Attitude Dynamics Fundamentals

Hanspeter Schaub

Simulated Reprint from Encyclopedia of Aerospace Engineering Chichester, UK, 2010, pp. 3181–3198 Article Number: eae295

Section 5.5.1 Attitude Dynamics Fundamentals

Hanspeter Schaub Aerospace Engineering Science Department, 431 UCB, University of Colorado, Boulder, CO 80309-0431, USA Phone: (303) 492-2767 Fax: (303) 492-2825 Email: [email protected]

File Name: attitude-fundamental.tex Software: LATEX Edited on: February 9, 2010

The rotational motion of a rigid body or system of rigid bodies is a fundamental ﬁeld for the study of spacecraft pointing problems. This Chapter presents the fundamental aspects of de- 1 Rigid Body Kinematics 1 1.1 Rotating Coordinate Frames ...... 1 scribing the orientation, angular momentum, energy and differ- 1.2 Attitude Parameters ...... 2 ential equations of motion of a rigid body. After discussing the 1.2.1 Direction Cosine Matrix ...... 2 rigid body kinematics, the inertia properties of a body about ar- 1.2.2 Euler Angles ...... 3 bitrary reference points are developed. The angular momentum 1.2.3 Principal Rotation Parameters .... 4 and energy of a rigid body are critical to discussing the equa- 1.2.4 Quaternions or Euler Parameters ... 4 1.2.5 Other Attitude Parameters ...... 5 tions of motion, as well as investigating the stability of torque- free spin solutions. Next, passive methods of attitude stabiliza- 2 Rigid Body Inertia 5 tion are discussed. Here the spacecraft angular momentum, or 2.1 Inertia Matrix Deﬁnition ...... 6 external torques such as the gravity gradient torque, are used 2.2 Principal Coordinate System ...... 6 to stabilize the orientation. These fundamental kinematics and 2.3 Parallel Axis Theorem ...... 6 kinetics properties form the foundation for follow-on Chapters 3 Angular Momentum 7 discussing more complex pointing scenarios including systems with multiple rigid bodies (reaction wheels, etc.), attitude esti- 4 Kinetic Energy 7 mation, as well active attitude pointing problems.

5 Rotational Equations of Motion 8 Keywords: Rigid body pointing, angular momentum, rotational 5.1 Euler’s Rotational Equations ...... 8 equations of motion, dual-spin spacecraft, gravity gradient torques 5.2 Principal Axis Spin Stability ...... 8 5.3 Numerical Simulation of Rigid Body Motion 8

6 Torque-Free Response 9 1 RIGID BODY KINEMATICS 6.1 Angular Velocity Solutions ...... 9 6.1.1 Axisymmetric Inertia Case ...... 9 This Chapter first discusses the angular rotation and ori- 6.1.2 General Inertia Case ...... 9 entation coordinates used to describe the motion of a rigid 6.2 Coning Motion ...... 9 body. By fixing a coordinate frame to a rigid body, under- 6.3 Polhodes ...... 10 standing how the coordinate frame orientation evolves over 7 Dual-Spin Stabilization 11 time is equivalent to understanding how the rigid body atti- 7.1 Equations of Motion ...... 11 tude behaves. 7.2 Linear Equilibrium Stability ...... 11 1.1 Rotating Coordinate Frames ˆ ˆ ˆ 8 External Torques for Passive Control 12 The coordinate frame : , b1, b2, b3 illustrated in B {O } 8.1 Gravity Gradient Torques ...... 12 Figure 1 is defined through its origin and the three mutu- O 8.2 Atmospheric Torques ...... 14 ally orthogonal unit direction vectors bˆ , bˆ , bˆ . A right- 8.3 Magnetic Torques ...... 14 { 1 2 3} handed coordinate frame satisfies bˆ b = bˆ . If this co- 1 ⇥ 2 3 ordinate frame is attached to a rigid body, then describ- 9 Conclusion 14 B ing the orientation of the rigid body is equivalent to studying References 15 the orientation of . For the study of attitude dynamics, the B translation of the rigid body is not of interest. As a result

1 Shuttle cockpit to tail, then this vector appears stationary as bˆ 2 seen by a Shuttle ﬁxed observer. However, this same r is ro- bˆ 3 !2 tating as seen by an earth-ﬁxed observer. A left super-script label is used on the time differential operator to denote the nˆ !3 3 B observer frame. The transport theorem is used to map a time N bˆ derivative seen by a frame into the equivalent derivative nˆ 2 1 B OB !1 seen by another frame [Likins, 1973, Schaub and Junkins, ON N nˆ 1 2010]:

Nd Bd (r)= (r)+! / r (4) dt dt B N ⇥ Figure 1: Coordinate Frame Illustration The vector r and frames and are only placeholders in B N Eq. (4). The transport theorem applies equally if any other coordinate frames are thus often defined through their unit vector or frames are substituted into this expression. direction vectors only. If no frame label is provided, then the time derivative is As with positions, orientations can only be described rel- assumed to be taken with respect to an inertial frame. ative to a reference orientation. Let be an inertial (non- N accelerating) coordinate frame. By defining the bî direction Nd vectors with respect to , we are able to describe the orien- x˙ (x) (5) N ⌘ dt tation of the body with respect to . B N The angular motion of relative to is described through This is by far the most common time derivative of a vector as B N the angular velocity vector ! / Newton’s and Euler’s laws require inertial derivatives to be B N taken. Further, let x be expressed in frame components as ˆ ˆ ˆ T B ! / = !1b1 + !2b2 + !3b3 (1) Bx =(x1,x2,x3) Notice that B N This vector is the instantaneous angular rotation vector of B x˙ 1 body relative to , and is typically expressed in body Bd B N x˙ 2 (x) x˙ (6) frame vector components. If only the and frames are 0 1 ) dt 6) B N x˙ 3 considered, the ! vector is often written simply as !. / @ A While we will seeB N that attitude coordinates sets are not vec- The time derivative of the angular velocity vector ! / tors, and do not abide by vector addition laws, amazingly has a special property worth noting: B N the angular velocity vector is truly a vector. Thus, if three frames , and are present, their angular velocities re- Bd Bd A B N !˙ / = ! / + ! / ! / = ! / (7) late through B N dt B N B N ⇥ B N dt B N ! / = ! / + ! / (2) In other words, if ! is the angular velocity vector between A N A B B N two particular frames, then the time derivative of ! as seen If we wish to express a vector ! as a 3 1 matrix, we ⇥ by either of these frames is the same. must specify with respect to which frame the vector components have been taken. If frame components are used as B 1.2 Attitude Parameters in Eq. (1), then the left super-script notation is used in this Having shown how a rigid body rotation can be described Chapter: through !, these next sections focus on how the rigid body orientation is described. To describe the three-dimensional B ! 1 orientation of rigid body a minimum of three coordinates B! = ! (3) 2 or attitude parameters are required. However, any minimal 0! 1 3 three-parameter set will encounter singularities that must be @ A These frame declarations can become cumbersome after a considered, while redundant sets of more than three parame- while if only 2 frames are present. If no label is made on a ters can avoid singularity issues at the cost of additional pa- matrix representation of a vector, then body frame compo- rameter constraints. nents are implied. 1.2.1 Direction Cosine Matrix Having defined a rotating frame , let us briefly discuss The 3 3 rotation matrix [BN] is a fundamental way to B ⇥ how to differentiate a vector r expressed in vector compo- express the orientation of with respect to . Besides this B B N nents. To discuss the time evolution of r, an observer frame 2-letter notation, T N is also used to represent the same ma- must be specified. For example, if r points from the Space trix, or [C] is oftenB used if only a single body frame is con-

