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From: Paul Hirose (no email)
Date: Fri Aug 05 2005 - 15:56:25 EDT
Marcel E. Tschudin wrote:
>
> 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 problem is similar to a rise/set calculation, isn't it? That is, you
want to know the altitude the object would have if refraction were
turned off and you could see through the Earth. That equals dip of the
horizon plus the total curvature due to refraction of the light from the
object to the observer.
Imagine a theodolite at the summit of say a 100 meter mountain
overlooking the sea. A star is precisely on the horizon. Dip of the
horizon at H meters high is about 1.75′√H, so in this example the
telescope must be 17.5′ below horizontal to center the star and horizon
in the crosshairs. That sets it parallel to the arriving light rays at
the theodolite.
However, it is not parallel to the rays at the distant point where
they're tangent to the sea. To make it so, the scope has to be tilted
down a little more, by the amount of refraction between the tangent
point and the observer. According to the Explanatory Supplement to the
Astronomical Almanac, it equals about .37′√H for H in meters, or 3.7′ in
this example.
Finally, you tilt the scope down still more to allow for the refraction
between the tangent point and the celestial body. This is simply the
horizontal refraction, about 34′ in standard conditions.
You end up with the telescope 55′ below horizontal. It is now parallel
to the rays from the star before they enter the atmosphere.
The 1.75′√H and .37′√H terms can be combined, in which case the
depression angle expression is -34′ - 2.12′√H.
I don't think that will be accurate at great height, though. For
example, the .37′√H term for refraction between horizon and observer can
increase without bound. In reality, it should never exceed the
horizontal refraction (34′).
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