DESIGN OF SURFACES 239Equation of the ruled surface B using two boundary curves is given by
rB(u,v) = (1 – v)rB(u, 0) + vrB(u, 1)
Since the boundary curves rB(0,v) and rB(1,v) are straight line edges of surface B, formulating
r 2 (u,v) and the correction Coon’s patch r 3 (u,v) as in Eq. (7.34) would not alter the above
equation since r 2 (u,v) and r 3 (u,v) would cancel each other due to rectilinear nature of
rB(0,v) and rB(1,v). Though rB(u,v) is computed as a ruled surface, it is actually a bilinear
Coon’s patch.
(iii) Patch D is similarly obtained. The corner points are (1.5, 0.5, 0), (1, 0.5, 1), (1.5, 1.5, 0),
(1, 1.5, 1) and the end tangents are {(0, 0, 1),(–1, 0, 0)}, {(0, 0, 1),( –1, 0, 0)}. The top boundary
of patch D coincides with patch A {(1, 0.5, 1), (1, 1.5, 1)} and the bottom boundary is a straight
line with end points {(1.5, 0.5, 0), (1.5, 1.5, 0)}. The boundary curves are
(iv) rD( , 0) = [uuuu 1]
2–2 1 1
–3 3 –2 –1
00 10
10 00(1.5, 0.5, 0)
(1, 0.5, 1)
(0, 0, 1)
(– 1, 0, 0)32⎡⎣⎢
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⎥rD( , 1) = [uuuu 1]2–2 1 1
–3 3 –2 –1
00 10
10 00(1.5, 1.5, 0)
(1, 1.5, 1)
(0, 0, 1)
(– 1, 0, 0)32⎡⎣⎢
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⎥Equation of the ruled surface D is given by
rD(u,v) = (1 – v)rD(u, 0) + vrD(u, 1)(v) The triangular Coon’s patch C has the corner points{(1, 0, 0), (1.5, 0.5, 0) and (1, 0.5, 1)}. Two
of its boundaries are coincident with right and left boundaries of patch B and D respectively. The
top boundary curve rC(u, 1) is a multiple point (1, 0.5, 1). Two side boundary curves are given by
rC(0, ) = [uuu 1]2–2 1 1
–3 3 –2 –1
00 10
10 00(1, 0, 0)
(1.5, 0.5, 1)
(0, 0, 1)
(0, 1, 0)v^32⎡⎣⎢
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⎥rC(1, ) = [uuu 1]2–2 1 1
–3 3 –2 –1
00 10
10 00(1.5, 0.5, 0)
(1, 0.5, 1)
(0, 0, 1)
(–1, 0, 0)v^32⎡⎣⎢
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⎥The bottom boundary is given byrC( , 0) = [uuuu 1]2–2 1 1
–3 3 –2 –1
00 10
10 00(1, 0, 0)
(1, 5, 0.5, 0)
(1, 0, 0)
(0, 1, 0)32⎡⎣⎢
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