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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

DIPapp[moa]=1.76*sqrt(H[m])

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
also
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
correspondingly
(n-1)=2.81622e-4
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
troposphere.

Marcel

   

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