Computer Aided Engineering Design

(backadmin) #1
DESIGN OF SURFACES 221

that the conditions in Eq. (7.37) are satisfied. In other words, using Eqs. (7.37) to determine the
correction surface would yield the same result as in Eq. (7.36). In matrix form, the linear Coon’s
patch can be expressed as


r

a
a

b
b

PP
PP

( , ) = [(1 – ) ]

()
()

+ [(1 – ) ]

()
()


  • [(1 – ) ]
    0 (1 – )
    1


0
1

00 01
10 11

uuu

u
u

v uu

v
v

vv

v
v
























⎥ (7.38a)

It is clear that [(1 –u)u] and [(1 –v)v] are the blending functions for the Coon’s Patch, and they
arebarycentric.
Any other set of functions φ 0 (u),φ 1 (u);ψ 0 (v),ψ 1 (v) may also qualify as a set of blending
functions so long as they satisfy the following properties:



  • Barycentric property: φ 0 (u) + φ 1 (u) = 1; ψ 0 (v) + ψ 1 (v) = 1

  • Corner conditions:


φ 0 (0) = 1; φ 1 (0) = 0; ψ 0 (0) = 1, ψ 1 (0) = 0
φ 0 (1) = 0; φ 1 (1) = 1; ψ 0 (1) = 0; ψ 1 (1) = 1
The Coon’s patch, in general, will be given by

r

a
a

b
b

pp
pp

( , ) = [ ( ) ( )]

()
()

+ [ ( ) ( )]

()
()


  • [ ( ) ( )]


()
()
01

0
1

01

0
1

01

00 01
10 11

0
1

uuu

u
u

v uu

v
v

vv

v
v

φφ ψψ φφ

ψ
ψ

























(7.38b)

If, in addition to the four boundary curves and corner points, the respective cross boundary tangents
s 0 (v),s 1 (v),t 0 (u) and t 1 (u) are also given as shown in Figure 7.15, a Hermite or bi-cubic Coon’s patch
can be created in a similar manner as discussed above.


s 0 (v) a^0 (v)

P 01
b 1 (u)

t 1 (u)

s 1 (v)
P 11

a 1 (v)

t 0 (u)

P (^00) b 0 (u)
u P
10
v
Figure 7.15 Schematic of a bi-cubic Coon’s patch
Blending boundary curves b 0 (u) and b 1 (u) using cross boundary tangents, t 0 (u) and t 1 (u) gives
r 1 (u,v) = φ 0 (v)b 0 (u) + φ 1 (v)b 1 (u) + φ 2 (v)t 0 (u) + φ 3 (v)t 1 (u) (7.39a)
Likewise, bi-cubic blending of a 0 (v) and a 1 (v) using cross boundary tangents, s 0 (v) and s 1 (v) gives
r 2 (u,v) = φ 0 (u)a 0 (v) + φ 1 (u)a 1 (v) + φ 2 (u)s 0 (v) + φ 3 (u)s 1 (v) (7.39b)
The bi- cubic Coon’s patch is expressed as in Eq. (7.34) with r 3 (u,v) as the correction surface so that
the boundary conditions are met. Now
r(u, 0) = φ 0 (0)b 0 (u) + φ 1 (0)b 1 (u) + φ 2 (0)t 0 (u) + φ 3 (0)t 1 (u)
+φ 0 (u)a 0 (0) + φ 1 (u)a 1 (0) + φ 2 (u)s 0 (0) + φ 3 (u)s 1 (0) –r 3 (u, 0) = b 0 (u)

Free download pdf