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    Re: Equation for dip?
    From: George Huxtable
    Date: 2006 Oct 1, 15:32 +0100

    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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