Concepts for Programming a Rigid Body Physics Engine Part 2

1. Concepts for Programming a Rigid Body Physics EnginePart 2 Presented by Scott Hawkins

2. source: Game Physics Engine Developmentby Ian Millington

3. Topics • Collision Resolution • Resolving Velocities • Resolving Positions • Friction

4. Collision Resolution • We have been given a list of contacts, each containing: • references to the two rigid bodies • contact point in world coordinates, q (Vector) • contact normal in world coordinates, n (Vector) • penetration depth, d (real) • coefficient of restituion, c (real) • coefficient of friction, μ (real)

5. Collision Resolution • After we detect collisions, what needs to be updated? • positions • orientations • velocities • angular velocities

6. Collision Resolution • Two steps • Resolve velocities (and angular velocities) • Resolve positions (and orientations)

7. Collision Resolution • Resolving velocities • Find the closing velocity for each contact • Choose the contact with greatest closing velocity • Resolve velocities for this contact (do not remove it from the list of contacts) • Repeat this until either • (1) all contacts are resolved, or • (2) we run out of iterations

8. Finding Closing Velocity • “Closing velocity” is the speed at which the two objects are moving toward each other at the point of contact. • We need this information to determine which contact to resolve first.

9. Finding Closing Velocity • How to find the closing velocity? • Find the velocity at the contact point for each object • Take the component in the direction of the contact • Sum them together

10. Finding Closing Velocity • How do you find the velocity of a point on a spinning object? • Start with the object’s overall velocity and add the component due to rotation • vpoint = p’ + θ’ x (ppoint - p) • where ppoint is the position of the point in world coordinates

11. Finding Closing Velocities • For a rigid body collision, some component of the closing velocity may result from rotation vs

12. Finding Closing Velocity • We want the velocity of each object at the contact point • We were given q, the contact point in world coordinates • vcontact,a = p’a + θ’a x (q - pa) • vcontact,b = p’b + θ’b x (q - pb)

13. Finding Closing Velocity • Find the component of this in the direction of the contact • We were given n, the contact normal, which is a unit vector pointing from object a to object b • vc,a = vcontact,a·n • vc,b = vcontact,b· -n • Use -n in the second equation because we want the component going from object b to object a

14. Finding Closing Velocity • Sum the components from each object to get the total closing velocity, vc • vc = vc,a + vc,b

15. Choosing the Greatest Closing Velocity • We now have the closing velocity for each contact • Choosing the greatest closing velocity will take O(n) time normally • use a max-heap?

16. Resolving Velocities • How to update velocity and angular velocity for a pair of colliding objects? • Accomplish this by applying some impulse, g (which we will find) • g represents a change in momentum • g = Δ(mv)

17. Resolving Velocities • This also creates an impulsive torque, u, given by: • u = (q - p) x g • Once again, q is the contact point in world coordinates

18. Resolving Velocities • To apply g to the objects: • Δp’a = ma-1g • Δθ’a = Ia-1ua= Ia-1[(q - pa) x g] • Δp’b = mb-1(-g) • Δθ’b = Ib-1ub= Ib-1[(q - pb) x (-g)] • Be sure to use -g for object b • I-1 in these equations is in world coordinates

19. Resolving Velocities • How to choose g?

20. Resolving Velocities • We want the objects to bounce off of each other according to the coefficient of restitution, c • vc,after = -cvc • At the contact point, and in the direction of the contact normal, we want a velocity change of Δv • Δv = -cvc - vc = -(1 + c)vc

21. Resolving Velocities • How can we choose g to generate this result? • We already know the direction of g • It must point in the direction of the contact normal, since it is like a normal force acting over an infinitely short time interval • Find the magnitude g by expressing g as • g = gn and plugging it into the previous equations

22. Resolving Velocities • If we applied the impulse gn, we would get: • Δp’a = ma-1gn • Δθ’a = Ia-1[(q - pa) x gn] • Δp’b = -mb-1gn • Δθ’b = -Ib-1[(q - pb) x gn]

23. Resolving Velocities • This causes a change in velocity at the contact point: • Δvcontact,a = Δp’a + Δθ’a x (q - pa)= ma-1gn + (Ia-1[(q - pa) x gn]) x (q - pa)= g[ma-1n + (Ia-1[(q - pa) x n]) x (q - pa)] • Δvcontact,b = Δp’b + Δθ’b x (q - pb)= -mb-1gn + (-Ib-1[(q - pb) x gn]) x (q - pb)= -g[mb-1n + (Ib-1[(q - pb) x n]) x (q - pb)]

24. Resolving Velocities • This contributes to a change in closing velocity: • Δvc,a = Δvcontact,a·n= g[ma-1n + (Ia-1[(q - pa) x n]) x (q - pa)]·n= g(ma-1 + [(Ia-1[(q - pa) x n]) x (q - pa)]·n) • Δvc,b = Δvcontact,b· -n= -g[mb-1n + (Ib-1[(q - pb) x n]) x (q - pb)]· -n= g(mb-1 + [(Ib-1[(q - pb) x n]) x (q - pb)]·n)

25. Resolving Velocities • Add these to get the total change in closing velocity: • Δvc = Δvc,a + Δvc,b= g(ma-1 + [(Ia-1[(q - pa) x n]) x (q - pa)]·n) +g(mb-1 + [(Ib-1[(q - pb) x n]) x (q - pb)]·n)= -(1 + c)vc

26. Resolving Velocities • Now we can isolate g • g = -(1 + c)vc / (ma-1 + [(Ia-1[(q - pa) x n]) x (q - pa)]·n +mb-1 + [(Ib-1[(q - pb) x n]) x (q - pb)]·n) • g = gn • Apply g to the objects as we discussed earlier

27. Resolving Velocities • After resolving the velocities for one contact, update its closing velocity to be • vc,after = -cvc • Update the closing velocities for every contact involving one of the two bodies that were changed These need to update vc This contact was resolved

28. Resolving Velocities • Continue resolving whichever contact has the greatest closing velocity until either • (1) all contacts are resolved (all closing velocities are less than or equal to zero) or • (2) we run out of iterations

29. Collision Resolution • Resolving positions • Choose the contact with greatest interpenetration depth • Resolve positions for this contact (do not remove it from the list of contacts) • Repeat this until either • (1) all contacts are resolved, or • (2) we run out of iterations

30. Choosing the Greatest Interpenetration Depth • The interpenetration depth d at each contact is given to us by the collision detector • Use another max-heap?

