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Jim van Zandt compared the "four-thirds earth model", for dip, used in
radar analysis, i.e.

    d = sqrt(2*h/(1.33*Re))

with the standard Nautical almanac formula

    d = sqrt(2*h/(1.2*Re))

which is a "six-fifths earth model", i.e. less refraction, and asked-

I wonder
whether that's because of the different wavelength, or different
typical heights (so different temperature and pressure profiles), or
what?
=====================

I'm not familiar with such radar calculations, but it seems reasonable
that there should be some important differences.

1. A significant part of the dip is the effect of refraction. No doubt
the refraction of radar depends on air density, in just the same way
as does the refraction of light, but the wavelengths are very
different. There's no reason why the refractive index of air should be
the same at radar wavelength as it is at optical wavelength. After
all, it varies significantly between red light and green, as we know.
However, I don't know the relative values; perhaps Jim does.

2. Because of the longer wavelength, no doubt diffraction effects come
into play, which allow wave energy to travel around the Earth's
curvature to some extent, whereas light has such a short wavelength
that is strictly limited to a direct (though curved) optical path
above the surface. So for radar, there will be a less-shrp cutoff, by
the horizon.

3. Dip depends on the bending of light in the very lowest layer of
atmosphere, between the observer's eye and sea level. Presumably the
standard value for dip assumes a reasonable value for temperature
gradient, for that layer. Radar signals are presumably coming from a
somewhat higher level, from a mast rather than from a ship's bridge,
and in the limiting case would then be skimming over a horizon to the
target, at the height of a ship's hull, say. However, the formula from
which standard optical dip values are taken covers a wide range of
heights-of-eye, so I would not expect that to be a relevant factor.
Assuming the effective temperature gradient to be similar, over those
different ranges, is probably good enough.

In addition, there's another difference.  Radar signals are having to
travel outwards, around the curve of the Earth, and then the
reflections have to return, again around the curve of the Earth. So
any horizon factor comes in both times, but I would expect the
resulting double attenuation to be taken care of elsewhere in the
maths.

===================

Frank Reed's restricted his comments to the effect of temperature
gradients at visible wavelengths; there will indeed be similar effects
at radar wavelegths, when the different refractive index is allowed
for.

When he related dip to the lapse rate (the rate of temperature
variation) in the lower atmosphere, he was of course quite correct.
However, it needs to be understood that the lower atmosphere being
considered is the EXTREME lower atmosphere, within a few tens of feet
of the sea surface; and not the lower atmosphere meteorologists
generally have in mind when speaking of lapse rates, which has a much
greater span of height.

George.


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