Oldest keyframe animation
Two conditions to make a moving image in 19th century
1. Higher than 10 frames per second
Keyframe animation
2. A period of blackness between images
Phenakistiscope
2D animation
Highly skilled animators draw the keyframes
Less skilled (lower paid) animators draw the inbetween frames
Difficult to create physically realistic animation
Time consuming process
Zoetrope
Praxinoscope
3D animation
Animators specify important keyframes in 3D
Computers generates the in-between frames
Some dynamic motion can be done by computers
(hair, clothes, etc)
Still time consuming; Pixar spent four years to
produce Toy Story
From the making of toy story
General pipeline
Story board
Keyframes
Inbetweens
Painting
Storyboards
Story boarding from “a bug’s life”
The film in outline form
specify the key scenes
specify the camera moves and edits
specify character gross motion
Typically paper and pencil sketches on individual
sheets taped on a wall
http://www.pixar.com/featurefilms/ab!behind_pop4.html
The process of keyframing
A keyframe
Specify the keyframes
Specify the type of interpolation
linear, cubic, parametric curves
Specify the speed profile of the interpolation
constant velocity, ease-in-ease-out, etc
In 2D animation, a keyframe is usually a single
image
In 3D animation, each keyframe is defined by a set
of parameters
Computer generates the in-between frames
Keyframe parameterization
in-between frames
What are the parameters?
Position and orientation in 3D
Joint angles of the hierarchy
Body deformation
Facial features
Hair
Clothing
Scene elements such as lights
and cameras
Linear interpolation
Cubic curve interpolation
Linear interpolation
Cubic curve interpolation
Linearly interpolate the parameters between keyframes
We can use three cubic functions to represent a 3D curve
Each function is defined with the range 0 ≤ t ≤ 1
t=5
Q(t) = [x(t) y(t) z(t)]
t = 10
or
3
t=0
Qx (t) = ax t + bx t2 + cx t + dx
t = 15
end key start key
Qy (t) = ay t3 + by t2 + cy t + dy
t − t0
(x1 − x0 )
x = x0 +
t1 − t0
end time
Qz (t) = az t3 + bz t2 + cz t + dz
start time
Compact representation

ax
 bx
C=
 cx
dx
ay
by
cy
dy

az
bz 

cz 
dz
T=

Q(t) =
!
t3
t2
ax
"  bx
t 1 
 cx
dx
!
ay
by
cy
dy
t3
t2
t 1

az
bz 
 = TC
cz 
dz
"
2t 1
0
matrices
A · B : multiplication
AB : multiplication
How many constraints do we need to determine a cubic
curve? 4
Redefine C as a product of the basis matrix M and the
geometry vector G
C=M·G
Q(t) =
"
vectors
a · b : inner product
a × b : outer product
ab : multiplication
Constraints on the cubics

!
d
Q̇ = Q(t) = 3t2
dt
bold: a vector or a matrix
italic: a scalar
!
t3
t2
C
=T·M·G
m11
"  m21
t 1 
 m31
m41
m12
m22
m32
m42
m13
m23
m33
m43

G1x
m14

m24 
  G2x
m34   G3x
G4x
m44
G1y
G2y
G3y
G4y

G1z
G2z 

G3z 
G4z
Hermite curves
Hermite basis matrix
A Hermite curve is determined by
Given desired constraints:
P4
R4
P1
1. endpoints P1 and P4
R1
2. tangent vectors R1 and R4 at the endpoints
Use these elements to construct the geometry vector

P1x
 P4x
Gh = 
 R1x
R4x
P1y
P4y
R1y
R4y

P1
0 0
 P4 
 1 1



 R1  = Gh =  0 0
R4
3 2
0
1
1
1
!
Q(1) =
!
0
1
0 0 1
"
· Mh · Gh = P1
1 1 1
"
· Mh · Gh = P4
2. tangent vectors meet R1 and R4
!
Q (1) =
!
!
0 0
3 2
1 0
"
· Mh · Gh = R1
1 0
"
· Mh · Gh = R4
Hermite Blending functions
We can solve for basis matrix Mh

Q(0) =
Q! (0) =

P1z
P4z 

R1z 
R4z
Hermite basis matrix

1. endpoints meet P1 and P4
Let’s define B as a product of T and M

1
1 
 · Mh · Gh
0 
0
Bh (t) =
!

t3
t2

2 −2 1
1
"  −3 3 −2 −1 

t 1 
 0
0
1
0 
1
0
0
0
Bh(t) indicates the weight of each element in Gh

0
 1
Mh = 
 0
3
0
1
0
2
0
1
1
1
−1
1
1 

0 
0

2
 −3
=
 0
1
−2
3
0
0

1
1
−2 −1 

1
0 
0
0
Bh (t)


P1
 P4 


Q(t) = Bh (t) · 
R1 
R4
P1
P4
R1
t
R4
Bézier curves
Bézier basis matrix
Indirectly specify tangent vectors by specifying two
intermediate points
R1
Establish the relation between Hermite and Bezier
geometry vectors
P3
R1 = 3(P2 − P1 )
P4
P2
R4 = 3(P4 − P3 )
P1
R4


