Where exactly is it?

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cikljamas

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Re: Where exactly is it?
« Reply #210 on: August 03, 2020, 03:34:18 PM »
The Average Motion of the Moon
Your math is wrong.
You are completely ignoring how it turns by 15 degrees, making it so you cannot just pretend it is linear motion.
Treating it as linear motion results in a very significant error.

My math is perfectly correct! What are you talking about? What linear motion? What is wrong with you? If you dug perfectly straight tunnel which connects two points (let's say at the equator) which are 1000 miles apart, for how much would you short that distance, what do you think?
"I can't breathe" George Floyd RIP

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JJA

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Re: Where exactly is it?
« Reply #211 on: August 03, 2020, 03:51:35 PM »
The Average Motion of the Moon
Your math is wrong.
You are completely ignoring how it turns by 15 degrees, making it so you cannot just pretend it is linear motion.
Treating it as linear motion results in a very significant error.

My math is perfectly correct! What are you talking about? What linear motion? What is wrong with you? If you dug perfectly straight tunnel which connects two points (let's say at the equator) which are 1000 miles apart, for how much would you short that distance, what do you think?

By asking that question you seem to already understand that a straight line is different from a curved one.

This is why your math is wrong, you're using straight lines when objects on the surface move in curves. They are not the same.

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JackBlack

  • 21560
Re: Where exactly is it?
« Reply #212 on: August 03, 2020, 03:58:24 PM »
My math is perfectly correct! What are you talking about? What linear motion?
The linear motion I have already pointed out.
You are pretending that over the course of an hour the moon travels in a perfectly straight line and likewise, so does the people on Earth.
This introduces very significant errors.

Here is an image to demonstrate it:

Purple is where you falsely place the moon/observer.
Green is where it should be. (Although that is just focusing on one error, you have 2 errors in opposite direction).

For your example you have the moon travelling at 100 000 km/hr.
If we start with the moon at (0,0), and have it initially moving to the left, then after 1 hour this places it at (-100000, 0).
But if you don't act like someone who failed math, and instead you actually do the calculation properly (but still pretending it is 15 degrees per hour) you end up with the moon at (-98998, -13033), and if you have the moon properly move ~14.45 degrees you end up with it at (-95477, -12108).
Notice the big difference?
You are out by over 1%. This then impacts the angles you calculate.


Like I said before, in your simple case of the velocity of the moon being based upon completing a circle in a day you end up with the angular velocity at the centre being 15 degrees per hour.
But that means the near side MUST have an angular velocity GREATER than 15 degrees per hour.
Instead you produce the nonsense of 14.74 degrees per hour.

Doing it "properly" for the GC fantasy such that after 27.5 days it circles Earth 26.5 times so it has a period of just over 1 day, it should be 14.45 degrees per hour, and again it MUST be faster for the point closer to the moon. But your nonsense produces 14.25.
The angles you calculate for the distance the moon travels is completely incorrect for your GC fantasy.

You have an error in your calculated angles of roughly 0.2 to 0.3 degrees.
That is the size of the angle you are trying to find.
As such your error completely swamps any difference.

As such, any conclusions drawn from it are likewise nonsense.

Like I said, do the math properly, without rounding pi to 3.14, without just saying it is 100 000 km, without switching 2541 km with 2500 km, and either for a much shorter period of time like 1 s, where then the moon only moves 15 arc seconds, or using the actual math required (law of cosines) to determine angles rather than pretending it is all just straight lines; or actually look at the math I have provided and tell me what is wrong with it.
Preferably this math here which is the general case which shows the 2 angles are identical:
https://www.theflatearthsociety.org/forum/index.php?topic=70921.msg1917997#msg1917997

If you can't stop pretending you are correct or that your math is perfect when in fact it is fundamentally flawed.
« Last Edit: August 03, 2020, 04:14:53 PM by JackBlack »

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cikljamas

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  • Ex nihilo nihil fit
Re: Where exactly is it?
« Reply #213 on: August 04, 2020, 02:30:01 AM »
The Average Motion of the Moon
Your math is wrong.
You are completely ignoring how it turns by 15 degrees, making it so you cannot just pretend it is linear motion.
Treating it as linear motion results in a very significant error.

My math is perfectly correct! What are you talking about? What linear motion? What is wrong with you? If you dug perfectly straight tunnel which connects two points (let's say at the equator) which are 1000 miles apart, for how much would you short that distance, what do you think?

