A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Equation for dip?
From: Frank Reed CT
Date: 2006 Oct 2, 21:28 EDT
From: Frank Reed CT
Date: 2006 Oct 2, 21:28 EDT
Lu, you wrote: "The surface of the earth, especially seawater, is a conductive medium and hence will bend electromagnetic waves." This is significant for long wavelength radio waves, but totally irrelevant to the behavior of visible light. "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.* The ground wave effect for visible light is a tad more than "almost negligible". It's nil, zilch, nada, certainly below one part in a trillion and I wouldn't be surprised if it's below one part in a trillion trillion (ok, even that's not zilch, but it's certainly irrelevant
). The distance to the visible horizon is different from the Euclidean geometric distance because of the variation in the density of the atmosphere. It gets thinner as you go up, so rays of light necessarily bend downwards allowing us to see beyond the distance to the geometric horizon. This variation in density depends somewhat on the weather. If the temperature in the air rises with height above the water, and rises steeply enough, the bending due to the density gradient can easily be much greater than normal and allow us to see many miles beyond the normal horizon. It's worth nothing that this sort of temperature gradient also promotes the formation of sea fog. So if you're in an area noted for sea fog, you can also expect unusual refraction including anomalous dip and the ability to see far beyond the normal horizon. And you wrote in a footnote: " * 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." Distance to the horizon can be calculated in the same way as dip, range of visibility, distance by angle to the horizon, etc. The last place in the multiplier does indeed jump around in the literature because different sources have made different assumptions about the rate of change in atmospheric density (which itself depends primarily on the lapse rate in temperature). They're trying to give a value for a mean state of the weather. Some mathy details: First: the curvature of the surface of the Earth is 1 minute of arc per nautical mile (this is a direct consequence of the definition of a nautical mile). What this means is if I place a vector on the ground at the equator perfectly horizontal at some point P pointing east and then travel 1 n.m. east and place a vector horizontally on the ground at some point Q, again pointing east, then the angle between those two vectors is 1 minute of arc. The second vector is tilted relative to the first vector by 1 minute of arc. Makes sense, right? Now suppose I shoot a light ray (aim a laser) horizontally at sea level from point P towards point Q. Under average conditions, the curvature of this light ray will be about 0.15 minutes of arc per nautical mile. Because it is proportional to horizontal distance traveled, this curvature can be accounted for by pretending that the light ray does not bend but the Earth is slightly less curved (bigger Earth radius). More generally, when you work out the refraction, you'll find that the curvature of the trajectory of a light ray is c = Q*(alpha0*Re/scaleHt)*(1-h/scaleHt) where alpha0 is the refraction constant for air, approximately 0.000281 at standard Temp/Pressure, Re is the actual radius of the Earth, h is height above sea level, scaleHt is the "scale height" of the atmosphere, and Q is the usual temperature pressure factor: Q=(P/P0)/(T/T0). The scale height is the distance over which the atmospheric density "e-folds" or decreases by a factor of e (base of the natural logs). The scale height is usually about 11 kilometers. A very reasonable range of values to expect for the low level scale height is from 7 to 13 km (corresponding to low-level lapse rates of +5 and -12 degrees C per km respectively). If the lapse rate should happen to be as low as -34.1 degrees C per km, the scale height is infinite and the curvature is zero. That's the case I mentioned in another message where the lowest layer of the atmosphere has constant density, not varying with height. In that case, the refraction is zero and it is just as if there is no air at all. If the lapse rate is positive and rather high, 129.6 degrees C per km, the scale height is about 1.7 km which implies a curvature of 1.0 minutes of arc per nautical mile --equal to the curvature of the Earth's surface. In that case, light rays remain parallel to the Earth's surface and it is "as if" the curvature of the Earth is zero. Flat-earthers rejoice! The dip is zero at all heights (assuming that lapse rate is maintained over the range of heights under consideration). Please note: I'm working from my notes from last January and some of the exact numbers here may be based on slightly different assumptions. The general conclusions are correct though, I think. -FER 42.0N 87.7W, or 41.4N 72.1W. www.HistoricalAtlas.com/lunars --~--~---------~--~----~------------~-------~--~----~ To post to this group, send email to NavList@fer3.com To unsubscribe, send email to NavListfirstname.lastname@example.org -~----------~----~----~----~------~----~------~--~---