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

**Re: Q: how to calculate refraction at higher altitudes on land?**

**From:**George Huxtable

**Date:**2002 Feb 28, 19:20 +0000

In reply to a question from Dan Allen, Dov Kruger said- >But at the lower angles, >the normal correction for pressure will presumably be too great, because >the reason the pressure is low is that you are high, not because your >whole region is experiencing low pressure. > >In the worst case, consider you are looking down at the horizon. Near >the horizon, your line of sight is passing through sea-level air. >Halfway, it is passing through air at half your altitude. Since the >correction is small in any case, why not just try to divide it in half >and use that? You know the upper bound (no pressure correction) and the >lower bound (full pressure correction) so you know exactly how bad your >assumption can be. ================== Comment from George Huxtable- I am not quite sure what Dov is getting at here, so if I have got the wrong end of the stick, I hope he will correct me. I take it that Dov is referring to the angle of dip that is relevant to the view of the horizon from a great height, such that the refraction at that height is significantly less than at sea level: and how to correct the dip to allow for that change in refraction. Have I understood him right? Dip is made up from two components, which both vary with the square root of the elevation of the observer. The dominant part is caused by the curvature of the surface of the sea. It is entirely geometrical and it is quite unaffected by refraction. Its magnitude is- 1.063 root h, in arc-minutes, if h is in feet. The other component arises from curvature of the light ray between the horizon and the eye, caused by refraction due to the density-gradient of the air along that path. If the air has its normal sea-level pressure all along that path, the magnitude of the refraction component of dip is- -.083 root h, in the same units. It is negative, working in the opposite direction to the curvature of the sea, so the net result is- total dip = 0.98 root h arc-minutes, h in feet. Dov Kruger is, I think, assessing the effect on the dip of the fact that, when the observer is high, the refraction is not constant all along the light-path, but varies significantly because of the changing altitude. It is true that there is such an effect, but it is a very small one. Take an observer's altitude of 1000 feet, for example, at which height the dip would be about 30 arc-minutes, and the pressure, and so the refraction, would be down to about 97% of the sea-level value. If the pressure changed at a uniform rate along the light-path (which it doesn't, but that would provide a worst-case value) then we could take, as Dov suggests, an average value of the pressure, and the refraction, at the mid-point of the light-path, which would show just half that reduction, to 98.5% of the sea-level value. So this would reduce the refraction contribution to the dip to 98.5% of its previous value, from -.083 root h, to -.0818 root h. And the total dip will then increase from .98 root h to .9812 root h, that is, by about 1 part in 1000 of the value given by the standard formula. Of course, that contribution would be greater at even higher altitudes. The point that I am making here is that because only a small part of the dip is refraction-dependent, the dip as a whole varies very little with the atmospheric pressure at the observer. If I have completely missed the point, I hope someone will put me right. George Huxtable. ------------------------------ george---.u-net.com George Huxtable, 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK. Tel. 01865 820222 or (int.) +44 1865 820222. ------------------------------