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