Spacecraft Dynamics and Control Matthew M. Peet Arizona State University Lecture 16: Euler s Equations
Attitude Dynamics In this Lecture we will cover: The Problem of Attitude Stabilization Actuators Newton s Laws Mi = d dth Fi = m d dt v Rotating Frames of Reference Equations of Motion in Body-Fixed Frame Often Confusing M. Peet Lecture 16: 2 / 39
Review: Coordinate Rotations Positive Directions If in doubt, use the right-hand rules. Figure: Positive Directions Figure: Positive Rotations M. Peet Lecture 16: 3 / 39
Review: Coordinate Rotations Roll-Pitch-Yaw There are 3 basic rotations a vehicle can make: Roll = Rotation about x-axis Pitch = Rotation about y-axis Yaw = Rotation about z-axis Each rotation is a one-dimensional transformation. Any two coordinate systems can be related by a sequence of 3 rotations. M. Peet Lecture 16: 4 / 39
Review: Forces and Moments Forces These forces and moments have standard labels. The Forces are: X Axial Force Net Force in the positive x-direction Y Side Force Net Force in the positive y-direction Z Normal Force Net Force in the positive z-direction M. Peet Lecture 16: 5 / 39
Review: Forces and Moments Moments The Moments are called, intuitively: L Rolling Moment Net Moment in the positive ω x -direction M Pitching Moment Net Moment in the positive ω y -direction N Yawing Moment Net Moment in the positive ω z -direction M. Peet Lecture 16: 6 / 39
6DOF: Newton s Laws Forces Newton s Second Law tells us that for a particle F = ma. In vector form: F = i F i = m d dt V That is, if F = [F x F y F z ] and V = [u v w], then F x = m du dt F x = m dv dt F z = m dw dt Definition 1. m V is referred to as Linear Momentum. Newton s Second Law is only valid if F and V are defined in an Inertial coordinate system. Definition 2. A coordinate system is Inertial if it is not accelerating or rotating. M. Peet Lecture 16: 7 / 39
Lecture 16 6DOF: Newton s Laws Forces Newton s Second Law tells us that for a particle F = ma. In vector form: 2018-04-19 6DOF: Newton s Laws F = Fi = m d V dt i That is, if F = [Fx Fy Fz] and V = [u v w], then Fx = m du dt Fx = m dv dt Fz = m dw dt Definition 1. mv is referred to as Linear Momentum. Newton s Second Law is only valid if F and V are defined in an Inertial coordinate system. Definition 2. A coordinate system is Inertial if it is not accelerating or rotating. We are not in an inertial frame because the Earth is rotating ECEF vs. ECI
Newton s Laws Moments Using Calculus, momentum can be extended to rigid bodies by integration over all particles. M = M i = d H dt i Definition 3. Where H = ( r c v c )dm is the angular momentum. Angular momentum of a rigid body can be found as H = I ω I where ω I = [p, q, r] T is the angular rotation vector of the body about the center of mass. p = ω x is rotation about the x-axis. q = ω y is rotation about the y-axis. r = ω z is rotation about the z-axis. ω I is defined in an Inertial Frame. The matrix I is the Moment of Inertia Matrix (Here also in inertial frame!). M. Peet Lecture 16: 8 / 39
2018-04-19 Lecture 16 Newton s Laws Newton s Laws Moments Using Calculus, momentum can be extended to rigid bodies by integration over all particles. M = Mi = d H dt i Definition 3. Where H = ( rc vc)dm is the angular momentum. Angular momentum of a rigid body can be found as H = I ωi where ωi = [p, q, r] T is the angular rotation vector of the body about the center of mass. p = ωx is rotation about the x-axis. q = ωy is rotation about the y-axis. r = ωz is rotation about the z-axis. ωi is defined in an Inertial Frame. The matrix I is the Moment of Inertia Matrix (Here also in inertial frame!). r c and v c are position and velocity vectors with respect to the centroid of the body.
Newton s Laws Moment of Inertia The moment of inertia matrix is defined as I xx I xy I xz I = I yx I yy I yz I zx I zy I zz (y I xy = I yx = xydm I xx = 2 + z 2) dm (x I xz = I zx = xzdm I yy = 2 + z 2) dm (x I yz = I zy = yzdm I zz = 2 + y 2) dm So H x I xx I xy I xz p I H y = I yx I yy I yz q I H z I zx I zy I zz r I where p I, q I and r I are the rotation vectors as expressed in the inertial frame corresponding to x-y-z. M. Peet Lecture 16: 9 / 39
Moment of Inertia Examples: Homogeneous Sphere I sphere = 2 1 0 0 5 mr2 0 1 0 0 0 1 Ring 1 I ring = mr 2 2 0 0 1 0 2 0 0 0 1 M. Peet Lecture 16: 10 / 39
Moment of Inertia Examples: Homogeneous Disk I disk = 1 1 + 1 h 3 r 0 0 4 mr2 2 0 1 + 1 h 3 r 0 2 1 0 0 2 F/A-18 23 0 2.97 I = 0 15.13 0 kslug ft 2 2.97 0 16.99 M. Peet Lecture 16: 11 / 39
Moment of Inertia Examples: Cube I cube = 2 3 l2 1 0 0 0 1 0 0 0 1 Box I box = b 2 +c 2 3 0 0 a 0 2 +c 2 3 0 0 0 a 2 +b 2 3 M. Peet Lecture 16: 12 / 39
Moment of Inertia Examples: Cassini 8655.2 144 132.1 I = 144 7922.7 192.1 kg m 2 132.1 192.1 4586.2 NEAR Shoemaker Figure 1: Sketch of NEAR Shoemaker spacecraft 3 473.924 0 0 I = 0 494.973 0 kg m 2 0 0 269.83 [4] C.D. Hall, P. Tsiotras, and H. Shen. Tracking Rigid Body Motion Using Thrusters and Momentum Wheels. The Journal of the Astronautical Sciences, 50(3):311 323, 2002. [5] C.D. Hall. Spacecraft Dynamics and Control, AOE 4140 Class Notes. http://www.aoe.vt.edu/ cdhall/courses/aoe4140/, February 5, 2003. [6] J.L. Meriam and L.G. Kraige. Engineering Mechanics: Dynamics. John Wiley & Sons M. Peet inc., 4 edition, Lecture 1997. 16: 13 / 39
Problem: The Body-Fixed Frame The moment of inertia matrix, I, is fixed in the body-fixed frame. However, Newton s law only applies for an inertial frame: M = i M i = d dt H Suppose the body-fixed frame is rotating with angular velocity vector ω. Then d for any vector, a, dt a in the inertial frame is d a = d a + ω a dt I dt B Specifically, for Newton s Second Law F = m d V + m ω V dt B and M = d H dt + ω H B M. Peet Lecture 16: 14 / 39
Equations of Motion Displacement The equation for acceleration (which we will ignore) is: F x F y = m d V +m ω V dt F B z u ˆx ŷ ẑ = m v + m det ω x ω y ω z ẇ u v w u + ω y w ω z v = m v + ω z u ω x w ẇ + ω x v ω y u As we will see, displacement and rotation in space are decoupled. no aerodynamic forces. M. Peet Lecture 16: 15 / 39
Equations of Motion The equations for rotation are: L M = d H + ω H dt N B I xx I xy I xz ω x I xx I xy I xz ω x = I yx I yy I yz ω y + ω I yx I yy I yz ω y I zx I zy I zz ω z I zx I zy I zz ω z I xx ω x I xy ω y I xz ω z ω x I xx ω y I xy ω z I xz = I xy ω x + I yy ω y I yz ω z + ω ω x I xy + ω y I yy ω z I yz I xz ω x I yz ω y + I zz ω z ω x I xz ω y I yz + ω z I zz I xx ω x I xy ω y I xz ω z +ω y (ω z I zz ω x I xz ω y I yz ) ω z (ω y I yy ω x I xy ω z I yz ) I yy ω y I xy ω x I yz ω z ω x (ω z I zz ω y I yz ω x I xz )+ω z (ω x I xx ω y I xy ω z I xz ) I zz ω z I xz ω x I yz ω y +ω x (ω y I yy ω x I xy ω z I yz ) ω y (ω x I xx ω y I xy ω z I xz ) Which is too much for any mortal. We simplify as: For spacecraft, we have I yz = I xy = I xz = 0 (two planes of symmetry). For aircraft, we have I yz = I xy = 0 (one plane of symmetry). M. Peet Lecture 16: 16 / 39
