Now, the math for the geocentric scenario :
Again, we have been over this all before.
The math ends up the same.
The difference in the ANGULAR velocity is the same.
And that angular velocity is what is observed.
Remember, the observer on Earth is not simply moving sideways, they are rotating.
And I didn't tell you to spam your same refuted nonsense.
I told you to show the problem with the math I provided, which clearly shows the angles are the same.
Remember this:
https://www.theflatearthsociety.org/forum/index.php?topic=70921.msg1917997#msg1917997Where I clearly went through the math showing how the angular velocities work out to be identical regardless of which is in motion? Where I demonstrated beyond any doubt that you cannot use a simple visual observation like this to determine which is moving?
The math you have repeatedly refused to address because you know there is no fault with it and it shows beyond any doubt that you are wrong?
Can you show any actual problem with that math?
If not, you argument is still pure nonsense.
Now, the math for the geocentric scenario :
Farther observer (on the opposite side of the Arctic circle) : 3639+665 = 4304 km/h
Closer observer moves wrt the Moon : 3639-665 = 2974 km/h
Why are you adding and subtracting speeds here?
For the GC model you are having Earth stationary.
For the GC model there is just one speed of the moon.
For some actual numbers with justification/explanation:
Taking Earth (in reality, it would be the sky in your GC fantasy) to rotate once every 24 hours (1 day) and the Moon to orbit at a distance of 400 000 km once every 28 days (or in your fantasy, still at 400 000 km but now it will do it 27 times in 28 days.
Also note that the moon orbits Earth in the same direction that Earth rotates. This means the angular motion due to the orbit of the moon is in the opposite direction to the angular motion due to the rotation of Earth.
Taking the Arctic circle to be at a radial distance of 2534 km.
This means the linear speed of Earth at this point is ~184 m/s or 663 km/hr.
The linear speed of the Moon in the model with Earth rotating is ~1039 m/s or 3740 km/hr
For the model in which Earth is magically held stationary and instead the moon circles Earth 27 times in 28 days we have an orbital period of 28*86400/27=89600, giving us linear velocity of ~28050m/s or 100 980 km/hr.
Now what angles would we expect.
For the simple GC model, where all the apparent motion is due to the moon, and notice there is no magical subtraction of velocities here, as Earth is stationary in this fantasy.
The 2 distances that we care about are the point when the Moon is closest to the observer at 400 000 - 2534 km = 397,466 km; and the one where it is the furthest at 400 000 + 2534 = 402,534 km.
Approximating the Moon's motion as a straight line, that means after 1 second the moon has moved 28050 m or 28.050 km.
For the near observer that amounts to an angle of 14.56 arc seconds.
For the far observer it amounts to 14.37 arc seconds.
This gives us a difference of 0.18 arc seconds.
Now the more complex and closer to reality model with a rotating Earth:
Now, we need to remember that we aren't just dealing with linear motion. Instead we are dealing with rotation.
Both observers are initially looking towards the moon (or are zero point is anyway).
But after the 1 second of rotation of Earth, this puts them looking 15 arc seconds rotated, which when projected to the moon places it (again, assuming a simple straight line) places it 29.089 km away from where the moon was.
Now adding in the real motion of the moon of 1.039 km, we end up with a change in position from our new rotated reference of 28.050 km.
This should seem familiar, it is the exact same as the GC fantasy.
This means the angular change in exactly the same.
For the near observer that amounts to an angle of 14.56 arc seconds.
For the far observer it amounts to 14.37 arc seconds.
This gives us a difference of 0.18 arc seconds.
So just like already shown countless times, the difference in angular velocity for the 2 positions is the same, regardless of if Earth rotates with the moon orbiting or if Earth is stationary with the moon circling much faster.
You cannot tell which is moving by such a simple visual observation.
However, there is no need for conducting such experiments because all we need is one fixed platform
You are right that there is no need for conducting such experiments, as no simple visual experiment can tell which is moving.
Instead now you are appealing to a "fixed platform",
But, as soon as we provided one fixed (in absolute terms) platform
Fat chance getting one of them. Regardless, that is an entirely separate argument.
Does this mean you accept your ZigZag argument is garabge?