Speed of the Sun & Setting Height

Just trying to work out how the sun is supposed to set in Flat Earth theory and can't seem to understand it.

Using FE, and basing the circumference of the equator at roughly 25,000m (or if a FEer has a different number Im happy to re-calculate), we can work out that a sun (which is supposedly about 2500 miles up) is travelling at about 1,042mph (25,000 miles in 24 hours).

On the equator at midday the sun will be direcly above you (at the right time of year), so effectively 90 degrees to the surface.

We can then work out that in 6 hours the sun travels about 6,250 miles. Basic trig can tell us that when the sun, 2,500 miles up and 6,250 miles away, will be at an angle of 21.8 degrees up in the air. But this is supposed to be when the sun sets and goes out of view? How does a sun that is 21.8 degrees up in the air drop below the line of sight (0 degrees)?

Additionally a sun that is (apparently) 32 miles in diameter, has a perceived width when directly above (midday). When it sets it still has the same perceived width. Apparently this is due to 'magnification' by the earths atmosphere (?), but for an object 6,250 miles away at 21.8 degrees angle in the air at 6pm (sunset), would have to be over 80 miles in diameter to have the same perceived width as the sun at midday. Thats a magnification of over 2.5 times... How come nothing else is affected by this magnification? Planes, stars etc....

Surely there are holes like this that are too big in FE Theory for people to still believe it?

Magic

Just trying to work out how the sun is supposed to set in Flat Earth theory and can't seem to understand it.

Using FE, and basing the circumference of the equator at roughly 25,000m (or if a FEer has a different number Im happy to re-calculate), we can work out that a sun (which is supposedly about 2500 miles up) is travelling at about 1,042mph (25,000 miles in 24 hours).

On the equator at midday the sun will be direcly above you (at the right time of year), so effectively 90 degrees to the surface.

We can then work out that in 6 hours the sun travels about 6,250 miles. Basic trig can tell us that when the sun, 2,500 miles up and 6,250 miles away, will be at an angle of 21.8 degrees up in the air. But this is supposed to be when the sun sets and goes out of view? How does a sun that is 21.8 degrees up in the air drop below the line of sight (0 degrees)?

Additionally a sun that is (apparently) 32 miles in diameter, has a perceived width when directly above (midday). When it sets it still has the same perceived width. Apparently this is due to 'magnification' by the earths atmosphere (?), but for an object 6,250 miles away at 21.8 degrees angle in the air at 6pm (sunset), would have to be over 80 miles in diameter to have the same perceived width as the sun at midday. Thats a magnification of over 2.5 times... How come nothing else is affected by this magnification? Planes, stars etc....

Surely there are holes like this that are too big in FE Theory for people to still believe it?

Quote from: Empirical on January 22, 2016, 11:23:19 AM

Magic

Fixed.

But yeah. While I will say I don't believe FEers accept that figure for the equator (it doesn't work on a disk, the interior would be far too small) , the question of sunset and why the Sun vanishes bottom-up rather than dwindling out is generally just responded to with a slightly handwaved reference to refraction.