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    Re: Dip: Astronomical Almanac vs. Nautical Almanac
    From: Brad Morris
    Date: 2013 Apr 21, 19:13 -0400

    Hi Marcel

    From my reading, D(H)=1.75sqrt(H,m) for the dip of the horizon.

    When computing the additional refraction of a celestial object on the horizon, for rise or set, one needs an additional refraction correction of R(H)=.37sqrt(H,m).

    Note the equation later on that shows A-D(H)-R(H), showing that the sign is negative.

    I believe they are showing how to compute astronomical refraction for a body on the horizon (rise/set) as compared to only the terrestrial refraction, dip.  If I am not mistaken, we discussed this point in a previous communication.

    Best Regards

    On Apr 21, 2013 5:48 PM, "Marcel Tschudin" <marcel.e.tschudin@gmail.com> wrote:

    While searching in the Internet something completely different I dropt
    on an excerpt of the "Explanatory Supplement to the Astronomical
    Almanac" containing "9.331 The Effects of Dip and Refraction". Reading
    its content here
    I notice that dip appears to differ to how I understand it is used in
    the Nautical Almanac.
    Nautical Almanac
    It mentions calculating the dip consisting (to my understanding) of
    the geometrical angle and *including* the effects of terrestrial
    refraction as
    Astronomical Almanac
    Here almost the same equation as in the Nautical Almanac is mentioned
    but for what appears to calculate only the geometrical dip:
    The terrestrial refraction appears then to be considered seperately by
    *adding* its contribution
    to the geometrical part and resulting in what appears to be the total
    (or apparent) value
    DIP[moa]=2.12*sqrt(H[m]) = DIPgeom + REFdip
    Where could the misunderstanding be for
    1) Almost the identical factor for calculating dip with and without
    terrestrial refraction? Do the two Almanac treat the dip indeed so
    much different?
    2) The terrestrial refraction being added whereas to my understanding
    it reduces the geometrical dip and should therefore be subtracted?
    Any ideas?

    View and reply to this message: http://fer3.com/arc/m2.aspx?i=123637

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