Computer Aided Engineering Design

DESIGN OF SURFACES 235

[r 00 r 01 r 02 r 03 ]II = [(0, 3, 0) (1, 3, 3) (2, 3, 3) (3, 3, 0)]

For slope continuity, Case I yields

rr 1 IIii = 0 II + (λ rr 3 Iii – 2 I), = 0, 1, 2, 3i

choosingλ = 2 results in

[r 10 r 11 r 12 r 13 ]II = [(0, 3, 0) (1, 3, 3) (2, 3, 3) (3, 3, 0)] + 2[(0, 1, 0) (0, 1, 1) (0, 1, 1) (0, 1, 0)]
= [(0, 5, 0) (1, 5, 5) (2, 5, 5) (3, 5, 0)]
The resultant composite surface is shown in Figure 7.26 with polyhedron I shown with thick linear
lines while polyhedron II is shown with thin lines.

4

2

0

z

8

6

4

2

(^00)
1
2
3
y x
Figure 7.26 A composite Bézier surface (Example 7.11) with gradient continuity (Case I)
With Case II, from Eq. (7.57), we have
[0 0 1 0] MGMVB BII BTT = [3 2 1 0] λ MGMVB IB TTB + ( + ) [1 1 1 1]
3
2
1
0
01 B B
I
B
T
2
μμv
v
v
MGM
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
Comparing the coefficients of V results in
[–3 3 0 0] = [3 2 1 0] + [1 1 1 1]
33 00
02 2 0
00
0000
B
II
B B
I
B B
I
B
T
10
10
10
B
GMGMGMλ M–T
μμ
μμ
μμ
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
Choosingλ = μ 0 =μ 1 = 1, which completely determines the right hand side, gives
3[(r 10 – r 00 ) (r 11 – r 01 ) (r 12 – r 02 ) (r 13 – r 03 )]II = [(3, 3, 9) (4, 3, 9) (5, 3, 0) (6, 3, –18)]