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 fromula from my old notes from school, this solves as: 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
