230 COMPUTER AIDED ENGINEERING DESIGN
Grr rr rr
rr rr rr
rr rr rrrr rrrr
rF =3( – ) 3( – )
3( – ) 3( – )
3( – ) 3( – ) 9( – – + ) 9( – – + )
3(00 03 01 00 03 02
30 33 31 30 33 32
10 00 13 03 00 10 01 11 02 12 03 13
3030 – r rr rrrr rrrr 20 ) 3( 33 – 23 ) 9( 20 – 30 – 21 + 31 ) 9( 22 – 32 – 23 + 33 )⎡⎣⎢
⎢
⎢
⎢⎤⎦⎥
⎥
⎥
⎥(7.51)which implies that the gradients and twist vectors at patch corners can be expressed in terms of the
characteristic Bézier polyhedron and thus the user may not need to specify higher order information.
Instead, we would be more comfortable maneuvering data points to implicitly control the slope and
mixed derivatives as opposed to specifying them. Figure 7.19 shows two adjacent bi-cubic Bézier
patches with corner points of their control polyhedra that lie on the respective patches. From
Eq. (7.28), the patches can be formulated as
rI(u,v)= UMBGBI MBTVTrII(u,v)=UMBGBIIMBTVT (7.52)For positional continuity across the common boundary, it is required that rI(1,v) = rII(0,v) for all
values of v, that is [1 1 1 1] MBGBI = [0 0 0 1]MBGBII or
[0001] = = [1000]00 01 02 03
10 11 12 13
20 21 22 23
30 31 32 33I
30
31
32
33I
00 01 02 03
10 11 12 13
20 21 22 23
30 31 32 33rrrr
rrrr
rrrr
rrrrr
r
r
rrrrr
rrrr
rrrr
rrrr⎡⎣⎢
⎢
⎢⎤⎦⎥
⎥
⎥⎡⎣⎢
⎢
⎢⎤⎦⎥
⎥
⎥⎡⎣⎢
⎢
⎢⎤⎦⎦⎥
⎥
⎥⎡⎣⎢
⎢
⎢⎤⎦⎥
⎥
⎥II
00
01
02
03II=r
r
r
rimplying that rI 3 j = rII 0 j,j = 0,... , 3, or in other words, the boundary polygon must be common
between the two patches.
Example 7.10.For blending two quadratic Bézier surfaces, a quadratic Bézier surface Sp has the
following control points:
ppp
ppp
ppp00 01 02
10 11 12
20 21 22= {0, 0, 0} = {0, 1, 2} = {0, 2, 2}
= {1, 0, 1} = {1, 1, 3} = {1, 3, 3}
= {2, 0, 0} = {2, 1, 2} = {2, 2, 2}⎡⎣⎢
⎢
⎢⎤⎦⎥
⎥
⎥Another quadratic Bézier surface Sq has the control points
vr 33 II
I IIr 00 I
uFigure 7.19 Adjacent Bézier patch boundariesr 03 I rr^33I = 03 IIr 30 II
rr 30 I = 00 II