2 sidered. The rotation matrix is determined through To map Br into equivalent vector components, the [NB] N rotation matrix is used: bˆ1 nˆ 1 bˆ1 nˆ 2 bˆ1 nˆ 3 · · · Nr =[NB] Br (16) [BN]= bˆ2 nˆ 1 bˆ2 nˆ 2 bˆ2 nˆ 3 (8) 2ˆ · ˆ · ˆ · 3 b3 nˆ 1 b3 nˆ 2 b3 nˆ 3 The inverse coordinate transformation is 4 · · · 5 1 T Let ↵ be the angle between bˆ and nˆ , then bˆ nˆ = Br =[NB] Nr =[NB] Nr =[BN] Nr (17) ij i j i · j cos ↵ij. The rotation matrix elements BNij are thus the cosines of the relative unit direction angles. As a result the 1.2.2 Euler Angles rotation matrix is also referred to as the Direction Cosine Ma- Euler angles are a minimal set of three attitude coordi- trix (DCM). nates. All minimal orientation description contain attitudes If the unit direction vectors bˆ are given in frame com- i N where the description or the associated differential kinematic ponents, or the unit direction vectors nˆ are expressed in i B equations become singular. The Euler angles describe the frame components, then the DCM is found through orientation of relative to through 3 sequential one-axis B N T rotations. There are 12 different sets of Euler angles which (Nbˆ1) T differ through the sequence of one-axis rotations. The popu- [BN]= ( bˆ ) = Bnˆ Bnˆ Bnˆ (9) 2 N 2 3 1 2 3 lar yaw , pitch ✓ and roll angles are a (3 2 1) Euler ( bˆ )T N 3 ⇥ ⇤ angle sequence as illustrated in Figure 2. The primed axis 4 5 The proper DCM (frame satisfy the right-hand rule) is or- labels denote the intermediate axes of the rotation sequence. Starting with frame orientation, the yaw axis is deﬁned as thogonal and has a determinant of +1. The inverse is simply N positive rotation about the current 3-axis (nˆ 3 or bˆ3), the pitch given by the transpose operator: ˆ ˆ is a rotation about the current 2-axis (b20 or b200), and the roll 1 T ˆ ˆ [BN] =[BN] =[NB] (10) is a rotation about the ﬁnal 1-axis (b100 or b1).

The time evolution of the DCM orientation measure is cap- ˆ ˆ b100 tured through the differential kinematic equation nˆ 1 ✓ b100 ˆ bˆ1 b20 ˆ ˆ b20 ˆ ˙ nˆ b10 ˆ b10 ˆ [BN]= [!˜ / ][BN] (11) 2 b200 b200 B N bˆ bˆ3 ˆ 2 ˆ bˆ bˆ b300 b300 where the tilde matrix notation is nˆ 3 30 30 0 ! ! 3 2 Figure 2: (3-2-1) Euler Angle Sequence [!˜ / ]= !3 0 !1 (12) B N 2 3 ! ! 0 th 2 1 The one-dimensional rotations about the i body axis can 4 5 be describe using rotation matrices [M (✓)] as: and represents a matrix representation of the vector cross i product through [!˜]a ! a. 10 0 ⌘ ⇥ Assume that the DCMs [AB] and [BN] are given. To add [M1(✓)] = 0 cos ✓ sin ✓ (18a) these 2 orientations (sequentially rotate ﬁrst from to , 20 sin ✓ cos ✓3 N B and then rotate from to ), and obtain the attitude of rel- B A A 4cos ✓ 0 sin ✓5 ative to , we simply multiply these DCM with each other: N [M2(✓)] = 01 0 (18b) 2 3 [AN]=[AB][BN] (13) sin ✓ 0 cos ✓ 4 cos ✓ sin ✓ 05 This simple DCM attitude addition property is a very fun- [M3(✓)] = sin ✓ cos ✓ 0 (18c) damental method to add successive orientations. To subtract 2 0013 [BN] from [AN], and get the relative attitude of relative A 4 5 to , we use Using the DCM addition property in Eq. (13), the (3-2-1) B yaw, pitch and roll rotations lead to 1 T [AB]=[AN][BN] =[AN][BN] =[AN][NB] (14) [BN]=[M1()][M2(✓)][M3( )] (19) A common use of the DCM is to perform three- or dimensional coordinate transformations. Assume a vector r c✓c c✓s s✓ is given in terms of frame coordinates as B [BN]= ss✓c cs ss✓s + cc sc✓ (20) 2 3 r = r bˆ + r bˆ + r bˆ (15) cs✓c + ss cs✓s sc cc✓ 1 1 2 2 3 3 4 5

3 The inverse transformations from the direction cosine matrix 1.2.3 Principal Rotation Parameters [BN] to the ( ,✓,) angles are While the Euler angles utilize three sequencial rotations to map from to , it is possible to rotate between two arbi- BN N B = arctan 12 (21a) trary frame using a single rotation about the principal axis eˆ BN ✓ 11 ◆ by the principal angle . This property is called Euler’s Prin- ✓ = arcsin (BN ) (21b) ciple Rotation theorem.[Whittaker, 1965 reprint] The princi- 13 BN pal rotation axis eˆ has the special property that = arctan 23 (21c) BN ✓ 33 ◆ Beˆ =[BN] Neˆ (25) A good description of alternate sets of Euler angles can be Thus eˆ is the eigenvector of [BN] corresponding to the +1 found in Junkins and Turner [1986]. eigenvalue. To add or subtract two orientations given in terms of Eu- The principal rotation parameters eˆ and are not unique. ler angles, these angles are mapped to the equivalent DCM An orientation can be represented through (using Eq. (19) for the (3-2-1) Euler angles) and then using (eˆ, )( eˆ, )(eˆ, 0)(eˆ, 0) (26) the simple DCM attitude addition and subtraction properties in Eqs. (13) and (14). Having obtained the desired DCM, the where 0 =2⇡ . While one principal angle describes equivalent (3 2 1) Euler angles can be extracted using the short rotation, say 30 degrees, the alternate principal an- Eq. 21. gle 0 describes the long rotation representation, say 330 de- The differential kinematic equations relate the (3-2-1) grees. Euler angle rates to the body angular velocity vector ! The DCM in terms of (eˆ, ) is given by[Shuster, 1993] through[Shuster, 1993] T [BN] = cos [I3 3]+(1 cos ) eêˆ sin [e˜] (27) ⇥ ˙ sin cos B ! 0 cos ✓ cos ✓ 1 The inverse transformation from the rotation matrix [BN] to ✓˙ = 0 cos sin !2 (22) 0 1 2 3 0 1 the principal rotation parameters is ˙ 1sin tan ✓ cos tan ✓ !3 1 @ A 4 5 @ A = arccos (trace([BN]) 1) (28a) [B( ,✓,)] ± 2 ✓ ◆ | {z } Note that the ! vector components here must be taken with BN23 BN32 1 respect to the frame because the Euler angles define the eˆ = BN31 BN13 (28b) B 2sin0 1 attitude of . Further, these kinematic differential equations BN12 BN21 B are singular if the 2nd rotation angle ✓ is 90 degrees. This @ A ± The four different principal rotation parameter sets are ob- singular behavior is true for all Euler angle sequences which tained by using either sign in Eq. (28a) and noting the alter- do not repeat a rotation axis (asymmetric Euler angles). If a nate 0 =2⇡ solution to the arccos function. rotation axis is repeated such as a (3-1-3) sequence, then the attitude coordinates are called symmetric Euler angles. In 1.2.4 Quaternions or Euler Parameters this case it is also the 2nd rotation angle ✓ which determines 2 The quaternions, also called the Euler Parameters (EPs), a singular orientation, but it must be ✓ = 0 or 180 degrees. 2 are a popular redundant attitude coordinate set which is sin- Besides the (3-2-1) yaw, pitch and roll Euler angles, the (3- gularity free in its attitude representation. The Euler param- 1-3) Euler angles are also popular to describe spacecraft or eter set =( , , , ) is defined in terms of the prin- orbit plane orientations. The DCM in terms of (3-1-3) Euler 0 1 2 3 cipal rotation parameters as angles (⌦,i,!) is 0 = cos (/2) (29a) c!c⌦ s!cis⌦ s!cic⌦+c!s⌦ s!si 1 = e1 sin (/2) (29b) [BN]= s!c⌦ c!cis⌦ c!cic⌦ s!s⌦ c!si (23) 2 sis⌦ sic⌦ ci 3 2 = e2 sin (/2) (29c) 4 5 3 = e3 sin (/2) (29d) while the differential kinematic equation is Note that 0 only depends on the principal rotation angle, ˙ sin ! cos ! B and is not influenced by the principal rotation axis, and is ⌦ sin i sin i 0 !1 thus called the scalar EP component. The remaining EP 1, i˙ = cos ! sin ! 0 !2 (24) 0 1 2 3 0 1 2 and 3 are referred to as the vector EP components. Alter- !˙ sin ! cot i cos ! cot i 1 !3 nate variable notations such as (q ,q ,q ,q ) are also popu- @ A 4 5 @ A 1 2 3 4 [B(⌦,i,!)] lar. Care should be taken how the scalar EP or quaternion | {z } component is labeled (q4 = 0 in this case).

