CSCE 441: Keyframe Animation/Smooth Curves (Cont.) Jinxiang Chai

Key-frame Interpolation Given parameter values at key frames, how to interpolate parameter values for inbetween frames. θ t

Key-frame Interpolation Given parameter values at key frames, how to interpolate parameter values for inbetween frames. t θ

Key-frame Interpolation Given parameter values at key frames, how to interpolate parameter values for inbetween frames. t θ Nonlinear interpolation

Review: Natural cubic cruves

Review: Natural cubic curves Properties: - Go through four control points - not good for local control

Review: Hermite Curves P1P1 R1R1 P4P4 R4R4 P 1 : start position P 4 : end position R 1 : start derivative R 4 : end derivative

Review: Hermite Curves P1P1 R1R1 P4P4 R4R4

P1P1 R1R1 P4P4 R4R4 Herminte basis matrix

Review: Hermite Curves P1P1 R1R1 P4P4 R4R4 Herminte basis matrix

Review: Hermite Curves P1P1 R1R1 P4P4 R4R4

P1P1 R1R1 P4P4 R4R4

P1P1 R1R1 P4P4 R4R4

P1P1 R1R1 P4P4 R4R4

P1P1 R1R1 P4P4 R4R4 Hermite basis functions

Review: Hermite Curves P1P1 R1R1 P4P4 R4R4 basis function 1basis function 2basis function 3 basis function 4

Review: Hermite Curves P1P1 R1R1 P4P4 R4R4 *P 1 *P 4 *R 1 *R =

Review: Bezier Curves

*v 0 *v 1 *v 2 *v =

Cubic curves: Hermite curves: Bezier curves: Review: Different basis functions

Complex curves Suppose we want to draw or interpolate a more complex curve

Complex curves Suppose we want to draw or interpolate a more complex curve How can we represent this curve?

Complex curves Suppose we want to draw a more complex curve Idea: we’ll splice together a curve from individual segments that are cubic Béziers

Complex curves Suppose we want to draw or interpolate a more complex curve Idea: we’ll splice together a curve from individual segments that are cubic Béziers

Splines A piecewise polynomial that has a locally very simple form, yet be globally flexible and smooth

Splines There are three nice properties of splines we’d like to have - Continuity - Local control - Interpolation

Continuity C 0 : points coincide, velocities don’t C 1 : points and velocities coincide What’s C 2 ? - points, velocities and accelerations coincide

Continuity Cubic curves are continuous and differentiable We only need to worry about the derivatives at the endpoints when two curves meet

Local control We’d like our spline to have local control - that is, have each control point affect some well-defined neighborhood around that point

Local control We’d like our spline to have local control - that is, have each control point affect some well-defined neighborhood around that point

Local control We’d like our spline to have local control - that is, have each control point affect some well-defined neighborhood around that point

Interpolation Bézier curves are approximating - The curve does not (necessarily) pass through all the control points - Each point pulls the curve toward it, but other points are pulling as well - the curve is always located within the convex hull based on control points. Instead, we may prefer a spline that is interpolating - That is, that always passes through every control point

B-splines We can join multiple Bezier curves to create B-splines Ensure C 2 continuity when two curves meet

Derivatives at end points

Continuity in B splines Suppose we want to join two Bezier curves (V 0, V 1, V 2,V 3 ) and (W 0, W 1, W 2, W 3 ) so that C 2 continuity is met at the joint

Continuity in B splines Suppose we want to join two Bezier curves (V 0, V 1, V 2,V 3 ) and (W 0, W 1, W 2, W 3 ) so that C 2 continuity is met at the joint

Continuity in B splines Suppose we want to join two Bezier curves (V 0, V 1, V 2,V 3 ) and (W 0, W 1, W 2, W 3 ) so that C 2 continuity is met at the joint

Continuity in B splines Suppose we want to join two Bezier curves (V 0, V 1, V 2,V 3 ) and (W 0, W 1, W 2, W 3 ) so that C 2 continuity is met at the joint

Continuity in B splines Suppose we want to join two Bezier curves (V 0, V 1, V 2,V 3 ) and (W 0, W 1, W 2, W 3 ) so that C 2 continuity is met at the joint

Continuity in B splines Suppose we want to join two Bezier curves (V 0, V 1, V 2,V 3 ) and (W 0, W 1, W 2, W 3 ) so that C 2 continuity is met at the joint

Continuity in B splines What does this derived equation mean geometrically? - What is the relationship between a, b and c, if a = 2b - c? b is the middle point of a and c. w 2 =v 1 +4v 3 -4v 2

de Boor points Instead of specifying the Bezier control points, let’s specify the corners of the frames that forms a B-spline These points are called de Boor points and the frames are called A-frames

de Boor points What is the relationship between Bezier control points and de Boor points? Verify this by yourself!

