3-3 The oblique frame of reference of the sphere 129
Solution (the third problem, direct transformation).
Such a problem can be immediately solved as outlined in Boxes 3.3, 3.4, 3.5, and 3.6. First, we have parameterized the transformation of reference frames {e 1 0 , e 2 0 , e 3 0 O} → {e 1 , e 2 , e 3 O} by means of the
pole position {λ 0 , φ 0 }. Second, the placement vector x ∈ S
2
r of a point of the reference sphere is represented in both the conventional equatorial frame of reference {e 1 , e 2 , e 3 O} and in the meta-equatorial
(oblique) frame of reference {e 1 0 , e 2 0 , e 3 0 O} at the origin O. Third, we substitute {e 1 , e 2 , e 3 O} in
favor of {e 1 0 , e 2 0 , e 3 0 O} by means of the backward transformation of reference frames, our first
setup. The final representation of the placement vector x(λ, φ; λ 0 , φ 0 ) is achieved in terms of (i) conventional equatorial coordinates {λ, φ} and (ii) equatorial coordinates {λ 0 , φ 0 } of the meta-North
Pole. The corresponding two coordinate transformations α(λ, φ; λ 0 , φ 0 ) and β(λ, φ; λ 0 , φ 0 ) are derived
in Box 3.5. The three identities for (i) cos α, (ii) sin α, and (iii) sin β or cos ψ are derived by representing (i) x
0 = r cos β cos α, (ii) y
0 = r cos β sin α, and (iii) z
0 = r sin β = r cos ψ in the oblique frame
of reference. The first identity is also called spherical sine lemma, the second identity is also called
spherical sine–cosine lemma, and the third identity is called spherical side cosine lemma. Indeed, we
have derived the collective formulae of Spherical Trigonometry, however, in a way to be used for other
reference surfaces, for instance, the ellipsoid-of-revolution – a surface with one Killing vector of symmetry. The direct transformation formulae {λ, φ; λ 0 , φ 0 } → {α, β} are presented in Box 3.6. First, by
dividing the second identity by the first identity, we arrive at tan α = f (λ, φ; λ 0 , φ 0 ). Alternatively,
we may chose cos α or sin α, which are additionally depending on cos β. Second, we repeat the third
identity sin β = cos ψ either for the meta-latitude β or the meta-colatitude ψ, also called meta-polar
distance – the space angle between the vectors x and x 0 .
End of Solution (the third problem, direct transformation).
Solution (the third problem, inverse transformation).
Such a problem can be immediately solved as outlined in Box 3.7. First, we have parameterized the
transformation of reference frames {e 1 , e 2 , e 3 O} → {e 1 0 , e 2 0 , e 3 0 O} by means of the pole position
{λ 0 , φ 0 }. Second, the placement vector x ∈ S
2
r of a point of the reference sphere is represented
in both the conventional equatorial frame of reference {e 1 , e 2 , e 3 O} and in the meta-equatorial
(oblique) frame of reference {e 1 0 , e 2 0 , e 3 0 O} at the origin O. Third, we substitute {e 1 0 , e 2 0 , e 3 0 O}
in favor of {e 1 , e 2 , e 3 O} by means of the forward transformation of reference frames, our first
setup. The final representation of the placement vector x(α, β; λ 0 , φ 0 ) is achieved in terms of (i)
meta- equatorial coordinates {α, β} and (ii) equatorial coordinates {λ 0 , φ 0 } of the meta-North Pole.
The corresponding two coordinate transformations λ(α, β; λ 0 , φ 0 ) and φ(α, β; λ 0 , φ 0 ) are derived in
Box 3.8. The three identities for (i) cos λ, (ii) sin λ, and (iii) sin φ are based on representing (i) x =
r cos φ cos λ, (ii) y = r cos φ sin λ, and (iii) z = r sin φ in the oblique frame of reference as derived in
the previous formulae. The inverse transformation formulae {α, β; λ 0 , φ 0 } → {λ, φ} are presented in
Box 3.9. First, by dividing the second identity by the first identity, we arrive at tan λ = f (α, β; λ 0 , φ 0 ).
Alternatively, we may take advantage of the elegant form tan(λ − λ 0 ), which is achieved as soon as
we implement the addition theorem for tan(α ± β). Indeed, another simple derivation is the following.
Take reference to the direct transformation formulae. Multiply, the third identity by cos φ 0 , namely
sin β cos φ 0 , as well as the first identity by sin φ 0 , namely cos β cos α sin φ 0 , and sum up. In this way, you
have found cos φ cos(λ − λ 0 ) = sin β cos φ 0 + cos β cos α sin φ 0 . From the second identity, you transfer
cos φ sin(λ−λ 0 ) = cos β sin α and divide to produce tan(λ−λ 0 ) = sin(λ−λ 0 )/ cos(λ−λ 0 ). Second, you
transfer the third identity of the inverse transformation to gain sin φ = g(α, β; λ 0 , φ 0 ). Alternatively,
you may multiply the first identity of the forward transformation by cos φ 0 , namely cos β cos α cos φ 0 ,
and replace cos β cos φ 0 cos(λ − λ 0 ) by the third identity, i. e. sin β − sin φ sin φ 0 = cos ψ − sin φ sin φ 0 .
Finally, solve for sin φ.
