Geoscience Reference
In-Depth Information
Table 5.10 Sample computation for the direct solution of the geodetic problem
Given data
Ellipsoidal parameters
Computational results
90
00
0
00.11
00
215
59
0
04.333
00
L
1
ﾼ
Krassowski Ellipsoid
L
2
ﾼ
35
00
0
00.22
00
30
29
0
20.964
00
B
1
ﾼ
B
2
ﾼ
100
00
0
00.33
00
290
32
0
53.389
00
A
1
ﾼ
A
2
ﾼ
215
59
0
13.059
00
S
ﾼ
15 000 000.2 m
GRS75 Ellipsoid
L
2
ﾼ
30
29
0
23.867
00
B
2
ﾼ
290
32
0
48.833
00
A
2
ﾼ
215
59
0
13.306
00
GRS80 Ellipsoid
L
2
ﾼ
30
29
0
23.947
00
A
2
ﾼ 290
32
0
48.708
00
B
2
ﾼ
Applying the cosine theorem to the triangle P
0
1
P
0
2
N
0
in Fig.
5.43
gives (l is used
to approximately replace
ʻ
, and hence the
˃
obtained is an estimated value
˃
0
):
cos
˃
0
ﾼ
sin u
1
sin u
2
þ
cos u
1
cos u
2
cos l
:
ð
5
:
103
Þ
From triangles P
0
1
P
0
2
Q
1
and P
0
1
P
0
2
N
0
, one obtains:
sin l
sin
sin m
0
ﾼ
cos u
1
sin A
1
ﾼ
cos u
1
cos u
2
˃
0
:
ð
5
:
104
Þ
The approximate estimated value m
0
of m is hereby obtained.
With (
5.92
), we take the approximate correction for
ʻ
:
0
ʔʻ ﾼ ʻ
l
ʱ
˃
0
sin m
0
ﾼ
0
:
003351
˃
0
sin m
0
:
ð
5
:
105
Þ
Hence, we have:
ʻ
0
ﾼ
l
þ ʔʻ:
Taking (
5.103
) and differentiating
˃
and l results in:
sin
˃
0
ʔ˃ ﾼ
cos u
1
cos u
2
sin
ʻ
0
ʔʻ
,
Or
sin
ʻ
0
ʔ˃ ﾼ
cos u
1
cos u
2
˃
0
ʔʻ ﾼ
sin m
0
ʔʻ:
ð
5
:
106
Þ
sin
Hence, the result is:
˃
1
ﾼ ˃
0
þ ʔ˃:
Substituting
ʻ
0
and
˃
1
into (
5.104
) produces:
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