C2 THE NAVIGATION TRIANGLE USING THE AGETON METHOD - 10/03/2004

In the nineteen thirties, Lt. Arthur A. Ageton, United States Navy,
proposed a method for solving the navigation triangle. He dropped
a perpendicular from one corner of the oblique spherical triangle
to the meridian of the observer (or point of departure) to form
two right triangles, and did a solution using secants and
cosecants only. This was an advantage because the logarithms are
always positive. To use this system, a special set of tables
called the Ageton tables are required. It was also published by
the U. S. Navy Hydrographic Office as publication "HO-211".

When sailing from point 1 (L1, LO1) to point 2 (L2, LO2)
t = DLO = difference in longitude (degrees)
R = length of the perpendicular
K = distance from the equator to the perpendicular as
measured along the meridian of point one
D = distance from point 1 to point 2 (degrees)
C = course angle at point 1
csc R = csc t * sec L2 ( R is less than 90 degrees )
csc K = csc L2 / sec R (note 1)
sec D = sec R * sec ( K-L1 )
csc C = csc R / csc D (note 2)

Turning the equations over and converting to the sine
cosine format, we get a set of formulas usable on
a calculator.

sin R = sin t * cos L2 ( R is less than 90 degrees )
sin K = sin L2 / cos R (note 1)(note 3)
cos D = cos R * cos (K-L1)
sin C = sin R / sin D (note 2)
(note 1) If t is greater than 90 degrees then
K is greater than 90 degrees.
(note 2) If L1 is greater than K then C is greater
than 90 degrees.
(note 3) If only the course angle is required,
Tan C = tan R / sin (K-L1)
If L1 is greater than K, C will be a negative angle
and the correct answer will be C+180 degrees.