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## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**Re: Refraction**

**From:**George Huxtable

**Date:**2005 Aug 7, 21:06 +0100

At 12:32 07/08/2005, Marcel wrote- >With homework I mean TRYING to calculate refraction for negative altitudes >using the following approach: > > > >1) Calculating for an observer at a given elevation (a.s.l.), a given >(negative) altitude (but e.g. still above the apparent horizon), a given >temperature and given pressure the tangent point on a line above and >parallel to the earth's curvature at sea level, this using calculation of >terrestrial refraction. I don' t really understand the distinction that Marcel makes between "terrestrial" and "astronomical" refraction. It's all refraction, integrated along different parts of the light path. But I agree, this is the first, and perhaps most difficult, step in the procedure. >2) Calculating for this tangent point the local temperature and pressure, >this using the standard atmosphere model. Yes. But the temperature and pressure are unimportant in themselves, all that's needed is the resulting air density at that altitude. >3) Adding to the terrestrial refraction the astronomical refraction for an >altitude of 0? corrected with temperature and pressure at the tangent point. I suggest doing step 3 rather differently, as follows- A ray of light passing horizontally through Marcel's tangent point undergoes a known refraction, on its way from outer space. If the tangent was at sea level this would be 29 arc-minutes if Saemundsson's formula is correct. At a higher altitude for the tangent point, it would need adjusting by the ratio of air density at that altitude to air density at sea level (call this factor D1). That adjustment factor should be roughly correct if the tangent point is somewhere in the lower atmosphere, but that may not be so at higher levels. If that ray of light then carries on to disappear into outer space again, then (from symmetry) it would undergo a further deflection of 29 D1 minutes, so in all 58 D1 minutes caused by the Earth. This is the total refraction the Earth would cause to that ray, as viewed by an observer far from the Earth. But Marcel's observer is not far from the Earth. He is at some altitude within the atmosphere. So the refraction he observes will be 58 D1 minutes, less the further refraction that would have occurred if that ray, instead of meeting the his eye, had carried on into space. If the observer is at a height where the density relative to sea level is D2, and he was looking at a negative angle of A degrees, what is needed is the refraction he would see if looking the other way, at a star at a POSITIVE apparent angle of 4 degrees, from a layer where the air density (relative to that at sea level) is D2. This is a familiar problem, with a well-known answer. Then that refraction needs to be subtracted from 58 D1 minutes. >Comparing the results obtained this way with table 6 of the Air Almanac may >lead to further, but now more specific questions. It seems that the table in the air almanac deals with very broad bands of observer's altitude, so is unlikely to provide very precise information. It may be good enough for Marcel's purposes, however. To get the answer to step 1, it may help if I quote from Young's "Sunset Science" article, referred to previously, that along the path of a light ray, the product nRsinz is invariant, where n is the local refractive index, R is the radius from the centre of the Earth, and z is the local zenith distance of the ray. I only quote, please don't ask me to justify or interpret that rule. George. =============================================================== Contact George at george---.u-net.com ,or by phone +44 1865 820222, or from within UK 01865 820222. Or by post- George Huxtable, 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.