4 Being a redundant 4-parameter set, the EPs must satisfy makes it straight-forward to subtract two orientations and the holonomic constraint solve for the relative orientation 0 given and 00 using Eq. (33). 2 2 2 2 0 + 1 + 2 + 3 =1 (30) The EP rates relate to the body angular velocity vector ! through If numerically integrating the EPs, this constraint must be ˙ enforced at each time step. As we had 4 sets of possible 0 0 !1 !2 !3 0 principal rotation parameters to represent an orientation, this ˙ 1 ! 0 ! ! 1 = 1 3 2 1 (35) results in 2 sets of feasible EPs and 0 = . If of 0 ˙ 1 2 3 0 1 0 2 2 !2 !3 0 !1 2 an EP set is greater than zero, then the EPs are describing a B˙3C 6!3 !2 !1 0 7 B3C short rotation from to with <180 degrees. If < 0 B C 6 7 B C N B 0 @ A 4 5 @ A then a long rotation is described. This duality is important or by transmutation when choosing which set of EPs is used in a feedback attitude ˙ control problem. 0 0 1 2 3 0 ˙ 1 ! The DCM is written in terms of the EPs as 0 11 = 2 1 0 3 23 0 11 (36) ˙2 2 2 3 0 1 !2 2+2 2 2 2( + ) 2( ) 0 1 2 3 1 2 0 3 1 3 0 2 B˙3C 63 2 1 07 B!3C 2 2 2 2 B C 6 7 B C [BN]= 2(12 03) 0 1+2 3 2(23 + 01) @ A 4 5 @ A 22( + ) 2( ) 2 2 2+2 3 The 4 4 matrix in Eq. (36) is again orthogonal and enjoys 1 3 0 2 2 3 0 1 0 1 2 3 ⇥ 4 (31)5 a simple inverse formulation. while the inverse mapping is 1.2.5 Other Attitude Parameters There are several other sets of attitude parameters such as 1 0 = trace([BN]) + 1 (32a) the (w, z) coordinates introduced by Tsiotras and Longuski ±2 [1996], or the Calyey-Klein parameters discussed by Whit- BNp23 BN32 1 = (32b) taker [1965 reprint]. A good survey of attitude coordinates 40 is published by Shuster [1993]. Besides using the quater- BN31 BN13 nions as a singularity free attitude measure, the Modiﬁed Ro- 2 = (32c) 40 drigues Parameters (MRPs) have become popular.[Wiener, BN12 BN21 1962] At the cost of a discontinuous attitude description, the 3 = (32d) 40 MRPs can represent any orientation without singularity using only 3 parameters.[Schaub and Junkins, 1996] As expected, this mapping yields 2 possible EP sets. Note that this algorithm in singular if 0 =0( = 180 degrees). A lengthier, but singularity free, mapping from DCM into bˆ equivalent EPs is given by Sheppard [1978]. 3 To add or subtract EP sets it is not necessary to map the ! EPs to and from DCMs. Instead the EP have a direct bi- r linear transformation. Let 0 and 00 represent to orientations bˆ2 which are to be added to yield the overall rotation as the EP dm set . Rc OB 00 00 00 00 0 0 0 1 2 3 0 B 1 00 00 00 00 0 N bˆ 0 1 = 2 1 0 3 2 3 0 11 (33) 1 00 00 00 00 0 2 2 3 0 1 2 B C 600 00 00 007 B0 C B 3C 6 3 2 1 0 7 B 3C O @ A 4 5 @ A By transmutation of Eq. (33) an alternate expression is found Figure 3: Discretized Continuous Body

0 0 0 0 00 0 0 1 2 3 0 0 0 0 0 00 0 11 = 2 1 0 3 23 0 1 1 (34) 0 0 0 0 00 2 RIGID BODY INERTIA 2 2 3 0 1 2 B C 60 0 0 0 7 B00C B 3C 6 3 2 1 07 B 3 C @ A 4 5 @ A Having covered how to describe the orientation of a rigid The 4 4 matrices in Eqs. (33) and (34) are orthogonal with body (kinematics), we now focus on how bodies with spe- ⇥ their inverse simply being the transpose of the matrix. This ciﬁc mass distributions will rotate (kinetics). This leads to

