# NavList:

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

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Re: Smart and Shearme Position Line Tables
From: David Pike
Date: 2020 May 5, 03:52 -0700

Lars Bergman you wrote:  Now a new try on the formula that David C asked about. How to find azimuth when knowing latitude, altitude and hour angle?

cos(azimuth - x) = cot(latitude)·tan(altitude)·cos(x),  where tan(x) = csc(latitude)·cot(hour angle)

or, using the more common trig functions

cos(azimuth - x) = tan(altitude)·cos(x) / tan(latitude),  where tan(x) = 1 / [sin(latitude)·tan(hour angle)]

That spherical triangle is uniquely defined.

Lars
That’s a very clever formula, which I’m struggling to make sense of at the moment, but I’m sure it’s correct.  Could you possibly provide a labelled diagram or explain a bit more about X.  Also, there’s still one point that’s troubling me.  Ignoring the maths, if you take a fixed length piece of string to represent zenith distance and a globe.  If you place one end of the string on the observers meridian at the appropriate latitude and swing it around the earth’s surface, the other end will cut the Sun/body’s meridian at two places, so the body will have two possible azimuths, unless the touch just happens to be tangential, when the angle opposite co-lat will be 90°.  One point might be highly unlikely, or impossible to observe, but there will be two points.  DaveP

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