5 ð3:146Þ In compact vector form ½CŠ is parameterized in terms of the MRP as4;20 ½CŠ ¼ ½I33Š þ 8½ ~ ssŠ2  4ð1  s2Þ½ ~ ssŠ ð1 þ s2Þ2 ð3:147Þ As is the case with the classical Rodrigues parameters, the simplest method to extract the MRP from a given direction cosine matrix is to first extract the Euler parameters and then use Eq. (3.137) to find the MRP vector s. If b0  0 is chosen when extracting the Euler parameters, then j sj  1. If b0 is chosen to be negative, then the alternate MRP vector corresponding to a larger princi- pal rotation angle is found. 120 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

A direct mapping from the direction cosine matrix ½CŠ to the MRP skew- symmetric matrix ½ ~ ssŠ has recently been provided by Hurtado in Ref. 33. ½ ~ ssŠ ¼ ½CŠT  ½CŠ zðz þ 2Þ ð3:148Þ where again the parameter z ¼ traceð½CŠÞ þ 1 p is used. This matrix transfor- mation will always provide an inner MRP set with j sj  1. However, although the MRPs are nonsingular for rotations up to 360 deg, this mapping contains a 0/0-type singularity for principal rotations of 180 deg, where z ¼ 1. These mathematical issues arise because of the two possible MRP answers that can occur with 180-deg rotations. If z ¼ 1, then an alternate method such as computing the Euler parameters, or extracting the principal rotation parameters as shown by Hughes in Ref. 32, can be employed. Using the tilde matrix defi- nition, the three MRP parameters can be directly computed using s ¼ s1 s2 s3 0 @ 1 A ¼ 1 zðz þ 2Þ C23  C32 C31  C13 C12  C21 0 @ 1 A ð3:149Þ The MRPs enjoy the same relative rotation identity as did the classical Rodrigues parameters: ½Cð sފT ¼ ½Cð sފ ð3:150Þ Given two MRP vectors s0 and s 00, let the overall MRP vector s be defined through ½FN ð sފ ¼ ½FBð s 00ފ½BNð s 0ފ ð3:151Þ Starting with the Euler parameter successive rotation property and using the MRP definitions in Eq. (3.137), the MRP successive rotation property is expressed as4 s ¼ ð1  j s0j2Þ s00 þ ð1  j s 00j2Þ s 0  2 s00  s0 1 þ j s 0j2j s 00j2  2 s 0  s00 ð3:152Þ Using Eq. (3.150), we are able to express the relative attitude vector s 00 in terms of s and s0 as s00 ¼ ð1  j s 0j2Þ s  ð1  j sj2Þ s 0 þ 2 s  s0 1 þ j s0j2j sj2 þ 2 s 0  s ð3:153Þ Although these expressions are more complicated than their Euler parameter or classical Rodrigues parameter counterparts, they do provide a numerically effi- cient method to compute the composition of two MRP vectors or find the rela- tive MRP attitude vector. RIGID BODY KINEMATICS 121D w U W M S m DO

Example 3.11 Given the Euler parameter vector b b ¼ ð0:961798; 0:14565; 0:202665; 0:112505ÞT the MRP vector s is found using Eq. (3.137) s1 ¼ 0:14565 1 þ 0:961798 ¼ 0:0742431 s2 ¼ 0:202665 1 þ 0:961798 ¼ 0:103306 s3 ¼ 0:112505 1 þ 0:961798 ¼ 0:0573479 The alternate shadow MRP vector sS can be found using  b instead of b in Eq. (3.137): sS 1 ¼ 0:14565 1  0:961798 ¼ 3:81263 sS 2 ¼ 0:202665 1  0:961798 ¼ 5:30509 sS 3 ¼ 0:112505 1  0:961798 ¼ 2:945 Note that if the direct mapping in Eq. (3.142) is used, the same vector sS is obtained. Because the vector j sj ¼ 0:139546  1, it represents the shorter principal rotation angle of F ¼ 7:94 deg. The vector j sSj ¼ 7:16611  1 repre- sents the longer principal rotation angle F0 ¼ F  360 deg ¼ 328:224 deg. The kinematic differential equation of the MRPs is found in a similar manner as the one for the classical Rodrigues parameters. The resulting matrix formulation is20;25 _ ss ¼ 1 4 1  s2 þ 2s2 1 2ðs1s2  s3Þ 2ðs1s3 þ s2Þ 2ðs2s1 þ s3Þ 1  s2 þ 2s2 2 2ðs2s3  s1Þ 2ðs3s1  s2Þ 2ðs3s2 þ s1Þ 1  s2 þ 2s2 3 2 6 6 4 3 7 7 5 B o1 o2 o3 0 B B @ 1 C C A ð3:154Þ The MRP kinematic differential equation in vector form is4;20 _ ss ¼ 1 4 ½ð1  s2Þ½I33Š þ 2½ ~ ssŠ þ 2 ssT Š Bx ¼ 1 4 ½Bð sފ B x ð3:155Þ Note that the MRPs retain a kinematic differential equation very similar to the classical Rodrigues parameters with only quadratic nonlinearities present. This 122 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

equation holds for either set of MRPs. However, the resulting vector _ ss will depend on which set of MRPs is being used. Just as a mapping exists betweens and sS, a direct mapping between _ ss and _ ssS is given by26 _ ssS ¼  _ ss s2 þ 1 2 1 þ s2 s4   ssT v ð3:156Þ Let the matrix ½BŠ transform v in Eqs. (3.154) and (3.155) into _ ss. This ½BŠ matrix is almost orthogonal except for a generally non-unit scaling factor. The inverse of ½BŠ can be written as ½BŠ 1 ¼ 1 ð1 þ s2Þ2 ½BŠT ð3:157Þ To prove Eq. (3.157), let’s study the expression ½BŠT ½BŠ. Using Eq. (3.155), this is written as ½BŠT ½BŠ ¼ 1  s2  ½I33Š  2½ ~ ssŠ þ 2 ss T   1  s2  ½I33Š þ 2½ ~ ssŠ þ 2 ssT   After carrying out all of the matrix multiplications, the ½BŠT ½BŠ expression is reduced to ½BŠT ½BŠ ¼ 1  s2  2½I33Š  4½ ~ ssŠ2 þ 4 ss T which can be further simplified using the identity ½ ~ ssŠ2 ¼ ss T  s2½I33Š to ½BŠT ½BŠ ¼ 1 þ s2  2½I33Š At this point it is trivial to verify that Eq. (3.157) must hold. The inverse trans- formation of Eqs. (3.154) and (3.155) then is in matrix notation B x ¼ 4 ð1 þ s2Þ2 ½BŠT _ ss ð3:158Þ and in vector form4 B x ¼ 4 ð1 þ s2Þ2 1  s2  ½I33Š  2½ ~ ssŠ þ 2 ss T _ ss ð3:159Þ Like the classical Rodrigues parameters, the MRPs can also be used to mini- mally parameterize higher-dimensional proper orthogonal matrix ½CŠ. Let the ½SŠ be a skew-symmetric matrix. The extended Cayley transform of ½CŠ in terms of ½SŠ is15;27 ½CŠ ¼ ð½I33Š  ½SŠÞ2ð1 þ ½SŠÞ 2 ¼ ð1 þ ½SŠÞ 2ð½I33Š  ½SŠÞ2 ð3:160Þ RIGID BODY KINEMATICS 123D w U W M S m DO

where the order of the matrix products is again irrelevant. For the case in which ½CŠ is a 3  3 matrix, then ½SŠ is the same as ½ ~ ssŠ. Therefore, Eq. (3.155) transforms a higher-dimensional proper orthogonal ½CŠ into higher-dimensional MRPs. For the special case of three-dimensional matrices, Hurtado demonstrates in Ref. 33 the remarkable Cayley-like transformations ½CŠ ¼ ðk½I33Š þ ½SŠÞ 1ðk½I33Š  ½SŠÞ ¼ ðk½I33Š  ½SŠÞðk½I33Š þ ½SŠÞ 1 ð3:161Þ ½SŠ ¼ kð½I33Š þ ½CŠÞ 1ð½I33Š  ½CŠÞ ¼ kð½I33Š  ½CŠÞð½I33Š þ ½CŠÞ 1 ð3:162Þ where in Eq. (3.161) kð sÞ ¼ ð1  s2Þ=2, while in Eq. (3.162) kð½CŠÞ ¼ z þ 1 p =ð z þ 1 p þ 2Þ with z ¼ traceð½CŠÞ. While the extended Cayley transform in Eq. (3.155) is valid for any orientations, the elegant Cayley-like transforma- tions in Eqs. (3.161) and (3.162) suffer from a mathematical singularity for 180- deg principal rotations as a result of the duality of MRP solutions that exist within j sj ¼ 1. Unfortunately, no direct inverse transformation exists like Eq. (3.134) for the higher order Cayley transforms.15 The transformation is achieved indirectly through the matrix ½W Š, where it is defined as the matrix square root of ½CŠ: ½CŠ ¼ ½W Š½W Š ð3:163Þ Because ½CŠ is orthogonal, it can be spectrally decomposed as ½CŠ ¼ ½V Š½DŠ½V Š* ð3:164Þ where ½V Š is the orthogonal eigenvector matrix and ½DŠ is the diagonal eigen- value matrix with entries of unit magnitude. The * operator stands for the adjoint operator that performs the complex conjugate transpose of a matrix. The matrix ½W Š can be computed as ½W Š ¼ ½V Š . . . 0 ½DŠii p 0 . . . 2 6 6 4 3 7 7 5½V ŠT ð3:165Þ The eigenvalues of ½CŠ are typically complex conjugate pairs. If the dimension of ½CŠ is odd, then the extra eigenvalue is real. For proper orthogonal matrices it is þ1 and its square root is also chosen to be þ1. The resulting ½W Š matrix will then itself also be a proper orthogonal matrix. As Ref. 15 shows, the geometric interpretation of ½W Š is that it represents the same higher-dimen- sional orientation as ½CŠ except that the corresponding principal rotation angles are halved. 124 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

The standard Cayley transforms in Eqs. (3.133) and (3.134) can be applied to map ½W Š into ½SŠ and back: ½W Š ¼ ð½I Š  ½SŠÞð½I Š þ ½SŠÞ 1 ¼ ð½I Š þ ½SŠÞ 1ð½I Š  ½SŠÞ ð3:166Þ ½SŠ ¼ ð½I Š  ½W ŠÞð½I Š þ ½W ŠÞ 1 ¼ ð½I Š þ ½W ŠÞ 1ð½I Š  ½W ŠÞ ð3:167Þ Therefore, to obtain a higher-dimensional MRP representation of ½CŠ, the matrix ½W Š must be found first and then substituted into Eq. (3.167). Note that by substituting Eq. (3.166) into Eq. (3.163), a direct forward transformation from ½SŠ to ½CŠ is found: ½CŠ ¼ ð½I Š  ½SŠÞ2ð½I Š þ ½SŠÞ 2 ¼ ð½I Š þ ½SŠÞ 2ð½I Š  ½SŠÞ2 ð3:168Þ The kinematic differential equations for ½SŠ are not written directly in terms of ½CŠ as they were for the classical Cayley transform. Instead the ½W Š matrix is used. Being an orthogonal matrix, its kinematic differential equation is of the same form as Eq. (3.27) ½ _WW Š ¼ ½ ~VVŠ½W Š ð3:169Þ where ½ ~VVŠ is the corresponding angular velocity matrix. It is related to the ½ ~ vvŠ matrix in Eq. (3.27) through ½ ~ vvŠ ¼ ½ ~VVŠ þ ½W Š½ ~VVŠ½W ŠT ð3:170Þ Analogously to Eq. (3.136), the kinematic differential equation of the ½SŠ matrix is given by ½ _SSŠ ¼ 1 2 ð½I Š þ ½SŠÞ½ ~VVŠð½I Š  ½SŠÞ ð3:171Þ Example 3.12 Consider the same orthogonal 4  4 matrix ½CŠ as is defined in Example 3.10. Using MATLAB1, its matrix square root ½W Š is found to be ½W Š ¼ 0:86416 0:35312 0:14580 0:32754 0:37209 0:69343 0:44177 0:43076 0:25488 0:50816 0:79065 0:22734 0:22320 0:36911 0:39807 0:80962 2 6 6 4 3 7 7 5 RIGID BODY KINEMATICS 125D w U W M S m DO

