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    Re: Determing the bearing off a celestial object.
    From: Bruce J. Pennino
    Date: 2015 Mar 31, 09:38 -0400
    Hello All:
    Greg, you raised a point which I never before considered. At any time, how  do you calculate the true bearing of the sun , or any celestial body, without H? So I went to my trusty “Celestial Navigation in the GPS Age” and found my answer. Maybe?
    Conceptually and visually  it seems  clear, time is known, your location is known and GP can be quickly found.   I would solve the triangle using 
                             Tan A = cos LHA/(cosLxtand – sinLxcosLHA)
    where A is true bearing, LHA of sun at the time of observation, d is declination of sun (or moon), L is latitude of observer. Is this correct? An easier way? When I try to make a sketch of the triangle something seems incorrect. Is A the true bearing or more correctly I must make a sketch and allow for quadrant......gotta think about this more.
    I’ll try this later today. Thanks in advance.


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