I have a dumb question. How do you know things like this?
For an observer at 22.5 degrees latitude, the sun should appear to have an angular elevation, relative to the horizon, of 67.5 degrees for the latitude shot to work, which it does. But, in the FE model, it's actual angular elevation is ~63.4 degrees. Okay, so maybe we can explain the 4.1 degree difference with bendy light. Let's continue.
For an observer at 45 degrees latitude, the sun appears to have an angular elevation of 45 degrees. Again, as it should be. Um, except that this means that light is bending back again from it's 4.1 degree deviation before. Hmm. Never mind, push on.
For an observer at 67.5 degrees latitude, the sun appears to have an angular elevation of 22.5 degrees above the horizon. In the FE model, it's actual angular elevation would be ~33.7 degrees above the horizon, so now we have an 11.2 degree deviation the other way. WTF? Anyway, let's finish the data set.
Is there a computer program you use, or a table in a book, or do you travel with a sextant a great deal?
G'day odes!
Okay, for the 'apparent' elevations of the sun, I'm going off the fact that sextants have been in use for centuries with great success, so we can safely say that these 'apparent' elevations are correct. For the 'flat earth' elevations, I am doing some trigonometry to work out the angles. Not too hard really, you just need the supposed altitude of the sun (3000 miles in this case), and your horizontal distance from a point directly beneath the sun, remembering that it is about 6000 miles from the equator to either pole, and 90 degrees for the same (statute miles that is, with about 66.7 miles per degree). To work out the angle it should be (at the equinox), use this equation:
angle = tan
-1 (altitude / horizontal distance)
So for my latitude (30 degrees South, or 2000 miles from the equator) you get:
angle = tan
-1 (3000 / 2000) = 56.3 degrees.
I actually went out on the equinox at local midday and measured the sun's apparent elevation, and it was exactly 60 degrees above the horizon, so I have personally verified for at least one latitude!
Note: for times of year other than the equinox, there is a little more maths involved in working out where the sun should appear to be. I can't remember how to work it out right now, but it's not overly complicated. It does only work for a round earth though.