It might just be much much more complicated than my understanding allows at this moment. I'm always learning. I'm not so sure it is flat anymore now that I can't figure this out. This is why I am asking for Physical Observer's help to explain.
Don't ask for his help. He has no interest in honest explanations. He will run away from any attempt to make an honest explanation of what is happening, dismissing it as goobly gloop or the like.
Reasons like the physics of flat and level water seeking the lowest point and staying calm on an allegedly 1000 mph spinning ball (yeah right). Try spinning a merry-go-round at 1000 mph with a bucket of water on there and see where the water goes.
It isn't quite the lowest point. It is the lowest apparent point, based upon the apparent gravitational potential which takes that spinning into consideration.
For a ball, the lowest point is the centre.
A sphere has each point being equally distant from the centre and thus equally high.
Physically, the equator is over 20 km higher than the poles. This is a result of the spinning.
As for spinning a merry go round, you don't spin it with the same tangential velocity to make it an honest comparison, nor would you do it with the same angular velocity.
The acceleration required to remain in a circular path around a point =omega^2*r=omega*v=v^2/r.
That means if you take Earth, with a radius of 6371 km, with a rotational speed of roughly 15 degrees an hour, you need an acceleration (at the equator) of 0.03 m/s^2, which is much much smaller than that provided by gravity, 9.8 m/s^2.
To scale this down keeping the acceleration constnat, based upon the acceleration being equal to v^2/r, this means r needs to be kept proportional to v^2.
So scaling it down to 6.371 m (the radius by a factor of 1 million (1000 squared)), the velocity needs to drop to 1 mile per hour (a factor of 1000).
So for an honest comparison, try spinning a merry go round with a radius of 6.371 m, such that the outside goes at a speed of 1 mile per hour. That is less than walking pace.
Besides, if the earth were round then you would see the curve of earth from side to side along the horizon.
No. You wouldn't.
You don't see the great circle of Earth as the horizon. Instead the horizon is a small circle centred on you with the same angle of dip all around.
Thus if you followed it all around, you would see a circle, at some slight (potentially undetectable) distance below you.
In order to see the curve of this circle, you need to be at a very great height.
As an example, try and find a surface that you can stick your head through, or get a camera and tripod.
Draw a large circle on this surface.
Now, position your eyes/camera in the centre of the circle, at a height such that the circle is just a half of a degree or so below level (in reality, the horizon is actually at a much smaller angle of depression for most cases, it is 5-10 km away from you, depending on your height, with you only being 1.5-2 m above the ground at your spot). Take a look at it and it should appear quite flat, especially with a low FOV like in a camera.
It is only when you get up higher and can see much more of the curve that it appears as a circle.
If Earth was flat, the horizon would be based upon your limit of vision to the atmosphere not allowing smooth passage of light and thus it would be a blur.
If you were to follow that curve while turning around you would end up at a point lower in height or altitude than where you started, therefore your elevation would decrease just by spinning your body around while following the curve of the horizon. My elevation does not decrease just by looking left or right or spinning around (how absurd is that?), therefore the earth cannot be round.
This actually argues against you and shows a flaw in your reasoning.
We know circular objects exist.
We know that you can look at the "horizon" from them.
We know that turning around doesn't magically make you higher.
As such, you conclusion about spherical objects contradicts reality, showing there should be some flaw in your reasoning.
Your key flaw is your assumption that it should curve, and that if you turn that curve should just continue.
The horizon should be at the same angle of depression all around, thus if you turn, it should still be at that angle, not curving down.