By asking that question you seem to already understand that a straight line is different from a curved one.

This is why your math is wrong, you're using straight lines when objects on the surface move in curves. They are not the same.

Original version of my ZIGZAG argument presumes mimicking linear motion of two observers on the opposite sides of the Arctic circle by using directional gyros (or by other means) for correction purposes. However, as soon as we use directional gyro for that purpose the game is over for a different reason. So, basically the practical application of my ZIGZAG argument depends on using gyros and because of that it boils down to some other kind of falsification of earth's rotation hypothesis by conducting much more simpler experiment (directional gyro experiment).

My math is perfectly correct! What are you talking about? What linear motion?
The linear motion I have already pointed out.
You are pretending that over the course of an hour the moon travels in a perfectly straight line and likewise, so does the people on Earth.
This introduces very significant errors.

Here is an image to demonstrate it:

Purple is where you falsely place the moon/observer.
Green is where it should be. (Although that is just focusing on one error, you have 2 errors in opposite direction).

For your example you have the moon travelling at 100 000 km/hr.
If we start with the moon at (0,0), and have it initially moving to the left, then after 1 hour this places it at (-100000, 0).
But if you don't act like someone who failed math, and instead you actually do the calculation properly (but still pretending it is 15 degrees per hour) you end up with the moon at (-98998, -13033), and if you have the moon properly move ~14.45 degrees you end up with it at (-95477, -12108).
Notice the big difference?
You are out by over 1%. This then impacts the angles you calculate.


Like I said before, in your simple case of the velocity of the moon being based upon completing a circle in a day you end up with the angular velocity at the centre being 15 degrees per hour.
But that means the near side MUST have an angular velocity GREATER than 15 degrees per hour.
Instead you produce the nonsense of 14.74 degrees per hour.

Doing it "properly" for the GC fantasy such that after 27.5 days it circles Earth 26.5 times so it has a period of just over 1 day, it should be 14.45 degrees per hour, and again it MUST be faster for the point closer to the moon. But your nonsense produces 14.25.
The angles you calculate for the distance the moon travels is completely incorrect for your GC fantasy.

You have an error in your calculated angles of roughly 0.2 to 0.3 degrees.
That is the size of the angle you are trying to find.
As such your error completely swamps any difference.

At the first glance it seems as if this mathematical method works pretty well :

If Earth is orbiting Sun, then we calculate relative to Sun:
Earth moves (2*Pi*149.6e6)/(365.25*24) = 107 232.5 km/h
Moon moves 107 232.5 ± (2*Pi*384 400)/(27.35*24) = 107 232.5 ± 3679.5 km/h
During solar eclipse it is minus, so we have 97 553 km/h.
Two observers in polar circle, one at closer end and another at farther end will travel 107 232.5 ± (2*Pi*2600)/24 = 107 232.5 ± 681 km/h
Closer observer 106 551.5 km/h, farther observer 107 913.5 km/h.
Now:
Closer observer: 106 551.5 - 97 553 = 8998.5 km/h ; ARCTAN(8998.5/381800) = 1.35 degrees per hour.
Farther observer: 107 913.5 - 97 553 = 10 360.5 km/h ; ARCTAN(10360.5/387000) = 1.53 degrees per hour
Angular speed difference between observers 0.18 degrees per hour.


If Space is orbiting Earth then we calculate relative to Earth:
Sun moves (2*Pi*149.6e6)/24 = 39 165 188.4 km/h
Moon moves (2*Pi*384 400)/24 - 3679.5 = 96 956.2 km/h
Now:
Closer observer: Sun ARCTAN(39165188.4 / 1496e6) = 14.67 degrees per hour ; Moon ARCTAN(96956.2 / 381800) = 14.245 degrees per hour ; difference 0.425
Farther observer: Sun ARCTAN(39165188.4 / 1496e6) = 14.67 degrees per hour ; Moon ARCTAN(96956.2 / 387000) = 14.065 degrees per hour ; difference 0.605
Angular speed difference between observers 0.18 degrees per hour.


So, i only made few modifications in order to make the math above simpler and more accurate. I got the different values (the slight overall differences) for HC and GC scenarios, and that is what i wanted to point out : even if we leave out the core of the original version of my ZIGZAG argument (mimicking linear motion by using directional gyros) we still get the difference between GC and HC model.