5 the basic rotational equations of motion dictating how a rigid Note that because the order, magnitude and sign of the eigen- body will accelerate rotationally due to external torques. vectors vi is not unique, the [FB] rotation is also not unique. The frame unit direction vectors fî are then given by 2.1 Inertia Matrix Definition F ˆ ˆ ˆ ˆ ˆ Assume a rigid body has a body-fixed coordinate frame Bf1 = v1 Bf2 = v2 Bf3 = Bf1 Bf2 (42) B ⇥ as illustrated in Figure 3. The origin is set to be equal OB ˆ to the center of mass location of the rigid body. Let r be the Determining Bf3 in this manner guarantees that ends up F position vector of a differential mass element dm. The 3 3 as a proper right-handed coordinate system. The frame ⇥ ˆ F inertia matrix of this rigid body about its center of mass point axes fi which diagonalize the rigid body inertia matrix are is defined as [Schaub and Junkins, 2010] called the principal axes. The frame then is a principal F coordinate frame of the body, with the eigenvalues i being [I ]= [r˜][r˜]dm (37) the principal inertias Ii of this body. c ZB For example, consider the inertia matrix expression with where represents the integral over the entire body . If r respect to a general body-fixed frame : B B is expressedB in frame components as in Eq. (15), then the R B B 28.700 2.279 2.340 symmetric inertia matrix is expressed in frame component B 2.279 24.400 1.585 2 form as B[I]= kg m (43) 22.340 1.585 21.9003 · B 2 2 r2 + r3 r1r2 r1r3 4 5 2 2 The principal inertias of this body are determined through the B[I ]= r r r + r r r dm (38) c 1 2 1 3 2 3 2 B 2 2 23 eigenvalues (30, 25, 20) kg m . The corresponding eigen- Z r1r3 r2r3 r + r 1 2 · 4 5 vectors are If the body contains a series of discrete mass elements, then the integral in Eq. (38) can also be replaced with a summation B 0.925 B 0.163 B 0.342 operator. v1 = 0.319 v2 = 0.823 v3 = 0.470 (44) 20.2053 20.5443 20.8143 2.2 Principal Coordinate System 4 5 4 5 4 5 Note that the inertia matrix calculation in Eq. (37) depends The desired rotation matrix [FB] to rotate the general body on the choice of the body frame used to represent r. What frame into a principal coordinate frame is B B if the inertia matrix is required with respect to another body 0.925 0.319 0.205 fixed frame ? Fortunately there exists a direct transforma- F [FB]= 0.163 0.823 0.544 (45) tion of B[Ic] into F [Ic] that maps one inertia matrix to any 2 0.342 0.470 0.8143 other frame without performing the body integral again. Let 4 5 [FB] be the DCM of the new body-fixed frame relative v F Here the eigenvector algorithm returned a set of vectors i to the previous body frame . The inertia matrix coordinate where v v = v . This is generally not the case and this B 1 ⇥ 2 3 frame transformation is then given by [Schaub and Junkins, condition must be checked. 2010]

T 2.3 Parallel Axis Theorem F[I]=[FB] B[I][FB] (39) If the inertia matrix is not required about the body center Generally the [I ] matrix is a fully populated 3 3 ma- of mass, but about another point O, then we would like to do c ⇥ trix. Studying Eq. (39) raises the following question. Is it this computation without reevaluating the body intergral in possible to pick such that the resulting inertia matrix F[I] Eq. (37). Let Rc be the position vector from the point O to F is diagonal with the body center of mass. The parallel axis theorem is used to map [Ic] into the body inertia matrix [IO] taken about O F I1 00 using [Schaub and Junkins, 2010] F[I]= 0 I2 0 (40) 2 3 ˜ ˜ T 00I3 [IO]=[Ic]+M[Rc][Rc] (46) 4 5 The answer, naturally, is yes. Let vi be unit eigenvectors of Note that this equation is written in a coordinate frame in- the inertia matrix B[I] and i be the corresponding eigenval- dependent manner. When evaluating the numerical values of ues. The desired coordinate transformation matrix [FB] is the inertias, care must be taken that both [Ic] and Rc are expressed with respect to the same coordinate frames. If not, T v1 the DCM is used to rotate either the inertia matrix compo- T [FB]= v2 (41) nents (see Eq. (39)) or the position vectors components (see 2 T 3 v3 Eq. (16)) into the appropriate frame. 4 5

6 Let us consider the following example. The inertia of a where [a˜]b a b and the inertia definition in Eq. (37) is ⌘ ⇥ body about its center of mass is given in Eq. (43). The body used. To study the rotational motion of a rigid body about is placed inside the space shuttle bay. The body mass is M = its center of mass we focus on Hc and do not consider the 20 kg. Let : sˆ , sˆ , sˆ be the space-shuttle fixed frame, spacecraft translation as a whole. If a principal coordinate S { 1 2 3} and frame is chosen, then H 2 is written as B | c| 2 2 2 2 2 2 2 2 r / =(1m)sˆ1 +(2m)sˆ2 (1 m)sˆ3 (47) H = Hc = I1 !1 + I2 !2 + I3 !3 (54) B S | | We wish to evaluate the the inertia the body about the ori- Let L be the total external torque acting on the rigid body. B gin of . Before we can use the parallel axis theorem, note This torque could be due to the spacecraft actuator thrusters, S that [I] is given in frame vector components, while r / or be environment torques such as the gravity gradient, atmo- B B S is given in terms of frame components. Let the relative spheric, or differential solar radiation torque. Euler’s equa- S orientation of with respect to be tion states that B S 00 1 H˙ = L (55) [BS]= 10 0 (48) 20103 This very simple and innocent looking differential equation 4 5 is the means to compute the rotational equations of motion If the final expression is desired in shuttle -frame compo- S of a rigid body, or even a system of rigid bodies. Thus, let nents, it is simplest to perform a coordinate transformation of us discuss Eq. (55) in more detail. First, the time derivative the body inertia [I] into -frame components using Eq. (39). S of the H vector must be taken as seen by an inertial coordi- T nate frame. Euler’s equation is not only valid for a system S[I]=[BS] B[I][BS] containing a single rigid body, it is also valid for a system N S 24.40 1.59 2.28 (49) containing rigid bodies. The angular momentum expres- 2 N = 1.59 21.90 2.34 kg m sion H = i=1 Hi must be the total angular momentum, 2 3 · 2.28 2.24 28.70 where Hi could represent the momentum of the the space- P 4 5 craft, a moving panel, or an attached fly-wheel. Finally, care Now the desired inertia matrix of about the origin O of must be taken in Eq. (55) both the momentum and the torque B S is evaluated using Eq. (46): vectors are taken about the identical reference point. This hinge point can be an inertial point, or the system center of T S[IO]=S[I]+M[r˜ / ][r˜ / ] B S B S mass. S 124.40 41.59 17.72 (50) = 41.59 61.90 37.66 kg m2 4 KINETIC ENERGY 217.72 37.66 128.703 · Referring again to Figure 3, the kinetic energy of a rigid 4 5 3 ANGULAR MOMENTUM body is 1 1 The total angular momentum of the continuous body of T = MR˙ R˙ + r˙ r˙dm = T + T (56) 2 c · c 2 · trans rot mass M shown in Figure 3 is written as ZB For attitude studies we focus again on the rotational energy ˙ HO = Rc MRc + r r˙dm (51) only, and ignore the translational motion of the body center ⇥ ⇥ ZB of mass. Using [a˜]b a b and the inertia definition in ⌘ ⇥ Here R MR˙ is the angular momentum of the body center Eq. (37), the rotational energy of a single rigid body is [Cur- c⇥ c of mass about point O, while tis, 2010]

1 1 T 1 H = r r˙dm (52) Trot = r˙ r˙dm = ! [Ic]! = ! Hc (57) c ⇥ 2 B · 2 2 · ZB Z If a principal coordinate system is chosen for , the energy is the angular momentum of the body about the center of B mass. The vector r is the position of each differential mass is written using the principal inertias Ii as element dm relative to the body center of mass. If the body I1 2 I2 2 I3 2 is rigid, then r˙ = ! r and Trot = ! + ! + ! (58) ⇥ 2 1 2 2 2 3 H = r (! r) dm =[I ]! (53) The power equation seeks to determine what the energy c ⇥ ⇥ c ˙ ZB rate T is for a dynamical system. Let Lc be the total external

