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IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?

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sandokhan

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #240 on: June 05, 2017, 03:10:10 PM »
CAN YOU SHOW ANYTHING WRONG WITH MY DERIVATION?

Your derivation stinks to high heaven.

It is a monumental piece of crap with no scientific value whatsoever.


How can you ask anybody to show what is wrong with your derivation when you made comments such as these:

Quote from: JackBlack on June 03, 2017, 11:47:37 PM
As such, for a given interferometer (with a fixed area of the light path), treating Earth's orbit as a rotation and Earth's daily rotation as a rotation where one revolution is 24 hours, the orbital Sagnac effect will be approximately 1/365 that of Earth's rotation, as the area is the same, the only thing that changes is the angular velocity which for the orbit is roughly 1/365 that of the day.

Quote from: JackBlack on June 04, 2017, 03:40:32 AM
No. It is easy to calculate that the orbital sagnac is a mere 1/365th that of the rotational.

That is because the 2 interferometers are the same and thus have the same R and the same A and thus it is entirely dependent upon ω

Again, the location of the centre of rotation is irrelevant.
All that matters is the area of the loop and its angular velocity.


Your derivation = a claim/conclusion that the orbital Sagnac = 1/365 of the rotational Sagnac.

As such any reference which proves otherwise, renders your derivation to be totally worthless.

Can you understand this much?


C.C. Su, "A Local-ether model of propagation of electromagnetic wave," in Bull. Am. Phys. Soc., vol. 45, no. 1, p. 637, Mar. 2000 (Minneapolis, Minnesota).

http://www.ee.nthu.edu.tw/ccsu/

Published in one of the most prestigious scientific journals in the world: Bulletin of the American Physical Society.

Both the rotational and the orbital motions of the earth together with the orbital
motion of the target planet contribute to the Sagnac
effect. But the orbital motion of the sun has no effects
on the interplanetary propagation. On the other hand, as
the unique propagation frame in GPS and intercontinental
links is a geocentric inertial frame, the rotational motion
of the earth contributes to the Sagnac effect. But the orbital
motion of the earth around the sun and that of the
sun have no effects on the earthbound propagation. By
comparing GPS with interplanetary radar, it is seen that
there is a common Sagnac effect due to earth’s rotation
and a common null effect of the orbital motion of the sun
on wave propagation. However, there is a discrepancy in
the Sagnac effect due to earth’s orbital motion. Moreover,
by comparing GPS with the widely accepted interpretation
of the Michelson–Morley experiment, it is seen that
there is a common null effect of the orbital motions on
wave propagation, whereas there is a discrepancy in the
Sagnac effect due to earth’s rotation.


Based on this characteristic of uniqueness and switchability of the propagation frame,
we propose in the following section the local-ether model
of wave propagation to solve the discrepancies in the in-
fluences of earth’s rotational and orbital motions on the
Sagnac effect and to account for a wide variety of propagation
phenomena.


Anyway, the interplanetary Sagnac effect is due to
earth’s orbital motion around the sun as well as earth’s
rotation. Further, for the interstellar propagation where
the source is located beyond the solar system, the orbital
motion of the sun contributes to the interstellar Sagnac
effect as well.

Evidently, as expected, the proposed local-ether model
accounts for the Sagnac effect due to earth’s rotation and
the null effect of earth’s orbital motion in the earthbound
propagations in GPS and intercontinental microwave link
experiments. Meanwhile, in the interplanetary radar, it accounts
for the Sagnac effect due both to earth’s rotation
and to earth’s orbital motion around the sun.


Based on the local-ether model, the propagation is entirely
independent of the earth’s orbital motion around
the sun or whatever and the velocity v for such an earthbound
experiment is referred to an ECI frame and hence
is due to earth’s rotation alone. In the original proposal,
the velocity v was supposed to incorporate earth’s orbital
motion around the sun. Thus, at least, v^2/c^2
=~ 10^-8. Then the amplitude of the phase-difference variation
could be as large as π/3, when the wavelength is
0.6 µm and the path length is 10 m. However, as the velocity
v is the linear velocity due to earth’s rotation alone,
the round-trip Sagnac effect is as small as v^2/c^2∼ 10^-12 which is merely 10^-4 times that due to the orbital motion.


The Sagnac effect is a FIRST ORDER effect in v/c.

Even in the round-trip nature of the Sagnac effect, as it was applied in the Michelson-Morley experiment, thus becoming a second order effect within that context, we can see that the ORBITAL SAGNAC IS 10,000 TIMES GREATER than the rotational Sagnac effect.


The orbital Sagnac is 10,000 times greater than the rotational Sagnac, a fact acknowledged by the Bulletin of American Physical Society, one of the top scientific journals in the world today.



You claimed, based on your derivation that the orbital Sagnac = 1/365 of the rotational Sagnac.


But actually the orbital Sagnac is 10,000 greater in magnitude than the rotational Sagnac.

The paper was published by the Bulletin of the American Physical Society.


Therefore, the claims/conclusions of your derivation have been shown to be totally erroneous, or to put it bluntly, they are totally worthless.
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rabinoz

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #241 on: June 05, 2017, 03:58:55 PM »
Quote from: sandokhan on June 05, 2017, 03:10:10 PM
CAN YOU SHOW ANYTHING WRONG WITH MY DERIVATION?

Your derivation stinks to high heaven.
Really! If there is anything wrong with JackBlack's derivation you would show how it should be done.

YOU can't do that, so just whose "derivation stinks to high heaven" - YOU CANNOT even do it, yet you complain.

Quote from: sandokhan
It is a monumental piece of crap with no scientific value whatsoever.

How can you ask anybody to show what is wrong with your derivation when you made comments such as these:


JackBlack's result: dt = 4.A.ω/c2
MathPages' resul: ", where A = πR2 is the area enclosed by the loop."

They look the same to my "untrained eye".

So YOU show in detail WHERE JackBlack's derivation is wrong!
You cannot do it because it is correct and we know that Sagnac does not depend on the shape of the loop!
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JackBlack

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #242 on: June 05, 2017, 04:15:54 PM »
Quote from: sandokhan on June 05, 2017, 03:03:52 PM
The peer reviewers at the Journal of Electromagnetic Waves and Applications agree that
They agree that that statement was used in a prior article which I assume is referenced and thus are happy to have it in the intro.

Also, if you read it in context, that is just for an aether model.

Quote from: Jane on June 05, 2017, 03:50:34 PM
If the orbital Sagnac had been 1/365 of the rotational Sagnac then nobody would have bothered with adopting a local aether model.
Except the nutters that want an aether model.

Quote from: Jane on June 05, 2017, 03:50:34 PM
It is only because the orbital Sagnac is much larger than the rotational Sagnac, and also missing, that the author and the peer reviewers agreed that a local aether model is needed.
Yes, for the nutters that want an aether model.
For those happy with reality, it isn't needed.

Quote from: sandokhan on June 05, 2017, 03:10:10 PM
CAN YOU SHOW ANYTHING WRONG WITH MY DERIVATION?
Your derivation stinks to high heaven.
It is a monumental piece of crap with no scientific value whatsoever.
Then why are you completely incapable of showing any flaw in the math or reasoning behind it?

Quote from: Jane on June 05, 2017, 03:50:34 PM
How can you ask anybody to show what is wrong with your derivation when you made comments such as these:
Because that is what the derivation shows.
Now can you show anything at all wrong with the derivation?
Can you show an error in the math?
Can you show a false assumption that was made?
If not, the derivation stands as correct.

Quote from: sandokhan on June 05, 2017, 03:10:10 PM
Your derivation = a claim/conclusion that the orbital Sagnac = 1/365 of the rotational Sagnac.
As such any reference which proves otherwise, renders your derivation to be totally worthless.
Can you understand this much?
Yes, which PROVES otherwise, not claims.
You are yet to provide one which does.
Instead all you can do is spout crap from articles using aether based models, which don't even use closed loops with counterpropogating waves, and thus have nothing to do with the Sagnac effect.
No where do they show anything at all wrong with my derivation or show that the orbital Sagnac effect is much larger.

Now then, CAN YOU SHOW ANYTHING WRONG WITH MY DERIVATION???
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Alpha2Omega

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #243 on: June 05, 2017, 08:15:45 PM »
Quote from: sandokhan on June 05, 2017, 03:10:10 PM
CAN YOU SHOW ANYTHING WRONG WITH MY DERIVATION?

Your derivation stinks to high heaven.

It is a monumental piece of crap with no scientific value whatsoever.

Well, that was informative. Thanks for taking the time to point out specific errors in such detail.

Quote
How can you ask anybody to show what is wrong with your derivation when you made comments such as these:

Quote from: JackBlack on June 03, 2017, 11:47:37 PM
As such, for a given interferometer (with a fixed area of the light path), treating Earth's orbit as a rotation and Earth's daily rotation as a rotation where one revolution is 24 hours, the orbital Sagnac effect will be approximately 1/365 that of Earth's rotation, as the area is the same, the only thing that changes is the angular velocity which for the orbit is roughly 1/365 that of the day.

