143. The Sun is like a spotlight

"People claim that if the Earth were flat, with the Sun circling over and around us, we should be able to see the Sun from everywhere all over the Earth, and there should be daylight even at night-time. Since the Sun is NOT 93 million miles away but rather just a few thousand and shining down like a spotlight, once it has moved significantly far enough away from your location it becomes invisible beyond the horizon and daylight slowly fades until it completely disappears. If the Sun were 93 million miles away and the Earth a spinning ball, the transition from daylight to night would instead be almost instantaneous as you passed the terminator line."

If you bother to do proper calculations the flat Earth explanation does not make sense.  Dubay and flat earthers in general rarely attempt to do proper calculations, preferring to hide behind vagueness.

"If the Sun were 93 million miles away and the Earth a spinning ball, the transition from daylight to night would instead be almost instantaneous ..."

The light from the Sun diffuses through the atmosphere.  This is why the sky is light during the day and not just the Sun. At sunset when the sun sets behind the horizon so an observer on the ground can no longer see it, the rays of the Sun are still lighting up the atmosphere above the observer.  As the Earth revolves and the sun sinks further below the horizon the amount of light reaching the atmosphere above the observer decreases until the sky is completely dark.  This process will obviously not occur almost instantaneously as Dubay claims.

http://earthsky.org/earth/twilight-2

How does the Sun disappear below the horizon?

In addition flat earthers cannot explain the Sun disappearing below the horizon bottom first.  If it was circling above the Earth and just moving away from the observer at sunset it would remain observably above the horizon and just become smaller and fainter.

Although flat earthers do not seem to have a definite distance for the Sun (Typically they know what they don't believe but they seem very vague about what they do believe), the current consensus places it at about 3000 miles above the plane of the flat Earth,  circling above the equator.  The circumference of the Earth is believed to be 24,900 miles.

At the equator there are 12 hours of daylight, so the torch like sun has to cover half of the circumference at the equator.

That works out to 6,225 miles edge to edge.  Assuming that the Sun is in the middle of this area this means that for an observer standing at the equator at sunset, the Sun must be 3113 miles further away along the equator.

If we assume that the Sun is 3000 miles above the flat Earth, that means that its angle from the ground at sunset must be about 45 degrees.

Oh dear flat earthers, that is nowhere near the horizon.  How do you explain that?

If you disagree with my figures for the circumference of the Earth, or the circumference of the flat Earth equator, please substitute your own.  The calculations that you need to use are included below for your convenience.

How to calculate the angle of the Sun at sunset/sunrise on a flat Earth

The radius of a circle = 2 π r where r is the radius
For the following right angle triangle Pythagoras's theorem states
c² = a² + b²

For the above right angle triangle the angle between a and c is = arcsin (b / c)
So:

Circumference of the flat Earth = 24,900
The radius of flat Earth is 24,900 / (2 π) = 3,963

The equator is halfway between the North Pole and the edge of the world, so the distance to the equator from the North Pole is 1,981 miles.

Therefore the circumference at the equator  = 2 π x 1,981 = 12,450 miles

At the equator the length of the day is 12 hours, so the sun must cover a width of half of the circumference at the equator = 6,225 miles.

If the Sun is above the centre of its area of illumination it must be 3113 miles from the edge of that illumination to the centre.

Using Pythagoras’ the hypotenuse of the triangle ( c ) formed by the triangle of the observer at sunset (i.e. the edge of the sunlit area, the sun and the point immediately below the sun) is =  √ ( the height of the Sun²  + observers distance from the point below the Sun²)
=  √ (30002 + 31132) = 4,423 miles

The angle of the Sun in the sky as seen by the observer at sunset is therefore
arcsin(3000/4423) = 42.7 degrees


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