Geography Reference
In-Depth Information
Example D.3 (Optimal universal transverse Mercator projection, biaxial ellipsoid
E
a,b
).
First, let us here establish the boundary condition for the universal transverse Mercator projection
modulo an unknown dilatation factor. Second, with respect to the d'Alembert-Euler equations
(Cauchy-Riemann equations), let us here solve the firstly formulated boundary value problem.
Finally, the unknown dilatation factor is optimally determined by an optimization of the total
distance distortion measure (Airy optimum)
or
of the total areal distortion.
The boundary condition.
r
), we
gener-
alize
the equidistant mapping of the
L
0
meta-equator, namely the
boundary condition
(
D.67
), by
an unknown
dilatation factor
, subject to later optimization. We begin with the solution to the
problem to express the boundary condition in the function space.
x
=
x{q
(
L
=
L
0
,B
)
,p
(
L
=
L
0
,B
)
}
=
a
(1
− e
2
)
B
0
With reference to the two examples (transverse Mercator projection of the sphere
S
d
B
e
2
sin
2
B
)
3
/
2
,
(D.67)
(1
−
y
=
y
{
q
(
L
=
L
0
,B
)
,p
(
L
=
L
0
,B
)
}
=0
.
Since the
boundary condition
(
D.67
) is given in terms of surface normal ellipsoidal latitude
B
,in
the
first step
, we have to introduce ellipsoidal isometric latitude
Q
(
B
):[
−π/
2
,
+
π/
2]
→
[0
,±∞
]
by (
D.68
), where
e
2
=(
a
2
− b
2
)
/a
2
=1
−
(
b
2
/a
2
) is the first numerical eccentricity.
Q
=ln
tan
π
e/
2
=lntan
π
1
4
+
B
e
sin
B
1+
e
sin
B
−
4
+
B
−
2
2
e
2
ln
1+
e
sin
B
e
sin
B
=
(D.68)
1
−
= artanh(sin
B
)
− e
artanh
e
sin
B.
In the
second step
,wesetupa
uniformly convergent series expansion
of the integral transformation
according to (
D.69
)bymeansofthe
recurrence relation
(
D.70
). The coecients
a
r
are collected
in (
D.71
). They are given as polynomials in 1,
e
2
,e
4
etc. and cos
B,
cos(2
B
)
,
cos(3
B
)etc.
x
=
∞
d
r
x
d
Q
r
(
Q
0
(
B
0
))
,q
:=
Q
x
0
:=
x
(
L
0
,B
0
)
,a
r
=
1
r
!
a
r
q
r
∀
−
Q
0
=
Q
−
Q
(
B
0
)
,
(D.69)
r
=0
d
r−
1
x
d
Q
r−
1
d
B
d
r
x
d
Q
r
=
e
2
sin
2
B
1
− e
2
1
d
B
d
B
d
Q
=cos
B
1
−
d
Q
∀
,
(D.70)
e
2
)
B
0
d
B
a
cos
B
1
− e
2
sin
2
B
,a
2
=
−
a
cos
B
sin
B
2
1
− e
2
sin
2
B
a
0
=
x
0
=
a
(1
−
e
2
sin
2
B
)
3
/
2
,a
1
=
,
(1
−
2s
in
2
B
+
e
2
sin
4
B
)
6(1
− e
2
)
1
− e
2
sin
2
B
a
3
=
−
a
cos
B
(1
−
,a
4
=
a
cos
B
sin
B
24(1
− e
2
)
2
1
− e
2
sin
2
B
×
×
[5
−
6sin
2
B − e
2
(1 + 6 sin
2
B −
9sin
4
B
)+
e
4
sin
4
B
(3
−
4sin
2
B
)]
,
(D.71)
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