Quote from: JackBlack on June 04, 2017, 03:40:32 AM
No. It is easy to calculate that the orbital sagnac is a mere 1/365th that of the rotational.

That is because the 2 interferometers are the same and thus have the same R and the same A and thus it is entirely dependent upon ω

Again, the location of the centre of rotation is irrelevant.
All that matters is the area of the loop and its angular velocity.

Your derivation = a claim/conclusion that the orbital Sagnac = 1/365 of the rotational Sagnac.

Can you give a meaningful explanation why you say this is wrong? Details, and show the math, please. Can you identify specific errors or provide a different derivation, or, even better, both?

Quote
As such any reference which proves otherwise, renders your derivation to be totally worthless.

Can you elaborate on this? Where, specifically, are the errors? What is the correct derivation?

Quote
Can you understand this much?

It's easy to understand. You're used to addressing an audience that is intimidated by doubletalk, bluster, and pontification. You need to do better now. Meaningless, evasive, and irrelevant comments and walls of text are not sufficient.

Quote
<wall of text>

Where is (are) the error(s) in JackBlack's derivation? Succinctly, please.
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"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan
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sandokhan

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #244 on: June 05, 2017, 09:00:44 PM »
If the claims/conclusions of a derivation are shown to be totally wrong, then the derivation itself is a piece of thrash.

It is as simple as this.

There is no need to go into any kind of other details, since the claims/conclusions were clearly stated as follows:

Quote from: JackBlack on June 03, 2017, 11:47:37 PM
As such, for a given interferometer (with a fixed area of the light path), treating Earth's orbit as a rotation and Earth's daily rotation as a rotation where one revolution is 24 hours, the orbital Sagnac effect will be approximately 1/365 that of Earth's rotation, as the area is the same, the only thing that changes is the angular velocity which for the orbit is roughly 1/365 that of the day.

Quote from: JackBlack on June 04, 2017, 03:40:32 AM
No. It is easy to calculate that the orbital sagnac is a mere 1/365th that of the rotational.

That is because the 2 interferometers are the same and thus have the same R and the same A and thus it is entirely dependent upon ω

Again, the location of the centre of rotation is irrelevant.
All that matters is the area of the loop and its angular velocity.


The claim/conclusion of the derivation is as follows: the orbital Sagnac = 1/365 of the rotational Sagnac and these have been highlighted in red: there is no going back now.

As such, if I can prove that this claim/conclusion is absolutely false, then the derivation is shown to be faulty, erroneous.

No need for any kind of wasteful analysis: would any of you even bother to analyze the details, when the clearly stated claim/conclusion is available for total and direct destruction? Certainly not, but that is exactly how you deviously ask of others to do.


The local aether model was adopted ONLY AFTER the author, the peer reviewers of the BAPS/IOP/EPL, proved that the orbital Sagnac is not only much larger than the rotational Sagnac, but that also it is missing.


If the claims/conclusions (as listed above) of the derivation had been true (orbital Sagnac = 1/365 rotational Sagnac), then nobody would have bothered to even investigate the subject.

No need to adopt a local aether model in order to explain something which is much smaller in magnitude than the rotational Sagnac.


Again, here is the claim/conclusion of the derivation: the orbital Sagnac is 1/365 of the rotational Sagnac.

A very specific claim, the most important part of the entire derivation.


Let us see how easy it is to put this claim to rest.


Dr. Daniel Gezari is a Post Doctoral Research Fellow at the California Institute of Technology, and Astrophysicist (Emeritus) - NASA/Goddard (2008 - present).

The lunar laser ranging experiment is an astronomical version of the Sagnac experiment.

However, G. Sagnac used the fringe-shift method to measure indirectly light travel time;
while Dr. Daniel Gezari uses clocks to measure directly light travel time in both directions.

Shooting light to the moon has to do with the behavior of light like GPS.

Dr. Daniel Gezari must calculate the Sagnac in order to fully account for the shooting of the laser to the mirror on the moon and back.

Calculations performed at the NASA Goddard Space Flight Center by Dr. Daniel Gezari:

https://arxiv.org/vc/arxiv/papers/0912/0912.3934v1.pdf

Please note the theoretical orbital sagnac shows up in these calculations, but is not picked up/registered/recorded by GPS satellites.

Motion of the Earth-Moon system in orbit around the Sun would average out in a two-way measurement, and only appear as a small (∼3 m/s) second-order residual.

Because of the two-way averaging, the orbital Sagnac effect registered is smaller than usual, however it is not 1/365 of the rotational Sagnac effect, in fact even in the diluted form permitted by the two-way averaging calculation, it represents a significant percentage of the rotational Sagnac effect.


These calculations were done by none other than Dr. Daniel Gezari (CalTech, Goddard Space Flight Center).

For instance, the Earth’s full 30 km/s orbital velocity along the line-of-sight would produce a second-order residual velocity of only ~3 m/s, so we cannot preclude the possibility that some part of the 8.4 m /s difference between co and c measured here is a real second-order residual due to motion of the Earth-Moon system relative to an absolute frame.

3/8.4 = 0.357

1/365 = 0.00274

0.357/0.00274 = 130.3


A total refutation of the claims, and thus of the derivation.


Any sane person would understand that the derivation has been rendered to be useless and worthless, but we are not dealing with such persons here.


Dr. Daniel Gezari, one of the highest rated astrophysicists in the world today, has performed the calculations which prove the claim/conclusion of the derivation to be absolutely wrong.


A second paper also showing that the claim/conclusion of the derivation is pure thrash can be found here:

https://www.theflatearthsociety.org/forum/index.php?topic=70614.msg1916512#msg1916512


It was published by an IOP journal.

Let us read the conclusions of the article:

http://qem.ee.nthu.edu.tw/f1b.pdf

On the other hand, the orbital
motion of the earth around the sun has a linear speed of
about 30 km/s which is about 100 times that of earth’s
rotation. Thus the present high-precision GPS would be
entirely impossible if the omitted correction due to orbital
motion is really necessary.


The author states that the orbital Sagnac is larger than the rotational Sagnac and that it is missing.


Again.

Meanwhile, as in GPS, no effects of earth’s orbital motion
are reported in these links, although they would be
easier to observe if they are in existence.

The orbital Sagnac would be easier to observe if and only if it was larger than the rotational Sagnac.


Then, and only then, the author adopts a local aether model in order to explain this extraordinary finding: the fact that not only is the orbital Sagnac larger in magnitude than the rotational Sagnac, but that also it is missing.


These are conclusions of a mainstream paper, accepted for publication by an IOP journal.

They totally refute the claim/conclusion that the orbital Sagnac is 1/365 of the rotational Sagnac.

They render the derivation to be a worthless piece of thrash.


If the conclusion of a derivation is shown to be wrong by such a huge margin, would it make any sense to go into any kind of further details? No, it would not.

The claim/conclusion of the derivation was clearly stated, there is no going back now.

But it was shown to be totally false.



Here is a third paper, this time it was published by the BULLETIN OF THE AMERICAN PHYSICAL SOCIETY, one of the most prestigious journals in the world today.


https://www.theflatearthsociety.org/forum/index.php?topic=70614.msg1916512#msg1916512


Let us read the conclusions of the article:

C.C. Su, "A Local-ether model of propagation of electromagnetic wave," in Bull. Am. Phys. Soc., vol. 45, no. 1, p. 637, Mar. 2000 (Minneapolis, Minnesota).

http://www.ee.nthu.edu.tw/ccsu/


Both the rotational and the orbital motions of the earth together with the orbital
motion of the target planet contribute to the Sagnac
effect.

Anyway, the interplanetary Sagnac effect is due to
earth’s orbital motion around the sun as well as earth’s
rotation.

Based on the local-ether model, the propagation is entirely
independent of the earth’s orbital motion around
the sun or whatever and the velocity v for such an earthbound
experiment is referred to an ECI frame and hence
is due to earth’s rotation alone. In the original proposal,
the velocity v was supposed to incorporate earth’s orbital
motion around the sun. Thus, at least, v^2/c^2
=~ 10^-8. Then the amplitude of the phase-difference variation
could be as large as π/3, when the wavelength is
0.6 µm and the path length is 10 m. However, as the velocity
v is the linear velocity due to earth’s rotation alone,
the round-trip Sagnac effect is as small as v^2/c^2∼ 10^-12 which is merely 10^-4 times that due to the orbital motion.


The Sagnac effect is a FIRST ORDER effect in v/c.

Even in the round-trip nature of the Sagnac effect, as it was applied in the Michelson-Morley experiment, thus becoming a second order effect within that context, we can see that the ORBITAL SAGNAC IS 10,000 TIMES GREATER than the rotational Sagnac effect.


Your claims are refuted for a third time, and your derivation is rendered to be worthless.


The author PROVED that the orbital Sagnac is 10,000 times greater than the rotational Sagnac, and the peer reviewers agree totally.


