Modeling and Solving Constraints Erin Catto Blizzard Entertainment - PowerPoint PPT Presentation

Modeling and Solving Constraints Erin Catto Blizzard Entertainment

Basic Idea � Constraints are used to simulate joints, contact, and collision. � We need to solve the constraints to stack boxes and to keep ragdoll limbs attached. � Constraint solvers do this by calculating impulse or forces, and applying them to the constrained bodies.

Overview � Constraint Formulas � Jacobians, Lagrange Multipliers � Modeling Constraints � Joints, Motors, Contact � Building a Constraint Solver � Sequential Impulses

Contact and Friction Constraint Types

Constraint Types Ragdolls

Particles and Cloth Constraint Types

Show Me the Demo!

Bead on a 2D Rigid Wire C x y = ( , ) 0 Implicit Curve Equation: This is the position constraint.

How does it move? The normal vector is perpendicular to the velocity. n v = n v dot( , ) 0

Enter The Calculus Position Constraint: ⎡ ⎤ x = ⎢ ⎥ x = x C ( ) 0 ⎣ ⎦ y If C is zero, then its time derivative is zero. & C = 0 Velocity Constraint:

Velocity Constraint & C = 0 � Velocity constraints define the allowed motion. � Next we’ll show that velocity constraints depend linearly on velocity.

The Jacobian Due to the chain rule the velocity constraint has a special structure: ⎡ ⎤ & x = ⎢ ⎥ & C = Jv v & ⎣ ⎦ y J is a row vector called the Jacobian. J depends on position. The velocity constraint is linear .

The Jacobian The Jacobian is perpendicular to the velocity. J T v & C = = Jv 0

Constraint Force Assume the wire is frictionless. v What is the force between the wire and the bead?

Lagrange Multiplier Intuitively the constraint force F c is parallel to the normal vector. F c v Direction known . = λ F J T Magnitude unknown . c implies

Lagrange Multiplier � The Lagrange Multiplier (lambda) is the constraint force signed magnitude. � We use a constraint solver to compute lambda. � More on this later.

Jacobian as a CoordinateTransform � Similar to a rotation matrix. � Except it is missing a couple rows. � So it projects some dimensions to zero. � The transpose is missing some columns, so some dimensions get added.

Velocity Transform & v J C Cartesian Constraint Space Space Velocity Velocity & C = Jv

Force Transform λ F J T c Constraint Cartesian Space Space Force Force = λ F J T c

Refresher: Work and Power Work = Force times Distance Work has units of Energy (Joules) Power = Force times Velocity (Watts) ( ) P = F V dot ,

Principle of Virtual Work Principle: constraint forces do no work. = λ F J T We can ensure this by using: c Proof (compute the power): ( ) T = = λ = λ = F v J v Jv T T P 0 c c The power is zero, so the constraint does no work.

Constraint Quantities C Position Constraint & C Velocity Constraint J Jacobian λ Lagrange Multiplier

Why all the Painful Abstraction? � We want to put all constraints into a common form for the solver. � This allows us to efficiently try different solution techniques.

Addendum: Modeling Time Dependence � Some constraints, like motors, have prescribed motion. � This is represented by time dependence. ( ) t = x C , 0 Position: & = + = Jv C b t ( ) 0 Velocity: velocity bias

Example: Distance Constraint = − x Position: C L x C = x v T & L Velocity: y x ⎡ ⎤ x = ⎢ ⎥ x Jacobian: ⎣ ⎦ y particle = x T J x b = λ is the tension Velocity Bias: 0

Gory Details ( ) dC d = + − 2 2 x y L dt dt ( ) 1 d dL = + − 2 2 x y + dt dt 2 2 2 x y ( ) + 2 xv yv = x y − 0 + 2 2 2 x y T ⎡ ⎤ ⎡ ⎤ x v v T x 1 = = x ⎢ ⎥ ⎢ ⎥ x + ⎣ ⎦ ⎣ v y ⎦ 2 2 x y y

Computing the Jacobian � At first, it is not easy to compute the Jacobian. � It gets easier with practice. � If you can define a position constraint, you can find its Jacobian. � Here’s how …

A Recipe for J � Use geometry to write C . � Differentiate C with respect to time. � Isolate v . � Identify J and b by inspection. & = + Jv C b

Constraint Potpourri � Joints � Motors � Contact � Restitution � Friction

T x = x J Joint: Distance Constraint g m λ = T J a x F = c F v y

Motors A motor is a constraint with limited force (torque). Example θ = θ − C sin t − ≤ λ ≤ 10 10 A Wheel Note: this constraint does work.

Velocity Only Motors Example ω & = ω − C 2 − ≤ λ ≤ 5 5 Usage: A wheel that spins at a constant rate. We don’t care about the angle.

