Thanks a lot for your remarks on generalizing the equations for an
oblique plane.
As a general rule, any plane being oblique (ie, inclining and declining)
where you
are is an horizontal one somewhere else. But where?
OK. You can consult Rohr's book or derive'em by yourself (it's pretty
easy), and
you'll get that the 'equivalent latitude' is given by
sin(LatEq) = cos(z)*sin(Lat)-sin(z)*cos(Lat)*cos(D)
and the 'equivalent longitude' is
LongitudeEq = Longitude + arctan( -sin(D) /
(sin(Lat)*cos(D)+cos(Lat)/tan(z)) )
being 'z' the zenithal distance of the plane (what you usually call
'inclination' in English, but
not in Spanish and French because it can be misleading, etc, etc, bla,
bla, bla) and 'D' is its
azimuth.
Now you have to be careful because the hour angle of the Sun in that
place is shifted an angle
given by the increase of longitude and the substyle line does not
coincide with the greatest slope
one, being the angle between them equal to:
SD = arctan(sin(D) / (cos(z)*cos(D)+sin(z)*tan(Lat)) )
Ah, and don't forget to carefully look at the sunrise and sunset times
over the plane and
over the horizon!
These are the classical formulae that perhaps everybody in the list
know, but maybe somebody
didn't have all together. Personally I prefer to use Fer's matricial
method which is not so straightforward
but much crisper and suitable for computers.
Eventually, we have to be VERY careful when computing the inverse
trigonometrical functions, because
we can get scrambled results. This is a general problem in gnomonics
which requires a deeper explanation
in some other day.
Cheers,
Anselmo
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