19-2 Special conic projections based upon the ellipsoid-of-revolution 407
19-22 Special conic projections of type conformal
Here, we depart from the postulate of conformality. A
fter integration- by- parts and after application
of the addition theorem, we are finally led to (19.15).
Λ 1 = Λ 2 ⇒
nf (∆)
N sin ∆
=
f
(∆)
M
⇒
f
f
=
nM
N sin ∆
⇔
df
f
= n
(1 − E
2 )d∆
(1 − E 2 cos 2 ∆) sin ∆
⇒ ln f = n(1 − E
2 )
d∆
(1 − E 2 cos 2 ∆) sin ∆
.
(19.10)
Here, let us substitute u := E cos ∆:
sin
2 ∆ =
E
2
− u
2
E 2
,
du
d∆
= E(− sin ∆) , ln f = −n(1 − E
2 )E
du
(1 − u 2 )(E 2 − u 2 )
. (19.11)
B y integration- by- parts, we find in detail:
du
(1 − u 2 )(E 2 − u 2 )
=
Adu
1 − u 2 +
Bdu
E 2 − u 2 ,
(19.12)
A(E
2
− u
2 ) + B(1 − u
2 ) = 1 , AE
2 + B − u
2 (A + B) = 1 ,
AE
2 + B = 1 , A + B = 0
⇒
A = −
1
1 − E 2 , B =
1
1 − E 2
⇒
du
(1 − u 2 )(E 2 − u 2 )
= −
1
1 − E 2
du
1 − u 2 −
du
E 2 − u 2
=
= −
1
1 − E 2
1
2
ln
1 + u
1 − u
−
1
2E
ln
E + u
E − u
=
= −
1
2(1 − E 2 )
ln
1 + E cos ∆
1 − E cos ∆
+
1
E
ln
1 − cos ∆
1 + cos ∆
.
(19.13)
We take advantage of the addition theorem:
d∆
(1 − E 2 cos 2 ∆) sin ∆
= −
1
2(1 − E 2 )
ln
1 + E cos ∆
1 − E cos ∆
+
1
E
ln tan
2 ∆
2
+ ln c
⇒
ln f =
nE
2
ln
1 + E cos ∆
1 − E cos ∆
tan
2/E ∆
2
+ ln c
= ln
1 + E cos ∆
1 − E cos ∆
En/2
tan
n ∆
2
c
⇒
(19.14)
f (∆) := c
1 + E cos ∆
1 − E cos ∆
E/2
tan
∆
2
n
.
(19.15)
19-22 Special conic projections of type conformal
Here, we depart from the postulate of conformality. A
fter integration- by- parts and after application
of the addition theorem, we are finally led to (19.15).
Λ 1 = Λ 2 ⇒
nf (∆)
N sin ∆
=
f
(∆)
M
⇒
f
f
=
nM
N sin ∆
⇔
df
f
= n
(1 − E
2 )d∆
(1 − E 2 cos 2 ∆) sin ∆
⇒ ln f = n(1 − E
2 )
d∆
(1 − E 2 cos 2 ∆) sin ∆
.
(19.10)
Here, let us substitute u := E cos ∆:
sin
2 ∆ =
E
2
− u
2
E 2
,
du
d∆
= E(− sin ∆) , ln f = −n(1 − E
2 )E
du
(1 − u 2 )(E 2 − u 2 )
. (19.11)
B y integration- by- parts, we find in detail:
du
(1 − u 2 )(E 2 − u 2 )
=
Adu
1 − u 2 +
Bdu
E 2 − u 2 ,
(19.12)
A(E
2
− u
2 ) + B(1 − u
2 ) = 1 , AE
2 + B − u
2 (A + B) = 1 ,
AE
2 + B = 1 , A + B = 0
⇒
A = −
1
1 − E 2 , B =
1
1 − E 2
⇒
du
(1 − u 2 )(E 2 − u 2 )
= −
1
1 − E 2
du
1 − u 2 −
du
E 2 − u 2
=
= −
1
1 − E 2
1
2
ln
1 + u
1 − u
−
1
2E
ln
E + u
E − u
=
= −
1
2(1 − E 2 )
ln
1 + E cos ∆
1 − E cos ∆
+
1
E
ln
1 − cos ∆
1 + cos ∆
.
(19.13)
We take advantage of the addition theorem:
d∆
(1 − E 2 cos 2 ∆) sin ∆
= −
1
2(1 − E 2 )
ln
1 + E cos ∆
1 − E cos ∆
+
1
E
ln tan
2 ∆
2
+ ln c
⇒
ln f =
nE
2
ln
1 + E cos ∆
1 − E cos ∆
tan
2/E ∆
2
+ ln c
= ln
1 + E cos ∆
1 − E cos ∆
En/2
tan
n ∆
2
c
⇒
(19.14)
f (∆) := c
1 + E cos ∆
1 − E cos ∆
E/2
tan
∆
2
n
.
(19.15)