It was published in one of the most respected journals in the world.


Thus the claim/conclusion that the orbital Sagnac is 1/365 of the rotational Sagnac was shown to be false.

As such, the derivation itself is worthless.


You know what you have to do now: please mail your derivation to both the BAPS and to the EPL and see if they will publish it.


Let us not forget that Journal of Electromagnetic Waves and Applications, one of the most famous mainstream scientific journals in the world, also agrees with me.


http://www.ee.nthu.edu.tw/ccsu/qem/f3c.pdf

For the interplanetary propagation, earth’s orbital
motion contributes to the Sagnac effect as well. This local-ether model
has been adopted to account for the Sagnac effect due to earth’s
motions in a wide variety of propagation phenomena, particularly the
global positioning system (GPS), the intercontinental microwave link,
and the interplanetary radar.


The peer reviewers at the Journal of Electromagnetic Waves and Applications agree that the orbital Sagnac is larger than the rotational Sagnac, that it is missing, and that a local-ether model has to be adopted in order to account for this fact.


A total refutation of the claim/conclusion/derivation.

 

Can you show an error in the math?

The error in math is humongous.

While you claimed that the orbital Sagnac is 1/365 of the rotational Sagnac, the very subject of the derivation, you were shown that actually the orbital Sagnac is 10,000 times larger than the rotational Sagnac.

A huge difference, which goes to show the catastrophic nature of the derivation.


The proofs provided by the above respected authors were accepted by some of the most prestigious journals in the world today.

You have to do the same now: please mail your derivation to those same journals and see if they will publish it.


I have mainstream science on my side, while you have nothing at all.


You claimed that the orbital Sagnac is 1/365 of the rotational Sagnac.


Here is how easy it is to put this claim to rest.



In the original proposal,
the velocity v was supposed to incorporate earth’s orbital
motion around the sun. Thus, at least, v^2/c^2
=~ 10^-8. Then the amplitude of the phase-difference variation
could be as large as π/3, when the wavelength is
0.6 µm and the path length is 10 m. However, as the velocity
v is the linear velocity due to earth’s rotation alone,
the round-trip Sagnac effect is as small as v^2/c^2∼ 10^-12 which is merely 10^-4 times that due to the orbital motion.


10,000 is a much larger figure than 1/365.

That is by how much your derivation is wrong.

Your derivation has been shown to be a total piece of worthless thrash.



« Last Edit: June 05, 2017, 11:04:51 PM by sandokhan »
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rabinoz

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #245 on: June 05, 2017, 10:53:54 PM »
Quote from: sandokhan on June 05, 2017, 09:00:44 PM
If the claims/conclusions of a derivation are shown to be totally wrong, then the derivation itself is a piece of thrash.

It is as simple as this.

There is no need to go into any kind of other details, since the claims/conclusions were clearly stated as follows:


YOU have shown nothing.

YOU have not shown where JackBlack's derivation might be in error and his results agree with the MathPages reference that YOU gave.

YOU only give some references, then drag what you like out of them.

So stop being a pig-headed idiot an show exactly where that analysis is incorrect.
All we asking is for you to show us how to do the correct derivation.
Surely for a person of your brilliance and mathematical prowess it should as easy as pie - is that π?

Addendum:  Here is a derivation from your favourite source of information, Wikipedia!
Quote from: Wikipedia
Sagnac Effect, Theories
The shift in interference fringes in a ring interferometer can be viewed intuitively as a consequence of the different distances that light travels due to the rotation of the ring.(Fig. 3) The simplest derivation is for a circular ring of radius R, with a refractive index of one, rotating at an angular velocity of ω, but the result is general for loop geometries with other shapes. If a light source emits in both directions from one point on the rotating ring, light traveling in the same direction as the rotation direction needs to travel more than one circumference around the ring before it catches up with the light source from behind. The time t1 that it takes to catch up with the light source is given by:
         
ΔL is the distance (black bold arrow in Fig. 3) that the mirror has moved in that same time:
         
Eliminating ΔL from the two equations above we get:
         
Likewise, the light traveling in the opposite direction of the rotation will travel less than one circumference before hitting the light source on the front side. So the time for this direction of light to reach the moving source again is:
         
               
Figure 3. Light traveling opposite
directions go different distances
before reaching the moving source

The time difference is:
         
For , this reduces to
         

where A is the area of the ring. The Sagnac area formula applies to any shape of loop.
Although this simple derivation is for a circular ring with an index of refraction of one, the result holds true for any shape of rotating loop with area A.

Wherever we go we seem stuck with and the result holds true for any shape of rotating loop with area, A
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sandokhan

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #246 on: June 05, 2017, 11:09:29 PM »
Here are the conclusions/claims of the derivation:

Quote from: JackBlack on June 03, 2017, 11:47:37 PM
As such, for a given interferometer (with a fixed area of the light path), treating Earth's orbit as a rotation and Earth's daily rotation as a rotation where one revolution is 24 hours, the orbital Sagnac effect will be approximately 1/365 that of Earth's rotation, as the area is the same, the only thing that changes is the angular velocity which for the orbit is roughly 1/365 that of the day.

Quote from: JackBlack on June 04, 2017, 03:40:32 AM
No. It is easy to calculate that the orbital sagnac is a mere 1/365th that of the rotational.

That is because the 2 interferometers are the same and thus have the same R and the same A and thus it is entirely dependent upon ω

Again, the location of the centre of rotation is irrelevant.
All that matters is the area of the loop and its angular velocity.


The derivation leads directly to these conclusions.


The claim is as follows: the orbital Sagnac is 1/365 of the rotational Sagnac.


Let us put this claim to the test.


These calculations were done by none other than Dr. Daniel Gezari (CalTech, Goddard Space Flight Center).

For instance, the Earth’s full 30 km/s orbital velocity along the line-of-sight would produce a second-order residual velocity of only ~3 m/s, so we cannot preclude the possibility that some part of the 8.4 m /s difference between co and c measured here is a real second-order residual due to motion of the Earth-Moon system relative to an absolute frame.

3/8.4 = 0.357

1/365 = 0.00274

0.357/0.00274 = 130.3


A total refutation of the claims, and thus of the derivation.


Any sane person would understand that the derivation has been rendered to be useless and worthless, but we are not dealing with such persons here.


Published by the BULLETIN OF THE AMERICAN PHYSICAL SOCIETY, one of the most prestigious journals in the world today.

C.C. Su, "A Local-ether model of propagation of electromagnetic wave," in Bull. Am. Phys. Soc., vol. 45, no. 1, p. 637, Mar. 2000 (Minneapolis, Minnesota).

http://www.ee.nthu.edu.tw/ccsu/


Anyway, the interplanetary Sagnac effect is due to
earth’s orbital motion around the sun as well as earth’s
rotation.

Based on the local-ether model, the propagation is entirely
independent of the earth’s orbital motion around
the sun or whatever and the velocity v for such an earthbound
experiment is referred to an ECI frame and hence
is due to earth’s rotation alone. In the original proposal,
the velocity v was supposed to incorporate earth’s orbital
motion around the sun. Thus, at least, v^2/c^2
=~ 10^-8. Then the amplitude of the phase-difference variation
could be as large as π/3, when the wavelength is
0.6 µm and the path length is 10 m. However, as the velocity
v is the linear velocity due to earth’s rotation alone,
the round-trip Sagnac effect is as small as v^2/c^2∼ 10^-12 which is merely 10^-4 times that due to the orbital motion.


The Sagnac effect is a FIRST ORDER effect in v/c.

Even in the round-trip nature of the Sagnac effect, as it was applied in the Michelson-Morley experiment, thus becoming a second order effect within that context, we can see that the ORBITAL SAGNAC IS 10,000 TIMES GREATER than the rotational Sagnac effect.


Your claims are refuted yet again, and your derivation is rendered to be worthless.


The author PROVED that the orbital Sagnac is 10,000 times greater than the rotational Sagnac, and the peer reviewers agree totally.


It was published in one of the most respected journals in the world.



Both claims cannot be true at the same time, right?


The derivation's conclusion is that the orbital Sagnac is 1/365 of the rotational Sagnac.

The mainstream papers published by these respected journals prove that the orbital Sagnac is 10,000 greater than the rotational Sagnac.


Even in the two way averaging calculation performed by Dr. Daniel Gezari, the orbital Sagnac is 130 times greater than the claim made by the derivation.


The claim/conclusion of the derivation has been shown to be totally worthless: as such, its very content is pure thrash, there no need to analyze any further details.

10,000 is a much larger figure than 1/365.

That is by how much the math in the derivation has been shown to be wrong.


As for idiotism, there can be no greater idiot than the person who can believe that four trillion billion gallons of water stay glued next to the exterior surface of a sphere.


Is Dr. Daniel Gezari an idiot?

Dr. Daniel Gezari is a Post Doctoral Research Fellow at the California Institute of Technology, and Astrophysicist (Emeritus) - NASA/Goddard (2008 - present).



