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Early in the morning I see the contrails of airplanes just above the horizon. How far away is the airplane? For this problem I assume that the airplane is flying at 30,000 feet. From my viewpoint the angle from the horizon is 30 degrees. To me the angle appears to be 30 degrees but with the curvature of earth, the angle will be greater. Is there an approximation I can use such as: If the angle I see is 30 degrees the plane is 13 miles, If the angle I see is 45 degrees the plane is 9 miles, etc.?

Yes there are analyses of apparent altitude vs. angle of observation above the horizon, taking into account the curvature of the Earth. However, things are not so simple at first sight because the atmosphere "bends" the light and another correction for the index of refraction of the atmosphere, in addition to the Earth's curvature, is needed. The derivations require some trigonometry which is too long to reproduce here, but you can find the details at the following web sites:

The topic of "real" vs. "apparent" altitude is problem treated in many contexts -- meteorology, ballistic calculations, astronomy, and navigation -- just to mention a few. Each approaches the analysis a bit differently depending upon the needs of each field, so there is no one simple formula, that I am aware of.

Vince Calder

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