All i can do for you Jack is to make a geometric (trigonometric) correctional allowance for Moon's circular motion

That would require you to make an allowance of 0.2 degrees for your hour.

And that isn't simply just adding it on to everything. It is letting the angular motion of the moon be unknown to that.

And that is much larger than the difference between the 2 locations, and thus you have no known difference.

Instead you have 2 very large values which overlap.

But that is not the case at all.

There is far more that you can do. You just don't want to because doing so would show you are wrong.

There is no reason at all for you to not do one of these options except that they both show you are wrong.

Why do you feel the need to cling to such a long period where a linear approximation would not hold? What is wrong with doing it for just 1 second?

Why do you feel the need to cling to such a faulty linear approximation where it doesn't hold? What is wrong with doing the math properly, using the actual angular motion?

Either way, if there actually was a difference, it would still show up. The only "difference" that would disappear is one based upon falsely applying a linear approximation where it does not hold.

There are 2 very simple things for you to do:

1 - STOP TREATING IT AS LINEAR MOTION!

Instead do the math properly, using the law of cosines to determine what the angles should be.

2 - Use a period of time where it actually approximates linear motion, like 1 second.

Either way do it properly, without just randomly rounding off numbers for later calculations.

That means it is 2541, not 2500. So the distances are 379959 and 385041, not 380000 and 385000

That means it is 3.1415926535897932385, not 3.14.

That means the moon has to travel 2403318 km in its orbit, not 2402100 km.

That means the moon's velocity in the GC fantasy is 96497 km / hr not the 100 000 km/hr you provided.

And so on.

Now, you might not think that is important.

But you have an error in your 4th significant figure. e.g. it should be 3.142 to 4 sig figs, not 3.140; it should be 2403..., not 2402.

This means while you say it is 14.36 degrees, it could be 14.34 or 14.39.

That means the difference could be 0.18, but it could also easily be 0.15 or 0.21.

So just like your poor linear approximation, your poor approximation for pi results in you having 2 very large ranges of values for GC and HC which overlap, and thus there is no difference between them which your horribly flawed math can see.

Once more, here is option 1 for the general case:

https://www.theflatearthsociety.org/forum/index.php?topic=70921.msg1917997#msg1917997It shows quite clearly that you end up with the same result regardless of which is moving.

Once more, here is option 2, for 1 second, this time with all the details you provided, and more, explaining each step. Note, I provide rounded numbers here, but leave them unrounded in excel for further calculation):

Taking the radius of the Arctic circle to be 2541 km.

Taking the distance of the moon from the centre of the Arctic circle to be 382500 km.

This means the distance to the near observer is = 382500 km - 2541 km = 379959 km.

This means the distance to the far observer is = 382500 km + 2541 km = 385041 km.

Taking the period of a moon's orbit (or the time to go from new moon to new moon) to be 27.5 days.

And taking the period of the rotation of Earth to be 24 hours, i.e. 86400 s

Then, for the rotating Earth reality:

The circumference of the Arctic circle is:

2541 km * 2 * pi = 15966 km.

A point on the Arctic circle must travel that distance in 86400 s, thus the speed is:

15966 km / 86400 s = 0.185 km/s

Now, the moon.

It has an orbital radius of 382500 km.

Its circumference = 382500 km * 2 * pi = 2403318 km.

That means the speed is 2403318 km / (27.5 day * 86400 s/day) = 1.011 km/s.

Now, that means the near observer, which travels in the same direction as the moon, has the speed of the moon relative to them as:

1.011 km/s - 0.185 km/s = 0.827 km/s

Now we focus on 1 second, so the moon has moved 0.827 km, relative to the observer.

The angle is calculated based upon the inverse tan.

atan(0.827 km /379959 km) = 0.45 arcseconds

The far observer, which travels in the opposite direction as the moon, has the speed of the moon relative to them as:

1.011 km/s + 0.185 km/s = 1.20 km/s

Now we focus on 1 second, so the moon has moved 1.196 km, relative to the observer.

The angle is calculated based upon the inverse tan.

atan(1.196 km /385041 km) = 0.64 arcseconds

This means in the rotating Earth reality, there is a difference of 0.64 arcseconds - 0.45 arcseconds = 0.19 arcseconds.

Note this difference is far-near. This will be used again below.

Now the stationary Earth fantasy.

The numbers above are quite similar, it is just instead of Earth rotating, you just have the sky magically rotate around Earth at a rate of 15 degrees per hour, in the opposite direction. This is 15 arcseconds per second, and thus 15 arcseconds for our 1 second (which will be used later). As this is in the opposite direction I will denote it as -15 arc seconds, to avoid any confusion that comes later.

But now, the moon, instead of orbitting once every 27.5 days, manages to complete 26.5 circles in 27.5 days.

So now its speed is given by the circumference, multiplied by 26.5 for the number of times it circles Earth, divided by the 27.5 days it takes.

i.e.

2403318 km * 26.5 / (27.5 day * 86400 s/day) = 26.805 km/s.

And again, as this is in the opposite direction, I will denote it as -26.805 km/s, to avoid confusion later.

This means for both observers, the moon will have moved -26.805 km in that 1 second.

So now the angles:

Near observer => atan(-26.805 km /379959 km) = -14.55 arcseconds.

To find it relative to the sky, we subtract the motion of the sky (and this is where the directionality matters):

-14.55 arcseconds - (-15 arcseconds) = 0.45 arcseconds.

Far observer => atan(-26.805 km /385041 km) = -14.36 arcseconds.

Relative to the sky

-14.36 arcseconds - (-15 arcseconds) = 0.64 arcseconds.

Note, the angles relative to the sky are identical to the angles in the rotating Earth case.

But the part you like harping on about, the difference, again as far - near:

-14.36 arcseconds - (-14.55 arcseconds) = 0.19 arcseconds.

Relative to the sky:

0.64 arcseconds - 0.45 arcseconds = 0.19 arcseconds.

So yet again, it has been conclusively shown that the rotating Earth reality and your stationary Earth fantasy produce the same simple visual observation.

Now stop trying to run away from the topic, stop spamming the same refuted nonsense and do one of those 2 options; and show just what problem there is with my math, or admit that there is no difference and such a simple experiment cannot determine which is in motion.

Either do the math using the law of cosines or the like rather than a linear approximation, or do the math for a period of 1 s.

Either way, you will then show that there is no difference between GC and HC.