2018-04-19 Lecture 16 Equations of Motion Equations of Motion The equations for rotation are: L M = d H + ω H dt N B Ixx Ixy Ixz = ωx Ixx Ixy Ixz Iyx Iyy Iyz ωy + ω ωx Iyx Iyy Iyz ωy Izx Izy Izz ωz Izx Izy Izz ωz Ixx ωx Ixy ωy Ixz ωz ωxixx ωyixy ωzixz = Ixy ωx + Iyy ωy Iyz ωz + ω ωxixy + ωyiyy ωziyz Ixz ωx Iyz ωy + Izz ωz ωxixz ωyiyz + ωzizz Ixx ωx Ixy ωy Ixz ωz+ωy(ωzizz ωxixz ωyiyz) ωz(ωyiyy ωxixy ωziyz) = Iyy ωy Ixy ωx Iyz ωz ωx(ωzizz ωyiyz ωxixz)+ωz(ωxixx ωyixy ωzixz) Izz ωz Ixz ωx Iyz ωy +ωx(ωyiyy ωxixy ωziyz) ωy(ωxixx ωyixy ωzixz) Which is too much for any mortal. We simplify as: For spacecraft, we have Iyz = Ixy = Ixz = 0 (two planes of symmetry). For aircraft, we have Iyz = Ixy = 0 (one plane of symmetry). If we use the matrix version of the cross-product, we can write M = I ω(t) + [ω(t)] Iω(t) Which is a much-simplified version of the dynamics! Recall x y z 0 z y = z 0 x y x 0
Equations of Motion Euler Moment Equations With I xy = I yz = I xz = 0, we get: Euler s Equations L M = I xx ω x + ω y ω z (I zz I yy ) I yy ω y + ω x ω z (I xx I zz ) N I zz ω z + ω x ω y (I yy I xx ) Thus: Rotational variables (ω x, ω y, ω z ) do not depend on translational variables (u,v,w). For spacecraft, Moment forces (L,M,N) do not depend on rotational and translational variables. Can be decoupled However, translational variables (u,v,w) depend on rotation (ω x, ω y, ω z ). But we don t care. M. Peet Lecture 16: 17 / 39
Euler Equations Torque-Free Motion Notice that even in the absence of external moments, the dynamics are still active: 0 I x ω x + ω y ω z (I z I y ) 0 = I y ω y + ω x ω z (I x I z ) 0 I z ω z + ω x ω y (I y I x ) which yield the 3-state nonlinear ODE: ω x = I z I y ω y (t)ω z (t) I x ω y = I x I z ω x (t)ω z (t) I y ω z = I y I x ω x (t)ω y (t) I z Thus even in the absence of external moments The axis of rotation ω will evolve Although the angular momentum vector h will NOT. occurs because tensor I changes in inertial frame. This can be problematic for spin-stabilization! M. Peet Lecture 16: 18 / 39
Euler Equations Spin Stabilization introduce products of inertia in the spacecraft inertia axis is illustrated in Figure 5. Conversely We can use Euler s equation to study Spin Stabilization. tensor. is spinning about its major axis, any nu Angular Momentum coincides will simply decay. There are twowith important Nominal Spin cases: Axis ASMOS can be used to investiga energy dissipation effects. With ASMO introduce various rates of internal energy the rigid body model by entering vi coefficients