From Riemann Manifolds to Riemann Manifolds - Map Projections: Cartographic Information Systems - page 107

y ) / [ R (1 + cos Φ )sin β ]. Next, add sin 2 Λ/ 2+cos 2 Λ/ 2=1
and you are done. Similarly, to solve the fourth problem, choose Λ = constant and eliminate
Φ from the direct equations of the mapping, for instance, by 1 + cos Φ = x/ [ R sin Λ/ 2] as Φ =
( x

as well as cos Λ/ 2=( R cos β sin Φ −

R sin Λ/ 2) / ( R sin Λ/ 2) and cos 2 Φ as well as by y/R +( x sin β ) / ( R tan Λ/ 2) = cos β sin Φ ,to
be squared to cos 2 β sin 2 Φ , leading to a quadratic form of type ax 2 + bxy + cy 2 + d = 0, indeed a
conic section.

−

End of Solution.

Fig. 1.32. Armadillo projection modified by Raisz : double projection, (i) sphere → torus, (ii) torus → plane,

obliquity β =20 ◦ , Tissot ellipses of distortion

Box 1.58 (Armadillo double projection. 1st: sphere to torus. 2nd: torus to oblique plane).

r ,
left coordinates
(spherical longitude Λ,
(spherical latitude Φ ):

Left manifold S

2
a,b

1
a

1

Right manifold T ∼ S × S

b ,

right coordinates
(toroidal longitude λ,
(toroidal latitude φ ):

2 2

S

r :=

T

a,t :=

x 2 + y 2

a 2

3

X 2 + Y 2 + Z 2 R 2 =0 , :=

3

:= { X ∈ R | − { x ∈ R | −

+ z 2 b 2 =0 , −

+ ,R> 0 + ,b + ,b

R ∈ R } , a ∈ R ∈ R ≤ a } ,

(1.343)

Φ − l : X ( Λ, Φ )= Φ − 1 : c ( λ, φ )=
= E 1 R cos Φ cos Λ + E 2 R cos Φ sin Λ += e 1 (( a + b cos φ )cos λ
+ e 2 (0 + b cos Φ )sin λ +

+ E 3 B sin Φ. + e 3 b sin Φ .