A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: David C
Date: 2020 Apr 29, 16:40 -0700
May I contribute with a Dreisonstok solution for completness? :-)
I am forced to calculate the azimuth with Table V in HO200 (-;
This method uses altitude, dec and hour angle, unlike most other tables wich use lat, dec and hour angle.
h 33° 03.7'
dec 14° 13'
With dec and t extract 9.39990
With h in the dec column look for the number closest to 9.9990 . Azimuth is the corresponding hour angle = 17°. With some very rough interpolation azimuth = 17.5.°. dec and lat are of contrary names so azimuthis named opposite of lat ----- N 17.5° E. This is the bearing method of naming azimuth. If the names are the same the rule is complicated and I have not attempted to decypher it.