On the first page in post #24 you said this :

If you were looking at buildings 4 km away, and just steps a few m to the left or right, do they appear to move? No.

That is because of just how small that is.

Using the maximum of 3.7 m, the angular displacement that causes is 0.05 degrees, for the 1.7 it is 0.02 degrees.

This linear motion is not causing the apparent motion of the moon.

Do you know what is? YOUR ROTATION. That is why it is important.

The motion, just due to translation would be a mere 0.05 degrees, at most.

But at the same time you have rotated 1.25 degrees.

You turning 1.25 degrees will make everything appear to move 1.25 degrees in the opposite direction.

Okay?

The translational motion of you and the moon combined will make it appear to move no more than 0.05 degrees. But your rotation will cause it to appear to move 1.25 degrees.

This video disproves you :

No it doesn't.

From all the shaking that is going on you cannot tell what kind of motion is happening.

At the start, it looks more like they are turning to the left, which would produce significant motion.

In fact, if you look closely, your footage actually provides a guide, the roof of the building in front.

If you notice, at the start, your straight line down the centre starts at the top right of a crest and ends at the bottom left. This indicates the camera is not aligned with it and instead is facing slightly to the right.

If you observe it at the end, it is the opposite, the top is at the left and the bottom is at the right, indicating the camera is facing to the left.

So by TURNING the camera from the right to the left, you have made the object (the distant mountain) appear to move from the left to the right.

So rather than refuting me you have backed me up, showing that rotation causes things to appear to move.

But I don't need your help.

Simple math proves I am correct.

Imagine you are lined up with an object 4 km (d) away so it is directly in the centre of your FOV.

Now, step 1 m to the left. (s)

This now forms a right angle triangle.

The object will now appear a slight angle off centre. This angle can be calculated as:

tan(a)=s/d

a=atan(s/d)=atan(1/4000)=0.014 degrees.

If your imagined rotation of the earth (which doesn't exist in reality)

No. It does exist, as has been shown by several experiments. You are yet to show it doesn't.

Every alleged proof of its non-rotation that you or anyone else has provided has been a massive failure and your zigzag argument is among them.

Your zigzag argument is completely unable to determine the difference between rotation and no-rotation as I showed above.

existed, and if the moon were so far away as you claim that it is, even then (with so distant moon) the moon would appear as if moves to the left at Noon scenario because your imagined rotation of the earth (which would be the main cause regarding the amount AND THE DIRECTION of moon's APPARENT motion in the sky) would determine not only the amount of moon's apparent displacement but it would also determine THE DIRECTION of moon's APPARENT motion.

Again, YOU ARE WRONG.

THE DIRECTION OF ROTATION WILL DETERMINE IT, NOT THE DIRECTION OF TRANSLATION!

As you are rotating CCW (to the left) the moon will appear to move CW (to the right).

There would be no change at noon.

Like I said, if you wish to spout such bullshit, do the math and prove it. I have done the math which proves the exact opposite.

You can't say CCW motion of earth's supposed rotation is the main factor which determines the amount of moon's apparent displacement in the sky, but then deny that CCW motion of earth's supposed rotation determines THE DIRECTION of moon's APPARENT motion in the sky, as well!

I don't. You are the one denying that.

Earth rotates CCW (to the left) both during the day and the night. That means the moon appears to move CW (to the right) during the day and the night.

You are the one that seems to want to ignore it/deny it.

You are the one acting like the rotation of Earth magically changes.

You could say something like this : >>>i don't deny that CCW motion of earth's supposed rotation is the main factor in both cases (1. the amount of moon's apparent displacement in the sky & 2. the direction of moon's apparent motion), only i claim that CCW motion of earth's supposed rotation determines moon's motion IN ONE SINGLE DIRECTION throughout 24 hours of one polar night during which we can observe the full moon full 24 hours!<<<

And that is what I claim and what I have claimed above. The main cause of the moons motion is Earth's CCW rotation (to the left) which makes the moon appear to move to the right.

This CCW rotation of Earth is constant and thus the apparent motion of the moon due to it will be constant and cause the moon to appear to go to the right.

The slight translational motion of a spot on Earth (which at the arctic circle is roughly 0.6% of the distance to the moon) will only contribute a small amount to the apparent motion of the moon.

You are the one claiming the opposite.

You are claiming that the rotation of Earth causes this 1.25 degree motion of the moon, and that magically it will change direction between midday and midnight, even thought Earth is still turning the same way.

Well, this claim would be simply wrong, because CCW direction of earth's supposed rotation wouldn't yield the same result regarding the direction of moon's apparent motion in Midnight and in Noon scenario, because in one scenario (Midnight scenario) the direction of earth's supposed motion would be to the left with respect to the moon, and in other scenario (Noon scenario) the direction of earth's supposed motion would be to the right with respect to the moon.

