Computer Aided Engineering Design

DESIGN OF CURVES 115

k = 0: DQnn^00 = ⇒qn = bn

k = 1: (1 – c)DQnn^1 –1 =^1 –1 ⇒ (1 – c) (bn – bn–1) = (qn – qn–1)

⇒qn–1 = (1 – c)bn–1 + cbn

k = 2: (1 – )c^2 DQ^2 nn–2 =^2 –2⇒ (1 – c)^2 (bn – 2bn–1 + bn–2) = (qn – 2qn–1 + qn–2)

⇒qn–2 = (1 – c)^2 bn–2 + 2c (1 – c)bn–1 + c^2 bn

k = 3: (1 – )c^3 DQ^3 nn–3 =^3 –3⇒ (1 – c)^3 (bn – 3bn–1 + 3bn–2 – bn–3) = (qn – 3qn–1 + 3qn–2 – qn–3)

⇒qn–3 = (1 – c)^3 bn–3 + 3c(1 – c)^2 bn–2 + 3c^2 (1 – c)bn–1 + c^3 bn

(4.55)

Eqs. (4.55) and (4.32) show that qbq bq bnnn =^0 , –1 =^1 n–1, n–2 = n^2 –2 and qbn–3 = n^3 –3 for u=c. In

general,qbnk– = nkk– , = 0,... ,kn that is, the control points for the second Bézier segment are the

intermediate de Casteljau points, bbn^0 ,^1 n–1,... , bn 0 for u = c which correspond to the bottom edge in
the triangular scheme in Figure 4.13. Figure 4.18 depicts the control polylines for the two subdivided
cubic segments. Note that b 03 is a common end point for both segments, and bb 02 03 and bb 0312 are
tangents to the respective segments.

Figure 4.18 Control polylines for the two subdivided curves (thick and

thicker solid lines) and subdivided curves (dashed lines)

b 0

b 1

b 2

b 3

b 01

b (^02) b
0
3
b 11
b 12
b 21
(u = 0)
(u = c)
(u = 1)

A procedure reverse to subdivision may also be employed to extend a curve. For given c between
0 and 1 and for given control points p 0 ,p 1 ,... , pn, the control polyline b 0 ,b 1 ,... , bn for the extended
curve (for 0 ≤u≤ 1) can be computed using Eqs. (4.51) by a series of forward substitutions.

Example 4.7. The equation for a Bézier curve with the control points P 0 ,P 1 ,P 2 ,P 3 is given by

rP() = ()

=0

3

uBu^3

i i i

Σ

The curve is required to be subdivided at u = 1/2. Develop a formulation for subdivision into two
Bézier segments (a) in the interval u∈ [0, 1/2] and (b) u∈ [1/2, 1].
Let the two segmentsbe represented as: