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.
θ
t
Key-frame Interpolation
• Given parameter values at key frames,
how to interpolate parameter values for
inbetween frames.
θ
t
Nonlinear interpolation
Review: Natural cubic cruves
a x t 3  bxt 2  cxt  d x  Qx (t )
Review: Natural cubic cruves
a x t 3  bxt 2  cxt  d x  Qx (t )
t
3
t2
 ax 
 
 bx 
t 1    Qx (t )
c
 x
d 
 x

Review: Natural cubic cruves
a x t 3  bxt 2  cxt  d x  Qx (t )
t
t

t

t
t

3
3
1
3
2
3
3
3
4
 ax 
 
 bx 
t 1    Qx (t )
c
 x
d 
 x

t2
t1
2
t2
2
t3
2
t4
2
t1 1  a x   Qx (t1 ) 
  

t 2 1 bx   Qx (t 2 ) 
 c    Q (t ) 
t3 1  x   x 3 
t 4 1 d x   Qx (t 4 ) 
 a x   t13 t12 t1
   3
2
 bx   t 2 t 2 t 2
c   t 3 t 2 t
 x   33 3 2 3
 d  t
 x   4 t4 t4
1

1

1
1
1
 Qx (t1 ) 


 Qx (t 2 ) 
 Q (t ) 
 x 3 
 Q (t ) 
 x 4 
Review: Natural cubic curves
Q(t )  t
3
t
2
 ax 
 
 bx 
t 1   t 3 t 2
c
 x
d 
 x
Properties:
- Go through four control points
- not good for local control
 t13
 3
t
t 1 2 3
 t3
t 3
4
2
t1
t2
2
t2
t3
2
t3
t4
2
t4
t1
1

1

1
1
1
 Q(t1 ) 


Q
(
t
)
 2 
 Q(t ) 
 3 
 Q(t ) 
 4 
Review: Hermite Curves
P1: start position
P4: end position
R1: start derivative
R4: end derivative
R1
P1
P4
R4
Review: Hermite Curves

Q (t )  t 3
t2
1  p1 
 2 2 1

 
  3 3  2  1 p4 
t 1
0
0
1
0  R1 

 
 1
0
0
0  R4 


R1
P1
P4
R4
Review: Hermite Curves

Q (t )  t 3
t2
1  p1 
 2 2 1

 
  3 3  2  1 p4 
t 1
0
0
1
0  R1 

 
 1
0
0
0  R4 


Herminte basis matrix
R1
P1
P4
R4
Review: Hermite Curves

Q (t )  t 3

t2
R1
1  p1 
 2 2 1

 
  3 3  2  1 p4 
t 1
0
0
1
0  R1 

 
 1
0
0
0  R4 


 P1 
 
P
3
2
3
2  4 
t  2t  t t  t  
R
 1
R 
 4
Herminte basis matrix
 2t 3  3t 2  1  2t 3  3t 2
P1

