review of interpolation. a method of constructing a function that crosses through a discrete set of...

Review of Interpolation

.

A method of constructing a function that crosses through a discrete set of known data points.

Spline (general case)

Given n+1 distinct knots xi such that:

with n+1 knot values yi find a spline function

with each Si(x) a polynomial of degree at most n.

Spline Interpolation

Linear, Quadratic, Cubic Preferred over other polynomial interpolation

More efficient High-degree polynomials are very computationally expensive Smaller error Interpolant is smoother

Natural Cubic Spline Interpolation

Si(x) = aix3 + bix2 + cix + di

4 Coefficients with n subintervals = 4n equations There are 4n-2 conditions

Interpolation conditions Continuity conditions Natural Conditions

S’’(x0) = 0

S’’(xn) = 0

Natural Conditions

Bezier Curves

A similar but different problem:

Controlling the shape of curves.

Problem: given some (control) points, produce and modify the shape of a curve passing through the first and last point.

http://www.ibiblio.org/e-notes/Splines/Bezier.htm

Bezier Curves

Idea: Build functions that are combinations of some basic and simpler functions.

Basic functions: B-splines

Bernstein polynomials

Bernstein Polynomials

Bernstein polynomials of degree n are defined by:

For v = 0, 1, 2, …, n, In general there are N+1 Bernstein Polynomials of degree N. For

example, the Bernstein Polynomials of degrees 1, 2, and 3 are: 1. B0,1(t) = 1-t, B1,1(t) = t;

2. B0,2(t) = (1-t)2, B1,2(t) = 2t(1-t), B2,2(t) = t2;

3. B0,3(t) = (1-t)3, B1,3(t) = 3t(1-t)2, B2,3(t)=3t2(1-t), B3,3(t) = t3;

Deriving the Basis Polynomials

Write 1 = t + (1-t) and raise 1 to the third power for cubic splines (2 for quadratic, 4 for quartic, etc…)

1^3 = (t + (1-t))^3 = t^3 + 3t^2(1-t) + 3t(1-t)^2 + (1-t)^3 The basis functions are the four terms:

B0(t) = (1-t)^3

B1(t) = 3t(1-t)^2

B2(t) = 3t^2(1-t)

B3(t) = t^3


Bernstein polynomials of degree 3

Using the Basis Functions

To use the Bernstein basis functions to make a spline we need 4 “control” points for a cubic. (3 for quadratic, 5 for quartic, etc…)

Let (x0,y0), (x1,y1), (x2,y2), (x3,y3) be the “control”points

The x and y coordinates of the cubic Bezier spline are given by:

x(t) = x0B0(t) + x1B1(t) +x2B2(t) + x3B3(t)

y(t) = y0B0(t) + y1B1(t) + y2B2(t) + y3B3(t)

Where 0<= t <= 1.


Given a set of control points {Pi}Ni=0, where Pi = (xi, yi). A Bezier curve of

degree N is:

P(t) = Ni=0 PiBi,N(t), Where Bi,N(t), for I = 0, 1, …, N, are the Bernstein polynomials of

degree N. P(t) is the Bezier curve

Since Pi = (xi, yi) x(t) = N

i=0xiBi,N(t) and y(t) = Ni=0yiBi,N(t)

The spline goes through (x0,y0) and (x3,y3)

The tangent at those points is the line connecting (x0,y0) to (x1,y1) and (x3,y3) to (x2,y2).

The positions of (x1,y1) and (x2,y2) determine the shape of the curve.

Bernstein Polynomials Example

Find the Bezier curve which has the control points (2,2), (1,1.5), (3.5,0), (4,1). Substituting the x- and y-coordinates of the control points and N=3 into the x(t) and y(t) formulas on the previous slide yields

x(t) = 2B0,3(t) + 1B1,3(t) + 3.5B2,3(t) + 4B3,3(t)

y(t) = 2B0,3(t) + 1.5B1,3(t) + 0B2,3(t) + 1B3,3(t)

Bernstein Polynomials Example

Substitute the Bernstein polynomials of degree three into these formulas to get

x(t) = 2(1 – t)3 + 3t(1 – t)2 + 10.5t2(1 – t) + 4t3

y(t) = 2(1-t)3 + 4.5t(1 – t)2 + t3

Simplify these formulas to yield the parametric equation:

P(t) = (2 – 3t + 10.5t2 – 5.5t3, 2 – 1.5t – 3t2 + 3.5t3)where 0 < t < 1

P(t) is the Bezier curvehttp://www.cs.brown.edu/exploratories/freeSoftware/repository/edu/brown/cs/exploratories/applets/bezierSplines/bezier_splines_java_browser.html

Composite Bezier Curves

In practice, shapes are often constructed of multiple Bezier curves combined together.

Allows for flexibility in constructing shapes. (Consecutive Bezier curves need not join smoothly.)


Linking curves:

Example: Find the composite Bezier curve for the four sets of control points:

{(-9,0), (-8,1), (-8,2.5), (-4,2.5)}, {(-4,2.5), (-3,3.5), (-1,4), (0,4)}{(0,4), (2,4), (3,4), (5,2)}, {(5,2), (6,2), (20,3), (18,0)}


Use the same process as earlier to yield these four parametric equations:

P1(t) = (-9 + 3t - 3t2 + 5t3, 3t + 1.5t2 - 2t3)

P2(t) = (-4 + 3t + 3t2 - 2t3, 2.5 + 3t - 1.5t2)

P3(t) = (6t - 3t2 + 2t3, 4 - 2t3)

P4(t) = (5 + 3t + 39t2 - 29t3, 2 + 3t2 - 5t3)


Linking Curves Smoothly To have two Bezier curves P(t) and Q(t) meet

smoothly would require: PN = Q0

P’(PN) = Q’(Q0)

It is sufficient to require that the control points PN-1, PN = Q0, and Q1 be collinear.


