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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.

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