86 COMPUTER AIDED ENGINEERING DESIGN
u
ddduj j2
12=1= 3 3
+
= 3.61 + 4.47
10.91
= 0.74, = 1Σ
Eq. (4.2) for the four data points becomes
1
1
1
1=03
02
0113
12
1123
22
2133
32
3133
22
11
0000
11
22
33uuu
uuu
uuu
uuuaa
aa
aa
aaxy
xy
xy
xyxy
xy
xy
xy⎡⎣⎢
⎢
⎢
⎢
⎢⎤⎦⎥
⎥
⎥
⎥
⎥⎡⎣⎢
⎢
⎢
⎢
⎢⎤⎦⎥
⎥
⎥
⎥
⎥⎡⎣⎣⎢
⎢
⎢
⎢
⎢⎤⎦⎥ ⎥ ⎥ ⎥ ⎥ ⇒⎡⎣⎢
⎢
⎢
⎢
⎢⎤⎦⎥
⎥
⎥
⎥
⎥⎡⎣⎢
⎢
⎢
⎢
⎢⎤⎦⎥
⎥
⎥
⎥
⎥⎡⎣⎢
⎢
⎢
⎢
⎢⎤⎦⎥
⎥
⎥
⎥
⎥0001
0.33 0.33 0.33 1
0.74 0.74 0.74 1
1111=–1 2
24
66
8432
3233
22
11
00aa
aa
aa
aaxy
xy
xy
xywhich gives
=–3.9 –17.15
5.15 16.74
7.83 2.4
–1 233
22
11
00aa
aa
aa
aaxy
xy
xy
xy⎡⎣⎢
⎢
⎢
⎢⎤⎦⎥
⎥
⎥
⎥⎡⎣⎢
⎢
⎢
⎢⎤⎦⎥
⎥
⎥
⎥The equation of the required curve is
r(u) = x(u)i + y(u)j = [–3.9u^3 + 5.15u^2 + 7.83u – 1]i+ [–17.15u^3 + 16.74u^2 + 2.4u + 2]jthe plots of which ae shown in Figure 4.1.
Figure 4.1 Parametric and Cartesian plots for Example 4.1•
•
•
•
7 6 5 4 3 2y(u)–20246810
x(u)0 0.2 0.4 0.6 0.8 1
uy(u)x(u)1086420–2P 1 (u 1 = 0.33)P 2 (u 2 = 0.74)P 1 (u 4 = 1)P 0 (u 0 = 0)