DESIGN OF SURFACES 237Considering the triple scalar product for the vectors on the left givespp p12 3 ( ) =113
103
0–10⋅× = 1(3) + 1(0) + 3 (–1) = 0while for those on the right gives
qq q12 3 ( ) =–1 0 3
0–10
21–6⋅× = – 1(6) + 0 (0) + 3 (2) = 0Example 7.12 (A Composite Surface with Coons Patches). Often, sharp corners in a machine
component is not desirable and need to be replaced by curved surfaces at the corners and edges. An
example of a composite surface is illustrated in Figure 7.28 for which the surface is divided into four
patches: A is the top flat patch, B and D are the adjacent ruled patches and C is the triangular Coon’s
patch.
(i) The top flat patch A is rectangular, parallel to horizontal (x-y) plane at a height ‘1’ along the z-
axes. The four corner points are (0, 0.5, 1), (0, 0.5, 1), (1, 1.5, 1) and (1, 1.5, 1). The equation
of the bilinear Coon’s patch is given by
rPP
A PP
( , ) = [1 – ] uuu
00 01 1 –
10 11vv
v⎡
⎣⎢⎤
⎦⎥⎡
⎣⎢⎤
⎦⎥+ [1 – ]1 –- [1 – ]
00 01 1 –
10 11
00 01
10 11vvv
vPP
PPPP
PP⎡
⎣
⎢⎤
⎦
⎥⎡
⎣
⎢⎤
⎦
⎥⎡
⎣
⎢⎤
⎦
⎥⎡
⎣
⎢⎤
⎦
⎥u
uuuor rA( , ) = [(1 – ) ]uuu
(0, 0.5, 1) (1, 0.5, 1)
(0, 1.5, 1) (1, 1.5, 1)(1 – )
v = { , (0.5 + ), 1}v
vvv⎡
⎣
⎢⎤
⎦
⎥⎡
⎣
⎢⎤
⎦
⎥(ii) Patch B has the top boundary a line rB(1,v), common with A, given by the end points
(0, 0.5, 1) and (1, 0.5, 1). The bottom boundary rB(0,v) is also a straight line with end points
(0, 0, 0) and (1, 0, 0). Let rB(u, 0) and rB(u, 1) represent the remaining two boundaries in terms
of Hermite curves. The end points of the left boundary curve rB(u, 0) are (0, 0, 0) and
(0, 0.5, 1), and the end tangents are unit vectors rBu(0, 0) = k= (0, 0, 1) and rBu(0, 1) = j=
(0, 1, 0). Similarly, the end points of the right boundary curve rB(u, 1) are (1, 0, 0), (1, 0.5, 1),
and the end tangents are unit vectors rBu(1, 0) = k= (0, 0, 1) and rBu(1, 1) =j=(0, 1, 0). Thus,
the boundary curves are given by
rB( , 0) = [uuuu 1]2–2 1 1
–3 3 –2 –1
00 10
10 00(0, 0, 0)
(0, 0.5, 1)
(0, 0, 1)
(0, 1, 0)32⎡⎣⎢
⎢
⎢
⎢⎤⎦⎥
⎥
⎥
⎥⎡⎣⎢
⎢
⎢
⎢⎤⎦⎥
⎥
⎥
⎥