ZIGZAG of the Full Moon - Irrefutable argument against the rotation of the earth

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Yes it can, do you know why?
Look :
NO! It can't.
Do you know why?
Look:
If Earth is rotating at -1.25 degrees per 5 minutes and the moon is orbiting at 400 000 km at -0.04 degrees per 5 minutes and you are at the arctic circle with a radius of 2500 km
[insert math from above, taken from onshape]
For midnight the moon moves 1.202 degrees to the right.
For midday, the moon moves 1.218 degrees to the right.

Now lets do it with a stationary Earth with the moon orbiting at 400 000 km at 1.21 degrees per 5 minutes and you are at the artic circle with a radius of 2500 km.
[insert math from above]
For midnight the moon moves 1.202 degrees to the right.
For midday, the moon moves 1.218 degrees to the right.

Notice how they are the same (if you like I can do all the math in full again and not even rely upon onshape)?

Because these are the same the argument cannot be used to tell the difference between the 2.

What this means is exactly what HC maniacs meant to say when pointing out that moon's apparent translation to the right is the DIRECT consequence of earth's CCW motion
Yes, its CCW ROTATION!!! Not its translation, it's rotation.
The big contributor is not moving (translating) to the left or right, it is rotating 1.25 degrees to the left which causes everything to appear to move 1.25 degrees to the right.
The motion of the other object (the moon in this case) can also be somewhat significant, with the moon contributing -0.04 degrees to the left.
And then your translation left or right contributes a pathetic 0.008 degrees.

So, if we cancel out our supposed (or real) 68 m high bulge in the middle of our 60 km (37,5 miles), and if our measurements (given such adjusted conditions) still show exactly the same results regarding the amount of angular displacement of the moon and the direction of an apparent displacement of the moon (at Noon and at Midnight), then all your excuses fall to pieces!
No it doesn't.
Again, the key contributor is not the translation, or the budge. The key contribution is the rotation, Earth turning 1.25 degrees to the left.

Yes, you seem to like indicating that this rotation is a tiny 1 degree so it shouldn't matter, BUT THAT IS HOW MUCH THE MOON MOVES (roughly)!
That is the dominant factor.
What you need to do to cancel out the dominant cause of the motion is not to try to move sideways, it is to rotate the camera 1.25 degrees  CW over 5 minutes, in a plane aligned with the equator.

The bulge contributes virtually nothing, even less than the 0.008 degrees. The sideways motion contributes a tiny amount, 0.008 degrees.
The ROTATION contributes a massive amount, 1.25 degrees, 2.5 angular diameters of the moon.


And the sweetest part is that we can compensate for these 68 meters very easily!
Go ahead, it wont help your case.

If you are honestly interested in putting this to the test 1 set of calculations is not enough.

Instead you need to do 2 sets (preferably with general variables)
One set shows the results for Earth rotating at -1.25 degrees over these 5 minutes (or rotating an angle thetaE), with the moon (at distance R) orbits at a rate of -0.04 degrees over 5 minutes (thetaM), with you observing it from a distance r from the centre of rotation and orbit, and seeing what result you get (again, general is good).

Then, you have another set, where instead of Earth rotating and the moon moving along its orbit, you have Earth stationary, with just the moon (at a distance R) obriting at a rate of 1.21 degrees over 5 minutes (thetaM-thetaE), with you observing it from a distance r from the centre of the orbit, and seeing what results you get (again, general is good).

I don't have the time to do all the general math now, but hopefully I will tonight and I will provide it for you, showing they are identical and thus you CANNOT use this to show Earth isn't rotating.

Remember, when you do this you need to note that your reference point (0 angle for your observation) is moving as well as it is on Earth and thus rotating with Earth.

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cikljamas

  • 2069
  • Ex nihilo nihil fit
The bulge contributes virtually nothing, even less than the 0.008 degrees. The sideways motion contributes a tiny amount, 0.008 degrees.
The ROTATION contributes a massive amount, 1.25 degrees, 2.5 angular diameters of the moon.
Rotation contributes nothing because there is no rotation.
We have to ensure three points (two at each end of our 60 km long Arctic line and one in the middle) and three cameras (with the same angle - with respect to the moon) which will be directed in the same direction (that is to say in the same direction with respect to the absolute (inertial) spatial frame of reference. And with all three cameras we have to take six pictures (one 2,5 min before Midnight, one at Midnight, one 2,5 min after Midnight, and then one 2,5 min before Noon, one at Noon, and one 2,5 min after Noon), and compare in which direction "apparent" motion of the moon occur 5 min around Midnight and 5 min around Noon. If there is no difference the earth is at rest!
Why?
Because of this :



Remember, when you do this you need to note that your reference point (0 angle for your observation) is moving as well as it is on Earth and thus rotating with Earth.
What reference point?
What is moving as well as it is on Earth and thus rotating with Earth?
"I can't breathe" George Floyd RIP

Okay, for this derivation, this is the key image needed:

It is not shown to scale and is shown with angles exaggerated.

O marks the centre of it all.
M0 marks the starting point of the moon, and M the position it has moved after some time.
N is the position of the observer at midnight after this time, D is the one for the day, and R is the reference point which is rotating with Earth.
θE is the angle that Earth has moved in the time.
θM is the angle that moon has moved along its orbit.

