Geography Reference
In-Depth Information
π/
2) = 4
Rπ/
√
6
π
=2
πc
1
, and
th
e length of the image of the pole
+
π/
2)
−
x
(
Λ
=0
,Φ
=
−
=
π/
2) = 4
Rπ/
√
6
π
=2
πc
1
. Obviously, the length of
image of the circular equator is twice the length of image of the central meridian or the pole.
x
(
Λ
=+
π,
|
Φ
|
=
π/
2)
−
x
(
Λ
=
−
π,
|
Φ
|
End of Solution (the first problem).
Solution (the second problem).
In order to derive the left Cauchy-Green deformation tensor, according to Box
1.10
,wedepart
from computing the left Jacobi matrix J
l
. First, the partial derivatives
D
Λ
x, D
Φ
x, D
Λ
y
,and
D
Φ
y
build up the left Jacobi matrix. Second, by means of the matrix product C
l
=J
l
G
r
J
l
,weareable
to compute the left Cauchy-Green matrix for the right matrix of the metric G
r
=I
2
. Indeed, the
chart
is covered by Cartesian coordinates whose metric is simply given by d
s
2
=d
x
2
+d
y
2
.
Though the special left Cauchy-Green matrix C
l
=J
l
J
l
looks simple, but is complicated in detail.
The elements
{c
11
,c
12
=
c
21
,c
22
}
document these features.
{
x, y
}
End of Solution (the second problem).
Solution (the third problem).
Box
1.11
outlines the solution of the third problem, namely the laborious analytical computation
of the left eigenvalues and the left eigencolumns. First, we refer to G
l
as the matrix of the
metric of the sphere
S
2
R
of radius
R
,andtoC
l
as the matrix of the left Cauchy-Green tensor, as
computed in Box
1.10
. The characteristic equation of the left general eigenvalue problem leads
to the solution
Λ
1
,
2
=
Λ
2
+
,−
as functions of the two
fundamental invariants
(i) tr [C
l
G
−
l
] and (ii)
det[C
l
G
−
l
]. While the elements of the matrix C
l
G
−
l
evoke simple, its trace is complicated. In
contrast, det[C
l
G
−
l
] = 1. Second, it is a straightforward proof that the product of eigenvalues
squared is identical to the second invariant, i.e.
Λ
1
Λ
2
=det[C
l
G
−
l
]. As proven, det[C
l
G
−
l
]=1(in
consequence
Λ
1
Λ
2
= 1) can be interpreted as the condition for an equiareal mapping. A detailed
computation of the left eigenvalues
is not useful due to the lengthy forms
involved. T
hi
rd, the same argument holds for the computed first eigencolumn, whic
h is
associated
{
Λ
1
,Λ
2
}
=
{
Λ
+
,Λ
−
}
to
Λ
1
=
Λ
1
∈
R
+
and for the second eigencolumn, which is associated to
Λ
2
=
Λ
2
∈
R
+
,and
lattice (for instance, 1
◦
×
1
◦
)
these are very lengthy. For practical use, a computation in a
{
Λ, Φ
}
is recommended.
End of Solution (the third problem).
Solution (the fourth problem).
}
is
not
an isometry. For an isometry,
Λ
1
=
Λ
2
= 1 is the postulate. If
Λ
1
=
Λ
2
, then it holds that
(tr[C
l
G
−
l
])
2
=4det[C
l
G
−
l
]. Since (tr[C
l
G
−
l
(
Λ
=0
,Φ
=0)])
2
=[(64+9
π
2
)
/
24
π
]
2
Box
1.12
collects the details of the proof that the “Eckert II mapping” of the point
{
Λ, Φ
}
=
{
0
,
0
= 4 due to
det[C
l
G
−
l
] = 1, it follows that
Λ
1
=
Λ
2
.
End of Solution (the fourth problem).
Example
1.5
documents that for various map projections it is practically impossible to analytically
compute the eigenspace which leads to the left and right Tissot ellipses. Numerically no problems
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