Here are the conclusions of his calculations:

For instance, the Earth’s full 30 km/s orbital velocity along the line-of-sight would produce a second-order residual velocity of only ~3 m/s, so we cannot preclude the possibility that some part of the 8.4 m /s difference between co and c measured here is a real second-order residual due to motion of the Earth-Moon system relative to an absolute frame.

3/8.4 = 0.357

1/365 = 0.00274

0.357/0.00274 = 130.3


A total refutation of the claims, and thus of the derivation.



Is Dr. C.C Su an idiot?

Here is a listing of his scientific papers:

http://www.ee.nthu.edu.tw/ccsu/


C.C. Su, "A Local-ether model of propagation of electromagnetic wave," in Bull. Am. Phys. Soc., vol. 45, no. 1, p. 637, Mar. 2000 (Minneapolis, Minnesota).

http://www.ee.nthu.edu.tw/ccsu/


Anyway, the interplanetary Sagnac effect is due to
earth’s orbital motion around the sun as well as earth’s
rotation.

Based on the local-ether model, the propagation is entirely
independent of the earth’s orbital motion around
the sun or whatever and the velocity v for such an earthbound
experiment is referred to an ECI frame and hence
is due to earth’s rotation alone. In the original proposal,
the velocity v was supposed to incorporate earth’s orbital
motion around the sun. Thus, at least, v^2/c^2
=~ 10^-8. Then the amplitude of the phase-difference variation
could be as large as π/3, when the wavelength is
0.6 µm and the path length is 10 m. However, as the velocity
v is the linear velocity due to earth’s rotation alone,
the round-trip Sagnac effect is as small as v^2/c^2∼ 10^-12 which is merely 10^-4 times that due to the orbital motion.


jack's derivation concluded that the orbital Sagnac is 1/365 of the rotational Sagnac.


The mainstream papers cited above prove the exact opposite: calculations performed by some of the greatest experts in the field.


It it as simple as this.

Go ahead and mail the derivation to the very same journals and see if they will accept it, only then you can come back here for a debate.


« Last Edit: June 05, 2017, 11:12:53 PM by sandokhan »
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Lonegranger

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #247 on: June 06, 2017, 01:11:50 AM »
Sorry old chap but one flight by Space-X renders all you arguments null and void.
The company employs more PhDs than you could shake a stick at, and by your reckoning that should really count for something.

Now these guys know about everything to do with orbital mechanics, they do build and fly rockets.

You can harp on about sub-quark that, or sagnac this but in the end a simple rocket flight, or should I say multiple filmed, publicly observed flights render all what you say obsolete.

Ps
How's your perpetual motion machine coming along?, you know the one you invented...

http://www.spacex.com

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rabinoz

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #248 on: June 06, 2017, 01:39:32 AM »
Quote from: sandokhan on June 05, 2017, 11:09:29 PM
Here are the conclusions/claims of the derivation:
No!
We don't want your conclusions of what JackBlack's derivation might or might not lead to! We want you to show what is wrong with that derivation
and of course along with that your derivation of what you claim is the correct result.

Every derivation I have seen ends up with the same result and many specifically state that:
  • The Sagnac delay is proportional to the projected area of the loop onto the plane of rotation.
  • The shape of the loop is irrelevant.
  • The location of the centre of rotation is irrelevant.
So stop being a pig-headed idiot an show exactly where those analyses are so incorrect.
All we asking is for you to show us how to do the correct derivation.
Surely for a person of your brilliance and mathematical prowess it should as easy as pie.

Of course, the reason you won't do it, is simple! You can't do it.

Lonegranger just reminded me of this.
I'm still waiting for information on your supplying these magnetic monopoles, a free energy machine and also
      the height of the sun, moon, planets and stars above the earth!
      If you have no idea, just admit it.

I guess you will just have to finally admit that your wondrous magnetic monopoles and free energy machines are hoaxes
and that you haven't the slightest of "the height of the sun, moon, planets and stars above the earth".
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sandokhan

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #249 on: June 06, 2017, 01:59:27 AM »
We don't want your conclusions of what JackBlack's derivation might or might not lead to!

rabinoz, do you really want to make an imbecile out of yourself, each and every time you post here?

Here are jack's own conclusions/claims: they are not mine.


Quote from: JackBlack on June 03, 2017, 11:47:37 PM
As such, for a given interferometer (with a fixed area of the light path), treating Earth's orbit as a rotation and Earth's daily rotation as a rotation where one revolution is 24 hours, the orbital Sagnac effect will be approximately 1/365 that of Earth's rotation, as the area is the same, the only thing that changes is the angular velocity which for the orbit is roughly 1/365 that of the day.

Quote from: JackBlack on June 04, 2017, 03:40:32 AM
No. It is easy to calculate that the orbital sagnac is a mere 1/365th that of the rotational.

Are you such a moron as to not know how to read plain English?


This is the basic claim made by jack right here in front of everybody, as evidenced/highlighted in red:

Orbital Sagnac is 1/365 of the rotational Sagnac.


Remember well that figure: 1/365.





Dr. Daniel Gezari emitted a pulse of photons from a point on earth, bounced those photons off a reflector on the moon, and then recorded the photons’ arrival time at that same point on earth.

Obviously, both the rotational and the orbital Sagnac effects have to be taken into account.

These calculations were done by none other than Dr. Daniel Gezari (CalTech, Goddard Space Flight Center).

For instance, the Earth’s full 30 km/s orbital velocity along the line-of-sight would produce a second-order residual velocity of only ~3 m/s, so we cannot preclude the possibility that some part of the 8.4 m /s difference between co and c measured here is a real second-order residual due to motion of the Earth-Moon system relative to an absolute frame.

3/8.4 = 0.357

1/365 = 0.00274

0.357/0.00274 = 130.3


Please note the orbital sagnac shows up in these measurements.

Now, because of the vast distance, if you are correct, you should see 1/365 of the rotational sagnac in the measurements and that will show up on this vast distance.

So, if you are correct, then we should see your 1/365 conclusions in the measurements. Guess what. We do not.

That means you are wrong.



jack claims the orbital Sagnac is 1/365 of the rotational Sagnac.

His derivation leads directly to this conclusion.


Dr. Daniel Gezari's calculations prove otherwise: even in the diluted two way averaging form, the orbital Sagnac amounts for a 3/8.4 = 0.357 (35.7%) percentage of the rotational Sagnac.



Both these figures cannot be right at the very same time.


Any sane person would understand that the derivation has been rendered to be useless and worthless, but we are not dealing with such persons here.


Published by the BULLETIN OF THE AMERICAN PHYSICAL SOCIETY, one of the most prestigious journals in the world today.

C.C. Su, "A Local-ether model of propagation of electromagnetic wave," in Bull. Am. Phys. Soc., vol. 45, no. 1, p. 637, Mar. 2000 (Minneapolis, Minnesota).

http://www.ee.nthu.edu.tw/ccsu/


Anyway, the interplanetary Sagnac effect is due to
earth’s orbital motion around the sun as well as earth’s
rotation.

Based on the local-ether model, the propagation is entirely
independent of the earth’s orbital motion around
the sun or whatever and the velocity v for such an earthbound
experiment is referred to an ECI frame and hence
is due to earth’s rotation alone. In the original proposal,
the velocity v was supposed to incorporate earth’s orbital
motion around the sun. Thus, at least, v^2/c^2
=~ 10^-8. Then the amplitude of the phase-difference variation
could be as large as π/3, when the wavelength is
0.6 µm and the path length is 10 m. However, as the velocity
v is the linear velocity due to earth’s rotation alone,
the round-trip Sagnac effect is as small as v^2/c^2∼ 10^-12 which is merely 10^-4 times that due to the orbital motion.


The Sagnac effect is a FIRST ORDER effect in v/c.

Even in the round-trip nature of the Sagnac effect, as it was applied in the Michelson-Morley experiment, thus becoming a second order effect within that context, we can see that the ORBITAL SAGNAC IS 10,000 TIMES GREATER than the rotational Sagnac effect.


Your claims are refuted yet again, and your derivation is rendered to be worthless.


The author PROVED that the orbital Sagnac is 10,000 times greater than the rotational Sagnac, and the peer reviewers agree totally.


It was published in one of the most respected journals in the world.


Both claims cannot be true at the same time, right?


The derivation's conclusion is that the orbital Sagnac is 1/365 of the rotational Sagnac.

The mainstream papers published by these respected journals prove that the orbital Sagnac is 10,000 greater than the rotational Sagnac.


There is no need to bother with the hapless derivation: its conclusions/claims have been shown to be totally erroneous.


rabinoz, do you suffer from some kind of a mental disorder which precludes you from understanding these basic things?


Dr. C.C. Su mailed his analysis to two of the most prestigious journals in the world: BAPS and EPL.

His article was peer reviewed and published.


