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    Dip and Temperature Gradient
    From: Marcel Tschudin
    Date: 2013 Apr 23, 18:23 +0300

    From Paul’s derivation of dip as a function of difference in air
    refractivity one can also determine the temperature gradient
    corresponding to a certain dip.
    The astronomical refraction values which are generally used compare
    well or may even have been calculated with the standard temperature
    gradient in the troposphere of -6.5 K/km. This may not apply for the
    dip, i.e. for the temperature gradient between height of eye, H, and
    sea level which is generally calculated with the empirical(?) formula
    One may therefore ask which temperature gradient the dip resulting
    from this formula corresponds to.
    The following calculation assumes that the height of eye is also 10 m
    (as in Paul’s examples) and that according to the above equation the
    dip is
    DIPapp = 5.57 moa
    Using the above result and Paul’s equation:
    DIPapp[moa] = 1.926 * sqrt(dn′ + H)
    one obtains
    dn’ = -1.648
    and further, by assuming for the earth the same radius of curvature as
    in Paul’s example with
    dn = dn’ / 6371000
    dn = -2.587e-7 = ((refractive index at eye) - (refractive index at sea))
    At this point it is assumed that the observer has (nautical) standard
    conditions, i.e. T=283.15 K and P=1010 hPa and that the refractivity
    of air (using the online calculator which Paul proposed) is
    Knowing that the refractivity of air is inversely proportional to the
    temperature allows calculating the temperature difference, dT, between
    height of eye and sea level as
    dT = dn/(n-1) * T = -0.260 K
    resulting finally in a temperature gradient, TG, between eye and sea level of:
    TG=-0.0260 K/m
    This means that the dip formula comprises a temperature gradient
    between height of eye and sea level which corresponds (surprisingly
    exact) to four times the standard temperature gradient in the

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