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Re: Derivation of the time-azimuth formula
From: Lars Bergman
Date: 2024 Oct 29, 06:58 -0700
From: Lars Bergman
Date: 2024 Oct 29, 06:58 -0700
Pat, you asked: "Is it derived from:
(1) Sin(Hc) =(Cos(LHA)*Cos(Lat)*Cos(Dec))+(Sin(Lat)*Sin(Dec))
(2) Cos (Z) = (Sin(Dec)-(Sin(Lat)*Sin(Hc))/(Cos(Lat)*Cos(Hc))"
It could be derived from those equations. Replace Sin(Hc) in (2) with the right-hand side of (1). Then, after some manipulation,
noting that Sin2(Lat)=1-Cos2(Lat), you divide by Cos(Lat)·cos(Dec) and noting that Cos(Hc)/Cos(Dec)=Sin(LHA)/Sin(Z), you get the formula.
I don't know anything of its origin.
Lars
