Yet the fact remains that the sun disappears from view by dropping below the horizon, not from fading to a blur through the "atmoplane".
Do you have any evidence to support your claim?
Really? You have never watched a sunset?
Do you need me to get you a video of one?
Then do you have an explanation for them?
An explanation for what?
Go back and read my comment, in context. It is clear what I am referring to.
Using trigonometry, that means it would come into view at above a 20 degree angle. Having seen the Sun descend to below zero degrees (below the horizon) over the ocean many times, I know that is false.
Please provide evidence for your claim.
What? Don't you understand simple trigonometry?
The numbers vary depending upon exactly where you are, so I will use the equator, on the equinox, with a FE model with the equator having a radius of 10 000 km and the sun being 5000 km high.
The sun rises/sets when the sun is roughly 1/4 of the way around Earth on the equator. (This is because it goes over the equator and there is roughly 12 hours of day and 12 hours of night with the sun directly overhead mid-way through the day, and the sun "moves" at a constant speed in the FE model. This means the sun takes 24 hours to do a complete circle and 6 hours to go from sunrise to directly overhead and then another 6 to go to sunset).
Using Pythagoras (you can use trig if you want), noting that 1/4 of a revolution is 90 degrees, we have a right angle triangle with the 2 short sides being 10 000 km long. The right angle is at the north pole (which the sun circles), one short side goes to the equator where the observer is, the other goes to where the sun is.
That places the sun at a distance of sqrt(10000^2+10000^2) km, or roughly 14 142 km.
Now, the sun is 5000 km above that point on the equator, and thus we set up another right angle triangle.
This time one side goes between the observer and the position directly below the sun, the other short side goes from that point to the sun (and the right angle is there).
Thus the angle of elevation follows the relation tan(theta)=5000 km/14142 km.
And thus solving for tan, you get theta=19.47 degrees, or roughly 20 degrees.
As such, if it was the "atomplane" restircting visibility, you would expect the sun to come in quite high in the sky, not rising from below the horizon.
(Prediction: you will simply contradict this or ignore entirely because it involves facts)
Incorrect. But you round earth logicians seem to be used to that.
Nope. He was completely correct, as you just showed by contradicting him.