Lecture 12 - Spline surfaces (slides)

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Unformatted text preview: 99 ·· ·· 11 10 ·· 11 10 14 10 ·· 14 10 11 ·· ·· 11 Corners Gradients in the x / μ direction " Gradients in the x / µ direction Example ! Example at µ, ⌫ = 0, 1 are defined directly in the question: The corners 01 1 P(0,@ P = (9, (10, 4, 13)T P8, , 0) T (10, 4, 13) 0) 4, 12) ( (14, 10) = @ A = P(0,@ µ = (9, 5, 11) 1) P(1, 1) = = , 5, 14) (10 0 2 (0,0) 1.5 7 8 x, µ 9 7 # 10 8 11 x, µ 9 # 10 Graphics Lecture 12: Slide 28! 11 2 · · 2 · · · · · · · · y, ⌫ ! 3 4 · · y, ⌫ ! · 10 3 4 14 12 · · 15 13 · 10 · 10 14 12 15 13 · 10 5 67 · ·· 9 ·· 5 67 11 10 · · ·· 14 10 · 9 ·· 11 ·· 11 10 · 14 10 · 11 ·· Gradients in the x / µ direction Gradients in the x / μ direction " @P @µ @P @µ @P @µ @P @µ = (10, 4, 13)T 2 (0,0) = (11, 4, 10)T (10, 5, 14)T (8, 5, 9)T 2 (0,1) = (9, 4, 12)T 2 (1,0) = (8, 4, 10)T (11, 5, 11)T (1,1) (9, 5, 11)T 2 = (1, 0, 1.5)T = (1, 0, 1)T = (1, 0, 2.5)T = (1, 0, 0)T Graphics Lecture 12: Slide 29! 29 / 35 Gradients in the y / ⌫ direction Gradients in the y / ν direction " @P @⌫ @P @⌫ @P @⌫ @P @⌫ = (9, 5, 11)T 2 (0,0) = (10, 5, 14)T (9, 6, 10)T (9, 4, 12)T 2 (0,1) = (10, 3, 15)T 2 (1,0) = (9, 3, 14)T (10, 6, 10)T (1,1) (10, 4, 13)T 2 = (0, 1, 1.5)T = (0, 1, 0.5)T = (0, 1, 1)T = (0, 1, 1.5)T Graphics Lecture 12: Slide 30! 30 / 35 Finding the boundary curves Finding the boundary curves " E.g. Finding curve P(µ, 0) P(µ, 0) = a3 µ +a2 µ +a1 µ+a0 3 2 01 0 a0 Ba1 C B B C=B @a2 A @ a3 1 0 3 2 0 1 2 1 0 0 3 2 10 1 0 P0 0 C BP0 C C B 0C 1 A @P1 A 1 P0 1 Cubic spline patch equation (Previous lecture) •  see cubic spline patch equation (previous lecture) ! Graphics Lecture 12: Slide 31! 31 / 35 Finding the boundary curves Finding the boundary curves" E.g. Finding curve P(µ, 0) P(µ, 0) = a3 µ3 +a2 µ2 +a1 µ+a0 01 0 a0 Ba1 C B B C=B @a2 A @ a3 1 0 3 2 After substituting in P(0, 0), 0 1 2 1 @P @µ 0 0 3 2 (0,0) 10 0 9 0 CB 1 CB 1 A @10 1 1 , P(1, 0), @P @µ 1 4 12 0 1.5C C 4 13 A 0 1 (1,0) Graphics Lecture 12: Slide 32! 32 / 35 Finding the boundary curvecurve 0) (µ, 0) Finding the boundary P(µ, P Finding the boundary curve P(µ, 0) Calculating the constant vectorsvectors, a 1, and a0 and a Calculating the constant a3 , a •  Calculating the constant vectors2 ! a3 a2 , a1 0 a0 = P0 = (9, 4, 12) a0 = P0 = (9, 4, 12) a1 = P0 = (10, 0, 1.5) 0 a1 = P0 = 0(1, 0, 1.5) 0 a2 = 3P0 2P0 3P1 P1 a2 = 3P0 2P0 3P1 P0 0 1 = 3 ⇥ (9, 4, 12) 2 ⇥ (1, 0, 1.5) + 3 ⇥ (10, 4, 13) (1, 0, 1) = 3 ⇥ (9, 4, 12) 2 ⇥ (1, 0, 1.5) + 3 ⇥ (10, 4, 13) (1, 0, 1) = (0, 0, 1) = (0,00, 1) a3 = 2P0 + P0 2P1 + P0 1 a3 = 2P0 + P0 2P1 + P0 1 = 2 ⇥ (9, 4, 12) +0(1, 0, 1.5) 2 ⇥ (10, 4, 13) + (1, 0, 1) = 2 ⇥ (9, 4, 12) + (1, 0, 1.5) 2 ⇥ (10, 4, 13) + (1, 0, 1) = (0, 0, 0.5) = (0, 0, 0.5) Graphics Lecture 12: Slide 33! 33 / 35 Finding the boundary curves P(µ, 1), P(0, ν), P(1, ν) •  These curves are found identically to P(µ, 0). ! •  We now have all the individual bits: ! P(µ, 0): a cubic polynomial in µ P(µ, 1): a cubic polynomial in µ P(0, ν): a cubic polynomial in ν P(1, ν): a cubic polynomial in ν
 P(0, 0), P(0, 1), P(1, 0) and P(1, 1): the corner points! •  Given values of µ and ν, we can calculate each of these eight points ! Graphics Lecture 12: Slide 34! So, for any given value for µ and ⌫ . . . So, for any given value for µ and ν …" . can evaluate the the coordinate Coons patch: . ... .wewe can evaluate coordinate on theon the Coons patch: ! P(µ, ⌫ ) = P(µ, 0)(1 ⌫ ) + P(µ, 1)⌫ + P(0, ⌫ )(1 P(0, 1)(1 µ)⌫ P(0, 0)(1 Graphics Lecture 12: Slide 35! µ) + P(1, ⌫ )µ µ)(1 P(1, 0)µ(1 ⌫) ⌫) P(1, 1)µ⌫ 35 / 35...
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This document was uploaded on 03/26/2014.

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