8-1 General mapping equations 223
8-1 General mapping equations
Setting up general equations of the mapping “ellipsoid-of-revolution to plane”: azimuthal projections in
the normal aspect (polar aspect).
There are again two basic postulates which govern the setup of general equations of mapping the
ellipsoid-of-revolution E
2
A 1 ,A 2
of semi-major axis A 1 and semi-minor axis A 2 , which are characterized
by A 1 > A 2 , to a tangential plane T E
2
A 1 ,A 2
attached to a point X ∈ T E
2
A 1 ,A 2
. Let the tangential plane
be covered by polar coordinates {α, r}. Then the following postulates are valid.
Postulate.
The polar coordinate α, which is also called azimuth, is identical to the ellipsoidal longitude, i. e.
α = Λ.
End of Postulate.
Postulate.
The polar coordinate r depends only on the ellipsoidal latitude Φ or on the ellipsoidal colatitude
∆ := π/2 − Φ, i. e. r =
x 2 + y 2 = f (∆) = f (π/2 − Φ). If Φ = π/2 or, equivalently, ∆ = 0, then
f (0) = 0 holds.
End of Postulate.
In last consequence, the general equations of an azimuthal mapping are provided by the following
vector equation:
x
y
=
r cos α
r sin α
=
f (∆) cos Λ
f (∆) sin Λ
.
(8.12)
Question.
Question: “How can we identify the images of the special coordinate lines Λ = constant
and Φ = constant, respectively?” Answer (y = x tan Λ, Λ = constant : elliptic meridian):
“The image of the elliptic meridian Λ = constant under an azimuthal mapping is the radial
straight line.” Answer (x
2 + y
2 = r
2 = f
2 (∆), ∆ = constant : parallel circle): “The image of
the parallel circle ∆ = constant (or Φ = constant) under an azimuthal mapping is the circle
S
1
r of radius r = f (∆). Such a mapping is called concircular.”
Proof (y = x tan Λ, Λ = constant : elliptic meridian).
Solve the first equation towards f (∆) = x/ cos Λ and substitute f (∆) in the second equation such
that y = f (∆) sin Λ = x sin Λ/ cos Λ = x tan Λ holds.
End of Proof (y = x tan Λ, Λ = constant : elliptic meridian).
Proof (x
2 + y
2 = r
2 = f
2 (∆), ∆ = constant : parallel circle).
Compute the terms x
2 and y
2 and add the two: x
2 + y
2 = f
2 (∆).
End of Proof (x
2 + y
2 = r
2 = f
2 (∆), ∆ = constant : parallel circle).
In summary, the images of the elliptic meridian and the parallel circle constitute the typical graticule
of an azimuthal mapping, i. e.
meridians (Λ = constant)
−→ radial straight lines ,
parallel circles
∆ = constant
Φ = constant
−→
equicentric circles .
(8.13)
8-1 General mapping equations
Setting up general equations of the mapping “ellipsoid-of-revolution to plane”: azimuthal projections in
the normal aspect (polar aspect).
There are again two basic postulates which govern the setup of general equations of mapping the
ellipsoid-of-revolution E
2
A 1 ,A 2
of semi-major axis A 1 and semi-minor axis A 2 , which are characterized
by A 1 > A 2 , to a tangential plane T E
2
A 1 ,A 2
attached to a point X ∈ T E
2
A 1 ,A 2
. Let the tangential plane
be covered by polar coordinates {α, r}. Then the following postulates are valid.
Postulate.
The polar coordinate α, which is also called azimuth, is identical to the ellipsoidal longitude, i. e.
α = Λ.
End of Postulate.
Postulate.
The polar coordinate r depends only on the ellipsoidal latitude Φ or on the ellipsoidal colatitude
∆ := π/2 − Φ, i. e. r =
x 2 + y 2 = f (∆) = f (π/2 − Φ). If Φ = π/2 or, equivalently, ∆ = 0, then
f (0) = 0 holds.
End of Postulate.
In last consequence, the general equations of an azimuthal mapping are provided by the following
vector equation:
x
y
=
r cos α
r sin α
=
f (∆) cos Λ
f (∆) sin Λ
.
(8.12)
Question.
Question: “How can we identify the images of the special coordinate lines Λ = constant
and Φ = constant, respectively?” Answer (y = x tan Λ, Λ = constant : elliptic meridian):
“The image of the elliptic meridian Λ = constant under an azimuthal mapping is the radial
straight line.” Answer (x
2 + y
2 = r
2 = f
2 (∆), ∆ = constant : parallel circle): “The image of
the parallel circle ∆ = constant (or Φ = constant) under an azimuthal mapping is the circle
S
1
r of radius r = f (∆). Such a mapping is called concircular.”
Proof (y = x tan Λ, Λ = constant : elliptic meridian).
Solve the first equation towards f (∆) = x/ cos Λ and substitute f (∆) in the second equation such
that y = f (∆) sin Λ = x sin Λ/ cos Λ = x tan Λ holds.
End of Proof (y = x tan Λ, Λ = constant : elliptic meridian).
Proof (x
2 + y
2 = r
2 = f
2 (∆), ∆ = constant : parallel circle).
Compute the terms x
2 and y
2 and add the two: x
2 + y
2 = f
2 (∆).
End of Proof (x
2 + y
2 = r
2 = f
2 (∆), ∆ = constant : parallel circle).
In summary, the images of the elliptic meridian and the parallel circle constitute the typical graticule
of an azimuthal mapping, i. e.
meridians (Λ = constant)
−→ radial straight lines ,
parallel circles
∆ = constant
Φ = constant
−→
equicentric circles .
(8.13)