134 COMPUTER AIDED ENGINEERING DESIGN

Φi′′

i i i

i

i i i

i

ii

i

i i

i

ii

i

ii

i

i ii

i

i i

tt

ttt

h

ttt

h

tt

h

tt

h

tt

h

tt

h

ttt

h

tt

( ) = [0 0 2 6 ]

(– 3) (3 – ) – –

6–6(2+ )(+ 2)

+1 +1

2

3

+1

2

3

+1

2

2

(^2) +1
2
+1
3
+1
3
+1 +1
2
+1tt
h
tt
h
tt
h
tt
h
tt
h
hhhh
y
y
s
s
i
i
i i
i
i i
i
i i
i
i i
i
iiii
i
i
i
i
2
+1
3
+1
3
+1
2
+1
2
3322
+1
+1
–3(+ ) 3(+ ) – (+2 ) –(2+ )
2 –2 11
⎡
⎣
⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥
⎡
⎣
⎢
⎢
⎢
⎢⎢
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ =
6(2 – – +1) – 6(2 – – ) 2(3 – – 2 ) 2(3 – 2 – )
3
+1
3
+1
2
+1
2
+1
+1
ttt
h
ttt
h
tt t
h
ttt
h
y
y
s
s
i i
i
i i
i
i i
i
i i
i
i
i
i
i
⎡
⎣
⎢
⎤
⎦
⎥
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
(5.8)
Likewise
′′
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
Φi ii
i
ii
i
ii
i
ii
i
i
i
i
i
t
tt t
h
tt t
h
tt t
h
tt t
h
y
y
s
s
–1
–1
–1
3
–1
–1
3
–1
–1
2
–1
–1
2
–1
–1
() =
6(2 – – ) – 6(2 – – ) 2(3 – – 2 ) 2(3 – 2 – )
⎦⎦
⎥
⎥
⎥
⎥
⎥
(5.9)
NowΦΦi′′–1() = ttiii′′() results in
6 –6 24

–6 6 –4 –2
–1
2
–1
(^2) –1 –1
–1
–1
22
+1
+1
hhhh
y
y
s
s
hhhh
y
y
s
s
iiii
i
i
i
i
iiii
i
i
i
i
⎡
⎣
⎢
⎤
⎦
⎥
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
⎡
⎣
⎢
⎤
⎦
⎥
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
or
s
h
s
hh
s
h
y
h
y
hh
y
h
i in
i
i
i i
i
i
i
i
i
i i
i
i
–1
–1 –1
+1 +1
2
–1
22
–1
–1

• 2 2
  ⎛ (^1) + (^1) + =^3 + 3 (^1) – (^1) –^3 , = 1, , – 1
  ⎝
  ⎜
  ⎞
  ⎠
  ⎟
  ⎛
  ⎝
  ⎜
  ⎞
  ⎠
  ⎟ ... (5.10)
  Note that Eqs. (5.5)-(5.10) describe a generalized Ferguson cubic composite curve for t 0 ≤t≤tn.
  Chapter 4 discusses a particular case wherein each knot span is normalized. That is, h 0 = h 1 = ... =
  hn–1 = 1 for which case Eq. (5.10) is identical to Eq. (4.17). Nevertheless, the exercise above suggests
  that a C^2 continuous Ferguson curve is a spline. Eqs. (5.10) are linear in n +1 unknowns,s 0 ,... , sn
  while the number of equations are only (n– 1). Thus, two additional conditions are needed to
  determine the unknown second derivatives siat each knot ti. These can be specified using one of the
  three possibilities:
  (i) Free end: Where there is no curvature at the end knots, that is, s 0 orsn= 0 at ti or tn, respectively.
  This gives a natural spline.