Thanks to Jorge Ramalho for pointing out a sign error in my last message. It
should have read as below. I still have the same questions, tho'
Steve
I'm trying to convert an azimuth value into an hour-angle, and would
appreciate some help with the formula and with my algebra. Using
dcl=declination, lat=latitude, azi=azimuth,ha=hour-angle, I started with a
standard formula for azimuth given hour-angle:
tan(azi) = sin(ha)/(sin(lat)*cos(ha) - tan(dcl)*cos(lat))
After multiplying out and rearranging, I got:
sin(lat)*tan(azi)*cos(ha) - sin(ha) = tan(azi)*tan(dcl)*cos(lat)
Substituting into a standard formula from my old notes from school
{solutions of a*cos(x)+b*sin(x)=c}, this means my solution is:
ha = arctan(-1 / sin(lat)*tan(azi)) + - arccos( tan(azi)*tan(dcl)*cos(lat)
/ sqrt( sin(lat)^2 * tan(azi)^2 +1 ))
where + - means plus or minus
Questions:
Is this the correct solution?
If so, how should I interpret the plus or minus element of the solution.
Surely for any given azimuth-latitude-declination there is only one possible
hour-angle, not the two implied by the plus or minus option?
Many thanks, Steve