And NP and DP are points used to draw lines parallel to O-M0.

Note: a positive angle is a turn CCW (to the left), and a negative one is CW (to the right), which is against the previous convention used by me in the posts above.

We are interested in finding out the angle at N and D which go to M, as this, along with a correction for the reference, will show us the angle the moon will appear, assuming it started at 0.

So, first what is the position of the moon?
Well, it is on a circle of radius R, and thus has a position R*(cos(θM), sin(θM)).
What is the position of N? Well, it is on a circle of radius r, and thus is at a position of r*(cos(θE), sin(θE))
With this, we can find the difference, or the position of the moon relative to N:
M-N=R*(cos(θM), sin(θM))-r*(cos(θE), sin(θE))
=(R*cos(θM)-r*cos(θE),R*sin(θM)-r*sin(θE)).

Then we want to know what angle this is, relative to the reference, and assuming it is within the range (-pi/2,pi/2) (which it is in this case), is given by:
aN=oN-θE
Thus oN=aN+θE
tan(oN)=(R*sin(θM)-r*sin(θE))/(R*cos(θM)-r*cos(θE))
Thus tan(aN+θE)=(R*sin(θM)-r*sin(θE))/(R*cos(θM)-r*cos(θE)) (which I will now call eq1).

M has a similar relation, where via inspection, the position of M is -N, and thus you end up with:
Thus tan(aM+θE)=(R*sin(θM)+r*sin(θE))/(R*cos(θM)+r*cos(θE))


Now lets look at the stationary Earth case.
This is effectively the same as above. The only difference is now θE=0, and θM=-1.21=0.04-1.25, or more generally, you get θM2=θM-θE.
The simplest way to explain it is by noting that you can simply rotate the system by -θE and you get the picture of that. But lets try the more mathematical way.
Lets stick the above into eq1.

tan(aN+0)=(R*sin(θM-θE)-r*sin(0))/(R*cos(θM-θE)-r*cos(0))
tan(aN)=R*sin(θM-θE)/(R*cos(θM-θE)-r)   (which I will call eq2)

Now, this looks different, but we can't say for sure as both sides are different and we cannot easily compare them, so lets see if we can simplify this at all:
We know that tan(a+b)=(tan(a)+tan(b))/(1-tan(a)*tan(b))
So lets try using that on eq1:
We have tan(aN+θE)=(R*sin(θM)-r*sin(θE))/(R*cos(θM)-r*cos(θE))

LHS=tan(aN+θE)
=(tan(aN)+tan(θE))/(1-tan(aN)*tan(θE))=RHS
Thus tan(aN)+tan(θE)=RHS-RHS*tan(aN)*tan(θE)
tan(aN)+tan(θE)=RHS-RHS*tan(aN)*tan(θE)
tan(aN)+RHS*tan(aN)*tan(θE)=RHS-tan(θE)
tan(aN)(1+RHS*tan(θE))=RHS-tan(θE)
tan(aN)=(RHS-tan(θE))/(1+RHS*tan(θE))
For simplicity, I will at first break this into the numerator (p) and the denominator (q)
p=RHS-tan(θE)
=(R*sin(θM)-r*sin(θE))/(R*cos(θM)-r*cos(θE))-tan(θE)

q=(1+RHS*tan(θE))
=1+((R*sin(θM)-r*sin(θE))/(R*cos(θM)-r*cos(θE)))*tan(θE)

Lets start with p:
p=((R*sin(θM)-r*sin(θE))/(R*cos(θM)-r*cos(θE))-tan(θE))
=((R*sin(θM)-r*sin(θE))/(R*cos(θM)-r*cos(θE))-sin(θE)/cos(θE))
=(R*sin(θM)-r*sin(θE))*cos(θE)/(cos(θE)*(R*cos(θM)-r*cos(θE)))-sin(θE)*(R*cos(θM)-r*cos(θE))/(cos(θE)*(R*cos(θM)-r*cos(θE)))
=((R*sin(θM)-r*sin(θE))*cos(θE)-sin(θE)*(R*cos(θM)-r*cos(θE)))/(cos(θE)*(R*cos(θM)-r*cos(θE)))
=((R*sin(θM)*cos(θE)-r*sin(θE)*cos(θE)-R*cos(θM)*sin(θE)+r*cos(θE)*sin(θE)))/(R*cos(θE)*cos(θM)-r*cos(θE)*cos(θE)))
=((R*(sin(θM)*cos(θE)-cos(θM)*sin(θE))))/(R*cos(θE)*cos(θM)-r*cos(θE)*cos(θE)))
=R*sin(θM-θE)/(R*cos(θE)*cos(θM)-r*cos(θE)*cos(θE))

Now q=1+((R*sin(θM)-r*sin(θE))/(R*cos(θM)-r*cos(θE)))*tan(θE)
=1+((R*sin(θM)-r*sin(θE))/(R*cos(θM)-r*cos(θE)))(sin(θE)/cos(θE))
=1+(R*sin(θM)-r*sin(θE))*sin(θE)/((R*cos(θM)-r*cos(θE))*cos(θE))
=(((R*cos(θM)-r*cos(θE))*cos(θE))+(R*sin(θM)-r*sin(θE))*sin(θE))/(R*cos(θM)*cos(θE)-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)*cos(θE)-r*cos(θE)*cos(θE))
=(R*cos(θM-θE)-r*(cos(θE)2+sin(θE)2))/(R*cos(θM)*cos(θE)-r*cos(θE)*cos(θE))
=(R*cos(θM-θE)-r)/(R*cos(θM)*cos(θE)-r*cos(θE)*cos(θE))

