I-1 The optimal Mercator projection (UM) 627
Furthermore, constitutional elements of the left Airy distortion energy are
Λ E
Λ W
dΛ
Φ E
Φ S
dΦ
(1 − E 2 sin
2 Φ) 3/2 =
Φ N
Φ S
(Λ E − Λ W )dΦ
(1 − E 2 sin
2 Φ) 3/2 ,
(I.15)
dx
(1 − E 2 sin
2 x) 3/2 =
= x
1 +
3
4
E
2 +
45
64
E
4
−
3
4
cos x sin x
1 +
15
16
E
2
E
2
−
15
32
E
4 cos x sin
3 x + O(E
6 )
⇒
c 1 (see first equation of (I.10)) ,
(I.16)
Λ E
Λ W
dΛ
Φ E
Φ S
dΦ
cos Φ(1 − E 2 sin
2 Φ)
=
Φ N
Φ S
(Λ E − Λ W )dΦ
cos Φ(1 − E 2 sin
2 Φ)
,
(I.17)
dx
cos x(1 − E 2 sin
2 x)
=
=
1 + E
2 + E
4
ln tan
π
4
+
x
2
− E
2
1 + E
2
sin x −
1
3
E
4 sin
3 x + O(E
6 )
⇒
c 2 (see second equation of (I.10)) .
(I.18)
Note that for the proof of Lemma I.4, we have collected all constitutional items in (I.10)–(I.18). Indeed,
as soon as we represent the left principal stretches Λ 1 = Λ 2 = Λ S according to (I.3) within the left Airy
distortion energy J lA , in particular (I.9), we are left with the quadratic polynomial of (I.11) which
constitutes the integrals of (I.12) and (I.13). First, the left principal stretch Λ S has to be integrated
over the area of interest. Second, the squared left principal stretch Λ
2
S has to be integrated over the area
of interest. In this way, we are led to the coefficients c 0 , c 1 , abd c 2 of type (I.14). (I.15)–(I.18) describe
the involved integrals which are computed by term-wise integration of the uniformly convergent kernel
series, namely by interchanging integration and summation. The integral series expansions are of the
order O(E
6 ) for (I.16) and (I.18).
Lemma I.5 (Minimal Airy distortion energy).
The Airy distortion energy (I.9) is minimal if the dilatation factor amounts to ˆ
ρ = c 1 /c 2 and
ˆ
ρ(Φ S , Φ N ) =
=
(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 sin Φ S )
−
15
32
E
4
cos Φ N sin
3 Φ N − cos Φ S sin
3 Φ S
/
ln
tan
π
4
+
Φ N
2
/ tan
π
4
+
Φ S
2
×
×
1 + E
2 + E
4
−E
2 (1 + E
2 )(sin Φ N − sin Φ S ) −
1
3
E
4 (sin
3 Φ N − sin
3 Φ S )
+ O(E
6 ) ,
ˆ
ρ(Φ S = −Φ N ) =
=
Φ N
1 +
3
4
E
2 +
45
64
E
4
−
3
4
1 +
15
16
E
2
E
2 cos Φ N sin Φ N −
15
32
E
4 cos Φ N sin
3 Φ N
/
1 + E
2 + E
4
ln tan
π
4
+
Φ N
2
− E
2 (1 + E
2 ) sin Φ N −
1
3
E
4 sin
3 Φ N
+ O(E
6 ) .
(I.19)
End of Lemma.
Furthermore, constitutional elements of the left Airy distortion energy are
Λ E
Λ W
dΛ
Φ E
Φ S
dΦ
(1 − E 2 sin
2 Φ) 3/2 =
Φ N
Φ S
(Λ E − Λ W )dΦ
(1 − E 2 sin
2 Φ) 3/2 ,
(I.15)
dx
(1 − E 2 sin
2 x) 3/2 =
= x
1 +
3
4
E
2 +
45
64
E
4
−
3
4
cos x sin x
1 +
15
16
E
2
E
2
−
15
32
E
4 cos x sin
3 x + O(E
6 )
⇒
c 1 (see first equation of (I.10)) ,
(I.16)
Λ E
Λ W
dΛ
Φ E
Φ S
dΦ
cos Φ(1 − E 2 sin
2 Φ)
=
Φ N
Φ S
(Λ E − Λ W )dΦ
cos Φ(1 − E 2 sin
2 Φ)
,
(I.17)
dx
cos x(1 − E 2 sin
2 x)
=
=
1 + E
2 + E
4
ln tan
π
4
+
x
2
− E
2
1 + E
2
sin x −
1
3
E
4 sin
3 x + O(E
6 )
⇒
c 2 (see second equation of (I.10)) .
(I.18)
Note that for the proof of Lemma I.4, we have collected all constitutional items in (I.10)–(I.18). Indeed,
as soon as we represent the left principal stretches Λ 1 = Λ 2 = Λ S according to (I.3) within the left Airy
distortion energy J lA , in particular (I.9), we are left with the quadratic polynomial of (I.11) which
constitutes the integrals of (I.12) and (I.13). First, the left principal stretch Λ S has to be integrated
over the area of interest. Second, the squared left principal stretch Λ
2
S has to be integrated over the area
of interest. In this way, we are led to the coefficients c 0 , c 1 , abd c 2 of type (I.14). (I.15)–(I.18) describe
the involved integrals which are computed by term-wise integration of the uniformly convergent kernel
series, namely by interchanging integration and summation. The integral series expansions are of the
order O(E
6 ) for (I.16) and (I.18).
Lemma I.5 (Minimal Airy distortion energy).
The Airy distortion energy (I.9) is minimal if the dilatation factor amounts to ˆ
ρ = c 1 /c 2 and
ˆ
ρ(Φ S , Φ N ) =
=
(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 sin Φ S )
−
15
32
E
4
cos Φ N sin
3 Φ N − cos Φ S sin
3 Φ S
/
ln
tan
π
4
+
Φ N
2
/ tan
π
4
+
Φ S
2
×
×
1 + E
2 + E
4
−E
2 (1 + E
2 )(sin Φ N − sin Φ S ) −
1
3
E
4 (sin
3 Φ N − sin
3 Φ S )
+ O(E
6 ) ,
ˆ
ρ(Φ S = −Φ N ) =
=
Φ N
1 +
3
4
E
2 +
45
64
E
4
−
3
4
1 +
15
16
E
2
E
2 cos Φ N sin Φ N −
15
32
E
4 cos Φ N sin
3 Φ N
/
1 + E
2 + E
4
ln tan
π
4
+
Φ N
2
− E
2 (1 + E
2 ) sin Φ N −
1
3
E
4 sin
3 Φ N
+ O(E
6 ) .
(I.19)
End of Lemma.