1. Designing Parametric Cubic Curves 1

2. 2 Objectives Introduce types of curves ­Interpolating ­Hermite ­Bezier ­B-spline Analyze their performance

3. 3 Matrix-Vector Form define then

4. 4 Interpolating Curve p0p0 p1p1 p2p2 p3p3 Given four data (control) points p 0, p 1, p 2, p 3 determine cubic p(u) which passes through them Must find c 0, c 1, c 2, c 3

5. 5 Interpolation Equations apply interpolating conditions at u = 0, 1/3, 2/3, 1 p 0 = p(0) = c 0 p 1 = p(1/3) = c 0 + (1/3)c 1 + (1/3) 2 c 2 + (1/3) 3 c 2 p 2 = p(2/3) = c 0 + (2/3)c 1 + (2/3) 2 c 2 + (2/3) 3 c 2 p 3 = p(1) = c 0 + c 1 + c 2 + c 2 or in matrix form with p = [p 0 p 1 p 2 p 3 ] T p = Ac

6. 6 Interpolation Matrix Solving for c find interpolation matrix c = M I p Note that M I does not depend on input data and can be used for each segment in x, y, and z

7. 7 Interpolating Multiple Segments use p = [p 0 p 1 p 2 p 3 ] T use p = [p 3 p 4 p 5 p 6 ] T Get continuity at join points but not continuity of derivatives

8. 8 Blending Functions Rewriting the equation for p(u) p(u) = u T c = u T M I p = b(u) T p where b(u) = [b 0 (u) b 1 (u) b 2 (u) b 3 (u)] T is an array of blending polynomials s.t. p(u) = b 0 (u)p 0 + b 1 (u)p 1 + b 2 (u)p 2 + b 3 (u)p 3 b 0 (u) = -4.5(u - 1/3)(u - 2/3)(u - 1) b 1 (u) = 13.5u (u - 2/3)(u - 1) b 2 (u) = -13.5u (u - 1/3)(u - 1) b 3 (u) = 4.5u (u - 1/3)(u - 2/3)

9. 9 Blending Functions functions are not smooth ­Hence interpolation polynomial not smooth

10. 10 Interpolating Patch Need 16 conditions to determine the 16 coefficients c ij Choose at u, v = 0, 1/3, 2/3, 1

11. 11 Matrix Form Define v = [1 v v 2 v 3 ] T C = [c ij ] P = [p ij ] p(u,v) = u T Cv If observe that for constant u (v), obtain interpolating curve in v (u), can show p(u,v) = u T M I PM I T v C = M I PM I

12. 12 Blending Patches Each b i (u)b j (v) is a blending patch Shows that can build and analyze surfaces from knowledge of curves

13. 13 Other Types of Curves and Surfaces How can get around limitations of interpolating form ­Lack of smoothness ­Discontinuous derivatives at join points Have four conditions (for cubics) that can apply to each segment ­Use them other than for interpolation ­Need only come close to data

14. 14 Hermite Form p(0)p(1) p’(0) p’(1) Use two interpolating conditions and two derivative conditions per segment Ensures continuity and first derivative continuity between segments

15. 15 Equations Interpolating conditions same at ends p(0) = p 0 = c 0 p(1) = p 3 = c 0 + c 1 + c 2 + c 3 Differentiate  p’(u) = c 1 + 2uc 2 + 3u 2 c 3 Evaluate at end points p’(0) = p’ 0 = c 1 p’(1) = p’ 3 = c 1 + 2c 2 + 3c 3

16. 16 Matrix Form Solve  c = M H q where M H = [Hermite matrix]

17. 17 Blending Polynomials p(u) = b(u) T q Although these functions are smooth, Hermite form not used directly in Computer Graphics and CAD because usually have control points but not derivatives However, Hermite form is basis of Bezier form

18. 18 Parametric and Geometric Continuity Can require derivatives of x, y, and z to each be continuous at join points (parametric continuity) Alternately, can only require that tangents of resulting curve be continuous (geometry continuity) The latter gives more flexibility as need satisfy only 2 conditions rather than 3 at each join point

19. 19 Example Here p and q have same tangents at ends of segment but different derivatives Generate different Hermite curves This techniques is used in drawing applications

20. 20 Higher Dimensional Approximations Techniques for both interpolating and Hermite curves can be used with higher dimensional parametric polynomials For interpolating form, resulting ­matrix becomes increasingly more ill- conditioned ­curves less smooth and more prone to numerical errors In both cases, more work in rendering resulting polynomial curves and surfaces