Geography Reference
In-Depth Information
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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