Bi-3 C 2 Polar Subdivision

1. Bi-3 C2 Polar Subdivision Ashish MylesUniversity of FloridaNew York University Jörg PetersUniversity of Florida

2. Overview: Bicubic C2 polar subdivision • What is polar subdivision?(increased artistic freedom) • Why is curvature continuity difficult?(and what makes it feasible?)

3. Uniform Bicubic Splines • Quad-grid mesh input • Degree (3, 3) • Curvature continuous (C2) • Well understood,explicit formulas • Mesh refinement

4. Generalize mesh connectivity valence ≠ 4 Catmull-Clark (C1, unboundedcurvature) polar Bicubic PolarSubdivision –Karciauskas, Peters '08 (C1, bounded curvature)

5. Why not Catmull-Clark on Polar? Catmull-Clark Bicubic PolarSubdivision

6. Subdivision on Polar Connectivity Catmull-Clark C1unbounded curvature Bicubic polarsubdivision C1boundedcurvature C2PS continuouscurvature Bicubic + C2 = Surprise!

7. Why a surprise? • Subdivision theory ([Peters, Reif '08] Subdivision book) • Our contribution: k = 2, q = 3: bicubic, simple masks, C2 at the poles Degree q = 6 needed for C2 at extraordinary points

8. Subdivision rules • Small masks that depend solely on the connectivity 1-link 2-link

9. Curvature comparison C2 Jet subdivision (degree (6,5)) C1 bicubic polar subdivision C2PS (degree (3,3))

10. Curvature comparison – finger C1 bicubic polar subdivision C2 Jet (6,5) subdivision C2PS (3,3)

11. Shape – C2PS

12. Part II • What is polar subdivision?(increased artistic freedom) • Why is curvature continuity difficult?(and what makes it feasible?)

13. z-coordinate quadratic in x and y coordinates need deg 6! need deg 2 deg 3 deg 1 Why deg 6 for C2 at the pole? • C2 & flexible ⇒ can reconstruct a paraboloid f(x, y) = (x, y, x2 + y2) E.g. polar valence = 6

14. curve C2PS need deg 2 curve deg 1 Infinite valence • Valence → ∞ ⇒ control points → curves⇒ polar coordinates: f(x, y) = (x, y, x2 + y2)f(r cos(t), r sin(t)) = (r cos(t), r sin(t), r2) valence =∞

15. Infinite valence ⇒ C2 f(x, y) = (x, y, x2 + y2) = (r cos(t), r sin(t), r2) f(x, y) = (x, y, x2 – y2) = (r cos(t), r sin(t), r cos(2t)) f(x, y) = (x, y, 2 x y) = (r cos(t), r sin(t), r sin(2t)) span{x2 + y2, x2 – y2, 2 x y} = span{x2, y2, x y} = span{x,y} × span{x,y}

16. ? ? ? ? ? ? ? ? ? Extraordinary neighborhood ... ...

17. A (5n+1) × (5n+1) matrix q0 q1 5n+1 vertices 5n+1 vertices Recall: Stationary subdivision analysis • q1 = A q0 • Radial-only subdivision ⇒ A is square • q∞ = A∞q0 → eigenanalysis of A

18. (5(2n)+1)×(5n+1) matrix A Standard theory not applicable to C2PS • q1 = A q0 • A is not square • q∞ = (...AAAA) q0 → Cannot use eigenanalysis • But: valence → ∞ q0 q1 5n+1 vertices 5(2n)+1 vertices

19. (5(2n)+1)×(5n+1) matrix A Trick – reformulate C2PS • Reformulate C2PS refinement in terms of 6 variables f(x, y) = p0 + p1x + p2y + p3 (x2 + y2) + p4 (x2 – y2) + p5 (2 x y) + o(x2 + y2) • C2PS = Diminishing perturbation of subdivision on infinite valence ⇒C2 q0 q1 5n+1 vertices 5(2n)+1 vertices

20. Future direction • C2 for polar and regular – how about extraordinary points? • Approx. double the extraordinary facet neighborhood • Can we make the subdivision masks easy too? Catmull-Clark non-stationary refinement

21. Summary • Bicubic C2 Polar subdivision • simple, curvature continuous • Analysis technique • compare with stationary algorithm(on curves) • Non-stationary valence/connectivity • allows for low degree + high smoothness

22. Acknowledgments • SIGGRAPHcommittee andreviewers • NationalScienceFoundation(0728797)

23. Thank you

24. Thank you