31. Resolving Positions • How to update position and orientation for a pair of colliding objects? • We apply changes in position and orientation to each object to make d = 0 • The changes we need to find are • Δpa, Δθa, Δpb, and Δθb

32. Resolving Positions • Once we have Δpa, Δθa, Δpb, and Δθb, we can perform the update • pa,after = pa + Δpa • θa,after = θa + ½(Δθa)(θa) • where Δθa = 0 + Δθa,x i + Δθa,y j + Δθa,z kand Δθa,x, Δθa,y, Δθa,z are components of Δθa • pb,after = pb + Δpb • θb,after = θb + ½(Δθb)(θb) • where Δθb = 0 + Δθb,x i + Δθb,y j + Δθb,z kand Δθb,x, Δθb,y, Δθb,z are components of Δθb

33. Resolving Positions • How do we decide what those changes should be?

34. Resolving Positions • Millington uses an approximation • Treat distance moved around a circle as if it were in a straight line

35. Resolving Positions • Interpenetration depth d resolved through a combination of linear and rotational movement from both objects • d = linear movement of a + angular movement of a + linear movement of b + angular movement of b

36. Resolving Positions • Each movement contributes some fraction of the total distance • linear movement of a = w1d • angular movement of a = w2d • linear movement of b = w3d • angular movement of b = w4d • where w1, w2, w3, and w4 are weights between 0 and 1 such that • w1 + w2 + w3 + w4 = 1

37. Resolving Positions • Impulse • To resolve velocities, the “impulse” we want is a force integrated over a short time • units are kg·m/s • To resolve positions, we need an “impulse” that is force integrated twice over a short time • units are kg·m

38. Resolving Positions • Inertia • normally, an object’s tendency to resist a change in motion • in this case, we say “inertia” is an object’s response to one kg·m of “impulse”

39. Resolving Positions • Choose w1, w2, w3, and w4 based on inertia • w1 = linear inertia of a / total inertia • w2 = angular inertia of a / total inertia • w3 = linear inertia of b / total inertia • w4 = angular inertia of b / total inertia • total inertia = linear inertia of a + angular inertia of a + linear inertia of b + angular inertia of b

40. Resolving Positions • Linear Inertia • How much linear movement occurs in response to a unit impulse? • (different kind of impulse than before) • linear inertia of a = m-1a • linear inertia of b = m-1b

41. Resolving Positions • Angular Inertia • How much angular movement occurs in response to a unit impulse? (in the direction of the contact) • angular inertia of a = [(I-1a[(q - pa) x n]) x (q - pa)] ·n • angular inertia of b = [(I-1b[(q - pb) x n]) x (q - pb)] ·n

42. Resolving Positions • We can now find the weights w1, w2, w3, and w4 • w1 = m-1a / total inertia • w2 = [(I-1a[(q - pa) x n]) x (q - pa)] ·n / total inertia • w3 = m-1b / total inertia • w4 = [(I-1b[(q - pb) x n]) x (q - pb)] ·n / total inertia • total inertia = m-1a +[(I-1a[(q - pa) x n]) x (q - pa)] ·n +m-1b +[(I-1b[(q - pb) x n]) x (q - pb)] ·n

43. Resolving Positions • Using these weights, we can find how much of each type of movement should occur • linear movement of a = w1d • angular movement of a = w2d • linear movement of b = w3d • angular movement of b = w4d

44. Resolving Positions • How can we generate this much movement?

45. Resolving Positions • Linear movement • Δpa = -n * linear movement of a • Δpb = n * linear movement of b

46. Resolving Positions • Angular Movement • How much rotation? • We are given an amount of movement • Find ratio of impulse per movement • Find ratio of rotation per impulse • rotation = (movement) * (impulse / movement) * (rotation / impulse)

47. Resolving Positions • How to find ratio of impulse per movement? • The angular inertia we used before was a ratio of movement per impulse • (impulse/movement)a = (angular inertia of a)-1 • (impulse/movement)b = (angular inertia of b)-1

48. Resolving Positions • How to find ratio of rotation per impulse? • (rotation/impulse)a = I-1a((q - pa) x n) • (rotation/impulse)b = I-1b((q - pb) x (-n)) • Use -n for object b, since it must receive an impulse in the opposite direction

49. Resolving Positions • Putting it all together • Δθa = rotationa= (movement)a * (impulse / movement)a * (rotation / impulse)a= w2d * (angular inertia of a)-1 * I-1a((q - pa) x n)= (angular inertia of a / total inertia) * d *(angular inertia of a)-1 * I-1a((q - pa) x n)

50. Resolving Positions • Putting it all together • Δθb = rotationb= (movement)b * (impulse / movement)b * (rotation / impulse)b= w4d * (angular inertia of b)-1 * I-1b((q - pb) x (-n))= (angular inertia of b / total inertia) * d *(angular inertia of b)-1 * I-1b((q - pb) x (-n))