Final exam date Final exam date has been announced: Dynamics III - PDF document

Final exam date  Final exam date has been announced: Dynamics III  Tuesday, February 27, 2007  2:45 - 4:45pm Numerical Integration  70-1435 Projects Assignments  Presentations:  Assignment 1 -- Framework  Most have been graded  Dates:  If not submitted please do so.  Week 9: Wed, Feb 14  Week 10: Mon, Feb 19  Assignment 2 -- Keyframing  Finals Week: Tues, Feb 27 (2:45-4:45)  Due Friday, Jan 12.  Questions?  15 minutes / presentation  Schedule now on Web  Assignment 3 -- Billiards  Please send me choice of time/day  To be given today Logistics Potential job opportunity  Prof Raj will be teaching a course on  Course Withdrawal deadline Saturdays to high school students.  Friday, January 26th  Looking for lab assistants  7 Saturdays thru May  Please contact rkr@cs.rit.edu 1

Plan Motivation Films  Physics 101 for rigid body animation  Computer Animated Feature Films  35 since (and including) Toy Story  Last Mon: translation and rotational dynamics  17 currently in Production  Last Wed: Forces, impacts, and collisions  http://www.boxofficemojo.com/genres/chart/?id=computeranimation.htm  Today: Numerical integration.  Major players  Pixar  But first…  PDI/Dreamworks  Blue Sky / FOX  Sony Imageworks  Disney Motivational Film Motivational Film  Short animations by Sony Imageworks  The ChubbChubbs (2002)  First animated short produced by Sony Imageworks  Previewed before Men in Black II.  Winner of the 2002 Academy Award for best animated short. Motivational Film Plan For Today  Early Bloomer (2003)  Topics  Started as an in-house training exercise  Numerical Integration  Shown at SIGGRAPH 2003.  Life Beyond Euler  Assignment #3  A Night at the Pool Hall 2

Laws of Motion Linear Motion Law I  For Linear Physical Motion  Every object in a state of uniform motion tends to remain Mass   in that state of motion unless an external force is applied  Measure of the amount of matter in a body to it. (Inertia)  From Law II: Measure of the a body’s resistance to motion Velocity Law II:    Change of motion with respect to time The acceleration of a body is proportional to the resulting  Acceleration force acting on the body, and this acceleration is in the   Change of velocity with respect to time same direction as the force. Force  Law III:   In short, force is what makes objects accelerate For every action there is an equal and opposite reaction. Momentum    mass x velocity  Another way of stating Law I: Momentum is conserved Rotational Motion Where we are Object properties  For Rotational Physical Motion Calculate forces Inertia Position, orientation   Measure of the amount/distribution of matter in a body Linear and angular velocity  From Law II: Measure of the a body’s resistance to motion Linear and angular momentum Angular Velocity  mass  Change of rotation with respect to time Angular Acceleration   Change of velocity with respect to time Torque   In short, torque is what makes objects rotate Angular Momentum  Update object properties Calculate accelerations  Inertia x velocity Using mass, momenta  Another way of stating Law I: Momentum is conserved Where we are Where we are  State of object at any given time  Derivative of object state position s ( t ) v ( t ) & � � � � s ( t ) � � � � � � & R ( t ) ( t ) R ( t ) rotation (in world coords) � � � � R ( t ) & � � � � S ( t ) momentum � � = = S ( t ) = & M ( t ) F ( t ) � � � � M ( t ) � � � � � � angular momentum � � & L ( t ) ( t ) � L ( t ) � � � � � � 3

Putting it all together Putting it all together  Step 1 Step 1  Calculate Forces, F(t), τ (t)  Calculate Forces, F(t), τ (t)  Step 2   Step 2 Integrate position/rotation   Integrate position/rotation s(t + Δ t) = s(t) + v(t) Δ t  q(t + Δ t) = q(t) + 0.5 ( ω (t)q(t)) Δ t /* normalize to avoid problems */  s(t + Δ t) = s(t) + v(t) Δ t  R(t + Δ t) = quatToRot (q(t + Δ t) )   R(t + Δ t) = R(t) + ( ω (t)*R(t)) Δ t /* CAREFUL HERE */  r(t + Δ t) = s(t + Δ t) + r body R(t + Δ t)  r(t + Δ t) = s(t + Δ t) + r body R(t + Δ t) Update Momentum (integrate accelleration)  Update Momentum (integrate accelleration)  M(t + Δ t) = M(t) + F(t) Δ t  M(t + Δ t) = M(t) + F(t) Δ t  L(t + Δ t) = L(t) + τ (t) Δ t   L(t + Δ t) = L(t) + τ (t) Δ t Putting It all together Numerical Integration  Step 3  Problems we are considering are first  Calculate velocities (for next step) order, initial value, ordinary differential  v(t + Δ t) = M(t + Δ t)/m + impulse equations (ODE)  I -1 (t + Δ t) = R(t + Δ t)I -1 body R(t + Δ t) T  We have derivative (acceleration, velocity)  ω (t + Δ t) = + I -1 (t + Δ t)L(t + Δ t) + impulse  We have initial values  Go to step 1  We need integral (velocity, position)  Questions? Numerical Integration Numerical Integration  Said another way  The derivative can be seen as forming a vector field in 2 dimensions dx v f ( x t , ) � = = dt  f’(t, x) = the derivate of our function at time t at point x. 4

