From Riemann Manifolds to Riemann Manifolds - Map Projections: Cartographic Information Systems - page 24

appear when we have a computer at hand. For a large number of map projections, there is no
problem to analytically compute the eigenspace. Such an example is considered after the boxes.

Box 1.9 (Eckert II, the first problem).

x = c 1 Λ 4

2 R

√ 6 π ,

− 3sin | Φ | , 1 :=

sign φ, c 2 := R 2 π
3

y = c 2 2 − 4 − 3sin |Φ|

= πc 1 . (1.74)

Meridians :

4

x
c 1 Λ ⇒

c 2
c 1

x
Λ =2 c 2 −

π x

− 3sin | Φ | = y =2 c 2 −

Λ ,

Λ = constant ⇒ y =2 c 2 − c 3 x, c 3 := π

Λ , L 1 ( Λ = constant)

2

:= {x ∈ R

|y =2 c 2 − c 3 x} .

(1.75)

Parallel circles:
Φ = constant ⇒ x = c 4 Λ, c 4 := c 1 4 − 3sin |Φ| ,y = c 5 ,
c 5 := 2 c 2 − c 2 4 − 3sin |Φ| , (1.76)

L 1 ( Φ = constant) :=

2

{ x ∈ R | x = c 4 Λ, y = c 5 } .

“Half”:
(i) length of the circular equator:
x ( Λ =+ π,Φ =0)

π,Φ =0)=8 Rπ/ √ 6 π,
(ii) length of the central meridian:
x ( Λ =0 ,Φ = π/ 2)

− x ( Λ = −

π/ 2) = 4 Rπ/ √ 6 π,

− x ( Λ =0 ,Φ = − (1.77)

(iii) length of the pole:

= π/ 2) = 4 Rπ/ √ 6 π. x ( Λ =+ π, | Φ | = π/ 2) − x ( Λ = − π, | Φ |

Box 1.10 (Eckert II, the second problem).
x = c 1 Λ 4

2 R

√ 6 π ,

− 3sin | Φ | ,c 1 :=

sign φ, c 2 := R 2 π
3

y = c 2 2

4

− − 3sin | Φ | = πc 1 . (1.78)

Left Jacobi matrix:

D Λ x = c 1 4

J l := D Λ xD Φ x

,

c 2 Λ 3cos Φ sign Φ

√ 4 − 3sin |Φ|

− 3sin | Φ | , D Φ x = − ,

(1.79)

D Λ y =0 ,D Φ y = 3 c 2

√ 4 − 3sin |Φ|

cos Φ

D Λ yD Φ y

.

Left Cauchy-Green matrix: