A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Lars Bergman
Date: 2015 May 16, 13:37 -0700
Greg wrote: "If Lars is watching then maybe he can work up something that will look good."
I think the following at least looks good:
hav LHA = (hav z - N)/(1 - Q)
where N=hav(lat-dec), P=hav(lat+dec) and Q=N+P
z is the zenith distance 90°-altitude
Like the azimuth formula, the division by 1-Q is the tricky part. If Q is small, then an approximation is to multiply with 1+Q instead.
Lars, 59N 18E