Lecture 11 - Spline curves (slides)

# See 34 38 reversing the de casteljau algorithm

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Unformatted text preview: = P3 3P2 + 3P1 P0 So So we get the directionsat the endpoints by using P P1 and.!P2 . •  we get the directions at the endpoints by using 1 and P2 Notehave shown the 3 are the equation is the same as a cubic patch and We that Pshown P blending endpointsis the same as a cubic patch ! •  We have 0 the blending equation Recall the matrix form used for a cubic spline patch 0 10 10 1 0 0 0 10 P 0 a0 1 0 1 10 0 0 0 C B PP1 3P0 C B a0 C B 1 a1 C B 0 1 0 0 C B 30 B C B a1 C = B 0 1 0 0 C B P0 C 3 @ A@ A@ A 2 3 1 CB 0 C P B a2 C = B 3 @ a2 A @ 3 2 3 1 A @ PP3A 3P2 a3 2 1 2 1 33 a3 2 1 2 1 P0 3 a1 a2 a2 a = = = = 1 Graphics Lecture 11: Slide 34! see 34 / 38 Reversing the de Casteljau algorithm &quot; Reversing the de Casteljau algorithm We start from the point P3,0 and work in reverse to express it in terms of its construction line. P3,0 = (1 = (1 = (1 = (1 µ)P2,0 + µP2,1 µ) {(1 µ)P1,0 + µP1,1 } + µ {(1 µ)2 P1,0 + 2µ(1 µ)2 {(1 +2µ(1 +µ2 {(1 µ)P1,1 + µ2 P1,2 µ)P0,0 + µP0,1 } µ) {(1 µ)P0,1 + µP0,2 } µ)P0,2 + µP0,3 } µ)P1,1 + µP1,2 } Reversing the de Casteljau algorithm Reversing the de Casteljau algorithm &quot; . . . continuing the expansion, we can drop the ﬁrst subscript . . . continuing the expansion, we get: (which indicates the recursion level) tocan drop the ﬁrst subscript (which indicates the recursion level) to get: ! P(µ) = (1 µ)2 {(1 +2µ(1 +µ2 {(1 = (1 µ)P0 + µP1 } µ) {(1 µ)P1 + µP2 } µ)P2 + µP3 } µ)3 P0 + 3µ(1 µ)2 P1 + 3µ2 (1 µ)P2 + µ3 P3 This the same as as expanded Bernstein blending polynomial This is is the same the the expanded Bernstein blending polynomial already shown is equivalent to a is equivalent to which we havewhich we have already shown cubic spline patcha cubic spline patch ! !see Graphics Lecture 11: Slide 36! 36 / 38 Control Points &quot; Control Points • We can summarise the four four point Bezier Curve by that it has We can summarise the point Bezier Curve by saying saying thattwohas ! that are interpolated (P0 , P3 ) I it points –  two points that are interpolated (P0, P3) two control points (P1 , P2 ) –  two control points (P1, P2) I • The curve starts at at0P0 and endsP3 P3 and shape can be The curve starts P and ends at at and its its shape can determined by moving control controlP1 , P2 . P , P .! be determined by moving points points 1 2 This could be done interactively using a mouse. •  This could be done interactively using a mouse. ! Graphics Lecture 11: Slide 37! In summary . . . …&quot; In summary In summary . . . •  simplest and most ↵ective way to way to draw smooth The The simplest andemost effective draw a smoothacurve curveset of points is to throughsimplestof pointssetto usewaycubic patch.a cubic patch. ! a through a is↵ective a to drawuse The and most e a smooth curve through a set of points is to use a cubic patch. No interaction needed? No interaction needed?! I No interaction gradients by the setting the needed? ! I setting the gradientsby the setting gradients by ! central di↵erence difference ! central 1 the centralerence (Pi+1 di↵i 1 ) is e↵ective. P 2 !1 2 (Pi+1 ! Pi 1) is e↵ective. is effective. ! ! ! ! User wants interactive shape User interactive shape User wants wants interactive ! adjustment? shape adjustment? adjustment? ! I ! the four point Bezier I the four point Bezier The four point is ideal! formulation Bezier formulation is ideal formulation is ideal ! ! ! Graphics Lecture 11: Slide 38! 38 / 38...
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## This document was uploaded on 03/26/2014.

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