A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Antoine Couëtte
Date: 2020 May 11, 03:34 -0700
Many thanks for your Kind and comprehensive reply.
It is interesting that you did visually check your initial result to be reasonable before publishing it .
I do agree that in most cases, problems similar to Dave's one generally yield solutions with Bodies LHA's inferior to 90°.
Still, for Observers close to the Poles, LHA's can be expected to easily exceed 90° even for reasonable altitudes.
As an example, with the following numbers rounded to integer values let's us start from :
- Tony's Position (i.e. from position N60° E030° , Hello Tony ! ) , and :
- Dave's Height ( 55° i.e. a sub-astral distance D = 2100 NM ), and :
- an Azimuth Az = 030°,
we get a subastral point at N73°17.4' E124°06.6' hence with LHA = 94°06.6'
Solving this with the direct method, the Sine rule "1st quadrant LHA" indicates 85°53.4' , with a sub-astral point at E115°53.4' .
The distance between both points is 8°13'2 of Longitude at a Mid-Latitude close to N80°, representing about 90 NM.
The final check on Distance seems a simple way here to pick up the correct "2 nd Quadrant LHA" here, i.e. 94°06.6' .
Best Regards to all