Gordon said: > > From Jean Meeus's "Astronomical Algorithms," 1951, p. 89: > > > > tan H = sin A / (cos A * cos f + tan h * cos f) > > > > where H is hour angle, A is azimuth, f is latitude and h is altitude, > > and H is obtained via the arctangent.
to which Bill said: > > But note that in Steve's problem the altitude (h) is not given. To calculate > the Altitude, you will need to account for declination and in the end that > will yeild 2 (and occasionally 1 or none) possible answers at a given azimuth. For practical purposes, lets say I know what day I'm doing my conversion so I know declination. If I then add a further constraint that I'm located outside the tropics, I believe it always comes down to a one-to-one relationship between azimuth and hour angle. My Northern Hemisphere thinking is that the sun starts in the East (approximately) at dawn, moves through South and on to the West for dusk, then sneaks on round past North during the night and back to the East for the start of the next day. And if I'm in the Arctic in the summer, it doesn't even bother to sink below the horison while it's passing North. Finally, Richard Koolish found a useful web page which gives a nice general background to the various relevant co-ordinate systems and converting between them http://www.seds.org/~spider/spider/ScholarX/coords.html I've had plenty of good information to answer my question, and to see that my own initial attempt at a solution gave the wrong answer. I learn as I go! Thanks, Steve