Example of collinear control points:

{(0,3), (1,5), (2,1), (3,3)} and {(3,3), (4,5), (5,1), (6,3)}

P(t) = (3t, 3 + 6t - 18t2 + 12t3)

Q(t) = (3 + 3t, 3 + 6t – 18t2 + 12t3) and

P’(t) = (3, 6 – 36t + 36t2)

Q’(t) = (3, 6 – 36t + 36t2)

Substituting t = 1 and t = 0 into P’(t) and Q’(t), yields

P’(1) = (3,6) = Q’(0)


It’s collinear!

Practical Applications


• 0.900 0.850 0.900 0.850 0.900 0.700 0.900 0.700• 0.900 0.850 0.850 0.930 0.800 0.950 0.666 0.950• 0.333 0.050 0.333 0.050 0.666 0.050 0.666 0.050• 0.100 0.800 0.120 0.910 0.200 0.950 0.333 0.950• 0.100 0.800 0.100 0.800 0.100 0.200 0.100 0.200• 0.100 0.200 0.120 0.090 0.200 0.050 0.333 0.050• 0.333 0.950 0.333 0.950 0.666 0.950 0.666 0.950• 0.900 0.150 0.850 0.070 0.800 0.050 0.666 0.050• 0.900 0.150 0.900 0.150 0.900 0.300 0.900 0.300• 0.900 0.700 0.900 0.700 0.780 0.700 0.780 0.700• 0.780 0.700 0.780 0.780 0.720 0.820 0.600 0.820• 0.900 0.300 0.900 0.300 0.780 0.300 0.780 0.300• 0.780 0.300 0.780 0.220 0.720 0.180 0.600 0.180• 0.200 0.700 0.200 0.700 0.200 0.300 0.200 0.300• 0.600 0.820 0.600 0.820 0.333 0.820 0.333 0.820• 0.600 0.180 0.600 0.180 0.333 0.180 0.333 0.180• 0.333 0.180 0.230 0.180 0.200 0.230 0.200 0.300• 0.333 0.820 0.230 0.820 0.200 0.770 0.200 0.700•

• 1.900 0.850 1.900 0.850 1.900 0.700 1.900 0.700• 1.900 0.850 1.850 0.930 1.800 0.950 1.666 0.950• 1.333 0.050 1.333 0.050 1.666 0.050 1.666 0.050• 1.100 0.800 1.120 0.910 1.200 0.950 1.333 0.950• 1.100 0.800 1.100 0.800 1.100 0.200 1.100 0.200• 1.100 0.200 1.120 0.090 1.200 0.050 1.333 0.050• 1.333 0.950 1.333 0.950 1.666 0.950 1.666 0.950• 1.900 0.150 1.850 0.070 1.800 0.050 1.666 0.050• 1.900 0.150 1.900 0.150 1.900 0.300 1.900 0.300• 1.900 0.700 1.900 0.700 1.780 0.700 1.780 0.700• 1.780 0.700 1.780 0.780 1.720 0.820 1.600 0.820• 1.780 0.300 1.780 0.220 1.720 0.180 1.600 0.180• 1.200 0.700 1.200 0.700 1.200 0.300 1.200 0.300• 1.600 0.820 1.600 0.820 1.333 0.820 1.333 0.820• 1.600 0.180 1.600 0.180 1.333 0.180 1.333 0.180• 1.333 0.180 1.230 0.180 1.200 0.230 1.200 0.300• 1.333 0.820 1.230 0.820 1.200 0.770 1.200 0.700• 1.780 0.300 1.780 0.300 1.600 0.300 1.600 0.300• 1.600 0.300 1.600 0.300 1.600 0.370 1.600 0.370• 1.600 0.370 1.600 0.370 1.950 0.370 1.950 0.370• 1.950 0.370 1.950 0.370 1.950 0.300 1.950 0.300• 1.950 0.300 1.950 0.300 1.900 0.300 1.900 0.300


0 1 2 3 4 5 60

1

2

3

4

5

6

7


4.35 0.00 4.35 0.00 4.35 0.19 4.35 0.19 4.35 0.19 3.53 0.23 3.39 0.36 3.39 1.09 3.39 1.09 3.39 1.08 3.39 6.20 3.39 6.20 3.39 6.20 3.39 6.20 3.93 6.20 3.93 6.20 3.93 6.20 5.07 6.20 5.29 6.02 5.52 4.92 5.52 4.92 5.52 4.92 5.76 4.92 5.76 4.92 5.76 4.92 5.76 4.92 5.70 6.62 5.70 6.62 5.70 6.62 5.70 6.62 0.30 6.62 0.30 6.62 0.24 4.92 0.24 4.92 0.30 6.62 0.30 6.62 0.48 4.92 0.48 4.92 0.24 4.92 0.24 4.92 2.07 6.20 0.93 6.20 0.71 6.02 0.48 4.92 2.61 6.20 2.61 6.20 2.07 6.20 2.07 6.20 2.61 1.09 2.61 1.08 2.61 6.20 2.61 6.20 1.65 0.19 2.47 0.23 2.61 0.36 2.61 1.09 1.65 0.00 1.65 0.00 1.65 0.19 1.65 0.19 1.65 0.00 1.65 0.00 4.35 0.00 4.35 0.00

Bézier surfaces in computer graphics

review of interpolation. a method of constructing a function that crosses through a discrete set of...

Documents