NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Equation for dip?
From: George Huxtable
Date: 2006 Oct 1, 15:32 +0100
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. --~--~---------~--~----~------------~-------~--~----~ To post to this group, send email to NavList@fer3.com To , send email to NavList-@fer3.com -~----------~----~----~----~------~----~------~--~---