However, i noticed two problems (after pondering about your objections) :

In GC scenario i have to use 98 600 km/h for the speed of the moon instead of 96 434 km/h to get 0,18 degrees difference.
In HC scenario we would have to swap values for closer and farther observer so to get greater angle for closer observer and smaller angle for farther observer.

So, be that as it may, even if it turns out that we can't give up the core of my ZIGZAG argument (mimicking linear motion of the observers on the edge of the Arctic circle) so to be able to compare the giveaway differences between two different scenarios (which might be revealed even by using this modified version of verification (of the existence of earth's rotation)), we can still (at any time) perform the simplest experiment of all (as Foucault did with his primitive gyro that vanished into thin air, just like so many other important scientific devices have vanished (and we know why they have been "lost")) by using directional gyros!!!

"I can't breathe" George Floyd RIP

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JackBlack

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Re: Where exactly is it?
« Reply #214 on: August 04, 2020, 05:54:50 AM »
Original version of my ZIGZAG argument presumes mimicking linear motion
So the original version of your argument is a pathetic strawman of reality, which in no way shows a difference between GC and HC.

If your argument relies upon it being linear motion, it fails.

So, basically the practical application of my ZIGZAG argument depends on using gyros and because of that it boils down to some other kind of falsification of earth's rotation
So what you are saying is that your zig-zag argument is pure nonsense and is completely incapable of determining if Earth is rotating with the moon stationary, Earth is stationary with the moon circling, and Earth rotating with the moon orbiting, i.e. exactly what I said before.

Care to just directly admit that?

So, i only made few modifications in order to make the math above simpler and more accurate.
No, you make it simpler and less accurate.
Like I said, your own numbers for the calculated angles are off by 0.2 degrees.
The difference in angle between the 2 locations is less than 0.2 degrees.
And the difference between the differences is smaller still.

The difference you are so desperately clinging to is entirely the result of those errors.

Like I said, DO IT PROPERLY

even if we leave out the core of the original version of my ZIGZAG argument (mimicking linear motion by using directional gyros) we still get the difference between GC and HC model.
NO, we don't.
If instead you treat it properly as angular motion you end up with identical values.
If you use a linear approximation which actually holds, you end up with values which are the same within error.

You do not get a difference between GC and HC.
It is only with significant mistakes that you pretend there is a difference.

Again, if you wish to assert that there is a difference between GC and HC, DO IT PROPERLY!
Either use a period of time where the linear approximation actually holds, e.g. 1 second, or do it properly using the law of cosines and the actual arcs involved.

All the work has already been done for you here:
https://www.theflatearthsociety.org/forum/index.php?topic=70921.msg1917997#msg1917997
It just shows that you are wrong.
It clearly shows, using the correct math based upon the points following circles rather than magically travelling off in a straight line, that the 2 scenarios produce the same difference in angle.

If you can't do it honestly, stop pretending there is a difference.

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cikljamas

  • 2432
  • Ex nihilo nihil fit
Re: Where exactly is it?
« Reply #215 on: August 04, 2020, 07:42:44 AM »
Jack, if you are right then Macarios is wrong, and i am still right :



So, if you are right (let's say that you are) by giving up the core of my ZIGZAG argument we have to give up this procedure, as well :

The math for the HELIOCENTRIC 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

4304/385000 = 0,01117922 --- (ctg)0,01117922 = 0,6404
2974/380000 = 0,00782631 ----(ctg)0,00782631 = 0,4484

However, the procedure above is perfectly correct if we retain the core of my ZIGZAG argument.

So, it seems that we have to go back to the core of my ZIGZAG argument which presumes mimicking linear motion of the observers on the edge of the Arctic circle which means that by pointing to a certain star within REAL (not mere alleged) HC reality for an hour of time the moon would really and apparently move 0,64 degrees to the left (for farther observer) and 0,44 degrees to the left (for closer observer). In REAL (not mere alleged) HC reality pointing to a certain star would coincide with directional gyro keeping rigidity in space, that is to say : the pointer of a directional gyro would be always aligned with the star we are following, as well.