7 torque acting on a rigid body about the body center of mass. Here two !i are zero initially resulting in zero !˙ . These pure This torque will cause the rotational energy to vary according principal axis spin conditions are common with many spin- to stabilized spacecraft. We next investigate the linear stability of such principal axis spins. Let be a principal body- ˙ B Trot = ! Lc (59) ﬁxed frame. Without loss of generality assume the body · ˆ ˆ is nominally spinning about b1 with !e = !e1 b1 and that The derivation of this power equation requires use of the rigid !e2 = !e3 =0. Next we assume the spin experiences small body rotational equations of motion in Eq. (61). Note that departures through !i = !ei + !i and linearize the equa- even though T is computed using the vectors ! and Hc in tions of motion in Eq. (62): Eq. (57), the answer is a scalar. The time derivative of scalar quantities is the same no matter what observer frame is used. !˙ 1 =0 (64a) Thus, the power equations in Eq. (59) can be obtain by dif- I3 I1 ferentiating the vectors with respect to any frame. !˙ 2 = !e1 !3 (64b) I2 I1 I2 5 ROTATIONAL EQUATIONS OF MOTION !˙ 3 = !e1 !2 (64c) I3 5.1 Euler’s Rotational Equations Immediately it is apparent that the spin departures !1 about Let us assume at ﬁrst the the rigid body has a fully pop- the nominal spin axis are marginally stable. Given an initial ulated inertia matrix [I]. Note that in this development all spin error !1(t0) the spin is constant: moments and torques are implied to be taken about the body center of mass, and the subscript “c” is dropped for conve- !1(t)=!e1 + !1(t0) (65) nience. The rotational equations of motion are obtained by evaluating Euler’s equation in Eq. (55) using the transport The spin about the other two axis are coupled. Differenti- theorem in Eq. (4): ating Eq. (64c), substituting into Eq. (64b), and simplifying yields the uncoupled result for !2. d H˙ = B (H)+! H = L I I I I dt ⇥ (60) 1 3 1 2 !¨2 + !e1 !e1 !2 =0 (66) Bd Bd I I = ([I]) ! +[I] (!)+[!˜][I]! = L ✓ 2 ◆✓ 3 ◆ dt dt k

Bd | {z } Because the body is rigid we ﬁnd that dt ([I]) is zero. Using Similarly the !3 motion is shown to satisfy Eq. (7) the famous Euler’s rotational equations of motion of a body with a general inertia matrix are found: !¨3 + k!3 =0 (67)

[I]!˙ = [˜!][I]! + L (61) ˆ Thus, k>0 for the spin about the equilibrium !e = !e1 b1 to be linearly stable. This is true if I1 is either the largest or Choosing a principal body ﬁxed coordinate system the inertia the smallest inertia. Any spin about the intermediate axis of [I] matrix is diagonal and Eq. (61) reduces to [Wiesel, 1989] inertia is guaranteed to be unstable. Note that the spin about the largest moment of inertia is always stable. However, as will be shown in the study of the polhodes, the spin about I !˙ = (I I )! ! + L (62a) 11 1 33 22 2 3 1 the least moment of inertia is only stable in the absence of I !˙ = (I I )! ! + L (62b) 22 2 11 33 3 1 2 energy dissipation. I33!˙ 3 = (I22 I11)!1!2 + L3 (62c) 5.3 Numerical Simulation of Rigid Body Motion Here L are the body frame vector components of the ex- i B The rotational equations of motion in Eq. (61) appear at ternal torque vector L. ﬁrst glance to be decoupled from any attitude coordinates. However, the orientation can couple into these equations if 5.2 Principal Axis Spin Stability the external torque L depends on the attitude. Such external Note that due to the gyroscopic cross coupling terms in torques include atmospheric torque, solar radiation torque, Eq. (62), the only spin equilibria (where !˙ =0) of a rigid gravity gradient torques, etc. Writing a numerical integration body occurs when a pure spin about a principal axis is per- routine for the spacecraft attitude, Euler’s rotational equa- formed. tions of motion in Eq. (61) need to be integrated simulate- neously with the appropriate kinematic differential equations ˆ !e = !ebi for i =1, 2, 3 (63) such as Eq. (22) or Eq. (36).

8 6 TORQUE-FREE RESPONSE Table 1: Rigid Body Duffing Analog Constants The remainder of this section studies the rotational motion of a rigid body in the absence of an external torque vec- i Ai Bi  +  2 tor. This is a common situation for spin-stabilized spacecraft. 1 12 3 13 2 12 13 I1I2I3 I2I3 Other Chapters in this Section explore both passive and ac-  +  2 tive attitude stabilization methods. 2 23 1 21 3 21 23 I1I2I3 I2I3 312 +321 23132 6.1 Angular Velocity Solutions 3 I1I2I3 I2I3 6.1.1 Axisymmetric Inertia Case Consider the special case where the spacecraft is axially symmetric. Without loss of generality, let bˆ3 be the axis of order differential equations of attitude and angular velocity. symmetry. Here I1 = I2 = IT and Eqs. (62) reduce to: For torque-free motion the angular momentum H is iner- IT !˙ 1 = (I3 IT )!2!3 (68a) tially constant. This result can be used to reduce the rotational motion to first order differential equations of the at- IT !˙ 2 =(I3 IT )!3!1 (68b) titude coordinates. Using an observation due to Jacobi, the I !˙ =0 (68c) 3 3 inertial frame is defined such that H = Hnˆ . Using a 3 principal coordinate frame the momentum is written as Eq. (68c) that !3 is constant in this case. Differentiating Eqs. (68a) and (68b) and sustituting the results into each B I ! N 0 yields 1 1 BH = I ! =[BN( ,✓,)] 0 (72) 0 2 21 0 1 2 I3!3 H !¨1 + !p!1 =0 (69a) 2 @ A @ A !¨2 + !p!2 =0 (69b) As a result of this inertial frame choice, the following coning and precession rates are taken relative to angular momentum I3 (3 2 1) with !p = 1 !3. Eqs. (69) are equivalent to spring- vector. Parameterizing the DCM using Euler IT angles in Eq. (20) and solving Eq. (72) for ! yields mass systems⇣ and have⌘ the analytical solution: ! H sin ✓ ! (t)=! (t ) cos ! t ! (t )sin! t (70a) 1 I1 1 1 0 p 2 0 p ! = H sin cos ✓ (73) 2 I2 ! (t)=! (t ) cos ! t + ! (t )sin! t (70b) 0 1 0H 1 2 2 0 p 1 0 p !3 cos cos ✓ I3 !3(t)=!3(t0) (70c) @ A @ A Substituting this !(✓,) result into the (3 2 1) Euler 6.1.2 General Inertia Case angle differential kinematic equation result in the first order For general rigid body principal inertias Junkins et al. attitude differential equations [1973] show that it is possible to write Eqs. (62) as three sin2 cos2 decoupled second order differential equations ˙ = H + (74a) I2 I3 3 ✓ ◆ !ï + Ai!i + Bi! =0 for i =1, 2, 3 (71) H 1 1 i ✓˙ = sin 2 cos ✓ (74b) 2 I3 I2 This homogenous, undamped Duffing equation appears of- ✓ ◆ 1 sin2 cos2 ten in structural mechanics problems. However, there the ˙ = H sin ✓ (74c) I I I Duffing equation is an approximation the full response. It is ✓ 1 2 3 ◆ amazing that for the rotation motion of a rigid body the angu- Note that the coning rate ˙ cannot be positive for the general lar velocity the Duffing formulation is the exact differential inertia case, while ✓˙ and ˙ can assume either sign. For a axi- equation. The coefficients A and B are defined in Table 1 i i symmetric body these attitude rates simplify greatly. Without where = I I and  =2I T H2. ij i j i i loss of generality assume that I = I and substitute into Note that the differential equations in Eq. (71) are uncou- 2 3 Eq. (74): pled, they are not independent. The constants Ai depend on the energy T and angular momentum magnitude H. These H ˙ = (75a) quantities depend on all three initial conditions of !i(t0). I2 ✓˙ =0 (75b) 6.2 Coning Motion nd I I Normally the rotation motion leads to either 2 order dif- ˙ = H 2 1 sin ✓ (75c) I I ferential equations of attitude coordinates, or two sets of first ✓ 1 2 ◆

