You claim that "The rest is easy". Would you then like to explain how, in the flat earth:
<< I'll put my points below under your answers >>
1. Please simplify question
- the angular size of the sun and moon can stay the same when the distance from the observer can vary by a factor of 3.
For simplicity let's look at the sun at an equinox for an observer on the equator.
The geometry is something like this. The diagram was for a slightly different purpose but is close enough:
I'm assuming that on the flat earth the sun is 50 km in diameter and 5000 km above the earth.
For an observer in South America, directly under the sun, the time would be solar noon and the sun 5000 km away, so the angular size is
50/5000 = 0.01 radians or 0.57°.
For the observer 90° further east in Africa the sun is setting and if the radius of the equator circle is 10,000 km so the horizontal distance from the observer to the point below the sun is
sqrt(100002 + 100002).
So the total distance to the sun is then
sqrt(100002 + 100002 + 50002) = 15,000 km.
Hence when the sun is setting the angular size is
50/15000 = 0.0033 radians or 0.19° - one third the size.
But in reality the setting sun is virtually the same size as it is when directly overhead.
This is, of course, worth a topic of its own and it has one, see
Observation of Sun Size During the Day, Started by Bobby Shafto.2. Research yourself to see visually
- the sun can rise almost exactly due east everywhere on earth, except close to either pole, on every equinox.
I've done all the research I need. I know that here near Brisbane at the equnox the sun does rise as near as I can measure to due east.
My "LunaSol App" also claims that on Wed March 21, 2018 the sun rose at 90.3° - that's near enough for me.
But when I look at the sun's location at sunrise there is no way it can appear in the east unless light from the sun can curve horizontally somehow.
This is how I picture the sun's position at sunrise in Brisbane on Wed March 21, 2018:
If that fits the flat earth model, then the sun would need to rise 37° east of North, not 90° as I know it to be.
If you can find places, not too close to either pole, where this is not true please let's know. I can readily show videos that the sun sets due west in Perth, Western Australia.
3. and 4. Same as q2
- both poles can receive 24 hour sunlight for a day or so either side of each equinox.
and
- the South Pole can receive 24 hour sunlight from the (Northern Hemisphere) Fall Equinox to the Spring Equinox.
.
No, it is not possible with the usual FE model for the sun, unless you postulate some very strange spotlight shape. So you really have some explaining to do here.
5.The fe map is projected from mercator map. Its like saying why is greenland so big when it isn't?
No the FE map is not "projected from mercator map". In fact the usual Ice-Wall map is a "North Polar Azimuthal Equidistant Projection" of the Globe earth.
If the earth really were flat then no projection of the flat earth would be needed to make a flat earth map. A flat earth map would be no more than a small scale representation of the full sized earth.
The scale-factor might be say, 1:1,000,000 fir a 40 × 40 cm map and that scale-factor would be constant over the whole map.
This is not possible on a projection of the Globe, where the scale-factor always varies over the map.