Computer Aided Engineering Design

DESIGN OF CURVES 113

DD D P P^2 i = i^11 +1 – i = ( ii i+ 2 – +1) – (P P P P P+1 – ) = i ii+ 2 – 2 +1 + i

DD D P^3 i = i^22 +1 – i = ( iii ii+3 – 2P P+2 + +1) – (P+2 – 2P P+1 + i)

=Pi+3 – 3Pi+2 + 3Pi+1 – Pi

using which the kth derivative of a Bézier segment can be written as

dkku du n n n n k B u

i

nk

i

nk

i

r( )/ = ( – 1) ( – 2)... ( – + 1) ( ) k

=0

• 
  

Σ – D (4.45)

4.4.2 Subdivision of a Bézier Segment

Subdivision may have many applications in curve design. We may desire to trim a curve at the
subdivision point retaining only a part, or, may subdivide a curve and design a segment separately
without changing the shape of the other segment thus gaining additional flexibility in design. Subdivision
may be performed as many times as desired. It involves partitioning a Bézier segment r(u) at some
pointu = c into two segments each of which by itself is a Bézier segment. The resulting segments
have their own control polylines and each are of the same degree as the parent curve. With n + 1
control points b 0 ,b 1 ,... bn, and a parameter value u=c, 0 < c < 1, two new sets of control points
p 0 ,p 1 ,... , pn and q 0 ,q 1 ,... , qn are required so that the two Bézier segments span the original
segment in the parameter range 0 ≤u≤c and c≤u≤ 1, respectively. To find the control points for
the first segment, the parent Bézier segment for 0 ≤u≤c may be re-parameterized with u′=u/c so
that when u = 0, u′ = 0 and when u = c,u′ = 1. The segment r 1 (u) for 0 ≤u≤c (0 ≤u′≤ 1) becomes

r 1 () = uBcu Bu′′ ′i=0 ( ) = =0 ()

n

i

n i

i

n

i

ΣΣbpn i (4.46)

Since the two curves are identical, so are their derivatives at u = 0.
Thus:

d

du

Bcu cn n k B cu

dcu

du

k

k i

n

i

n i

i

nk

i

nk

i

k

k

′

′

⎡

⎣

⎢

⎤

⎦

⎥ ... ′

⎧

⎨

⎩

⎫

⎬

⎭

′

′

⎧

⎨

⎩

⎫

⎬

⎭

( ) = ( )( – 1) ( – + 1) ( )

()

=0 =0

• –

ΣΣb D

= ( – 1) ( – + 1) ( )

=0

• cnnk n k B– cu
  i

nk

i

nk

i

... Σ ′Dk (4.47)

while

d

du

Bu nn n k B cu

k

k i

n

i

n i

i

nk

i

nk

i

k

′

′

⎡

⎣

⎢

⎤

⎦

(^) =0 ( ) ⎥ = ( – 1) ( – + 1) ... =0 ( ′)

• 
  

ΣΣp – P

(4.48)

wherePik are the differences in control points Pi related in a manner similar to Eq. (4.44d) as

PP Pij = ij+1–1 – ij–1, = 1,... , ; = 0,... , – , with jninjPi^0 = pi (4.49)

Note that Di^0 = bi using which Dik can be computed accordingly from Eq. (4.44d).

Comparing Eqs. (4.47a) and (4.47b) for u′ = 0 gives.

cknkkDP 00 = k, = 0,... , (4.50)