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End of Corollary.
In practice, points on the biaxial ellipsoid are given in the chart of surface normal coordinates
of type { longitude L , latitude B} , but not in the chart of isometric coordinates {p, q} .Thus,
similarly to the second example “second choice”, we take advantage of the Taylor series expansion
of Q ( B ) around Q 0 ( B 0 ) as given by ( D.79 ) according to standard series inversion . Additionally,
implementing the remove step b
q ( b ), we finally arrive at Corollary D.6 .
c r b r
d r q r .
q =
b =
(D.79)
r =1
r =1
a,b , universal transverse Mercator projection modulo unknown dilatation param-
eter, c:c:cha-cha-cha).
Corollary D.6 (
E
The solution of the boundary value problem, where the L 0 meta-equator is modulo a dilatation
parameter equidistantly mapped, is given by ( D.80 ), ( D.81 ), and the a r of Tables D.2 and D.3 .
x ( l,b )= a 0 + a 10 b + a 20 b 2 + a 02 l 2 + a 30 b 3 + a 12 bl 2 + a 40 b 4 +
+ a 22 b 2 l 2 + a 04 l 4 + a 50 b 5 + a 32 b 3 l 2 + a 14 bl 4 + o x ( b 6 ,l 6 ) ,
(D.80)
y ( l,b )= a 01 l + a 11 bl + a 21 b 2 l + a 03 l 3 + a 31 b 3 l + a 13 bl 3 +
+ a 41 b 4 l + a 23 b 2 l 3 + a 05 l 5 + o y ( b 6 ,l 6 ) .
(D.81)
End of Corollary.
Tab l e D. 1 E a,b , the transverse Mercator projection, the series expansion q ( b )= r =1
c r b r for B = B 0 ,the
coecients c r are given up to N =5,upto N = 10 available from the author
− e 2
1
c 1 = Q ( B )=
e 2 sin 2 B ) ,
cos B (1
3sin 2 B )+ e 4 (
2+3sin 2 B )]
2! Q ( B )= sin B [1 + e 2 (1
c 2 = 1
,
e 2 sin 2 B ) 2
2cos 3 B (1
c 3 = 1
1
[1+sin 2 B
1+5sin 2 B +2sin 4 B )
3! Q ( B )=
e 2 (
e 2 sin 2 B ) 3
6cos 3 B (1
10 sin 2 B +11sin 4 B
9sin 6 B )
e 6 sin 2 B (6
13 sin 2 B +9sin 4 B )] ,
e 4 (2
c 4 = 1
sin B
24 cos 4 B (1
4! Q IV ( B )=
− e 2 sin 2 B ) 4 ×
[5 + sin 2 B + e 2 (
18 sin 2 B
5sin 4 B )+ e 4 (20
63 sin 2 B +96sin 4 B
17 sin 6 B )+
×
1
24 + 104 sin 2 B
159 sin 4 B +82sin 6 B
27 sin 8 B )+
+ e 6 (
+ e 8 sin 2 B (
24 + 68 sin 2 B
65 sin 4 B +27sin 6 B )] ,
c 5 = 1
1
5! Q V ( B )=
− e 2 sin 2 B ) 5 ×
120 cos 5 B (1
[5 + 18 sin 2 B +sin 4 B
e 2 (1 + 22 sin 2 B +93sin 4 B +4sin 6 B )
×
e 4 (20 + 136 sin 2 B
338 sin 4 B +68sin 6 B
86 sin 8 B )
e 6 (24
380 sin 2 B + 1226 sin 4 B
1588 sin 6 B + 1178 sin 8 B
220 sin 10 B )
−e 8 sin 2 B (240
1100 sin 2 B + 1936 sin 4 B −
1633 sin 6 B + 518 sin 8 B −
81 sin 10 B )
e 10 sin 4 B (120
420 sin 2 B + 541 sin 4 B
298 sin 6 B +81sin 8 B )] .
 
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