P4
R4
Review: Hermite Curves

Q (t )  t 3

t2
R1
1  p1 
 2 2 1

 
  3 3  2  1 p4 
t 1
0
0
1
0  R1 

 
 1
0
0
0  R4 


 2t 3  3t 2  1  2t 3  3t 2
P1
 P1 
 
P
3
2
3
2  4 
t  2t  t t  t  
R
 1
R 
 4

P4
R4
Review: Hermite Curves

Q (t )  t 3

t2
R1
1  p1 
 2 2 1

 
  3 3  2  1 p4 
t 1
0
0
1
0  R1 

 
 1
0
0
0  R4 


 2t 3  3t 2  1  2t 3  3t 2
P1
 P1 
 
P
3
2
3
2  4 
t  2t  t t  t  
R
 1
R 
 4

P4
R4
Review: Hermite Curves

Q (t )  t 3

t2
R1
1  p1 
 2 2 1

 
  3 3  2  1 p4 
t 1
0
0
1
0  R1 

 
 1
0
0
0  R4 


 2t 3  3t 2  1  2t 3  3t 2
P1
 P1 
 
P
3
2
3
2  4 
t  2t  t t  t  
R
 1
R 
 4

P4
R4
Review: Hermite Curves

Q (t )  t 3

t2
R1
1  p1 
 2 2 1

 
  3 3  2  1 p4 
t 1
0
0
1
0  R1 

 
 1
0
0
0  R4 


 2t 3  3t 2  1  2t 3  3t 2
P1
 P1 
 
P
3
2
3
2  4 
t  2t  t t  t  
R
 1
R 
 4

P4
R4
Review: Hermite Curves

Q (t )  t 3

t2
R1
1  p1 
 2 2 1

 
  3 3  2  1 p4 
t 1
0
0
1
0  R1 

 
 1
0
0
0  R4 


 2t 3  3t 2  1  2t 3  3t 2
P1
 P1 
 
P
3
2
3
2  4 
t  2t  t t  t  
R
 1
R 
 4
Hermite basis functions

P4
R4
Review: Hermite Curves

Q (t )  t 3
t2


basis function 1
1
1
0.9
0.8
0.8
0.7
0.7
0.6
0.6
0.5
0.5
0.4
0.4
0.3
0.3
R4
P1

basis function 2
0.9
P4
 P1 
 
P
3
2
3
2  4 
t  2t  t t  t  
R
 1
R 
 4
 2t 3  3t 2  1  2t 3  3t 2
basis function 3
basis function 4
0.16
0
0.14
-0.02
0.12
-0.04
0.1
-0.06
0.08
-0.08
0.06
-0.1
0.04
-0.12
0.2
0.2
0.02
0.1
0.1
0
R1
1  p1 
 2 2 1

 
  3 3  2  1 p4 
t 1
0
0
1
0  R1 

 
 1
0
0
0  R4 

0
20
40
60
80
100
120
0
0
20
40
60
80
100
120
0
-0.14
0
20
40
60
80
100
120
-0.16
0
20
40
60
80
100
120
Review: Hermite Curves

Q (t )  t 3
t2
R1
1  p1 
 2 2 1

 
  3 3  2  1 p4 
t 1
0
0
1
0  R1 

 
 1
0
0
0  R4 


P4
R4
P1
 P1 
 
P
3
2
3
2  4 
t  2t  t t  t  
R
 1
R 
 4


 2t 3  3t 2  1  2t 3  3t 2
 P1  (2t 3  3t 2  1)  P4  (2t 3  3t 2 )  R1  (t 3  2t 2  t )  R4  (t 3  t 2 )
1
1
0.9
0.9
*P1
=
0.8
0.7
+
*P4 +
0.8
0.7
0.6
0.6
0.5
0.5
0.4
0.4
0.3
0.3
0
*R1
0.14
0.12
+
*R4
-0.02
-0.04
0.1
-0.06
0.08
-0.08
0.06
-0.1
0.04
-0.12
0.2
0.2
0.02
0.1
0.1
0
0.16
0
20
40
60
80
100
120
0
0
20
40
60
80
100
120
0
-0.14
0
20
40
60
80
100
120
-0.16
0
20
40
60
80
100
120
Review: Bezier Curves
R1  3(v1  v0 )
R2  3(v3  v2 )
P1  v0
P4  v3
Review: Bezier Curves

Q (t )  t 3
t2
 1 3  3

 3 6 3
t 1
3 3
0

 1
0
0


1  v0 
 
0  v1 
0  v2 
 
0  v3 
Review: Bezier Curves

Q (t )  t 3

t2
 1 3  3

 3 6 3
t 1
3 3
0

 1
0
0


  t 3  3t 2  3t  1 3t 3  6t 2  3t
1  v0 
 
0  v1 
0  v2 
 
0  v3 
 3t 3  3t 2
 v0 
 
v
3  1
t  
v
 2
v 
 3

Review: Bezier Curves

Q (t )  t 3
t2
 1 3  3

 3 6 3
t 1
3 3
0

 1
0
0



  t 3  3t 2  3t  1 3t 3  6t 2  3t
1  v0 
 
0  v1 
0  v2 
 
0  v3 
 3t 3  3t 2
 v0 
 
v
3  1
t  
v
 2
v 
 3

1
0.45
0.45
1
0.9
0.4
0.4
0.9
0.35
0.35
0.8
0.8
0.7
0.7
0.3
0.3
0.25
0.25
0.2
0.2
0.15
0.15
0.1
0.1
0.05
0.05
0.6
0.6
0.5
0.5
0.4
0.4
0.3
0.3
0.2
0.1
0
0
20
40
60
80
100
120
0
0
20
40
60
80
100
120
0
0.2
0.1
0
20
40
60
80
100
120
0
0
20
40
60
80
100
120
Review: Bezier Curves

