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Proof (( 16.48 ): l 2 ).
d α 2 = 2 A 1 A 2 cos i ( A 2 cos 2 i A 1 )sin α cos α
d 2 L
,
( A 1 cos 2 α + A 2 cos 2 i sin 2 α ) 2
(16.63)
2 l 2 := d 2 L
d α 2 ( α 0 ) .
End of Proof.
Proof (( 16.52 )-( 16.54 )).
In a first step, ( 16.44 ) is specified by
Δx (meta-equator) =
= α 1 Δq + β 1 Δl + α 2 ( Δq 2
Δl 2 )+ β 2 2 ΔqΔl +O x 3 ,
(16.64)
Δy (meta-equator) =
= β 1 Δq − α 1 Δl + β 2 ( Δq 2
− Δl 2 ) − α 2 2 ΔqΔl +O y 3 =0 .
Implementation of ( 16.50 ) constitutes the second step:
Δx (meta-equator) =
= α 1 q 1 Δb + α 1 q 2 Δb 2 + β 1 Δl + α 2 ( q 1 Δb 2
− Δl 2 )+ β 2 2 q 1 ΔbΔl +O x 3 ,
(16.65)
Δy (meta-equator) =
= β 1 q 1 Δb + β 1 q 2 Δb 2
− Δl 2 ) − α 2 2 q 1 ΔbΔl +O y 3 =0 .
In a third step, the boundary condition in the above form is represented
in the meta-longitude dependence by means of ( 16.47 )-( 16.49 )
[where in a fourth step we identify ( 16.45 )by( 16.46 )]:
− α 1 Δl + β 2 ( q 1 Δb 2
Δx (meta-equator) =
= α 1 ( q 1 b 1 Δα + q 1 b 2 Δα 2 + q 2 b 1 Δα 2 )+ β 1 ( l 1 Δα + l 2 Δα 2 )+
+ α 2 ( q 1 b 1 Δα 2
l 1 Δα 2 )+ β 2 2 q 1 b 1 l 1 Δα 2 +O x 3 =
= s 1 Δα + s 2 Δα 2 ,
(16.66)
Δy (meta-equator) =
α 1 ( l 1 Δα + l 2 Δα 2 )+ β 1 ( q 1 b 1 Δα + q 1 b 2 Δα 2 + q 2 b 1 Δα 2 )
α 2 2 q 1 b 1 l 1 Δα 2 + β 2 ( q 1 b 1 Δα 2
l 1 Δα 2 )+O y 3 =0 .
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