Re: Query about solar azimuth formula

[brian whatcott <inet@intellisys.net>]

Paul's hour angle explicitly enters the calculation of Sun Altitude.
Sun Altitude enters his azimuth calculation.

The Sun has the same altitude at two azimuths, except at local noon,
so I fancy Paul's azimuth calculation is in error.
 But I could be corrected by even a single worked counter-example.
Say latitude N45 deg, t=14 hours, declination = 0?

Humbly
Brian

At 14:26 11/7/97 EST, you wrote:
> That would be accounted for by the hour angle of the Sun.
>
> > 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.

brian whatcott <inet@intellisys.net>
Altus OK