Mercator Projection and Polycylindric Projection - Map Projections: Cartographic Information Systems - page 811

d x

E 2 sin 2 x ) 3 / 2 =

(1 −

= x 1+ 3

64 E 4 4 cos x sin x 1+ 15 16 E 2 E 2 4 E 2 + 45

3 15

32 E 4 cos x sin 3 x +O( E 6 )

− − (I.16)

⇒
c 1 (see first equation of ( I.10 )) ,

Λ E

d Λ Φ E
Φ S E 2 sin 2 Φ ) = Φ N

d Φ ( Λ E − Λ W )d Φ

E 2 sin 2 Φ ) ,

(I.17)

cos Φ (1 − cos Φ (1 − Λ W Φ S

d x

E 2 sin 2 x ) = cos x (1 −

=(1+ E 2 + E 4 )lntan π

2

4 + x

1

E 2 (1 + E 2 )sin x

3 E 4 sin 3 x +O( E 6 )

− − (I.18)

⇒
c 2 (see second equation of ( I.10 )) .

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 accordingto( I.3 ) within
the left Airy distortion energy J l A , 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 S has
to be integrated over the area of interest. In this way, we are led to the coecients 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 ρ = c 1 /c 2 and

ρ ( Φ S ,Φ N )=

= ( Φ N − Φ S ) 1+ 3

64 E 4

1+ 15

16 E 2 E 2 (cos Φ N sin Φ N − cos Φ S sin Φ S ) 4 E 2 + 45

3
4

−

cos Φ S sin 3 Φ S ) / ln tan π

/ tan π

15

4 + Φ N 4 + Φ S 32 E 4 (cos Φ N sin 3 Φ N −

− ×

2 2

sin 3 Φ S ) +O( E 6 ) ,

1

(1 + E 2 + E 4 ) E 2 (1 + E 2 )(sin Φ N

3 E 4 (sin 3 Φ N

× − − sin Φ S ) − − (I.19)

ρ ( Φ S = − Φ N )=

= Φ N 1+ 3

64 E 4

1+ 15

16 E 2 E 2 cos Φ N sin Φ N − 32 E 4 cos Φ N sin 3 Φ N / 4 E 2 + 45

3
4

15

−

(1 + E 2 + E 4 )lntan π

3 E 4 sin 3 Φ N +O( E 6 ) .

4 + Φ N

1

E 2 (1 + E 2 )sin Φ N −

−

2

End of Lemma.