Here are his own words again:


In the original proposal,
the velocity v was supposed to incorporate earth’s orbital
motion around the sun. Thus, at least, v^2/c^2
=~ 10^-8. Then the amplitude of the phase-difference variation
could be as large as π/3, when the wavelength is
0.6 µm and the path length is 10 m. However, as the velocity
v is the linear velocity due to earth’s rotation alone,
the round-trip Sagnac effect is as small as v^2/c^2∼ 10^-12 which is merely 10^-4 times that due to the orbital motion.


So, jack's own claim, that the orbital Sagnac is 1/365 of the rotational Sagnac has been totally refuted.


Have jack mail his derivation to these same two journals and see if they will publish his thoughts on the matter.

Until then, this thread is over, there nothing to discuss further here.


jack's derivation claims the figure 1/365.

Dr. C.C. Su informs us that the figure is 10,000.

Both these figures cannot be true at the same time.

Please mail the derivation to the same scientific journals which published Dr. Su's paper and see if they will publish it.


IOP published Dr. C.C. Su findings, again peer reviewed.

Let us read the conclusions of the article:

http://qem.ee.nthu.edu.tw/f1b.pdf

On the other hand, the orbital
motion of the earth around the sun has a linear speed of
about 30 km/s which is about 100 times that of earth’s
rotation. Thus the present high-precision GPS would be
entirely impossible if the omitted correction due to orbital
motion is really necessary.


The author states that the orbital Sagnac is larger than the rotational Sagnac and that it is missing.


Again.

Meanwhile, as in GPS, no effects of earth’s orbital motion
are reported in these links, although they would be
easier to observe if they are in existence.

The orbital Sagnac would be easier to observe if and only if it was larger than the rotational Sagnac.


One of the top journals in the world, EUROPHYSICS LETTERS JOURNAL, published Dr. Su's figures exactly as quoted here.


Dr. Su tells us that the orbital Sagnac IS LARGER than the rotational Sagnac.

jack's derivation leads to the conclusion that the orbital Sagnac is 1/365 of the rotational Sagnac.

Both these conclusions cannot be true at the same time.


Dr. Su mailed his paper to the IOP and his findings were published in the EPL journal.

Now, jack has to do the very same thing: to mail his derivation to the EPL.

Until he does so, I am right and you are wrong: the missing orbital Sagnac is larger than the rotational Sagnac.
« Last Edit: June 06, 2017, 02:19:21 AM by sandokhan »
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sandokhan

  • Flat Earth Sultan
  • Flat Earth Scientist
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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #250 on: June 06, 2017, 02:36:13 AM »
All we asking is for you to show us how to do the correct derivation.

Very easy.

Earth's radius = 6357 km; r² = 40411449

Earth's orbital radius = 150,000,000 km r² = 22500000000000000

∆t = 4πR²ω/(c²-v²)
or

I use the linear velocity.

∆t = 4πRv/( c² - v² ), where v is the linear velocity.

For the earth's rotation, it is 0.4638333 km/ sec and the orbit v = 30km/sec.

∆t = 0.62831852628 for the earth's orbit.
Total path of the orbit is 2πr=2π(150,000,000 km) = 942,477,780km

Hence, the sagnac effect for a 1 km path, that means light source in the center and two receivers placed at .5km is:
0.62831852628 / 942,477,780km = 6.6666667 e-10 sec / km

Now, for the earth's rotation.
∆t = 4.1170061 e-7 seconds
Total path of the rotation is 2πr=2π(6357 km) = 39942.21 km


4.1170061 e-7 seconds / 39942.21 km = 1.0307407 e-11 sec / km


The sagnac effect for the earth's orbit is greater than that of the rotation.



The orbital Sagnac, though much larger than the rotational Sagnac, is not being registered by GPS satellites.



The BULLETIN OF THE AMERICAN PHYSICAL SOCIETY, EUROPHYSICS LETTERS JOURNAL, and the JOURNAL OF ELECTROMAGNETIC WAVES AND APPLICATIONS all agree with me.


Published by the BULLETIN OF THE AMERICAN PHYSICAL SOCIETY, one of the most prestigious journals in the world today.

C.C. Su, "A Local-ether model of propagation of electromagnetic wave," in Bull. Am. Phys. Soc., vol. 45, no. 1, p. 637, Mar. 2000 (Minneapolis, Minnesota).

http://www.ee.nthu.edu.tw/ccsu/




Anyway, the interplanetary Sagnac effect is due to
earth’s orbital motion around the sun as well as earth’s
rotation.

Based on the local-ether model, the propagation is entirely
independent of the earth’s orbital motion around
the sun or whatever and the velocity v for such an earthbound
experiment is referred to an ECI frame and hence
is due to earth’s rotation alone. In the original proposal,
the velocity v was supposed to incorporate earth’s orbital
motion around the sun. Thus, at least, v^2/c^2
=~ 10^-8. Then the amplitude of the phase-difference variation
could be as large as π/3, when the wavelength is
0.6 µm and the path length is 10 m. However, as the velocity
v is the linear velocity due to earth’s rotation alone,
the round-trip Sagnac effect is as small as v^2/c^2∼ 10^-12 which is merely 10^-4 times that due to the orbital motion.


The Sagnac effect is a FIRST ORDER effect in v/c.

Even in the round-trip nature of the Sagnac effect, as it was applied in the Michelson-Morley experiment, thus becoming a second order effect within that context, we can see that the ORBITAL SAGNAC IS 10,000 TIMES GREATER than the rotational Sagnac effect.


« Last Edit: June 06, 2017, 02:38:31 AM by sandokhan »
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JackBlack

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #251 on: June 06, 2017, 04:23:49 AM »
Quote from: sandokhan on June 05, 2017, 09:00:44 PM
If the claims/conclusions of a derivation are shown to be totally wrong, then the derivation itself is a piece of thrash.
It is as simple as this.
Not quite.
Yes, if the conclusions are shown to be totally wrong, the derivation must be wrong, but you need to be able to show the conclusions are wrong and that means there must be an error in the derivation which you should be able to find..
A simple claim, even in a peer reviewed publication does not show it is wrong, especially one based upon the assumption of aether does not show the conclusion is wrong.

Similarly, math can also be used to show the claim is wrong.

Do you know the only way to really show this conclusion is wrong (other than pointing out an error in the derivation)? By recording an orbital Sagnac effect which is much larger than the rotational one, but then you wouldn't have a problem.

Quote from: sandokhan on June 05, 2017, 09:00:44 PM
There is no need to go into any kind of other details, since the claims/conclusions were clearly stated as follows:
Yes there is, as you are unable to show anything wrong with the claims or provide a source which actually refutes the claims.

So again, CAN YOU SHOW ANYTHING WRONG WITH MY DERIVATION???


Quote from: sandokhan on June 05, 2017, 09:00:44 PM
As such, if I can prove that this claim/conclusion is absolutely false, then the derivation is shown to be faulty, erroneous.
Yes, ABSOLUTELY FALSE!!
You haven't even come close to that.

Quote from: sandokhan on June 05, 2017, 09:00:44 PM
No need for any kind of wasteful analysis: would any of you even bother to analyze the details, when the clearly stated claim/conclusion is available for total and direct destruction? Certainly not, but that is exactly how you deviously ask of others to do.
That depends, are you open minded enough to admit that you may be wrong about the conclusions?
If so, yes, you look at the derivation and try to find something wrong with it.

However, if you are brainwashed or close minded you just find whatever excuse you can to dismiss it.

If the conclusion is wrong there MUST be something wrong with the derivation and thus you should be able to show what is wrong with it.

If you can't, then you are just being closed minded to reject the conclusion.

Quote from: sandokhan on June 05, 2017, 09:00:44 PM
The local aether model was adopted ONLY AFTER the author, the peer reviewers of the BAPS/IOP/EPL, proved that the orbital Sagnac is not only much larger than the rotational Sagnac, but that also it is missing.
NO!!
No local aether model has been adopted by mainstream science.
Mainstream science does not use aether at all.
It is only the fringe that still wants to use aether that was forced to adopt it after it was shown that the effects of Earth's orbit are much larger than that of Earth's rotation in an aether based model.

In a relativistic model, the orbital Sagnac effect is MUCH SMALLER than the rotational one.

Quote from: sandokhan on June 05, 2017, 09:00:44 PM
If the claims/conclusions (as listed above) of the derivation had been true (orbital Sagnac = 1/365 rotational Sagnac), then nobody would have bothered to even investigate the subject.
Except the people that want to keep looking at aether.

Quote from: sandokhan on June 05, 2017, 09:00:44 PM
No need to adopt a local aether model in order to explain something which is much smaller in magnitude than the rotational Sagnac.
And there is no need. It is only those wanting to use aether that need to.

Quote from: sandokhan on June 05, 2017, 09:00:44 PM
A very specific claim, the most important part of the entire derivation.
No, one of the least important parts of the derivation, it is merely a conclusion from it.

Quote from: sandokhan on June 05, 2017, 09:00:44 PM
Let us see how easy it is to put this claim to rest.
You have repeatedly tried and repeatedly failed.

