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    Re: Equation for dip?
    From: D Winchurch
    Date: 2006 Oct 02, 08:45 -0400

    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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