B spline basis matrix

Building complex splines Constraining a Bezier curve made of many segments to be C 2 continuous is a lot of work - for each new segment we have to add 3 new control point - only one of the control points is really free B-splines are easier (and C 2 ) - First specify 4 vertices (de Boor points), then one per segment

B splines properties √ Continuity √ Local control x Interpolation

Catmull-Rom splines If we are willing to sacrifice C 2 continuity, we can get interpolation and local control. If we set each derivative to be a constant multiple of the vector between the previous and the next control points, we get a Catmull-Rom spline

Catmull-Rom splines

The segment is controlled by p 1,p 2,p 3,p 4 0.5

Catmull-Rom splines The segment is controlled by p 1,p 2,p 3,p 4

Catmull-Rom splines The effect of t: how sharply the curve bends at the control points

Catmull-Rom splines The segment is controlled by p 1,p 2,p 3,p 4

Catmull-Rom splines From Hermite curves

Catmull-Rom splines From Hermite curves

Catmull-Rom splines From Hermite curves

Catmull-Rom splines

What do we miss?

Catmull-Rom splines ? ?

Catmull-Rom splines have C 1 continuity (not C 2 continuity) Do not lie within the convex hull of their control points.

Catmull-Rom splines Catmull-Rom splines have C 1 continuity (not C 2 continuity) Do not lie within the convex hull of their control points.

Catmull-Rom splines properties X Continuity (C 2 ) √ Local control √ Interpolation

Keyframe Interpolation t=0 t=50ms

Curves in the animation Interpolate each parameter separately Each animation parameter is described by a 2D curve Q(u) = (x(u), y(u)) Treat this curve as θ = y(u) t = x(u) where θ is a variable you want to animate and t is a time index. We can think of the results as a function: θ(t) u u θ t

Keyframing interpolation is a two-step algorithm: - Compute the spline parameter u corresponding to the time t: u t=x(u)

Keyframing interpolation is a two-step algorithm: - Compute the spline parameter u corresponding to the time t: how to evaluate the inverse function?

Keyframing interpolation is a two-step algorithm: Compute the spline parameter u corresponding to the time t: how to evaluate the inverse function? - linear function: easy (t=u) - complex function: finite difference or numerically root finding

Keyframing interpolation is a two-step algorithm: Compute the spline parameter u corresponding to the time t: how to evaluate the inverse function? - linear function: easy - complex function: finite difference or numerically root finding Compute the θ corresponding to the spline parameter u: u θ = y(u)

Outline Process of keyframing Key frame interpolation Hermite and bezier curve Splines Speed control

Spline parameterization The spline is parameterized by u (0<=u<=1) and not by time t.

Spline parameterization The spline is parameterized by u (0<=u<=1) and not by time t. Hard to deal with the questions like -What is the location of point p at time t -What is the velocity of point p at time t -What is the acceleration of point p at time t

Spline parameterization Solution: reparameterize the curve in terms of t - express spline as a function of the arc length s: u=g(s) - express the arc length s as a function of t: s=f(t)

Speed control Time warping function to control speed - positive - monotonic; you cannot reverse the time s=f(t)

Speed control Simplest form is to have constant velocity along the path s=f(t)

Speed control Assume that the motion slows down at the beginning and end of the motion curve s=f(t)

Arc-length reparameterization Now we want a way to find u given a particular arclength s: u=g(f(t)) Not possible analytically for most curves (e.g. B- splines)

Finite difference Sample the curve at small intervals of the parameter and determine the distance between samples Use theses distances to build a table of arclength for this particular curve: (u i,s i ) What’s parametric value for the arc length.08?

Finite difference Sample the curve at small intervals of the parameter and determine the distance between samples Use theses distances to build a table of arclength for this particular curve: (u i,s i )

Sample the curve at small intervals of the parameter and determine the distance between samples Use theses distances to build a table of arclength for this particular curve: (u i,s i ) Finite difference What’s parametric value for the arc length.08?

Finite difference Sample the curve at small intervals of the parameter and determine the distance between samples Use theses distances to build a table of arclength for this particular curve: (u i,s i ) What’s parametric value for the arc length.08?

Finite difference Sample the curve at small intervals of the parameter and determine the distance between samples Use theses distances to build a table of arclength for this particular curve: (u i,s i ) What’s parametric value for the arc length.05? A linear interpolation of 0.00 and 0.05

Speed control recipe Given a time t, lookup the corresponding arclength S in the speed curve For S, look up the corresponding value of u in the reparameterization table Evaluate the curve at u to obtain the correct interpolated position for the animated object for the given time t