Quote from: sandokhan on June 05, 2017, 09:00:44 PM
The lunar laser ranging experiment is an astronomical version of the Sagnac experiment.
No it isn't.
Let me remind you, the Sagnac effect is a fringe shift due to 2 counterpropogating light pulses around a rotating ring.
The lunar ranging experiment does not have that and thus is not the Sagnac experiment.

As such, any crap you say about it is irrelevant.

Quote from: sandokhan on June 05, 2017, 09:00:44 PM
A total refutation of the claims, and thus of the derivation.
No. Not a refutation of the claims as it has nothing to do with the Sagnac effect, and thus it has no bearing on the derivation.
Regardless, all this would be is a claim. It has no proof backing it.

Quote from: sandokhan on June 05, 2017, 09:00:44 PM
Any sane person would understand that the derivation has been rendered to be useless and worthless, but we are not dealing with such persons here.
No. Sane people would realise there is nothing wrong with the derivation. You are the insane person here.

Quote from: sandokhan on June 05, 2017, 09:00:44 PM
Dr. Daniel Gezari, one of the highest rated astrophysicists in the world today, has performed the calculations which prove the claim/conclusion of the derivation to be absolutely wrong.
Yet you are unable to provide anything other than baseless bullshit to back up this claim of yours.

Quote from: sandokhan on June 05, 2017, 09:00:44 PM
A second paper also showing that the claim/conclusion of the derivation is pure thrash can be found here:
https://www.theflatearthsociety.org/forum/index.php?topic=70614.msg1916512#msg1916512
That is your insane ramblings, not a paper. Try again.

Quote from: sandokhan on June 05, 2017, 09:00:44 PM
Let us read the conclusions of the article:
How about we focus on what the article was about? An aether based model of light, so nothing at all to do with the discussion at hand. Try again.

Quote from: sandokhan on June 05, 2017, 09:00:44 PM
The author states that the orbital Sagnac is larger than the rotational Sagnac and that it is missing.
It isn't really the Sagnac effect, and it is for an aether based model.

Quote from: sandokhan on June 05, 2017, 09:00:44 PM
These are conclusions of a mainstream paper, accepted for publication by an IOP journal.
No they aren't. They are your lies about the papers.

Quote from: sandokhan on June 05, 2017, 09:00:44 PM
They totally refute the claim/conclusion that the orbital Sagnac is 1/365 of the rotational Sagnac.
No. They don't even come close.

Quote from: sandokhan on June 05, 2017, 09:00:44 PM
They render the derivation to be a worthless piece of thrash.
Not in the slightest. They don't show a single error in it.

Quote from: sandokhan on June 05, 2017, 09:00:44 PM
If the conclusion of a derivation is shown to be wrong by such a huge margin, would it make any sense to go into any kind of further details? No, it would not.
Yes it would, to show where the derivation is wrong, something you are completely unable to do.

Quote from: sandokhan on June 05, 2017, 09:00:44 PM
But it was shown to be totally false.
No it wasn't.

Quote from: sandokhan on June 05, 2017, 09:00:44 PM
Here is a third paper, this time it was published by the BULLETIN OF THE AMERICAN PHYSICAL SOCIETY, one of the most prestigious journals in the world today.
https://www.theflatearthsociety.org/forum/index.php?topic=70614.msg1916512#msg1916512
Again, that isn't a paper, that is a link to your bullshit.

I've had enough of dealing with your irrelavent spam for now.


So I ask again:
CAN YOU POINT OUT A SINGLE THING WRONG WITH MY DERIVATION???

If not, then either shut up, or accept it is correct.

Quote from: sandokhan on June 06, 2017, 02:36:13 AM
All we asking is for you to show us how to do the correct derivation.
Very easy.
Earth's radius = 6357 km; r² = 40411449
Earth's orbital radius = 150,000,000 km r² = 22500000000000000
∆t = 4πR²ω/(c²-v²)
or
I use the linear velocity.
∆t = 4πRv/( c² - v² ), where v is the linear velocity.
For the earth's rotation, it is 0.4638333 km/ sec and the orbit v = 30km/sec.
∆t = 0.62831852628 for the earth's orbit.
Total path of the orbit is 2πr=2π(150,000,000 km) = 942,477,780km
I take it you don't understand what a derivation is?

This is you just making up a formula and sticking in some numbers.

Guess what? YOU ARE COMPLETELY WRONG!

As shown by numerous sources and my derivation the A is the area of the Sagnac loop, the R is the radius of the loop.
As such the R is equal for both.

This has already been pointed out to you numerous times.

As I said before, if you wish to spout such stupid crap you need to derive that formula (as I did) showing that the R in question is earth's orbital radius or whatever.
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sandokhan

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #252 on: June 06, 2017, 05:47:07 AM »
In a relativistic model, the orbital Sagnac effect is MUCH SMALLER than the rotational one.

There is no such thing as the relativistic model.

See how your cognitive dissonance shows up again?

It is very easy to demolish your relativistic model.

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg769750#msg769750

There is no such thing as space-time geometry. Here is the step by step demonstration.

G.F. Riemann introduced the additional variables as a supporting theory for his logarithm branch cuts, NOT ever to present time as a new variable.





http://www.maths.tcd.ie/pub/HistMath/People/Riemann/Geom/WKCGeom.html

the abstract concept of n-dimensional geometry to facilitate the geometric representation of functions of a complex variable (especially logarithm branch cut). 'Such researches have become a necessity for many parts of mathematics, e.g., for the treatment of many-valued analytical functions.'

Never did he think to introduce TIME as a separate dimension or variable.

How was this done?

In contrast Riemann’s original non-Euclidian geometry dealt solely with space and was therefore an “amorphous continuum.” Einstein and Minkowski made it metric.

Minkowski's four-dimensional space was transformed by using an imaginary (√-1.ct ) term in place of the real time ( t ). So the coordinates of Minkowski's Four-Dimensional Continuum, ( x1, x2, x3, x4 ) are all treated as space coordinates, but were in fact originally ( x1, x2, x3, t ) or rather ( x1, x2, x3,√-1.ct ), therefore the 4th space dimension x4 is in fact the imaginary √-1.ct substitute. This imaginary 4-dimensional union of time and space was termed by Minkowski as 'world'. Einstein called it 'Spacetime Continuum'. In fact, Minkowski never meant it to be used in curved space. His 4th dimension was meant to be Euclidean dimensions (straight), because it was well before the introduction of General Relativity. Einstein forcibly adopted it for 'curved' or 'None Euclidean' measurements without giving a word of explanations why he could do it. In fact, if there was an explanation Einstein would have given it. Yet, this was how 'Time' became 'Space' or '4th dimensional space' for mathematical purpose, which was then used in 'Spacetime Curvature', 'Ripples of Spacetime' and other applications in General Relativity, relativistic gravitation, which then went on to become Black Hole, etc., ...



EINSTEIN HIMSELF ON THE ABSURDITY OF THE SPACE TIME CONTINUUM CONCEPT:

Einstein, following Minkowski, welded space and time together into what critics have called ‘the monstrosity called space-time’. In this abstract, four-dimensional continuum, time is treated as a negative length, and metres and seconds are added together to obtain one ‘event’. Every point in the spacetime continuum is assigned four coordinates, which, according to Einstein, ‘have not the least direct physical significance’. He says that his field equations, whose derivation requires many pages of abstract mathematical operations, deprive space and time of ‘the last trace of objective reality’.


EINSTEIN FALLACIES:

http://web.archive.org/web/20090309113407/http://ourworld.compuserve.com/homepages/dp5/relativ.htm


REASONS WHY EINSTEIN WAS WRONG:

http://web.archive.org/web/20120205135201/http://www.kevin.harkess.btinternet.co.uk/reasons_einstein_wrong/reasons_einstein_wrong.html (one of the best works on the variability of light)


EINSTEIN'S THEORY OF RELATIVITY: SCIENTIFIC THEORY OR ILLUSION? by Milan Pavlovic

http://web.archive.org/web/20080705084812/http://users.net.yu/~mrp/chapter5.html


“it is difficult to find a theory so popular, and yet so unclear, incomplete, paradoxical
and contradictory, as is the theory of relativity…. The special theory of relativity can be said to be, in essence, a sum of deceptions.”



ALBERT IN RELATIVITYLAND

http://www.gsjournal.net/old/ntham/amesbury.pdf

However, space-time as a fourth dimension is nothing more than the product of professor Minkowski's cerebral and mathematical imagination.



Your knowledge of science amounts to a big zero jack.


The most extraordinary proofs on HOW EINSTEIN FAKED HIS 1919/1922 DATA FOR THE SO CALLED EINSTEIN SHIFT:

http://einstein52.tripod.com/alberteinsteinprophetorplagiarist/id9.html


http://www.ekkehard-friebe.de/dishones.htm (scroll down to the section: With regard to the politics that led to Einstein's fame Dr. S. Chandrasekhar's article [46] states...)


http://web.archive.org/web/20070202201854/http://www.nexusmagazine.com/articles/einstein.html



HOW EINSTEIN MODIFIED HIS FORMULA RELATING TO MERCURY'S ORBIT IN ORDER TO FIT THE RESULTS:

http://www.gravitywarpdrive.com/Rethinking_Relativity.htm (scroll down to The advance of the perihelion of Mercury’s orbit, another famous confirmation of General Relativity, is worth a closer look...)


Dr. F. Schmeidler of the Munich University Observatory has published a paper  titled "The Einstein Shift An Unsettled Problem," and a plot of shifts for 92 stars for the 1922 eclipse shows shifts going in all directions, many of them going the wrong way by as large a deflection as those shifted in the predicted direction! Further examination of the 1919 and 1922 data originally interpreted as confirming relativity, tended to favor a larger shift, the results depended very strongly on the manner for reducing the measurements and the effect of omitting individual stars.


Moreover, Einstein made a terrible blunder.

Einstein, 1905:

"The principle of the constancy of the velocity of light is of course contained in Maxwell's equations”

We can infer immediately that Einstein had no knowledge whatsoever of the original ether equations derived by Maxwell, and based his false/erroneous conclusions on the MODIFIED/CENSORED Heaviside-Lorentz equations.


"Einstein claims that “The principle of the constancy of the velocityof light is of course contained in Maxwell's equations”.

If the Lorentz force had still been included as one of Maxwell’s equations, they could
have been written in total time derivative format (see Appendix A in ‘The Double
Helix Theory of the Magnetic Field’) and Einstein would not have been able to make
this claim. A total time derivative electromagnetic wave equation would allow the
electromagnetic wave speed to alter from the perspective of a moving observer."


Here are the censored Heaviside-Lorentz equations, USED BY EINSTEIN to justify his erronous claim regarding the speed of light:



HERE IS THE ORIGINAL SET OF JAMES CLERK MAXWELL'S EQUATIONS: THE EXISTENCE OF ETHER, AETHER AND THE VARIABILITY OF THE SPEED OF LIGHT:




http://theflatearthsociety.org/forum/index.php?topic=30499.msg1608815#msg1608815 (more information on the set of original Maxwell equations)


http://web.archive.org/web/20071006083222/http://www.wbabin.net/science/tombe4.pdf
(also includes the appendix called Maxwell's Minor Errors discussing the wrong minus sign in equation D)

E = vXB − ∂Α/dt +gradψ

The most important scientific paper ever published: ON PHYSICAL LINES OF FORCE, by JAMES CLERK MAXWELL - the original set of ether equations, which are almost unknown to modern physics.

http://vacuum-physics.com/Maxwell/maxwell_oplf.pdf


"A solution to the original/corrected Maxwell equations indicates that these equations are invariant under the Galilean transformation. Velocity vectors are additive, which means that the speed of light can be exceeded."

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1848776#msg1848776




The Sagnac effect is far larger than the effect forecast by relativity theory.

STR has no possible function in explaining the Sagnac effect.

The Sagnac effect is a non-relativistic effect.

COMPARISON OF THE SAGNAC EFFECT WITH SPECIAL RELATIVITY, starts on page 7, calculations/formulas on page 8

http://www.naturalphilosophy.org/pdf/ebooks/Kelly-TimeandtheSpeedofLight.pdf

page 8

Because many investigators claim that the
Sagnac effect is made explicable by using the
Theory of Special Relativity, a comparison of
that theory with the actual test results is given
below. It will be shown that the effects
calculated under these two theories are of very
different orders of magnitude, and that
therefore the Special Theory is of no value in
trying to explain the effect.


Thus the Sagnac effect is far larger than any
purely Relativistic effect. For example,
considering the data in the Pogany test (8 ),
where the rim of the disc was moving with a
velocity of 25 m/s, the ratio dtS/dtR is about
1.5 x 10^7. Any attempt to explain the Sagnac
as a Relativistic effect is thus useless, as it is
smaller by a factor of 10^7.


Referring back to equation (I), consider a disc
of radius one kilometre. In this case a fringe
shift of one fringe is achieved with a velocity
at the perimeter of the disc of 0.013m/s. This
is an extremely low velocity, being less than
lm per minute. In this case the Sagnac effect
would be 50 billion times larger than the
calculated effect under the Relativity Theory.


Post (1967) shows that the two (Sagnac and STR) are of very different orders of magnitude. He says that the dilation factor to be applied under SR is “indistinguishable with presently available equipment” and “is still one order smaller than the Doppler correction, which occurs when observing fringe shifts” in the Sagnac tests. He also points out that the Doppler effect “is v/c times smaller than the effect one wants to observe." Here Post states that the effect forecast by SR, for the time dilation aboard a moving object, is far smaller than the effect to be observed in a Sagnac test.
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sandokhan

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #253 on: June 06, 2017, 05:53:46 AM »
Let me remind you, the Sagnac effect is a fringe shift due to 2 counterpropogating light pulses around a rotating ring.
The lunar ranging experiment does not have that and thus is not the Sagnac experiment.

jack, people are beginning to see through your miserable denial of reality.


The lunar laser ranging experiment is an astronomical version of the Sagnac experiment.

However, G. Sagnac used the fringe-shift method to measure indirectly light travel time;
while Dr. Daniel Gezari uses clocks to measure directly light travel time in both directions.

Shooting light to the moon has to do with the behavior of light like GPS.

The arrival time of light to a receptor is influenced by the motion of
the receptor relative to the earth: this is the basic discovery of G. Sagnac.

This fact has to be incorporated into the lunar laser ranging calculations.

Here is a basic reference which confirms this fact:

Ring-laser tests of fundamental physics and geophysics, G.E. Steadman, 1997, pg 15




You need both the orbital and rotational Sagnac to calculate the correct timing, there is no way around that.


Dr. Daniel Gezari emitted a pulse of photons from a point on earth, bounced those photons off a reflector on the moon, and then recorded the photons’ arrival time at that same point on earth.


Please note the theoretical orbital sagnac shows up in these calculations, but is not picked up/registered/recorded by GPS satellites.

Motion of the Earth-Moon system in orbit around the Sun would average out in a two-way measurement, and only appear as a small (∼3 m/s) second-order residual.

Because of the two-way averaging, the orbital Sagnac effect registered is smaller than usual, however it is not 1/365 of the rotational Sagnac effect, in fact even in the diluted form permitted by the two-way averaging calculation, it represents a significant percentage of the rotational Sagnac effect.


THE SMALL (~3M/S) SECOND ORDER RESIDUAL IS THE ORBITAL SAGNAC.


For instance, the Earth’s full 30 km/s orbital velocity along the line-of-sight would produce a second-order residual velocity of only ~3 m/s, so we cannot preclude the possibility that some part of the 8.4 m /s difference between co and c measured here is a real second-order residual due to motion of the Earth-Moon system relative to an absolute frame.

THE 8.4 M/S DIFFERENCE IS THE ROTATIONAL SAGNAC.


Listen to what Dr. Daniel Gezari says:


For instance, the Earth’s full 30 km/s orbital velocity along the line-of-sight would produce a second-order residual velocity of only ~3 m/s, so we cannot preclude the possibility that some part of the 8.4 m /s difference


3/8.4 = 0.357

1/365 = 0.00274

0.357/0.00274 = 130.3


A total refutation of the claims, and thus of the derivation.


Now, because of the vast distance, if you are correct, you should see 1/365 of the rotational sagnac in the measurements and that will show up on this vast distance.

So, if you are correct, then we should see your 1/365 conclusions in the measurements. Guess what. We do not.

That means you are wrong.



jack claims the orbital Sagnac is 1/365 of the rotational Sagnac.

His derivation leads directly to this conclusion.


Dr. Daniel Gezari's calculations prove otherwise: even in the diluted two way averaging form, the orbital Sagnac amounts for a 3/8.4 = 0.357 (35.7%) percentage of the rotational Sagnac.



Both these figures cannot be right at the very same time.


Any sane person would understand that the derivation has been rendered to be useless and worthless, but we are not dealing with such persons here.


« Last Edit: June 06, 2017, 10:09:09 AM by sandokhan »
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rabinoz

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #254 on: June 06, 2017, 06:01:25 AM »
Quote from: sandokhan on June 06, 2017, 05:53:46 AM
Let me remind you, the Sagnac effect is a fringe shift due to 2 counterpropogating light pulses around a rotating ring.
The lunar ranging experiment does not have that and thus is not the Sagnac experiment.
Show your derivation of the correct Sagnac time delay!
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sandokhan

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #255 on: June 06, 2017, 06:01:44 AM »
How about we focus on what the article was about? An aether based model of light, so nothing at all to do with the discussion at hand. Try again.

jack, you are embarrassing yourself.