In our GC reality by strictly following a certain star we would get the opposite result (the angle would be greater for closer observer and smaller for farther observer), and while keeping rigidity in space our directional gyro would (apparently) drift 15 degrees/hour (wrt the stars), better to say the stars would move 15 degrees/hour with respect to a rigid pointer of directional gyro which would be directed let's say towards North, all the time.
"I can't breathe" George Floyd RIP

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JackBlack

  • 21560
Re: Where exactly is it?
« Reply #216 on: August 04, 2020, 02:29:04 PM »
Jack, if you are right then Macarios is wrong, and i am still right
If he is treating 15 degrees as still being linear, then he is wrong.
Either way, YOU ARE WRONG!

Again, is you wish to disagree, then either do the math properly, without pretending it is linear, using the law of cosines; or by using a much shorter period of time where the linear approximation holds, e.g. 1 second.
And either way, you still need to deal with my math which refutes you.

The reason I keep bringing up the option of 1 second is quite simple.
If there actually was a difference which changes by which is moving, then the ratio of these differences (as difference HC/difference GC) will only grow as you get to a shorter and shorter time period as it maximises when the difference occurs (e.g. focusing on half a day would give no difference, while focusing on only the tiny fraction of time when the difference is greatest should produce the largest ratio).
If instead the difference is purely a fabrication based upon you making a massive error with your linear approximation not holding, then the better the linear approximation becomes the smaller the ratio becomes, until it reaches unity.

And if you do the math properly (i.e. without rounding off prematurely other than to the limit of Excel's precision or saying it is roughly 100 000 km), then you get the following ratios (note: the latter ones are shown to more decimal places to show were the first point of difference is):
3600 s - 1.064
1800 s - 1.016
600 s - 1.002
300 s - 1.0004
60 s - 1.00002
30 s - 1.000004
1 s - 1.000000005

And do you know what that indicates?
That this difference is entirely a fabrication due to your incorrect linear approximation for such a large range.

The other thing to note, as pointed out before, with your 27.5 day period for the moon's orbit, this gives us an angular velocity of the moon (at the north pole) of 14.45 degrees per hour. This equates to 14.45 arc seconds per second.
Using the GC model in the case of 1 s, you get angular displacements of 14.55 and 14.36 arc seconds. One above and one below, just like you would expect.
And if you look at the angle covered at the north pole, you get 14.45 arc seconds, just like you would expect instead of the 14.16 degrees you get from 1 hour.

Once again, this shows the difference is entirely due to you using a linear approximation.

So, it seems that we have to go back to the core of my ZIGZAG argument
The core of your zigzag argument is treating a clearly non-linear path as a linear path to get a significant error which falsely indicates there is a difference when there is none.

So no, we don't have to go back to that dishonesty at all.

So, it seems that we have to go back to the core of my ZIGZAG argument which presumes mimicking linear motion of the observers on the edge of the Arctic circle which means that by pointing to a certain star within REAL (not mere alleged) HC reality for an hour of time the moon would really and apparently move 0,64 degrees to the left (for farther observer) and 0,44 degrees to the left (for closer observer). In REAL (not mere alleged) HC reality pointing to a certain star would coincide with directional gyro keeping rigidity in space, that is to say : the pointer of a directional gyro would be always aligned with the star we are following, as well.

In our GC reality by strictly following a certain star we would get the opposite result (the angle would be greater for closer observer and smaller for farther observer), and while keeping rigidity in space our directional gyro would (apparently) drift 15 degrees/hour (wrt the stars), better to say the stars would move 15 degrees/hour with respect to a rigid pointer of directional gyro which would be directed let's say towards North, all the time.
Still wrong.
Even using your faulty linear approximation, you are right that in the HC reality, the moon moves 0.45 degrees to the left for the near observer and 0.64 degrees for the far observer.
But for the GC fantasy, the star moves 15 degrees to the right, while the moon for the near observer moves 14.25 degrees to the right for the near observer which is 0.75 degrees left of the star, while for the far observer it moves 14.07 degrees right, which is 0.93 degrees left of the star.

If instead of using a completely broken linear approximation we limit the time to 1 second so the linear approximation holds, then in reality, the moon has moved 0.45 arc seconds for the near observer and 0.64 arc seconds for the far observer, still both to the left, while in the GC fantasy the stars move 15 arc seconds right, while the moon moves 14.55 arc seconds right, and thus is 0.45 arc seconds left of the star, while the far observer sees the moon move 14.36 arc seconds to the right and thus the moon is 0.64 arc seconds left of the star.