Using Eq. (3.167), the higher-dimensional, skew-symmetric MRP matrix ½SŠ representing ½CŠ is found: ½SŠ ¼ 0 0:20952 0:10114 0:14383 0:20952 0 0:28309 0:24040 0:10114 0:28309 0 0:17471 0:14383 0:24040 0:17471 0 2 6 6 4 3 7 7 5 By back substitution of this ½SŠ into Eq. (3.160), it can be verified that it does indeed parameterize ½CŠ. 3.8 Other Attitude Parameters There exists a multitude of other attitude parameters sets in addition to those discussed so far. This section will briefly outline a selected few. 3.8.1 Stereographic Orientation Parameters The stereographic orientation parameters (SOPs) are introduced in Ref. 20. They are formed by projecting the Euler parameter constraint surface, a four- dimensional unit hypersphere, onto a three-dimensional hyperplane. The pro- jection point can be anywhere on or within the constraint hypersphere, while the mapping hyperplane is chosen to be a unit distance away from the projec- tion point. There are two types of SOPs, the symmetric and asymmetric sets. The symmetric sets have a mapping hyperplane that is perpendicular to the b0 axis. Because b0 ¼ cos F=2 contains information only about the principal rotation angle, the resulting sets will all have a geometric singularity at a specific prin- cipal rotation angle F only, regardless of the corresponding principal rotation axis ^ee. The classical and modified Rodrigues parameters are examples of symmetric SOPs. Figure 3.12 illustrates the stereographic projection of the symmetric stereo- graphic orientation parameters (SSOP). The projection point is set at b0 ¼ a, and the projection plane is defined through b0 ¼ a þ 1. The mapping from Euler parameters b to a SSOP set g is Zi ¼ bi b0  a i ¼ 1; 2; 3 ð3:172Þ However, whereas the CRP and MRP represent a unique attitude, the generalized SSOP g do not. Note how in Fig. 3.12 there are two sets of b that project to the same g values. One possible set of Euler parameters has b0 > a, while the alternate set has b0 < a.31 This nonuniqueness of the attitude representation of g must be carefully considered before using these coordinates. The typical use of the general SSOP is for a situation where the craft should not depart from the reference motion by more than a fixed principal rotation angle F. Reference 31 uses the SSOP to develop elegant constrained attitude control laws where 126 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

the pointing departure constraint is enforced through the use of SSOP attitude coordinates. The inverse mapping from SSOP to Euler parameters is31 b0 ¼ S1 1 þ Z2 ð3:173aÞ bi ¼ Zi S2 1 þ Z2 for i ¼ 1; 2; 3 ð3:173bÞ with S1 ¼ aZ2  1 þ ð1  a2ÞZ2 p ð3:174aÞ S2 ¼ a  1 þ Z2 1  a2ð Þ p ð3:174bÞ The ‘‘þ’’ sign corresponds to the case where b0 > 0 (shorter rotation), and the ‘‘7’’ sign is for the alternate case with b0 < a (longer rotation). Fig. 3.12 Stereographic projection of the Euler parameters to symmetric stereo- graphic parameters. RIGID BODY KINEMATICS 127D w U W M S m DO

The direction cosine matrix is compactly expressed in terms of the SSOP as31 C½ Š ¼ 1 1 þ Z2ð Þ2 S2 2 ~ZZ½ Š2 þ ggT    2 ~ZZ½ ŠS1S2 þ S2 1 I½ Š33 ð3:175Þ where the differential kinematic equation is _gg ¼ 1 2 S1 S2 ½I33Š þ ½~ggŠ þ ggT   Bx ð3:176Þ Note that Eqs. (3.175) and (3.176) reduce to the expected formulations for the CRPs and the MRPs if either a ¼ 0 or 71 is chosen. Asymmetric SOPs have a mapping hyperplane that is not perpendicular to the b0 axis. The condition for a geometric singularity will now depend on both the principal rotation axis ^ee and the angle F. As an example, consider the asymmetric SOP vector g. It is formed by having a projection point at b1 ¼ 1 and having a mapping hyperplane at b1 ¼ 0. In terms of the Euler parameters, it is defined as Z1 ¼ b0 1 þ b1 Z2 ¼ b2 1 þ b1 Z3 ¼ b3 1 þ b1 ð3:177Þ with the inverse transformation being b0 ¼ 2Z1 1 þ Z2 b1 ¼ 1  Z2 1 þ Z2 b2 ¼ 2Z2 1 þ Z2 b3 ¼ 2Z3 1 þ Z2 ð3:178Þ where Z2 ¼ gT g. From Eq. (3.177) it is evident that g has a geometric singularity whenever b1 ! 1. This means that g goes singular whenever it represents a pure single axis rotation about the first body axis by the principal angles F1 ¼ 180 deg or F2 ¼ þ540 deg. This type of asymmetric principal angle rotation range is typical for all asymmetric SOPs. However, because the g vector has a distinct shadow counterpart, any geometric singularities can be avoided by switching between the two sets through the mapping gS ¼  g Z2 ð3:179Þ The direction cosine matrix is written in terms of the g vector components as ½CŠ ¼ 1 ð1 þ Z2Þ2  4 Z2 1  Z2 2  Z2 3   þ ð1  Z2Þ2 8Z1Z3 þ 4Z2ð1  Z2Þ 8Z1Z3 þ 4Z2ð1  Z2Þ 4ðZ2 1 þ Z2 2  Z2 3Þ  ð1  Z2Þ2    8Z1Z2 þ 4Z3ð1  Z2Þ 8Z2Z3  4Z1ð1  Z2Þ 2 6 4 8Z1Z2 þ 4Z3ð1  Z2Þ    8Z2Z3 þ 4Z1ð1  Z2Þ 4ðZ2 1  Z2 2 þ Z2 3Þ  ð1  Z2Þ2 3 7 5 ð3:180Þ 128 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

The kinematic differential equation of the g vector is _gg ¼ 1 4 1  2Z2 1 þ Z2 2 Z1Z3  Z2   2 Z1Z2 þ Z3   2 Z3  Z1Z2   2 Z2Z3 þ Z1   1  2Z2 2 þ Z2 2 Z1Z3 þ Z2   1 þ 2Z2 3  Z2 2 Z1  Z2Z3   2 6 4 3 7 5 Bx ð3:181Þ Having a projection point on the constraint surface provides for the largest possible range of singularity free rotations. This is evident when comparing the classical and the modified Rodrigues parameters. The classical Rodrigues para- meters have a projection point within the constraint hypersphere at b0 ¼ 0. Their principal rotation range is half of that of the MRPs whose projection is on the constraint surface at b0 ¼ 1. 3.8.2 Higher Order Rodrigues Parameters The higher order Rodrigues parameters (HORP) are introduced in Ref. 27. The classical Cayley transform in Eq. (3.133) is expanded such that it parame- terizes ½n  nŠ orthogonal matrices through a skew-symmetric, higher order Rodrigues parameter matrix X : ½CŠ ¼ ½I33Š  ½X Š  m ½I33Š þ ½X Š   m ð3:182Þ The corresponding attitude vector x is given by x ¼ tan F 2m   ^ee ð3:183Þ For m ¼ 1 the vector x is the classical Rodrigues vector and for m ¼ 2 it is the MRP vector. Note that the domain of validity of the x vector is jFj < mp. The HORP sets are generally also not unique as is the case with the MRPs. Corre- sponding ‘‘shadow’’ sets can also be used here to avoid any geometric singula- rities. Note that for a given m there are typically m sets of possible HORPs. A particular set of HORP is the t vector where m ¼ 4. In terms of the Euler parameters, the first two HORP vectors t are defined through ti ¼ bi 1 þ b0  2ð1 þ b0Þ p i ¼ 1; 2; 3 ð3:184Þ with the inverse transformation being b0 ¼ 2 1  t2 1 þ t2  2 1 bi ¼ 4tið1  t2Þ ð1 þ t2Þ2 i ¼ 1; 2; 3 ð3:185Þ RIGID BODY KINEMATICS 129D w U W M S m DO

where t2n ¼ ð tT tÞn. Each vector t defined in Eq. (3.184) can be mapped to the corresponding shadow vector tS through tS ¼  t 1  t2 2t2 þ 1 þ t2ð Þt   ð3:186Þ where t ¼ t2 p . Combined Eqs. (3.184) and (3.186) yield the four possible HORP vectors for m ¼ 4. In terms of the principal rotation elements, the four sets can be expressed as t ¼ tan F  2kp 8   k ¼ 0; 1; 2; 3 ð3:187Þ Therefore, it will always be possible to switch from one t vector to another to avoid geometric singularities. The kinematic differential equation of the t vector is _ tt ¼ 1 8ð1  t2Þ 2ð3  t2Þ ttT þ 4ð1  t2Þ½ ~ ttŠ þ ð1  6t2 þ t4Þ½I33Š Bx ð3:188Þ Note that the kinematic differential equations of the HORP lose the simple second-order polynomial form that is present for the classical and modified Rodrigues parameters. Also, although the t vector itself is defined for rotations up to F ¼ mp, the kinematic differential equations encounter mathematical singularities of the type 0=0 whenever t2 ! 0. This corresponds to F ! 360 deg. By using the mapping in Eq. (3.186) to transform a t vector to an alternate set whenever j tj  tanðF=8Þ, any geometrical and mathematical singularities are avoided altogether. 3.8.3 The (w , z) Coordinates The ðw; zÞ attitude coordinates were introduced by Tsiotras and Longuski in Ref. 28. They are a minimal coordinate set and lend themselves well to be used in control problems of under-actuated axially symmetric spacecraft.29 The complex coordinate w describes the heading of one of the body axes, typically the spin axis. The coordinate z is the relative rotation angle about this axis defined by w. Let the heading of the chosen body axis be given by the vector ^bbi ¼ ða; b; cÞT . Because the vector ^bbi is a unit vector, the three components a; b, and c are not independent. They must satisfy the constraint sphere equa- tion a2 þ b2 þ c2 ¼ 1 ð3:189Þ 130 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

By performing a stereographic projection of the constraint sphere from the projection point ð0; 0; 1Þ onto the complex ðw1; w2Þ plane, the three redun- dant axis heading coordinates ða; b; cÞ are reduced to the complex variable w: w ¼ w1 þ iw2 ¼ b  ia 1 þ c ð3:190Þ The inverse transformation from w to ða; b; cÞ is given by a ¼ iðw  wwÞ 1 þ jwj2 b ¼ w þ ww 1 þ jwj2 c ¼ 1  jwj2 1 þ jwj2 ð3:191Þ Let’s assume that the spin axis is the third body axis; then the direction cosine matrix in terms of ðw; zÞ is given by ½CŠ ¼ 1 1 þ jwj2 Re½ð1 þ w2ÞeizŠ Im½ð1 þ w2ÞeizŠ 2 ImðwÞ Im½ð1  ww2Þe izŠ Re½ð1  ww2Þe izŠ 2 ReðwÞ 2 ImðweizÞ 2 ReðweizÞ 1  jwj2 2 6 4 3 7 5 ð3:192Þ The kinematic differential equations of the ðw; zÞ coordinates are given by _ww1 ¼ o3w2 þ o2w1w2 þ o1 2 1 þ w2 1  w2 2   ð3:193aÞ _ww2 ¼ o3w1 þ o1w1w2 þ o2 2 1 þ w2 2  w2 1   ð3:193bÞ _zz ¼ o3  o1w2o2w1 ð3:193cÞ 3.8.4 Cayley–Klein Parameters The Cayley–Klein parameters are a set of four complex parameters that are closely related to the Euler parameter vector b. They form a once-redundant, nonsingular set of attitude parameters. Let i ¼ 1 p , then they are defined in terms of b as14 a ¼ b0 þ ib3 b ¼ b2 þ ib1 g ¼ b2 þ ib1 d ¼ b0  ib3 ð3:194Þ The inverse transformation from the Euler parameters to the Cayley–Klein parameters is b0 ¼ a þ dð Þ=2 b1 ¼ i b þ gð Þ=2 b2 ¼  b  gð Þ=2 b3 ¼ i a  dð Þ=2 ð3:195Þ RIGID BODY KINEMATICS 131D w U W M S m DO

The direction cosine matrix is parameterized by the Cayley–Klein parameters as ½CŠ ¼ ða2  b2  g2 þ d2Þ=2 iða2 þ b2  g2 þ d2Þ=2 ðbd  agÞ iða2 þ b2  g2  d2Þ=2 ða2 þ b2 þ g2 þ d2Þ=2 iðag þ bdÞ ðgd  abÞ iðab þ gdÞ ðad þ bgÞ 2 6 6 4 3 7 7 5 ð3:196Þ 3.9 Homogeneous Transformations All previous sections in this chapter deal with methods to describe the rela- tive orientation of one coordinate frame to another. In particular, the direction cosine matrix is a convenient tool to map a vector with components taken in one reference frame to a vector with components taken in another. However, one underlying assumption here is that both reference frames have the same origin. In other words, any translational differences between the two frames in question are not taken into account when the vector components are mapped from one frame to another. Figure 3.13 shows an illustration in which two coordinate frames differ both in orientation and in their origins. Let us define the following two reference frames N and B: N : fON ; ^nn1; ^nn2; ^nn3g B: fOB; ^bb1; ^bb2; ^bb3g Fig. 3.13 Illustration of two coordinate frames with different origins and orienta- tions. 132 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

