Computer Aided Engineering Design

232 COMPUTER AIDED ENGINEERING DESIGN

Further, for Sp remaining unaltered while Sq changed to Sq1 having different control points, say

qqq

qqq

qqq

00 01 02

10 11 12

20 21 22

= {0, 2, 2} = {0, 3, 2} = {0, 4, 1}

= {1, 3, 3} = {1, 4, 2} = {1, 5, 2}

= {2, 2, 2} = {2, 3, 2} = {2, 4, 1}

⎡

⎣

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

the surface Sq1 and the resulting composite surface is shown in Figure 7.22.

Figure 7.22 (a) Surface Sq1 and (b) the composite surface with patches Sp and Sq1

2.5

2

1.5

1

0

0.5

1

1.5

2

2

3

4

2

1

0

0

0.5

1

1.5

2

0

1

2

3

4

(a) Surface Sq1 (b) Composite surface

In Figure 7.23, for gradient continuity, the tangent plane of patch I at u = 1 must coincide with that
of patch II at u = 0 for v∈ [0, 1]. This implies that the direction of surface normal at the boundary
must be unique. Or,

∂

∂

× ∂

∂

∂

∂

× ∂

uu∂

(0, ) (0, ) = ( ) (1, ) (1, )rrII v II rrI I

v

vvv

v

λ v (7.53)

whereλ(v) is a scalar function that takes into account the discontinuities in magnitudes of the surface

normals. Since the positional continuity ensures that ∂
∂

∂

v ∂

v

v

(0, ) = (0, )rrII I v, a solution for Eq.

(7.53) is to have:

Case I

∂

∂

∂

uu∂

(0, ) = ( ) (1, )rrII vvvλ I (7.54)

⇒ [0 0 1 0] MBGBIIMBTVT = λ(v)[3 2 1 0] MBGBI MBTVT

Since the left hand side is cubic in v, the right hand side should be such that λ(v) = λ, a constant to
match the degree in v. Further, equating coefficients of V and post-multiplying with MB–T results in

rr 1 IIii – 0 II = (λ r r 3 Iii – 2 I), = 0, 1, 2, 3i (7.55)

which means that the four pairs of polyhedron edges meeting at the boundary must be collinear as
shown in Figure 7.23 (thick lines).