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9
=
S
M cos A
ʔ
B
S
N sin A sec B
ʔ
L
:
ð
5
:
52
Þ
;
S
N sin A tan B
ʔ
A
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
dS,
M
with ( 5.49 ), dS
c os A dB, we obtain:
sin A
cos A
M sin BdB
N cos B
dA
,
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
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
dr,
with r
N cos B, substituting into the above equation gives:
sin A
cos A
dr
r :
dA
We obtain:
r cos AdA
þ
sin Adr
0,
dr
r :
cot AdA
After integration, it can be rewritten as:
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