and wheel inertias. The user resulting motion. This motion can also further analysis. Figure 3. Spin Stabilized Spacecraft Angular Momentum Vector & New Spin Ax Axisymmetric: I x = I y I = I x 0 0 0 I x 0 0 0 I z Angular Momentum Vector Non-Axisymmetric: Figure 1: Sketch of NEAR Shoemaker I x spacecraft I y 3 [4] C.D. Hall, P. Tsiotras, and H. Shen. Tracking Rigid Body Motion Using Thrusters and Momentum Wheels. The Journal of the Astronautical Sciences, 50(3):311 323, 2002. I [5] C.D. = Hall. 473.924 0 0 Spacecraft 0 Dynamics and 494.973 Control, AOE 4140 Class Notes. 0 kg m 2 http://www.aoe.vt.edu/ cdhall/courses/aoe4140/, February 5, 2003. 0 0 269.83 [6] J.L. Meriam and L.G. Kraige. Engineering Mechanics: Dynamics. John Wiley & Sons, inc., 4 edition, 1997. Old Sp M. Peet Lecture 16: 19 / 39
Angular Momentum coincides with Nominal Spin Axis Figure 1: Sketch of NEAR Shoemaker spacecraft 3 [4] C.D. Hall, P. Tsiotras, and H. Shen. Tracking Rigid Body Motion Using Thrusters and Momentum Wheels. The Journal of the Astronautical Sciences, 50(3):311 323, 2002. [5] C.D. Hall. Spacecraft Dynamics and Control, AOE 4140 Class Notes. http://www.aoe.vt.edu/ cdhall/courses/aoe4140/, February 5, 2003. [6] J.L. Meriam and L.G. Kraige. Engineering Mechanics: Dynamics. John Wiley & Sons, inc., 4 edition, 1997. 11 2018-04-19 Lecture 16 Euler Equations Note we say a body is axisymmetric if I x = I y. We don t need rotational symmetry... Euler Equations Spin Stabilization introduce products of inertia in the spacecraft inertia axis is illustrated in Figure 5. Conversely We can use Euler s equation to study Spin Stabilization. tensor. is spinning about its major axis, any nu will simply decay. There are two important cases: Figure 3. Spin Stabilized Spacecraft Axisymmetric: Ix = Iy Ix 0 0 I = 0 Angular Momentum Ix 0 Vector 0 0 Iz Nominal Spin Axis Nutation Angle Figure 4. Spacecraft Nutational Motion In the absence of energy dissipation, nutational motion is stable about the axis of either the maximum or minimum moment of inertia. This implies that the amplitude of motion is bounded by initial conditions. However, all real spacecraft experience some form of energy dissipation. In this case, nutational motion is only stable about the axis of maximum moment of inertia. The axes of minimum and maximum moments of inertia are referred to as minor and major axes, respectively. Thus, if a spacecraft is spinning about its minor axis, nutational motion will grow until the spacecraft tumbles and eventually reorients itself spinning about its major axis. Reorientation of the spin ASMOS can be used to investiga energy dissipation effects. With ASMO introduce various rates of internal energy the rigid body model by entering vi coefficients and wheel inertias. The user resulting motion. This motion can also further analysis. Angular Momentum Vector & New Spin Ax