Now, subbing that back in we get:
tan(aN)=p/q
=(R*sin(θM-θE)/(R*cos(θE)*cos(θM)-r*cos(θE)*cos(θE)))/((R*cos(θM-θE)-r)/(R*cos(θM)*cos(θE)-r*cos(θE)*cos(θE)))
=R*sin(θM-θE)/(R*cos(θM-θE)-r)

Thus:
tan(aN)=R*sin(θM-θE)/(R*cos(θM-θE)-r)

This looks familiar doesn't it?
If you recall, this is exactly the same as before (eq2)

So we get the exact same result for midnight. In both cases, the apparent angle of the moon after a period of time is given by the formula:
tan(aN)=R*sin(θM-θE)/(R*cos(θM-θE)-r)
Which can also be expressed as:
tan(aN+θE)=(R*sin(θM)-r*sin(θE))/(R*cos(θM)-r*cos(θE))

And you get basically the same for midday, just:
tan(aM)=R*sin(θM-θE)/(R*cos(θM-θE)+r)
or
tan(aN+θE)=(R*sin(θM)+r*sin(θE))/(R*cos(θM)+r*cos(θE))

So no, YOU CANNOT USE THIS ARGUMENT TO TELL IF EARTH IS STATIONARY!
You get the EXACT SAME RESULT if Earth is rotating with the moon orbiting, or if Earth is stationary with the moon orbiting around us.

If you don't agree, point out exactly what you think is wrong.

Rotation contributes nothing because there is no rotation.
That is what you are trying to refute. You can't refute it by simply asserting it doesn't exist.
If Earth rotate 1.25 degrees over these 5 minutes, that will contribute a 1.25 degree change in apparent position.

We have to ensure three points (two at each end of our 60 km long Arctic line and one in the middle) and three cameras (with the same angle - with respect to the moon) which will be directed in the same direction (that is to say in the same direction with respect to the absolute (inertial) spatial frame of reference. And with all three cameras we have to take six pictures (one 2,5 min before Midnight, one at Midnight, one 2,5 min after Midnight, and then one 2,5 min before Noon, one at Noon, and one 2,5 min after Noon), and compare in which direction "apparent" motion of the moon occur 5 min around Midnight and 5 min around Noon. If there is no difference the earth is at rest!
Why?
Again, how much motion are you expecting here? Because it works out to be roughly 0.032 or 0.048 degrees.
That is no where near the 2.5 moon diameters you have been claiming.

Again, this wont tell if Earth is at rest or moving. Both produce the same result.

Remember, when you do this you need to note that your reference point (0 angle for your observation) is moving as well as it is on Earth and thus rotating with Earth.
What reference point?
What is moving as well as it is on Earth and thus rotating with Earth?
You are standing still on Earth.
If you are standing still and looking towards the north pole, after Earth rotates 1.25 degrees, you are still facing the north pole and thus your view has moved 1.25 degrees.

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rabinoz

  • 26296
  • Real Earth Believer

What you are doing in that video is quite unrealistic. You are have a photo taken fron one point and are sliding it along.
Of course everything moves by the equivalent of your 40 cm. This is not at all the same as the observer moving 40 cm.

Do a proper test and move the camera along the 40 cm and see the result in proper perspective.
The way you have tried to show it is totally invalid!

You don't seem to have the slightest appreciation for perspective, where nearby objects appear to move much more than do distant objects.

It's just as unrealistic as your "ZIG ZAG ARGUMENT IS THE WINNER" video.
At 2:23 you show the sun very close, so you get your "zig-zag".
Please be honest with the heliocentric model and put the sun at the equivalent of 150,000,000 km away!
Then that angle is only about 0.005° and completely imperceptible.

All of your "irrefutable proofs" boil down to your complete ignorance of either physics, the globe or simple perspective.

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cikljamas

  • 2069
  • Ex nihilo nihil fit
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 :



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

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!

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!<<<

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.

And we can test it within our scaled down model.

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.

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

4km distance and 60 cm lateral translation is proportional ratio regarding the supposed distance of the moon of 400 000 km and 5550 km diameter of the Arctic circle.

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.

ON TOP OF THAT :





« Last Edit: June 11, 2017, 06:05:51 AM by cikljamas »
"I can't breathe" George Floyd RIP

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Mikey T.

  • 2419
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 :



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

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!

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!<<<

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.

And we can test it within our scaled down model.

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.

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

4km distance and 60 cm lateral translation is proportional ratio regarding the supposed distance of the moon of 400 000 km and 5550 km diameter of the Arctic circle.

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.



