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In-Depth Information
2
4
3
5
cos h ðx 1
a
y 1 ¼ c þ a
2
4
3
5
cos h ðx 2
a
y 2 ¼ c þ a
2
3
5 a
2
3
ðx 2
a
ðx 1
a
L ¼ a
4
4
5
sin h
sin h
2
3
q
L
sin h ðx 2 x 1 Þ
2
a
2
4
5 ða can be solved for numerically at this point
2
ðy 2 y 1 Þ
¼
2
Þ
a
0
@
1
A
0
@
1
A
sin h x 2
a
x 1
a
M ¼
sin h
0
@
1
A
0
@
1
A
cos h x 2
a
x 1
a
N ¼
cos h
(7.87)
0
1
M
N
@
A
tan h 1
if N > M m ¼
M
sin h
N
cos h
Q ¼
ðmÞ ¼
ðmÞ
2
4
0
@
1
A
3
5
L
Q a
b ¼ a
sin h 1
m
0
@
1
A
N
M
tan h 1
i f M > N
m ¼
N
sin h
M
cos h
Q ¼
ðmÞ ¼
ðmÞ
2
0
1
3
L
Q a
b ¼ a
4
@
A
5
cos h 1
m
A relaxation process is used as the second and final step in positioning the vertices. The effect of
gravity is ignored; it has been used implicitly in the formation of the catenary curves using the interior
vertices. The exterior vertices are initially positioned to affect a downward pull. The relaxation pro-
cedure repositions each vertex to satisfy unit distance from each of its neighbors. For a given vertex,
displacement vectors are formed to each of its neighbors. The vectors are added to determine the direc-
tion of the repositioning. The magnitude of the repositioning is determined by taking the square root of
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