Q (t )  t 3
t2
 1 3  3

 3 6 3
t 1
3 3
0

 1
0
0


1  v0 
 
0  v1 
0  v2 
 
0  v3 

  t 3  3t 2  3t  1 3t 3  6t 2  3t
 3t 3  3t 2
 v0 
 
v
3  1
t  
v
 2
v 
 3

 v0  (t 3  3t 2  3t  1)  v1  (3t 3  6t 2  3t )  v2  (3t 3  3t 2 )  v3  t 3
1
0.45
0.45
1
0.9
0.4
0.4
0.9
0.35
0.35
0.8
0.7
=
*v0
0.6
0.5
+
*v1
0.3
0.25
+
0.8
*v2
0.3
0.25
0.2
0.2
0.15
0.15
0.1
0.1
0.05
0.05
+
0.4
0.6
0.5
0.4
0.3
0.3
0.2
0.1
0
*v3
0.7
0
20
40
60
80
100
120
0
0
20
40
60
80
100
120
0
0.2
0.1
0
20
40
60
80
100
120
0
0
20
40
60
80
100
120
Review: Different basis functions
• Cubic curves:
Q(t )  t 3 t 2
 ax 
 
b 
t 1 x   t 3 t 2
c
 x
d 
 x
• Hermite curves:

Q (t )  t 3

t1
2
t2
2
t3
2
t4
2
t1 1 

t 2 1

t3 1 
t 4 1
1
 Q(t1 ) 


 Q(t 2 ) 
 Q(t ) 
 3 
 Q(t ) 
 4 
t2
1  P1 
 2 2 1

 

3
3

2

1

 P4 
t 1
0
0
1
0  R1 

 
 1
 R 
0
0
0

 4 
t2
 1 3  3

 3 6 3
t 1
3 3
0

 1
0
0

• Bezier curves:
Q (t )  t 3
 t13
 3
t
t 1 2 3
 t3
t 3
4


1  v0 
 
0  v1 
0  v2 
 
0  v3 
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
• C0: points coincide, velocities don’t
• C1: points and velocities coincide
• What’s C2?
- 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 C2 continuity when two curves
meet
Derivatives at end points

