Paper Summaries Any takers? Interpolation Papers for interpolation - - PDF document
Paper Summaries Any takers? Interpolation Papers for interpolation Projects Logistics Proposals due Wednesday. E-mail. Who didnt get any? Jobs, Jobs, Jobs Plan for today and I dont mean Steve!
Interpolation
Any takers? Papers for interpolation…
Proposals due Wednesday.
E-mail….
Who didn’t get any?
…and I don’t mean Steve! Intel is looking for Graphics people
2D MPEG/video 3D For those graduating soon. Resumes to Raun Krisch (by Friday)
Keyframing and Interpolation But first…
Two early attempts of significant use of
CG elements in film
TRON
Disney (1982) Clips shown at SIGGRAPH 82
Question from SIGGRAPH Bowl
Q: Which special effects company was
responsible for the effects of TRON?
A: ALL OF THEM
MAGI Triple I Robert Abel & Associates Digital Effects
Interesting TRON Factoid
No polygons! All objects were defined using
mathematical formula.
Lesson Learned from TRON
Great effects Bad movie Overestimated the “Geek Factor”
The Last Starfighter (1985)
Effects by Digital Productions Universal Pictures
Last Starfighter factoids
First movie to do all special effects (except
makeup) on a computer.
All shots of spacecraft, space, etc. were generated
Lesson Learned from Last Starfighter
Great effects Bad movie Overestimated the “Geek Factor”
Life imitating art imitating life…
Sort of… A prototype for A Last Starfighter video
game was produced by Atari (remember them?) but was never distributed.
Note that there WAS a TRON video game
(made by Bally/Midway – 1983)
Key framing and interpolation
action
The two main approaches to hand drawn
animation.
Straight ahead action
“the animator works straight ahead from his first drawing
in the scene.”
Pose-to-pose (keyframing)
the animator plans his actions, figures out just what
drawings will be needed to animate the business, makes the drawings concentrating on the poses, ... and then draws the inbetweens.”
Link
The simplest form of animating an object. Object has:
A beginning state An end state A number of intermediate “key” states
It is up to the keyframe system to determine
the states for inbetween frames
P1 P2 P3 P4 P5 P6
Time = 0 Time = 10 Time = 35 Time = 50 Time = 55 Time = 60
60 sec animation
Video: 30 frames / sec = 1800 frames We have 6 key frames Need to create position of objects for the
remaining 1794 frames.
According to webster.com
in·ter·po·late (n) - to estimate values of
(a function) between two known values
Note: “state” can be defined as a function
F(u)
Things to consider:
Interpolation or approximation?
Interpolation – Control points fall on the curve Approximation – Control points fall outside the
curve.
Things to consider:
Continuity – defines “smoothness” of curve. Determined by how many derivatives are
continuous
0th order – are the curves attached at the points
themselves (positional continuity)
1st order – is the first derivative continuous (tangential
continuity)
2nd order – is the 2nd derivative continuous (curvature
continuity)
No positional continuity Order 1 Continuity Order 0 continuity 2nd order continuity For animation purposes, Tangential (1st Order) continuity is usually sufficient
Things to consider:
Local vs. Global control
Does changing one point change the entire
function?
Questions?
in·ter·po·late (n) - to estimate values of (a
function) between two known values
Function f (u)
u = interpolation control variable
Two known values
P1 = f(0) P2 = f(1)
Interpolation scheme will provide value of the
function for u between 0 and 1
Some interpolation schemes will require additional points to
perform interpolation
The simplest solution
P(0) = P0 P(1) = P1 P(u)
1
) 1 ( ) ( P u P u u P ⋅ + ⋅ − =
1
) ( ) ( P u P P u P + ⋅ − =
Also can be expressed as
Pros
Straight-forward Easy to implement
Cons
Lacks tangential continuity Why is this important?
Remember our principles of animation?
“The visual path of action from one
extreme to another is always described by an arc.”
