Next message: Fred Hebard: "Re: Refraction"
Marcel,
I am still wondering under what conditions you would get a negative
elevation.
If it's for locating yourself on land, it seems hopeless: If you're at
an unknown location in a valley, and trying to use surrounding
mountains as an indication of horizon, you wouldn't know your height
above sea level to begin with, so wouldn't know how much higher the
mountains were than you, not to mention not knowing how far away they
were. Also, the height of mountains varies, so you would have to know
the precise spot on the mountain below the celestial object, to
determine its height above sea level at that point. It seems hopeless
to me. Better to use an artificial horizon or a plumb bob.
Fred
On Aug 4, 2005, at 12:20 PM, Marcel E. Tschudin wrote:
> Fred,
>
> Yes, how do I get in this? Just trying to cover in a self-made program
> the
> situation from an object at the horizon (over sea level) as seen from a
> mountain or air craft.
>
> The "real" calculation is done via integration. But since this is not
> very
> practical one uses approximative formulae like e.g. the one from
> Bennett
> which Meeus mentions in his book Astronomical Algorithms. All tables on
> refraction I found so far do end at 0° elevation and for none of the
> approximative formulae I could find an indication that they also would
> be
> valid for negative elevations.
>
> I also was wandering whether the approximate formulae could be used by
> calculating the Refraction R for e.g. -2° the follwing way:
>
> R(-2°) = R(0°) + ( R(0°) - R(+2°) )
>
> If this would be correct then one would not need separate formula for
> negative elevations.
>
> Greetings from Marcel
>