FRAME OF REFERENCE.  What did this tower move left and right in reference to?  Your shaky camera? The Building close to you? 
No your video does nothing to disprove Jack, it does add more proof to the case for your basic misunderstanding of frames of reference.  You clearly used something further away than what your scaled down model was discussing moved less in an attempt to use the "look further away with less movement" to fool naive idiots.  You moved left and right with a shaky camera, made a claim with no reference point.  Ok, I will make the reference point.  The sky behind the tower.  I zoomed in, watched pixel by pixel in slow motion.  There was no movement between the tower and the sky behind it.  It was hard to follow, had to frame by frame it since your camera was so damn shaky, but I watched 6 pixels that were in a color transition area so they were easier to track, on either side of the tower that were 110 pixels away, 6 on Left 5 on Right pixels away and 1 pixel touching the tower.  None of them moved in reference to the tower.  ZIGZAG DEAD.

 the United States would be the laughing stock of history for the next five hundred years. That would be the epitaph for that nation in every history book; "Aren't you the guys who think the Earth is a snowglobe?" The US would necessarily shrink from world importance and self-implode into a theocratic nightmare of science-denying and modern-day witch-hunting and book-burning... anything to ensure the flat Earth cult doesn't have to look itself in the eye and realise the human cost of their YouTube clickbait empire.

But don't worry. Too many people's actual real jobs depend on the Earth's spherical and rotating shape. A flat Earth America would have no pilots, no artillerymen, no naval officers and no engineers beyond automotive and electrical.

You can't train an artilleryman that the Earth doesn't rotate. He will miss his target.

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cikljamas

  • 2069
  • Ex nihilo nihil fit
"I can't breathe" George Floyd RIP

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.


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


You should go to a gun range and talk to someone who does long distance shooting.  They will will tell you the affect the earths rotation has on fired projectiles.

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Mikey T.

  • 2419
ZIGZAG DEAD.
Your brain is dead!
Do you have anything other than insults?  Or was it too disheartening for someone to point out your BS?  Did it hurt your feelings that bad that you decided to just insult me as a response rather than the content of what I showed?  Guess it must be hard when someone shows you that they are not naive enough to just take your word for it.

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cikljamas

  • 2069
  • Ex nihilo nihil fit
These three videos DEFINITELY confirm veracity of my ZIGZAG argument (watch them all CAREFULLY - in a given order - before you decide to discard them on the basis of your prejudices) :

GYROCOMPASS DRIFT LIE DEBUNKED :



DIRECTIONAL GYRO AND ZIGZAG ARGUMENT  :



ZIGZAG ARGUMENT IS THE WINNER :

"I can't breathe" George Floyd RIP

These three videos DEFINITELY confirm veracity of my ZIGZAG argument (watch them all CAREFULLY - in a given order - before you decide to discard them on the basis of your prejudices) :
No. I'm sick of you trying to distract with pathetic bullshit.
Those videos likely just repeat the same refuted crap.

I have already put in more than an effort to refute your bullshit, and you just completely ignore it and link to shitty videos.
Why should I watch them just to refute what has likely already been refuted? Especially when you will likely just ignore the refutation.


Go and read what I have provided, the mathematical proof that your zig-zag argument is pure BS, and if you wish to object, point out exactly what you think is wrong, why it is wrong, and what the correct thing would be.

If you are unable to do that, your zig-zag argument remains as a pile of refuted bovine excrement.

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cikljamas

  • 2069
  • Ex nihilo nihil fit
These three videos DEFINITELY confirm veracity of my ZIGZAG argument (watch them all CAREFULLY - in a given order - before you decide to discard them on the basis of your prejudices) :
No. I'm sick of you trying to distract with pathetic bullshit.
Those videos likely just repeat the same refuted crap.

I have already put in more than an effort to refute your bullshit, and you just completely ignore it and link to shitty videos.
Why should I watch them just to refute what has likely already been refuted? Especially when you will likely just ignore the refutation.


Go and read what I have provided, the mathematical proof that your zig-zag argument is pure BS, and if you wish to object, point out exactly what you think is wrong, why it is wrong, and what the correct thing would be.

If you are unable to do that, your zig-zag argument remains as a pile of refuted bovine excrement.

All i had to say in relation to your stupid argumentation against my argument (which boils down to a wrong understanding of how would CCW and CW alleged motion of the earth work (regarding the apparent motion of celestial lights) - if the earth would really spin on it's axis) i have already said in the post #66 (on this very page)!

My video (which you commented - in your previous post - with a load of bullshit) is experimental proof that you don't know what you are talking about. So, math doesn't help if your logic is wrong. You should finally learn that lesson.
As we say here in Croatia : "paper can withstand everything", so you can continue to load here and elsewhere as much meaningless mathematical equations as you want, it won't help your cause.

Just a reminder : https://www.theflatearthsociety.org/forum/index.php?topic=70614.msg1910504#msg1910504

"I can't breathe" George Floyd RIP

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cikljamas

  • 2069
  • Ex nihilo nihil fit
Jack, show me your mathematical justification (in the light of your fraudulent HC claims which are in dissonance with the experimental (scaled down) truth which doesn't match HC theory) for what you can see in my new experiment (which is presented in the last part of this video) :

"I can't breathe" George Floyd RIP

Jack, show me your mathematical justification (in the light of your fraudulent HC claims which are in dissonance with the experimental (scaled down) truth which doesn't match HC theory) for what you can see in my new experiment (which is presented in the last part of this video) :



as we can clearly see with your attempt to make a lateral movement you also introduced rotational movements.
clearly to see that the view does not shift to left and right in a smooth movement.

that shows that you do not do a correct setup of that experiment.
for a correct experiment you would also have to show how you did the lateral and rotational movements.

you proven only that you do not have any knowledge how to even set up an experiment.

do you really think that somebody would believe this video as a proof?