So you are wrong, yet again.
In both cases, the HC reality and the GC fantasy, the moon appears to move move relative to the background stars for the far observer than for the near observer.
If you limit yourself to the time where the linear approximation actually holds, this difference in angle is the same for HC reality and GC fantasy.

Again, this cannot allow you to distinguish between the 2.

Again, the 2 situations are literally just related by a simple rotation.
If you take the HC system (centred on Earth) and rotate it by roughly 15 degrees per hour, you end up with the GC system.
This means any simple visual observation will be identical for both, i.e. they cannot distinguish between the 2.

The simple fact is such a simple visual observation cannot distinguish between Earth rotating with the moon stationary, the moon circling with Earth stationary, or both moving.


Now going to actually try to address the arguments that have been raised? Or will you just continue to pretend they haven't been made and spam the same refuted nonsense again?
« Last Edit: August 04, 2020, 03:13:47 PM by JackBlack »

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cikljamas

  • 2432
  • Ex nihilo nihil fit
Re: Where exactly is it?
« Reply #217 on: August 05, 2020, 03:21:43 AM »
The Average Motion of the Moon
Your math is wrong.
You are completely ignoring how it turns by 15 degrees, making it so you cannot just pretend it is linear motion.
Treating it as linear motion results in a very significant error.

My math is perfectly correct! What are you talking about? What linear motion? What is wrong with you? If you dug perfectly straight tunnel which connects two points (let's say at the equator) which are 1000 miles apart, for how much would you short that distance, what do you think?

By asking that question you seem to already understand that a straight line is different from a curved one.

This is why your math is wrong, you're using straight lines when objects on the surface move in curves. They are not the same.

Original version of my ZIGZAG argument presumes mimicking linear motion of two observers on the opposite sides of the Arctic circle by using directional gyros (or by other means) for correction purposes. However, as soon as we use directional gyro for that purpose the game is over for a different reason. So, basically the practical application of my ZIGZAG argument depends on using gyros and because of that it boils down to some other kind of falsification of earth's rotation hypothesis by conducting much more simpler experiment (directional gyro experiment).

My math is perfectly correct! What are you talking about? What linear motion?
The linear motion I have already pointed out.
You are pretending that over the course of an hour the moon travels in a perfectly straight line and likewise, so does the people on Earth.
This introduces very significant errors.

Here is an image to demonstrate it:

Purple is where you falsely place the moon/observer.
Green is where it should be. (Although that is just focusing on one error, you have 2 errors in opposite direction).

For your example you have the moon travelling at 100 000 km/hr.
If we start with the moon at (0,0), and have it initially moving to the left, then after 1 hour this places it at (-100000, 0).
But if you don't act like someone who failed math, and instead you actually do the calculation properly (but still pretending it is 15 degrees per hour) you end up with the moon at (-98998, -13033), and if you have the moon properly move ~14.45 degrees you end up with it at (-95477, -12108).
Notice the big difference?
You are out by over 1%. This then impacts the angles you calculate.


Like I said before, in your simple case of the velocity of the moon being based upon completing a circle in a day you end up with the angular velocity at the centre being 15 degrees per hour.
But that means the near side MUST have an angular velocity GREATER than 15 degrees per hour.
Instead you produce the nonsense of 14.74 degrees per hour.

Doing it "properly" for the GC fantasy such that after 27.5 days it circles Earth 26.5 times so it has a period of just over 1 day, it should be 14.45 degrees per hour, and again it MUST be faster for the point closer to the moon. But your nonsense produces 14.25.
The angles you calculate for the distance the moon travels is completely incorrect for your GC fantasy.

You have an error in your calculated angles of roughly 0.2 to 0.3 degrees.
That is the size of the angle you are trying to find.
As such your error completely swamps any difference.

At the first glance it seems as if this mathematical method works pretty well :

If Earth is orbiting Sun, then we calculate relative to Sun:
Earth moves (2*Pi*149.6e6)/(365.25*24) = 107 232.5 km/h
Moon moves 107 232.5 ± (2*Pi*384 400)/(27.35*24) = 107 232.5 ± 3679.5 km/h
During solar eclipse it is minus, so we have 97 553 km/h.
Two observers in polar circle, one at closer end and another at farther end will travel 107 232.5 ± (2*Pi*2600)/24 = 107 232.5 ± 681 km/h
Closer observer 106 551.5 km/h, farther observer 107 913.5 km/h.
Now:
Closer observer: 106 551.5 - 97 553 = 8998.5 km/h ; ARCTAN(8998.5/381800) = 1.35 degrees per hour.
Farther observer: 107 913.5 - 97 553 = 10 360.5 km/h ; ARCTAN(10360.5/387000) = 1.53 degrees per hour
Angular speed difference between observers 0.18 degrees per hour.


