On March 21-22 the sun is directly overhead at the equator and appears 45 degrees above the horizon at 45 degrees north and south latitude. As the angle of sun above the earth at the equator is 90 degrees while it is 45 degrees at 45 degrees north or south latitude, it follows that the angle at the sun between the vertical from the horizon and the line from the observers at 45 degrees north and south must also be 45 degrees. The result is two right angled triangles with legs of equal length. The distance between the equator and the points at 45 degrees north or south is approximately 3,000 miles. Ergo, the sun would be an equal distance above the equator.
How do you refute that?
Easy. Watch me.
According to the FAQ, the diameter of the flat earth is given as 24,900 miles. Divide that by 2 to get a radius of 12,450 miles. Now, I'm guessing is that lines of latitude work pretty much the same in FET as in RET. That is, there are 180 degrees of latitude in the form of concentric circles uniformly spaced (please let me know if I'm wrong about this). Anyway, a 12,450 mile radius divided by 180 degrees of latitude gives us 69.166666667 miles per degree of latitude. If the sun is directly over the equator at noon then an observer at 45 degrees latitude will observe the sun to be said 45 degrees above the horizon forming a right triangle. The distance from the equator to 45 degrees latitude is 45 degrees times 69.166666667 miles per degree giving us 3112.5 miles. Close to enough to Tom's 3000 mile figure that I'm not going to worry about it. Now, we we multiply that 3112.5 mile distance by the tangent of the 45 degree angle to the sun and ... Drum roll please.... we get a distance to the sun of 3112.5 miles. Woo hoo!! Another win for FE!!
Or is it?
So, what do you suppose happens if the observer is at some other latitude? Should be very close to the same result, shouldn't it? Well, let's try it and see what happens.
At 50 degrees latitude, we are 50 degrees times 69.166666667 miles per degree putting us 3458.3333333 miles from the equator and we get a 40 degree angle to the sun. Do the math and the distance to the sun comes out to... Hmmm... That can't be right. It comes out to 2901.9 miles.
Let's try that again at 60 degrees latitude. Hmmm, 60 degrees is 4150 miles times tangent of 30 degrees comes out to... 2396 miles? How can this be?
OK, how about 30 degees? 2075 miles times tangent of 60 degrees is 3594 miles.
As it turns out, the farther from the equator you go, the closer the sun is to the flat earth. In fact, the sun is only 107.5 miles away when you are at 89 degrees latitude. On the other hand, at 1 degree from the equator, the sun is 3962.5 miles above the flat earth.
So Tom, are you sure that this is the right way of measuring the distance to the sun?