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E
2
sin
2
B
0
)
7
/
2
b
23
=
−
(1
−
E
2
)
4
cos
4
B
0
×
12
A
1
(1
−
15
E
2
+4sin
2
B
0
+13
E
2
sin
2
B
0
+ 210
E
4
sin
2
B
0
−
32
E
2
sin
4
B
0
×
(2
−
315
E
6
sin
4
B
0
+
+16
E
2
sin
6
B
0
+ 520
E
4
sin
6
B
0
+ 881
E
6
sin
6
B
0
−
176
E
4
sin
8
B
0
−
536
E
4
sin
4
B
0
−
−
852
E
6
sin
8
B
0
+ 280
E
6
sin
10
B
0
)
,
b
05
=
E
2
(1
E
2
sin
2
B
0
)
7
/
2
40
A
1
(1
− E
2
)
5
−
×
×
(4 + 15
E
2
−
8sin
2
B
0
−
172
E
2
sin
2
B
0
−
210
E
4
sin
2
B
0
+ 184
E
2
sin
4
B
0
+872
E
4
sin
4
B
0
+
+315
E
6
sin
4
B
0
−
704
E
4
sin
6
B
0
−
944
E
6
sin
6
B
0
+ 648
E
6
sin
8
B
0
)
.
15-612 The Second Step: Polynomial Representation of Conformal Coordinates in the Second
Strip Replaced by the Conformal Coordinates in the First Strip
The standard polynomial representation of conformal coordinates of type Gauss-Krueger or UTM
in the
L
02
-strip is given by (
15.117
)and(
15.118
) subject to the longitude/latitude differences
l
2
:=
L
B
02
with respect to the longitude
L
02
of the reference meridian and
the latitude
B
02
of the reference point
−
L
02
and
b
2
:=
B
−
{
L
02
,B
02
}
of series expansion.
x
1
=
ρ
(
x
10
l
2
+
x
11
l
2
b
2
+
x
30
l
2
+
x
12
l
2
b
2
+O
4
x
)
.
Easting :
(15.117)
y
1
=
ρ
(
y
0
+
y
01
b
2
+
y
20
l
2
+
y
02
b
2
+
y
21
l
2
b
2
+
y
03
b
2
+O
4
y
)
.
Northing :
(15.118)
y
0
denotes the length of the meridian arc from zero ellipsoidal latitude to the ellipsoidal latitude
B
02
of the reference point
{
L
02
,B
02
}
chosen by the identity
B
01
=
B
02
=
B
0
for operational
reasons. The coecients
can be taken from Boxes
15.4
and
15.5
. In addition, the optimal
dilatation factor
ρ
has been set identical in the
L
01
-strip and the
L
02
-strip of the same strip width.
The longitude/latitude differences
{
x
ij
,y
ij
}
{
l
1
,b
1
}
as well as
{
l
2
,b
2
}
are related by
l
2
=(
L
01
−
L
02
)+
l
1
B
02
)+
b
1
being derived from the invariance
L
=
L
01
+
l
1
=
L
02
+
l
2
and
B
=
B
01
+
b
1
=
B
02
+
b
2
. Now we are on duty to replace
and
b
2
=(
B
01
−
L
02
)+
l
1
and
b
2
=(
B
01
−B
02
)+
b
1
by
{
(
L
01
−L
02
)+
l
1
,b
1
}
within (
15.117
)and(
15.118
), which leads us to
{
l
2
,b
2
}
by means of
l
2
=(
L
01
−
x
2
=
ρ
[
x
10
(
L
01
− L
02
)+
x
10
l
1
+
x
11
(
L
01
− L
02
)
b
1
+
x
11
l
1
b
1
+
x
30
(
L
01
−
L
02
)
3
+
(15.119)
L
02
)
2
l
1
+3
x
30
(
L
01
−
L
02
)
l
1
+
x
30
l
1
+
x
12
(
L
01
−
L
02
)
b
1
+
x
12
l
1
b
1
+O
4
x
]
,
+3
x
30
(
L
01
−
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