By Les Piegl, Wayne Tiller (auth.)

The booklet covers all features of non-uniform rational B-splines essential to layout geometry in a computer-aided surroundings. uncomplicated B-spline beneficial properties, curve and floor algorithms, and cutting-edge geometry instruments are all mentioned. distinct code for layout algorithms and computational methods are lined too, in a lucid, easy-to-understand sort, with no less than arithmetic and utilizing various labored examples. The e-book might be a needs to for college kids, researchers, and implementors whose paintings contains using splines.

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22 Curve and Surface Basics ALGORITHM A1. 4 PointOnBezierCurve(P,n,u,C) { 1* Compute point on Bezier curve. 0; for (k=O; k<=n; k++) C = C + B[k]*P[k]; } Using property Pl. 7, it is easy to derive the general expression for the derivative of a Bezier curve d (tBi,n(U)Pi) C'(u) = i=O n = ~ B~ (u) p. 9) = n LBi,n-I(U)(Pi+1 - Pi) i=O From Eq. g. 10) Notice from Eqs. 10) that • the derivative of an nth-degree Bezier curve is an (n - 1)th-degree Bezier curve; • the expressions for the end derivatives at U = 0 and U = 1 are symmetric (due, of course, to the symmetry of the basis functions); • the kth derivative at an endpoint depends (in a geometrically very intuitive manner) solely on the k + 1 control points at that end.

1 shown with the polynomial segments represented in Bezier form. such curves. First, redundant data must be stored: twelve coefficients, where only eight are required for C l continuous cubic curves, and only six for C 2 continuous cubic curves. Second, for the Bezier form the continuity of C(u) depends on the positions of the control points, hence there is little flexibility in positioning control points while maintaining continuity. If a designer wants C l continuity and is satisfied with the segments Cl(u) and C 3 (u), but wants to modify the shape of C 2 (u), he is out of luck: none of C 2 (u)'s control points can be modified.

22. 15) compute the control points Qj (uo) for the isocurve C;:'O=1/3{V). 23. Let n = 3 and m = 2. Consider the nonrational Bezier surface defined by the control net {Pi,O} = {(O, 0, 0), (3, 0, 3), (6, 0, 3), (9, 0, O)} {Pi,I} = {(O, 2, 2), (3, 2, 5), (6, 2, 5), (9, 2, 2)} {Pi,2} = {(O, 4, 0), (3, 4, 3), (6,4,3), (9, 4, O)} a. sketch this surface; b. use the deCasteljau algorithm to compute the surface point S(1/a, 1M; c. 24. Let =L n S(u, v) L Bi,n (u)Bj,m (v) P;,j m ;=0 j=O and assume that Po,o = Pl,O = ...