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    Re: Position from altitude and azimuth.
    From: Antoine Couëtte
    Date: 2020 May 11, 03:34 -0700

    Dear Peter,

    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


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