No it would't.

Again, you are thinking translation.

The ROTATION is still the same direction. Earth is still turning CCW or to the left, and thus the moon will still appear to move CW or to the right.

The direction of Earth's rotation doesn't magically change between midday and midnight. It is still going CCW.

That tower (in my video) is 9 miles = 14,4 km away from my house, and after ONLY 40 cm of lateral translation of the camera to the left, that tower obviously apparently moved to the right.

No. After turning the camera to the left, the tower has appeared to the right.

That is not just translation. Try again.

But you said that we wouldn't notice any apparent translation of an object at 4 km distance if we moved laterally 60 cm.

No. I said the translation will have no effect. If you turn the camera at the same time then it will appear to move, based almost solely on the rotation of the camera.

So what is shown in my video is more than generous towards you, since 14,4 is 3,6 times greater distance than 4 km, and 40 cm lateral displacement is 5,4 times shorter length than 216 cm which would be the proper length (within our scaled down model) for an object which is 14,4 away.

And this is just more dishonest presentation for you.

You act like you are giving out 2 bonuses here.

You are not.

If you are going to correct the sideways motion for the distance, then you just use that.

ON TOP OF THAT :

Deal with your zigzag BS before moving on to more BS, and if you want to make an argument, don't just link to a few shitty pictures, provide it yourself, in text form here.

Providing it in text form allows me to easily quote sections without having to rewrite it and point out exactly what is wrong.

Now then, if you want to keep up this BS, do a valid test, and refute the above math showing you get the same result with a rotating Earth and a stationary one.

Or did you need the extra stuff for the day case?

If so:

Recall, for the rotating Earth case you have: (yes, I know, I accidently used M instead of D. It is fixed here)

tan(aD+θE)=(R*sin(θM)+r*sin(θE))/(R*cos(θM)+r*cos(θE)) (eq 3)

Again, the stationary Earth case is a special case of the above where you have θE2=0, and θM2=θM-θE.

Subbing that in gives us:

tan(aD+0)=(R*sin(θM-θE)+r*sin(0))/(R*cos(θM-θE)+r*cos(0))

tan(aD)=R*sin(θM-θE)/(R*cos(θM-θE)+r) (eq 4)

Again, it looks different to eq3 so we need to simplify it.

tan(aD+θE)=(R*sin(θM)+r*sin(θE))/(R*cos(θM)+r*cos(θE))

Remember, from above, how we expanded tan? The same works for aD

tan(aN)=(RHS-tan(θE))/(1+RHS*tan(θE))

tan(aD)=(RHS-tan(θE))/(1+RHS*tan(θE))

and now RHS is (R*sin(θM)+r*sin(θE))/(R*cos(θM)+r*cos(θE))

So just like before, breaking it into p and q: (and I will combine some steps for brevity)

p=RHS-tan(θE)

=(R*sin(θM)+r*sin(θE))/(R*cos(θM)+r*cos(θE))-sin(θE)/cos(θE)

=(R*sin(θM)*cos(θE)+r*sin(θE)*cos(θE)-R*cos(θM)*sin(θE)-r*sin(θE)*cos(θE))/((R*cos(θM)+r*cos(θE))*cos(θE))

=R*sin(θM-θE)/((R*cos(θM)+r*cos(θE))*cos(θE))

And q=1+RHS*tan(θE)

=1+((R*sin(θM)+r*sin(θE))/(R*cos(θM)+r*cos(θE)))*(sin(θE)/cos(θE))

=1+(R*sin(θM)*sin(θE)+r*sin(θE)*sin(θE))/((R*cos(θM)+r*cos(θE))*cos(θE))

=(R*cos(θM)*cos(θE)+r*cos(θE)*cos(θE)+R*sin(θM)*sin(θE)+r*sin(θE)*sin(θE))/((R*cos(θM)+r*cos(θE))*cos(θE))

=(R*cos(θM-θE)+r)/((R*cos(θM)+r*cos(θE))*cos(θE))

Now to find the solution:

tan(aD)=p/q=(R*sin(θM-θE)/((R*cos(θM)+r*cos(θE))*cos(θE)))/((R*cos(θM-θE)+r)/((R*cos(θM)+r*cos(θE))*cos(θE)))

tan(aD)=R*sin(θM-θE)/(R*cos(θM-θE)+r)

which is exactly the same as eq4.

As such, the rotating Earth case gives the EXACT SAME RESULT as the stationary Earth case.

As such, this argument is unable to tell the difference between a rotating Earth and a stationary one.

If you wish to claim otherwise, you need to show what is wrong with the math.