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Re: Sunrise to sunset azimuth
From: Antoine Couëtte
Date: 2018 Feb 26, 10:24 -0800
From: Antoine Couëtte
Date: 2018 Feb 26, 10:24 -0800
Ah ! Thank you Frank ! I had more or less - rather "less" actually - guessed that you wanted to compute the way the "offset" depends on both Declination and Latitude.
Starting from your equation : sin(Amp) = sin(Dec) / cos(Lat), and with d(U/V) = (V*dU - U*dV)/V², I am getting the following result:
cos(Amp)dAmp = [cos(Dec)dDec*cos(Lat) - sin(Dec)*{-sin(Lat)dLat}]/cos²(Lat). Hence, in order to compute dAmp as a function of dLat, get:
dAmp = [cos(Dec)dDec*cos(Lat) + sin(Dec)*sin(Lat)dLat]/[cos²(Lat)*cos(Amp)]
I have not closely checked whether our results exactly match, but we should be close enough now.
Antoine
