Geography Reference
In-Depth Information
Box 1.37 (Representation of
Φ
−
l
and
Φ
−
r
in terms of conformal coordinates:
E
A
1
,A
1
,A
2
→
S
r
)
.
Φ
−
l
:
X
(
Λ, Φ
)=
Φ
−
r
:
x
(
λ, φ
)=
=
E
1
A
1
cos
Φ
cos
Λ
1
+
=
e
1
r
cos
φ
cos
λ
+
E
2
sin
2
Φ
−
+
E
2
A
1
cos
Φ
sin
Λ
1
− E
2
sin
2
Φ
+
+
e
2
r
cos
φ
sin
λ
+
E
2
)sin
Φ
1
−
E
2
s
in
2
Φ
+
E
3
A
1
(1
−
=
+
e
3
r
sin
φ
=
A
1
cos
f
−
1
(
P
2
+
Q
2
)
P
p
4
r
2
+
p
2
+
q
2
+
P
2
+
Q
2
+=
e
1
4
r
2
=
E
1
1
(1.256)
E
2
sin
2
f
−
1
P
2
+
Q
2
−
A
1
cos
f
−
1
(
P
2
+
Q
2
)
Q
q
4
r
2
+
p
2
+
q
2
+
P
2
+
Q
2
++
e
2
4
r
2
=
E
2
1
− E
2
sin
2
f
−
1
P
2
+
Q
2
E
2
)sin
f
−
1
(
P
2
+
Q
2
)
+
e
3
r
4
r
2
(
p
2
+
q
2
)
4
r
2
+
p
2
+
q
2
.
+
E
3
A
1
(1
−
−
1
.
E
2
sin
2
f
−
1
(
P
2
+
Q
2
)
−
Isoparametric mapping:
p
=
P, q
=
Q.
(1.257)
Solution (the third problem).
By means of the two mapping equations “left”
{
P
=
f
(
Φ
)cos
Λ, Q
=
f
(
Φ
)sin
Λ
}
and of the two
mapping equations “right”
{
p
=
g
(
φ
)cos
λ, q
=
g
(
φ
)sin
λ
}
, we are able to compute the factors
Λ
l
,λ
l
}
Λ
r
,λ
r
}
of type “right”. The detailed formulae are
reviewed in Box
1.34
. If we specify “left”
Φ
=
π/
2 (ellipsoidal North Pole) or “right”
φ
=
π/
2
(spherical North Pole), we are led to
l
Λ
1
(
π/
2) =
l
Λ
2
(
π/
2) =
Λ
l
(
π/
2) = 1 and
r
Λ
1
(
π/
2) =
r
Λ
2
(
π/
2) =
Λ
r
(
π/
2) = 1. Obviously, at the North Pole, “left UPS” and “right UPS” are an
isometry. We shall see later that this is a built-in constraint for any UPS.
of conformality
{
of type “left” and
{
End of Solution (the third problem).
Solution (the fourth problem).
A “simple conformal mapping” of
E
A
1
,A
1
,A
2
→
S
r
is the isoparametric mapping, which is conve-
niently characterized by
p
=
P, q
=
Q
or
2
r
tan
4
−
2
cos
λ
=
2
)
1+
E
sin
Φ
1
−E
sin
Φ
E/
2
cos
Λ,
(1
−
E
)
E/
2
φ
2
A
1
√
1
−E
2
(1+
E
)
E/
2
tan(
4
−
Φ
(1.258)
2
r
tan
4
−
2
sin
λ
=
2
)
1+
E
sin
Φ
1
−E
sin
Φ
E/
2
sin
Λ.
(1
−
E
)
E/
2
φ
2
A
1
(1+
E
)
E/
2
tan(
4
−
Φ
√
1
−E
2
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