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−A 1 sin B 0
x 31 =
E 2 sin 2 B 0 ) 3 / 2 ×
6(1
E 2 )(1
E 2
6sin 2 B 0 (1 + E 2 )+3 E 2 sin 4 B 0 (3 + E 2 )
4 E 4 sin 6 B 0 ] ,
×
[5
E 2 )sin B 0
x 13 = A 1 (1
E 2 B 0 ) 7 / 2 ×
6(1
9 E 2 +2 E 2 sin 2 B 0 (5
3 E 2 )+4 E 4 sin 4 B 0 ] ,
×
[1
(15.89)
A 1 cos B 0
x 30 =
E 2 sin 2 B 0 ) 1 / 2 ×
120(1
E 2 ) 3 (1
4sin 2 B 0 (7 + 4 E 2 )+2sin 4 B 0 (12 + 43 E 2 +13 E 4 )
4 E 2 sin 6 B 0 (18 + 25 E 2 +3 E 4 )+ E 4 sin(77 + 39 E 2 ) 28 E 6 sin 10 B 0 ] ,
× [5 − E 2
A 1 cos B 0
120(1 − E 2 )(1 − E 2 sin 2 B 0 ) 5 / 2 ×
x 32 =
× [5 − E 2
2sin 2 B 0 (9 + 4 E 2 + E 4 )+15 E 2 sin 4 B 0 (3 + E 2 )
2 E 4 sin 6 B 0 (23 + 3 E 2 )+16 E 6 sin 8 B 0 ] ,
A 1 (1 − E 2 )cos B 0
24(1
x 14 =
E 2 sin 2 B 0 ) 9 / 2 ×
2 E 2 )+
+12 E 4 sin 4 B 0 (5 2 E 2 )+8 E 6 sin 6 B 0 ] .
9 E 2 +36 E 2 sin 2 B 0 (1
×
[1
Box 15.5 (A representation of the non-vanishing coecients in a polynomial setup of a
conformal mapping of type Gauss-Krueger or UTM).
y ( l,b )=
= y 01 b + y 20 l 2 + y 02 b 2 + y 21 l 2 b + y 03 b 3 + y 40 l 4 + y 22 l 2 b 2 + y 04 b 4
+ y 41 l 4 b + y 23 l 2 b 3 +
(15.90)
+ y 05 b 5 +O(6) ,
E 2 )
A 1 (1
E 2 sin 2 B 0 ) 3 / 2 ,
y 20 = A 1 cos B 0 sin B 0
y 01 =
(1
2(1 − E 2 sin 2 B 0 ) 1 / 2 ,
y 02 = 3 A 1 E 2 (1
E 2 )cos B 0 sin B 0
E 2 sin 2 B 0 ) 5 / 2
2(1
 
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