Interpolating values
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Transcript of Interpolating values
Interpolating values
CSE 3541Matt 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
2. Construct an arc-length-parametric-value function for the curve
3. Construct time-arc-length function according to given constraints
p=P(u)
u=U(s)
s=S(t)
p=P(U(S(t)))
Lab 5
See specification
Choosing an interpolating functionTradeoffs 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
Therefore, choose a cubic polynomial
With a point of inflection, the curvecan match arbitrary tangents at end points
Continuity ordersC-1
discontinuous C0 continuous
C1 first derivative is 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
jk
k kj
kjj xx
xxyxP
1
)(
See http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html for more details
Lagrange Polynomial
x
jk
k kj
kjj xx
xxyxP
1
)(
Written explicitly term by term is:
Lagrange Polynomial
Results are not always good for animation
Polynomial curve formation
• Given by the equationP(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 FormulationMatrix multiplication
MBUP T
123 uuuGeometric information (i.e the points or tangents)
Coefficient matrix (given by which type of curve you are using)
Where u is a scalar value in [0,1]
The point at “time” u
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
• Use with cubic equation
Note:- c values correspond to a, b, c, and d in the earlier cubic equation- c values are vectors
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
1
1
2
23 *
0010
05.005.0
5.025.21
5.05.15.15.0
*1)(
i
i
i
i
p
p
p
p
uuuuP
Blended Parabolas/Catmull-Rom*Visual example
1
1
2
23 *
0010
05.005.0
5.025.21
5.05.15.15.0
*1)(
i
i
i
i
p
p
p
p
uuuuP
* End conditions are handled differently(or assume curve starts at P1 and stops at Pn-2)
Bezier Curve
p0
p1
p2
p3
• Find the point x on the curve as a function of parameter u:
x(u)•
de Casteljau Algorithm
• Describe the curve as a recursive series of linear interpolations
• Intuitive, but not the most efficient form
de Casteljau Algorithm
p0
p1
p2
p3
• Cubic Bezier curve has four points (though the algorithm works with any number of points)
de Casteljau Algorithm
p0
q0
p1
p2
p3
q2
q1
322
211
100
,,
,,
,,
ppq
ppq
ppq
uLerp
uLerp
uLerp
Lerp = linear interpolation
de Casteljau Algorithm
q0
q2
q1
r1
r0
211
100
,,
,,
qqr
qqr
uLerp
uLerp
de Casteljau Algorithm
r1x
r0•
10 ,, rrx uLerp
Bezier Curve
x•
p0
p1
p2
p3
Cubic Bezier
3
2
123
0.00.00.00.1
0.00.00.30.3
0.00.30.60.3
0.10.30.30.1
1
i
i
i
i
p
p
p
p
uuuP
Curve runs through Pi and Pi+3
with starting tangent PiPi+1 and ending tangent Pi+2Pi+3
Controlling Motion along p=P(u)
Step 2. Reparameterization by arc lengthu = U(s) where s is distance along the curve
p=P(u)But segments may nothave equal length
Step 1. vary u from 0 to 1create points on the curve
Step 3. Speed control
s = ease(t) where t is timefor example, ease-in / ease-out
Reparameterizing by Arc Length
AnalyticForward differencing
SupersamplingAdaptive approach
NumericallyAdaptive Gaussian
Reparameterizing by Arc Length - supersample
1.Calculate a bunch of points at small increments in u2.Compute summed linear distances as approximation to arc
length3.Build table of (parametric value, arc length) pairs
Notes1.Often useful to normalize total distance to 1.02.Often useful to normalize parametric value for multi-
segment curve to 1.0
index u Arc Length
(s)
0 0.00 0.000
1 0.05 0.080
2 0.10 0.150
3 0.15 0.230
... ... ...
20 1.00 1.000
Build table of approx. lengths
Speed Control
Time-distance functionEase-in Ease-out
SinusoidalCubic polynomialConstant acceleration
General distance-time functions
time
distance
Time Distance Function
s = S(t)
s
t the global time variable
S
Ease-in/Ease-out Function
s = S(t)
s
t
S
0.0
0.01.0
1.0
Normalize distance and time to 1.0 to facilitate reuse
Ease-in: Sinusoidal
2/)12/sin()( tteases
Ease-in: Single Cubic
23 32)( ttteases
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
2. Construct an arc-length-parametric-value function for the curve
3. Construct time-arc-length function according to given constraints
p=P(u)
u=U(s)
s=S(t)
p=P(U(S(t)))
Arbitrary Speed ControlAnimators 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/LagrangeInterpolatingPolynomial.html