A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Antoine Couëtte
Date: 2015 Feb 16, 02:11 -0800
Feb 16th, 2015
Nice job and nice display/drawing of yours as Hewitt earlier noticed, alotgether with your good, solid and sturdy differential computations for corrections.
Just one question as regards your Azimuths determinations.
If I am not mistaken, the Azimuths seem to be derived from their SINE values through the (SINE)-1 / or ARCSIN functions. No ?
If such is the case (i.e. Azimuths determined from their Sine Values), I wish to get some clarification here because of the following facts detailed here-under.
Generally speaking, one should be [very] careful when attempting to recover an Azimuth value through only its SINE value, simply because sin (90°+&°) equals sin (90°-&°). Accordingly you have no immediate means to sort out the right answer when attempting to recover an Azimuth value from ONLY its SINE value when this value is close from 90° or 270°. A "correct recovery" procedure for such cases does exist, but it is not straightforward.
In other words, and even with your "strict" limitations, i.e.: " This paper will consider only the case of the Local Hour Angle from zero to ninety degrees, both the observer and the celestial body having the "same name" (in the Northern Hemisphere), and the declination is less than the observer's latitude. ", there are many cases of bodies with declinations smaller than the Observer's Latitude and with LHA's less than 90° which - in addition to the 180°-270° / 090°-180° Azimuths quadrants - also ly in the 270°-360° / 000°-090° Azimuths quadrants.
2 examples hereunder for Bodies with no appreciable parallax (i.e. stars cases):
Example 1 : Observer's Latitude 45°00'0 N , Body Declination : 44°00'0 N and Local Hour Angle : 89°00'0 yields : Geocentric height = 30°00'3 and Azimuth = 303.8°
Example 2 : Observer's Latitude 45°00'0 N , Body Declination : 02°10'0 N and Local Hour Angle : 45°00'0 yields : Geocentric height = 31°45'7 and Azimuth = 236.2°
Both examples fit into your constraints. Their azimuts differ by 67.6° while they nonetheless have the same SINE values. If you are using only the ARC SINE function to determine the bodies azimuths, how are you actually proceeding ?
On the other hand, recovering Azimuths through the (COSINE)-1 or through the (COTANGENT)-1 functions is not prone to such traps. Solving for Azimuths through these functions still gives you as earlier 2 different values over the 360° Azimuths Range, but one of these values lies East of the Observer and the second one lies West of the Observer. One always knows (or should know) whether the Body LHA lies in the [000°-180°] range - yielding an Azimuth in the West of the Observer - or in the [180°-360°] range - yielding an Azimut in the East of the Observer. Accordingly the Azimuths [un]determination is easily solved when using the (COSINE)-1 or the (COTANGENT)-1 functions.
Thank you for your Kind Attention, and for your subsequent clarification on this "Azimuths" topic.
Best "Navigational" Regards
Antoine M. "Kermit" Couëtte