Animation Lecture 10 Slide Fall 2003

Transcription

1 Animation Lecture 10 Slide Fall 2003

2 Conventional Animation Draw each frame of the animation great control tedious Reduce burden with cel animation layer keyframe inbetween cel panoramas (Disney s Pinocchio)... Lecture 10 Slide Fall 2003

3 Keyframing Computer-Assisted Animation automate the inbetweening good control less tedious creating a good animation still requires considerable skill and talent Procedural animation describes the motion algorithmically express animation as a function of small number of parameteres Example: a clock with second, minute and hour hands hands should rotate together express the clock motions in terms of a seconds variable the clock is animated by varying the seconds parameter Example 2: A bouncing ball Abs(sin(ωt+θ 0 ))*e -kt Image adapted from: Lasseter, John. "Principles of Traditional Animation applied to 3D Computer Animation." ACM SIGGRAPH Computer Graphics 21, no. 4 (July 1987): Lecture 10 Slide Fall 2003

4 Computer-Assisted Animation Physically Based Animation Assign physical properties to objects (masses, forces, inertial properties) Simulate physics by solving equations Realistic but difficult to control Image removed due to copyright considerations. Motion Capture Captures style, subtle nuances and realism You must observe someone do something Lecture 10 Slide Fall 2003

5 Overview Hermite Splines Keyframing Traditional Principles Squash & Stretch in Luxo Jr.'s Hop Articulated Models Forward Kinematics Inverse Kinematics Optimization Differential Constraints Image adapted from: Lasseter, John. "Principles of Traditional Animation applied to 3D Computer Animation." ACM SIGGRAPH Computer Graphics 21, no. 4 (July 1987): Lecture 10 Slide Fall 2003

6 Keyframing Image adapted from: Lasseter, John. "Principles of Traditional Animation applied to 3D Computer Animation." ACM SIGGRAPH Computer Graphics 21, no. 4 (July 1987): Describe motion of objects as a function of time from a set of key object positions. In short, compute the inbetween frames. Lecture 10 Slide Fall 2003

7 Interpolating Positions Given positions: ( x, y, t ), i = 0, K, n i i i x() t find curve C ( t ) = such that y ( t ) C ( t ) i x i = y i ( x2, y 2, t2) C ( t ) u ( x, y, t ) 0 ( x1, y1, t1) Lecture 10 Slide Fall 2003

8 Linear Interpolation ( x, y, t ) ( x0, y 0, t0) ( x1, y1, t1) Simple problem: linear interpolation between first two points assuming t 0=0 and t 1=1: x( t) = x 0( 1 t ) + x1t The x-coordinate for the complete curve in the figure: x() t t1 t t t0 ) x 0 + x 1, t t 0, t 1 t1 t0 t1 t 0 = t2 t t t1 x1 + x2, t t1, t2 t2 t1 t2 t 1 Lecture 10 Slide Fall 2003 Derivation?

9 Polynomial Interpolation ( x2, y 2, t2) ( x0, y 0, t0) ( x1, y1, t1) parabola An n-degree polynomial can interpolate any n+1 points. The Lagrange formula gives the n+1 coefficients of an n-degree polynomial that interpolates n+1 points. The resulting interpolating polynomials are called Lagrange polynomials. On the previous slide, we saw the Lagrange formula for n = 1. Lecture 10 Slide Fall 2003

10 Spline Interpolation Lagrange polynomials of small degree are fine but high degree polynomials are too wiggly. t t t x ( t ) 8-degree polynomial spline spline vs. polynomial How many n-degree polynomials interpolate n+1 points? Lecture 10 Slide Fall 2003

11 Spline Interpolation Lagrange polynomials of small degree are fine but high degree polynomials are too wiggly. Spline (piecewise cubic polynomial) interpolation produces nicer interpolation. t t t x ( t ) 8-degree polynomial spline spline vs. polynomial Lecture 10 Slide Fall 2003

12 Spline Interpolation A cubic polynomial between each pair of points: 2 3 x() t = c + c t + c t + c t Four parameters (degrees of freedom) for each spline segment. Number of parameters: n+1 points n cubic polynomials 4n degrees of freedom Number of constraints: interpolation constraints n+1 points (n-1) = 2n interpolation constraints endpoints + each side of an internal point rest by requiring smooth velocity, acceleration, etc. Lecture 10 Slide Fall 2003

13 Hermite Splines We want to support general constraints: not just smooth velocity and acceleration. For example, a bouncing ball does not always have continuous velocity: Solution: specify position AND velocity at each point Derivation? c, c, c, c =? for x, v : t = t and x, v : t = t Lecture 10 Slide Fall 2003

14 Keyframing Given keyframes ( 0 1 K ) 0 p t 1 find curves K () t = p ( t ) such that K ( t i ) = K M 0 1 What are parameters p, p, K? K = p, p,, t, i = 0, K, n i i i i ( ) position, orientation, size, visibility, i i i Interpolate each curve separately Lecture 10 Slide Fall 2003

