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Re: Determing the bearing off a celestial object.
From: Bruce J. Pennino
Date: 2015 Mar 31, 09:38 -0400
Bruce
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.
Bruce
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