P1
 P2 

Gb = 
 P3 
P4
Bézier basis matrix
Q(t) = T · Mh · Gh = T · Mh · (Mhb · Gb )

 
P1
1
 P4   0
 
Gh = 
 R1  =  −3
R4
0
0 0
0 0
3 0
0 −3
Mb = Mh Mhb
3 −3
−6 3
3
0
0
0

1
0 

0 
0
http://www.cs.unc.edu/~mantler/research/bezier/index.html

 = Mhb · Gb

Bezier blending functions are also called Bernstein
polynomials
Bb (t) =
−1
 3
=
 −3
1

Bézier blending functions

= T · (Mh · Mhb ) · Gb = T · Mb · Gb


0
P1
 P2
1 

0   P3
3
P4
!
t3
t2
t

−1 3 −3 1

"
3 −6 3 0 

1 
 −3 3
0 0 
1
0
0 0
Bb (t)


P1
 P2 


Q(t) = Bb (t) · 
P3 
P4
P1
P4
P2
P3
t
Complex curves
What if we want to model a curve that passes through
these points?
Splines
A piecewise polynomial function that has locally
very simple form, yet be globally flexible and
smooth
There are three nice properties of splines we’d like
to have
Problem with higher order polynomials:
Continuity
Wiggly curves
Local control
No local control
Interpolation
Continuity
Continuity
C0: points coincide, velocities don’t
Cubic curves are continuous and differentiable
C1: points and velocities coincide
We only need to worry about the derivatives at the
endpoints when two curves meet
What’s C2? points, velocities and accelerations coincide
Local control
Bezier and Hermite curves don’t have local control;
moving a single control point affects the whole
curve
We’d like to have each control point on the spline
only affect some well-defined neighborhood around
that point
Interpolation
Bezier curves do not necessarily pass through all
the control points
We’d like to have a spline interpolating the control
points so that the spline always passes through
every control points
B-splines
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
V1
We can join multiple Bezier curves to create Bsplines
Ensure C2 continuity when two curves meet
V0
V3
Qv (1) = Qw (0)
Q!v (1)
=
Q!w (0)
Q!!v (1) = Q!!w (0)
W1
V2
W0
W2
W3
V3 = W0
V3 − V2 = W1 − W0
V1 − 2V2 + V3 = W0 − 2W1 + W2
W2 = V1 + 4V3 − 4V2
Continuity in B-splines
What does this derived equation mean geometrically?
W2 = V1 + 4V3 − 4V2
de Boor points
Instead of specifying the Bezier control points, let’s
specify the corners of the frames that forms a B-spline
What is the relationship between
a, b and c, if a = 2b - c?
V1 V
2
V0
B1
W0 = V3
W1 = 2V3 − V2
V3 = W0
W2 = V1 + 4V3 − 4V2
= 2(2V3 − V2 ) − (2V2 − V1 )
W1
W3
W2
= 2W1 − B1
What is B2?
de boor points
What is the relationship between Bezier control points and
de Boor points?
These points are called de Boor points and the frames are
called A-frames
B-spline basis matrix

Q(t) =
"
!
1
1
2
V0 =
B0 + (B1 − B0 ) + B1 + (B2 − B1 )
2
3
3
1
V1 = B1 + (B2 − B1 )
3
2
V2 = B1 + (B2 − B1 )
3
"
!
1
1
2
V3 =
B1 + (B2 − B1 ) + B2 + (B3 − B2 )
2
3
3
Verify this by yourself
!
t3

Mbs
t2
t
−1 3
1
3 −6

= 
6 −3 0
1
4

B1
 B2 
"

1 Mbs 
 B3 
B4
−3
3
3
1

1
0 

0 
0
http://www.engin.umd.umich.edu/CIS/course.des/cis577/projects/BSP/welcome.html
B-Spline properties
B-Spline properties
C2 continuity?
C2 continuity?
Local control?
Local control?
Interpolation?
Interpolation?
Catmull-Rom splines
Catmull-Rom splines
D0 = C1 − C0
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
controls, we get a Catmull-Rom spline
C3
1
D1 = (C2 − C0 )
2
D2 =
1
(C3 − C1 )
2
..
.
Dn = Cn − Cn−1
C1
C0
D1
D0
D2
C2
Catmull-Rom Basis matrix

