Bezier Curves and Surfaces (Redbook ch12)

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Interpolating Cubic Polynomials Simplest case, but rarely used in practice Four control points p 0, ..., p 3 Fit a cubic polynomial through them● Space u evenly: p 0= p(0), p 1=[= p(1/3), ...1p 0c 2 2 p 1]3c3 3 p 21 2 pp 3= p1=c 0c 1c 2c 33 2 23 3p 0 = p0=c 0p 1= p 1 3 =c 0 1 3 c 1 1 3 2c 2 1 3 3c 3p 2 = p 2 3 =c 0 2 3 c 1 2 3 2[0 0 01 1 3 1 3 2 1 3 3 2 3 31 1 1 1][c0c 1]c 2c 3CMPT370: <strong>Bezier</strong> curves 27 Mar 20076
Geometry matrix Invert this matrix to get the geometry matrix● Multiply the geometry matrix by the fourcontrol points to get the coefficients[c 0c 1] =[1 0 0 0] [p 0−5.5 9 −4.5 1 p 1]c 29 −22.5 18 −4.5 p 2c 3−4.5 13.5 −13.5 4.5 p 3 The coefficients define the cubic polynomial thatinterpolates these control points● Can render, e.g., by using many small linesegments (GL_LINE_STRIP)CMPT370: <strong>Bezier</strong> curves 27 Mar 20077
Page 1 and 2: Bezier Curves and Surfaces(Redbook
Page 3 and 4: What's on for today Polynomial curv
Page 5: Polynomial curves Restrict the func
Page 9 and 10: Hermite polynomial curves Another w
Page 11 and 12: Bezier curves Widely used, provided
Page 13 and 14: PR EFACIO.cuentran continuamente en

Bezier Curves and Surfaces (Redbook ch12)

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