# 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: Bruce J. Pennino
Date: 2015 Apr 2, 23:18 -0400
Hello All:

Frank , something has been missed here.  Let me repeat. I was curious to learn the technique to calculate Z(true) without knowing H. Then be able to check a  hand compass, which was the topic. I saw the Tan A  equation and thought that  it was the simplest  single step manner to directly calculate A, of course knowing my location. I never mentioned “old methods”. Use one equation , simple, I thought. As you wrote, I incorrectly submitted the  equation (should have morning coffee before writing trig, with or without x). Sorry about equation error. I wrote  “Easier way?” because there usually is an easier way.  After I did several  trials with TanA  equation I realize there seems to be a “ quadrant” or sign issue, even if a sketch is made.  I’m sure someone will explain or correct me.

My conclusion is that you are correct.  With a known location, time, declination, etc.,  simply calculate H via

Sin H = sinLsind + cosLcosdcosLHA then calculate Cos A  with standard equation .

Use fool  proof rule “ If LHA LT 180     Zn = 360 – A or LHA GT 180    Zn = A “.  For compass check, must add or subtract local magnetic variation, staying away from transmission lines, pipelines,etc.  So , using an app, or equations, or other methods, now I and others know how to do it.

Bruce

From: Frank Reed
Sent: Tuesday, March 31, 2015 2:31 PM
Subject: [NavList] Re: Determing the bearing off a celestial object.

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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