Computer Aided Engineering Design

DESIGN OF SURFACES 227

t

PP

ij i PP

ij ij

ij ij

= D

• | – |

+1 –1

+1 –1

, where Di = min (| Pij – Pij–1 |, | Pij+1 – Pij |) (7.46)

The twist vectors can be assumed to be zero. The geometric matrix for this Ferguson’s patch would
then be

G

PP t t

PP t t

ss

ss

=

|

|

––|––

|0 0

|0 0

+1 +1

+1 +1 +1 +1 +1 +1

+1

+1 +1 +1

ij ij ij ij

ij ij ij ij

ij ij

ij ij

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥

(7.47)

Note that for i = 0 or i = m,P– 1 j and Pm+1j, respectively, are not known and so the user will have to
specifys 0 jandsmj for all j = 0,... , n. Similarly, slopes ti 0 andtin for i = 0,... , mwill also need to
be specified. In other words, slopes along u and v are to be specified on the boundaries of the
composite surface. The so-called FMILL method to generate a composite surface using Ferguson’s
patches described above works well for evenly spaced data points. However, local flatness or bulging
for unevenly spaced data points producing unnatural surface normals is often seen. This is primarily
due to the assumption of zero twist vectors.

Example 7.9. For given control points

P 00 = {0, 0, 0}, P 10 = {1, 0, 0}, P 20 = {2, 0, 0}

P 01 = {0, 1, 0}, P 11 = {1, 1, 0}, P 21 = {2, 1, 0}

P 02 = {0, 1, 2}, P 12 = {1, 1, 2}, P 22 = {2, 1, 4}

determine Ferguson’s patches using the FMILL method to get the composite surface. Take the end
slopess 0 j = smj = ti 0 =tin = {0, 0, 0} for all i = 0,... , 2; j = 0,... , 2.
The intermediate slopes sijcan be computed using Eq. (7.46) as

s 10 = [min (| P 10 – P 00 |, | P 20 – P 10 |)]
PP
PP

20 00

20 00

• | – |

= [min (| (1, 0, 0) – (0, 0, 0) |, | (2, 0, 0) – (1, 0, 0)]

(2, 0, 0) – (0, 0, 0)

| (2, 0, 0) – (0, 0, 0) |

= (1, 0, 0)

s 11 = [min (| P 11 – P 01 |, | P 21 – P 11 |)]

PP

PP

21 01

21 01

• | – |

= [min (| (1, 1, 0) – (0, 1, 0) |, | (2, 1, 0) – (1, 1, 0)]

(2, 1, 0) – (0, 1, 0)

| (2, 1, 0) – (0, 1, 0) |

= (1, 0, 0)

s 12 = [min (| P 12 – P 02 |,| P 22 – P 12 |)]
PP
PP

22 02

22 02

• | – |

= [min (| (1, 1, 2) – (0, 1, 2) |, | (2, 1, 4) – (1, 1, 2)]

(2, 1, 4) – (0, 1, 2)

| (2, 1, 4) – (0, 1, 2) | =

1

2

, 0,^1

2

⎛

⎝

⎜

⎞

⎠

⎟