1 / 16

Chap. 14 Curves Mathematics for Computer Graphics Applications

Chap. 14 Curves Mathematics for Computer Graphics Applications. Seminar for Beginner Summer 2002 Jang Su-Mi 2002-08-07. Parametric Equations of Curve. x = x(u) y = y(u) z = z(u) x(u) = au 2 + bu + c p = p (u) p (u) = [x(u) y(u) z(u)]. Plane Curves(1).

Télécharger la présentation

Chap. 14 Curves Mathematics for Computer Graphics Applications

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Chap. 14 CurvesMathematics for Computer Graphics Applications Seminar for Beginner Summer 2002 Jang Su-Mi 2002-08-07

  2. Parametric Equations of Curve x = x(u) y = y(u) z = z(u) x(u) = au2 + bu + c p = p(u) p(u) = [x(u) y(u) z(u)]

  3. Plane Curves(1) x(u) = axu2 + bxu + cx y(u) = ayu2 + byu + cy z(u) = azu2 + bzu + cz p(u) = au2 + bu + c  Algebraic form

  4. Plane Curve(2) 3 Pointare needed. p0=[x0 y0 z0] ; u = 0 p0.5=[x0.5 y0.5 z0.5] ; u = 0.5 p1=[x1 y1 z1] ; u = 1  Algebraic form에 대입 x0 = cx x0.5 = 0.25ax + 0.5bx + cx x1 = ax + bx + cx  y, z에 대해서도 비슷한 결과

  5. Plane Curve(3) ax = 2x0 - 4x0.5 + 2x1 bx = -3x0 +x0.5 - x1 cx = x0 ax bx cx에 대하여 푼 것 x(u) = (2x0 - 4x0.5 + 2x1)u2 +(-3x0 +x0.5 - x1)u + x0 y(u), z(u)도 비슷한 결과 x(u) = (2u2 – 3u +1)x0 + (-4u2 + 4u) x0.5 +(2u2 – u) x1 • x0 x0.5 x1에 대하여 정리 p(u) = (2u2-3u+1)p0+(-4u2+4u)p0.5+(2u2–u)p1  Geometric form

  6. Plane Curves(4) • Matrix Algebra (Algebraic form) p(u) = au2 + bu + c  a [u2 u 1] b = au2 + bu + c c U = [u2 u 1] A = [a b c]T = ax ay az  p(u) = UA bx by bz cx cy cz Algebraic coefficients

  7. Space Curves(5) • Matrix Algebra (Geometric form) p(u) = (2u2-3u+1)p0+(-4u2+4u)p0.5+(2u2–u)p1  p(u) =[(2u2-3u+1) (-4u2+4u) (2u2–u)] [p0 p0.5 p1]T F = [(2u2-3u+1) (-4u2+4u) (2u2–u)] P = [p0 p0.5 p1]T = x0 y0 z0 x0.5 y0.5 z0.5 x1 y1 z1 p(u)=FP Blending function matrix Control Point matrix Geometric Coefficients

  8. Plane Curves(6) FP = UA F = [u2 u 1] 2 -4 2 -3 4 -1 M 1 0 0 F = UM UMP = UA MP = A A = MP P = M-1A Basis transformation matrix

  9. Space Curve • Cubic Polynomials : x(u) y(u) z(u), p(u) • 4 Points are needed : p0 p1/3 p2/3 p1 • Same process with the Plane curve p(u) = UA Algebraic form p(u) = GP Geometric form G = UN N : basis transformation matrix GP = UA UNP = UA A = NP

  10. The Tangent Vector • Use 2 end point, 2 tangents instead of 4 point. (p0 p1 pu0 pu1 ) • Tangent vector pu(u) = [ dx(u)i/dudy(u)j/du dz(u)k/du] pu = [xu yu zu] x(u) = axu3 + bxu2 + cxu + dx xu = 3axu2 + 2bxu + cx

  11. The Tangent Vector u=0, u=1대입  x0 x1 xu0 xu1에 대하여 정리 • ax bx cx dx 에 대하여 정리  치환대입 정리 x(u) = (2x0-2x1+xu0 +xu1 )u3+(-3x0 +3x1-2xu0-xu1 ) u2+xu0u +x0 • x0 x1 xu0 xu1에 대하여 정리 x(u) = (2u3-3u2+1)x0 +(-2u3 +3u2) x1 +(u3 -2u2 +u)xu0 +(u3-u2)xu1 p(u) = (2u3-3u2+1)p0 +(-2u3 +3u2)p1 +(u3 -2u2 +u)pu0 +(u3-u2)pu1 F B

  12. The Tangent Vector p(u) = UA p(u) = FB F = UM UMB = UA A = MB (magnitude of the tangent vector account into) pu0 = m0t0 pu1 = m1t1 p(u) = (2u3-3u2+1)p0 +(-2u3 +3u2)p1 +(u3 -2u2 +u)m0t0 +(u3-u2)m1t1

  13. F blending Function G blending Function Blending Function

  14. Reparameterization • reverse direction

  15. Continuity and Composit Curves • Parametric Continuity : Cn • Geometric Continuity : Gn

  16. Approximating a Conic Curve • Conic Curves • Hyperbola • Parabola • Ellipse

More Related