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Show me the fucking money!
Look at it this way. Find a piece of flat land somewhere, say, Kansas. Now, you will have to prepare this ahead of time.
Pick a spot where you know the sun will be directly overhead, say, at about noon. Since the sun travels in an east-west direction, get somebody to stand 16 nm east of the spot where the sun will be directly overhead, and get another person to stand 16 nm directly west of the spot. Stand them so that if you draw a straight line between the two people, the line would pass through the spot where the sun is directly overhead, and this spot would be right in the middle of the line, kinda like this >>> ·+· . (The two dots are the people. The vertical line shows the center of the horizontal distance the people are apart. The intersect is where the sun is directly overhead.)
If the two people look straight up, they should each be looking directly at opposite edges of the sun. But they aren't. Even transits or telescopes will show that they are still looking at the center of the sun. This means that the sun must be far wider than you suggest, or that there is some contradiction in you calculations somewhere.
Because of the size of the earth, any atmospheric distortion would be too small to notice looking straight up. Perspective isn't included, because the guys are looking directly at the sun, even with the telescopes, not at some width where they need to focus perspectively.
^^^ I know the Sun's exact size from the 90 degree measurement, it's simply a matter of proving/showing that the 1 minute per mile ratio used on the ground also applies to celestial objects.
The refractive layer is key here, I now think odolvlobo could have been on to something when he mentioned it. I just need to take step back and approach the problem from another (no pun intended) angle.
Well, of course we know the sun's exact size. My calc, above, repeated in this quote shows how you and I both calculated it absolutely using the 90 degree measurement.
On a FE, 32 degrees equals the sun's size. You have said this often in the past. You have shown how 32 degrees equals 32 nm. My calc simply shows the most logical way to apply the 32 degrees/nm to the sun's size. Since the earth is flat (right?), simply look straight up from opposite ends of the 32 nm. This will give you the 90 degree angle, because the flat earth is 90 degrees with respect to looking straight up.
If the sun were off in the distance somewhere, at other angles to both people in my calculations, then there might be some other calc that goes into it. But since:
1. The earth is flat;
2. 32 degrees = 32 nm;
3. The people are looking straight up (90 degrees to the FE) at opposite edges of the sun;
4. Simple trig shows that the sun is 32 nm wide, no matter how high it is above.
If the sun is very far away, you might need some kind of super transit with each of the people, so that they can be sure that they are looking absolutely, perfectly straight up. But there you have it. A directly straight up view that determines the sun's width.
If the sun is wider or narrower, we have to change the 90 degrees to the FE that the people look up. The 90 degrees is no longer applicable.
So, take your pick. Either the sun is 32 nm wide, or the angles are different than 90 degrees. Even sextant measurements will show this when the sun is directly overhead.
If you understand the sun to be some 3,000 nm across, on the FE, get a couple of people to stand 3,000 miles apart, and use their super-transits to look exactly straight up. The transits will see the opposite edges of the sun. It's the way trig and math work.
If your people can't detect the edges of the sun, it's because the sun is so far away that a super-transit isn't able to tell the difference, because it isn't sensitive/accurate enough for such distance. OR, the sun is so small (32 nm) that it isn't visible in the transits at all when the transits are 3,000 nm apart looking straight up.
You can't have it both ways.