Let the position vector from the N frame origin to the B frame origin be given by rB=N. The position vector of point P is expressed in B frame components as Brp. These vector components are mapped into N frame components by pre- multiplying by the direction cosine matrix ½NBŠ. Although this provides the correct N frame components of the vector rP, it does not provide the correct position vector of point P as seen by the N frame because the two frames have different origins. To obtain these vector components, we compute N rp ¼ NrB=N þ ½NBŠ Brp ð3:197Þ By defining the 4  4 homogeneous transformation30 ½NBŠ ¼ NB N rB=N 013 1 " # ð3:198Þ it is possible to transform the position vector taken in B frame components directly into the corresponding position vector in N frame components. In robotics literature, this transformation is typically referred to as N B T. To accom- plish this, we define the 4  1 position vector Bp ¼ Brp 1 " # ð3:199Þ Observing Eq. (3.197), it is clear that N p ¼ ½NBŠ Bp ð3:200Þ This formula is very convenient when computing the position coordinate of a chain of bodies such as are typically found in robotics applications. However, care must be taken when considering the order of the translational and rota- tional differences between the two frames. The homogeneous transformation, as shown in Eq. (3.198), performs the translation first and the rotation second. This order is important. Assume a rotational joint has a telescoping member attached to it. To compute the homogeneous transformation from the joint to the telescoping member tip, a rotation must be performed first and a translation second. Note that this homogeneous transformation matrix abides by the same successive transformation property as the direction cosine matrix does. Con- sider the two vectors Ap ¼ ½ABŠ Bp ð3:201Þ Np ¼ ½NAŠ Ap ð3:202Þ RIGID BODY KINEMATICS 133D w U W M S m DO

Substituting Eq. (3.201) into (3.202), we find that Np ¼ ½NAŠ½ABŠ Bp ¼ ½NBŠ Bp ð3:203Þ Thus, two successive transformations are combined through ½NBŠ ¼ ½NAŠ½ABŠ ð3:204Þ However, the inverse matrix formula for the homogeneous transformation is not quite as elegant as the matrix inverse of the orthogonal direction cosine matrix. The following partitioned matrix inverse is convenient to compute the inverse of the ½NBŠ. Let ½M Š be defined as3 ½M Š ¼ A B C D   ð3:205Þ Then the inverse is given by ½M Š 1 ¼ A 1 þ A 1BD 1CA 1 A 1BD 1 D 1CA 1 D 1 " # ð3:206Þ with the Schur complement being defined as D ¼ D  CA 1B ð3:207Þ Substituting ½AŠ ¼ ½NBŠ ½BŠ ¼ ½NrB=N Š ½CŠ ¼ ½013Š ½DŠ ¼ ½1Š the Schur complement is given by D ¼ ½1Š ð3:208Þ and the inverse of the homogeneous transformation is the remarkably simple formula: ½NBŠ 1 ¼ ½NBŠT ½NBŠT N rB=N 013 1   ð3:209Þ Here the fact was used that ½NBŠ is orthogonal and that ½NBŠ 1 ¼ ½NBŠT . 134 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

References 1Junkins, J. L., and Turner, J. D., Optimal Spacecraft Rotational Maneuvers, Elsevier, Amsterdam, 1986. 2Kaplan, W., Advanced Calculus, 4th ed., Addison Wesley, New York, 1991. 3Junkins, J. L., and Kim, Y., Introduction to Dynamics and Control of Flexible Structures, AIAA Education Series, AIAA, Washington DC, 1993. 4Shuster, M. D., ‘‘A Survey of Attitude Representations,’’ Journal of Astronautical Sciences, Vol. 41, No. 4, 1993, pp. 439 517. 5Rugh, W. J.., Linear System Theory, Prentice Hall, Englewood Cliffs, NJ, 1993. 6Bowen, R. M., and Wang, C. C., Introduction to Vectors and Tensors, Vol. 1, Plenum, New York, 1976. 7Goldstein, H., Classical Mechanics, Addison Wesley, 1950. 8Likins, P. W., Elements of Engineering Mechanics, McGraw Hill, New York, 1973. 9Bar Itzhack, I. Y., and Markley, F. L., ‘‘Minimal Parameter Solution of the Orthogonal Matrix Differential Equation,’’ IEEE Transactions on Automatic Control, Vol. 35, No. 3, March 1990, pp. 314 317. 10Nelson, R. C., Flight Stability and Automatic Control, McGraw Hill, New York, 1989. 11Battin, R. H., An Introduction to the Mathematics and Methods of Astrodynamics, AIAA Education Series, AIAA, New York, 1987. 12Junkins, J. L., and Shuster, M. D., ‘‘The Geometry of Euler Angles,’’ Journal of Astronautical Sciences, Vol. 41, No. 4, 1993, pp. 531 543. 13Rodrigues, O., ‘‘Des Lois Geometriques qui Regissent Les Deplacements D’Un Systeme Solide Dans l’Espace, et de la Variation des Coordonnes Provenants de ces Deplacements Considers Independamment des Causes Qui Peuvent les Preduire,’’ LIOUV, Vol. III, 1840, pp. 380 440. 14Whittaker, E. T., Analytical Dynamics of Particles and Rigid Bodies, Cambridge Univ. Press, 1965 reprint, pp. 2 16. 15Schaub, H., Tsiotras, P., and Junkins, J. L., ‘‘Principal Rotation Representations of Proper NxN Orthogonal Matrices,’’ International Journal of Engineering Science, Vol. 33, No. 15, 1995, pp. 2277 2295. 16Nazaroff, G. J., ‘‘The Orientation Vector Differential Equation,’’ Journal of Guidance and Control, Vol. 2, 1979, pp. 351 352. 17Jiang, Y. F., and Lin, Y. P., ‘‘On the Rotation Vector Differential Equation,’’ IEEE Transactions on Aerospace and Electronic Systems, Vol. AES 27, 1991, pp. 181 183. 18Bharadwaj, S., Osipchuk, M., Mease, K. D., and Park, F. C., ‘‘Geometry and Optimality in Global Attitude Stabilization,’’ Journal of Guidance, Control, and Dynamics, Vol. 21, No. 6, 1998, pp. 930 939. 19Sheppard, S. W., ‘‘Quaternion from Rotation Matrix,’’ Journal of Guidance and Control, Vol. I, No. 3, 1978, pp. 223 224. 20Schaub, H., and Junkins, J. L., ‘‘Stereographic Orientation Parameters for Attitude Dynamics: A Generalization of the Rodrigues Parameters,’’ Journal of Astronautical Sciences, Vol. 44, No. 1, 1996, pp. 1 19. 21Federov, F., The Lorentz Group, Nauka, Moscow, 1979. 22Cayley, A., ‘‘On the Motion of Rotation of a Solid Body,’’ Cambridge Mathematics Journal, Vol. 3, 1843, pp. 224 232. RIGID BODY KINEMATICS 135D w U W M S m DO

23Wiener, T. F., Theoretical Analysis of Gimballess Inertial Reference Equipment Using Delta Modulated Instruments, Ph.D. Dissertation, Dept. of Aeronautics and Astronautics, Massachusetts Inst. of Technology, Cambridge, MA, 1962. 24Marandi, S. R., and Modi, V. J., ‘‘A Preferred Coordinate System and the Associated Orientation Representation in Attitude Dynamics,’’ Acta Astronautica, Vol. 15, No. 11, 1987, pp. 833 843. 25Tsiotras, P., ‘‘Stabilization and Optimality Results for the Attitude Control Problem,’’ Journal of Guidance, Control, and Dynamics, Vol. 19, No. 4, 1996, pp. 772 779. 26Schaub, H., Robinett, R. D., and Junkins, J. L., ‘‘New Penalty Functions for Optimal Control Formulation for Spacecraft Attitude Control Problems,’’ Journal of Guidance, Control, and Dynamics, Vol. 20, No. 3, 1997, pp. 428 434. 27Tsiotras, P., Junkins, J. L., and Schaub, H., ‘‘Higher Order Cayley Transforms with Applications to Attitude Representations,’’ Journal of Guidance, Control, and Dynamics, Vol. 20, No. 3, 1997, pp. 528 534. 28Tsiotras, P., and Longuski, J. M., ‘‘A New Parameterization of the Attitude Kine matics,’’ Journal of Astronautical Sciences, Vol. 43, No. 3, 1996, pp. 342 262. 29Tsiotras, P., ‘‘On the Choice of Coordinates for Control Problems on SO(3),’’ March 20 22, 1996. 30Craig, J. J., Introduction to Robotics Mechanics and Control, Addison Wesley, 1989. 31Southward, C. M., Ellis, J., and Schaub, H., ‘‘Spacecraft Attitude Control Using Symmetric Stereographic Orientation Parameters,’’ Journal of the Astronautical Sciences, Vol. 55, No. 3, July Sept. 2007, pp. 389 405. 32Hughes, P. C., Spacecraft Attitude Dynamics, Wiley, New York, 1986. 33Hurtado, J. E., ‘‘Interior Parameters, Exterior Parameters, and a Cayley Like Trans form,’’ Journal of Guidance, Control, and Dynamics, Vol. 32, No. 2, 2009, pp. 653 657. Problems 3.1 Given three reference frames N , B, and F, let the unit base vectors of the reference frames B and F be ^bb1 ¼ 1 3 1 2 2 0 @ 1 A ^bb2 ¼ 1 2 p 0 1 1 0 @ 1 A ^bb3 ¼ 1 3 2 p 4 1 1 0 @ 1 A and ^ff1 ¼ 1 4 3 2 3 p 0 B @ 1 C A ^ff2 ¼ 1 2 1 0 3 p 0 B @ 1 C A ^ff3 ¼ 1 4 3 p 2 3 p 1 0 B @ 1 C A 136 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

where the base vector components are written in the N frame. Find the direction cosine matrices ½BFŠ that describe the orientation of the B frame relative to the F frame, along with the direction cosine matrices ½BN Š and ½FNŠ that map vectors in the N frame into respective B or F frame vectors. 3.2 Assume a body frame B has its orientation defined through f^bb1; ^bb2; ^bb3g. Further, assume that these three ^bbi unit vectors are given with vector components taken in the N frame. Show that the direction cosine matrix [BN] can be expressed as ½BN Š ¼ N ^bb1  T N ^bb2  T N ^bb3  T 2 6 6 6 6 4 3 7 7 7 7 5 ¼ B ^nn1   B ^nn2   B ^nn3   3.3 Given the following 3  3 matrices, numerically verify if these are proper orthogonal rotation matrices by checking (a) What their determinant is. (b) If all column vectors have unit length. (c) If the three column vectors are mutually orthogonal. (d) If the three column vectors abide by the right-hand rule. (e) Comment on which matrices are proper, orthogonal rotation matrices. ½BNŠ ¼ 0:6 0 0:8 0:8 0 0:6 0 1 0 2 6 4 3 7 5 ½FNŠ ¼ 0:6 0 0:8 0:8 0 0:6 0 1 0 2 6 4 3 7 5 ½GN Š ¼ 0 0:8 0:6 0 0:6 0:8 1:1 0 0 2 6 4 3 7 5 3.4 Let the vector v be written in B frame components as Bv ¼ 1^bb1 þ 2^bb2  3^bb3 RIGID BODY KINEMATICS 137D w U W M S m DO

The orientation of the B frame relative to the N frame is given through the direction cosine matrix ½BN Š ¼ 0:87097 0:45161 0:19355 0:19355 0:67742 0:70968 0:45161 0:58065 0:67742 2 4 3 5 (a) Find the direction cosine matrix ½NBŠ that maps vectors with compo- nents in the B frame into a vector with N frame components. (b) Find the N frame components of the vector v. 3.5 Using the direction cosine matrix ½BN Š in problem 3.4, find its real eigen- value and corresponding eigenvector. 3.6 The angular velocity vectors of a spacecraft B and a reference frame motion R relative to the inertial frame N are given by v B=N and v R=N . The vector vR=N is given in R frame components, while vB=N is given in B frame components. The error angular velocity vector of the space- craft relative to the reference motion is then given by d v ¼ v B=N v R=N . Find the relative error angular acceleration vector d _ vv with components expressed in the B frame. (a) Find d _ vv by only assigning coordinate frames at the last step. (b) Find d _ vv by first expressing d v in B frame components as Bd v ¼ B v B=N  ½BRŠR v R=N and then performing an inertial derivative. 3.7 The reference frames N : f ^nn1; ^nn2; ^nn3g and B: f^bbL; ^bby; ^bbrg are shown in Fig. P3.7. (a) Find the direction cosine matrix ½BN Š in terms of the angle f. (b) Given the vector Bv ¼ 1^bbr þ 1^bby þ 2^bbL, find the vector N v. 3.8 Starting with Eqs. (3.21) and (3.22), verify Eq. (3.24). Fig. P3.7 Disk rolling on circular ring. 138 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

