Computer Aided Engineering Design

112 COMPUTER AIDED ENGINEERING DESIGN

Thus, the derivative of a Bézier segment is a degree n–1 Bézier segment with control points
n(P 1 – P 0 ),n(P 2 – P 1 ), ..., n(Pn – Pn–1). Alternatively, it is the difference of two Bézier segments of
degreen–1, times n. This derivative is usually referred to as the hodographof the original Bézier
segment. Note that n(Pi – Pi–1) are no longer position vectors but are free vectors instead. It is when
the tails of the vectors are made to coincide with the origin that they may be termed as the control
points for the hodograph. If a is any vector along which the original control points, P 0 ,P 1 ,... , Pn
are displaced, the original Bézier segment gets displaced by a (Eq. 4.42). However, its hodograph
remains unchanged.

(j ) Higher Order Derivatives
From Eq. (4.43)

dudu n dB udu

i

n

i

n

i i

22

=0

–1

–1

rPP()/ = Σ (^) []()/ (+1 – )
= ( – 1)[ ( ) – ( )] ( – )
=0
–1
–1
–2 –2
nn BuBui +1
n
i
n
i
n

Σ PPi i

= ( – 1) ( )( – ) – ( – 1) ( )( – )

=0

–1

–1

–2

+1 =0

–1

–2

nn i Bu nn Bu+1

n

i

n

i i i

n

i

n

ΣΣPP PPi i

= ( – 1) ( )( – ) – ( – 1) ( )( – )

=0

–2

–2 +2 +1

=0

–2

nn Bu nn Bu–2 +1

i

n

i

n ii

i

n

i

ΣΣPP n PPi i

(BuBu–1n–2( ) = nn–1–2( ) = 0; index i shifted accordingly)

= ( – 1) ( ) [( – ) – ( – )]

=0

–2 –2

nn i B u + 2 +1 +1

n

i

n

Σ PP PPii ii (4.44a)

= ( – 1) ( ) [ – 2 + ]

=0

–2 –2

nn i B u +2 +1

n

i

n

Σ PPPiii (4.44b)

We may find by inspection, from Eqs. (4.44a and b) that

dudu nn n i B u P

n

i

n

iii iii

33

=0

–3 –3

r( )/ = ( – 1)( – 2) Σ ( )[(PPP+3 – 2 +2 + +1) – ( +2 – 2PP+1 + )}

= ( – 1)( – 2) =0 ( )[ – 3 + 3 – ]

–3

–3

nn n i B u +3 +2 +1

n

i

n

Σ PP PPii ii (4.44c)

To express higher order derivatives more concisely, a finite difference scheme may be employed.

DD Dij = ij+1–1 – ij–1, = 1,... , ; = 0,... , – jninj

DPi^0 = i (4.44d)

Thus

DD D P P^1 i = i^00 +1 – i = i+1 – i