9 Note that in this special inertia case all three Euler angle rates Momentum H3 Energy have constant rates. Sphere Ellipsoid

6.3 Polhodes

Let H = H1bˆ1 +H2bˆ2 +H3bˆ3 be the angular momentum vector in frame components. With torque-free motion the B angular momentum magnitude H is constant. Assuming a principal coordinate frame H is written as

2 2 2 2 2 2 2 2 2 2 H2 H = H1 + H2 + H3 = I1 !1 + I2 !2 + I3 !3 (76)

In terms of !i the momentum constraint describes the surface of an ellipsoid. Using the momentum coordinates Hi the constraint describes the surface of a sphere. Without external H1 torque the power equation in Eq. (59) shows that the energy T is also a constant. (a) Maximum Inertia Axis Spin (Minimum Energy)

1 2 1 2 1 2 T = I1!1 + I2!2 + I3!3 (77) 2 2 2 Energy Momentum Ellipsoid Using Hi = Ii!i the constant energy constrain is written in Sphere the energy ellipsoid form:

H2 H2 H2 1= 1 + 2 + 3 (78) 2I1T 2I2T 2I3T

Polhodes are three-dimensional line plots of the !i evo- lutions. They are a convenient method to study torque-free H2 rigid body motion. Even with internal friction the angular momentum of a body remains ﬁxed in the absence of external forces. However, the energy can vary. As a result it is common to plot the rigid body motion that result for a ﬁxed H1 H and let the energy levels vary. The momentum constraint in terms of Hi is a sphere in Eq. (76), while the energy constraint in Eq. (78) is an ellipsoid. The actual rotation motion is at the intersection of the momentum sphere and energy el- (b) Intermediate Inertia Axis Spin (Intermediate En- ergy) lipsoid. Without loss of generality, let us assume the inertia or- dering I I I . Figure 4 illustrates the polhodes for 1 2 3 the three possible principle axis spin cases. The minimum Momentum energy state in Eq. (77) is a pure spin about the maximum Sphere principal inertia axis bˆ1. The maximum energy state is the spin about the axis of least inertia.

H2 H2 H2 Tmax = Tint = Tmin = (79) 2I1 2I2 2I3 H2 The Tmax state represents the smallest energy ellipsoid which still touches the momentum sphere, while Tmin is the largest ellipsoid to ﬁt inside the sphere and still make contact. The polhode curves for the energy levels T T T min   max are illustrated in Figure 5. Small spin departures about the H1 Energy bˆ and bˆ axis result in bounded neighboring motion, while 1 3 Ellipsoid any departure from the intermediate axis spin about bˆ2 results in unstable motion as predicted be linear stability analysis. Internal friction in a body can reduce the energy level (c) Minimum Inertia Axis Spin (Maximum Energy) while maintaining a ﬁxed angular momentum vector. The Figure 4: Polhodes of Principal Axis Spin Cases. 10 2 bˆ2 H3 H Tmax = 2I3 bˆ1

⌦ sepratrix

bˆ3 H2 H1 Figure 6: Illustration of a dual-spin spacecraft with the wheel H2 rotation axis aligned with a spacecraft principal axis. T = 2 min 2I H 1 Tint = 2I2 the body angular velocity of the main craft, while ! / = W B ⌦bˆ1 is the angular velocity of the fly wheel relative to the spacecraft. The total angular momentum is then given by Figure 5: Rigid body Polhode for various energy states H =[Is]! +[IW ](⌦bˆ1 + !) (80) polhodes in Figure 5 illustrate how such an energy loss in a where [Is] is the inertia matrix of the main spacecraft system, pure min-inertia spin will result in unstable motion because while [IW ] is the inertia of the fly wheel component. Note the !i eventually intersects with the intermediate axis pol- that in the case where the dual-spin craft is rotating an entire hode (septratrix) before settling down to a stable minimum segment of the main craft, then [IW ] would be the equivalent energy spin about the axis of largest inertia. spacecraft component inertia matrix. Next, let us define the combined inertia matrix [I] as 7 DUAL-SPIN STABILIZATION [I]=[Is]+[IW ] (81) While for a single rigid body only the principal axes spin Assuming a principal coordinate frame , no external torque about the axes of largest inertia is passively stable in the B vector, and a constant wheel spin rate with ⌦˙ =0, we arrive presence of energy loss, in many applications (geostationary at the three scalar dual-spin spacecraft equations of motion satellites rotating to point an antenna at a fixed Earth point) using Euler’s equation H˙ = 0: it would be of great advantage to be able to stabilize a spacecraft spin about any principal axes, regardless of the corre- I1!˙ 1 =(I2 I3)!2!3 (82a) sponding axes inertia. The dual-spin spacecraft is a simple I !˙ =(I I )! ! I ! ⌦ (82b) system where passive attitude stability is achieved by adding 2 2 3 1 1 3 Ws 3 I !˙ =(I I )! ! + I ! ⌦ (82c) a single fly-wheel to the rigid spacecraft. Beyond the geo- 3 3 1 2 1 2 Ws 2 stationary communication satellites, another application of Because ⌦ =0, the only dual-spin equilibrium configuration the dual-spin concept is the interplanetary Galileo spacecraft 6 where !˙ remains zero is which traveled to Jupiter. Here the main antenna needed to continuously point back at the Earth. To stabilize this orien- ˆ !e = !e1 b1 (83) tation the half of the body rotated at 3 revolutions per minute, while the sensor components rotated very slowly to align the If the wheel spin axis is aligned with another principal body communication antenna to the slowly changing spacecraft- axis, then the dual-spin equilibria would be about this new to-Earth vector. However, this spin rate magnitude must be axis. chosen very carefully. While the dual-spin concept can stabilize any principal axis spacecraft spin, if used incorrectly it 7.2 Linear Equilibrium Stability can also be the cause of instability. Next, let us examine the dual-spin spacecraft equilibria stability if the fly wheel is a constantly rotating compo- 7.1 Equations of Motion nent, and assuming no system energy loss is present. We To develop the dual-spin system equations of motion, linearize the attitude motion about the equilibrium rotation ˆ without loss of generality, assume that the rotating fly-wheel !e = !e1 b1. Let the actual angular velocity be given by is aligned with the first principal axis bˆ1 of the main space- ! = !e + ! (84) craft component as illustrated in Figure 6. Let ! = ! / be B N

11 T where ! =(!1,!2,!3) is the departure motion. Sub- ⌦ˆ 1 ⌦ˆ 2 0 stituting this ! into the equations of motion in Eq. (82) and ⌦ˆ dropping higher order terms leads to the linearized departure equations of motion: (a) Case where I1 >I2 >I3.

!˙ 1 =0 (85a) ⌦ˆ 1 0 ⌦ˆ 2 I I I 3 1 Ws ⌦ˆ !˙ 2 = !e1 !3 (85b) I2 I2 ✓ ◆ (b) Case where I2 >I1 >I3. I1 I2 IWs !˙ 3 = !e1 + !2 (85c) I3 I3 ✓ ◆ 0 ⌦ˆ 2 ⌦ˆ 1 Note that because !˙ 1 =0, the departure angular velocity ⌦ˆ about bˆ1 (i.e. the ﬂy wheel spin axis in this case), is only (c) Case where I3 >I2 >I1. marginally stable with !1(t)=!e1 +!1(t0) being constant. To determine the stability of the ! (t) and ! (t) mo- 2 3 ˆ tions, we differentiate Eq. (85b) and substitute Eq. (85c) to Figure 7: Stabilizing ⌦ range (shaded) illustration. ﬁnd

!¨2 + k! =0 (86) is illustrated in Figure 7(b). As expected, note that here the origin is not included in the stabilizing ⌦ˆ range. Without the where the stiffness-like parameter k is stabilizing effect of the fly wheel, the single rigid body spin !2 about the intermediate axis of inertia is unstable. k = e1 I I + I ⌦ˆ I I + I ⌦ˆ 1 3 Ws 1 2 Ws (87) Lastly, we explore the stabilizing wheel speed range for I2I3 ⇣ ⌘⇣ ⌘ a minimum axis of inertia spin where I3 >I2 >I1. The and ⌦ˆ = ⌦ is the non-dimensional fly-wheel spin rate. For admissible wheel speeds are illustrated in Figure 7(c). Be- !e1 the non-spin axis departure velocities to be stable we require cause the minimum inertia spin case is linearly stable in the absence of a fly wheel, the origin is once again included in k>0 (88) the admissible range. However, there is still finite range of positive ⌦ˆ values which would lead to an unstable system. Given the principal inertias I , I and I , we would like to 1 2 3 If the system undergoes energy loss (wheel bearing fric- determine for what range of values ⌦ˆ the dual-spin space- tion), then the above stability analysis must be modified. craft motion is passively stable. There are two critical wheel Dual-spin spacecraft stability analysis with energy loss is speeds which cause the inequality conditions to be either true discussed in further detail in both Hughes [1986] or Curtis or false: [2010]. I I ⌦ˆ = 3 1 (89a) 1 I Ws 8 EXTERNAL TORQUES FOR PASSIVE I2 I1 ⌦ˆ 2 = (89b) CONTROL IWs The dual-spin stability condition in Eq. (88) is satisfied if: Next, let us consider what external torques can be used to passively stabilize a spacecraft without resorting to a feed- ˆ ˆ ˆ ˆ Condition 1: ⌦ > ⌦1 and ⌦ > ⌦2 (90a) back control strategy.

Condition 2: ⌦ˆ < ⌦ˆ 1 and ⌦ˆ < ⌦ˆ 2 (90b) 8.1 Gravity Gradient Torques Let us examine these 2 stability conditions for three types of principal axis rotations. First, consider the maximum in- Not all parts of a rigid body will experience the same gravitational attraction to the planet they are orbiting due to dif- ertia spin scenario where I1 >I2 >I3. Because I1 is fering separation distances. While position difference are the largest inertia, both ⌦ˆ 1 and ⌦ˆ 2 are negative values with very small, they due lead to a noticeable net torque being ⌦ˆ 1 < ⌦ˆ 2. The resulting ranges of stabilizing ⌦ˆ values are graphically illustrated in Figure 7(a). Note that because the applied to the spacecraft called gravity gradient torque. maximum inertia spin is stable in absence of the ﬂy wheel, Assume the center of mass of object Earth’s center is the B the feasible ⌦ˆ range must include the origin. inertial position vector Rc. Let the vector LG be the external The stabilizing ﬂy wheel spin rates for the case where the gravity gradient torque experienced by a rigid object mea- spacecraft is to rotate about an intermediate axis of inertia sured about its center of mass. For a solid body this torque is

12 defined through in a gravity gradient equilibrium orientation if the orbit frame axes are also principal axes of the body inertia matrix [I]. L = r dF (91) To analyze the stability of the spacecraft attitude motion G ⇥ G ZB about, let us assume small angular departures (linearized stability analysis), and use the 3 2 1 Euler angles ( ,✓,) where the vector r is the position vector of an infinitesimal to describe the departure orientations. Assuming a circular body element relative to the center of mass and FG is the gravitational attraction experienced by this element. Using orbit with a mean orbit rate ⌦, the linearized pitch equation Newton’s Gravitational Law this force is written as becomes: I I GMe ¨ 2 11 33 dFG = Rdm (92) ✓ + 3⌦ ✓ =0 (97) R 3 I22 | | ✓ ◆ where Me is Earth’s mass, dm is the body element mass and where I11 is the principal inertia about the oˆtheta orbit along- R is its inertial position vector measured from Earth’s center. track axis, I22 is the principal inertia about oˆh, and I33 is the principal inertia about the orbit radial axis oˆr. These pitch R = Rc + r (93) equations are the dynamical equivalent of a simple spring- mass system. It is immediately clear from linear control the- Substituting Eq. (92) into the LG expression, and expand- ory that for the pitch mode to be stable ing use a binomial expansion to first order, yields the general gravity gradient torque expression: ([Junkins and Turner, I11 I33 (98) 1986, Greenwood, 1988, Schaub and Junkins, 2010]) must be true. Thus, for pitch stability, the spacecraft orien- 3GM e ˜ tation must be chosen such that the inertia about the along- LG = 5 [Rc][I]Rc (94) Rc track axis must be larger than the inertia about the orbit radial Note that no particular frame has been chosen in Eq. (94) axis. to express this torque. For example, if orbit frame : The linearized roll and yaw motions are given through the O coupled equations: oˆ✓, oˆh, oˆr vector components are chosen, the ORc = { T } [0, 0,Rc] and O[I] is a fully populated matrix for general ¨ 0⌦(1k ) ˙ spacecraft orientations. Here the gravity gradient torque ex- + Y ¨ ⌦(kR 1) 0 ˙ pression in Eq. (94) reduces to:[Schaub and Junkins, 2010] ✓ ◆  ✓ ◆ 2 4⌦ kY 0 3GMe + 2 =0 (99) OLG = 3 ( I23oˆr + I13oˆh) (95) 0⌦kR Rc  ✓ ◆ k k Note that the gravity gradient torque will never produce a where the inertia ratios R and Y are defined as torque about the orbit radial axis. In contrast, assume is I22 I11 T B k = (100) a principal body frame. Here BRc =[Rc ,Rc ,Rc ] and R 1 2 3 I33 B[I]=diag(I1,I2,I3), leading to the simple expression: I I k = 22 33 (101) Y I B 11 Rc2 Rc3 (I33 I22) 3GMe LG = Rc1 Rc3 (I11 I33) (96) To guarantee stability it is necessary and sufficient that R5 c 0R R (I I )1 c1 c2 22 11 @ A 1+3kY + kY kR > 4 kY kR (102) Studying Eq. (96) it is clear that the gravity gradient torque k k > 0 (103) can be zero for several scenarios. For example, if the space- R Y p craft body shape is such that I1 = I2 = I3 (spherical sym- The two possible regions where we have marginal stability metry for example) leads to a zero torque. Further, if Rc is on all three degrees of freedom are illustrated in Figure 8. aligned with one of the principal body axes, then one R ci Region I requires that vector component is non-zero, while the remaining two are zero. Here too the torque LG is zero. I22 I11 I33 (104) To determine all possible gravity gradient equilibrium orientations, one of the body fixed axes must align with the orbit This gravity gradient stabilized configuration requires that nadir axis oˆr. To ensure that !˙ (equilibrium condition), the the inertia about the orbit normal axis oˆh is the largest in- gyroscopic term [!˜][I]! in Eq. (61) requires that the other ertia. The configurations in Region II are not typically em- two body axes are aligned with either the along-track direc- ployed because they are unstable if damping and energy loss tion oˆ✓ or the orbit normal direction oˆh. Thus, a spacecraft is is considered.