15 Interpolating Key Frames Interpolation is not fool proof. The splines may undershoot and cause interpenetration. The animator must also keep an eye out for these types of side-effects. Image adapted from: Lasseter, John. "Principles of Traditional Animation applied to 3D Computer Animation." ACM SIGGRAPH Computer Graphics 21, no. 4 (July 1987): Lecture 10 Slide Fall 2003

16 Traditional Animation Principles The in-betweening, was once a job for apprentice animators. We described the automatic interpolation techniques that accomplish these tasks automatically. However, the animator still has to draw the key frames. This is an art form and precisely why the experienced animators were spared the inbetweening work even before automatic techniques. The classical paper on animation by John Lasseter from Pixar surveys some the standard animation techniques: "Principles of Traditional Animation Applied to 3D Computer Graphics, SIGGRAPH'87, pp Lecture 10 Slide Fall 2003

17 Squash and stretch Squash: flatten an object or character by pressure or by its own power Stretch: used to increase the sense of speed and emphasize the squash by contrast Squash & Stretch in Luxo Jr.'s Hop Squash & Stretch In Bouncing Ball Image adapted from: Lasseter, John. "Principles of Traditional Animation applied to 3D Computer Animation." ACM SIGGRAPH Computer Graphics 21, no. 4 (July 1987): Image adapted from: Lasseter, John. "Principles of Traditional Animation applied to 3D Computer Animation." ACM SIGGRAPH Computer Graphics 21, no. 4 (July 1987): Lecture 10 Slide Fall 2003

18 Timing Timing affects weight: Light object move quickly Heavier objects move slower Timing completely changes the interpretation of the motion. Because the timing is critical, the animators used the draw a time scale next to the keyframe to indicate how to generate the inbetween frames. TIMING CHART FOR BALL BOUNCE EXTREME EXTREME Image adapted from: Lasseter, John. "Principles of Traditional Animation applied to 3D Computer Animation." ACM SIGGRAPH Computer Graphics 21, no. 4 (July 1987): Lecture 10 Slide Fall 2003

19 Anticipation An action breaks down into: Anticipation Action Reaction Anatomical motivation: a muscle must extend before it can contract. Prepares audience for action so they know what to expect. Directs audience s attention. Amount of anticipation can affect perception of speed and weight. Squash & Stretch in Luxo Jr.'s Hop Image adapted from: Lasseter, John. "Principles of Traditional Animation applied to 3D Computer Animation." ACM SIGGRAPH Computer Graphics 21, no. 4 (July 1987): Lecture 10 Slide Fall 2003

20 Articulated Models Articulated models: rigid parts connected by joints They can be animated by specifying the joint angles as functions of time. q i q t i () t 1 t 2 t 1 t 2 Lecture 10 Slide Fall 2003

21 Forward Kinematics Describes the positions of the body parts as a function of the joint angles. 1 DOF: knee 2 DOF: wrist 3 DOF: arm Lecture 10 Slide Fall 2003

22 Skeleton Hierarchy Each bone transformation described relative to the parent in the hierarchy: xh, yh, zh, θh, φh, σh θt, φt, σt hips left-leg... r-thigh θ c y r-calf θ, φ v s f f x r-foot z Derive world coordinates for an effecter with local coordinates? v w v s Lecture 10 Slide Fall 2003

23 Forward Kinematics xh, yh, zh, θh, φh, σh θt, φt, σt Transformation matrix for an effecter v s is a matrix composition of all joint transformation between the effecter and the root of the hierarchy. θ c y θ f, φ f v s v w = T( x, y, z ) R( θ, φ, σ ) z v s h h h h h h x TR( θφσ,, t t t )TR( θ ) TR ( θ, φ c ) v θ φ σ θ φ σ θ θ φ w=s x h, y, z,,,,,,,,, v =S( p) v p h h h h h t t t c f f s s Lecture 10 Slide Fall 2003 f f v s

24 Inverse Kinematics Forward Kinematics Given the skeleton parameters (position of the root and the joint angles) p and the position of the effecter in local coordinates v s, what is the position of the sensor in the world coordinates v w? Not too hard, we can solve it by evaluating Inverse Kinematics Given the position of the effecter in local coordinates v s and the desired position ṽ w in world coordinates, what are the skeleton parameters p? S ( p) v s xh, yh, zh, θh, φh, σh θt, φt, σt θ c Much harder requires solving the inverse ( ) = % of the non-linear function: p? such that S p v s vw Underdetermined problem with many solutions θ f, φ f v s v% w Lecture 10 Slide Fall 2003