Non-Axisymmetric: Ix Iy 473.924 0 0 0 494.973 0 kg m 2 I = 0 0 269.83 Conclusion Old Sp Figure 5. Reorientation of the Sp ASMOS is a simulation tool th animated 3-D computer graphics to visu attitude motion. The program runs personal computers and features pull do dialog boxes making the program accessi use. The program is capable of animating a wide range of rigid body a The rigid body model includes an e investigating stability and energy dissipat References 1. Hughes, P.C., "Spacecraft Attitude D Wiley & Sons, Inc., 1986. 2. Kaplan, M.H., "Modern Spacecraf Control," John Wiley & Sons, Inc., 1986. 3. Wertz, J.R, "Spacecraft Attitude Det Control," D. Reidel Publishing Co., 1985. 4. Lampton, C., "Flights of Fantasy / Pr Video Games in C++", Waite Group Pres
Spin Stabilization Axisymmetric Case An important case is spin-stabilization of an axisymmetric spacecraft. Assume symmetry about z-axis (I x = I y ) Then recall Thus ω z = constant. ω z = I y I x I z ω x (t)ω y (t) = 0 The equations for ω x and ω y are now [ ] [ ωx 0 Iz Iy I = x ω y Ix Iz I y ω z 0 Which is a linear ODE. ω z introduce products of inertia in th tensor. Angular Momentum coi with Nominal Spin A ] [ωx ] (t) ω y (t) Figure 3. Spin Stabilized S Angular M Vec M. Peet Lecture 16: 20 / 39
Spin Stabilization Axisymmetric Case Fortunately, linear systems have closed-form solutions. let λ = Iz Ix I x ω z. Then Combining, we get which has solution Differentiating, we get ω x (t) = λω y (t) ω y (t) = λω x (t) ω x (t) = λ 2 ω x (t) ω x (t) = ω x (0) cos(λt) + ω x(0) λ sin(λt) ω y (t) = ω x(t) = ω x (0) sin(λt) ω x(0) cos(λt) λ λ = ω x (0) sin(λt) + ω y (0) cos(λt) ω x (t) = ω x (0) cos(λt) ω y (0) sin(λt) M. Peet Lecture 16: 21 / 39
Spin Stabilization Axisymmetric Case Define ω xy = ωx 2 + ωy. 2 ω 2 xy = (ω x (0) sin(λt) + ω y (0) cos(λt)) 2 + (ω x (0) cos(λt) ω y (0) sin(λt)) 2 = ω x (0) 2 sin 2 (λt) + ω y (0) 2 cos 2 (λt) + 2ω x (0)ω y (0) cos(λt) sin(λt) + ω x (0) 2 cos 2 (λt) + ω y (0) 2 sin 2 (λt) 2ω x (0)ω y (0) cos(λt) sin(λt) = ω x (0) 2 (sin 2 (λt) + cos 2 (λt)) + ω y (0) 2 (cos 2 (λt) + sin 2 (λt)) = ω x (0) 2 + ω y (0) 2 Thus ω z is constant rotation about axis of symmetry ωx 2 + ωy 2 is constant rotation perpendicular to axis of symmetry This type of motion is often called Precession! M. Peet Lecture 16: 22 / 39
Circular Motion in the Body-Fixed Frame Thus ω(t) = ω x(t) cos(λt) sin(λt) 0 ω y (t) = sin(λt) cos(λt) 0 ω x(0) ω y (0) = R 3 (λt) ω x(0) ω y (0) ω z (t) 0 0 1 ω z (0) ω z (0) M. Peet Lecture 16: 23 / 39
Lecture 16 Circular Motion in the Body-Fixed Frame 2018-04-19 Circular Motion in the Body-Fixed Frame Thus ωx(t) cos(λt) sin(λt) 0 ωx(0) ωx(0) ω(t) = ωy(t) = sin(λt) cos(λt) 0 ωy(0) = R3(λt) ωy(0) ωz(t) 0 0 1 ωz(0) ωz(0) For λ > 0, this is a Positive (counterclockwise) rotation, about the z-axis, of the angular velocity vector ω as expressed in the body-fixed coordinates!
