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 x i such that: with n+1 knot values y i find a spline function with each S i (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 S i (x) = a i x 3 + b i x 2 + c i x + d i 4 Coefficients with n subintervals = 4n equations There are 4 n-2 conditions Interpolation conditions Continuity conditions Natural Conditions S’’(x 0 ) = 0 S’’(x n ) = 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.

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. B 0,1 (t) = 1-t, B 1,1 (t) = t; 2. B 0,2 (t) = (1-t) 2, B 1,2 (t) = 2t(1-t), B 2,2 (t) = t 2 ; 3. B 0,3 (t) = (1-t) 3, B 1,3 (t) = 3t(1-t) 2, B 2,3 (t)=3t 2 (1-t), B 3,3 (t) = t 3 ;

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: B 0 (t) = (1-t)^3 B 1 (t) = 3t(1-t)^2 B 2 (t) = 3t^2(1-t) B 3 (t) = t^3

Bernstein Polynomials 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 (x 0,y 0 ), (x 1,y 1 ), (x 2,y 2 ), (x 3,y 3 ) be the “control”points The x and y coordinates of the cubic Bezier spline are given by: x(t) = x 0 B 0 (t) + x 1 B 1 (t) +x 2 B 2 (t) + x 3 B 3 (t) y(t) = y 0 B 0 (t) + y 1 B 1 (t) + y 2 B 2 (t) + y 3 B 3 (t) Where 0<= t <= 1.

Bernstein Polynomials Given a set of control points {P i } N i=0, where P i = (x i, y i ). A Bezier curve of degree N is: P(t) = N i=0 P i B i,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 P i = (x i, y i ) x(t) = N i=0 x i B i,N (t) and y(t) = N i=0 y i B i,N (t) The spline goes through (x 0,y 0 ) and (x 3,y 3 ) The tangent at those points is the line connecting (x 0,y 0 ) to (x 1,y 1 ) and (x 3,y 3 ) to (x 2,y 2 ). The positions of (x 1,y 1 ) and (x 2,y 2 ) 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) = 2B 0,3 (t) + 1B 1,3 (t) + 3.5B 2,3 (t) + 4B 3,3 (t) y(t) = 2B 0,3 (t) + 1.5B 1,3 (t) + 0B 2,3 (t) + 1B 3,3 (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) t 2 (1 – t) + 4t 3 y(t) = 2(1-t) t(1 – t) 2 + t 3 Simplify these formulas to yield the parametric equation: P(t) = (2 – 3t t 2 – 5.5t 3, 2 – 1.5t – 3t t 3 ) where 0 < t < 1 P(t) is the Bezier curve r.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.)

Composite Bezier Curves 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)}

Composite Bezier Curves Use the same process as earlier to yield these four parametric equations: P 1 (t) = (-9 + 3t - 3t 2 + 5t 3, 3t + 1.5t 2 - 2t 3 ) P 2 (t) = (-4 + 3t + 3t 2 - 2t 3, t - 1.5t 2 ) P 3 (t) = (6t - 3t 2 + 2t 3, 4 - 2t 3 ) P 4 (t) = (5 + 3t + 39t t 3, 2 + 3t 2 - 5t 3 )

Composite Bezier Curves Linking Curves Smoothly To have two Bezier curves P(t) and Q(t) meet smoothly would require: P N = Q 0 P’(P N ) = Q’(Q 0 ) It is sufficient to require that the control points P N-1, P N = Q 0, and Q 1 be collinear.

Composite Bezier Curves 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 - 18t t 3 ) Q(t) = (3 + 3t, 3 + 6t – 18t t 3 ) and P’(t) = (3, 6 – 36t + 36t 2 ) Q’(t) = (3, 6 – 36t + 36t 2 ) Substituting t = 1 and t = 0 into P’(t) and Q’(t), yields P’(1) = (3,6) = Q’(0)

Composite Bezier Curves It’s collinear!

Practical Applications

Practical Applications

Bézier surfaces in computer graphics