8-2 Special mapping equations 227
We depart from the representation of the meridian arc length as a function of the polar distance ∆,
the complement of the surface normal latitude Φ. Let us transform the integral kernel (which is a
function of Φ) to Φ
∗ (which is the circular reduced latitude). Such a polar coordinate is generated by
projecting a meridianal point P vertically onto a circle S
1
A 1
of radius A 1 . Note that the geometrical
situation is illustrated in Fig. 8.2 and Fig. 8.3. Furthermore, note that the relations ∆ := π/2 − Φ and
∆
∗ := π/2 − Φ
∗ hold, and
f (∆) = A 1 (1 − E
2 )
∆
0
d∆
(1 − E 2 cos 2 ∆ ) 3/2
⇔
f (∆
∗ ) = A 1
∆
∗
0
1 − E 2 sin
2 ∆ d∆ .
(8.25)
The transformation formulae Φ → Φ
∗ and Φ
∗
→ Φ, respectively, are summarized in Box 8.4, originating
from
√
X 2 + Y 2 = A 1 cos Φ
∗ and Z = A 2 sin Φ
∗ , taking reference to the semi-major axis A 1 and the
semi-minor axis A 2 . Here, we refer to
sin Φ =
sin Φ
∗
√
1 − E 2 cos 2 Φ ∗
or
cos ∆ =
cos ∆
∗
1 − E 2 sin
2 ∆ ∗
,
and
A 1 (1 − E
2 )
(1 − E 2 cos 2 ∆) 3/2 =
A 1
√
1 − E 2
(1 − E
2 sin
2 ∆
∗ )
3/2 ,
and
sin ∆ =
√
1 − E 2
1 − E 2 sin
2 ∆ ∗
sin ∆
∗
⇒
cos ∆d∆ =
√
1 − E 2
(1 − E 2 sin
2 ∆ ∗ ) 3/2 cos ∆
∗ d∆
∗ ,
⇒
d∆ =
cos ∆
∗
cos ∆
√
1 − E 2
(1 − E 2 sin
2 ∆ ∗ ) 3/2 d∆
∗ =
√
1 − E 2
1 − E 2 sin
2 ∆ ∗ d∆
∗ ,
(8.26)
in order to have derived
A 1 (1 − E
2 )
(1 − E 2 cos 2 ∆) 3/2 d∆ =
A 1
√
1 − E 2
(1 − E
2 sin
2 ∆
∗ )
3/2
√
1 − E 2
1 − E 2 sin
2 ∆ ∗ d∆
∗ ,
A 1 (1 − E
2 )
(1 − E 2 cos 2 ∆) 3/2 d∆ = A 1
1 − E 2 sin
2 ∆ ∗ d∆
∗ .
(8.27)