NavList:

Rather than solving it and spoiling the fun, I'll post some steps that I would take to analyze this question. First, let's put an upper limit on the distance. Suppose I have an aircraft at 42,000 feet altitude. What is the maximum distance at which I could see this aircraft if it were sitting right on the visible horizon and assuming my height of eye is negligible? This is just a visibility range question, and it is equivalent to asking how far away is the observed horizon as seen from that high flying aircraft (under the unphysical assumption that we actually could see the horizon from the lower stratosphere!). The basic relationship for those of us who like distances in nautical miles and heights in feet would be:
   distance (n.m.) = k · sqrt[height (feet)].
Note that this relationship can be derived directly from the geometry of a plane right triangle with the additional assumption that the central angle is small which allows us to replace cos(x) by 1-x²/2. In this relationship, k is a constant which is just about 1.065 for the Euclidean approximation which ignores refraction and treats light rays as straight lines. Switching from pure geometry to physics, k is about 1.16 (roughly) for real light rays which are refracted and somewhat "follow" the curvature of the Earth as they travel. But the 1.16 is for light rays that stay quite low in the atmosphere (lowest mile is fair), so we should probably split the difference and use 1.11 for light that is travelling from the lower stratosphere down to an observer on the ground. Doing the math, the upper limit on range would be 218 nautical miles for the unrefracted case. For the refracted case, assuming k=1.11 as a reasonable guess, the upper limit on range would be 227 n.m. It's quite a distance! Note that this problem scales with the square root of the height above ground. So if we had an object nine times higher --maybe a meteor burning up-- the ranges would be three times larger. And that's a bit amazing: when you see a bright meteor low near the horizon that distinctly appears to be falling just beyond the next hill or in your neighbor's backyard, it is probably closer to 500 miles away!

The problem that you have described, Antoine, places the aircraft 0.76° or 45.6' above the true horizon which means that the aircraft is closer to us. Nonetheless we can count on the problem scaling approximately with the square root of the aircraft's altitude so the simple upper limit calculation, above, should be about right. We need a name for this distance when the aircraft is at an arbitrary angular altitude, and there is, in fact, a common name, covering both aircraft and satellites in orbit. It's called the "slant height" or "slant range". So we ask Google... And we find that there are a number of online calculators for this sort of problem in the unrefracted case. Here's one.

I'll stop there.

By the way, I was inspired to think about your puzzle by an impressive contrail at sunset tonight. I didn't see the aircraft that generated it, but it was aligned along the usual track for flights from Europe and points east heading to JFK or sometimes beyond (e.g. Dulles --IAD). I looked up about ten minutes after local sunset, and a dense contrail stretched straight across the sky, just beginning to show signs of decay. It spread slowly, but even 35 minutes after sunset it was still clearly a contrail.

Frank Reed