Diameter of Flat Earth: 24,900 miles (Radius: 12,450 miles)
Solar path Radius at equinox: 6,225 miles
Approximate deviation between solstices: +/- 1615.6 miles
Height of Sun: 3,000 miles
Diameter of Sun: 32 miles (Radius: 16 miles)
Scenario: During the equinox, when an observer stands upon the equator, at precisely 6am and 6pm solar time, the observer will be exactly halfway through either a sunrise or a sunset at sea level.
Objective: Using the above information, determine the upward curvature of light to observe this phenomena.
Linear distance from observer to the point at sea level on the earth that the sun is directly over at this time:
Since this forms a right isosceles triangle with the north pole, this distance is merely the square root of twice the Solar path Radius at equinox squared.
D12 = (6,225 mi)2 + (6,225 mi)2
D12 = 77,501,250 mi2
D1 = 8803.4794258 mi
The Height of the sun is fixed at: D2 = 3000 mi
The linear distance to the center-point of the sun would then be:
D32 = D12 + D22
D32 = 77,501,250 mi2 + 9.000.000 mi2
D32 = 86,501,250 mi2
D3 = 9300.6048190 mi
The center-point light from the sun should be intersecting and almost horizontal with the earth at the point of the observer at sea level.
It should be noted that there should be increasing upward curvature with increasing elevation of the observer above sea level, allowing a fuller view of the sun at the horizon, while significantly moving horizontally closer or further from the sun will likewise impact the view.
Based on these angles, the sun should appear directly in the NE on sunrise and NW on sunset at the equator on the equinox. Since we know this is not true and the sun is instead viewed directly in the E and W respectively, there would also need to be some horizontal curvature of light to the observer as well. This curvature would be less pronounced at the North Pole and more exaggerated at the rim in order to maintain this effect world wide.
I'll have to consider this some more when I have more time...