Keyframing. CS 448D: Character Animation Prof. Vladlen Koltun Stanford University

Transcription

1 Keyframing CS 448D: Character Animation Prof. Vladlen Koltun Stanford University

2 Keyframing in traditional animation Master animator draws key frames Apprentice fills in the in-between frames

3 Keyframing in computer animation Animator specifies object state for time t i, for all i State for intermediate frames is computed by interpolation State can include: Position Orientation Material properties Many other things

4 Key values Not all parameters are specified for all key frames A key frame is only key for a subset of parameters

5 How do we interpolate? Depends on type of parameter This lecture: Position Orientation has issues, will be covered later

6 Polynomial interpolation Theorem: Any n+1) distinct points can be interpolated by a polynomial of degree n. Given x 0,y 0 ), x,y ),..., x n,y n ) there is a polynomial px) =a 0 x n + a 1 x n 1 + a x n a n such that px i )=y i

7 Polynomial interpolation y 0 = a 0 x 0 n + a 1 x 0 n 1 + a x 0 n a n y 1 = a 0 x 1 n + a 1 x 1 n 1 + a x 1 n a n y n = a 0 x n n + a 1 x n n 1 + a x n n a n

8 Polynomial interpolation x 0 n x 0 n x n 1 x n a 0 a 1... = y 0 y 1... x n n x n n a n y n Linear system. Solve Gaussian elimination, LU decomposition). Gives the desired polynomial. px) =a 0 x n + a 1 x n 1 + a x n a n

9 Polynomial interpolation What happens in three dimensions? Express x 1,y 1,z 1 ),..., x n,y n,z n ) as xt 1 ),yt 1 ),zt 1 )),..., xt n ),yt n ),zt n )) Compute the polynomials xt), yt), and zt) In dealing with position interpolation, we will sometimes discuss only the univariate case, knowing that all methods generalize to interpolating position in higher dimensions.

10 Lagrange interpolation Need to interpolate x 0,y 0 ), x,y ),..., x n,y n ) Express px) as a linear combination of n+1) basis polynomials L i, such that L i x i ) = 1 and L i x j ) = 0 for all j i n If we can find such L i, we can set px) = y i L i x) i=0 Set L i x) = 0 j n, j i x x j x i x j

11 Global versus local interpolation These were global interpolation methods Computationally expensive. Potentially unstable numerically. A local change of an input point triggers a complete re-computation. Unweildy for animators, who want to be able to make local manipulations. Local interpolation methods connect input points with polynomial arcs

12 Linear interpolation x 5,y 5 ) x 3,y 3 ) x 1,y 1 ) x 4,y 4 ) x,y ) x 0,y 0 ) Interpolate between x i,y i ) and x i+1,y i+1 ) with p i x) =y i + x x i x i+1 x i y i+1 y i )

13 C n Orders of continuity continuity: The n-th derivative is continuous. Linear interpolation provides but potentially jerky motion. C 0 continuity. Continuous Want to achieve at least C 1, and sometimes C continuity.

14 Hermite interpolation How do we achieve C 1 continuity and local control? We enforce shared tangents at control points and connect consecutive input points with polynomial arcs subject to the positional and tangential constraints at the endpoints.

15 Hermite interpolation p 0) p0) pt) p 1) p1) Four linear equations that constrain the coefficients of p. How many coefficients do we need? Four. What is the degree of p? It s a cubic. pt) = a 0 t 3 + a 1 t + a t + a 3 p t) = 3a 0 t +a 1 t + a

16 Hermite interpolation p 0) p0) pt) p 1) p1) a 3 = p0) a = p 0) a 0 + a 1 + a + a 3 = p1) 3a 0 +a 1 + a = p 1) Solve to obtain the coefficients.

17 Hermite interpolation pt) = t 3 t t 1 ) p0) p1) p 0) p 1) pt) = 3t t 10 ) p0) p1) p 0) p 1)

18 Hermite interpolation pt) =a 0 t 3 + a 1 t + a t + a 3 pt) =T T MB T = t 3 t t 1 ) M = B = p0) p1) p 0) p 1)

19 Catmull-Rom spline How do we get the tangents? Can be specified by the animator along with the control points, but this can be tedious and time-consuming. The Catmull-Rom idea: p t i )= 1 pt i+1 ) pt i 1 ) ) pt i 1 ) pt i+1 ) pt i ) p t i )

20 Bezier interpolation x 1 x 0 x x 3 With two control points it s equivalent to Hermite interpolation. M = pt) =T T MB T = t 3 t t 1 ) B = x 0 x 1 x x 3

21 Diversion: Bezier curves x 1 x 4 x 0 x 3 A Bezier curve can have any number of control points. pt) = n i=0 x ) n 1 t) n i t i x i i Bernstein polynomials: Wolfram Research, Inc.

22 Kochanek-Bartels spline Hermite lets us specify the tangents directly. Catmull-Rom completely automates the shape of the spline at the input points. Can we have some degree of control over the spline, but in a more intuitive way than direct tangent specification? Yes. Kochanek-Bartels gives us three intuitive degrees of freedom for the tangents: tension, continuity, and bias. tension continuity bias

23 Kochanek-Bartels spline Tension p leftt i ) = 1 T p rightt i ) = 1 T ) pt i ) pt i 1 ) ) pt i ) pt i 1 ) + 1 T + 1 T ) pt i+1 ) pt i ) ) pt i+1 ) pt i )

24 Kochanek-Bartels spline Continuity p leftt i ) = 1 C p rightt i ) = 1+C ) pt i ) pt i 1 ) ) pt i ) pt i 1 ) + 1+C + 1 C ) pt i+1 ) pt i ) ) pt i+1 ) pt i )

25 Kochanek-Bartels spline Bias p leftt i ) = 1+B p rightt i ) = 1+B ) pt i ) pt i 1 ) ) pt i ) pt i 1 ) + 1 B + 1 B ) pt i+1 ) pt i ) ) pt i+1 ) pt i )

26 Kochanek-Bartels spline p leftt i ) = p rightt i ) = 1 T )1 C)1+B) 1 T )1+C)1+B) pti ) pt i 1 ) ) + pti ) pt i 1 ) ) + 1 T )1+C)1 B) 1 T )1 C)1 B) pti+1 ) pt i ) ) pti+1 ) pt i ) )

27 Velocity Control We now have a parametric curve pt) that smoothly interpolates keys. But we still have a problem: uncontrolled velocity of movement.

28 Reparameterization We want to be able to control the distance covered along the curve per unit of time. Need a function T that maps from normalized distance covered, s, to appropriate parameter value, t. Then pts)) will move at uniform velocity. We approximate T by approximating its inverse S that maps from parameter values, t, to distance covered, s.

29 Finite differencing Sample t uniformly and approximate pt) by piecewise linear segments. Approximate S by normalized distance covered along the approximating curve.

30 Adaptive finite differencing Maintain a set of candidate curve segments. For each such segment pa),pb)), if pa)+pb) pa) + pb) pa)+pb) pb) pa) > ε pa), pb)) pa), pa)+pb) ) pa)+pb) then replace with and and iterate until no segments need to be broken up. ), pb)

31 Velocity control We can now produce uniform velocity motion along the curve St) pt s)) T s) =S 1 t) by approximating, computing the resulting, and moving along as s increases uniformly from 0 to 1. s time

32 Velocity control We can also drive the motion along the curve in more general ways, with the distance covered being a non-uniform function pt στ))) στ) of time. Then the motion can be expressed as. One example is the ease-in/ease-out behavior: στ) στ) = sin τπ π ) τ