120 COMPUTER AIDED ENGINEERING DESIGN

Using position and slope continuity conditions in Eq. (4.60) yields

tr (1) = tr (0)

2

1

2 1

2

1

(^22)
2
× 2 2
⎛
⎝
⎞
⎠
d ×
du
d
du
α
α (4.61)
Equation that satisfies the condition above is
d
du
d
du
d
du
2
1
2 1
2
1
(^22)
2
(1) = rrr 2 2 (0) + 2 2 (0)
α
α
⎛ μ
⎝
⎞
⎠

()
(0) + (0)
2
2
2
2
2 2 2 2
m
n
d
du
d
λ du
rrμ (4.62)
for some scalar μ. Using Eq. (4.44b),
d
du
mm B mm
i
m
i
(^2) m iiimmm
1
(^21) =0
–2

(1) = ( – 1) r Σ –2(1)[ppp+2 – 2 +1 + ] = ( – 1)[ppp – 2 –1 + –2] (4.63a)

while d
du

nn B nn

i

n

i

(^2) n iii
2
(^21) =0
–2

(0) = ( – 1) r Σ –2(0)[qqq+2 – 2 +1 + ] = ( – 1)[qqq 2 – 2 + 1 0 ] (4.63b)

Substituting Eq. (4.63) into (4.62) and also using d

du

n

2

(0) = (r 210 qq – ) from Eq. (4.43), we get

mm m

n

( – 1)[ mm m – 2 + ] = nn n

()

–1 – 2 ( – 1) [ – 2 + ] + ( – )

2

pp p λ 2 qqq 210 μ qq 10

or mm mm

mn

n

( – 1)( mm – ) – ( – 1)( m m – ) + n

( – 1)

–1 –1 – 2 – ( – )

2

pp p p λ 2 μ qq 10

⎡

⎣

⎢

⎤

⎦

⎥

=

( – 1)

( – )

2

2 21

mn

nλ

⎡

⎣

⎢

⎤

⎦

⎥qq

UsingC^1 continuity (Eq. 4.59), we gets

mn

n

nmm mmmm m m

2

2 –1 –1 – 2

( – 1)

• 
  
  
  
  • ( – 1) ( – )– ( – 1)( – )
    λ
  
  
  
  

μλ

⎡

⎣

⎢

⎤

⎦

⎥ pp p p

=

( – 1)

( – )

2

2 21

mn

nλ

⎡

⎣

⎢

⎤

⎦

⎥qq (4.64)

This implies that (q 2 – q 1 ) expressed as a linear combination of vectors (pm – pm–1) and
(pm–1 – pm–2) lies in the plane containing the latter two. In other words, pm–2,pm–1,pm = q 0 ,q 1 and q 2
arecoplanar. Note that for a composite, C^1 continuous planar Bézier curve, this condition is inherently
satisfied. However, for a spatial, C^2 continuous composite curve, q 2 is constrained to lie in the same
plane as pm–2,pm–1,pm = q 0 and q 1. The foregoing generalized analysis was for two Bézier segments
of degrees m and n. To design a C^2 continuous composite Bézier curve with cubic segments, the first