These three videos DEFINITELY confirm veracity of my ZIGZAG argument (watch them all CAREFULLY - in a given order - before you decide to discard them on the basis of your prejudices) :
No. I'm sick of you trying to distract with pathetic bullshit.
Those videos likely just repeat the same refuted crap.

I have already put in more than an effort to refute your bullshit, and you just completely ignore it and link to shitty videos.
Why should I watch them just to refute what has likely already been refuted? Especially when you will likely just ignore the refutation.


Go and read what I have provided, the mathematical proof that your zig-zag argument is pure BS, and if you wish to object, point out exactly what you think is wrong, why it is wrong, and what the correct thing would be.

If you are unable to do that, your zig-zag argument remains as a pile of refuted bovine excrement.

All i had to say in relation to your stupid argumentation against my argument (which boils down to a wrong understanding of how would CCW and CW alleged motion of the earth work (regarding the apparent motion of celestial lights) - if the earth would really spin on it's axis) i have already said in the post #66 (on this very page)!

My video (which you commented - in your previous post - with a load of bullshit) is experimental proof that you don't know what you are talking about. So, math doesn't help if your logic is wrong. You should finally learn that lesson.
As we say here in Croatia : "paper can withstand everything", so you can continue to load here and elsewhere as much meaningless mathematical equations as you want, it won't help your cause.

Just a reminder : https://www.theflatearthsociety.org/forum/index.php?topic=70614.msg1910504#msg1910504


Off course math nothing to do with reality off course, shhhh there are no such thing as predictive model, nooo its not possible to visualize math, no don't look here

https://celestiaproject.net/

This software is not real,

Its obviously cant show ALL the angle, distances, speeds and trajectories above...


Cause there are no such thing as math model predict something

All i had to say in relation to your stupid argumentation against my argument (which boils down to a wrong understanding of how would CCW and CW alleged motion of the earth work (regarding the apparent motion of celestial lights) - if the earth would really spin on it's axis) i have already said in the post #66 (on this very page)!
Except you are yet to demonstrate a single thing wrong with it.
Instead you just assert the same refuted bullshit.

You might also want to double check your count. As far as I can tell, post #66 is by Rab, not you.

Can you provide any rational argument against what I have said?
Can you show anything wrong with the math at all?


My video (which you commented - in your previous post - with a load of bullshit) is experimental proof that you don't know what you are talking about. So, math doesn't help if your logic is wrong. You should finally learn that lesson.
No, I didn't comment with a load of bullshit.
I commented with a load of reason, something you seem completely incapable of refuting.
Your video did not show a mere translation. It showed a translation with a rotation.

There is not a single thing wrong with my logic. Again, if there was you would be able to point it out.
Instead you just repeat the same refuted crap.

As we say here in Croatia : "paper can withstand everything", so you can continue to load here and elsewhere as much meaningless mathematical equations as you want, it won't help your cause.
I don't particularly care what you say in Croatia. The simple fact is paper cannot withstand anything.

If there is a flaw (which is required for it to not match reality) you should be able to point it out.

For example, I destroyed your zig-zag argument, pointing out exactly what was wrong with it. So no, it cannot withstand anything. Bullshit can be refuted.
Do you know the only kind of things that can withstand anything? The truth; rational, sound arguments.

If you are unable to find a single thing wrong with my argument, that means it stands as correct and your zig-zag BS remains as a pile of refuted bullshit.

Just a reminder : https://www.theflatearthsociety.org/forum/index.php?topic=70614.msg1910504#msg1910504
I don't need a reminder of your prior ignorance.


Jack, show me your mathematical justification (in the light of your fraudulent HC claims which are in dissonance with the experimental (scaled down) truth which doesn't match HC theory) for what you can see in my new experiment (which is presented in the last part of this video) :


I already pointed out what was wrong with this, but like usual, you ignored it because it shows you to be full of shit.
You are not just translating the camera. You are rotating it as well.
This is quite obvious when you look at the roof near the bottom of the camera. In the rightmost position the vertical line does not go straight down the roof. Instead it at the top (distant) it is on the right of a crest, at the bottom (near) it is on the left of the same crest. This indicates the camera is facing right.
Then, at the leftmost position, you get the opposite.
The line at the top is on the left of a crest and at the bottom at the right of the crest.
This means the camera is facing to the left.
What this means is instead of merely translating the camera, you turned it.

And just like a rational person would expect, rotating the camera makes things appear to move.

If you want to do something like this, then control it properly.
Set it up on a straight track and slide it along. Make sure the track is secure and if you like record it as well.
If I get time, I will do something similar with a motorised track.
Holding it with your hand and massively shaking the camera as you move it left and right is not a valid, controlled experiment.
If you want something close to that, use a theodolite app which shows your bearing accurately and use that. That way you can at least see if you have turned (assuming it is accurate enough). You can also intentionally rotate the camera to get the bearing to various objects.


Or you can speak to navigators or surveyors or the like and see how they measure bearings and angles and see if it matches the BS you are spouting, where barely existing motion to the left or right will cause something 400 000 km away to appear to move 1.25 degrees.

