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Interesting discussion.
On the topic, sort of, at the campus where I teach there sis a very
large dish antenna.  I notice that when I turn of my GPS it occasionally
give erroneous bearings to  a way point.  The cell phone also does not
always connect.

Would these phenomena be related to interference from the dish antenna?

Lu Abel wrote:
> Electromagnetic waves, whether radio, radar, or light, tend to "cling"
> to conductive surfaces.   The effect is often likened to two wheels on
> an axle, where the one wheel is on a hard surface and the other in mud
> or sand.  As the axle makes forward progress it also rotates towards the
> "clingy" medium.
>
> The surface of the earth, especially seawater, is a conductive medium
> and hence will bend electromagnetic waves.
>
> This effect is inversely proportional to the frequency of the
> electromagnetic wave.  Thus light bends very little and the visual
> horizon is an almost negligible amount beyond the geometric horizon.*
> Radar bends more, hence the "radar horizon" is farther away than the
> visual horizon.  Visual horizon is usually given as 1.17 * sqrt (he),
> radar horizon as 1.22 * sqrt (he).**   At sufficiently low frequencies,
> radio waves cling well enough that they can travel long distances
> hugging the surface of the earth.  Examples are the 100KHz frequency of
> Loran-C and the even lower 10KHz frequency of the discontinued Omega
> system, both of which rely on these "ground waves" for accurate positioning.
>
> Lu Abel
>
> * Before one of the more erudite people on the list jumps on me, I'm
> fully aware that light IS refracted near the horizon and we DO apply
> refraction corrections when taking low-angle sights; distance to the
> horizon is a much cruder calculation.  In fact, I believe the last place
> in the 1.17 multiplier has jumped around as people have argued about the
> effect of atmospheric refraction.
>
> ** The four percent greater distance to the radar horizon (1.17 vs 1.22)
> is reasonably consistent with Jim van Zandt's equations for dip
> (remembering that there is a square root involved in the dip equations)
>
> George Huxtable wrote:
>> 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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