3.9 Parameterize the direction cosine matrix ½CŠ in terms of (2-3-2) Euler angles. Also, find appropriate inverse transformations from ½CŠ back to the (2-3-2) Euler angles. 3.10 Find the kinematic differential equations of the (2-3-2) Euler angles. What is the geometric condition for which these equations will encounter a mathematical singularity? 3.11 Given the (3-2-1) Euler angles c ¼ 10 deg; y ¼ 15 deg, and f ¼ 20 deg and their rates _cc ¼ 2 deg=s; _yy ¼ 1 deg=s, and _ff ¼ 0 deg=s, find the vectors B v and N v. 3.12 The initial yaw, pitch, and roll angles of a vehicle are ðc; y; fÞ ¼ ð40; 30; 80Þ deg. Assume that the angular velocity vector of the craft is given in body frame components as x ¼ sinð0:1tÞ 0:01 cosð0:1tÞ 0 @ 1 A20 deg =s Write a program to numerically integrate the yaw, pitch, and roll angles over a simulation time of 1 min. Note that you must integrate these equa- tions using radians as the angular units. However, show the results in terms of degrees. 3.13 The orientation of an object is given in terms of the (3-1-3) Euler angles ð30 deg; 40 deg; 20 degÞ, (a) Find the corresponding principal rotation axis ^ee. (b) Find the two principal rotation angles F and F0. (c) Find the corresponding Euler parameters. (d) Find the corresponding classical Rodrigues parameters. (e) Find the corresponding modified Rodrigues parameters. 3.14 A spacecraft performs a 45-deg single axis rotation about ^ee ¼ 1= 3 p 1; 1; 1ð ÞT . Find the corresponding (3-2-1) yaw, pitch, and roll angles that relate the final attitude to the original attitude. 3.15 Verify that the exponential matrix mapping ½CŠ ¼ e F½~eeŠ does have the finite form given in Eq. (3.81). 3.16 Verify that Eq. (3.89) is indeed the inverse mapping of the differential kinematic equation of _ gg given in Eq. (3.88). 3.17 Starting from the direction cosine matrix ½CŠ in Eq. (3.71) written in terms of the principal rotation elements, derive the parameterization of ½CŠ in terms of the Euler parameters. RIGID BODY KINEMATICS 139D w U W M S m DO

3.18 Starting with Eq. (3.96), verify the composite rotation property of the Euler parameter vector given in Eqs. (3.97) and (3.98). 3.19 Derive the kinematic differential equations for the second, third, and fourth Euler parameter. 3.20 Verify the transformation in Eq. (3.114) that maps a classical Rodrigues parameter vector into an Euler parameter vector. 3.21 Show the details of transforming the classical Rodrigues parameter defini- tion in terms of the Euler parameters qi ¼ bi=b0 into the expression qi ¼ tan ðF=2Þ^eei which is in terms of the principal rotation elements. 3.22 Show that the classical Rodrigues parameters are indeed a stereographic projection of the Euler parameter constraint surface (a four-dimensional unit hypersphere) onto the three-dimensional hyperplane defined through b0 ¼ 1 with the projection point being b ¼ ð0; 0; 0; 0ÞT . 3.23 Given the classical Rodrigues parameter vector q ¼ ð0:5; 0:2; 0:8ÞT, use the Cayley transform in Eq. (3.133) to find the corresponding direction cosine matrix ½CŠ. Also, verify that this ½CŠ is the same as is obtained through the mapping in Eqs. (3.119) or (3.120). 3.24 Verify the transformation in Eq. (3.138) that maps a MRP vector into an Euler parameter vector. 3.25 Show the details of transforming the MRP definition in terms of the Euler parameters si ¼ bi=ð1 þ b0Þ into the expression si ¼ tan ðF=4Þ^eei, which is in terms of the principal rotation elements. 3.26 Show that the MRPs are a stereographic projection of the Euler parameter constraint surface (a four-dimensional unit hypersphere) onto the three- dimensional hyperplane defined through b0 ¼ 0 with the projection point being b ¼ ð1; 0; 0; 0ÞT . 3.27 Derive the MRP parameterization of the direction cosine matrix ½CŠ given in Eq. (3.146). 3.28 Let the initial attitude vector be given through the MRP vectorsðt0Þ ¼ ð0; 0; 0ÞT . The body angular velocity vector vðtÞ is given as ð1; 0:5; 0:7ÞT rad=s. Integrate the resulting rotation for 5 s and use the mapping between original and shadow MRPs in Eq. (3.142) to enforce j sj  1. 3.29 Derive the mapping between _ ss and its shadow counterpart _ ss S in Eq. (3.156) starting with the kinematic differential equation of the MRP in Eqs. (3.155) and (3.142). 140 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

3.30 Given the MRP vector s ¼ ð0:25; 0:4; 0:3ÞT, use the Cayley trans- form in Eq. (3.168) to find the corresponding direction cosine matrix ½CŠ. Also, verify that this ½CŠ is the same as is obtained through the mapping in Eqs. (3.146) or (3.147). 3.31 You are flying in a lunar transit spacecraft and your current attitude relative to an inertial frame is given in terms of 3-2-1 Euler angles as (230, 70, and 103 deg). You are interested in docking onto the space station, which attitude relative to the same inertial frame is given through the Euler parameters ðb0; b1; b2; b3Þ ¼ ð0:328474; 0:437966; 0:801059; 0:242062Þ. What is the attitude of your spacecraft relative to the space station? Express your answer using the principal rotation angle f and principal rotation axis ^ee. 3.32 The initial 3-2-1 Euler angles yaw, pitch, and roll of a vehicle are ðc0; y0; f0Þ ¼ ð40; 30; 80Þ deg at time t0. Assume that the angular velo- city vector of the craft is given in body frame components as x ¼ sinð0:1tÞ 0:01 cosð0:1tÞ 0 @ 1 A20 deg=s The time t is assumed to be given here in units of seconds. (a) Translate this initial attitude description into the corresponding Euler parameters. (b) Write a program to numerically integrate the Euler parameters over a simulation time of 1 min. Note that you must integrate these equations using radians as the angular units. Plot the four Euler parameter time histories. (c) Given the result of the numerical integration, plot the Euler parameter constraint b2 0 þ b2 1 þ b2 2 þ b2 3; comment on these values. RIGID BODY KINEMATICS 141D w U W M S m DO

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4 Eulerian Mechanics 4.1 Introduction The dynamics of a continuous body, as presented in Chapter 2, Newtonian Mechanics, are specialized in this chapter for the case of rigid body dynamics. This means that all continuous bodies studied will have a constant shape. This is a common case for many applications. Systems such as satellites, aircraft, or robots are all typically modeled as sets of rigid bodies. The rotational dynamics of a rigid body are often referred to as Eulerian mechanics, because Euler’s equation _HH ¼ L and Euler’s rotational equation of motion generally govern this field. Unlike Chapter 2, this chapter first investigates the rigid-body angular momentum vector H and its derivative, along with the kinetic energy, before developing the rotational equations of motion. Next the rigid body dynamics in a torque-free environment are studied in more detail. Simple passive spin stabi- lization is investigated through the dual-spin spacecraft concept. Further, the dynamics of a rigid body is studied when a set of reaction wheels, control moment gyroscopes, or variable speed control moment gyroscopes are present, or the body is under the influence of gravity-gradient torques. 4.2 Rigid Body Dynamics 4.2.1 Angular Momentum The following discussions parallel the development in Section 2.5 for the case in which no body deformations were allowed. Let the moment be taken either about the center of mass or the inertial coordinate frame origin. In either case Euler’s equation reduces to _HH ¼ L ð4:1Þ Let R be the inertial position vector of an infinitesimal mass element dm. Let’s choose the moment to be defined about the coordinate frame origin O. It turns out this case will include the case of having the moment defined about the center of mass Rc. The angular momentum vector in Eq. (2.95) is reduced to H O ¼ ð B R  _RR dm ð4:2Þ 143D w U W M S m DO

Because R ¼ Rc þ r, this is rewritten as H O ¼ ð B ðRc þ rÞ  ð _RRc þ _rrÞ dm ð4:3Þ which is then expanded to H O ¼ ð B Rc  _RRc dm þ ð B r dm  _RRc þ Rc  ð B _rr dm þ ð B r  _rr dm ð4:4Þ Noting that the mass of the rigid body is constant and using the definition of the center of mass in Eq. (2.77), the angular momentum vector about the coor- dinate frame origin O is reduced to the expression H O ¼ Rc  M _RRc þ ð B r  _rr dm ð4:5Þ Equation (4.5) is written for the general case in which the rigid body B is rotating about its center of mass and the center of mass is moving indepen- dently at an inertial velocity _RRc; as shown in Fig. 4.1. The first term of H O is the angular momentum of the mass center about the origin, and its behavior was studied when discussing the dynamics of a single particle. The second term is more interesting because it contains the angular momentum vector H c of the rigid body B about its mass center Rc. From here on we will be discussing mainly H c and not the more general H O: H c ¼ ð B r  _rr dm ð4:6Þ Fig. 4.1 General rigid body rotation. 144 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

At this point we will make use of some of the kinematics results from the previous chapter. By definition, the vector _rr is defined to be an inertial deriva- tive, therefore _rr ¼ N d dt rð Þ ¼ Bd dt rð Þ þ x  r ð4:7Þ where the vector x is the instantaneous angular velocity vector of the rigid body frame B relative to the inertial frame N . Because B is a rigid body, the term Bd=dtðrÞ is zero. Thus _rr reduces to _rr ¼ x  r ð4:8Þ The angular momentum vector about the center of mass is then defined as H c ¼ ð B r  x  rð Þ dm ¼ ð B ½~rrŠ½~rrŠ dm   x ð4:9Þ Let the vectors ^bbi be the B frame unit direction vectors, then the vector r, x, and H c are written in B frame coordinates as r ¼ r1 ^bb1 þ r2 ^bb2 þ r3 ^bb3 ð4:10Þ x ¼ o1 ^bb1 þ o2 ^bb2 þ o3 ^bb3 ð4:11Þ H c ¼ Hc1 ^bb1 þ Hc2 ^bb2 þ Hc3 ^bb3 ð4:12Þ After carrying out the triple cross product and collecting all of the terms, the angular momentum vector H c is expressed as H c ¼ B Hc1 Hc2 Hc3 0 @ 1 A ¼ ð B B r2 2 þ r2 3 r1r2 r1r3 r1r2 r2 1 þ r2 3 r2r3 r1r3 r2r3 r2 1 þ r2 2 2 4 3 5 B o1 o2 o3 0 @ 1 A dm ð4:13Þ The entries in the 3  3 matrix are the moments and products of inertia of the rigid body B about its center of mass. Note that because the r vector components were taken in the B frame, the corresponding matrix components are also taken in the B frame. If a different coordinate system were assigned to the rigid body, the corresponding inertia matrix would be different, too. Let this symmetric inertia matrix be called ½IcŠ, where the subscript c indicates about which point the moments and products of inertia were taken. If this letter is omitted, then it is understood that this inertia matrix is defined about the center of mass: B½IcŠ ¼ ð B  ½~rrŠ½~rrŠ dm ¼ ð B B r2 2 þ r2 3 r1r2 r1r3 r1r2 r2 1 þ r2 3 r2r3 r1r3 r2r3 r2 1 þ r2 2 2 4 3 5 dm ð4:14Þ EULERIAN MECHANICS 145D w U W M S m DO

Because x does not vary over the volume, it can be taken outside the integral. Unless noted otherwise, from here on it will be assumed that the vectors and inertia matrices are written in the B frame and the superscript B will be dropped. The angular momentum vector of a rigid body about its center of mass can then be written in its simplest form as1 H c ¼ ½IcŠx ð4:15Þ Example 4.1 Consider the inertia matrix [I] of the simple spacecraft model illustrated in Fig. 4.2. The cylindrical body has four massless rods with point masses mi attached to it in a symmetrical fashion. The body fixed frame is given through B: f^bb1; ^bb2; ^bb3g. Assume the inertia of the main cylindrical hub is given through B½IhŠ ¼ B I1 0 0 0 I2 0 0 0 I3 2 4 3 5 Note that the inertia matrix definition in Eq. (4.14) requires the integration of all mass elements of the body. The integration of a continuous body is equiva- lent to the summation of discrete body elements. Thus, to account for the iner- tia caused by the four tip masses, we can compute ½I Š ¼ ½IhŠ  P4 i¼1 mi½~rriŠ½~rriŠ Using the tip position vector definitions with respect to the spacecraft hub mass center r1=3 ¼ r ^bb1 r2=4 ¼ r ^bb2 leads to B½I Š ¼ I1 0 0 0 I2 0 0 0 I3 2 4 3 5 þ ðm1 þ m3Þ 0 0 0 0 r2 0 0 0 r2 2 4 3 5 þ ðm2 þ m4Þ 0 0 0 r2 0 0 0 0 r2 2 4 3 5 Fig. 4.2 Illustration of spacecraft with four symmetric beams. 146 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

