From my above post, I will take the suns form as round, and explore the maths involved.
If the sun is 3000 miles above the plane earth, then when the observers lattitude is at about the same declination as the sun the sun will be overhead, or if on the equator about march 21st it will be overhead.
When i took my observations yesterday, i measured the suns diameter as 32 minutes of an arc. Or 0.53333333333333 of a degree.
If the suns half diameter was 16 minutes of the arc, (0.266666) then simple trig will give us the diameter.
Sorry, i am using an iPad, and i am no good at diagrams, but it is easy enough to follow;
The angle i measured was 16 minutes of arc for half the diameter, with the suns distance being the adjacent side and half the diameter being opposite, so
Tangent 0.266666666 multiplied by the distance will give half the diameter, which works out as 13.96 miles or a diameter of 27.92miles, so lets say 28 miles for round figures.
Now that works for noonday if the sun is 3000 miles away and 28 miles in diameter, all fits well so far. I am taking them to be statute miles, even though nautical miles would be easier to calculate.
Now retaking the sun at intervals uses plane trigonometry if the earth is a plane.
Where at noon, the position of the sun over the earth was the same as mine, more or less, there was no horizontal distance separating us, but there will be later. (Suns position over the earth relative to the observer is is referred to as Local Hour angle, and declination)
Next at about 15:00 lt i retook the suns diameter and it had not changed., however its distance from me had changed.
It’s lattitude (declination) had not changed much, but in 3 hours it had travelled 3 x 15 = 45 degrees, as it is not disputed that the sun travels 360 degrees in 24 hours.
Therefore the sun was 45 degrees of longitude horizontally away from me, and 3,000 miles vertically above me.
The 45 degrees of longitude at 15 degrees north is equal to 2700 minutes of longitude, multiplied by Cos 15 to get nautical miles. Which is 2608 (rounded up) converted to statute miles is 3,000 miles. (Coincidentally)
Now using Pythagoras the hypotenuse (straight line distance) is the square root of (3,000 squared plus 3,000 squared) which equals 4,42 miles.
Given the measured arc was constant at .5666666 of a degree or semi diameter of .266666666 degree we can use the same formula as above to get the diameter in miles of the sun.
Tan 0.26666666 multiplied by distance (4,242) equals 19.74 miles for half diameter or 39.5 miles across, which is NOT the same as the first calculation at noon. (41% bigger than the noons diameter) the apparent altitude was 45 degrees
Using the next observation as above at 17:00 the sun had travelled a total of 5 x 15 = 75 degrees of longitude, or 4,500 minutes of arc or 4346 Nmiles or 5,000 statute miles and was still 300 miles above the plane, therefore the actual diameter should be, Tan 0.26666666 x 5,000 = which is equal to 23.27 miles or 46.5 miles which is 66% bigger than at noon. The apparent altitude was 10 degrees.
As can be seen either the sun changes diameter (not possible) or should get smaller. I dont see Enag giving any form of proof other than a sketch to show how much bigger it gets! So i am forced to believe through my own eyes, and own empirical observations that the sun cannot be where the FE theory say it is.