# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

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Re: Determing the bearing off a celestial object.
From: Frank Reed
Date: 2015 Mar 31, 11:23 -0700

Bruce, you wrote:
"At any time, how do you calculate the true bearing of the sun , or any celestial body, without H?"

But you don't have to worry about that! If you know your position and time, which you certainly do in the modern world, then you can instantly calculate the altitude, H, and then use the standard equation, which is dependent on H, to calculate the azimuth. The problem with most thinking about celestial azimuths today is that we forget this simple reality --we know our position. You don't have to go back to 1950 methodologies where calculation is expensive and position is uncertain (let alone go back to 1800 methodologies where "amplitude sights" are required) in order to use the Sun for azimuth. Whenever it is low in the sky, the Sun's azimuth can be instantly available for an exact compass check via an app or some other live software tool. The azimuth can be calculated for any altitude, low or not, but as folks have already pointed out, it's difficult to use the azimuth when the Sun is high without a specially designed instrument. The azimuths of the Sun and other celestial bodies are the ultimate gold standard for azimuth.

Regarding the equation you gave as
"Tan A = cos LHA/(cosLxtand – sinLxcosLHA)",

tan A = sin LHA / (cos L · tan d - sin L · cos LHA)

or for a more "computer-ese" version:
tan(A)=sin(LHA)/[cos(L)*tan(d)-sin(L)*cos(LHA)].

Note that there's an easy way to see that this has to be the case by considering an observer at the north pole.
If latitude is 90°, then the equation reduces to:
tan A = sin LHA / (0 · tan d - 1 · cos LHA) or
tan A = sin LHA / cos LHA = tan LHA or
A = LHA
which makes sense, right?

Frank Reed
Conanicut Island USA

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