This is not an arc!
x, y, z given as function of u Approximated by cubic polynomials x x x x
d u c u b u a u x + + + =
2 3
) (
y y y y
d u c u b u a u y + + + =
2 3
) (
z z z z
d u c u b u a u z + + + =
2 3
) (
1
2 3
u u u U =
⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ =
z z z z y y y y x x x x
d c b a d c b a d c b a C
[ ]
) ( ) ( ) ( ) ( u z u y u x u Q =
Separate C matrix into basis +
geometry 4 4 3 4 4 2 1 4 4 4 4 3 4 4 4 4 2 1
points control 4 3 2 1 4 3 2 1 4 3 2 1 matrix basis 44 43 42 41 34 33 32 31 24 23 22 21 14 13 12 11
⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ =
z z z z y y y y x x x x
G G G G G G G G G G G G m m m m m m m m m m m m m m m m C
Basis & Control points distinguishes types of
curves
control basis
Hermite Curves
specified by endpoints and tangent vectors
from each endpoint
Hermite Curves
⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − − − = 1 1 1 2 3 3 1 1 2 2
H
M ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ′ ′ =
+ + 1 1 i i i i H
P P P P G
Let’s go to the video tape
Problem with Hermite Curves
Animator generally will not enjoy specifying
tangent vectors
Solution: The Catmull-Rom spline
Catmull-Rom spline
Uses previous and next control point to determine
tangent vector
Spline goes through all control points!
Catmull-Rom
⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − − − − = 2 1 1 1 4 5 2 1 3 3 1 2 1
CR
M ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ =
+ + − 2 1 1 i i i i CR
P P P P G Note: for interpolation between Pi and Pi+1 you need 4 points
Let’s go to the video tape
Bezier Curves
specified by endpoints and intermediate
points, not on the curve, that approximate the tangent vectors
Bezier Curves
⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − − − − = 1 3 3 3 6 3 1 3 3 1
B
M ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ =
3 2 1
P P P P GH
Let’s go to the video tape
Bezier Curves suffer from same problem as
Hermite Curves
Animator generally will not enjoy specifying
tangent vectors…even from control points
Exception: Provide animator with an interactive curve
editor
DeCasteljau Construction
Algorithm to automatically define Bezier control points Described in Parent Appendix and Shomake Quat. paper
B-Splines
Control points are outside of the curve Local control
Moving one control point only affects a portion
Individual curve segments share control
points.
Knots - points connecting individual curve
segments.
B-Splines
B-Splines - local control
Segment Qi uses control points Pi-3, Pi-2, Pi-1 and Pi
B-Splines
⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − − − − = 1 4 1 3 3 3 6 3 1 3 3 1 6 1
B
M ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ =
− − − i i i i H
P P P P G
1 2 3
Let’s go to the video tape
Uniform Nonrational B-Splines
uniform - knots are equally spaced nonrational - does not use homogeneous
coordinates (more on this later)
B - basis -- curves are weighted sums of a
basis function
Rational Splines
expressed using homogeneous coordinates
[ ]
) ( ) ( ) ( ) ( ) ( u W u Z u Y u X u Q = ) ( ) ( ) ( , ) ( ) ( ) ( , ) ( ) ( ) ( u W u Z u z u W u Y u y u W u X u x = = =
Rational splines -- why bother?
Invariant under rotation, scaling,
translation, and perspective transforms
Can define precisely any of the conic
etc)
NURBS
Non-uniform Rational B-Splines Non-uniform - knots need not be equally
spaced.
Most useful and used method for
representing surface curves in CG.
Probably overkill for cubic interpolation
Pros
We now have our tangential continuity More complex than linear but not too
complex
Cons
May require an interactive “curve editor”
tool.
