[Peter Winfield]

At the Equinox (coming soon) on the Equator, the Sun is vertically overhead. At the North Pole the Sun is at a constant angle of about 23.4 degrees above the horizon at the North Pole. I think this is consistent with both RE and FE models, but do correct me if this is wrong.

On a flat Earth, simple trigonometry says that the height of the Sun is about 0.43 of the distance from the Pole to the Equator (tan(23.4)~=0.43). If that distance is 10,000Km then the height of the Sun would be about 4300Km.

Is this roughly correct? I am clearly taking this angle from data generated from an RE model of the Earth, so that may be the problem, but a similar calculation could be done at any latitude as long as the measurement is done at the highest point (noon).

I am aware that the theory of Electromagnetic Acceleration would give different figures but (according to the WiKi) this is not a complete theory and there is no experimental evidence for it. When there is an experiment that can measure light curving upwards then we can use those results to adjust the figure I calculated.

[Longtitube]

At the Equinox (coming soon) on the Equator, the Sun is vertically overhead. At the North Pole the Sun is at a constant angle of about 23.4 degrees above the horizon at the North Pole..

You should check that figure for the Pole, it’s wrong for the equinox. Probably correct for noon at the summer solstice.

[Peter Winfield]

At the Equinox (coming soon) on the Equator, the Sun is vertically overhead. At the North Pole the Sun is at a constant angle of about 23.4 degrees above the horizon at the North Pole..

You should check that figure for the Pole, it’s wrong for the equinox. Probably correct for noon at the summer solstice.

You are right, of course! At the equinox it is horizontal, and 23.4degrees at the summer solstice. So replace 10,000 with 7,400 Km and you get around 3200 Km high.

[Longtitube]

Ah, but you haven’t taken account of Electromagnetic Acceleration or EA, according to which light bends up over large distances. This means your ordinary trigonometry doesn’t work to calculate the height of the sun above a flat earth. You’ll find this discussed in the wiki, as well as other attempts to calculate the height of the sun above a flat earth: they don’t work too because for some reason they don’t take account of EA either.

There’s an equation for EA given in the wiki but it’s not complete: one of its terms doesn’t yet have a value, so you can’t calculate the effect of EA to correct the trigonometry.

[Peter Winfield]

Ah, but you haven’t taken account of Electromagnetic Acceleration

Ah, yes, Electromagnetic Acceleration, the theory which is not complete, cannot be measured experimentally, and doesn't solve problems with horizontal curvature.

[Iceman]

Ah, yes, Electromagnetic Acceleration, the theory which is not complete, cannot be measured experimentally, and doesn't solve problems with horizontal curvature.

To be fair to EA proponents, it's their contention that natural phenomena like sunsets and objects disappearing behind the horizon are evidence and the experimental measurement for EA.

The observation is that things disappear bottom to top behind the horizon. There are multiple interpretations to explain that. Earth's curvature being the accepted interpretation, upward bending of light rays away from a flat earth being another.