Now I ask again, do you have any rational objection to the arguments I have provided?
Can you point out a single flaw in any of the math I provided which shows you get the exact same result for a stationary Earth and a rotating Earth?

If not, my argument stands correct and your's stands refuted.

Until you can do the math and show what you would expect for a rotating and non-rotating Earth, showing that they are different (rather than the real math which shows them to be identical, as has been pointed out several times), your zigzag "argument" remains as a pile of refuted bullshit.

And don't bother bringing in other crap such as using gyros or the like to remove the rotation as that is a completely different argument based upon being able to directly detect and compensate for the rotation of Earth, which most gyros cannot do.


My video (which you commented - in your previous post - with a load of bullshit) is experimental proof that you don't know what you are talking about. So, math doesn't help if your logic is wrong. You should finally learn that lesson.
As we say here in Croatia : "paper can withstand everything", so you can continue to load here and elsewhere as much meaningless mathematical equations as you want, it won't help your cause.

If the logic was wrong the math would not work out.  Go look up ballistics, go talk to someone who fires long distance, talk to a meteorologist.  They will all tell you the same thing.  You dont even have to do the math, it is already done for you.

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Mikey T.

  • 2419
The ZIGZAG zombie died again.  sickly pajamas provided the video that removes all doubt that it is absolutely false.  He just doesn't want to admit it.  His widdle feewings are hurt.

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Mikey T.

  • 2419


This is how I feel every time I "discuss" things with a FE hoaxer

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cikljamas

  • 2069
  • Ex nihilo nihil fit
One question for Jack :

Since the parallax of the moon is allegedly 400 times greater than the parallax of the sun due to the "fact" that the sun is 400 times farther away than the moon, it means that the rotational effect of the earth is 400 times greater in relation to the moon, then in relation to the sun.

So, right after the total solar eclipse is over, when observing it from the northern hemisphere (we are facing the south) the sun is coming out (ceases to be obscured by the moon) TO THE RIGHT (as if the sun apparently moves faster - in it's "apparent" motion TO THE RIGHT - than the moon).

In the southern hemisphere, the situation is totally opposite. We are facing north, and the sun is coming out of the moon's eclipse TO THE LEFT (as if the sun apparently moves faster - in it's "apparent" motion TO THE LEFT - than the moon).

Due to the fact that the rotational effect is 400 times greater in relation to the moon (than in relation to the sun), right after the total solar eclipse is over, when observing it from the northern hemisphere the sun should come out (ceases to be obscured by the moon) TO THE LEFT (contrary from what happens in reality), and when observing it from the southern hemisphere the sun should come out TO THE RIGHT (contrary from what occurs in reality, also)!!!

This is the reason why HC theory needed to reverse moon's motion to the left (in an opposite direction from everyone who has ever lived seen it go).

The alleged speed of the moon (to the LEFT) is 3660 km/h
The sun is allegedly stationary.
However, earth-moon system allegedly travels (to the RIGHT) 108 000 km/h

Now, you could say, 108 000 km/h is ONLY 30 times greater than moon's alleged motion in an opposite direction from the orbital motion (to the right) of the earth-moon system, but since the parallax of the moon is 400 times greater than sun's parallax, it means that the alleged speed of the moon could be 400 times lesser than the orbital motion of the earth-moon system in order to cancel it out.

But such an answer doesn't add up, since the earth's rotational effect is already 400 times greater in relation to the moon, than in relation to the sun.

So, if moon's motion (to the left) is 13,5 times greater than the speed needed to overcome orbital motion of earth-moon system (to the right) it is not enough, since the rotational effect of the earth is still 400 times greater in relation to the moon, than in relation to the sun.

What if we applied this situation to the stars?

The shapes of "constellations" don't change over thousands of years.

Let's take a closer look at above sentence and the meaning of these words.

If the Earth moved in it's 300 000 000 km wide orbit, and even if we allowed that HC bull shit assumption (regarding "endless" distances of the stars) is true, what kind of change we would be able to observe in the sky, anyway?

1. This information comes from World Almanac. The distance to the main stars of the Big Dipper ranges from about 68 light-years (ly) to about 210 light-years...

2. Scientists studying the North Star Polaris found that it is about 323 light-years from the sun and Earth, substantially closer than a previous estimate of 434 light-years

3. Beta Ursae Minoris, traditionally called Kochab, is only slightly less bright than Polaris with its apparent magnitude of 2.08. Located around 131 light years away from Earth

4. At magnitude 4.95. the dimmest of the seven stars of the Little Dipper is Eta Ursae Minoris. An F-type main sequence star of spectral type F5V, it is 97 light-years distant. It is double the Sun's diameter and is 1.4 times as massive and shines with 7.4 times its luminosity. Nearby Zeta lies 5.00-magnitude Theta Ursae Minoris. Located around 855 light-years distant.

5. The Orion nebula is located at a distance of 1,344 ± 20 light years

6. Casiopeia is approximately 11,000 light-years (3.4 kpc) away from us in the Milky Way.

7. The Andromeda Galaxy /ænˈdrɒmɨdə/ is a spiral galaxy approximately 2.5 million light-years (2.4×1019 km) from Earth in the Andromeda constellation.

