Interpolating values
CSE 3541
Matt Boggus
Problem: paths in grids or graphs are jagged
Path smoothing
Path smoothing example
Problem statement
• How can we construct smooth paths?
– Define smooth in terms of geometry
– What is the input?
– Where does the input come from?
• Pathfinding data
• Animator specified
Interpolation
• Interpolation: The process of inserting in a
series an intermediate number or quantity
ascertained by calculation from those already
known.
• Examples
– Computing midpoints
– Curve fitting
Interpolation terms
• Order (of the polynomial)
– Linear
– Quadratic
– Cubic
• Dimensions
– Bilinear (data in a 2D grid)
– Trilinear (data in a 3D grid)
Linear interpolation
• Given two points, P0 and P1 in 2D
• Parametric line equation:
P = P0 + t (P1 – P0) ; OR
X = P0.x + t (P1.x – P0.x)
Y = P0.y + t (P1.y – P0.y)
• t = 0  Beginning point P0
• t = 1  End point P1
Linear interpolation
• Rewrite the parametric equation
P = (1-t)P0 + t P1 ; OR
X = (1-t)P0.x + t P1.x
Y = (1-t)P0.y + t P1.y
• Formula is equivalent to a weighted average
• t is the weight (or percent) applied to P1
• 1 – t is the weight (or percent) applied to P0
• t = 0.5  Midpoint between P0 and P1 = Pmid
• t = 0.25  1st quartile = midpoint between P0 and Pmid
• t = 0.75  3rd quartile = midpoint between Pmid and P1
Bilinear interpolation process
• Given 4 points (Q’s)
• Interpolate in one
dimension
– Q11 and Q21 give R1
– Q12 and Q22 give R2
• Interpolate with the
results
– R1 and R2 give P
Bilinear interpolation application
• Resizing an image
Bilinear vs. bicubic interpolation
Higher order interpolation requires more sample points
Curve fitting
• Given a set of points
– Smoothly (in time and space) move an object
through the set of points
• Example of additional temporal constraints
– From zero velocity at first point, smoothly accelerate
until time t1, hold a constant velocity until time t2,
then smoothly decelerate to a stop at the last point
at time t3
Example
Observations:
• Order of labels (A,B,C,D)
is computed based on
time values
• Time differences do not
have to correspond to
space or distance
• The last specified time is
arbitrary
Solution steps
1. Construct a space curve that
interpolates the given points with
piecewise first order continuity
p=P(u)
2. Construct an arc-length-parametricvalue function for the curve
u=U(s)
3. Construct time-arc-length function
according to given constraints
s=S(t)
p=P(U(S(t)))
Lab 5
See specification
Choosing an interpolating function
Tradeoffs and commonly chosen options
Interpolation vs. approximation
Polynomial complexity: cubic
Continuity: first degree (tangential)
Local vs. global control: local
Information requirements: tangents needed?
Interpolation vs. Approximation
Polynomial complexity
Low complexity
reduced computational cost
With a point of inflection, the curve
can match arbitrary tangents at end points
Therefore, choose a cubic polynomial
Continuity orders
C-1 discontinuous
C1 first derivative is continuous
C0 continuous
C2 first and second derivatives are continuous
Local vs. Global control
Information requirements
just the points
tangents
interior control points
just beginning and ending
tangents
Curve Formulations
• General solution
– Lagrange Polynomial
• Piecewise cubic polynomials
– Catmull-Rom
– Bezier
Lagrange
Polynomial
x  xk
Pj ( x)  y j 
k 1 x j  xk
x
k j
See http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html for more details
Lagrange
Polynomial
x  xk
Pj ( x)  y j 
k 1 x j  xk
x
k j
Written explicitly term by term is:
Lagrange
Polynomial
Results are not always good for animation
Polynomial curve formation
• Given by the equation
P(u) = au3 + bu2 + cu + d
• u is a parameter that varies from 0 to 1
– u = 0 is the first point on the curve
– u = 1 is the last point on the curve
• Remember this is a parametric equation!
– a, b, c, and d are not scalar values, but vectors
Polynomial curve formation
P(u) = au3 + bu2 + cu + d
• The point P(u) = (x(u), y(u), z(u))
where
x(u) = axu3 + bxu2 + cxu + dx
y(u) = ayu3 + byu2 + cyu + dy
z(u) = azu3 + bzu2 + czu + dz
Polynomial Curve Formulation
Matrix multiplication
The point at “time” u
P  U MB
T
u
3
u2

u 1
Where u is a scalar
value in [0,1]
Geometric information
(i.e the points or tangents)
Coefficient matrix
(given by which type of curve you are using)
So how do we set a, b, c, and d?
Or the matrices M and B?
Catmull-Rom spline
• Passes through control points
• Tangent at each point pi is based on the
previous and next points: (pi+1 − pi−1)
– Not properly defined at start and end
• Use a toroidal mapping
• Duplicate the first and last points
Catmull-Rom spline derivation
τ is a parameter for tension
(most implementations set it to 0.5)
Catmull-Rom spline derivation
Catmull-Rom spline derivation
Note:
- c values correspond to a, b, c, and d
in the earlier cubic equation
- c values are vectors
• Use with cubic equation
Catmull-Rom spline derivation
• Solve for ci in terms of control points
Catmull-Rom matrix form
• Set τ to 0.5 and put into matrix form