Q (t )  t 3
t2
 1 3  3

 3 6 3
t 1
3 3
0

 1
0
0


1  v0 
 
0  v1 
0  v2 
 
0  v3 
t=1
t=0
Derivatives at end points

Q (t )  t 3
t2
 1 3  3

 3 6 3
t 1
3 3
0

 1
0
0


1  v0 
 
0  v1 
0  v2 
 
0  v3 
t=1

Q (t )  3 t 2
 1 3  3

 3 6 3
2t 1 0 
3 3
0

 1
0
0


1  v0 
 
0  v1 
0  v2 
 
0  v3 
t=0
Derivatives at end points

Q (t )  t 3
t2
 1 3  3

 3 6 3
t 1
3 3
0

 1
0
0


1  v0 
 
0  v1 
0  v2 
 
0  v3 
t=1
Q (t )  3 t 2
 1 3  3

 3 6 3
2t 1 0 
3 3
0

 1
0
0

1  v0 
 
0  v1 
0  v2 
 
0  v3 
(t )  6 t
Q
 1 3  3

 3 6 3
2 0 0 
3 3
0

 1
0
0

1  v0 
 
0  v1 
0  v2 
 
0  v3 


t=0
Derivatives at end points

Q (t )  t 3
t2
 1 3  3

 3 6 3
t 1
3 3
0

 1
0
0


1  v0 
 
0  v1 
0  v2 
 
0  v3 
t=1
Q (t )  3 t 2
 1 3  3

 3 6 3
2t 1 0 
3 3
0

 1
0
0

1  v0 
 
0  v1 
0  v2 
 
0  v3 
(t )  6t
Q
 1 3  3

 3 6 3
2 0 0 
3 3
0

 1
0
0

1  v0 
 
0  v1 
0  v2 
 
0  v3 


t=0
 1 3  3

 3 6 3
Q (0)  3* 0 2 2 * 0 1 0
3 3
0

0
0
 1
 3(v1  v0 )
 1 3  3

(0)  6 * 0 2 0 0  3  6 3
Q
 3 3
0

0
0
 1
 6(v0  2v1  v2 )
1  v 0 
 
0  v1 
0  v 2 
 
0  v3 
1  v0 
 
0  v1 
0  v2 
 
0  v3 
Derivatives at end points

Q (t )  t 3
t2
 1 3  3

 3 6 3
t 1
3 3
0

 1
0
0


1  v0 
 
0  v1 
0  v2 
 
0  v3 
t=1

Q (t )  3 t 2
(t )  6 t
Q
 1 3  3

 3 6 3
2t 1 0 
3 3
0

 1
0
0

1  v0 
 
0  v1 
0  v2 
 
0  v3 
 1 3  3

 3 6 3
2 0 0 
3 3
0

 1
0
0

1  v0 
 
0  v1 
0  v2 
 
0  v3 

t=0
Q (0)  3(v1  v0 )
Q (1)  3(v  v )
3
2
(0)  6(v  2v  v )
Q
0
1
2
(1)  6(v  2v  v )
Q
3
2
1
Continuity in B splines
• Suppose we want to join two Bezier curves (V0,
V1, V2,V3) and (W0, W1, W2, W3) so that C2
continuity is met at the joint
Continuity in B splines
• Suppose we want to join two Bezier curves (V0,
V1, V2,V3) and (W0, W1, W2, W3) so that C2
continuity is met at the joint
Qv (1)  Qw (0)
Continuity in B splines
• Suppose we want to join two Bezier curves (V0,
V1, V2,V3) and (W0, W1, W2, W3) so that C2
continuity is met at the joint
Qv (1)  Qw (0)
Q (1)  Q (0)
v
w
 (1)  Q
 (0)
Q
v
w
Continuity in B splines
• Suppose we want to join two Bezier curves (V0,
V1, V2,V3) and (W0, W1, W2, W3) so that C2
continuity is met at the joint
v3  w0
Qv (1)  Qw (0)
Q (1)  Q (0)
3(v3  v2 )  3( w2  w1 )
 (1)  Q
 (0)
Q
v
w
6(v3  2v2  v1 )  6( w0  2w1  w2 )
v
w
Continuity in B splines
• Suppose we want to join two Bezier curves (V0,
V1, V2,V3) and (W0, W1, W2, W3) so that C2
continuity is met at the joint
v3  w0
3(v3  v2 )  3( w2  w1 )
6(v3  2v2  v1 )  6( w0  2w1  w2 )
Continuity in B splines
• Suppose we want to join two Bezier curves (V0,
V1, V2,V3) and (W0, W1, W2, W3) so that C2
continuity is met at the joint
v3  w0
3(v3  v2 )  3( w2  w1 )
6(v3  2v2  v1 )  6( w0  2w1  w2 )
w0  v3
w1  2v3  v2
w2  v1  4v3  4v2
Continuity in B splines
What does this derived equation mean
geometrically? w2=v1+4v3-4v2
- What is the relationship between a, b and c, if a = 2b - c?
b is the middle point of a and c.
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!
Building complex splines
• Constraining a Bezier curve made of many
segments to be C2 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 C2)
- 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 C2 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
Catmull-Rom splines
 ( p3  p1 )
 ( p4  p2 )
The segment is controlled by p1,p2,p3,p4
Q ( 0)  p 2
Q (1)  p3
Q (0)   ( p3  p1 )
Q (1)   ( p  p )
4
0.5
2
Catmull-Rom splines
 ( p3  p1 )
 ( p4  p2 )
The segment is controlled by p1,p2,p3,p4
Q ( 0)  p 2
Q (1)  p3
Q (0)   ( p3  p1 )
Q (1)   ( p  p )
4
2
Catmull-Rom Splines
The effect of t: how sharply the curve bends
at the control points
Catmull-Rom splines
 ( p3  p1 )
 ( p4  p2 )
The segment is controlled by p1,p2,p3,p4
Q ( 0)  p 2
Q (1)  p3
Q (0)   ( p3  p1 )
Q (1)   ( p  p )
4
2
 Q ( 0)   0

 
Q
(
1
)

  0
 Q (0)     

 
 Q (1)   0

 
0 0  p1 
 
0 1 0  p2 
0  0  p3 
 
  0   p4 
1
Catmull-Rom splines
 ( p3  p1 )
 ( p4  p2 )
From Hermite curves

Q(t )  u 3 u 2
1  Q(0) 
 2 2 1



  3 3  2  1 Q(1) 
u 1
0
0
1
0  Q (0) 



 1
0
0
0  Q (1) 


Catmull-Rom splines
 ( p3  p1 )
 ( p4  p2 )
From Hermite curves

Q(t )  u 3 u 2
1  Q(0) 
 2 2 1



  3 3  2  1 Q(1) 
u 1
0
0
1
0  Q (0) 



 1
0
0
0  Q (1) 


 Q ( 0)   0

 
 Q (1)   0
 Q (0)     

