Reference Ellipsoid and the Geodetic Coordinate System - Geodesy: Introduction to Geodetic Datum and Geodetic Systems - page 248

Solution of Spherical Triangles

1. Find A 2
Since A 2 ﾼ

A 0 2 + 180 , from the right-angled spherical triangle P 0 0 Q 2 P 0 2 , one

obtains:

tan m
cos M

tan A 0 2 ﾼ

tan A 2 ﾼ

Þ :

ð 5 : 96 Þ

ð þ ˃

2. Find u 2
Again from the right-angled spherical triangle P 0 0 Q 2 P 0 2 , we have:

cos A 0 2 tan M tan u 2 ﾼ ð þ ˃ Þﾼ cos A 2 tan M ð þ ˃ Þ: ð 5 : 97 Þ

3. Compute ʻ

From the right-angled spherical triangles P 0 0 Q 1 P 0 1 and P 0 0 Q 2 P 0 2 , we get:

9
=
; :

tan ʻ 1 ﾼ sin m tanM ﾼ sin u 1 tan A 1

tan ʻ 2 ﾼ sin m tan M ð þ ˃ Þﾼ sin u 2 tan A 2 ð 5 : 98 Þ

ʻ ﾼ ʻ 2 ʻ 1

So far, the three unknowns A 2 , u 2 , and ʻ on the spherical surface have all been

obtained.

Project Elements from the Spherical Surface onto the Ellipsoidal Surface

1. Given u 2 , find B 2 :

q

1 e 0 2 tan B 2 ﾼ þ tan u 2 : ð 5 : 99 Þ

2. Convert ʻ into l and find L 2 :

h i ,

0 0 sin 0 sin 2

l ﾼ ʻ sin m ʱ ˃ þ ʲ ˃ cos 2M ð þ ˃ Þ þ ʳ ˃ cos 4M ð þ 2 ˃ Þ ð 5 : 100 Þ

where