138 3 Coordinates
Box 3.17 (The backward problem of transforming spherical frames of reference: the transverse aspect. Input
variables: A, B, Ω, I = π/2. Output variables: Λ, Φ).
(i) Longitude:
tan(Λ − Ω) = −
tan B
cos A
.
(3.86)
(ii) Latitude:
sin Φ = cos B sin A .
(3.87)
(iii) Substitutions:
sin(Λ − Ω) = −
sin B
cos Φ
,
cos(Λ − Ω) = +
tan Φ
tan A
,
tan(Λ − Ω) = −
sin B
sin Φ
sin A
cos A
= −
tan B
cos A
.
(3.88)
3-34 The transverse frame of reference of the sphere: part two
The transverse case is a special case of an oblique frame of reference. Since it has gained great interest
in map projections, we devote another special section to the transverse aspect. In short, for such
a peculiar aspect, the meta-North Pole is chosen to be located in the conventional equator of the
reference sphere S
2
r . In short, the spherical latitude φ 0 = 0
◦ of the meta-North Pole is fixed to zero.
Accordingly, we specialize the forward and backward transformation formulae according to Boxes 3.6
and 3.9 to such a special pole configuration. In order to understand better the conventional choice of
the transverse frame of reference, we additionally consider the following example.
Example 3.7 (On the transverse frame of refererence).
Let us choose λ 0 = 270
◦ and φ 0 = 0
◦ , namely a placement of the meta-North Pole in the West Pole.
For such a configuration, the meta-equator (then called transverse equator) agrees with the Greenwich
meridian of reference. e 1 0 = e S is directed to the South, e 2 0 = e E is directed to the East. Such an
oblique frame of reference has not found the support of traditional map projectors. They prefer a
transverse frame of reference
e 1 ∗ , e 2 ∗ , e 3 ∗
O
, namely a right-handed orthonormal frame of reference
that is oriented “East, North, Vertical” and relates to
e 1 0 , e 2 0 , e 3 0
O
by
e 1 ∗ = +e 2 0 = e E (“Easting”) ,
e 2 ∗ = −e 1 0 = e S (“Northing”) ,
e 3 ∗ = +e 3 0 = e V (“Vertical”) .
(3.89)
In terms of this example, we may alternatively choose λ
0 = λ 0 − 270
◦ = 90
◦ + λ 0 or λ 0 = 270
◦ + λ
0 ,
and α = α S = 90
◦ + α E or α E = α S − 90
◦ = 270
◦ + α S , such that {λ
0 = 0
◦ , φ 0 = 0
◦
} identifies the
Greenwich meridian of reference. Indeed, we have shifted both λ 0 and α E for 3π/2 ∼ 270
◦ .
End of Example.
Box 3.17 (The backward problem of transforming spherical frames of reference: the transverse aspect. Input
variables: A, B, Ω, I = π/2. Output variables: Λ, Φ).
(i) Longitude:
tan(Λ − Ω) = −
tan B
cos A
.
(3.86)
(ii) Latitude:
sin Φ = cos B sin A .
(3.87)
(iii) Substitutions:
sin(Λ − Ω) = −
sin B
cos Φ
,
cos(Λ − Ω) = +
tan Φ
tan A
,
tan(Λ − Ω) = −
sin B
sin Φ
sin A
cos A
= −
tan B
cos A
.
(3.88)
3-34 The transverse frame of reference of the sphere: part two
The transverse case is a special case of an oblique frame of reference. Since it has gained great interest
in map projections, we devote another special section to the transverse aspect. In short, for such
a peculiar aspect, the meta-North Pole is chosen to be located in the conventional equator of the
reference sphere S
2
r . In short, the spherical latitude φ 0 = 0
◦ of the meta-North Pole is fixed to zero.
Accordingly, we specialize the forward and backward transformation formulae according to Boxes 3.6
and 3.9 to such a special pole configuration. In order to understand better the conventional choice of
the transverse frame of reference, we additionally consider the following example.
Example 3.7 (On the transverse frame of refererence).
Let us choose λ 0 = 270
◦ and φ 0 = 0
◦ , namely a placement of the meta-North Pole in the West Pole.
For such a configuration, the meta-equator (then called transverse equator) agrees with the Greenwich
meridian of reference. e 1 0 = e S is directed to the South, e 2 0 = e E is directed to the East. Such an
oblique frame of reference has not found the support of traditional map projectors. They prefer a
transverse frame of reference
e 1 ∗ , e 2 ∗ , e 3 ∗
O
, namely a right-handed orthonormal frame of reference
that is oriented “East, North, Vertical” and relates to
e 1 0 , e 2 0 , e 3 0
O
by
e 1 ∗ = +e 2 0 = e E (“Easting”) ,
e 2 ∗ = −e 1 0 = e S (“Northing”) ,
e 3 ∗ = +e 3 0 = e V (“Vertical”) .
(3.89)
In terms of this example, we may alternatively choose λ
0 = λ 0 − 270
◦ = 90
◦ + λ 0 or λ 0 = 270
◦ + λ
0 ,
and α = α S = 90
◦ + α E or α E = α S − 90
◦ = 270
◦ + α S , such that {λ
0 = 0
◦ , φ 0 = 0
◦
} identifies the
Greenwich meridian of reference. Indeed, we have shifted both λ 0 and α E for 3π/2 ∼ 270
◦ .
End of Example.