Prolate vs. Oblate The speed of the precession is given by the natural frequency: λ = I z I x ω z I x with period T = 2π λ = 2πIx I z I x ωz 1. Direction of Precession: There are two cases Definition 4 (Direct). An axisymmetric (about z-axis) rigid body is Prolate if I z < I x = I y. Definition 5 (Retrograde). An axisymmetric (about z-axis) rigid body is Oblate if I z > I x = I y. Thus we have two cases: λ > 0 if object is Oblate (CCW rotation) λ < 0 if object is Prolate (CW rotation) Note that these are rotations of ω, as expressed in the Body-Fixed Frame. M. Peet Lecture 16: 24 / 39
Pay Attention to the Body-Fixed Axes Figure: Prolate Precession Figure: Oblate Precession The black arrow is ω. The body-fixed x and y axes are indicated with red and green dots. Notice the direction of rotation of ω with respect to these dots. The angular momentum vector is the inertial z axis. M. Peet Lecture 16: 25 / 39
Motion in the Inertial Frame As these videos illustrate, we are typically interested in motion in the Inertial Frame. Use of Rotation Matrices is complicated. Which coordinate system to use??? Lets consider motion relative to h. Which is fixed in inertial space. We know that in Body-Fixed coordinates, I x ω x h = I ω = I y ω y I z ω z Now lets find the orientation of ω and ẑ with respect to this fixed vector. M. Peet Lecture 16: 26 / 39
Motion in the Inertial Frame Let ˆx, ŷ and ẑ define the body-fixed unit vectors. We first note that since I x = I y and h = Ix ω xˆx + I y ω y ŷ + I z ω z ẑ = I x (ω xˆx + ω y ŷ + ω z ẑ) + (I z I x )ω z ẑ = I x ω + (I z I x )ω z ẑ we have that ω = 1 I x I z h + ẑ I x I x ω z which implies that ω lies in the ẑ h plane. M. Peet Lecture 16: 27 / 39
Motion in the Inertial Frame We now focus on two constants of motion Since θ - The angle h makes with the body-fixed ẑ axis. γ - The angle ω makes with the body-fixed ẑ axis. h x I x ω x h = h y = I y ω y h z I z ω z The angle θ is defined by h 2 x + h 2 y I x ω x 2 + ωy 2 tan θ = = h z I z ω z = I x I z ω xy ω z Since ω xy and ω z are fixed, θ is a constant of motion. M. Peet Lecture 16: 28 / 39
2018-04-19 Lecture 16 Motion in the Inertial Frame Motion in the Inertial Frame We now focus on two constants of motion θ - The angle h makes with the body-fixed ẑ axis. γ - The angle ω makes with the body-fixed ẑ axis. Since h = hx hy = Ixωx Iyωy hz Izωz The angle θ is defined by h 2 x + h 2 y Ix ωx 2 + ωy 2 tan θ = = = Ix ωxy hz Izωz Iz ωz Since ωxy and ωz are fixed, θ is a constant of motion. Again, h here is in the body-fixed frame This is why it changes over time.
Motion in the Inertial Frame The second angle to consider is γ - The angle ω makes with the body-fixed ẑ axis. As before, the angle γ is defined by ωx 2 + ωy 2 tan γ = = ω xy ω z ω z Since ω xy and ω z are fixed, γ is a constant of motion. We have the relationship tan θ = I x I z ω xy ω z Thus we have two cases: 1. I x > I z - Then θ > γ = I x I z tan γ 2. I x < I z - Then θ < γ (As Illustrated) M. Peet Lecture 16: 29 / 39
Motion in the Inertial Frame Figure: The case of I x > I z (θ > γ) Figure: The case of I z > I x (γ > θ) M. Peet Lecture 16: 30 / 39
Lecture 16 Motion in the Inertial Frame 2018-04-19 Motion in the Inertial Frame Figure: The case of Ix > Iz (θ > γ) Figure: The case of Iz > Ix (γ > θ) We illustrate the motion using the Space Cone and Body Cone The space cone is fixed in inertial space (doesn t move) The space cone has width ω θ The body cone is centered around the z-axis of the body. In body-fixed coordinates, the space cone rolls around the body cone (which is fixed) In inertial coordinates, the body cone rolls around the space cone (which is fixed)