End of Solution (the third problem, inverse transformation).
Solution (the third problem, direct transformation).
Such a problem can be immediately solved as outlined in Boxes 3.3, 3.4, 3.5, and 3.6. First, we have parameterized the transformation of reference frames {e 1 0 , e 2 0 , e 3 0 O} → {e 1 , e 2 , e 3 O} by means of the
pole position {λ 0 , φ 0 }. Second, the placement vector x ∈ S
2
r of a point of the reference sphere is represented in both the conventional equatorial frame of reference {e 1 , e 2 , e 3 O} and in the meta-equatorial
(oblique) frame of reference {e 1 0 , e 2 0 , e 3 0 O} at the origin O. Third, we substitute {e 1 , e 2 , e 3 O} in
favor of {e 1 0 , e 2 0 , e 3 0 O} by means of the backward transformation of reference frames, our first
setup. The final representation of the placement vector x(λ, φ; λ 0 , φ 0 ) is achieved in terms of (i) conventional equatorial coordinates {λ, φ} and (ii) equatorial coordinates {λ 0 , φ 0 } of the meta-North
Pole. The corresponding two coordinate transformations α(λ, φ; λ 0 , φ 0 ) and β(λ, φ; λ 0 , φ 0 ) are derived
in Box 3.5. The three identities for (i) cos α, (ii) sin α, and (iii) sin β or cos ψ are derived by representing (i) x
0 = r cos β cos α, (ii) y
0 = r cos β sin α, and (iii) z
0 = r sin β = r cos ψ in the oblique frame
of reference. The first identity is also called spherical sine lemma, the second identity is also called
spherical sine–cosine lemma, and the third identity is called spherical side cosine lemma. Indeed, we
have derived the collective formulae of Spherical Trigonometry, however, in a way to be used for other
reference surfaces, for instance, the ellipsoid-of-revolution – a surface with one Killing vector of symmetry. The direct transformation formulae {λ, φ; λ 0 , φ 0 } → {α, β} are presented in Box 3.6. First, by
dividing the second identity by the first identity, we arrive at tan α = f (λ, φ; λ 0 , φ 0 ). Alternatively,
we may chose cos α or sin α, which are additionally depending on cos β. Second, we repeat the third
identity sin β = cos ψ either for the meta-latitude β or the meta-colatitude ψ, also called meta-polar
distance – the space angle between the vectors x and x 0 .
End of Solution (the third problem, direct transformation).
Solution (the third problem, inverse transformation).
Such a problem can be immediately solved as outlined in Box 3.7. First, we have parameterized the
transformation of reference frames {e 1 , e 2 , e 3 O} → {e 1 0 , e 2 0 , e 3 0 O} by means of the pole position
{λ 0 , φ 0 }. Second, the placement vector x ∈ S
2
r of a point of the reference sphere is represented
in both the conventional equatorial frame of reference {e 1 , e 2 , e 3 O} and in the meta-equatorial
(oblique) frame of reference {e 1 0 , e 2 0 , e 3 0 O} at the origin O. Third, we substitute {e 1 0 , e 2 0 , e 3 0 O}
in favor of {e 1 , e 2 , e 3 O} by means of the forward transformation of reference frames, our first
setup. The final representation of the placement vector x(α, β; λ 0 , φ 0 ) is achieved in terms of (i)
meta- equatorial coordinates {α, β} and (ii) equatorial coordinates {λ 0 , φ 0 } of the meta-North Pole.
The corresponding two coordinate transformations λ(α, β; λ 0 , φ 0 ) and φ(α, β; λ 0 , φ 0 ) are derived in
Box 3.8. The three identities for (i) cos λ, (ii) sin λ, and (iii) sin φ are based on representing (i) x =
r cos φ cos λ, (ii) y = r cos φ sin λ, and (iii) z = r sin φ in the oblique frame of reference as derived in
the previous formulae. The inverse transformation formulae {α, β; λ 0 , φ 0 } → {λ, φ} are presented in
Box 3.9. First, by dividing the second identity by the first identity, we arrive at tan λ = f (α, β; λ 0 , φ 0 ).
Alternatively, we may take advantage of the elegant form tan(λ − λ 0 ), which is achieved as soon as
we implement the addition theorem for tan(α ± β). Indeed, another simple derivation is the following.
Take reference to the direct transformation formulae. Multiply, the third identity by cos φ 0 , namely
sin β cos φ 0 , as well as the first identity by sin φ 0 , namely cos β cos α sin φ 0 , and sum up. In this way, you
have found cos φ cos(λ − λ 0 ) = sin β cos φ 0 + cos β cos α sin φ 0 . From the second identity, you transfer
cos φ sin(λ−λ 0 ) = cos β sin α and divide to produce tan(λ−λ 0 ) = sin(λ−λ 0 )/ cos(λ−λ 0 ). Second, you
transfer the third identity of the inverse transformation to gain sin φ = g(α, β; λ 0 , φ 0 ). Alternatively,
you may multiply the first identity of the forward transformation by cos φ 0 , namely cos β cos α cos φ 0 ,
and replace cos β cos φ 0 cos(λ − λ 0 ) by the third identity, i. e. sin β − sin φ sin φ 0 = cos ψ − sin φ sin φ 0 .
Finally, solve for sin φ.
End of Solution (the third problem, inverse transformation).