210 COMPUTER AIDED ENGINEERING DESIGN
Figure 7.5 Bi-cubic surface patch in 16 point formr(0, 1)r(1/3, 1) r(2/3, 1)
r(1, 1)r(0, 2/3)r(0, 1/3)r(0, 0) r(1, 0)r(1/3, 0) r(2/3, 0)r(1, 1/3)r(1, 2/3)r(1/3, 2/3) r(2/3, 2/3)r(1/3, 1/3) r(2/3, 1/3)( – 3.625) – 7.75 0( – 3.625) 7.75 0( – 3) 0 6= 0xyz46.5 (x– 3.625) + 46.5(y– 3.625) = 0, that is x + y = 7.25The surface area of the patch can be obtained by Eq. (7.19) as
ru(u,v)×rv(u,v) = 6 (–21u^2 + 16u + 5)i– 6(21u^2 – 26u)j|ru(u,v)×rv(u,v) | = 6[(–21u^2 + 16u + 5)^2 + (21uu^22 – 26 ) ]1(^2) = f (u)
Sfudud
u
= ( ) = 54.64
=0
1
=0
1
∫∫v
v
7.1.3 Sixteen Point Form Surface Patch
For 16 uniformly spaced points on a surface patch, to fit a bi-cubic tensor product surface of the form
given in Eq. (7.5) with m = n = 3,
rDDDDDDDDDDDDDDDDD( , ) = = [ 1]1=03
=03
3233 32 31 3023 22 21 2013 12 11 1003 02 01 0032
uuuuu
ji ij
vvijvvvΣΣ
⎡⎣⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎤⎦⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎡⎣⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎤⎦⎥ ⎥ ⎥ ⎥ ⎥ ⎥(7.22)has 16 unknowns Dij to be determined. Let u∈ [0, 1], v∈ [0, 1] and each interval be subdivided as
[0, 1/3, 2/3, 1]. The given sixteen points on the surface are (each rij is a triplet) such that