If Space is orbiting Earth then we calculate relative to Earth:
Sun moves (2*Pi*149.6e6)/24 = 39 165 188.4 km/h
Moon moves (2*Pi*384 400)/24 - 3679.5 = 96 956.2 km/h
Now:
Closer observer: Sun ARCTAN(39165188.4 / 1496e6) = 14.67 degrees per hour ; Moon ARCTAN(96956.2 / 381800) = 14.245 degrees per hour ; difference 0.425
Farther observer: Sun ARCTAN(39165188.4 / 1496e6) = 14.67 degrees per hour ; Moon ARCTAN(96956.2 / 387000) = 14.065 degrees per hour ; difference 0.605
Angular speed difference between observers 0.18 degrees per hour.


So, i only made few modifications in order to make the math above simpler and more accurate. I got the different values (the slight overall differences) for HC and GC scenarios, and that is what i wanted to point out : even if we leave out the core of the original version of my ZIGZAG argument (mimicking linear motion by using directional gyros) we still get the difference between GC and HC model.

However, i noticed two problems (after pondering about your objections) :

1. In GC scenario i have to use 98 600 km/h for the speed of the moon instead of 96 434 km/h to get 0,18 degrees difference.
2. In HC scenario we would have to swap values for closer and farther observer so to get greater angle for closer observer and smaller angle for farther observer.



This is a geometrical explanation for the problem No 1 :





So, now we've got a physical corroboration for the correctness of my math :

CASE 2 (GC scenario) :

Quote
The GEOCENTRIC scenario :

100 000/385 000 = 0,259740259 --- (ctg)0,259740259 = 14,5602
100 000/380 000 = 0,263157894 --- (ctg)0,263157894 = 14,7435

14,7435-14,5602 = 0,183

96 500/385 000 = after the same procedure as above we get 14,0712 degrees
96 500/380 000 = after same procedure as above we get 14,2489 degrees

14,2489-14,0712 = 0,177 which makes even greater difference between geocentric (0,192 ) scenario and heliocentric scenario...

We are looking for the difference which is somewhere in between 0,183 and 0,177 :

98600/385000 = 0,256103896 --- (ctg) 0,256103896 = 14,36
98600/380000 = 0,259473684 --- (ctg) 0,259473684 = 14,54

14,54-14,36 = 0,18
0,18/2= 0,09

14,36 + 0,09 = 14,45
14,54 - 0,09 = 14,45

14,45 * 24 = 346,8 (exact match : 13,2 degrees less than 360 degrees (Moon's average motion relative to the stars)

This is a geometrical explanation for the problem No 2 :



Since the numbers are much smaller than in GC scenario (Moon moves only 3639 km instead of 96 500 km, and we have had to add or subtract from that number (3639 km) even smaller number (the distance (665 km) our observers travel in 1 hour)), then i don't think we have to check in a similar (physical) manner (as above (GC case)) whether there is some geometrical issue with the following math :

Now, the math for the HELIOCENTRIC 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

4304/385000 = 0,01117922 --- (ctg)0,01117922 = 0,6404
2974/380000 = 0,00782631 ----(ctg)0,00782631 = 0,4484

0,6404-0,4484 = 0,192 [/b][/size][/quote]

0,192 / 2 = 0,096

0,6404 -  0,096 = 0,5444
0,4484 + 0,096 = 0,5444

15 degrees - 0,544 = 14,456
14,456*24 = 346,944 which is very, very close to 13,2 degrees less than 360 degrees (Moon's average motion relative to the stars).

CONCLUSION : Calculated HC difference 0,6404-0,4484 = 0,192 [/b][/size][/quote] is quite correct!!!
« Last Edit: August 05, 2020, 03:30:15 AM by cikljamas »
"I can't breathe" George Floyd RIP

Re: Where exactly is it?
« Reply #218 on: August 05, 2020, 03:54:50 AM »
Bruh, your writing looks more like arabic than latin

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JackBlack

  • 21560
Re: Where exactly is it?
« Reply #219 on: August 05, 2020, 04:20:19 AM »
You ignoring all the refutations of your claims wont' magically make it go away.

