108 COMPUTER AIDED ENGINEERING DESIGN

For a composite curve, individual segments need to be of lower order, preferably cubic. Thus, a cubic
Bézier segment in algebraic and matrix forms for data points P 0 ,P 1 ,P 2 and P 3 is given by

r(u) = (1 – u)^3 P 0 + 3u(1 – u)^2 P 1 + 3u^2 (1 – u)P 2 + u^3 P 3

= (1 – 3u + 3u^2 – u^3 )P 0 + (3u – 6u^2 + 3u^3 )P 1 + (3u^2 – 3u^3 )P 2 + u^3 P 3

=u^3 (–P 0 + 3P 1 – 3P 2 + P 3 ) + u^2 (3P 0 – 6P 1 + 3P 2 ) + u(–3P 0 + 3P 1 ) + P 0

= [ 1]

–1 3 –3 1

3–630

–3 3 0 0

1000

(^32) =
0
1
2
3
uuu B
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
P
P
P
P
UM G (4.40b)
Eq. (4.40b) is similar to the Hermite cubic segment (Eq. (4.7)) with the parametric vector U, the
4 × 4 Bézier matrix MB and the geometric matrix G of size 4 × 3which is an array of data points.
The geometric matrix G is to be defined by the user (U and Mg remaining the same for all cubic
Bézier curves). Note that the curve does not pass through the points P 1 and P 2. To change the curve’s
shape, the user may relocate any of the control point P 0 ,P 1 ,P 2 or P 3. Recall that for Fergulon’s segments,
the user had to specify the end slopes for a particular shape which is difficult to speculate a peiori.
A Bézier curve more or lees mimics the shape of the control polyline, which is easier to specify.
Example 4.6. A set of control points is given by P 0 = (4, 4), P 1 = (6, 8), P 2 = (8, 9) and P 3 = (10, 3).
Compute the Bézier curve. Let the coordinate axes be moved to (2, 2) and then rotated by 30°
counter-clockwise. What is the effect on the shape of the curve? Observe the shape change when:
(a)P 2 is moved to (12, 12) and (b) when P 1 is located at (3, 10).
From Eq. (4.40b),
[ ( ) ( )] = [ 1]
–1 3 –3 1
3–630
–3 3 0 0
1000
44
68
89
10 3
xu yu u u u^32
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
= [6u + 4, – 4u^3 – 9u^2 + 12u + 4]
plot of which is given in Figure 4.16(a). The transformation given is equivalent to moving the curve
towards the origin by (–2, –2) and then rotating the curve by –30° clockwise. The net transformation
is (see Chapter 3).
T =
cos (–30 ) – sin (– 30 ) 0
sin (–30 ) cos (–30 ) 0
001
10–2
01–2
00 1

0.866 0.500 – 2.732

• 0.500 0.866 – 0.732
  001

°°

°°

⎡

⎣

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

⎡

⎣

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

⎡

⎣

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

which when applied to the original segment expressed in homogenous coordinates gives

()

()

1

=

0.866 0.500 –2.732

• 0.500 0.866 – 0.732
  001

6 + 4

–4 – 9 + 12 + 4

1

32

xu

yu

u

uu u

′

′

⎡

⎣

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

⎡

⎣

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

⎡

⎣

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