Thus, the final spacecraft inertia matrix is expressed in body frame B compo- nents as B½I Š ¼ B I1 þ ðm2 þ m4Þr2 0 0 0 I2 þ ðm1 þ m3Þr2 0 0 0 I3 þ Mr2 2 4 3 5 where M ¼ m1 þ m2 þ m3 þ m4 is the total tip mass. 4.2.2 Inertia Matrix Properties Developing Eq. (4.15), it was assumed that the rigid body B was free to rotate in space. Now it is assumed that the rigid body is no longer rotating independently from the center of mass motion, but instead it is orbiting a fixed point O such that it always keeps the same side facing this point, as shown in Fig. 4.3. Examples of this type of rotation would be the moon orbiting Earth or a rigid body swinging back and forth at the end of a suspended rope. The center of mass position vector Rc with this type of rotation is fixed in the B frame and therefore has the following inertial derivative: _RRc ¼ Bd dt ðRcÞ þ x  Rc ¼ x  Rc ð4:16Þ Substituting Eq. (4.16) into Eq. (4.5) and making use of Eq. (4.15), the angular momentum vector about the origin O is written as H O ¼ M Rc  ðx  RcÞ þ ½IcŠx ð4:17Þ After making use of the skew-symmetric tilde operator defined in Eq. (3.23), the vector H O is written as H O ¼ ð½IcŠ  M ½ ~RRcŠ½ ~RRcŠÞx ð4:18Þ Fig. 4.3 Rigid body rotation about origin. EULERIAN MECHANICS 147D w U W M S m DO

This leads to the famous parallel axis theorem. Given the moment of inertia matrix ½IcŠ of a rigid body B about its center of mass and the position vector Rc of this center of mass relative to some fixed point O, then the inertia matrix B about O is given through the transformation ½IOŠ ¼ ½IcŠ þ M ½ ~RRcŠ½ ~RRcŠT ð4:19Þ Note that the fixed point O does not have to be the origin, but it can be any inertially fixed location. Also, note that when expressing ½IcŠ and Rc in compo- nent form, for the matrix subtraction in Eq. (4.19) to be meaningful, both ½IcŠ and Rc must be expressed in the same coordinate frame. The resulting matrix ½IOŠ will also have components taken in the same frame. Example 4.2 Consider the oblate disk of mass m and radius r rolling on the level surface as shown in Fig. 4.4. The disk is attached to a vertical shaft through a massless rod of length L. This horizontal rod is clamped to the center of the disk and pinned to the vertical shaft, which is rotating at a constant rate _ff. What is the normal force N that the surface is exerting onto the disk? Let the coordinate frame B: f^bbL; ^bby; ^bbrg be attached to the rolling disk, E: f^eeL; ^eef; ^ee3g be attached to the rotating support rod, and N : f ^nn1; ^nn2; ^nn3g be an inertial frame. Because the disk is rolling without slip, the angular rate _yy can be related to the shaft rotating rate _ff through _yy ¼ L r _ff Fig. 4.4 Oblate disk rolling on level surface. 148 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

The angular velocity vectors between the respective frames are xE=N ¼ _ff ^nn3 ¼ _ff^ee3 ¼ E 0 0 _ff 0 B @ 1 C A xB=E ¼ _yy^eeL ¼  L r _ff^bbL ¼  L r B _ff 0 0 0 B @ 1 C A Let Is be the disk inertia about its spin axis ^bbL and It the transverse inertias, then the disk inertia matrix about its center of mass is given in B frame components by the diagonal matrix ½IcŠ ¼ B Is 0 0 0 It 0 0 0 It 2 4 3 5 Because the disk is axisymmetric about the ^bbL ¼ ^eeL axis, for this example B½IcŠ ¼ E ½IcŠ must hold. Because the disk is rotating at an offset distance L about the ^nn3 axis, to find the disk inertia matrix about the point O we must use the parallel axis theorem in Eq. (4.19). The position vector of the disk center of mass is Rc ¼ L^eeL ¼ E L 0 0 0 @ 1 A The disk inertia matrix ½IOŠ about point O is then given in E frame components by ½IOŠ ¼ ½IcŠ þ m½ ~RRcŠ½ ~RRcŠT ¼ E Is 0 0 0 It þ mL2 0 0 0 It þ mL2 2 4 3 5 The angular momentum vector H O of the disk about the point O is the sum of the angular momentum due to the shaft rotation about the ^nn3 direction and the rolling about the ^bbL direction: H O ¼ ½IcŠxB=E þ ½IOŠxE=N ¼ IS L r _ff^eeL þ ðIt þ mL2Þ _ff ^nn3 EULERIAN MECHANICS 149D w U W M S m DO

The inertial angular momentum vector rate _HHO is found using the transport theorem. _HHO ¼  Ed dt Is L r _ff^eeL    xE=N  Is L r _ff^eeL þ Nd dt ððIt þ mL2Þ _ff ^nn3Þ ¼  IsL r _ff2 ^eef The normal force is defined as N ¼ N ^ee3, the surface friction force is Ff ¼ Ff ^eef, and the gravity force is given by Fg ¼ mg^ee3. Note that at the hinge point only the torques tf ¼ tf ^eef and t3 ¼ t3 ^ee3 are applied. The exter- nal torque about point O is LO ¼ Rc  ðRc þ NÞ þ tf þ t3 þ ðRc  r^ee3Þ  Ff ¼ rFf ^eeL þ Lðmg  N Þ þ tf  ^eef þ ðt3 þ RcFf Þ^ee3 Note that by taking the moments about point O the reaction forces of the pin joint O do not appear. Using Euler’s equation _HH O ¼ LO and equating vector components, we find N ¼ mg þ Is r _ff2 þ tf L t3 ¼ 0 Ff ¼ 0 To express tf, we compute all moments and torque about the disk center of mass and find H c ¼ ½IcŠxB=N ¼ Is L r _ff^eeL þ It _ff^ee3 Lc ¼ Ff r^eeL  tf ^eef  t3 ^ee3 Using t3 ¼ Ff ¼ 0 and _HH c ¼ Lc, we find tf ¼ Is L r _ff2 Note that if _ff were not 0, then we would have a nonzero Ff term. Finally, the normal force magnitude is expressed as N ¼ mg þ 2 Is r _ff2 The polar moment of inertia of a circular disk of mass m and radius r is Is ¼ m 2 r2 150 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

which allows N to be written as N ¼ m g þ r _ff2   Equation (4.15) is valid for any choice of body-fixed coordinate axes with their origin at the body center of mass. Note that the inertia matrix ½I Š is calcu- lated for a specific coordinate system. Let the reference frames B and F both be proper body-fixed coordinate systems. All angular velocities are measured relative to an inertial reference frame N . Let the direction cosine matrix ½FBŠ transform vectors written in the B frame into vectors expressed in the F frame. Therefore, using Eq. (3.17), we can write FH c ¼ ½FBŠ BH c ð4:20Þ Fx ¼ ½FBŠ Bx ð4:21Þ Let us use the following notation. The matrix F ½I Š is the inertia matrix written in the respective F frame and B½I Š is the inertia matrix in the B frame. Equa- tion (4.15) can then be written as BH c ¼ B ½I Š Bx ð4:22Þ which is expanded using Eqs. (3.18), (4.20), and (4.21) to FH c ¼ ½FBŠ B½I Š½FBŠT Fx ¼ F ½I Š Fx ð4:23Þ Thus, an inertia matrix written in the B frame is rewritten into the F frame through the similarity transformation F ½I Š ¼ ½FBŠ B½I Š½FBŠT ð4:24Þ Whereas Eq. (3.17) maps a vector written in one frame into a vector expressed in another frame, Eq. (4.24) performs the analogous operation for matrices. It allows matrices with components taken in one frame to be expressed with components taken in another frame through the use of the corresponding direc- tion cosine matrix between the two frames. Given this similarity transformation, the following question arises. Is there a judicious rotation matrix ½CŠ that will rotate the current coordinate frame B into a new frame F such that the inertia matrix in this F frame is diagonal? The answer to this is yes, this is always possible. Let’s define F ½I Š to be diago- nal. Then Eq. (4.24) can be rewritten as C11 C12 C13 C21 C22 C23 C31 C32 C33 2 4 3 5 B I11 I12 I13 I12 I22 I23 I13 I23 I33 2 4 3 5 ¼ F I1 0 0 0 I2 0 0 0 I3 2 4 3 5 C11 C12 C13 C21 C22 C23 C31 C32 C33 2 4 3 5 ð4:25Þ EULERIAN MECHANICS 151D w U W M S m DO

After carrying out the algebra and equating the proper components, Eq. (4.25) can be reduced to B I11 I12 I13 I12 I22 I23 I13 I23 I33 2 4 3 5 Ci1 Ci2 Ci3 0 @ 1 A ¼ IiiF Ci1 Ci2 Ci3 0 @ 1 A ð4:26Þ for i ¼ 1; 2; 3. Studying Eq. (4.26), it is evident that each row of the desired ½CŠ matrix is an eigenvector of the B½I Š inertia matrix. Assuming that vi are the eigenvectors of B½I Š, we have ½CŠ ¼ ½V ŠT ¼ vT 1 vT 2 vT 3 2 6 4 3 7 5 ð4:27Þ The diagonal entries of the new F ½I Š; called principal inertias, are the eigen- values of the old B½I Š matrix. Note that the eigenvectors will always be ortho- gonal because ½CŠ is an orthogonal rotation matrix. The new set of body-fixed coordinate axes whose inertia matrix is diagonal are called the principal axes. Many analytical problems consider only the simpler case of diagonal inertia matrices because they assume that the appropriate coordinate transformation has already been done. However, in practice it is often difficult to find the exact prin- cipal axes of a given body. Here a set of coordinate axes are typically chosen that are close, but not perfectly aligned with the principal axes. The resulting inertia matrix will have dominant diagonal and small off-diagonal terms. Example 4.3 Find the rotation matrix ½CŠ that will transform the current coordinate frame to a new frame F that diagonalizes the inertia matrix ½I Š ¼ 3 1 1 1 5 2 1 2 4 2 4 3 5 Using MATLAB1, the eigenvector matrix ½V Š and eigenvalue vector L are found to be ½V Š ¼ 0:32799 0:59101 0:73698 0:73698 0:32799 0:59101 0:59101 0:73698 0:32799 2 4 3 5 L ¼ 7:04892 2:30798 2:64310 0 @ 1 A Note that numerical software packages will not necessarily return eigenvectors of unit length. If they are not unit length, they would have to be normalized at this point. In our case the eigenvectors returned are already of unit length. Secondly, we must verify that the set of eigenvectors fv1; v2; v3g form a right-hand set. By 152 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

inspection it is clear that our first eigenvector v1 crossed with the second eigen- vector v2 does not yield the third eigenvector v3, but rather v3. To correct this, we change the sign of v3 by simply reversing the sign of each element of the third column of ½V Š. The proper, orthogonal, right-hand ½V Š matrix is then ½V Š ¼ 0:32799 0:59101 0:73698 0:73698 0:32799 0:59101 0:59101 0:73698 0:32799 2 4 3 5 Because each row of the desired ½CŠ matrix is an eigenvector of ½I Š, then ½CŠ ¼ ½V ŠT ¼ 0:32799 0:73698 0:59101 0:59101 0:32799 0:73698 0:73698 0:59101 0:32799 2 4 3 5 The new principal inertia matrix components are the eigenvalues of ½I Š: F I1 ¼ 7:04892 F I2 ¼ 2:30798 F I3 ¼ 2:64310 4.2.3 Euler’s Rotational Equations of Motion Given the previous results, the equations of motion of a rigid body can be developed in a very straightforward fashion. Using the transport theorem, Euler’s equation is expressed as _HHc ¼ Bd dt ðH cÞ þ x  H c ¼ Lc ð4:28Þ Using Eqs. (4.15) and the fact that ½I Š is constant as seen from the B frame for a rigid body, the derivative of the angular momentum vector H c as seen in the B frame is written as Bd dt H c  ¼ Bd dt ð½I ŠÞx þ ½I Š Bd dt ðxÞ ¼ ½I Š _xx ð4:29Þ The last step in Eq. (4.29) is true because the derivative of the body angular velocity vector x is the same as seen in the B and the N frame: _xx ¼ N d dt xð Þ ¼ Bd dt xð Þ þ x  x ¼ Bd dt ðxÞ ð4:30Þ Substituting Eqs. (4.15) and (4.29) into Eq. (4.28) yields Lc ¼ ½I Š _xx þ x  ð½I ŠxÞ ð4:31Þ EULERIAN MECHANICS 153D w U W M S m DO

