Computer Aided Engineering Design

DESIGN OF SURFACES 229

Solving yields, four relations which can be summarized to the following two

(A) sij + 4si+ 1 j + si+ 2 j= 3(Pi+ 2 j – Pij)

(B) χχχχχij + 4χχχχχi+ 1 j + χχχχχi+ 2 j = 3(ti+ 2 j–tij) i = 0,... , m– 2 for fixed j (7.49a)

Similarly, for the boundary between patches I and III

∂

∂

∂

∂

2

2

I^2

2

( , 1) = III( , 0)

vv

rruu

UMGIMT [6 2 0 0 ]T = UMGIIIMT [0 2 0 0]T

or UMGI [6 –6 2 4]T = UMGIII [– 6 6 – 4 –2]T

or GI [6 –6 2 4]T = GIII [– 6 6 – 4 – 2]T

which yields the two relations as

(C) tij+ 4tij+ 1 + tij+ 2 = 3(Pij+ 2 – Pij)

(D) χχχχχij+ 4χχχχχij+ 1 + χχχχχij+ 2 = 3(sij+ 2 – sij)j = 0,... , n– 2 for fixed i (7.49b)

Thus, for given information

Data points: Pij,i = 0,... , m and j = 0,... , n

Boundary slopes: s 0 j,smj for all j = 0,... , n

ti 0 ,tin for all i = 0,... , m

Twist vectors: χ χ χ χ χ 0 j,χχχχχmj for all j = 0,... , n

χχχχχi 0 ,χχχχχin for all i = 0,... , m

We need to solve for

sij with (A)

tijwith (C)

χχχχχij using (B) and (D) for i = 1,... , m – 1, j = 1,... , n – 1

Eqs. (7.49) are all tri-diagonal and can be solved efficiently with algorithms available to get the
Ferguson’s geometric matrix for each patch. We can realize that the higher order slopes and twist
vectors are still needed to be specified which is a drawback with Ferguson’s composite patches.

7.3.2 Composite Bézier Surface

Both bi-cubic Ferguson and Bézier patches being tensor products, their equivalence is stated by the
relation

r(u,v) = UMFGFMFTVT = UMBGBMBTVT

or MFGFMFT = MBGBMBT

or GF = (MF–1MB)GB(MF–1MB)T (7.50)

where the subscript F refers to the Ferguson’s patch and B relates to the Bézier’s patch. Using
Eq. (7.6) for MF and Eq. (7.28) for MB and GB, we realize that