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    Re: refraction
    From: Marcel Tschudin
    Date: 2005 Aug 7, 22:43 +0300

    While doing some home work on terrestrial refraction, I found the following
    which might be of interest to you:
    Paul wrote::
    > 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.
    The following web page
    explains how these approximate formulae are derived. (This very good
    explanation is unfotunatly in German, but I think  twith the help of a
    translation machine one should be able to understand it.) The important to
    note is that the above approximations contain a factor k for terrestrial
    refraction. I do not know whether the Astronomical Almanac mentions that
    this terrestrial factor varies during the day and depending on the whether
    and that therefore the above approximations are only correct for certain
    situations. The above web page indicates the source for the following
    k=0.125 during the day, undisturbed atmosphere, i.e. clear sky and normal
    sight conditions
    k=0.20 for covered sky (day and night) or all year round
    k=0.30 during clear nights
    The web page here http://www.iol.ie/~geniet/eng/refract.htm#Terrestrial
    (this time in English!) mentions the following values for terrestrial
    refraction, this time as K=1/k
    K=4.91 at noon,
    K=10.64 during sun set/rise and night (equinox, wind speed [4 m/sec] (at 10
    m height) and latitude 53°)
    The one of you who used the expression "black magic" was probably wondering
    how one can select the correct value for k ;-)

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