626 I Mercator projection and polycylindric projection
Lemma I.4 (Generalized Mercator projection, Airy distortion energy).
In case of the generalized Mercator projection, the left Airy distortion energy J lA is the quadratic
form in terms of the dilatation factor ρ, in particular
J lA (ρ) = c 0 − 2c 1 ρ + c 2 ρ
2 ,
(I.9)
such that
c 0 = 1 ,
c 1 =
(Φ N − Φ S )
1 +
3
4
E
2 +
45
64
E
4
−
3
4
1 +
15
16
E
2
E
2 (cos Φ N sin Φ N − cos Φ S cos Φ S )−
−
15
32
E
4
cos Φ N sin
3 Φ N − cos Φ S sin
3 Φ S
/
sin Φ N +
2
3
E
2 sin
3 Φ N +
3
5
E
4 sin
5 Φ N − sin Φ S −
2
3
E
2 sin
3 Φ S −
3
5
E
4 sin
5 Φ S
+ O(E
6 ) ,
c 2 =
1 + E
2 + E
4
ln
tan
π
4
+
Φ N
2
/ tan
π
4
+
Φ S
2
− E
2
1 + E
2
(sin Φ N − sin Φ S ) −
−
1
3
E
3 (sin
3 Φ N − sin
3 Φ S )
/
sin Φ N +
2
3
E
2 sin
3 Φ N +
3
5
E
4 sin
5 Φ N − sin Φ S −
2
3
E
2 sin
3 Φ S −
3
5
E
4 sin
5 Φ S
+ O(E
6 )
(I.10)
hold.
End of Lemma.
Constitutional elements of the left Airy distortion energy are
J lA =
1
S
S
(Λ S − 1)
2 dS =
1
S
S
(Λ
2
S − 2Λ S + 1)dS = 1 −
2
S
S
Λ S dS +
1
S
S
Λ
2
S dS ,
(I.11)
S
Λ S dS = ρ
S
1 − E 2 sin
2 Φ
cos Φ
A
2
1 (1 − E
2 ) cos Φ
(1 − E 2 sin
2 Φ) 2 dΛdΦ ,
(I.12)
S
Λ
2
S dS = ρ
2
S
1 − E
2 sin
2 Φ
cos 2 Φ
A
2
1 (1 − E
2 ) cos Φ
(1 − E 2 sin
2 Φ) 2 dΛdΦ ,
(I.13)
J lA (ρ) = c 0 − 2c 1 ρ + c 2 ρ
2
⇔
c 0 := 1
⇔
c 1 :=
1
S
S
Λ S dS =
1
S
A
2
1 (1 − E
2 )
S
dΛdΦ
(1 − E 2 sin
2 Φ) 3/2 ,
c 2 :=
1
S
S
Λ
2
S dS =
1
S
A
2
1 (1 − E
2 )
S
dΛdΦ
cos Φ(1 − E 2 sin
2 Φ)
.
(I.14)
Lemma I.4 (Generalized Mercator projection, Airy distortion energy).
In case of the generalized Mercator projection, the left Airy distortion energy J lA is the quadratic
form in terms of the dilatation factor ρ, in particular
J lA (ρ) = c 0 − 2c 1 ρ + c 2 ρ
2 ,
(I.9)
such that
c 0 = 1 ,
c 1 =
(Φ N − Φ S )
1 +
3
4
E
2 +
45
64
E
4
−
3
4
1 +
15
16
E
2
E
2 (cos Φ N sin Φ N − cos Φ S cos Φ S )−
−
15
32
E
4
cos Φ N sin
3 Φ N − cos Φ S sin
3 Φ S
/
sin Φ N +
2
3
E
2 sin
3 Φ N +
3
5
E
4 sin
5 Φ N − sin Φ S −
2
3
E
2 sin
3 Φ S −
3
5
E
4 sin
5 Φ S
+ O(E
6 ) ,
c 2 =
1 + E
2 + E
4
ln
tan
π
4
+
Φ N
2
/ tan
π
4
+
Φ S
2
− E
2
1 + E
2
(sin Φ N − sin Φ S ) −
−
1
3
E
3 (sin
3 Φ N − sin
3 Φ S )
/
sin Φ N +
2
3
E
2 sin
3 Φ N +
3
5
E
4 sin
5 Φ N − sin Φ S −
2
3
E
2 sin
3 Φ S −
3
5
E
4 sin
5 Φ S
+ O(E
6 )
(I.10)
hold.
End of Lemma.
Constitutional elements of the left Airy distortion energy are
J lA =
1
S
S
(Λ S − 1)
2 dS =
1
S
S
(Λ
2
S − 2Λ S + 1)dS = 1 −
2
S
S
Λ S dS +
1
S
S
Λ
2
S dS ,
(I.11)
S
Λ S dS = ρ
S
1 − E 2 sin
2 Φ
cos Φ
A
2
1 (1 − E
2 ) cos Φ
(1 − E 2 sin
2 Φ) 2 dΛdΦ ,
(I.12)
S
Λ
2
S dS = ρ
2
S
1 − E
2 sin
2 Φ
cos 2 Φ
A
2
1 (1 − E
2 ) cos Φ
(1 − E 2 sin
2 Φ) 2 dΛdΦ ,
(I.13)
J lA (ρ) = c 0 − 2c 1 ρ + c 2 ρ
2
⇔
c 0 := 1
⇔
c 1 :=
1
S
S
Λ S dS =
1
S
A
2
1 (1 − E
2 )
S
dΛdΦ
(1 − E 2 sin
2 Φ) 3/2 ,
c 2 :=
1
S
S
Λ
2
S dS =
1
S
A
2
1 (1 − E
2 )
S
dΛdΦ
cos Φ(1 − E 2 sin
2 Φ)
.
(I.14)