 
 Q (1)   0

 
0 0  p1 
 
0 1 0  p2 
0  0  p3 
 
  0   p4 
1
Catmull-Rom splines
 ( p3  p1 )
 ( p4  p2 )
From Hermite curves

Q(t )  u 3 u 2
1  Q(0) 
 2 2 1



  3 3  2  1 Q(1) 
u 1
0
0
1
0  Q (0) 



 1
0
0
0  Q (1) 



Q (t )  u 3
u2
 Q ( 0)   0

 
 Q (1)   0
 Q (0)     

 
 Q (1)   0

 
1  0
 2 2 1



3
3

2

1

 0
u 1
0
0
1
0   


 1
0
0
0  0


0 0  p1 
 
0 1 0  p2 
0  0  p3 
 
  0   p4 
1
0 0  p1 
 
0 1 0  p2 
0  0  p3 
 
  0   p4 
1
Catmull-Rom splines
 ( p3  p1 )
 ( p4  p2 )

Q (t )  u 3

 u3
u2
u2
1  0
 2 2 1


  3 3  2  1 0
u 1
0
0
1
0   


 1
 0
0
0
0



 

 2
u 1


 0


2 
 2
  3 3  2
0

1
0
0 0  p1 
 
0 1 0  p2 
0  0  p3 
 
  0   p4 
1
  p1 
 
   p2 
0  p3 
 
0  p4 
Catmull-Rom splines
 ( p3  p1 )
 ( p4  p2 )

Q (t )  u 3

 u3
u2
u2
1  0
 2 2 1


  3 3  2  1 0
u 1
0
0
1
0   


 1
 0
0
0
0



 

 2
u 1


 0


2 
 2
  3 3  2
0

1
0
0 0  p1 
 
0 1 0  p2 
0  0  p3 
 
  0   p4 
1
  p1 
 
   p2 
0  p3 
 
0  p4 
Catmull-Rom splines
 ( p3  p1 )
 ( p4  p2 )

Q (t )  u 3

 u3
u2
u2
1  0
 2 2 1


  3 3  2  1 0
u 1
0
0
1
0   


 1
 0
0
0
0



 

 2
u 1


 0


2 
 2
  3 3  2
0

1
0
0 0  p1 
 
0 1 0  p2 
0  0  p3 
 
  0   p4 
1
  p1 
 
   p2 
0  p3 
 
0  p4 
What do we miss?
Catmull-Rom splines
?
?
Catmull-Rom splines
( p 4  p3 )
( p1  p0 )
Catmull-Rom splines
• Catmull-Rom splines have C1 continuity
(not C2 continuity)
• Do not lie within the convex hull of their
control points.
Catmull-Rom splines
• Catmull-Rom splines have C1 continuity
(not C2 continuity)
• Do not lie within the convex hull of their
control points.
Catmull-Rom splines properties
X Continuity (C2)
√ Local control
√ Interpolation
Catmull-Rome Splines for
Kyeframe Interpolation
• Given a set of control points, interpolation splines are obtained by
fitting the input points with a piecewise cubic polynomial curve that
passes through every control point
Catmull-Rome Splines for
Kyeframe Interpolation
•
Fit the interpolation function between each pair of control points with a cubic
function
pk
Pk+1
Pk+2
pk-1
Catmull-Rome Splines for
Kyeframe Interpolation
•
Fit the interpolation function between each pair of control points with the following
cubic function:

p (u )  u 3
u2
 

 2
u 1


 0


2 
 2
  3 3  2
0

1
0
  pk 1 


   pk 
0  pk 1 




0  pk  2 
pk
Pk+1
Pk+2
pk-1
0<=u<=1
Outline
•
•
•
•
•
Process of keyframing
Key frame interpolation
Hermite and bezier curve
Splines
Speed control
Speed Control
• Click here
Speed control
• Time warping function to control speed
(mapping from the original time line to the
output time line)
- positive
- monotonic; you cannot reverse the time
s=f(t)
new timeline
Original timeline
Speed control
• Simplest form is to have constant velocity
along the path
s=f(t)
Speed control
• Simplest form is to have constant velocity
along the path
s=f(t)
Which one is “speed up” or which one is “slow down”
Speed control
• Simplest form is to have constant velocity
along the path
Speed up
s=f(t)
Slow down
Speed control
• Assume that the motion slows down at the
beginning and end of the motion curve
s=f(t)
Speed control
• Assume that the motion slows down at the
beginning and end of the motion curve
s=f(t)
Adjust timing for each key frame