10 keyframe-ik-opt

Lecture 10
Slide 12
6.837 Fall 2003
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:
Keyframing
automate the inbetweening good control less tedious creating a good animation still requires considerable skill and talent

Procedural animation
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


ACM © 1987 “Principles of traditional animation
hands should rotate together express the clock motions in terms of a “seconds” variable the clock is animated by varying the seconds parameter
(x 1 , y 1 , t 1 ) Simple problem: linear interpolation between first two points assuming t 0 =0 and t 1=1: x (t ) x 0 1 t x 1t The x-coordinate for the complete curve in the figure:
x t
8-degree polynomial
lynomial
Lecture 10
Slide 11
6.837 Fall 2003
Spline Interpolation
A cubic polynomial between each pair of points: x (t ) c 0 c 1t c 2t 2 c 3t 3
s(t )
Lecture 10
Slide 6
6.837 Fall 2003
Interpolating Positions
Given positions: (x i , y i ,t i ), i 0, , n
x (t ) find curve C (t ) such that C (t i ) y (t )
Lecture 10
Slide 9
6.837 Fall 2003
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 10 6.837 Fall 2003
Spline Interpolation
Overview
Hermite Splines Keyframing Traditional Principles Articulated Models Forward Kinematics Inverse Kinematics Optimization Differential Constraints
Lecture 10
Slide 5
6.837 Fall 2003
Keyframing
ACM © 1987 “Principles of traditional animation applied to 3D computer animation”
Describe motion of objects as a function of time from a set of key object positions. In short, compute the inbetween frames.
interpolation constraints n+1 points 2 + 2 (n-1) = 2n interpolation constraints

“endpoints” + “each side of an internal point” rest by requiring smooth velocity, acceleration, etc.
Lecture 10 Slide 8
Derivation?
6.837 Fall 2003
Polynomial Interpolation
(x 2 , y 2 , t 2 )
(x 0 , y 0 ,t 0 )
(x 1 , y 1 , t 1 )
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.
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.

describes the motion algorithmically applied to 3D computer animation” express animation as a function of small number of parameteres Example: a clock with second, minute and hour hands
t t0 t1 t t t x 0 t t x 1 , t t 0 , t 1 1 0 x (t ) 1 0 t 2 t x t t 1 x , t t , t 1 2 t 2 t 1 1 t 2 t 1 2
What are parameters pi0 , pi1 ,?

position, orientation, size, visibility, …
Interpolate each curve separately
Lecture 10
Slide 14
6.837 Fall 2003
Interpolating Key Frames
6.837 Fall 2003
Keyframing
0 1 Given keyframes K i pi , pi , ,t i , i 0, , n


p 0 t find curves K (t ) p 1 t such that K (t i ) K i
x i y i
(x 2 , y 2 , t 2 )
C (t )
( xu, y 0 , t 0 ) 0
0
(x 1 , y 1 , t 1 )
Lecture 10
Slide 7
6.837 Fall 2003
Linear Interpolation
(x 2 , y 2 , t 2 )
(x 0 , y 0 ,t 0 )
Reduce burden with cel animation

ACM © 1997 “Multiperspective panoramas for cel animation”
Lecture 10
Slide 2
6.837 Fall 2003
Computer-Assisted Animation
Solution: specify position AND velocity at each point Derivation? c 0 , c 1 , c 2 , c 3 ? for x 0 ,v 0 : t t 0 and x 1 ,v 1 : t t 1
Lecture 10
Slide 13
Abs(sin(wt+q0))*e-kt

Example 2: A bouncing ball

Lecture 10
Slide 3
6.837 Fall 2003
Computer-Assisted Animation