15-3 Constraints to the Korn–Lichtenstein equations (Gauss–Krueger/UTM mappings) 325
15-3 Constraints to the Korn–Lichtenstein equations
(Gauss–Krueger/UTM mappings)
The constraints to the Korn–Lichtenstein equations generating the Gauss–Krueger conformal mapping or
the UTM conformal mapping.
The equidistant mapping of a meridian of reference L 0 immediately establishes the proper constraint
to the Korn–Lichtenstein equations which leads to the standard Gauss–Krueger conformal mapping
or universal transverse Mercator projection conformal mapping. The arc length of the coordinate line
L 0 = const., namely the meridian, between latitude B 0 and B is computed by (15.74) as soon as we
set up uniformly convergent Taylor series of type (15.75) and integrate term-wise.
y(0, b) =
B
B 0
G 22 (B ∗ )dB
∗ =
B
B 0
M (B
∗ )dB
∗ =
∞
n=1
y 0n b
n ,
(15.74)
G 22 (B) = M (B) =
A 1 (1 − E
2 )
(1 − E 2 sin
2 B) 3/2 =
∞
n=1
1
n!
G
(n)
22 (B 0 )b
n .
(15.75)
Box 15.2, which follows subsequently, contains a list of resulting coefficients y 0n , which establish the
setup of the constraints defined in Definition 15.3.
Definition 15.3 (Constraints to the Korn–Lichtenstein equations of conformal mapping).
Let there be given the ellipsoidal Korn–Lichtenstein equations (15.76), subject to the integrability condition, the Laplace–Beltrami equations (15.77), which generate a conformal mapping via a polynomial
representation of type (15.29)–(15.32) and the coefficient constraints given by (15.69)–(15.73).
y l = −
G 11 /G 22 x b , y b =
G 22 /G 11 x l ,
(15.76)
sx ll + (rx b ) b = 0 , sy ll + (ry b ) b = 0 .
(15.77)
The equidistant mapping of the meridian of reference L 0 establishes by means of constraints of type
(15.78) the conformal mapping of type Gauss–Krueger or UTM.
x(0, b) = 0 , y(0, b) =
∞
n=1
y 0n b
n .
(15.78)
End of Definition.
Box 15.2 (The equidistant mapping of the meridian of reference L 0 , y(0, b) =
P ∞
n=1 y 0n b
n , coefficients
y 01 , . . . , y 04 ).
y 01 =
√
G 22
˛
˛
˛
B 0
=
A 1 (1 − E
2 )
(1 − E 2 sin
2 B 0 ) 3/2 ,
y 02 =
1
2
[
√
G 22 ]
=
1
4
G
22 /
√
G 22
˛
˛
˛
B 0
=
3
2
A 1 E
2 (1 − E
2 ) cos B 0 sin B 0
(1 − E 2 sin
2 B 0 ) 5/2
,
y 03 =
1
24
[2G 22 G
22 − G
22
2 ]/G
3/2
22
˛
˛
˛
B 0
=
1
2
A 1 E
2 (1 − E
2 )
(1 − E 2 sin
2 B 0 ) 7/2 (1 − 2 sin
2 B 0 + 4E
2 sin
2 B 0 − 3E
2 sin
2 B 0 ) ,
y 04 =
1
192
[4G
2
22 G
22 − 6G 22 G
22 G
22 + 3 G
22
3 ]/G
5/2
22
˛
˛
˛
B 0
=
1
8
A 1 E
2 (1 − E
2 )
(1 − E 2 sin
2 B 0 ) 9/2 cos B 0 sin B 0 (4 − 15E
2 + 22E
2 sin
2 B 0 − 20E
4 sin
2 B 0 + 9E
4 sin
4 B 0 ) .
(15.79)
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