P(u )  u 3 u 2
 0.5 1.5  1.5 0.5   pi  2 
 1  2 .5 2  0 .5   p 
i 1 



u 1*
*
  0 .5 0
0 .5
0   pi 

  
1
0
0   pi 1 
 0

Blended Parabolas/Catmull-Rom*
Visual example

P (u )  u 3
u2
 0.5 1.5  1.5 0.5   pi  2 
 1
 p 

2
.
5
2

0
.
5
 *  i 1 
u 1 *
  0 .5
0
0 .5
0   pi 


 
1
0
0   pi 1 
 0

* End conditions are handled differently
(or assume curve starts at P1 and stops at Pn-2)
Bezier Curve
p1
p0
•
x(u)
p2
• Find the point x on
the curve as a
function of
parameter u:
p3
de Casteljau Algorithm
• Describe the curve as a recursive series of
linear interpolations
• Intuitive, but not the most efficient form
de Casteljau Algorithm
p1
p0
p2
• Cubic Bezier curve has four points
(though the algorithm works with
any number of points)
p3
de Casteljau Algorithm
p1
q1
q0
p0
p2
q2
q 0  Lerp u , p 0 , p1 
q1  Lerp u , p1 , p 2 
q 2  Lerp u , p 2 , p 3 
Lerp = linear interpolation
p3
de Casteljau Algorithm
q0
r0
q1
r1
r0  Lerp u , q 0 , q1 
r1  Lerp u , q1 , q 2 
q2
de Casteljau Algorithm
r0
•
x
x  Lerp u , r0 , r1 
r1
Bezier Curve
p1
•
p0
x
p2
p3
Cubic Bezier
P  u
3
u
2
  1 .0 3 .0  3 .0
 3 .0  6 .0 3 .0
u 1
  3 .0 3 .0
0 .0

0 .0
0 .0
 1 .0

Curve runs through Pi and Pi+3
with starting tangent PiPi+1 and
ending tangent Pi+2Pi+3
1.0   pi 
0.0  pi 1 
0.0  pi  2 


0.0  pi 3 
Controlling Motion
along p=P(u)
Step 1. vary u from 0 to 1
create points on the curve
p=P(u)
But segments may not
have equal length
Step 2. Reparameterization by arc length
u = U(s)
where s is distance along the curve
Step 3. Speed control
for example, ease-in / ease-out
s = ease(t)
where t is time
Reparameterizing by Arc Length
Analytic
Forward differencing
Supersampling
Adaptive approach
Numerically
Adaptive Gaussian
Reparameterizing by Arc
Length - supersample
1.Calculate a bunch of points at small increments in u
2.Compute summed linear distances as approximation to arc
length
3.Build table of (parametric value, arc length) pairs
Notes
1.Often useful to normalize total distance to 1.0
2.Often useful to normalize parametric value for multisegment curve to 1.0
Build
table of
approx.
lengths
index
u
0
0.00
Arc
Length
(s)
0.000
1
0.05
0.080
2
0.10
0.150
3
0.15
0.230
...
...
...
20
1.00
1.000
Speed Control
distance
Time-distance function
Ease-in Ease-out
Sinusoidal
Cubic polynomial
Constant acceleration
General distance-time functions
time
Time Distance Function
S
s
s = S(t)
t  the global
time variable
Ease-in/Ease-out Function
s
S
1.0
s = S(t)
0.0
0.0
1.0
t
Normalize distance and time to 1.0 to facilitate reuse
Ease-in: Sinusoidal
s  ease(t )  sin( t   / 2  1) / 2
Ease-in: Single Cubic
s  ease(t )  2t  3t
3
2
Ease-in: Constant Acceleration
Ease-in: Constant Acceleration
Ease-in: Constant Acceleration
Ease-in: Constant Acceleration
Motion on a curve – solution steps
1. Construct a space curve that interpolates the
given points with piecewise first order continuity
p=P(u)
2. Construct an arc-length-parametric-value function
for the curve
u=U(s)
3. Construct time-arc-length function according to
given constraints
s=S(t)
p=P(U(S(t)))
Arbitrary Speed Control
Animators can work in:
Distance-time space curves
Velocity-time space curves
Acceleration-time space curves
Set time-distance constraints
References
• Images from
• http://www.cambridgeincolour.com/tutorials/
image-interpolation.htm
• http://mathworld.wolfram.com/LagrangeInter
polatingPolynomial.html