211
32
.
0
2
80
⋅
⋅
=
E
....)
7560
1320
42
3
32
.
0
(
)
32
.
0
(
)
32
.
0
(
)
32
.
0
(
7
5
3
−
+
−
)
0
0000025
.
0
00078019
.
0
10666667
.
0
(
64
−
+
−
=
=
6.
777
This example is symmetry spiral transition N1=N2
,
E1=E2
⑷
Calculation of shift value
R
∆
)
cos
1
(
τ
−
−
=
∆
R
E
R
18
cos
1
(
100
777
.
6
−
−
=
∆
R
°20
′
06
″
)
=
1.700
Symmetry spiral transition
R
R
2
1
∆
=
∆
⑸
Calculation of Spiral Transition coordinate
τ
sin
R
N
N
m
−
=
=63.348-100sin18°20
′
06
″
=31.891
Symmetry spiral transition
N
N
m
m
2
1
=
⑹
Calculation of Tangent Distance
N
R
R
D
m
LA
LA
ec
LA
R
1
1
2
1
)
cot(
)
(
cos
)
2
tan(
+
∆
−
∆
+
=
+
=
LA
111°55
′
47
″
,
sin
1
cos
=
ec
,
tan
1
cot
=
=
D
1
100 * tan(111°55
′
47
″
/ 2) +1.7(1 / sin111°55
′
47
″
)
–1.7(1 / tan 111°55
′
47
″
) +31.891
=148.06015 + 1.8326 + 0.6844 +31.891
=182.468
D
D
2
1
=
⑺
Calculation of the coordinate KA1
α
1
1
1
1
cos
⋅
−
=
D
N
N
IP
KA
α
1
1
1
1
sin
⋅
−
=
D
E
E
IP
KA
Bearing from BP to IP1
⇒
=
α
1
74°03
′
16.6
″
=
N
KA
1
1300 –182.468 * cos 74°03
′
16.6
″
=1249.872 m
=
E
KA
1
1750 –182.468 * sin 74°03
′
16.6
″
=1574.553 m
⑻
Calculation of Arc Length
)
(
2
1
τ
τ
+
−
=
LA
R
L
=
R
(111°55
′
47
″
-2 * 18°20
′
06
″
)
=100( 75°15
′
35
″
180
o
π
)
=131.353 m
Summary of Contents for STS-750
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