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9
=
S
M
cos A
ʔ
B
ᄐ
S
N
sin A sec B
ʔ
L
ᄐ
:
ð
5
:
52
Þ
;
S
N
sin A tan B
ʔ
A
ᄐ
Clairaut's Equation for the Geodesic
Integrating the differential equations of geodesics gives Clairaut's equation. It
provides the basis for solving the geodetic problems over long distance.
From (
5.51
) we can get:
sin A sin B
N cos B
dA
ᄐ
dS,
M
with (
5.49
), dS
ᄐ
c
os
A
dB, we obtain:
sin A
cos A
M sin BdB
N cos B
dA
ᄐ
,
where, as shown in Fig.
5.27
, P is a point on the surface of the ellipsoid, PP
0
is the
arc element of the meridian, and PP
0
ᄐ
MdB. The difference in the radius of the
parallel at point P and P
0
is dr. When P
0
moves to P, the latitude increases and the
radius of parallel decreases. Let PP
0
P
00
be a small right-angled plane triangle; we
get:
M sin BdB
ᄐ
dr,
with r
ᄐ
N cos B, substituting into the above equation gives:
sin A
cos A
dr
r
:
dA
ᄐ
We obtain:
r cos AdA
þ
sin Adr
ᄐ
0,
dr
r
:
cot AdA
ᄐ
After integration, it can be rewritten as:
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