Accurate Simulation of Rigid Body Rotation and Extensions to Lie Groups Sam Buss Dept. of Mathematics U.C. San Diego Visible Luncheon Gwyddor Cyfrifiador University of Swansea March 3, 2011 Topics: ◮ Algorithms for simulating rotating rigid bodies. ◮ All algorithms preserve angular momentum. ◮ Algorithms can be made energy preserving. ◮ Generalization to Lie group setting. Talk outline: 1. Rigid body rotations. 1st thru 4th order algorithms. Unexpected terms. 2. Generalization to Taylor series methods over Lie groups/Lie algebras. 3. Energy preservation based on Poinsot ellipsoid. 4. Numerical simulations and efficiency. 1 Part I: The simple rotating, rigid body I = Inertia matrix (tensor). ω L = Angular momentum. ω = Rotation axis & rate, L L = Iω (Euler’s equation) ω = I−1L ω˙ = I−1(L˙ ω Iω) − × ω¨ = ω ω˙ + I−1(L¨ ω˙ L 2ω L˙ + ω (ω L)) × − × − × × × ... ... ω = 2ω ω¨ ω (ω ω˙ ) + I−1[L 3ω L¨ 3ω˙ L˙ ω¨ L × − × × − × − × − × + ω˙ (ω L) + 2ω (ω˙ L) + 3ω (ω L˙ ) ω (ω (ω L))] × × × × × × − × × × Wobble: ω˙ = 0 even when no applied torque (L˙ = 0). 6 vL˙ = Rate of change of momentum = Applied Torque. 2 Framework for simulating rigid body motion We assume the rigid body has a known angular momentum, and the external torques are completely known. The orientation (and hence the angular velocity) is updated in discrete time steps, at times t0, t1, t2,.... Update Step: At a given time t , let h = ∆t = t t , and assume i i+1 − i orientation Ωi at time ti is known, and that momentum is known (at all times). Update step calculates a net rotation rate vector, ω¯, and sets Ωi+1 = Rhω¯ Ωi, where Rν performs a rotation around axis ν of angle ν . || || Nearly every rigid body simulation method fits this framework. 3 First-order update method Use ω = I−1L as the estimate for ω¯. First-order algorithm: Set := −1 . ω¯ ωi = Ii Li Set Ωi+1 := Rhω¯ Ωi. This first-order method performs poorly. A wobbling, spinning object quickly gains energy and soon ends up spinning on a principal axis. “Good enough for computer games” (?) 4 Second-order update method h Use ω and ω˙ to estimate ω¯ as ω¯ = ω + 2ω˙ . Second-order algorithm: Set := −1 . ωi Ii Li Set := −1 ˙ . ω˙ i Ii (Li ωi Li) Set := h −. × ω¯ ωi + 2ω˙ i Set Ωi+1 := Rhω¯ Ωi. The second-order method performs substantially better. However, a wobbling, spinning object still steadily gains energy and ends up spinning on a principal axis. 5 False third-order update method h h2 Try using ω, ω˙ and ω¨ to estimate ω¯ as ω¯ = ω + 2ω˙ + 3! ω¨. False third-order algorithm: Set := −1 . ωi Ii Li Set := −1 ˙ . ω˙ i Ii (Li ωi Li) Set := − −1ר ˙ . ω¨ i ωi ω˙i + Ii (Li ω˙ i Li 2ωi Li + ωi (ωi Li)) Set := × 1 1 2−. × − × × × ω¯ ωi + 2ω˙ ih + 6ω¨h Set Ωi+1 := Rhω¯ Ωi. Surprisingly, this however turns out to be slightly worse than the second- order method! In fact, the Taylor series estimate for ω¯ is not second-order accurate. 6 The new third-order term - Motivation A rolling disk - Total rotation ϕ in time t0 (or tall skinny cone): b a′ ϕ ϕ c c′ a b′ 1 p p + 2vt0 p + vt0 ω - at point p, pointing into figure, i.e., away from viewer. ω˙ - pointing to the right. 7 The new third-order term - Motivation A rolling disk - Total rotation ϕ in time t0 (or tall skinny cone): ϕ q b a′ c ′ ϕ/2 ϕ/2 c a b′ 1 p p + 2vt0 p + vt0 ω - at point p, pointing into figure, i.e., away from viewer. ω˙ - pointing to the right. ω¯ - correct value is the point q, since a rotation around q yields the correct net motion. 8 ϕ ω¯ ∞ - ideal rotation center ω¯ ∞ ω¯ 2+ ω¯ 1 ω¯ 2 ω¯ 1 - 1st order update ω¯ 2 - 2nd order update ω¯ 2+ - augmented 2nd order. 2 ω¯ = ω¯ + h ω˙ ω. 2+ 2 12 × 9 Augmented second-order update method 2 Second-order approximation: ω¯ = ω¯ + h ω˙ ω. 2+ 2 12 × Augmented second-order algorithm: Set := −1 . ωi Ii Li Set := −1 ˙ . ω˙ i Ii (Li ωi Li) − ×2 Set ω¯ := ω + hω˙ + h (ω˙ ω ). i 2 i 12 i× i Set Ωi+1 := Rhω¯ Ωi. The augmented second-order method performs substantially better than the second-order method, and has more energy stability, although the energy does drift steadily. Extra computation cost: only one more cross-product than the second- order method. 