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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 l }
Λ r 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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