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Yes. That's how it works. It's worth drawing yourself a triangle to see that the angles would be identical if light travelled in Euclidean straight lines. Try it out:

• draw a circle for the earth with a 4-inch radius (implying a scale of roughly one inch to 1000 miles). Label its center, C.
• Place a dot marking a point, P, at an altitude of 500 miles or half an inch above the circle.
• Draw a line, CP, from the center of the earth to that dot.
• Mark the point where the line CP crosses the Earth's surface (the circle) as P'. It's the point on the ground directly below the obsserver at P.
• Draw another line passing through point P and perpendicular to CP. This represents the true horizon at that point. It's 90° from the zenith for the observer at P. Mark the end of that line H₀.
• Now draw a horizon-skimming line from P that just barely touches the surface of the earth. This represents the viewline of an observer at the dot seeing the sea horizon far off in the distance.
• Mark the point where this last line just touches the earth as H. It represents the "visible" or "sea" horizon.
• Draw a line CH from the center of the earth to the point H.
• Key geometry: notice that the line CH is perpendicular to the line PH. We have a right triangle!
• Now notice: the angle PCH between the line CP and the line CH is exactly proportional to distance across the surface of the Earth from P' to H. In fact, if we measure distances on the earth's surface in terms of angles at the earth's center, then they are identical: angle PCH = distance P'H. And of course that's just what we do: if the angle is in minutes of arc, then the distance is in nautical miles.
• Next the dip. The dip angle in this diagram is the angle between the true horizon line PH₀ and the visual horizon linne PH. We can call that angle HPH₀.
• More key geometry: dig out your Euclidean theorems and convince yourself that the anglle at the earth's center, PCH is identical to the dip angle HPH₀.

With that simple diagram, you have proved that dip and horizon distance are identical (that is the horizon distance in nautical miles is identical to the dip in minutes of arc). And if you apply some simple trig to the right-angled triangle, and thus use a small angle approximation [cos(x) = 1 - (1/2)x²], you get the dip equation. Oh, but rats! It doesn't work. Where's that 20% difference? We know they shouldn't be identical!

The 20% difference between dip and horizon distance is caused by refraction. Light rays don't travel in straight lines. Fortunately, nearly horizontal rays of light do something very simple in the lower atmosphere (this does not apply to altitudes more than a few thousand feet). They bend downward slowly as they travel. A light ray that starts out exactly horizontal will curve slowly downward at a rate k which is conveniently measured in minutes of arc of bending per nautical mile. On average, at sea, the downward rotation of light rays, k, is about 0.15' per nautical mile. This is completely identical mathematically (in these circumstances of low altitude nearly horizontal light rays) to a change in the effective radius of curvature of the earth. So we can fix up our equations just by changing the radius of the earth. And that's exactly what we do to get the proper dip and horizon distance equations. You replace R, the actual radius of the Earth, with R/(1-k). And then everything works out beautifully. You find that dip is reduced and distance to the horizon is increased by a proportional amount leading to that 20% separation between the two.

But what if the weather conditions are funny? That rotation of light rays, k, is only an average value. It is really is the norm, but if the temperature profile of the air near sea level changes a bit, it's quite possible for k to be different. It's not unusual for k to 0.3' per nautical mile. In that case, the dip is reduced a bit more and the distance to the horizon is extended ever further. It's also possible, with a temperature inversion, for that refraction number to reach unity: k=1.0' per nautical mile. Under those conditions, light rays follow the earth's surface, neither rising nor following, just orbiting around until they are absorbed by the "haze" of distance. Under those conditions, the dip is reduced to near zero, and the distance to the horizon become undefined --infinite. And thus the earth is flat (but only visually!).

Frank Reed

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