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    Re: Equation for dip?
    From: Lu Abel
    Date: 2006 Oct 01, 09:50 -0700

    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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