“Sphere to Cone”: Polar Aspect - Map Projections: Cartographic Information Systems - page 438

n

2 R n sin Φ 1 + c
R cos Φ 1

−

Λ 1 | Φ = Φ 1 = =1 . (17.54)

We solve this equation in order to determine the integration constant c and obtain ( 17.55 ). Since
c can also be computed from Λ 1 | Φ = Φ 2 := 1, we obtain ( 17.56 )and( 17.57 ). Resubstituting this
result into ( 17.56 ) gives the final form of c ,namely( 17.58 ).

c = R 2 cos 2 Φ 1 +2 n sin Φ 1
n 2

, (17.55)

R 2 cos 2 Φ 1 +2 n sin Φ 1
n 2

= R 2 cos 2 Φ 2 +2 n sin Φ 2
n 2

⇒ (17.56)

cos 2 Φ 1 +2 n sin Φ 1 =cos 2 Φ 2 +2 n sin Φ 2 ,

n = cos 2 Φ 1 −

cos 2 Φ 2 sin 2 Φ 1 − 1+sin 2 Φ 2 sin 2 Φ 2 − sin 2 Φ 1

sin Φ 1 ) = 1

−

=

sin Φ 1 ) = 2(sin Φ 2 − 2(sin Φ 2 − sin Φ 1 ) 2(sin Φ 2 −

= (sin Φ 2 −

sin Φ 1 )(sin Φ 2 +sin Φ 1 )
2(sin Φ 2 − sin Φ 1 )

= (17.57)

= 1

2 (sin Φ 1 +sin Φ 2 ) ,
c =4 R 2 1+sin Φ 1 sin Φ 2

(sin Φ 1 +sin Φ 2 ) 2 .

(17.58)

As a matter of course, the cone constant and the integration constant are symmetric in Φ 1 and Φ 2 .
The final mapping equations are given by ( 17.59 )or( 17.60 ), and the final left principal stretches
are provided by ( 17.61 ). Indeed, the requirements Λ 1 | Φ = Φ 1 = Λ 2 | Φ = Φ 1 = Λ 1 | Φ = Φ 2 = Λ 2 | Φ = Φ 2 =1
are met and the inverse mapping equations are defined through ( 17.62 ).
α
r

=

=

,

nΛ

n √ cos 2 Φ 1 +2 n sin Φ 1 −

(17.59)

R

2 n sin Φ

x
y

=

2 n sin Φ cos( nΛ )

,

n cos 2 Φ 1 +2 n sin Φ 1 −

= R

(17.60)

sin( nΛ )

= 1+sin Φ 1 sin Φ 2 Λ 1 = √ cos 2 Φ 1 +2 n sin Φ 1 −

2 n sin Φ − (sin Φ 1 +sin Φ 2 )sin Φ

cos Φ

, (17.61)

cos Φ

Λ 2 = Λ − 1 .

cos 2 Φ 1 +2 n sin Φ 1 −

R 2 .

nr

Λ = α

n , Φ =arcsin 1

(17.62)

2 n