John & Richard: Whatr a sloppy mess my posting was, where I told the solution for h.
This is a better way to present it, and a neat, convenient way to evaluate the solution: The quadratic formula: [-b +/- sqrt (b^2 - 4ac] / 2a -------------------------- a,b, & c, for this solution: a = sin^2 Lat + 1/tan^2 Az b = 2 tan dec cos Lat / tan Az c = tan^2dec cos^2 Lat - sin^2 Lat ----------------------------- By the quadratic formula: sin h = [-b +/- sqrt(b^2 - 4ac] / 2a -------------------------------- Constraints on h: If intended azimuth is west of the north-south line, then h and its sine are positive If intended aziuth is east of the north-south line, then h and its sine are negative. If intended azimuth is south of the east-west line, then cos h > tan dec / tan Lat. If intended azimuth is north of the east-west line, then cos h < tan dec / tan Lat. -------------------------------------- Michael Ossipoff If intened azimuth is north of the east-west line, then cos h < tan dec
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