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Animation CS 551 / 651

Baraff, 1991. Animation CS 551 / 651. Lecture 6 Rigid-body Simulation. Assignment 1 Rigid-body Simulator. Goal: To demonstrate understanding of rigid-body dynamics by writing a simulator 3ds file loader (given) Compute COM and MOI (given) User-selected force vector (how?)

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Animation CS 551 / 651

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  1. Baraff, 1991 AnimationCS 551 / 651 Lecture 6 Rigid-body Simulation

  2. Assignment 1Rigid-body Simulator • Goal: To demonstrate understanding of rigid-body dynamics by writing a simulator • 3ds file loader (given) • Compute COM and MOI (given) • User-selected force vector (how?) • Compute location of impact (ray / polygon intersection) • Compute change in linear and angular accelerations (Hecker) • Animate…

  3. Rigid-body Simulator • Not permitted resources • Physical simulation source code • Not from web, books, bathroom wall… • Permitted resources • All books (physics, graphics, C++, game design, …) • Graphics interface libraries • File loader and MOI libraries

  4. Rigid-body Simulator • Deliverables • Load any legal (closed) 3ds model • an aside…

  5. Closed objects • AKA “solid” • More rigorously: closed, orientable manifolds • Local neighborhood of all points isomorphic to disc • Boundary partitions space into interior & exterior No Yes

  6. Manifold • Examples of manifold objects: • Sphere • Torus • Well-formedCAD part

  7. Back-Face Culling • Examples of non-manifold objects: • A single polygon • A terrain or height field • polyhedron w/ missing face • Anything with cracks or holes in boundary • one-polygon thick lampshade

  8. Rigid-body Simulator • Deliverables (basic) • Load any legal (closed) 3ds model • Define an arbitrary force vector • Animate the model as is moves

  9. Rigid-body Simulator • Deliverables (intermediate) • Add gravity and drop object to the ground. Use constraint forces to make it rest there • Weld a spring to a vertex of the model and suspend model by spring from ceiling • Why springs? Because it is much more complicated to force the model to hang from a string (or bar) of fixed length… more constraints to worry about

  10. Rigid-body Simulator • Deliverables (advanced) • Add more objects and springs • Newton’s Cradle • Join two objects with springs • Suspend one object by spring and drop other objects on it • Implement constrained dynamics • Replace springs with rods

  11. Rigid-body Simulator • Grading will take place during a demo you’ll provide for the TA • Time TBA, but as early as 6:15 Wed, Sep 19th • Either bring your machine to him • Or make sure it works on Windows machine with VisualStudio .net • Machines in Stacks and Small Hall should be set up this way

  12. Reading for Monday • Prepare three questions for each paper • Timewarp Rigid Body Simulation, B. Mirtich, SIGGRAPH 2000. • Stephen Chenney and D.A.Forsyth, "Sampling Plausible Solutions to Multi-Body Constraint Problems". SIGGRAPH 2000 Conference Proceedings, pages 219-228, July 2000.

  13. Collisions • We’re in 2D… Collisions are: • Vertex / edge • Vertex / vertex • Edge / edge • Compute normal to collision • v/e: perpendicular to edge (pointing towards A by convention) • e/e: perpendicular to edge • v/v: something reasonable (perhaps use an edge-perp)

  14. Collisions • Compute relative normal velocity • f • Must be negative for a collision to take place • If equal to 0… resting contact (special case)

  15. After collision is detected • Consider applying a force to both bodies • This is how nature “simulates” collisions • Already interpenetrating objects will remain in this state for at least one more time step • Cannot instantaneously change velocity • Forces need time to be integrated to accelerations and velocities

  16. After collision is detected • We need instantaneous change in velocity • Impulse • This is a hack that gets us out of the jam we created when assuming impenetrable bodies exist • Generalization of subtle surface properties • Like simulating a large force for a small time step • Will change velocity like we need

  17. Calculating impulse • Duration of impulse is “no time” • This is a small amount of time • All other forces are ignored during this period • No friction • Coefficient of resitution • Models complicated compression and restitution of impacting bodies • Models dissipation of energy

  18. Coefficient of restitution • e = 1  superball (perfectly elastic) • e = 0  clay (perfectly inelastic)

  19. Calculating an impulse • Solve for one number, j • Apply j in direction of n to A • Apply j in direction of –n to B equal and opposite

  20. Computing Impulse • 1st: Assume objects cannot rotate • 2nd: Use definition of coeff. of rest. to derive a second set of v+ equations • 3rd: Use substitution to solve for j

  21. Computing impulse • Things to note: • n doesn’t have to be normalized • A or B can be fixed by setting mass to infinity • If MA = 1, MB = inf, vB = 0, e=1 • Computes reflection of vA about n

  22. Accounting for rotation • Consider velocity of collision point P after collision • Derived from two equations

  23. Accounting for rotation • Returning to elasticity • Start substituting • Solve for j

  24. Notes • Time of impact • Must apply impulse exactly at time of impact • After detecting interpenetration, use binary search (or more sophisticated) to fine tune • Beware of “tunneling” when dt is so large collisions are missed • Edge/edge collisions are modeled as point/point in this system • Only two colliding bodies at a time • 3D is harder because of variety of collision types

  25. Numerical Integration (from Dr. Tom Hobbs Systems Ecology course at Colorado State) Y = f(t), unknown f(t+t), unknown Y known t, specified t

  26. Example point to estimate Analytical solution to dy/dt Y0 = 10  t = 0.5

  27. Euler (pronounced “oiler”) y0 = 10 analytical y k1 = dy/dt at y0 k1 = 6*10-.007*(10)2 y = k1*t yest=y0 + y y estimated y y  t = 0.5

  28. Runge-Kutta (pronounced Run-gah Kut-tah) point to estimate Problem: estimate the slope to calculate y y  t = 0.5

  29. Runge-Kutta (4th order) slope = k1 y t/2 t

  30. Step 1: Evaluate slope at current value of state variable. y0 = 10 k1 = dy/dt at y0 k1 = 6*10-.007*(10)2 k1 = 59.3 k1=slope 1

  31. Step 2: Calculate y1at t +t/2 using k1. Evaluate slope at y1. y1 = y0 + k1* t /2 y1 = 24.82 k2 = dy/dx at y1 k2 = 6*24.8-.007*(24.8)2 k2 = 144.63 k2=slope 2 y1  t = 0.5/2

  32. Step 3: Calculate y2 at t +t/2 using k2. Evaluate slope at y2. y2 = y0 + k2* t /2 y2 = 46.2 k3 = slope 3 k3 = dy/dt at y2 k3 = 6*46.2-.007*(46.2)2 k3 = 263.0 y2  t = 0.5/2

  33. Step 4: Calculate y3 at t +t using k3. Evaluate slope at y3. y3 = y0 + k3* t y3 =141.0 y3 k4 = slope 4 k4 = dy/dt at y2 k4 = 6*141.0-.007*(141.0)2 k4 = 706.9 y2  t = 0.5

  34. Step 5: Calculate weighted slope. Use weighted slope to estimate y at t +t weighted slope = true value weighted slope estimated value  t = 0.5

  35. Conclusions • 4th order Runge-Kutta offers substantial improvement over Eulers. • Both techniques provide estimates, not “true” values. • The accuracy of the estimate depends on the size of the step used in the algorithm. Analytical Runge-Kutta Euler

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