Splines II – Interpolating Curves

1. Splines II – Interpolating Curves based on: Michael Gleicher: Curves, chapter 15 in Fundamentals of Computer Graphics, 3rd ed. (Shirley & Marschner) Slides by Marc van Kreveld

2. Polynomial pieces Canonical form of a polynomial of degree n defined with vector coefficients ai Generalized form of a polynomial defined with vector coefficients cithat are blended by the m polynomials bi(t) The degree is the max of the degrees of the bi(t) Using blending polynomials is the way to make splines

3. Recall: Basis and constraint matrices • Specifications of a curve give a constraint matrixp0 = f(0) = a0 + 0 a1 + 02a2p1 = f(0.5) = a0 + 0.5 a1 + 0.52a2p2 = f(1) = a0 + 1 a1 + 12a2 • Its inverse B = C–1 is the basis matrix (quadratic curve)

4. Blending functions • Blending functions (or basis functions)are functions of u and specify how to “mix” the specified constraints (points to pass through, derivatives, …) • Let u = [ 1 uu2u3 … un ] be the powers of u • b(u) = uB, a vector whose elements are the blending functions

5. Blending functions • u = [ 1 uu2u3 … un ] • b(u) = uB so we obtain for the usual quadric example with three points specified:

6. Blending functions • We can now blend the control points

7. Blending functions We see the contributions of each point depending on u For fixed u, we linearly interpolate the three points

8. Blending functions 1 0 1 0.5 u

9. Blending functions • Note the sum of the contributions +

10. Polynomials for interpolation • Given points p = (p0, p1, … , pn) and increasing parameter values t = (t0, t1, … , tn), we can make a polynomial of degree n that passes through pi exactly at parameter value ti so f(ti) = pi p1 p2 p5 p4 p0 p3

11. Polynomials for interpolation • Given points p = (p0, p1, … , pn) and increasing parameter values t = (t0, t1, … , tn), we can make a polynomial of degree n that passes through pi exactly at parameter value ti so f(ti) = pi p1 p2 p5 p4 p0 p3 the brown curve has t1’ > t1 and t4’ < t4

12. Polynomials for interpolation • Method • Set up constraint matrix as before • Invert to get basis matrix, giving the n+1 basis functions bi(t) and the polynomial f(t) : • Alternative method (Lagrange form)

13. Why not use polynomials to interpolate 5 or more points • Polynomials of higher degree have • extra wiggles • overshoots • non-locality: moving the point pn changes the curve even near p0; also when we add a point at the end

14. Why not use polynomials to interpolate 5 or more points • This gets worse with higher degree (more points)

15. Blending again • A piecewise-linear curve (polygonal line) can also be defined using blending functions 1 0 u 0.5 0.5  u  1 p1 p0 p2 0 1 u 0.5

16. Piecewise cubic polynomials • Allows position and derivative at each end • Allows C2 continuity • Are in a sense the most smooth curve: minimum curvature over its length (curvature: second derivative) curvature low curvature 0 curvature high

17. Cubic polynomials • f(u) = a0 + a1u + a2u2 + a3u3 in canonical form • Four control points (points or derivatives on curve) needed • For piecewise cubic curves: n pieces require 4n control points, but C0 continuity already fixes n – 1 of them

18. Cubic polynomials • The ideal properties: • Each piece is cubic • The curve interpolates the control points • The curve has local control • The curve has C2 continuity • We can have any three but not all four properties: • Natural cubics do not have local control • Cubic B-splines do not interpolate the control points • Cardinal splines are not C2

19. Natural cubics • The first piece specifies start and end positions, and the first and second derivative at the start • For each other piece, the position, first and second derivative match with the piece before it only the endpoint can be specified freely • A curve with n pieces has n+3 control “points” in total (n+1 points and 2 derivative specifications) p1 p2 p0 p3

20. Natural cubics • We get (with f(u) = a0 + u a1 + u2a2+ u3 a3):p0 = f (0) = a0 + 0 a1 + 02a2 + 03a3p1 = f’(0) = 1 a1 + 20 a2 + 302a3p2 = f’’(0) = 2 a2 + 60 a3p3 = f (1) = a0 + 1 a1 + 12a2 + 13a3 constraint matrix

21. Natural cubics • When you modify your natural cubic, for instance by changing the derivative at the start of the first piece, then the whole curve changes not local

22. Hermite cubics • Specifies positions of start and end, and the derivatives at these points • C1 continuous since the start position of a piece must be the same as the end position of the previous piece, and the same is true for the derivatives • A curve with n pieces has 2n+2 control “points” in total p1 p2 p3 p0

23. Hermite cubics • Local control: changing the position or derivative of any point influences only the piece before and the piece after that point • See slides 34-35 of Curves I lecture

24. Cardinal cubics • Specifies positions only; the derivatives at each point are determined by the points before and after it • C1 continuous • A curve with n pieces has n+3 control points in total p1 p2 p3 p0 p1p3 p0p2

25. Cardinal cubics • A tension parameter t [0,1) determines the “strength” of bending • Each derivative vector pi-1pi+1 is scaled by (1 – t)/2 p1 p1 p2 p2 p0 p0 p3 p3 t = 0.5 t = 0 (Catmull-Rom splines)

26. Cardinal cubics ??? scaling factor related,not the tension t

27. p1 p2 p0 p3 Cardinal cubics • We can set up the constraints:f(0) = p1f(1) = p2f’(0) = ½ (1 – t) (p2 – p0)f’(1) = ½ (1 – t) (p3 – p1) • Rewrite to get expressions for p0, p1, p2 and p3 • The first and last control points are not interpolated; no derivative would be specified • Therefore, n pieces require n+3 points

28. Summary • Piecewise cubic curves are popular for modeling • We want C2 continuity, locality, and interpolation; we can get only two of these with cubics • Natural cubics do not have locality • Hermitecubics are only C1 continuous • Cardinal cubics are only C1 continuous • The splines that follow next (Bezier and B-spline) will not interpolate but approximate the control points

29. Questions • Assume piecewise linear interpolation by blending functions, where the curve passes through points p0, p1, p2, p3,and p4 at u = 0, 0.25, 0.5, 0.75, and 1. Define the five blending functions bi(u) so that • Assume that someone defines quadratic blending functions as on slides 5-9, except that b2(u) = u2. Does the resulting curve pass through its defining points. Is the resulting curve translation-invariant?