Q(t) =
!
t3
t2
−1
"1 2

t 1
2  −1
0
3 −3
−5 4
0
1
2
0

1
P1
 P2
−1 

0   P3
0
P4
CATmull-Rom properties




C2 continuity?
Local control?
Interpolation?
Derive Catmull-Rom basis matrix by yourself
CATmull-Rom properties
C2 interpolating splines
How can we keep the C2 continuity of B-splines but
get interpolation property as well?
C2 continuity?
Local control?
Suppose we have a set of Bezier control points, our
goal is to find a C2 spline that that passes through
all the points
Interpolation?
C1
C2
C0
C3
C2 interpolating splines
Let’s assume we know the first derivative of the spline at
these control points
C1
D1
C0
C2
D2
C3
D0
D3
Derivatives of splines
V0 = C0
W0 = C1
1
V1 = C0 + D0
3
1
V2 = C1 − D1
3
1
W1 = C1 + D1
3
1
W2 = C2 − D2
3
V3 = C1
W3 = C2
To solve for D’s we apply C2 continuity requirement
If we want to create a Bezier curve between each pair of
these points, what are the V’s and W’s control points in
terms of C’s and D’s?
6(V1 − 2V2 + V3 ) = 6(W0 − 2W1 + W2 )
D0 + 4D1 + D2 = 3(C2 − C0 )
Derivatives of splines
Boundary conditions
D0 + 4D1 + D2 = 3(C2 − C0 )
We can impose more conditions on the spline to solve the
two extra degrees of freedom
D1 + 4D2 + D3 = 3(C3 − C1 )
..
.
Dm−2 + 4Dm−1 + Dm = 3(Cm − Cm−2 )
How many equations do we have?
How many variables are we trying to solve?
Natural C2 interpolating splines require second derivative
to be zero at the endpoints
6(V0 − 2V1 + V2 ) = 0
Linear system
Choice of splines
Collect m+1 equations into a linear system

2 1
 1 4


1





1
4

1
..
.
1
4
1
D0
  D1

  D2


..

.

1   Dm−1
2
Dm

Spline types

3(C1 − C0 )
  3(C2 − C0 )
 
  3(C3 − C1 )
 
=
..
 
.
 
  3(Cm − Cm−2 )
3(Cm − Cm−1 )
Continuity Interpolation Local control









Use forward elimination to zero out every thing below
the diagonal, then back substitute to compute D’s
Displaying Bezier curves
B-Splines
C2
No
Yes
Catmull-Rom
Splines
C1
Yes
Yes
C2 interpolating
splines
C2
Yes
No
Subdivide and Conquer
V1
V0!!
DisplayBezier(V0, V1, V2, V3)
V0!
begin
V1
if (FlatEnough(V0, V1, V2, V3))
Line(V0, V3)
V1!
V2
V1!!
V2
V2!
V0
V0
else
do something
It would be nice if we had an adaptive algorithm that
would take into account flatness
V3
DisplayBezier(V0, V1, V2, V3)
begin
if (FlatEnough(V0, V1, V2, V3))
! Line(V0, V3)
else
! Subdivide(V) -> L, R
! DisplayBezier(L0, L1, L2, L3)
! DisplayBezier(R0, R1, R2, R3)
V3
Flatness Test
Endpoints of B-splines
V1
V2
We can see that B-splines don’t interpolate the de
Boor points
V0
V3
Compare total length of control polygon to length of line
connecting endpoints:
It would be nice if we could at least control the
endpoints of the splines explicitly
There’s a trick to make the spline begin and end at
the de Boor points by repeating them
|V0 − V1 | + |V1 − V2 | + |V2 − V3 |
<1+!
|V0 − V3 |
Curves in the animation
Each animation parameter is determined by a 2D curve
Q(u) = (x(u), y(u))
Treat this curve as
θ(u) = y(u)
t(u) = x(u)
where θ is a variable you want to animate. We can think
of the results as a function: θ(t)
Apply constraints to make sure that θ(t) actually is a
function
Wrapping the curves
Wrapping is an important feature that makes the
animation restart smoothly when looping back to
the beginning
Create “phantom” control points before and after
the first and the last control points
t
C−1
Cn
C1
C0
C2
Cn+1
Speed control
Arclength-Time function
Simplest form is to have constant velocity along the path
In the linear interpolation case, we can use time (t)
as the parameter, increment it a constant amount,
and evaluate the new position
s
But there is a problem in the cubic polynomial case
t
Non-Uniformity in the
parameter
Arclength reparameterization
The problem is that uniform steps in the parameter (u)
don’t correspond to uniform path distances
We have a function Lengh(u) which evaluates the
arclength for a given u, but we also want a way to
find u given a particular arclength S
Not possible analytically for most curves (e.g. Bsplines)
We need a relationship between the parameters (u) and
the distance traveled
Finite Differences
Finite differences
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
index
parametric value
arclength
0
1
2
3
4
0.00
0.05
0.10
0.15
0.20
0.000
0.080
0.150
0.230
0.320
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
If we want to know S at a particular value of the
parameter u = 0.73, spacing of the table is 0.05
Find the entry in the table closest to this u
Or take the u before and after it and interpolate
linearly
Similarly, we can find u for a given S
Ease-in Ease-out curve
Assume that the motion slows down at the
beginning and end of the motion curve
s
t
Do you know this?
Open questions
What should the key values be?
When should the key values occur?
Why does a movie run at 24 frames per second
while a TV runs at 30 frames per second?
Are there better ways to specify key values?
What kind of bad things can occur from
interpolation? How do we prevent them?
Invalid configurations (pass through walls)
Unnatural motions (painful twists)
What’s next?
What about rotations?
Can we interpolate rotations using the these same
techniques?