Numerical Integration Numerical Integration  And  Instead  we start at an initial point x(t 0 ) t  Step along the field (using f’) to determine x ( t ) x ( t ) f ( t , x ) dt � = + � value at subsequent time steps 0 t 0  For most interesting cases, this integral cannot be calculate analytically Numerical Integration Numerical Integration  Euler method  Euler method  Let  Assumes average gradient over h is the gradient at time t  t i+1 = t i + h  then x x h f ( t , x ) = + � � i + 1 i i i Numerical Integration Numerical Integration  Euler Method  Taylor series expansion of a function  Pros dx h 2 d 2 x h 3 d 3 x  Easy x ( t h ) x ( t ) h ( t ) L + = + + + +  Intuitive 2 3 dt 2 ! dt 3 ! dt  Cons  Error prone  Can be unstable Euler All this is method stops error here 5

Numerical Integration Numerical Integration  Euler is said to be a 1 st order method  Note that we can improve on Euler method by reducing h dx h 2 d 2 x h 3 d 3 x x ( t h ) x ( t ) h ( t ) L + = + + + + 2 3 dt 2 ! dt 3 ! dt dx x ( t h ) x ( t ) h ( t ) O ( h 2 ) + = + + dt Numerical Integration Numerical Integration  To improve, consider 2 nd order terms  Problem with direct 2 nd order solutions  Need both f’(t) and f’’(t) at each time to calculate f(t) 2 2 3 3 dx h d x h d x x ( t h ) x ( t ) h ( t )  Which we don’t have + = + + + + L dt 2 ! dt 2 3 ! dt 3  Recall, we are calculating v(t) by applying a first order to a(t) then calculating x(t) by applying first order to v(t)  There is a way to get around this (Adams-Bashforth method), but would require several values for f(t - Δ t), f’(t - 2 Δ t), etc to calculate f(t). 2 nd order All this is method stops  I.e. Not self-starting error here Numerical Integration Numerical Integration  Midpoint Method  Midpoint Method  Assumes average gradient over h is the  Let gradient at the midpoint of h  t i+1 = t i + h  then h x � x x h f ( t , x ) = + � � + + i 1 i i i + 2 2 6

Numerical Integration Numerical Integration  Midpoint method  Find midpoint  But how does one calculate?  x mid = x i + (h/2) f' (t i, x i ) h x  We know x i � f ( t , x ) � + + i i 2 2  Derivative is Δ x / Δ t  Use Euler h x x x � � f ( t , x ) mid i + + = i i h 2 2 2 Numerical Integration Numerical Integration  The Midpoint method can be shown to  Midpoint Method have 2 nd order accuracy  Pros  2 nd Order accuracy  Self starting h x � 3 x x h f ( t , x ) O ( h ) = + � + + +  Cons i 1 i i i + 2 2  Additional application of Euler required at each step. Numerical Integration Numerical Integration  Runge-Kutta Method  4 th order Runge-Kutta Method  Family of methods symmetrical w.r.t. the k 1 = h � f ( t i , x i ) � interval h k  Midpoint method is a 2 nd order Runge- k h f ( t , x 1 ) = � � + + 2 i i 2 2 Kutta method h k  Fourth order Runge-Kutta k h f ( t , x 2 ) � = � + + 3 i i 2 2  Uses gradient at 4 points to estimate gradient over h. k h f ( t h , x k ) = � � + + 4 i i 3  Has 4 th order accuracy 1 x x ( k 2 k 2 k k ) = + + + + i + 1 i 6 1 2 3 4 7