Using Eq. (3.23), the famous Euler rotational equations of motion are1 ½I Š _xx ¼ ½ ~xxŠ½I Šx þ Lc ð4:32Þ Note that these equations were developed assuming that the angular momentum vector H c and torque vector Lc are taken about the body center of mass. However, these equations are also valid if the angular momentum and torque are taken about an arbitrary inertial point. By choosing a body-fixed coordinate system that is aligned with the principal body axes, the inertia matrix ½I Š will be diagonal and Eq. (4.32) reduces to2 I11 _oo1 ¼ ðI33  I22Þo2o3 þ L1 ð4:33aÞ I22 _oo2 ¼ ðI11  I33Þo3o1 þ L2 ð4:33bÞ I33 _oo3 ¼ ðI22  I11Þo1o2 þ L3 ð4:33cÞ For the special case in which the body is axially symmetric and no external torques are present, the rotational equations of motion in Eq. (4.33) are reduced to IT _oo1 ¼ ðI33  IT Þo2o3 ð4:34aÞ IT _oo2 ¼ ðI33  IT Þo3o1 ð4:34bÞ I33 _oo3 ¼ 0 ð4:34cÞ where the transverse inertia IT is given by IT ¼ I11 ¼ I22 ð4:35Þ From Eq. (4.34c) it is clear that body angular velocity component o3 along the axis of symmetry will remain constant. Using this fact while differentiating Eq. (4.34a), and substituting Eq. (4.34b), a second-order differential equation for o1 is found: €oo1 þ I33 IT  1  2 o2 3o1 ¼ 0 ð4:36Þ Similarly, we can find the second-order differential equation of o2 to be €oo2 þ I33 IT  1  2 o2 3o2 ¼ 0 ð4:37Þ 154 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

Note that these differential equations have the standard form of undamped oscillators. Therefore, the solution of o1ðtÞ and o2ðtÞ are given by o1ðtÞ ¼ A1 cos opt þ B1 sin opt ð4:38aÞ o2ðtÞ ¼ A2 cos opt þ B2 sin opt ð4:38bÞ where op is defined as op ¼ I33 IT  1   o3 ð4:39Þ Let oi0 be the initial body angular velocity components; then the constants A1 and A2 must be A1 ¼ o10 A2 ¼ o20 ð4:40Þ Differentiating Eqs. (4.38a) and (4.38b) and substituting them into Eqs. (4.34a) and (4.34b), the constants B1 and B2 are found to be B1 ¼ A2 B2 ¼ A1 ð4:41Þ The closed-form solution of the body angular velocity components are given for this axially symmetric, torque-free case through o1ðtÞ ¼ o10 cos opt  o20 sin opt ð4:42aÞ o2ðtÞ ¼ o20 cos opt þ o10 sin opt ð4:42bÞ o3ðtÞ ¼ o30 ð4:42cÞ Note that for this axisymmetric case, we find o2 12 ¼ o2 1 þ o2 2 ¼ o2 10 þ o2 20 ¼ constant ð4:43Þ The o1 and o2 time histories evolve on a circle of magnitude o12, while o3 remains constant. The equations of motion in Eqs. (4.34a) and (4.34b) can be written using Eq. (4.39) as _oo1 ¼ opo2 ð4:44Þ _oo2 ¼ opo1 ð4:45Þ Thus, if the body is prolate (I3 < IT ), then op < 0, and the vector x12 ¼ o1 ^bb1 þ o2 ^bb2 ð4:46Þ rotates about ^bb3 in the opposite direction of the symmetric axis spin o3 ^bb3. If the body is oblate (I3 > IT ), then x12 rotates about ^bb3 in the same direction as EULERIAN MECHANICS 155D w U W M S m DO

o3. The cylindrical motion of the x body frame vector components for an axisymmetric body is illustrated in Fig. 4.5. Example 4.4 Let us develop the equations of motion of the dual-gimbal gyroscope illustrated in Fig. 4.6. A disk is spinning at a rate os about the disk normal axis ^ww3. The rotor can be gimbaled about the axes ^gg1 at a nutation rate on and ^{{3 at a precession rate op. The inertial frame is given by I : f^{{1; ^{{2; ^{{3g. The gimbal frame G: f ^gg1; ^gg2; ^gg3g differs from the inertial frame by the angle f through a single-axis rotation about ^{{3. The wheel frame W: f ^ww1; ^ww2; ^ww3g is attached to the rotor and shares the third frame vector with the gimbal frame because ^ww3 ¼ ^gg3. The wheel and gimbal frame differ through the angle y about the axis ^gg1. The angular velocity of the gimbal frame relative to the inertial frame is given as the sum of the precession and nutation angular velocities xG=I ¼ op ^{{3 þ on ^gg1 The angular velocity of the wheel frame relative to the gimbal frame is simply given by xW =G ¼ os ^ww3 ¼ os ^gg3 Let us express all vectors using the gimbal frame G unit direction vectors. This requires the vector ^{{3 to be expressed in G frame vector components. Note that the rotation matrix [GI] is given by ½GI Š ¼ ½M1ðfފ½M3ðyފ ˆb1 ˆb2 ˆb3 ω ω3 ω2 ω1 ω12 ω3 Fig. 4.5 Illustration of the (x1; x2; x3) components for an axisymmetric body. 156 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

which leads to G ^{{3 ¼ ½GI Š I 0 0 1 0 @ 1 A ¼ G 0 sin y cos y 0 @ 1 A Thus xG=I is written in vector form as xG=I ¼ on ^gg1 þ op sin G ^gg2 þ op cos y ^gg3 The angular velocity of the wheel frame relative to the inertial frame is xW=I ¼ xW=G þ xG=I ¼ on ^gg1 þ op sin y ^gg2 þ ðos þ op cos yÞ ^gg3 Next, let us assume that the gimbal frame is massless. Because of axial symmetry about ^ww3, the wheel inertia is expressed in gimbal frame components as G½IW Š ¼ G I1 0 0 0 I2 0 0 0 I3 2 4 3 5 To develop the rotor differential equations of motion using Euler’s equation _HH ¼ L, the angular momentum is written as H ¼ ½IW ŠxW=I Fig. 4.6 Dual-gimbal gyroscope illustration. EULERIAN MECHANICS 157D w U W M S m DO

Using Euler’s equation and the transport theorem yields _HH ¼ Gd dt ½IW ŠxG=I  þ xG=I  ½IW ŠxW=I  ¼ L where L is the torque acting on the rotor. To evaluate the time derivatives as seen by the G frame, note that Gd dt xW=I  ¼ _oon ^gg1 þ ð _oop sin y þ op cos y _yyÞ ^gg2 þ ð _oos þ _oop cos y  op sin y _yyÞ ^gg3 Gd dt ½IW Š  ¼ 0 Carrying out the remaining vector algebra and equating G-frame vector compo- nents finally lead to the desired dual-gimbal gyroscope equations of motion: I1 _oon þ ðI3  I2Þo2 p sin y cos y þ I3opos sin y ¼ L1 I2 _oop sin y þ ðI1 þ I2  I3Þopon cos y  I3onos ¼ L2 I3 _oos þ I3 _oop cos y þ ðI2  I1  I3Þopon sin y ¼ L3 4.2.4 Kinetic Energy The kinetic energy of a continuous system is shown in Eq. (2.88) to be T ¼ 1 2 M _RRc  _RRc þ 1 2 ð B _rr  _rr dm ¼ Ttrans þ Trot ð4:47Þ Because we are now dealing only with nondeformable rigid bodies, the kinetic energy component Trot describes only the rotational energy of the rigid body B: Trot ¼ 1 2 ð B _rr  _rr dm ð4:48Þ After substituting Eq. (4.8), the rotational kinetic energy of a rigid body is expressed as Trot ¼ 1 2 ð B x  rð Þ  x  rð Þ dm ð4:49Þ After making use of the trigonometric identity a  bÞ  c  a  b  cð Þð ð4:50Þ the rotational kinetic energy is rewritten as Trot ¼ 1 2 x  ð B r  x  rð Þ dm ð4:51Þ 158 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

Note that the integral is exactly equal to H c in Eq. (4.9). Therefore, after using Eqs. (4.9) and (4.15), the rigid body rotational kinetic energy expression is simplified to the form Trot ¼ 1 2 x  H c ¼ 1 2 xT ½I Šx ð4:52Þ The total kinetic energy of a rigid body B is the sum of translational and rota- tional energy as shown in Eq. (4.47) and is given by T ¼ 1 2 M _RRc  _RRc þ 1 2 xT ½I Šx ð4:53Þ To find the work done onto a rigid body B, let us find the derivative of Eq. (4.52): _TTrot ¼ 1 2 _xx  H c þ 1 2 x  _HHc ð4:54Þ Using Eqs. (4.1) and (4.15), this is rewritten as _TTrot ¼ 1 2 _xxT ½I Šx þ 1 2 x  Lc ð4:55Þ After substituting Eq. (4.32) and simplifying the resulting expression, the rota- tional kinetic energy rate for a rigid body is found to be _TTrot ¼ x  Lc ð4:56Þ Using Eq. (2.91), the total kinetic energy rate or power is then given by _TT ¼ F  _RRc þ Lc  x ð4:57Þ If the force vector F is conservative and due to a potential function VcðRcÞ and the torque vector Lc is also conservative and due to a potential function VT , then Eq. (4.57) can be written as dT dt þ dVc dt þ dVT dt ¼ 0 ð4:58Þ which states that the total system energy E ¼ T þ Vc þ VT is conserved in this case. To find the work W done onto the rigid body B between two time steps, Eq. (4.57) is integrated once to yield W ¼ Tðt2Þ  T ðt1Þ ¼ ðt2 t1 F  _RRc dt þ ðt2 t1 Lc  x dt ð4:59Þ EULERIAN MECHANICS 159D w U W M S m DO

Example 4.5 Let us investigate the dynamical system shown in Fig. 4.7 where one solid disk of radius r and mass m is rolling off another disk of radius R without slip. The coordinate frame N ¼ f ^nn1; ^nn2; ^nn3g is an inertial frame with its origin O attached to the center of the stationary disk of radius R. A second coordinate from E ¼ f^eer; ^eey; ^ee3g has the same origin O. Note that er tracks the heading of the disk center O0 relative to O. The angle y specifies the angular position of the disk center O0, while the angle f defines the orientation of the rolling disk relative to the inertial ^nn2 axis. Since the disk rolls without slip, the angles y and f must be related through ðR þ rÞy ¼ rf First, let’s find an expression for the normal force that the lower disk exerts onto the rolling disk. The center of mass position vector of the rolling disk is given through rc ¼ ðR þ rÞ^eer The angular velocity of the disk center of mass to the N frame is xE=N ¼ _yy^ee3 Upon differentiating ^rrc, the inertial velocity and acceleration of the disk center of mass are found to be _rrc ¼ ðR þ rÞ_yy^eey €rrc ¼ ðR þ rÞ_yy2 ^eer þ ðR þ rÞ€yy^eey Fig. 4.7 Disk rolling off another disk. 160 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

Let N be the normal force component acting along the ^eer direction and Ff be the frictional force component acting along the ^eey direction. Considering the constant gravity field case, the total force vector acting on the rolling disk is given by F ¼ N  mg cos yð Þ^eer þ mg sin y  Ff  ^eey The super particle theorem for a continuous body states that m€rrc ¼ F which leads to mðR þ rÞ_yy2 ^eer þ mðR þ rÞ€yy^eey ¼ N  mg cos yð Þ^eer þ mg sin y  Ff  ^eey Equating ^eer and ^eey components, expressions are found for the normal force component N and the friction component Ff : N ¼ mg cos y  mðR þ rÞ_yy2 Ff ¼ mg sin y  mðR þ rÞ€yy To write Ff purely in terms of y and not €yy, we study the rotational motion of the disk about its center of mass. The torque LO0 experienced by the disk is LO0 ¼ rFf For this simple planar disk, Euler’s rotational equations of motion simplify to Ic €ff ¼ LO0 ¼ rFf where Ic is the polar mass moment of inertia of the disk given by Ic ¼ m 2 r2 Using the relationship €ff ¼ R þ r r €yy the angular acceleration €yy is expressed as €yy ¼ 2Ff mðR þ rÞ The friction force component Ff can now be expressed as Ff ¼ 1 3 mg sin y EULERIAN MECHANICS 161D w U W M S m DO

