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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). - PowerPoint PPT Presentation
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Chap. 14 CurvesMathematics for Computer Graphics Applications
Seminar for Beginner Summer 2002
Jang Su-Mi2002-08-07
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)]
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
Plane Curve(2)
3 Point are needed.
p0=[x0 y0 z0] ; u = 0p0.5 =[x0.5 y0.5 z0.5] ; u = 0.5p1=[x1 y1 z1] ; u = 1
Algebraic form 에 대입x0 = cx
x0.5 = 0.25ax + 0.5bx + cx
x1 = ax + bx + cx
y, z 에 대해서도 비슷한 결과
Plane Curve(3)
ax = 2x0 - 4x0.5 + 2x1
bx = -3x0 + x0.5 - x1
cx = x0 ax bx cx 에 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 x1 에 대하여 정리
p(u) = (2u2-3u+1)p0+(-4u2+4u)p0.5+(2u2–u)p1
Geometric form
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) = UAbx by bz
cx cy cz
Algebraic coefficients
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
Control Point matrix
Geometric Coefficients
Blending function matrix
Plane Curves(6)
FP = UA
F = [u2 u 1] 2 -4 2-3 4 -1 M 1 0 0
F = UMUMP = UAMP = AA = MPP = M-1A
Basis transformation matrix
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 curvep(u) = UA Algebraic formp(u) = GP Geometric formG = UN N : basis transformation m
atrixGP = UAUNP = UAA = NP
The Tangent Vector
• Use 2 end point, 2 tangents instead of 4 point. (p0 p1 pu
0 pu1 )
• Tangent vectorpu(u) = [ dx(u)i/du dy(u)j/du dz(u)k/du]pu = [xu yu zu]
x(u) = axu3 + bxu2 + cxu + dx
xu = 3axu2 + 2bxu + cx
The Tangent Vector
u=0, u=1 대입 x0 x1 xu0 xu
1 에 대하여 정리ax bx cx dx 에 대하여 정리 치환대입 정리
x(u) = (2x0-2x1+ xu0 + xu
1 )u3
+(-3x0 +3 x1-2xu0- xu
1 ) u2
+ xu0u
+ x0
x0 x1 xu0 xu
1 에 대하여 정리x(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
The Tangent Vector
p(u) = UAp(u) = FBF = UMUMB = UAA = MB
(magnitude of the tangent vector account into)pu
0 = m0t0
pu1 = m1t1
p(u) = (2u3-3u2+1)p0 +(-2u3 +3u2)p1
+(u3 -2u2 +u)m0t0
+(u3-u2)m1t1
Blending Function
• F blending Function • G blending Function
Reparameterization
• reverse direction
Continuity and Composit Curves
• Parametric Continuity : Cn
• Geometric Continuity : Gn
Approximating a Conic Curve
• Conic Curves– Hyperbola– Parabola– Ellipse