So, now we've got a physical corroboration for the correctness of my math :
No, we have a few pictures with no explanation which do not help your case at all.
Your argument still relies treating a clearly non-linear situation as linear. It is this false approximation which causes the difference in values between your faulty GC model and your faulty HC model.

Once again, the moon doesn't travel in a straight line.

If your argument really held and didn't rely upon this false approximation you would have done the much shorter period of time (1 second), rather than continuing with these pathetic deflections.

Like I said, actually address the arguments raised.
Either do the math properly without any linear approximations; or do a linear approximation for a much shorter period of time.
Either way, refute the math I provided which clearly shows the 2 produce identical results.

Once more, here is a link to the math:
https://www.theflatearthsociety.org/forum/index.php?topic=70921.msg1917997#msg1917997

Once more, what the ratio of the difference approach 1 as the linear approximation becomes more and more correct (due to the shorter distance), clearly showing that the difference you are focusing on between GC and HC is entirely due to your faulty linear approximation rather than an actual difference between the systems:
If you do the math properly (i.e. without rounding off prematurely other than to the limit of Excel's precision or saying it is roughly 100 000 km), then you get the following ratios (note: the latter ones are shown to more decimal places to show were the first point of difference is):
3600 s - 1.064
1800 s - 1.016
600 s - 1.002
300 s - 1.0004
60 s - 1.00002
30 s - 1.000004
1 s - 1.000000005

So will you actually honestly address this, or will you continue to bury your head in the sand?
« Last Edit: August 05, 2020, 04:25:14 AM by JackBlack »

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cikljamas

  • 2432
  • Ex nihilo nihil fit
Re: Where exactly is it?
« Reply #220 on: August 05, 2020, 06:36:39 AM »
All i can do for you Jack is to make a geometric (trigonometric) correctional allowance for Moon's circular motion (within HC paradigm, that is to say for one Moon's diameter/hour), just like i did for Moon's circular motion within GC reality (see my previous post). Even if i allowed 20 % allowance (which is more than anyone could ever dream about or dare to ask for) you will end up with 0,191 difference for HC scenario...You see, as i said in my previous post, HC numbers (for one of the cathetus in HC triangles) are much smaller, and that is why there is no room for correction here, no matter how generous i would be allowing you certain extensions (for Moon's motion within HC scenario) so to provide for you more precise  trigonometrical calculations (corrections)...So, it seems that you lost just one among many battles that you had already lost. However, no one will dare to fight this (ZIGZAG) battle on the ground (verifying our numbers on the battlefield (in reality)), because all experts already know that the earth is stationary (including you). After all, directional gyros in GC reality drift wrt the stars (the stars actually drift wrt DG), and in HC fairy tale DG would drift relative to the earth, which is pure fiction and utter fantasy!!!

No experiment has ever been performed with such excruciating persistence and meticulous precision, and in every conceivable manner, than that of trying to detect and measure the motion of the Earth. Yet they have all consistently and continually yielded a velocity for the Earth of exactly ZERO mph.

The toil of thousands of exasperated researchers, in the extremely varied experiments of Arago, De Coudre's induction, Fizeau, Fresnell drag, Hoek, Jaseja's lasers, Jenkins, Klinkerfuess, Michelson-Morley interferometry, Lord Rayleigh's polarimetry, Troughton-Noble torque, and the famous 'Airy's Failure' experiment, all conclusively failed to show any rotational or translational movement for the earth, whatsoever."

You can search all 29 volumes of this final authority on all scientific matters (Encyclopedia Britannica) but you will look in vain for any PROOF for this revolution of the earth around the sun and its spinning on its axis every 24 hours. It is simply stated as DOGMA and to doubt is to be damned to a spinning hell forever by the "scientific" community.

Many "astronomers" cite the Foucault pendulum experiment that was carried out in Paris in 1851 as PROOF that the earth turns. It's a pity that the NASA space program has not provided them with more recent proof.