10 True third-order update method 2 Now include the new h ω˙ ω term in ω¯. 12 × True third-order algorithm: Set := −1 . ωi Ii Li Set := −1 ˙ . ω˙ i Ii (Li ωi Li) Set := − −1ר ˙ . ω¨ i ωi ω˙i + Ii (Li ω˙ i Li 2ωi Li + ωi (ωi Li)) × 2 − 2 × − × × × Set ω¯ := ω + hω˙ + h ω¨ + h ω˙ ω . i 2 i 6 i 12 i× i Set Ωi+1 := Rhω¯ Ωi. As expected, this is third-order correct, and performs better than the augmented second-order method. 11 True fourth-order update method 3 Additional new term: h ω¨ ω . (Pattern does not continue) 24 i× i True fourth-order algorithm: Set := −1 . ωi Ii Li Set ω˙ := I−1(L˙ ω L ). i i i − i× i Set ω¨ := ω ω˙ + I−1(L¨ ω˙ L 2ω L˙ + ω (ω L )). ...i i× i i i − i× i − i× i i× i× i Set ωi := ( equation on earlier slide #2 ). ··· 2 2 3 ... ···3 Set ω¯ := ω + hω˙ + h ω¨ + h ω˙ ω + h ω + h ω¨ ω . i 2 i 6 i 12 i× i 24 i 24 i× i Set Ωi+1 := Rhω¯ Ωi. Performs better than the true third-order method. Experiments confirm fourth-order accuracy. 12 Part II: Generalize to Lie groups/Lie algebras The extra third- and fourth-order terms can be generalized to the Lie group / Lie algebra setting. This gives Taylor series methods over Lie groups. Related to: Runge-Kutta methods on Lie groups by Crouch & Grossman ’93; Marthinsen & Owren ’98; Munthe-Kaas ’98,’99; who give higher order corrector terms for Runge-Kutta algorithms. We write [u, v] for u v. Also, [u, v, w] for [u, [v, w]]. u, v,... are Lie group elements, and [×, ] is a Lie group product. · · We now use “W ” instead of “ω”, etc. These are elements of the associated Lie algebra. For Z is in the Lie algebra, exp(Z) is in the Lie group. z = exp(Z) is analogous to the rotation operation represented by rotation vector Z. So, Z ω¯ and exp(Z) Rω. ∼ ∼ ¯ 13 Suppose W (t) is a time-varying Lie algebra element. Let h > 0. We want to find a Z = Z(h) which is equivalent to applying W (t) over the time interval 0 to h: 0 h ih exp(h Z) = lim exp W . · N→∞ N · N i=YN−1 Analogy: W (t) is time varying instantaneous rotation vector. Z is ω¯. Goal: Find power series for Z in terms of W (0), W˙ (0), W¨ (0),.... Let Y = Y (h)= h Z. Let y(h) = exp(Y ). Now, by defn of Z, Y , y, · y′(t) = W (t). Also, taking first derivative of y(h) = exp(Y ), ′ ′ y (t) = (d exp)Y (t)(Y (t)). 14 Power Series expansions: (W0 = W (0), W˙ 0 = W˙ (0), etc.) 1 1 1 (d exp) = 1+ ad(Y )+ (ad(Y ))2 + (ad(Y ))3 + Y 2 3! 4! ··· (recall (ad(A))(B)=[A, B].) 1 2 1 3 ... W (t) = W0 + tW˙ 0 + t W¨ 0 + t W 0 + , 2 3! ··· Y (t) = tY + 1t2Y + 1 t3Y + 1 t4Y + , 0 2 1 3! 2 4! 3 ··· Y ′(t) = Y + Y t + 1Y t2 + 1 Y t3 + . 0 1 2 2 3! 3 ··· Equating coefficients of powers of t and solving for Yi’s gives: Y0 = W0. Y1 = W˙ 0. 1 Y = W¨ + [W˙ ,W ]. analogous to the “2+” term. 2 0 2 0 0 ... Y3 = W 0 +[W¨ 0,W0]. analogous to the “3+” term. 15 .... 3 ... 1 1 Y4 = W 0 + [W 0,W0]+[W¨ 0, W˙ 0]+ [W˙ 0, W˙ 0,W0] [W0, W¨ 0,W0] 2 2 − 6 1 ˙ 6[W0,W0, W0,W0]. .....− ... .... 5 ˙ ¨ ˙ 1 ˙ ¨ Y5 = W 0 + 2[W 0, W0]+2[W 0,W0]+2[W0, W0,W0]+ 2[W0, W0,W0] 1 ... 1 [W0, W 0,W0] [W0,W0, W¨ 0,W0] [W0, W˙ 0, W˙ 0,W0]. −2 − 2 − Fifth-order accurate formula for Z(h)= h−1Y (h): 2 3 ... 3 h ˙ h ¨ 1 2 ˙ h h ¨ Z(h) = W0 + 2W0 + 6 W0 + 12h [W0,W0]+ 24W 0 + 24[W0,W0] 4 .... 4 ... 4 4 h h h ¨ ˙ h ˙ ˙ +120W 0 + 80[W 0,W0]+ 120[W0, W0]+ 240[W0, W0,W0] 4 4 h [W , W¨ ,W ] h [W ,W , W˙ ,W ] −720 0 0 0 − 720 0 0 0 0 5 ..... 5 ... 5 .... 5 h h ˙ h h ¨ ˙ +720 W 0 + 288[W 0, W0]+ 360[W 0,W0]+ 360[W0, W0,W0] h5 h5 ... h5 + [W˙ 0, W¨ 0,W0] [W0, W 0,W0] [W0,W0, W¨ 0,W0] 1440 − 1440 − 1440 5 5 h [W ,W , W¨ ,W ] h [W , W˙ , W˙ ,W ]+ O(h6). −1440 0 0 0 0 − 720 0 0 0 0 16 Part III: Back to rigid body: Poinsot inertial ellipsoid Inertial ellipsoid Polhode Invariable plane Herpolhode Inertial ellipsoid is attached to the rigid body, and rolls on the plane.