Computer Aided Engineering Design

DESIGN OF CURVES 117

for Bézier curve of degree n defined by control points b 0 ,b 1 ,... , bn, to raise its degree by one
requires finding a new set of n+ 2 control points q 0 ,q 1 ,... , qn+1. Since the two segments are identical

ΣΣ Σ

i

n

n

i

nii

i i

n

n

i

ni i

i i

n

n

i

ni i

=0 Cuu Cuu uuCuui

+1

+1 +1–

=0

• =0
  (1 – ) qb = (1 – ) = (1 – + ) (1 – )– b

= (1 – ) + (1 – )

=0

–+1

=0

ΣΣ–+1

i

n

n

i

ni i

i i

n

n

i

ni i

CuubbCuui

Comparing the coefficients of (1 – u)n+1–iui yields

n+1C

iqi =

nC

ibi +

nC

i–1bi–1

or qbbiii

i

n

i

n

= 1 – in

+ 1

+

+ 1

⎛ –1, = 0,... , + 1

⎝

⎞

⎠

⎛

⎝

⎞

⎠ (4.56)

Note that for i = 0, q 0 = b 0 and when i = n + 1, qn+1 = bn. Even though expressions involving b–1 and
bn+1 may appear, they may not be required as the respective coefficients are 0 at i = 0 and i = n + 1.
Eq. (4.56) suggests that qi is the weighted linear combination of bi–1 and bi with non-negative weights
that add to 1. Thus, qi lies in the convex hull of bi–1 and bi. More precisely, the new control polyline
lies within the convex hull of the old polyline and the Bézier segment lies within the convex hulls of
both polylines. The process of degree elevation may be repeated as many times as desired. Each time
the degree elevation is performed, the resultant control polyline moves closer to the Bézier segment,
adding one control point at a time. When the number of times the degree elevation is performed
approaches infinity, the control polyline approaches the Bézier segment.

Example 4.8 Given data points A(0, 0), B(1, 2), C(3, 2) and D(6, –1), elevate the degree of this cubic
Bézier segment to four and five and show the new control polylines.
Using Eq. (4.56), data points for the degree 4 segment can be computed as

Polyline,n = 4

Polyline,n = 5

Polyline,n = 3

Bézier curve of degree, = 3, 4, 5

012345 6

x

2.5

2

1.5

1

0.5

0

–0.5

–1

–1.5

y

Figure 4.19 Degree elevation of a Bézier segment

q 0 = b 0 = (0, 0)

qbb 110 = 1 –

1

4

+

1

4

⎛ = (0.75, 1.50)

⎝

⎞

⎠

⎛

⎝

⎞

⎠

qbb 221 = 1 –

2

4

+

2

4

⎛ = (2.00, 2.00)

⎝

⎞

⎠

⎛

⎝

⎞

⎠

qbb 332 = 1 –^3

4

+^3

4

⎛ = (3.75, 1.25)

⎝

⎞

⎠

⎛

⎝

⎞

⎠

q 4 = b 3 = (6, –1)

Similarly, using qi,i = 0,... , 4 above, data points

can be computed for degree 5 segment. Figure

4.19 shows the control polylines for degree 3, 4

and 5 Bézier segments.

4.4.4 Relationship between Bézier and Ferguson Segments

That Bézier and Ferguson cubic segments have similar matrix forms (Eqs. (4.7) and (4.40)), we may
realize that the two geometric matrices may be related. In other words, a Ferguson’s segment may be