Btw, we celebrate today (here in Croatia) one important anniversary (victory), an event that took place 25 years ago :
"I can't breathe" George Floyd RIP

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Stash

  • Ethical Stash
  • 13398
  • I am car!
Re: Where exactly is it?
« Reply #221 on: August 05, 2020, 10:43:02 AM »
No experiment has ever been performed with such excruciating persistence and meticulous precision, and in every conceivable manner, than that of trying to detect and measure the motion of the Earth. Yet they have all consistently and continually yielded a velocity for the Earth of exactly ZERO mph.

The toil of thousands of exasperated researchers, in the extremely varied experiments of Arago, De Coudre's induction, Fizeau, Fresnell drag, Hoek, Jaseja's lasers, Jenkins, Klinkerfuess, Michelson-Morley interferometry, Lord Rayleigh's polarimetry, Troughton-Noble torque, and the famous 'Airy's Failure' experiment, all conclusively failed to show any rotational or translational movement for the earth, whatsoever."

Funny, you stole and posted the same exact thing back in 2017:

No experiment has ever been performed with such excruciating persistence and meticulous precision, and in every conceivable manner, than that of trying to detect and measure the motion of the Earth. Yet they have all consistently and continually yielded a velocity for the Earth of exactly ZERO mph.

The toil of thousands of exasperated researchers, in the extremely varied experiments of Arago, De Coudre's induction, Fizeau, Fresnell drag, Hoek, Jaseja's lasers, Jenkins, Klinkerfuess, Michelson-Morley interferometry, Lord Rayleigh's polarimetry, Troughton-Noble torque, and the famous 'Airy's Failure' experiment, all conclusively failed to show any rotational or translational movement for the earth, whatsoever."


There too you did not cite the author of these words. Why is that? Why do you continue to plagiarize people? That's probably not the funniest part though. The funny part is that back in 2017 when you copy and pasted someone elses words the above was prefaced with this:

If the earth is at rest, can she still be round? If yes then why, if no then why?

The earth is motionless, there isn't the slightest doubt about that, now since we know for the fact that the earth is at rest, round-earth hypothesis isn't sustainable any more for many reasons.

What happened? You were stealing other people's words back then in support of a flat earth. Then fast forward 3 years and you're still stealing the exact words to support a motionless globe earth. Which is it?

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cikljamas

  • 2432
  • Ex nihilo nihil fit
Re: Where exactly is it?
« Reply #222 on: August 05, 2020, 11:14:11 AM »
What happened?

I had done my job, that is what happened...

ONLY A FOOL NEVER CHANGES HIS MIND!!!

5. Jon McIntyre - Truth Seeker says :
Hey Odiupicku I've got a couple of more flat earth tests I've done that seem to show curvature. I've got four videos up now that all seem to show curvature. I came at this debate from a completely neutral perspective and in truth I actually preferred to find that the earth was flat. That's because if it was and it could be proven the whole corrupt system running the world would collapse. To my disappointment I keep finding what appears to be curvature but the  truth is that is what I'm finding. I've actually got another test in the can and will be uploading that one soon as well. It is called "The Floating Levels Test" and it shows the surface of a lake to be convex or at least it clearly appears that way. Could you please mirror my new videos and give a link back to my channel. I ask you mostly because I believe that spreading truthful inquiry and experiments is valuable. I also feel that you have shown yourself to have the character to admit you are wrong and pursue the truth just like I did. Thanks for all of your work. I'll also let you know when my latest test is uploaded. Thanks!?

OBJECT LESSON?

ONLY A FOOL NEVER CHANGES HIS MIND!!!

READ MORE : https://www.theflatearthsociety.org/forum/index.php?topic=78821.msg2172534#msg2172534
READ THIS ALSO (i suggest you to read this whole thread, it is going to be very interesting to read it, indeed) : https://forum.tfes.org/index.php?topic=15449.msg200225#msg200225

Btw, one of the main reasons for believing in flat earth (at that time) was the problem of superluminal motion of the stars...since you are so curious malevolent vulture...

EDIT : Stash, i forgot to play a violin for you (sorry) : Relax Stash, just relax, you can't change the truth, so relax and keep singing while i am playing this nice melody for you...
« Last Edit: August 05, 2020, 11:20:58 AM by cikljamas »
"I can't breathe" George Floyd RIP

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JackBlack

  • 21560
Re: Where exactly is it?
« Reply #223 on: August 05, 2020, 03:14:22 PM »
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#msg1917997
It 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.