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⇒
(16.42)
A
2
sin
i
(1
sin
α
A
1
cos
2
α
+
A
2
cos
2
i
sin
2
α
tan
B
=
.
−
E
2
)
End of Proof.
Proof.
x
3
=
z
=0
⇒
tan
α
=
y
x
A
1
A
2
=
=
y
x
1
√
1
− E
2
=
=
√
1
−
E
2
cos
2
i
−
x
1
sin
Ω
cos
i
+
x
2
cos
Ω
cos
i
+
x
3
sin
i
x
1
cos
Ω
+
x
2
sin
Ω
=
√
1
(16.43)
−
E
2
=
√
1
cos
B
sin(
L
E
2
)sin
B
sin
i
−
E
2
cos
2
i
−
Ω
)cos
i
+(1
−
√
1
cos
B
cos(
L
−
Ω
)
−
E
2
⇒
tan
α
=
√
1
−
E
2
cos
2
i
1
cos(
L
√
1
− E
2
E
2
)sin
i
tan
B
+cos
i
sin(
L
Ω
)
[(1
−
−
Ω
)]
.
−
End of Proof.
16-3 The Equations of the Oblique Mercator Projection
Universal oblique Mercator projection. D'Alembert-Euler equations (Cauchy-Riemann equa-
tions), oblique elliptic meta-equator.
The fundamental solution (
16.16
) of the d'Alembert-Euler equations (Cauchy-Riemann equa-
tions) here are specified by
}
UMP
of type (
16.10
) and by the boundary condition of an
equidistant mapping of the oblique elliptic meta-equator illustrated by Figs.
16.3
and
16.4
.In
particular, we depart from (
16.16
)and(
16.10
), conventionally written as (
16.44
), here only given
up to degree three.
{
p, q
Δx
:=
x
−
α
0
=
=
α
1
Δq
+
β
1
Δl
+
α
2
(
Δq
2
Δl
2
)+
β
2
2
ΔqΔl
+O
x
3
,
−
(16.44)
Δy
:=
y
−
β
0
=
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