Re: Query about solar azimuth formula

[brian whatcott <inet@intellisys.net>]

I see that Paul's Sun Altitude agrees with Margaret's.
But his Sun Azimuth does not account for time of day.
Perhaps a term is missing?

Brian

At 10:05 11/3/97 EST, Paul Camp wrote:
> This is the coordinate system transformation taken from Peter 
> Duffet-Smith's book Practical Astronomy with your Calculator. To go 
> from equatorial to horizon coordinates:
>
> sinA = sinD sinL + cosD cosL cosh
> cosB = (sinD - sinL sinA)/(cosL cosA
>
> This gives you both solar altitude and solar azimuth (I think I 
> translated from Duffet-Smith's notation correctly)
>
> > Paul J. Camp 

[Margaret]
 calculate A from -
> >         sinA = sinLsinD + cosLcosDcosh
> >
> > where   L = latitude of observer
> >         D = solar declination
> >         h = solar hour angle = (t-12) x pi/12  (in radians)
> >         and t = mean solar time in hours.
> >
> >
> >         (1)  B = -arcsin(-cosDsinh/cosA)
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brian whatcott <inet@intellisys.net>
Altus OK