13 1.0 Equilibrium orientations subject to the atmospheric torque are such orientations where La is zero. This can be achieved kRkY < 0 kY < kR by having symmetrical atmospheric pressure distribution rel- 0.5 ative to the spacecraft velocity vector. Such symmetry will cause the atmospheric torque about the spacecraft compo- 2 (1 + 3kY + kRkY ) < 16kY kR nents to mutually cancel, even though a net atmospheric Region I kR 0.0 force (atmospheric drag) is applied. Another particular attitude is the Torque Equilibrium Atti- Region II tude (TEA). This orientation ensures that all external torques -0.5 acting on the craft mutually cancel each other. For example, if the spacecraft shape causes a small atmospheric torque

kRkY < 0 which cause the craft to begin to rotate, it is possible to orient the craft such that the gravity gradient torque perfectly can- -1.0 -1.0 -0.5 0.0 0.5 1.0 cels the atmospheric torque. For example, the space station kY often flies in TEA’s to avoid having external torques cause excessive momentum build-up. Figure 8: Linearized Gravity Gradient Spacecraft Stability Regions 8.3 Magnetic Torques If the spacecraft contains a magnetic field due to the pres- 8.2 Atmospheric Torques ence of magnetic torque rods, then these can also produce an external torque L due to the interaction with Earth’s mag- In the rarefied atmosphere of low Earth orbits, a small m netic field. If m is the magnetic moment of the spacecraft, amount of gas molecules can hit the spacecraft and impart and B is the geocentric magnetic flux density, the resulting a force. Because the local gas dynamics behaves like a free magnetic torque is given by[Wertz and Larson, 1999] molecular flow, the deflected gas particles have a negligible influence on the gas particles. This non-interaction of incom- L = m B (107) ing and deflected gas particles allows the net aerodynamic m ⇥ force or torque to be computed by summing the contributions of each of the spacecraft components individually. This al- For a spinning spacecraft motion the induced eddy cur- lows the vehicle shape to be dissected into simple sub-shapes rents can cause a small magnetic torque. Such torques [Wertz to easy the aerodynamic force and torque computation. and Larson, 1999] can cause the spin axis to precess, and also A conservative estimate of the total aerodynamic force Fa result in nutation damping. acting on the spacecraft can be obtained through the following approximation: 9 CONCLUSION 1 F = C ⇢v2A (105) This Chapter develops the fundamental kinematics and a 2 D kinetics of a single rigid body. The orientation of the body is where ⇢ is the local atmospheric density, A is the projected tracked by studying the rotation of a body-fixed coordinate frame . Analytical solutions to the body angular velocities area of the spacecraft normal to the incident flow (spacecraft B velocity direction), v is the velocity of the spacecraft relative are explored for the axi-symmetric and general inertia case. to the local atmosphere, and CD is the drag coefficient. To Assuming no external torque, the linear spacecraft spin obtain a conservative estimate of the atmospheric force, a stability about the principal inertia axis is explored. Using conservative value of CD should be used. polhode plots, the torque free response is illustrated for a The magnitude of the atmospheric torque is estimated us- range of spin conditions, including the case where the body ing is losing energy due to internal damping. Passive attitude stability can be achieved using dual-spin configurations, or La = lFa (106) external influences such as the gravity gradient of atmospheric torques. The following Chapters explore the where is the moment arm of the center of pressure rela- attitude response of a spacecraft using active control tive to the spacecraft center of mass. For conservative esti- methods (applying external torques through thrusters or mates, this moment arm is assumed to be at least one-third of using internal momentum exchange devices), as well as how the spacecraft’s maximum dimension. This includes all ap- to estimate the orientation parameters using both external pendages, and should be done even if the spacecraft is sym- observations (sun or star sensor) and internal rate metric. measurements.

14 REFERENCES Wie B. Space Vehicle Dynamics and Control. AIAA Education Series, Reston, VA, 2nd edition, 2008. Curtis HD. Orbital Mechanics for Engineering Students. 2nd Edition, Elsevier Aerospace Engineering Series, Wiener TF. Theoretical Analysis of Gimballess Inertial Burlington, MA, 2010. Reference Equipment Using Delta-Modulated Instruments. Ph.D. dissertation, Department of Greenwood DT Principles of Dynamics. Prentice-Hall, Inc, Aeronautics and Astronautics, Massachusetts Institute of Englewood Cliffs, New Jersey, 2nd edition, 1988. Technology, March 1962. Hughes PC. Spacecraft Attitude Dynamics. John Wiley & Wiesel WE. Spaceﬂight Dynamics. McGraw-Hill, Inc., New Sons, Inc., New York, 1986. York, 1989. Junkins JL, Jacobson ID, and Blanton JN. a nonlinear oscillator analog of rigid body dynamics. Celestial Mechanics and Dynamical Astronomy, Vol. 7, No. 4, 1973, pp. 398–407.

Junkins JL and Turner JD. Optimal Spacecraft Rotational Maneuvers. Elsevier Science Publishers, Amsterdam, Netherlands, 1986.

Likinsj PW. Elements of Engineering Mechanics. McGraw-Hill, New York, 1973.

Schaub H and Junkins JL. Stereographic orientation parameters for attitude dynamics: A generalization of the rodrigues parameters. Journal of the Astronautical Sciences, Vol. 44, No. 1, 1996, pp. 1–19.

Schaub H and Junkins JL. Analytical Mechanics of Space Systems. 2nd Edition, AIAA Education Series, Reston, VA, 2009.

Sheppard SW. Quaternion from rotation matrix. AIAA Journal of Guidance, Control, and Dynamics, Vol. 1, No. 3, 1978, pp. 223–224.

Shuster MD. A survey of attitude representations. Journal of the Astronautical Sciences, Vol. 41, No. 4, 1993, pp. 439–517.

Sidi MJ. Spacecraft Dynamics and Control: A Practical Engineering Approach. Cambridge University Press, Cambridge, England, 1997.

Thomson WT. Introduction to Space Dynamics. Dover Publications, Toronto, Canada, 1986.

Tsiotras P and Longuski JM. A new parameterization of the attitude kinematics. Journal of the Astronautical Sciences, Vol. 43, No. 3, 1996, pp. 342–262.

Wertz, JR and Larson WJ. Space Mission Analysis and Design Microcosm, Inc., El Segundo, CA, 3rd edition, 1999.

Whittaker ET. Analytical Dynamics of Particles and Rigid Bodies. Cambridge University Press, 1965 reprint. pp. 2–16.