Note that the friction component depends only on the angle y, and not on the disk radius r. The only assumption made here is that the disk inertia satisfies the formula used for Ic. To find at what angle y the rolling disk will leave the lower disk, the normal force component N is set to zero. This leads to the first condition that y must satisfy when the disk leaves the surface: mg cos y ¼ mðR þ rÞ_yy2 Let the scalar function V ðyÞ be the potential function of the rolling disk: V ðyÞ ¼ mgðR þ rÞ cos y The kinetic energy of the disk is given by T ¼ 1 2 m_rrc  _rrc þ 1 2 Ic _ff2 ¼ 3 4 m R þ rð Þ2 _yy2 Recall that f is measured relative to an inertial axis. Because the dynamical system is conservative, the total energy is conserved. The initial energy E0 is E0 ¼ mgðR þ rÞ The total energy at y is E ¼ mgðR þ rÞ cos y þ 3 4 m R þ rð Þ2 _yy2 Equating the two energy states leads to the expression _yy2 ¼ 4 3 g ð1  cos yÞ ðR þ rÞ Substituting this _yy2 into the first condition on y (setting the normal force component N equal to zero) leads to y ¼ cos 1 4 7   ¼ 55:15 deg Thus the normal force component becomes zero at the same angle of y regard- less of the disk mass m and radius r. Again, the underlying assumption is that Ic ¼ ðm=2Þr2 is satisfied. 4.3 Torque-Free Rigid Body Rotation 4.3.1 Energy and Momentum Integrals If no external torques are acting on a system, then Eqs. (2.38), (2.72), and (2.105) show that the total angular momentum vector H is constant. This truth does not depend on whether the system is a single particle, a collection of 162 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

particles, or a continuous body. If no external forces are present, then the rigid body rotational kinetic energy will also be a constant, as seen in Eq. (4.56). Let us write the angular momentum vector H in terms of body frame B components as H ¼ BH ¼ H1 ^bb1 þ H2 ^bb2 þ H3 ^bb3 ð4:60Þ Note that _HH is a derivative taken relative to the inertial reference frame N . Because _HH ¼ 0, the angular momentum vector will appear constant only when seen from the inertial N frame. Relative to the body-fixed B reference frame, the vector H will generally not appear to be constant but rotating. Therefore, the B frame Hi vector components will be time varying. However, the magni- tude of H will be constant in all frames. The present discussion will assume that the body-fixed coordinate axes are all aligned with principal inertia axes; therefore, the rigid body inertia matrix is diagonal. For notational compactness, let us use the shorthand notation Ii ¼ Iii. The angular momentum vector is then written as H ¼ BH ¼ B H1 H2 H3 0 @ 1 A ¼ B I1o1 I2o2 I3o3 0 @ 1 A ð4:61Þ Because the angular momentum magnitude is constant, all possible angular velocities must lie on the surface of the following momentum ellipsoid: H 2 ¼ H T H ¼ I 2 1 o2 1 þ I 2 2 o2 2 þ I 2 3 o2 3 ð4:62Þ Because the kinetic energy is constant, too, the angular velocities must also lie on the surface following energy ellipsoid: T ¼ 1 2 I1o2 1 þ 1 2 I2o2 2 þ 1 2 I3o2 3 ð4:63Þ Thus, the dynamical torque-free rotation of a rigid body must be such that the corresponding body angular velocity vector xðtÞ satisfies both Eqs. (4.62) and (4.63). The geometric interpretation of this is that xðtÞ must lie on the intersec- tion of the momentum and energy ellipsoid surfaces. To more easily visualize the intersection of these two ellipsoids, they are written in terms of the B frame angular momentum vector components Hi instead of the body angular velocity vector components oi. Using Hi as inde- pendent coordinates, the momentum ellipsoid becomes the momentum sphere H 2 ¼ H 2 1 þ H 2 2 þ H 2 3 ð4:64Þ and the energy ellipsoid is written as 1 ¼ H 2 1 2I1T þ H 2 2 2I2T þ H 2 3 2I3T ð4:65Þ EULERIAN MECHANICS 163D w U W M S m DO

where 2IiT p are the corresponding semi-axes. For the torque-free rotation to satisfy both Eqs. (4.64) and (4.65), the energy ellipsoid and the momentum sphere must intersect. The intersection forms a trajectory of feasible xðtÞ, as illustrated in Fig. 4.8. This geometrical interpretation is very useful to make qualitative studies on the nature and limiting properties of large rotations. Clearly, for a given jHj, only a certain range of kinetic energy is possible. For the current discussion, let us hold the angular momentum vector magnitude constant and sweep the kinetic energy through its two extrema. Also, assume that the inertia matrix entries Ii are ordered such that I1  I2  I3 ð4:66Þ With this ordering of inertias, the largest kinetic energy ellipsoid semi-axis 2I1T p occurs about the ^bb1 axis as shown in Fig. 4.8, and the smallest semi- axis is about the ^bb3 axis. Equation (4.65) shows that varying T will only uniformly scale the corresponding kinetic energy ellipsoid. The overall shape and aspect ratio of the ellipsoid will remain the same for each choice in T. Three special energy cases are shown in Fig. 4.9. Because the kinetic energy ellipsoid and the momentum sphere must intersect, the smallest possible T would scale the energy ellipsoid such that its largest semi-axis is equal to H ¼ jHj. The momentum sphere perfectly envelops the energy ellipsoid, as shown in Fig. 4.9a. he only points of intersection are BH ¼ H ^bb1 ð4:67Þ Fig. 4.8 General intersection of the momentum sphere and the energy ellipsoid. 164 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

Fig. 4.9 Special cases of kinetic energy ellipsoid and momentum sphere inter- sections. EULERIAN MECHANICS 165D w U W M S m DO

Therefore, for this minimum kinetic energy case, the rigid body B is spinning purely about its axis of maximum inertia ^bb1, and the corresponding kinetic energy is Tmin ¼ H 2 2I1 ð4:68Þ As the kinetic energy T is enlarged, the next special case arises when the intermediate energy ellipsoid semi-axis is equal to H as shown in Fig. 4.9b. The intersection curve between the momentum sphere and the energy ellipsoid is called the sepratrix. The kinetic energy for any motion along the sepratrix is given by Tint ¼ H 2 2I2 ð4:69Þ Note that any small departure from the pure spin case about the intermediate inertia axis ^bb2 will result in general ‘‘tumbling’’ motion. This result agrees well with the common experience that it is very difficult to throw an object into the air and have it spin purely about the intermediate inertia axis without starting to turn and twist about the other axes. As the kinetic energy is enlarged to its largest possible value, the corre- sponding kinetic energy ellipsoid perfectly envelops the momentum sphere, as shown in Fig. 4.9c. This maximum kinetic energy case Tmax ¼ H 2 2I3 ð4:70Þ corresponds to a pure spin about the smallest axis of inertia ^bb3 because the only intersection point is at BH ¼ H ^bb3 ð4:71Þ For a general rigid body motion as shown in Fig. 4.8, once the initial kinetic energy T and angular momentum vector H are established, the angular velocity vector x will theoretically trace out a particular intersection curve forever. The assumption here is that the body B is perfectly rigid and that no energy is lost (i.e., no internal dampening, heat loss, etc.). However, this assumption is highly idealistic. No body is perfectly rigid and devoid of internal damping. Therefore, real rigid bodies spinning in a torque-free environment do actually lose energy, though typically at a slow rate. Figure 4.10 shows a family of energy ellipsoid and momentum sphere inter- sections for varying levels of kinetic energy. Note that except for the sepratrix case, all feasible xðtÞ paths form closed trajectories. A typical example of a torque-free rigid body rotation would be a rigid satellite launched into an Earth orbit. Once the satellite is spun up about a particular axis and the thrusters are shut down, the satellite will not experience any external torques and the H 166 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

vector will remain constant. We are ignoring here the affects of atmospheric and solar drag. Let’s study what happens if a satellite is spun up about the axis of least inertia ^bb3. For a given angular momentum, this corresponds to the maximum kinetic energy case. Because any real rigid body will lose energy over time simply due to internal damping, this satellite’s energy is expected to decrease over time. Figure 4.10 shows how the satellite will start to ‘‘wobble’’ about the ^bb3 axis as the energy ellipsoid is reduced. After some time the xðtÞ curves will cross the sepratrix and the satellite will start to ‘‘wobble’’ about the axis of maximum inertia ^bb1. Ultimately, as the energy approaches the minimum energy ellipsoid, the satellite will be spinning purely about the ^bb1 axis. There- fore, under the presence of a negative energy rate, only the spin about the axis of maximum inertia is a stable spin. The pure spin case about ^bb3 will become unstable over time. This behavior is demonstrated in nature in that all planets are essentially spinning about their axis of maximum inertia. This fact was rediscovered during early space explorations when Explorer 1 was launched into orbit spin- ning about its axis of least inertia. It took less than a fraction of an orbit before it started to tumble. If a body is performing a near spin about the axis of maximum inertia ^bb1, then the x body frame vector components will rotate about the ^bb1 or H1 axis in a counterclockwise direction. For near-spin scenarios about the axis of mini- mum inertia ^bb3, the x vector components will revolve about H3 in the clock- wise direction. Fig. 4.10 Family of energy ellipsoid and momentum sphere intersections. EULERIAN MECHANICS 167D w U W M S m DO

Example 4.6 This example explores the polhode illustration for the two particular body geometries where the body is either axisymmetric with I1 ¼ I2, or the body principal inertias are all equal with I1 ¼ I2 ¼ I3. This last condition occurs if the body is a homogeneous cube and the body frame B is a principal coordinate frame, or the body is a homogeneous sphere with any body fixed frame B. We first consider the torque-free motion of an axisymmetric spacecraft. Without loss of generality, we assume that I1 ¼ I2. Using Eq. (4.33), this results in _oo3 ¼ 0 and o3ðtÞ ¼ o3ðt0Þ being a constant value. Using Eq. (4.65), the kinetic energy ellipsoid has identical semi-axis in the H1 and H2 direction, with a unique semi-axis in the H3 direction. Intersecting this energy constraint surface with the spherical momentum constraint surface leads to circular intersec- tion trajectories about the H3 as illustrated in Fig. 4.11a. Because I1 ¼ I2 > I3 in this case, the resulting x trajectories orbit the H3 axis in a clockwise fashion. On the momentum sphere (H1, H2) equatorial plane an interesting motion occurs. Note that here H3 ¼ o3 ¼ 0, and for this axisymmetric body we know that o3ðtÞ ¼ o3ðt0Þ ¼ 0 Studying Euler’s rotational equations of motion in Eq. (4.33), we find that _oo1 ¼ _oo2 ¼ 0 in this case. This, if o3 ¼ 0, then o1 and o2 will have constant values. In the polhode plot this constant angular rate condition is illustrated by having dots on the momentum sphere equator instead of trajectories. Next, let us examine Fig. 4.11 Polhode plot illustration for particular principal inertia configurations. 168 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

the condition where all principal inertias are equal. This could be caused by a spherically symmetric body or a homogeneous cube with a principal coordinate frame B. In this case, Euler’s rotational equations of motion in Eq. (4.33) indicated that _oo1 ¼ €oo2 ¼ €oo3 ¼ 0 and that all of the oiðtÞ will be constant. Geometrically, this condition indicates that the energy ellipsoid has become a sphere that is identical to the momentum sphere. Thus, every point on the sphere is an intersection of the momentum and energy constraints. The polhode plot in Fig. 4.11b illustrates this behavior by plotting a discrete set of solutions as points. 4.3.2 General Free Rigid Body Motion In this section we would like to derive the general rotational equations of motion for a rigid body free of any external torques. The attitude coordinates are chosen to be the (3-2-1) Euler angles, also known as the yaw, pitch, and roll angles ðc; y; fÞ. However, the method used here to derive the equations of motion could be used for any set of attitude coordinates presented in Chapter 3. Assume the rigid body has a coordinate system B attached to it that is aligned with the principal inertia axes, and let N be an inertial reference frame. For the free motion of a rigid body, the angular momentum vector H will remain constant. Using a trick due to Jacobi, we can therefore always align our inertial space unit axes ^nni such that ^nn3 is aligned with H: H ¼ NH ¼ H ^nn3 ¼ N 0 0 H 0 @ 1 A ð4:72Þ The direction cosine matrix ½BN Š translates any vector written in N frame components into a vector in B frame components as shown in Eq. (3.17): BH ¼ ½BN Š NH ð4:73Þ The direction cosine matrix in terms of the (3-2-1) Euler angles is given in Eq. (3.34). After using Eq. (4.72) to carry out the matrix multiplication and equat- ing the resulting B frame components to Eq. (4.60), we obtain H1 ¼ H sin y ¼ I1o1 ð4:74aÞ H2 ¼ H sin f cos y ¼ I2o2 ð4:74bÞ H3 ¼ H cos f cos y ¼ I3o3 ð4:74cÞ EULERIAN MECHANICS 169D w U W M S m DO

which can be solved for the body angular velocity vector x as H I1 sin y  H I2 sin f cos y  H I3 cos f cos y 0 B B B B B B B @ 1 C C C C C C C A ¼ o1 o2 o3 0 B @ 1 C A ð4:75Þ To find an expression for the individual Euler angle rates ð _cc; _yy; _ffÞ, we substi- tute Eq. (3.56) into Eq. (4.75) and obtain the following equations of motion for a torque-free rigid body: _cc ¼ H sin2 f I2 þ cos2 f I3   ð4:76aÞ _yy ¼ H 2 1 I3  1 I2   sin 2f cos y ð4:76bÞ _ff ¼ H 1 I1  sin2 f I2  cos2 f I3   sin y ð4:76cÞ Note that the precession rate _cc in Eq. (4.76a) cannot be positive, while the nutation rate _yy and _ff can have either sign. 4.3.3 Axisymmetric Rigid Body Motion The equations of motion in Eqs. (4.76) are valid for a general rigid body with the body-fixed axes aligned with the principal inertia axes. Now we would like to study a particular case of these equations in which the rigid body is axisymmetric. Without loss of generality, assume that I2 ¼ I3. Then the yaw, pitch, and roll angle rates in Eqs. (4.76) simplify to _cc ¼  H I2 ð4:77aÞ _yy ¼ 0 ð4:77bÞ _ff ¼ H I2  I1 I1I2   sin y ð4:77cÞ Having chosen the inertial angular momentum vector NH to be in the positive ^nn3 direction, the precession rate _cc will be a positive constant for an axisym- metric rigid body. The relative spin rate _ff is also a constant-like precession rate. However, the sign of _ff in Eq. (4.77c) depends on the relative size of I1 and I2 and on the pitch angle y. On the other hand, the pitch rate _yy is zero for axisymmetric rigid body rotations; therefore, the pitch angle y will remain constant throughout the motion. 170 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