25 Real IK Problem Find a natural skeleton configuration for a given collection of pose constraints. Definition: A scalar objective function g(p) measures the quality of a pose. The objective g(p) reaches its minimum for the most natural skeleton configurations p. Definition: A vector constraint function C(p) = 0 collects all pose constraints: S ( ) % 0 p v 0 v 0 S ( ) % 1 p v1 v1 = 0 M C ( p) Lecture 10 Slide Fall 2003

26 Optimization Compute the optimal parameters p * that satisfy pose constraints and maximize the natural quality of skeleton configuration: Example objective functions g(p): deviation from natural pose: g ( p) = p p T M( p p) joint stiffness power consumption * p = argmin g( p) p s.t. C ( p) = 0 ( ) Lecture 10 Slide Fall 2003

27 Unconstrained Optimization Define an objective function f(p) that penalizes violation of pose constraints: Necessary condition: f ( p) = g ( p) + wi Ci( p) * p = argmin f ( p) p i * * * f ( p + p) f ( p ) 0 ( p is local minimum) * T f ( p ) p + L 0 (Taylor series) * f ( p ) = 0 ( p is arbitrary) 2 Lecture 10 Slide Fall 2003

28 Numerical Solution Gradient methods Guess initial solution x 0 T Iterate xk+ 1 = xk + αkdk, αk > 0, f ( xk) dk < 0 Until f ( x k ) = 0 T The conditions α k > 0, f ( xk) dk < 0 guarantee that each new iterate is more optimal f ( x ) f ( x ). Derive? k+ 1 < k Some choices for direction d k : Steepest descent g C i dk = f ( xk) = 2 wici p p 2 Newton s method d k = f ( xk) f ( xk) Quasi-Newton methods dk = Dk f ( xk) 1 i Lecture 10 Slide Fall 2003

29 Gradient Computation Requires computation of constraint derivatives: Compute derivatives of each transformation primitive Apply chain rule Example: C ( p) T( x, y, z ) R( θ, φ, σ ) TR( θ, φ, σ ) TR( θ ) TR( θ, φ ) v v h h h h h h t t t c f f s w C R( θ ) = c T( x, y, z ) R( θ, φ, σ ) TR( θ, φ, σ ) T TR( θ, φ ) v θ θ c h h h h h h t t t f f s c R( θc ) Derive if R is a rotation around z-axis? θ c Lecture 10 Slide Fall 2003

30 Constrained Optimization Unconstrained formulation has drawbacks: Sloppy constraints The setting of penalty weights w i must balance the constraints and the natural quality of the pose Necessary condition for equality constraints: Lagrange multiplier theorem: f ( p * ) + λ ( * i Ci p ) = 0 λ 0, λ 1, are scalars called Lagrange multipliers Interpretations: i Cost gradient (direction of improving the cost) belongs to the subspace spanned by constraint gradients (normals to the constraints surface). Cost gradient is orthogonal to subspace of feasible variations. Lecture 10 Slide Fall 2003

31 Example * p = argmin p + p p s.t. p + p = f ( p ) = 1 * 1 * 2 C( p ) = 2 Lecture 10 Slide Fall 2003

32 Nonlinear Programming Use Lagrange multipliers and nonlinear programming techniques to solve the IK problem: * p = argmin g( p) nonlinear objective p } s.t. C ( p) = 0 nonlinear constraints In general, slow for interactive use! Lecture 10 Slide Fall 2003

33 Differential Constraints Differential constraints linearize original pose constraints. Rewrite constraints by pulling the desired effecter locations to the right hand side Construct linear approximation around the current parameter p. Derive with Taylor series. ( ) ( ) S % 0 p v 0 v 0 = S % 1 p v1 v 1 M M C ( p) C( p) p = v% p Lecture 10 Slide Fall 2003

34 IK with Differential Constraints Interactive Inverse Kinematics User interface assembles desired effecter variations Solve quadratic program with Lagrange multipliers: * T p = argmin p M p p C( p) s.t. p = v% p % v Update current pose: Objective function is quadratic, differential constraints are linear. Some choices for matrix M: p + = + α * k pk k p Identity: minimizes parameter variations Diagonal: minimizes scaled parameter variations 1 Lecture 10 Slide Fall 2003

35 Quadratic Program Elimination procedure Apply Lagrange multiplier theorem and convert to vector notation: general form in scalar notation: f ( p * ) + λ ( * i Ci p ) = 0 Rewrite to expose p: i Use this expression to replace p in the differential constraint: Solve for Lagrange multipliers and compute p * C( p) M p + λ = 0 p 1 C( p) T p = M p C( p) 1 C( p) T M p p T λ λ = v% Lecture 10 Slide Fall 2003

36 Kinematics Kinematics vs. Dynamics Describes the positions of body parts as a function of skeleton parameters. Dynamics Describes the positions of body parts as a function of applied forces. Lecture 10 Slide Fall 2003

37 Next Dynamics ACM 1988 Spacetime Constraints Lecture 10 Slide Fall 2003