So, for example, Casiopeia constellation is 11 000 ly away from us, and the distance to the main stars of the Big Dipper ranges from about 68 light-years (ly) to about 210 light-years, and while the Earth whirls in it's 300 000 000 km wide orbit around the Sun there is no change in relative position of the stars (for the observer on the Earth) due to their very different distances from the Earth???
« Last Edit: June 16, 2017, 05:17:27 AM by cikljamas »
"I can't breathe" George Floyd RIP

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sokarul

  • 17555
  • Discount Chemist
No no no no
Sokarul

ANNIHILATOR OF  SHIFTER

Run Sandokhan run

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rabinoz

  • 26296
  • Real Earth Believer
One question for Jack :
Sorry, Jack's busy at the moment helping another client, may I be of assistance?

Quote from: cikljamas
Since the parallax of the moon is allegedly 400 times greater than the parallax of the sun due to the "fact" that the sun is 400 times farther away than the moon, it means that the rotational effect of the earth is 400 times greater in relation to the moon, then in relation to the sun.

So, right after the total solar eclipse is over, when observing it from the northern hemisphere (we are facing the south) the sun is coming out (ceases to be obscured by the moon) TO THE RIGHT (as if the sun apparently moves faster - in it's "apparent" motion TO THE RIGHT - than the moon).

In the southern hemisphere, the situation is totally opposite. We are facing north, and the sun is coming out of the moon's eclipse TO THE LEFT (as if the sun apparently moves faster - in it's "apparent" motion TO THE LEFT - than the moon).
Incorrect and totally irrelevant. Parallax is a measure of how much the angle to the object varies due to a change in the viewing location.
On earth the largest possible baseline is the earth's diameter, not that parallax measurements have been made from that baseline).
So the maximum parallax
           of the moon would be about 12,742/381,600 radians or 1.9° and
           of the sun would be about 12,742/149,000,000 radians or 0.0049°.
But these do not bear the slightest relation to how much they appear to move as the earth rotates and the moon orbits.

Quote from: cikljamas
Due to the fact that the rotational effect is 400 times greater in relation to the moon (than in relation to the sun), right after the total solar eclipse is over, when observing it from the northern hemisphere the sun should come out (ceases to be obscured by the moon) TO THE LEFT (contrary from what happens in reality), and when observing it from the southern hemisphere the sun should come out TO THE RIGHT (contrary from what occurs in reality, also)!!!
Incorrect and totally irrelevant.
The rotational effect in relation to the moon is not is 400 times greater than that in relation to the the sun.
The lunar and solar parallax are totally irrelevant!

Try again!

Quote from: cikljamas
This is the reason why HC theory needed to reverse moon's motion to the left (in an opposite direction from everyone who has ever lived seen it go).
The HC system never needed to "reverse moon's motion". As soon as it was accepted that the earth rotated on its axis once in about 23.9344699 hours (tat precision was not known back there),

Quote from: cikljamas
The alleged speed of the moon (to the left) is 3660 km/h
The sun is allegedly stationary.
However, earth-moon system allegedly travels (to the left) 108 000 km/h
So?

Quote from: cikljamas
Now, you could say, 108 000 km/h is ONLY 30 times greater than moon's alleged motion in an opposite direction from the orbital motion (to the right) of the earth-moon system, but since the parallax of the moon is 400 times greater than sun's parallax, it means that the alleged speed of the moon should be 400 times greater than the orbital motion of the earth-moon system in order to cancel it out.
Once again! The parallax of the moon and sun are totally irrelevant, other that as an indication of their distances.

Quote from: cikljamas
But such an answer doesn't add up, since the earth's rotational effect is already 400 times greater in relation to the moon, than in relation to the sun.

So, if moon's motion (to the left) is 13,5 times greater than the speed needed to overcome orbital motion of earth-moon system (to the right) it is not enough, since the rotational effect of the earth is 400 times greater in relation to the moon, than in relation to the sun.
Of course it doesn't add up! You claim "the rotational effect of the earth is 400 times greater in relation to the moon, than in relation to the sun" is total hogwash!

Quote from: cikljamas
What if we applied this situation to the stars?
<<< more utterly irrelevant claims >>>
If you are going yo argue against the Heliocentric Globe, please go and learn a little bit about the subject and stop wasting everyone's time!

I finally see the relevance of Mikey T's post!
This is how I feel every time I "discuss" things with an FE hoaxer ignoramus.

Maybe when JackBlack reads your post he might have kinder words to say, but I seriously doubt it!

You're always using the typical Straw-man arguments. You stae that the HC claims something totally false and ridiculous then spend pages proving that your own totally false and ridiculous claims are totally false and ridiculous.
How smart and amazing you are, proving that your own total garbage is total garbage!
If such deception is what it takes to prop up your Pathetic Pepperoni Pizza Planet, then maybe you should toss it all in the trash-can!

Yes, your ZIG-ZAG trash is irrefutable proof that you have not the slightest understanding of the Heliocentric system.

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cikljamas

  • 2069
  • Ex nihilo nihil fit
« Last Edit: June 16, 2017, 05:46:08 AM by cikljamas »
"I can't breathe" George Floyd RIP

One question for Jack :

Since the parallax of the moon is allegedly 400 times greater than the parallax of the sun due to the "fact" that the sun is 400 times farther away than the moon, it means that the rotational effect of the earth is 400 times greater in relation to the moon, then in relation to the sun.
No it doesn't.
While the approximation that the parallax is 400 times greater as it is 1/400 times the distance is effectively correct (as the parallax is effectively given by atan(s/d), where s is the sideways displacement (causing parallax) and d is the distance, where s/d is very small and thus the angle is small, and for small x, tan(x)=x noting that x is in radians), the rest is not.

