If you can't support this claim (bolded above) discretely, then you have no business claiming it as true. I can construct many meaningful geometries as counter cases that the first 4 do not apply to. I can probably go grab a textbook or two and find a few more, especially historical ones. What do you mean by 'meaningful'?

Care to provide such an example?

Remember, if the first 2 don't apply then the last one can't apply.

Regardless, that is entirely irrelevant.

You are trying to go off on a tangent to avoid admitting your claims are entirely baseless.

The simple fact is you are using axioms of Euclidean geometry in non-Euclidean geometry.

The burden is on you to establish that these axioms still hold in this non-Euclidean geometry.

Without that the Ferrari effect is entirely baseless.

Your disproof is trash and doesn't apply to topic at hand.

No, my disproof is rock solid, as shown by you needing to continually avoid the main issue.

As the discoverer of the non-euclidean flat earth, I should know what my theory states and that your proof is against an entirely different idea.

This is not about your non-flat flat Earth in general.

This is about one specific claim which is used to try and support it.

Knowing that I know my disproof is against it, not some other idea.

A claim which uses axioms from Euclidean geometry and pretend they apply in non-Euclidean geometry.

Here is a key part of it:

Let us again venture into thought experiment: eject some pods towards the earth from one such of our imaginary satellites at regular intervals along our orbit such that they are in free fall. Again, we can assume these are straight lines extending below to a translatable location on the surface of the earth, its geolocation. We can say these lines are normal to the trajectory of the satellite and they are normal to the ground, thus making the lines parallel.

Notice how you are relying upon properties of parallel straight lines, which hold in Euclidean space, and not necessarily in non-Euclidean space?

Also note:

It requires the first axiom, as it has straight lines.

It requires the second axiom, as it extends these straight lines.

I would say it requires I think the third, the one on right angles, as it is discussing normals. If this didn't need it then you need to specifies these normals in a less ambiguous way.

So the only 2 axioms left are regarding circles and parallel lines.

Regardless, as it is not Euclidean geometry, you can't use the axioms of Euclidean geometry to claim things about lines being parallel.

As I pointed out in another thread, the surface of a ball is a wonderful example of this (but other orbits and/or trajectories are as well).

Note: this is discussing a ball, not specifically Earth, however it is describing its surface in terms of spherical geometry and thus will use terms like equator, latitude and longitude, and other terms typically associated with a round Earth.

The equator (or any great circle), is a straight line.

If you project lines from this equator, normal to it (i.e. at right angles), at equal increments along the lines, you will have lines of longitude drawn which are also great circles which wrap around the ball.

If you now draw lines of constant curvature (where a straight line would be one where the curvature is 0) that remain normal to these lines of longitude, you have the various lines of latitude.

But these lines of latitude are not straight lines.

If you were to go north 10 degrees from the equator (at 0 E) and draw a straight line, it would be a great circle.

This great circle does not remain 10 degrees north of the equator. Instead, at 90 E and 90 W, it crosses the equator.

At 180 E, it is 10 S of the equator.

I was even nice and provided a pictorial representation of that:

The green and blue are straight.

The others are not, yet your method indicates the other 2 are straight as the green is straight.

Even your picture shows this problem:

That gravitationally affected path (i.e. the path of the ball), is a straight line as per your definition.

The curvature of space makes this a straight line.

Notice how it doesn't remain with lines connecting it to the orbital path being normal to both?

And as for your "non-gravitationally affected path", sure it appears straight here and parallel to the orbital path, but that is because of the small scale used.

Orbital paths are gravitationally affected to. A non-gravitationally affected does not remain the same distance from Earth.