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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