Let O ¼ o1 be the body spin rate about its axis of symmetry ^bb1 as shown in Fig. 4.12. Using Eq. (4.74a), it is expressed as O ¼ H I1 sin y ð4:78Þ For a positive pitch angle 0  y  p=2, the body spin rate O about the symme- try axis must be positive. Instead of being written in terms of the angular momentum magnitude H, the precession rate _cc and the relative spin rate _ff can now be expressed in terms of O as _cc ¼  I1 I2 O sin y ð4:79Þ _ff ¼ I2  I1 I2 O ð4:80Þ The angular momentum vector along the axis of symmetry ^bb1 is labeled in Fig. 4.12 as H a. It is defined as H a ¼ I1O^bb1 ð4:81Þ As is easily seen in Fig. 4.12, for positive pitch angles y and I2 > I1, the axisymmetric rigid body B will have a positive spin rate about ^bb1. Because the pitch angle y is shown to remain constant during this torque- free rotation, the resulting motion can be visualized by two cones rolling on each other. Figure 4.13 shows the two cases in which either I2 > I1 or I2 < I1. Fig. 4.12 Angular velocity and momentum vector relationship for the case I2 > I1. EULERIAN MECHANICS 171D w U W M S m DO

The space cone is fixed in space and its cone axis is always aligned with the angular momentum vector H. The cone angle b is defined as the angle between the vectors H and x. The body cone axis is aligned with the body axis ^bb1 and has the cone angle a; which is the angle between x and ^bb1. If I2 > I1; then the body cone will roll on the outside of the space cone as shown in Fig. 4.13a, and the resulting motion is called a direct precession. If I2 < I1, then the space cone will lie on the inside of the body cone, and the resulting motion is called a retrograde precession. 4.4 Dual-Spin Spacecraft Although for a single rigid body only the principal axes spin about the axes of largest inertia is passively stable in the presence of energy loss, in many applications it would be of great advantage to be able to stabilize a spacecraft spin about any principal axes, regardless of the corresponding axes inertia. The dual-spin spacecraft is a simple system where passive attitude stability is achieved by adding a single fly-wheel to the rigid spacecraft. For example, consider the typical cylindrical geostationary communications satellite illu- strated in Fig. 4.14. To have the sensor or communications antenna of compo- nent 2 to continuously point at the Earth, it would require a spin rate of one revolution per orbit. However, spinning about the least axis of inertia is unstable if energy losses are considered. Instead, the lower component 1 can spin at a different rate x1 to stabilize the system. However, this spin rate magnitude must be chosen very carefully. Although the dual-spin concept can stabilize any principal axis spacecraft spin, if used incorrectly it can also be the cause of instability. Beyond the geostationary communication satellites, another application of the dual-spin concept is the interplanetary Galileo spacecraft, which traveled to Fig. 4.13 Conic illustration of direct and retrograde precession of a freely rotating rigid body. 172 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

Jupiter. Here the main antenna needed to continuously point back at the Earth. To stabilize this orientation, half of the body rotated at three revolutions per minute, while the sensor components rotated very slowly to align the communication antenna to the slowly changing spacecraft-to-Earth vector. Thus, the dual-spin concept is not restricted to systems where the auxiliary rotating component is a spacecraft internal fly-wheel; it is also possible to have part of the main spacecraft structure rotate relative to the remaining structure to achieve an equivalent stabilizing effect. 4.4.1 System Equations of Motion To develop the dual-spin system equations of motion, assume that the rotat- ing fly-wheel is aligned with the first principal axis ^bb1 of the main spacecraft component as illustrated in Fig. 4.15. Let x ¼ xB=N be the body angular velo- city of the main craft, while xW=B ¼ O^bb1 is the angular velocity of the fly- wheel relative to the spacecraft. The total angular momentum is then given by H ¼ ½IsŠx þ ½IW ŠðO^bb1 þ xÞ ð4:82Þ where [Is] is the inertia matrix of the main spacecraft system, while [IW] is the inertia of the fly-wheel component. Note that in the case where the dual-spin craft is rotating an entire segment of the main craft, then [IW] would be the equivalent spacecraft component inertia matrix. Using Euler’s equation _HH ¼ L, where L is the external torque acting on this system, we find _HH ¼ L ¼ ð½IsŠ þ ½IW ŠÞ _xx þ ½ ~xxŠð½IsŠ þ ½IW ŠÞx þ ½IFŠð _OO^bb1Þ þ ½ ~xxŠ½IW ŠðO^bb1Þ ð4:83Þ Fig. 4.14 Geostationary dual-spin communication spacecraft illustration. EULERIAN MECHANICS 173D w U W M S m DO

Next, let us define the combined inertia matrix [I] as ½I Š ¼ ½IsŠ þ ½IW Š ð4:84Þ Assuming no external torque is present, and that the body frame B is a princi- pal frame of both the main body and the fly-wheel component, the equations of motion are written as I1 _oo1 ¼ ðI2  I3Þo2o3  IWs _OO ð4:85aÞ I2 _oo2 ¼ ðI3  I1Þo1o3  IWso3O ð4:85bÞ I3 _oo3 ¼ ðI1  I2Þo1o2 þ IWso2O ð4:85cÞ where IWs is the fly-wheel inertia about the spin axis ^bb1. For the dual-spin spacecraft concept, the spin rate O is typically held at a constant value. The equations of motion thus simplify to I1 _oo1 ¼ ðI2  I3Þo2o3 ð4:86aÞ I2 _oo2 ¼ ðI3  I1Þo1o3  IWso3O ð4:86bÞ I3 _oo3 ¼ ðI1  I2Þo1o2 þ IWso2O ð4:86cÞ Next, let us examine the equilibrium spin configurations. This requires determining what xe ¼ oe1 ^bb1 þ oe2 ^bb2 þ oe3 ^bb3 and O 6 ¼ 0 results in _xxe ¼ 0. Studying Eq. (4.86), this is only feasible if both oe2 and oe3 are zero. Thus, the dual-spin spacecraft equilibrium condition is xe ¼ oe1 ^bb1 ð4:87Þ All of the preceding developments assume that the fly-wheel spin axis is aligned with ^bb1. If it is aligned with another axis, the condition of Eq. (4.87) would change accordingly. Fig. 4.15 Illustration of a dual-spin spacecraft with a rotating component aligned with the first spacecraft principal axis. 174 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

4.4.2 Linear Stability Conditions To examine the stability of a dual-spin spacecraft with a constantly rotating component, we linearize the attitude motion about the equilibrium rotation xe ¼ oe1 ^bb1. Let the actual angular velocity be given by x ¼ xe þ dx ð4:88Þ where dx ¼ ðdo1; do2; do3ÞT is the departure motion. Substituting this x into the equations of motion in Eq. (4.86) and dropping higher-order terms lead to the linearized departure equations of motion: d _oo1 ¼ 0 ð4:89aÞ d _oo2 ¼ I3  I1 I2 oe1  IWs I2   do3 ð4:89bÞ d _oo3 ¼ I1  I2 I3 oe1 þ IWs I3   do2 ð4:89cÞ Because do1 is constant from Eq. (4.89a), the actual angular rate about the fly-wheel spin axis ^bb1 is also constant o1ðtÞ ¼ oe1 þ do1ðt0Þ ¼ constant ð4:90Þ and the o1ðtÞ behavior is thus marginally stable. Determining the stability of do2 and do3 is slightly more complex because their first-order differential equations are coupled. However, note that the do1 departure motion does not couple to the do2 and do3 motion, and that d _oo2 only depends on do3, while d _oo3 solely depends on the state do2. This allows us to rewrite the two coupled first-order differential equations into two uncoupled second-order differential equations. To do this, let us differentiate Eq. (4.89b) to yield d €oo2 ¼ I3  I1 I2 oe1  IWs I2   d _oo3 ð4:91Þ Next we substitute Eq. (4.89c) to obtain the uncoupled differential equation of do2: d €oo2 þ I1  I3 I2 oe1 þ IWs I2 O   I1  I2 I3 oe1 þ IWs I3 O   |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} k do2 ¼ 0 ð4:92Þ For the departure motion do2 to be neutrally stable, the inertias and spin rate O must be chosen such that k > 0. The same process can be performed to solve for the uncoupled second-order differential equation of do3. However, this process leads to exactly the same stability condition (i.e., d €oo3 þ kdo3 ¼ 0). EULERIAN MECHANICS 175D w U W M S m DO

Either both do2 and do3 departure motions are neutrally stable, or they are both unstable. Using the nondimensional spin rate ^OO ¼ O oe1 ð4:93Þ the dual-spin spacecraft stability condition (necessary for bounded oscillatory motion) is finally rewritten as k ¼ o2 e1 I2I3 I1  I3 þ IWs ^OO   I1  I2 þ IWs ^OO   > 0 ð4:94Þ Let us first investigate the condition in Eq. (4.94) for the special case where the fly-wheel is fixed relative to the main spacecraft, and thus ^OO ¼ 0. This case represents a situation where the entire system is acting as a single rigid body, and thus we should obtain the preceding stability conditions of a rigid craft rotating about a principal axis. With ^OO ¼ 0 the parameter k simplifies to k ¼ o2 e1 I2I3 I1  I3  I1  I2ð Þ > 0 ð4:95Þ This condition is true if either I1 > I3 I1 > I2 ) spin about axis of maximum inertia or I1 < I3 I1 < I2 ) spin about axis of maximum inertia Note that this linear stability analysis does not consider stability in the presence of energy dissipation. A spin about the intermediate axis of inertia is always unstable. Next we study further the general dual-spin stability condition in Eq. (4.94) for the typical case where ^OO 6 ¼ 0. Given the principal inertias I1, I2, and I3, we would like to determine what range of values ^OO leads to a stable system. There are two critical wheel speeds that cause the inequality conditions to be either true or false: ^OO1 ¼ I3  I1 IWs ð4:96aÞ ^OO2 ¼ I2  I1 IWs ð4:96bÞ 176 ANALYTICAL MECHANICS OF SPACE SYSTEMSD w U W M S m DO

The dual-spin stability conditions are then simplified to: Condition 1: ^OO > ^OO1 and ^OO > ^OO2 ð4:97aÞ Condition 2: ^OO < ^OO1 and ^OO < ^OO2 ð4:97bÞ Let us examine these two stability conditions for three types of principal axis rotations. First, consider the maximum inertia spin scenario where I1 > I2 > I3. Because I1 is the largest inertia, both ^OO1 and ^OO2 are negative values with ^OO1 < ^OO2. Condition 1 thus results in the requirement that ^OO > O2, whereas condition 2 requires that ^OO < ^OO1. The resulting range of stabilizing ^OO values is graphically illustrated in Fig. 4.16a. Note that because the maximum inertia spin is stable in the absence of the fly-wheel, the feasible ^OO range must include the origin. Any positive ^OO actually strengthens the stability. If the wheel speed is negative, then beyond the critical speed ^OO2 the system becomes unstable. Once the speed ^OO is less than the ^OO1, the fly-wheel stability once again is dominant, and the system is stable. The stabilizing fly-wheel spin rates for the case where the spacecraft is to rotate about an intermediate axis of inertia are illustrated in Fig. 4.16b. As expected, note that here the origin is not included in the stabilizing ^OO range. Without the stabilizing effect of the fly-wheel, the single rigid body spin about the intermediate axis of inertia is unstable. Once the magnitude of ^OO exceeds the critical speeds, then the intermediate axis spin becomes stable. Last we explore the stabilizing wheel speed range for a minimum axis of inertia spin where I3 > I2 > I1. The admissible wheel speeds are illustrated in Fig. 4.16c. Because the minimum inertia spin case is linearly stable in the Fig. 4.16 Stabilizing ^OO range (shaded) illustration. EULERIAN MECHANICS 177D w U W M S m DO