118 COMPUTER AIDED ENGINEERING DESIGN
converted to the cubic Bézier form and vice-versa. Given control points Pi,i = 0,... , 3, and realizing
that a Ferguson’s segment would pass through P 0 and P 3 , equating the two forms results in
r
P
P
T
T
( ) = [ 1]
2–21 1
–3 3 –2 –1
0010
1000
32
0
3
0
3
uuuu
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
= [ 1]
–1 3 –3 1
3–630
–3 3 0 0
1000
32
0
1
2
3
uuu
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
P
P
P
P
(4.57a)
or[ 1]
2 – 2 + +
–3 + 3 – 2 –
= [ 1]
- 3 – 3 +
3 – 6 + 3
–3 + 3
- 3 – 3 +
32
0303
03 03
0
3
32
0123
012
01
uuu uuu
PPTT
PPTT
T
T
PPPP
PPP
PP
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥ PP 0
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
Comparing the coefficients of u gives T 0 = 3(P 1 – P 0 ) while comparing those of u^2 results in
–3P 0 + 3P 3 – 2T 0 – T 3 = 3P 0 – 6P 1 + 3P 2
or T 3 = –6P 0 + 6P 1 – 3P 2 + 3P 3 – 2T 0
= 6(P 1 – P 0 ) + 3(P 3 – P 2 ) – 6(P 1 – P 0 ) = 3(P 3 – P 2 )
Equating the coefficients of u^3 thereafter becomes redundant. Thus, given control points Pi,
i = 0,... , 3, the geometric matrix for the Ferguson’s segment can be written as
G = [P 0 P 3 3(P 1 – P 0 ) 3(P 3 – P 2 ]T.
Likewise, for given two end points Pi and Pi+1, and end tangents, Ti and Ti+1 for Ferguson’s model,
the geometric matrix for the Bézier segment can be constructed as
GP
T
PP
T
= P
3
+ –
+1 3
+1
i +1
i
i i
i
i
T
⎛
⎝
⎞
⎠
⎛
⎝
⎞
⎠
⎡
⎣
⎢
⎤
⎦
⎥ (4.57c)
4.5 Composite Bézier Curves
In foregoing sections, Bézier segment of a generic degree was considered and its properties were
discussed in detail. Consider, in a composite curve, any two contiguous Bézier segments, r 1 (u 1 ) of
degreem with data points p 0 ,p 1 ,... , pm, and r 2 (u 2 ) of degree n with data points q 0 ,q 1 ,... , qn. For
position (C^0 ) continuity, since the segments pass through the end points, the last point in r 1 (u 1 ) should
coincide with the first point in r 2 (u 2 ), that is (Figure 4.20)
r 1 (u 1 = 1) = r 2 (u 2 = 0)
or pm = q 0 (4.58)