Q(s, t) = S M = S M [ G 1 (t) G 2 (t) G 3 1(t) G 4 (t) ] T

Transcription

1 Curves an Surfaces: Parametric Bicubic Surfaces - Intro Surfaces are generalizations of curves Use s in place of t in parametric equation: Q(s) = S M G where S equivalent to T in Q(t) = T M G If G is parameterize in terms of t, have surface representation: Q(s, t) = S M G(t) = S M [ G 1 (t) G 2 (t) G 3 1(t) G 4 (t) ] T For fie t, Q(s, t) is a curve because G(t) is constant Family of curves - a surface - efine by 0 t 1 If G i (t) are cubic, resulting surface is parametric bicubic surface where G i = [ g i1 g i2 g i3 g i4 ] T G i (t) = T M G i G i (t) = G T i M T T T = [ g i1 g i2 g i3 g i4 ] M T T T So Q(s, t) = S M g 11 g 12 g 13 g 14 g 21 g 22 g 23 g 24 g 31 g 32 g 33 g 34 g 41 g 42 g 43 g 44 M T T T for 0 s, t 1 = S M G M T T T Above can be written separately for, y, an z 1

2 Curves an Surfaces: Hermite Surfaces Formulation of G H parallels formulation of general case alreay iscusse (s) = S M H G H (s, t) = S M H G H (t) = S M H [ P 1 (t) P 4 (t) R 1 (t) R 4 (t) ] T For every value of t, have 2 en points an 2 tangent vectors P 1 (t) = T M H [ g 11 g 12 g 13 g 14 ] T P 4 (t) = T M H [ g 21 g 22 g 23 g 24 ] T R 1 (t) = T M H [ g 31 g 32 g 33 g 34 ] T R 4 (t) = T M H [ g 41 g 42 g 43 g 44 ] T [ P1 (t) P 4 (t) R 1 (t) R 4 (t) ] = T M H G T H where Therefore, G H = [ P1 (t) P 4 (t) R 1 (t) R 4 (t) ] T g 11 g 12 g 13 g 14 g 21 g 22 g 23 g 24 g 31 g 32 g 33 g 34 g 41 g 42 g 43 g 44 = g 11 g 12 g 13 g 14 g 21 g 22 g 23 g 24 g 31 g 32 g 33 g 34 g 41 g 42 g 43 g 44 = G H M T H T T M T H T T So, (s, t) = S M H G H M T H T T 2

3 Curves an Surfaces: Hermite Surfaces (2) In G H g 11 represents (0, 0), the starting point for P 1 (t), which starts (s, 0) g 12 represents (0, 1), the ening point for P 1 (t), which starts (s, 1) g 13 represents t (0, 0), the starting tangent vector at P 1(t), which starts (s, 0) g 14 represents 2 t 2 (0, 0), the starting tangent vector of R 1 (t) G H = (0, 0) (0, 1) t(0, 0) (1, 0) (1, 1) t(1, 0) s(0, 0) s(1, 0) t(0, 1) t(1, 1) s (0, 1) 2 st (0, 0) 2 st(0, 1) s (1, 1) 2 st (1, 0) 2 st(1, 1) Upper-left 2 2 quarant represents 4 corners of patch Upper-right an lower-left 2 2 quarants represent coorinates of tangent vectors along each parametric irection Lower-right 2 2 quarant represents partial erivatives wrt s an t This represents amount of twist at corners 3

4 Curves an Surfaces: Hermite Surfaces (3) Hermite bicubic patch allows C 1 an G 1 continuity across patches To have C 0 continuity across ege, 2 ege curves must be the same I.e., have same CPs To have C 1 continuity across ege, twist an tangent vectors must be the same To have G 1 continuity across ege, tangent vectors must have the same irection 4

5 Curves an Surfaces: Bézier Surfaces Formulation same as for Hermite (s, t) = S M B G B MB T T T Has 16 CPs Continuity conitions: To have C 0 an G 0 continuity across ege, 4 common CPs must be the same To have G 1 continuity across ege, 4 CPs on either sie of the ege must be colinear with the ege CPs 5

6 Curves an Surfaces: B-spline Surfaces Formulation: (s, t) = S M BS G BS M T BS T T Continuity conitions: C 2 across ege automatic 6

7 Curves an Surfaces: Surface Normals Surface efine by Q(s, t) = S M G M T T T, 0 s, t 1 s tangent vector efine by t tangent vector efine by s Q(s, t) = s (S M G M T T T ) = s (S) M G M T T T = [ 3s 2 2s 1 0 ] M G M T T T t Q(s, t) = t (S M G M T T T ) = S M G M T t (T T ) = S M G M T [ 3t 2 2t 1 0 ] T Normal is the Cartesian prouct of the above 2 tangent vectors Normal is biquinitic: A fifth egree polynomial with 2 variables 7

8 Curves an Surfaces: Drawing Surfaces Approaches use to rawing curves apply to surfaces as well 1. Iterative evaluation Draw curves of fie s along t Draw curves of fie t along s (a) Brute force O(n 2 ), where n is number of increments in s, t (b) Forwar ifferences Nee to apply forwar ifference to generate curve i + 1 from curve i From curve case, have D = E(δ) C DD = E(δ s ) A E(δ t ) T where DD is a 4 4 matri, an A is a 4 4 matri of coefficients for (s, t) First row of DD consists of (0, 0), t (0, 0), 2 t (0, 0), an 3 t (0, 0) where t is forwar ifference base on t This can be use to compute (0, t) Nee to compute (δ s, t) in terms of forwar ifferences Row 2 of DD consists of first forwar ifference on s of R 1 Row 3 of DD consists of secon forwar ifference on s of R 1 Row 4 of DD consists of thir forwar ifference on s of R 1 Using Row 1 = row 1 + row 2 Row 2 = row 2 + row 3 Row 3 = row 3 + row 4 generates (δs, 0), t (δs, 0), 2 t (δs, 0), an 3 t (δs, 0) in first row These are use to compute (δs, t) using forwar ifferences This is iteratively applie to generate (2δs, t), (3δs, t),... These are curves of constant s To generate curves of constant t, iterate on columns: Column 1 = column 1 + column 2 Column 2 = column 2 + column 3 Column 3 = column 3 + column 4 which results in (s, δt), s (s, δt), 2 s(s, δt), an 3 s(s, δt) in first column 8

9 Curves an Surfaces: Drawing Surfaces (3) 2. Recursive subivision (using Bézier) Principle: (a) Split surface along one parameter (b) Then split along the other Surface ivie into 4 patches: Base case etermine by either (a) Flatness Compute plane through 3 of the 4 corner vertices Distance of other 13 must be less than some ɛ (b) Depth of recursion Problems: (a) Gaps may appear between patches Result of ifferent levels of subivision in ajacent patches Solution: Recurse to fie epth or use small ɛ 9

10 Quarics are surfaces efine by Curves an Surfaces: Quarics f(, y, z) = a 2 + by 2 + cz 2 + 2y + 2eyz + 2fz + 2g + 2hy + 2iz + j = 0 Can also be represente as P T Q P = 0 where Q = a f g b e h f e c i g h i j an P = [ y z 1 ] T Normal to surface at (, y, z) given by [ f/ f/y f/z ] T Given a transform M, where M is translation or scale Q = (M 1 ) T Q M 1 Avantages of quarics: 1. Ease of surface normal computation 2. Ease of testing whether point lies on surface 3. z computable from known an y 4. Ease of computation of surface intersection 10