By David F. Rogers

For one week after receiving this e-book I agreed with an previous very severe evaluation. I replaced my brain. the topic isn't really effortless yet written through a person who understands his company. Having obtained used to his notation i locate this publication progressively more necessary and refer again to it every time an issue arises and typically find the solution or a few pointer to the reply.

**Read Online or Download An Introduction to NURBS: With Historical Perspective (The Morgan Kaufmann Series in Computer Graphics) PDF**

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**Additional info for An Introduction to NURBS: With Historical Perspective (The Morgan Kaufmann Series in Computer Graphics)**

**Sample text**

A) P ( t ) ; (b) P ' ( t ) ; (c) P"(t). Hence, the first derivative is P'(t) = BoJ~,o(t) + B1J~,l(t) + B2J~,2(t) + BaJJ,3(t) Recalling Ex. 1 and differentiating the basis functions directly yields J3,o(t) = (1 - t) 3 => J~,o(t) = - 3 ( 1 - t) 2 Ja, l(t) = 3t(1 - t ) 2 =~ J3,1(t) = 3(1-t) 2 -6t(1-t) J3,2(t) = 3t~(1 - t) J3,3(t) = t 3 =~ J~,2(t) = 6t(1 - t) - 3t 2 =~ J~,3(t) = 3t 2 E v a l u a t i n g these results at t = 0 gives J~,0(t)(0) = - 3 J~,l (t)(0) = 3 J~,2(t)(0) --- 0 J3,3(t)(0) ~---0 S u b s t i t u t i n g these values into the e q u a t i o n for the first derivative yields P ' ( 0 ) = - 3 B 0 + 3B1 = 3(B1 - B0) Thus, t h e direction of the t a n g e n t vector at the beginning of the curve is the same as t h a t of the first polygon span (see Fig.

11) yields t h e first derivative at any p o i n t on t h e curve. For e x a m p l e , at t = 1/2 P ' ( 1 / 2 ) -- - 3 ( 1 - 1/2)2Bo -[- 3(1 - 3/2)(1 - 1/2)B1 q- (3/2)(2 - 3/2)B2 -[- 3/4B3 -- -3/4 B0 - 3/4 81 + 3/4 B2 + 3/4 B3 = - 3 / 4 (B0 + B1 - B2 - B3) C o m p l e t e results for B0, B1, B2, B3 given in Ex. 1 are s h o w n in Fig. 6. Similarly, t h e second derivatives are J~:0(t) = { ( - 3 t ) ~ - 3t~} (1 - t)~ = 6(1 - t) t 2 ( 1 - t) 2 J~',l(t) = {(1 - 3t) 2 - 3t 2 - (1 - 2t)} (3t)(1 - t) 2 = - 6 ( 2 - 3t) t2(1 - t)2 J~',2(t) = {(2 j, {(a a,a(t) = - - 3t) 2 3t 2 2(1 t2(1 - t ) 2 - - at) ~ 3t ~ a(1 t2(1 _ t ) 2 - - - - 2t)} (3t2)( 1 2 t ) } t ~ = 6t _ t) = 6(1 - 3t) 31 Using Eq.

Partly through that summer, Elaine Cohen and I visited Pierre B~zier at Renault. His intuitive way of controlling the shape of a curve was quite the topic of discussion in CAD research communities, because the curve, defined by a user-specified "control polygon," gracefully mimicked its shape. This was a completely novel idea that had set the CAD world abuzz trying to understand what was fundamentally going on. In this first of many meetings and communications, B~zier was gracious and kind and demonstrated his new system on the graphics output of the time, a large flatbed plotter.