I'm not saying the sun needs to be far away and close. I'm pointing out that the flat earth model requires a sun that is much closer, and you have no direct proof that the sun is as far away as is required for your calculations to work. I also have no proof that the sun is closer for the flat earth model to work. I'm not proving one or the other, I'm simply pointing out that the same evidence can be used in either model. For all practical purposes it really doesn't matter which model you prefer.
I was tempted to answer this point because I DID already answer this for BOTH a near and far sun. And I explained it several times ad nauseum . I just verbally explained the proof. Here's what I said for all to witness (with emboldened text to point out the critical points):
If we find a point on earth where the sun is overhead, it's altitude is in direct proportion with the distance traveled, as in the angular measure on a protractor. For example, if we travel 690 miles and the sun is now 80 degrees elevation, another 690 it will be 70 degrees--- linear. Now, if the earth was flat and 690 miles gave us 80 degrees, We'd have closer to 71 degrees (70.6). The error gets greater further out. A circular distance of 2760 miles yields an altitude of 50 degrees. On a flat earth this distance yields 54.8 degrees altitude, almost 5 degrees error which is quite noticeable. The sun's elevation is then a function of the cotangent of the distance rather than a linear arc distance. Of course this does not happen. BTW, to get these figures, the sun needs to be only 3913 miles. If the sun is much further, even just a million miles, the difference in elevation would not be great at all -- the sun's elevation would be virtually the same on all points of the flat earth. To prove all this is not deep scientific theory. It's simple geometry/trigonometry to indisputably prove the earth is round.
SO we can see I addressed quite adequately a close sun and a distant sun and NEITHER will work for a flat earth model. And this is the conclusion drawn:
You aren't presenting any proofs whatsoever. You're simply presenting some assertions that you assume no one understands. Fine, if that's what you think, then the burden of proof is upon you to show us that you understand them yourself. You quite obviously have no idea what you're talking about in the first place. Present your arguments, proofs etc.
Seriously? I just did the calculations and gave the verbal answer. You really have to learn the difference between assertion and calculation-- nobody "asserts" specific numbers. They are calculated. And that's right, the burden of proof is on me, so now here's the calculation (and a funny dialog between hypothetical people just to lighten things up):
The cotangent in Trigonometry is defined as the ratio between the leg adjacent to the angle when it is considered part of a right triangle and the leg opposite for an acute angle of a right triangle. In a picture it is shown as thus:
And the tangent is defined as the reciprocal of cotangent.
Now, I'm on a point on a flat earth, in the tropics sipping a cocktail, with the sun directly overhead when I call my astronomer friend 690 miles away. "What is the elevation of the sun there?". 80 degrees he says. Thus, How far is the sun? I ask myself. The sun, flat earth with me under it form a right triangle. So the sun's distance:
This is where I derived the solar distance from. So now I call my friend twice the distance and I say to myself, I know what the answer will be. It will now be:
since cot(θ)=1380/3913 we take the inverse function
cot-1 (1380/3913) = 70.6º (almost 71º )
"Hello Ralph, what is the solar altitude at your location--- "70 degrees" he says. "Hmmm, that's off a tad. I know. I'll call my friend on this vast plane 6215 miles from me and the answer will be:
cot-1 (6215/3913) = 32.2º
"Sorry, it's not 32.2º, the sun is setting" (about zero degrees)
Aha! I just noticed the altitude decreases linearly with the distance It sinks 10º every 690 miles.
So now with this very realistic scenario, this is what I was trying to verbally convey to you sparing the simple calculations. At this point we may not know the shape of the earth yet, but we know that a flat earth model does not work because mathematically you cannot make a (co)tangential function work in a linear fashion. In reality, the degrees are related to the distance, a sure give-away that we are talking circular surfaces likened to radians (the distance traveled) and degrees (compared to solar altitude).
Now with all this, I can show harder math but that is not the point, and indeed one needs no calculus to show the flat earth model does not work.
You are not off the hook yet showing an electro-magnetic proof and I'll show my mathematical model displaying the seasons using Kepler's Law and spherical trig to describe the primary motions.