Motion in the Inertial Frame The orientation of the body in the inertial frame is defined by the sequence of Euler rotations ψ - R 3 rotation about h. Aligns hx perpendicular to ẑ. θ - R 1 rotation by angle θ about h x. Rotate hz-axis to body-fixed ẑ vector We have shown that this angle is fixed! θ = 0. φ - R 3 rotation about body-fixed ẑ vector. Aligns hx to ˆx. The Euler angles are related to the angular velocity vector as ω x ψ sin θ sin φ ω y = ψ sin θ cos φ ω z φ + ψ cos θ = constant M. Peet Lecture 16: 31 / 39
Lecture 16 Motion in the Inertial Frame The orientation of the body in the inertial frame is defined by the sequence of Euler rotations 2018-04-19 Motion in the Inertial Frame ψ - R3 rotation about h. Aligns hx perpendicular to ẑ. θ - R1 rotation by angle θ about hx. Rotate hz-axis to body-fixed ẑ vector We have shown that this angle is fixed! θ = 0. φ - R3 rotation about body-fixed ẑ vector. Aligns hx to ˆx. The Euler angles are related to the angular velocity vector as ωx ψ sin θ sin φ ωy = ψ sin θ cos φ ωz φ + ψ cos θ = constant This comes from 0 0 0 ω = R 3(φ)R 1(θ)R 3(ψ) 0 + R 3(φ)R 1(θ) θ + R 3(φ) 0 ψ 0 φ ψ sin θ sin φ 0 0 = ψ sin θ cos φ + 0 + 0 ψ cos θ 0 φ
Motion in the Inertial Frame To find the motion of ω, we differentiate ω x ψ φ sin θ cos φ ω y = ψ φ sin θ sin φ ω z 0 Now, substituting into the Euler equations yields ψ = I z (I x I z ) cos θ φ There are two cases here: I x > I z - Direct precession ψ and φ aligned. I y > I x - Retrograde precession ψ and φ are opposite. M. Peet Lecture 16: 32 / 39
2018-04-19 Lecture 16 Motion in the Inertial Frame Motion in the Inertial Frame To find the motion of ω, we differentiate ωx ψ φ sin θ cos φ ωy = ψ φ sin θ sin φ ωz 0 Now, substituting into the Euler equations yields Iz ψ = (Ix Iz) cos θ φ There are two cases here: Ix > Iz - Direct precession ψ and φ aligned. Iy > Ix - Retrograde precession ψ and φ are opposite. Recall ω x and ω y can be expressed in terms of ω x and ω y
Motion in the Inertial Frame Figure: Retrograde Precession (I z > I x) Figure: Direct Precession (I z < I x) M. Peet Lecture 16: 33 / 39
Mathematica Demonstrations Mathematica Precession Demonstration M. Peet Lecture 16: 34 / 39
Prolate and Oblate Spinning Objects Figure: Prolate Object: Ix = Iy = 4 and Iz = 1 M. Peet Figure: Oblate Object: Vesta Lecture 16: 35 / 39
Next Lecture Note Bene: Precession of a spacecraft is often called nutation (θ is called the nutation angle). By most common definitions, for torque-free motions, N = 0 Free rotation has NO nutation. This is confusing M. Peet Lecture 16: 36 / 39
Precession Example: Chandler Wobble Problem: The earth is 42.72 km wider than it is tall. How quickly will the rotational axis of the earth precess due to this effect? Solution: for an axisymmetric ellipsoid with height a and width b, we have I x = I y = 1 5 m(a2 + b 2 ) and I z = 2 5 mb2. Thus b = 6378km, a = 6352km and we have (m e = 5.974 10 24 kg) I z = 9.68 10 37kg m2, I x = I y = 9.72 10 37kg m2 If we take ω z = 2π T = 2πday 1, then we have λ = I z I x ω z =.0041day 1 I x That gives a period of T = 2π λ as the Chandler Wobble. Note: This is only the Torque-free precession. = 243.5days. This motion of the earth is known M. Peet Lecture 16: 37 / 39
Precession Actual period is 434 days Actual Ix = I y = 8.008 10 37 kg m 2. Actual Iz = 8.034 10 37 kg m 2. Which would predict T = 306days M. Peet Lecture 16: 38 / 39
Lecture 16 Precession 2018-04-19 Precession Actual period is 434 days Actual Ix = Iy = 8.008 10 37 kg m 2. Actual Iz = 8.034 10 37 kg m 2. Which would predict T = 306days The precession of the earth was first notices by Euler, D Alembert and Lagrange as slight variations in lattitude. Error partially due to fact Earth is not a rigid body(chandler + Newcomb). Magnitude of around 9m Previous plot scale is milli-arc-seconds (mas)
Next Lecture In the next lecture we will cover Non-Axisymmetric rotation Linearized Equations of Motion Stability Energy Dissipation The effect on stability of rotation M. Peet Lecture 16: 39 / 39