Questions
Normalized Quaternions are defined on the
unit 4D sphere
Linear interpolation generates unequal
spacing when projected onto circle
Spherical Linear Interpolation
where
θ θ θ θ sin ) sin( sin ) ) 1 sin(( ) , ( slerp
2 1 , 2 1
u q u q u q q + − =
2 1 2 1 2 1
Note:
Spherical Linear Interpolation has the same
problem as linear interpolation (no tangential continuity)
For smooth interpolation, Shomake recommends:
Bezier curves using DeCasteljau construction of control
points
Use slerp in constructing these points See SHOMAKE paper or text for details.
Linear Interpolation Cubic Interpolation
Hermite Catmull-Rom Bezier B-Splines
Interpolation of quaternions. Break.
Position / Orientation isn’t the only
thing that can be interpolated.
Frightening face demo Any value that changes over time is a
candidate.
u=1/3 u=2/3 u=0.0 u=1.0 u does not correspond to t.
Our problem
Spline control does not correspond to time. What we have
P(u) – position w.r.t. spline control
What we need
P(t) – position w.r.t. time (note we have this
for control points)
Solution
Determine:
u(t) – spline control w.r.t time
Then
P(t) = P (u(t))
Solution
Determine:
U(t) – spline control w.r.t time
Then
P(t) = P (U(t))
Can use interpolation on u(t) using t at control
points
Linear Cubic
Questions?
P1 P2 P3 P4 P5 P6
Time = 0 Time = 10 Time = 35 u = 0.0 Time = 50 u=1.0 Time = 55 Time = 60 at t = 42.5 u = 0.5
Refers to the spacing of the inbetween
frames at maximum positions.
Example:
a bouncing ball moves faster as it approaches
approaches leaves its maximum position
Suppose animator knows path but
wants more control over speed?
What we have
P(u) – position w.r.t. spline control
Note that this is NOT
S(u) – length traveled w.r.t. spline control.
And we need to give animator control of
S(t) – length traveled at a given time.
Suppose animator knows path but wants
more control over speed?
What we have
P(u) – position w.r.t. spline control
What we need
U(S) – length w.r.t. spline control
And let animator define
S(t) – length w.r.t. time (velocity/speed)
Then
P(t) = P (U (S (t)))
Arc Length = distance traveled along a
curve.
Mathematically defined as:
u u
2 1
2 1
So, arc length is an integral
Ways to solve
Analytic – integral not always solvable Numerical Integration – more when we talk
about physics.
Forward Differencing
Forward Differencing
Sample the curve at equally spaced u values. Estimate arc length by calculating linear distance
between sampled points.
For example,
Given P(u) calculate at u= 0.00, 0.05, 0.10, 0.15, … Define arc length s(u) s (0.0) = 0.0 s (0.05) = distance between P(0.05) and P(0.0) s (0.1) = s (0.05) + distance between P(0.10) and P(0.05) Etc.
Forward Differencing
This will give us a mapping from P(u) to
s(u) for sampled u values.
Perform interpolation (linear or cubic)
between sampled values
s(u) u
Idx
0.720 0.40
8
0.600 0.35
7
0.500 0.30
6
0.400 0.25
5
0.320 0.20
4
0.230 0.15
3
0.150 0.10
2
0.080 0.05
1
0.000 0.00
Find u corresponding to s = 0.25 Find closest idx to s = 0.25 Find relative distance to s for next idx
222 . 23 . 32 . 23 . 25 . = − −
Linearly interpolate u 161 . ) 15 . 20 . ( 22 . 15 . ) 25 . ( = − + = u
Animator can define s(t)
Length traveled w.r.t time Speed
Basic ease-in/ease out
Animator defines speed
Arc length given time
Spline control determined by forward
differencing
222 . 23 . 32 . 23 . 25 . = − −
Linearly interpolate u 161 . ) 15 . 20 . ( 22 . 15 . ) 25 . ( = − + = u Find u corresponding to s = 0.25 Find closest idx to s = 0.25 Find relative distance to s for next idx
Position determined by spline
P = P (0.161) Questions?