It's reasonably well know that you can determine your latitude by taking a measurement off the sun. There's even

a page about it in the wiki. Thing is though, for this to work in a flat earth model, light has to do some pretty funky bending.

I'm going to use the circular, north pole centered FE model, with the sun at an altitude of 3,000 miles, as this seems to be the most commonly used model, and all measurements are based on the position of the sun at local midday on the equinox unless stated otherwise.

For an observer on the equator, the sun appears directly overhead. This is fine, it fits in with everything.

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.

For an observer at 90 degrees latitude (the North pole, or the Southern equivalent), the sun appears to be setting (or rising) for the entire day, as it is partially obscured by the horizon. But in the FE model, it's actual position would be ~26.6 degrees above the horizon. Now we have a 26.6 degree shift!

Like I said, some pretty funky bending! Of course, the straight line distances to the sun may have some influence, so I'll leave you with some calculated FE data (for the model used in this example) which someone more mathematically minded can have some fun with.

Latitude: 0, Apparent Angular elevation of sun (AAE for the rest of this data): 90, Actual: 90, Distance: 3,000 miles

Latitude: 22.5, AAE: 67.5, Actual: 63.4, Distance: 3,354.1 miles

Latitude: 45, AAE: 45, Actual: 45, Distance: 4,242.6 miles

Latitude: 67.5, AAE: 22.5, Actual: 33.7, Distance: 5,408.3 miles

Latitude: 90, AAE: 0, Actual: 26.6, Distance: 6,708.2 miles

Oh, wait, that's not all! It's not just latitude shots at midday, but sunsets as well. For someone at the equator, the sun would be about 8485.3 miles away (line of sight) when it appears to set, with an actual angular elevation of 19.5 degrees, which means that light has to change direction a second time! If you head down to 45 degrees South latitude, the sun is 10,816.7 miles away with an actual angular elevation of 15.5 degrees when it appears to set.

If anyone can explain how that works in FET, I look forward to it! Especially if the explanation is supported with working formulas.

EDIT: I just realised, it gets even worse! If you were at the South pole (or the FE equivalent thereof) at midnight on the summer solstice, the sun would be nearly 20,000 miles away, at an actual elevation of 8.8 degrees, but an apparent elevation of 23.5, AND on the opposite side of the North pole, which should be in darkness!

Of course, RET covers all of this quite gracefully and without any convoluted BS explanations.