The parallax is a deviation from the rotational effects.
The rotational effect is the same, the moon will have a greater deviation due to parallax (but there is a more important point that will come up in a second)

So, right after the total solar eclipse is over, when observing it from the northern hemisphere (we are facing the south) the sun is coming out (ceases to be obscured by the moon) TO THE RIGHT (as if the sun apparently moves faster - in it's "apparent" motion TO THE RIGHT - than the moon).
Parallax and the rotation of Earth have effectively nothing to do with this. Remember, the parallax is quite small, and the rotation is the same. So what will matter is motion in addition to that.
For the prior discussion we have been using relative motion based upon solar days. So for the sun, that means there is no additional motion. But remember, the moon is orbiting Earth to the left, such that in 5 minutes it would move 0.04 degrees to the left.
This is the dominant cause of that apparent motion.

(note: apparent motion):
The sun is moving to the right at a rate of 1.25 degrees per 5 minutes. The moon is moving to the right at a rate of 1.21 degrees per 5 minutes. At most, at mid day for an eclipse, it would be 1.218 degrees to the right.
That means the apparent motion of the sun relative to the moon would be 0.04 degrees over 5 minutes, TO THE RIGHT!

So what is the problem here?
The sun has an apparent motion to the right that is faster than the moon, so you would expect the sun to appear to move to the right relative to the moon after an eclipse.


In the southern hemisphere, the situation is totally opposite. We are facing north, and the sun is coming out of the moon's eclipse TO THE LEFT (as if the sun apparently moves faster - in it's "apparent" motion TO THE LEFT - than the moon).
Yes, it is opposite due to the change in reference.
In the northern hemisphere, you are looking due south, with west to your right and east to your left.
In the southern hemisphere you are looking due north, with west to your left and east to your right.
So it appears opposite as you are effectively upside down.
Don't worry you can easily simulate this without going to the other hemisphere by standing on your head.

Due to the fact that the rotational effect is 400 times greater in relation to the moon (than in relation to the sun), right after the total solar eclipse is over, when observing it from the northern hemisphere the sun should come out (ceases to be obscured by the moon) TO THE LEFT (contrary from what happens in reality), and when observing it from the southern hemisphere the sun should come out TO THE RIGHT (contrary from what occurs in reality, also)!!!
That isn't a fact. That is a baseless claim that goes directly against HC theory.
Due to the fact that the sun is moving at a faster relative rate to the right than the moon it should appear to move to the right of the moon after an eclipse.
This model matches what happens in reality.

This is the reason why HC theory needed to reverse moon's motion to the left (in an opposite direction from everyone who has ever lived seen it go).
Yes, in opposite direction to that which it appears to be going. The same can happen with someone walking around a merry go round. This isn't a flaw in the model.

What if we applied this situation to the stars?
How about we stick to the simple system of the moon and now the sun to make sure you understand it first?
I will respond this time, but if you keep up ignoring the main issues I will stop going on with the more complex stuff.

If the Earth moved in it's 300 000 000 km wide orbit, and even if we allowed that HC bull shit assumption (regarding "endless" distances of the stars) is true, what kind of change we would be able to observe in the sky, anyway?
No BS. The near stars are measured by their parallax.
Proxima Centuri is only 4.2 light years away. Alpha Centuri is 4.4

So, for example, Casiopeia constellation is 11 000 ly away from us, and the distance to the main stars of the Big Dipper ranges from about 68 light-years (ly) to about 210 light-years, and while the Earth whirls in it's 300 000 000 km wide orbit around the Sun there is no change in relative position of the stars (for the observer on the Earth) due to their very different distances from the Earth???
Yes, that is right, due to how small the parallax is. Why don't we just use a single unit, light years. The orbit of Earth is 0.0000317 ly in diameter and 0.0000159 ly in radius
The closest star in the big dipper is 68 ly away.
So, the maximum change in its angular position is given by atan(0.0000159/68). That is roughly 0.000014 degrees or less than 0.05 seconds of arc.

That means from one point in the year to half a year later, it will change position by 0.05 seconds of arc (as opposed to the sun, with a parallax of roughly 180 degrees).
Compare that to the separation between the stars. I'm not sure which star you are using. I assume you are using delta Ursae Majoris, at a distance of 58 ly (not going back to correct the math, the difference is tiny).
The closest one to that is gamma Ursae Majoris.
Lets just look at the declination (which will underestimate the difference). delta is at 57° 01′ 57.4156″. gamma is at 53° 41′ 41.1350″
That is a difference of 3° 20′ 16.2806″.
Or, to get it all in the same units, that is 12016.2806 arc seconds.
You are looking for a difference of less than 0.05 out of 12016.2806. That works out to be 1 part in 240000. Do you really think that would be noticeable with the naked eye?
I don't.
There is only 1 star with a parallax that great (talking about parallax for half a year), and that is sol which appears to move 180 degrees relative to the background stars over the course of half a year.

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