Inequality Constraints � So far we’ve looked at equality constraints (because they are simpler). � Inequality constraints are needed for contact and joint limits. � We put all inequality position constraints into this form: t ≥ x C ( , ) 0

Inequality Constraints The corresponding velocity constraint: C ≤ 0 If & C ≥ 0 enforce: Else skip constraint

Inequality Constraints ≤ λ ≤ ∞ 0 Force Limits: Inequality constraints don’t suck .

Contact Constraint � Non-penetration. � Restitution: bounce � Friction: sliding, sticking, and rolling

Non-Penetration Constraint body 2 n = δ C δ p (separation) body 1

Non-Penetration Constraint & C = − ⋅ v v n ( ) p 2 p 1 ( ) ( ) = ⎡ + × − − − × − ⎤ ⋅ v ω p x v ω p x n ⎣ ⎦ 2 2 2 1 1 1 − T ⎡ ⎤ ⎡ n ⎤ v 1 ⎢ ⎥ ⎢ ⎥ ( ) − − × p x n ω ⎢ ⎥ ⎢ ⎥ = ⎢ 1 1 ⎥ ⎢ ⎥ Handy Identities n v ⎢ ⎥ ⎢ 2 ⎥ ( ) ( ) − × p x n ⋅ × = ω A B C ⎣ ⎦ ⎣ ⎦ 2 2 ( ) ⋅ × = C A B ( ) J ⋅ × B C A

Restitution Relative normal velocity − ⋅ v v n ฀ v ( ) n p 2 p 1 Velocity Reflection + − ≥ − v ev n n Adding bounce as a velocity bias & + − ev − = + ≥ = 0 C v ev b n n n

Friction Constraint Friction is like a velocity-only motor. The target velocity is zero. & C = ⋅ v t p ( ) = ⎡ + × − ⎤ ⋅ v ω p x t ⎣ ⎦ T ⎡ ⎤ ⎡ ⎤ t v = ⎢ ⎥ ⎢ ⎥ ( ) − × p x t ω ⎣ ⎦ ⎣ ⎦ p t J

Friction Constraint The friction force is limited by the normal force. λ ≤ μλ Coulomb’s Law: t n − μλ ≤ λ ≤ μλ In 2D: n t n 3D is a bit more complicated. See the references.

Constraints Solvers � We have a bunch of constraints. � We have unknown constraint forces. � We need to solve for these constraint forces. � There are many ways different ways to compute constraint forces.

Constraint Solver Types � Global Solvers (slow) � Iterative Solvers (fast)

Solving a Chain Global: λ 1 solve for λ 1, λ 2, and λ 3 simultaneously. m 1 λ 2 Iterative: m 2 while !done solve for λ 1 λ 3 solve for λ 2 m 3 solve for λ 3

Sequential Impulses (SI) � An iterative solver. � SI applies impulses at each constraint to correct the velocity error. � SI is fast and stable. � Converges to a global solution.

Why Impulses? � Easier to deal with friction and collision. � Lets us work with velocity rather than acceleration. � Given the time step, impulse and force are interchangeable. = P F h

Sequential Impulses Step1: Integrate applied forces, yielding tentative velocities. Step2: Apply impulses sequentially for all constraints, to correct the velocity errors. Step3: Use the new velocities to update the positions.

We separate applied forces and mass matrix c F Step 1: Newton’s Law + a F = constraint forces. Mv &

Step 1: Mass Matrix ⎡ ⎤ Particle m 0 0 ⎢ ⎥ = ⎢ M 0 m 0 ⎥ ⎢ ⎥ ⎣ ⎦ 0 0 m ⎡ E 0 ⎤ Rigid Body m = ⎢ M ⎥ 0 I ⎣ ⎦ May involve multiple particles/bodies.

Step 1: Applied Forces � Applied forces are computed according to some law. � Gravity: F = mg � Spring: F = -kx � Air resistance: F = -cv 2

Step 1 : Integrate Applied Forces Euler’s Method for all bodies. − = + v v M F 1 h 2 1 a This new velocity tends to violate the velocity constraints.

Step 2: Constraint Impulse The constraint impulse is just the time step times the constraint force. = P F h c c

Step 2: Impulse-Momentum Newton’s Law for impulses: Δ = M v P c In other words: − = + v v M P 1 2 2 c

Step 2: Computing Lambda For each constraint, solve these for λ : − = + v v M P 1 Newton’s Law: 2 2 c = λ P J T Virtual Work: c + = Jv b 0 Velocity Constraint: 2 Note: this usually involves one or two bodies.

Step 2: Impulse Solution ( ) λ = − + Jv m b C 2 1 = m − C JM J 1 T The scalar m C is the effective mass seen by the constraint impulse: & Δ = λ m C C