Computer Aided Engineering Design

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208 COMPUTER AIDED ENGINEERING DESIGN


Suudud
u

= |u( , ) ( , ) |
,v
∫∫ rrvvv× v (7.19)

For Ferguson’s bi-cubic patch, from Eq. (7.12)


r UM GM V r UMG M V
r UM M V r UM G M V
r UMG M V

u
uTT TT

uu
uu T T
u
uTT
TT

uu
uu
u

(, ) = (, ) = ( )
(, ) = (, ) = ( )
(, ) = ( )

vv
vv
v

v
v

v
v

vv
vv

(7.20)

whereMu and Muu are M 1 and M 2 , respectively in Eq. (4.9). Muu and Mvv are identical with the
difference that they are used with their respective parameter matrices U and V.


MMuuu = = M M

0000
6–6 3 3
–6 6 –4 –2
00 10

, = =

0000
0000
12 – 12 6 6
–6 6 –4 –2

vvv

























(7.21)

Example 7.2. A Ferguson surface patch has the following geometric coefficients:


G =

(6, 0, 0) (6, 0, 6) (0, 0, 6) (0, 0, 6)
(0, 6, 0) (0, 6, 6) (0, 0, 6) (0, 0, 6)
(0, 5, 0) (0, 5, 0) 0 0
(– 5, 0, 0) (–5, 0, 0) 0 0















Determine the tangents, normal, Gaussian curvature, mean curvature, principal curvatures, and equation
of the tangent plane at r(u = 0.5, v = 0.5). Also determine whether the surface is developable and find
the surface area of the patch.
The patch is given by r(u,v) = UMGMTVT = [xyz] = [(7u^3 – 13u^2 + 6) (–7u^3 + 8u^2 + 5u) (6v)]
whose plot is shown in Figure 7.3. At (u = 0.5, v = 0.5), the co-ordinates are (3.625, 3.625, 3). Using
Eq. (7.20), the slopes at any point on the surface are given by


ru(u,v) = [(21u^2 – 26u) (–21u^2 + 16u + 5) 0]
rv(u,v) = [0 0 6]

In particular, at (u = 0.5, v = 0.5),


ru(0.5, 0.5) = [–7.75 7.75 0]

rv(0.5, 0.5) = [0 0 6]

The unit normal can be determined using


ijk


  • 7.75 7.75 0 ij
    006


= 46.5 + 46.5
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