C.C. Su, "A Local-ether model of propagation of electromagnetic wave," in Bull. Am. Phys. Soc., vol. 45, no. 1, p. 637, Mar. 2000 (Minneapolis, Minnesota).

http://www.ee.nthu.edu.tw/ccsu/










Both the rotational and the orbital motions of the earth together with the orbital
motion of the target planet contribute to the Sagnac
effect. But the orbital motion of the sun has no effects
on the interplanetary propagation. On the other hand, as
the unique propagation frame in GPS and intercontinental
links is a geocentric inertial frame, the rotational motion
of the earth contributes to the Sagnac effect. But the orbital
motion of the earth around the sun and that of the
sun have no effects on the earthbound propagation. By
comparing GPS with interplanetary radar, it is seen that
there is a common Sagnac effect due to earth’s rotation
and a common null effect of the orbital motion of the sun
on wave propagation. However, there is a discrepancy in
the Sagnac effect due to earth’s orbital motion. Moreover,
by comparing GPS with the widely accepted interpretation
of the Michelson–Morley experiment, it is seen that
there is a common null effect of the orbital motions on
wave propagation, whereas there is a discrepancy in the
Sagnac effect due to earth’s rotation.


Based on this characteristic of uniqueness and switchability of the propagation frame,
we propose in the following section the local-ether model
of wave propagation to solve the discrepancies in the in-
fluences of earth’s rotational and orbital motions on the
Sagnac effect and to account for a wide variety of propagation
phenomena.


Anyway, the interplanetary Sagnac effect is due to
earth’s orbital motion around the sun as well as earth’s
rotation. Further, for the interstellar propagation where
the source is located beyond the solar system, the orbital
motion of the sun contributes to the interstellar Sagnac
effect as well.

Evidently, as expected, the proposed local-ether model
accounts for the Sagnac effect due to earth’s rotation and
the null effect of earth’s orbital motion in the earthbound
propagations in GPS and intercontinental microwave link
experiments. Meanwhile, in the interplanetary radar, it accounts
for the Sagnac effect due both to earth’s rotation
and to earth’s orbital motion around the sun.


Based on the local-ether model, the propagation is entirely
independent of the earth’s orbital motion around
the sun or whatever and the velocity v for such an earthbound
experiment is referred to an ECI frame and hence
is due to earth’s rotation alone. In the original proposal,
the velocity v was supposed to incorporate earth’s orbital
motion around the sun. Thus, at least, v2/c2
=~ 10-8. Then the amplitude of the phase-difference variation
could be as large as π/3, when the wavelength is
0.6 µm and the path length is 10 m. However, as the velocity
v is the linear velocity due to earth’s rotation alone,
the round-trip Sagnac effect is as small as v2/c2∼ 10-12 which is merely 10-4 times that due to the orbital motion.


The Sagnac effect is a FIRST ORDER effect in v/c.

Even in the round-trip nature of the Sagnac effect, as it was applied in the Michelson-Morley experiment, thus becoming a second order effect within that context, we can see that the ORBITAL SAGNAC IS 10,000 TIMES GREATER than the rotational Sagnac effect.



Three papers written by some of the best physicists in the world today, Dr. Daniel Gezari and Dr. C.C. Su, which directly contradict your failed/miserable derivation.

It is as simple as this.


Please read:

In the original proposal,
the velocity v was supposed to incorporate earth’s orbital
motion around the sun. Thus, at least, v2/c2
=~ 10-8. Then the amplitude of the phase-difference variation
could be as large as π/3, when the wavelength is
0.6 µm and the path length is 10 m. However, as the velocity
v is the linear velocity due to earth’s rotation alone,
the round-trip Sagnac effect is as small as v2/c2∼ 10-12 which is merely 10-4 times that due to the orbital motion.


THE ORBITAL SAGNAC, EVEN IN THE CONTEXT OF THE ROUND TRIP MM EXPERIMENT, IS 10,000 TIMES GREATER THAN THE ROTATIONAL SAGNAC.



A total refutation of your failed piece of garbage analysis.


Both the IOP journal and the APS agree with me.

They published the fact that the orbital Sagnac is much larger than the rotational Sagnac, and that it is missing.

Then, they accepted the local aether model, since the missing orbital Sagnac has to be explained.

It is as simple as this.



There is no need to go into any kind of other details, since the claims/conclusions were clearly stated as follows:

Quote from: JackBlack on June 03, 2017, 11:47:37 PM
As such, for a given interferometer (with a fixed area of the light path), treating Earth's orbit as a rotation and Earth's daily rotation as a rotation where one revolution is 24 hours, the orbital Sagnac effect will be approximately 1/365 that of Earth's rotation, as the area is the same, the only thing that changes is the angular velocity which for the orbit is roughly 1/365 that of the day.

Quote from: JackBlack on June 04, 2017, 03:40:32 AM
No. It is easy to calculate that the orbital sagnac is a mere 1/365th that of the rotational.

That is because the 2 interferometers are the same and thus have the same R and the same A and thus it is entirely dependent upon ω

Again, the location of the centre of rotation is irrelevant.
All that matters is the area of the loop and its angular velocity.


The claim/conclusion of the derivation is as follows: the orbital Sagnac = 1/365 of the rotational Sagnac and these have been highlighted in red: there is no going back now.



You say 1/365.


Dr. C.C. Su says 10,000.


Both these figures cannot be true at the same time, right?


The conclusion/claim of your failed derivation is as follows, per your own very words listed above:

the orbital Sagnac = 1/365 of the rotational Sagnac



In the paper published by the BULLETIN OF THE AMERICAN PHYSICAL SOCIETY, one of the most prestigious journals in the world today, Dr. C.C. Su says that the orbital Sagnac is 10,000 times the magnitude of the rotational Sagnac.


He agrees with me.


You have to mail your derivation to both the BAPS and EPL to see if they will publish it.

It is as simple as this.


Until then I win.


Mainstream science agrees with me.


Published by an IOP journal.

Let us read the conclusions of the article:

http://qem.ee.nthu.edu.tw/f1b.pdf

On the other hand, the orbital
motion of the earth around the sun has a linear speed of
about 30 km/s which is about 100 times that of earth’s
rotation. Thus the present high-precision GPS would be
entirely impossible if the omitted correction due to orbital
motion is really necessary.


The author states that the orbital Sagnac is larger than the rotational Sagnac and that it is missing.


Again.

Meanwhile, as in GPS, no effects of earth’s orbital motion
are reported in these links, although they would be
easier to observe if they are in existence.

The orbital Sagnac would be easier to observe if and only if it was larger than the rotational Sagnac.


Then, and only then, the author adopts a local aether model in order to explain this extraordinary finding: the fact that not only is the orbital Sagnac larger in magnitude than the rotational Sagnac, but that also it is missing.


These are conclusions of a mainstream paper, accepted for publication by an IOP journal.

They totally refute the claim/conclusion that the orbital Sagnac is 1/365 of the rotational Sagnac.

They render the derivation to be a worthless piece of thrash.


If the conclusion of a derivation is shown to be wrong by such a huge margin, would it make any sense to go into any kind of further details? No, it would not.

The claim/conclusion of the derivation was clearly stated, there is no going back now.

But it was shown to be totally false.



You are done here jack.

You lost again, big time.


Please mail your derivation to the above mentioned journals and see if they publish your derivation.

« Last Edit: June 06, 2017, 09:39:15 AM by sandokhan »
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sandokhan

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #256 on: June 06, 2017, 06:06:31 AM »
Show your derivation of the correct Sagnac time delay!

There is nothing to derive you numskull.

You simply plug in the numbers in the correct formula.

That is what is missing from your derivation: the correct application of the formula.


Your derivation leads to the following conclusion/claim:

Here are jack's own conclusions/claims: they are not mine.


Quote from: JackBlack on June 03, 2017, 11:47:37 PM
As such, for a given interferometer (with a fixed area of the light path), treating Earth's orbit as a rotation and Earth's daily rotation as a rotation where one revolution is 24 hours, the orbital Sagnac effect will be approximately 1/365 that of Earth's rotation, as the area is the same, the only thing that changes is the angular velocity which for the orbit is roughly 1/365 that of the day.

Quote from: JackBlack on June 04, 2017, 03:40:32 AM
No. It is easy to calculate that the orbital sagnac is a mere 1/365th that of the rotational.

Are you such a moron as to not know how to read plain English?


This is the basic claim made by jack right here in front of everybody, as evidenced/highlighted in red:

Orbital Sagnac is 1/365 of the rotational Sagnac.


Remember well that figure: 1/365.


These calculations were done by none other than Dr. Daniel Gezari (CalTech, Goddard Space Flight Center).

For instance, the Earth’s full 30 km/s orbital velocity along the line-of-sight would produce a second-order residual velocity of only ~3 m/s, so we cannot preclude the possibility that some part of the 8.4 m /s difference between co and c measured here is a real second-order residual due to motion of the Earth-Moon system relative to an absolute frame.

3/8.4 = 0.357

1/365 = 0.00274

0.357/0.00274 = 130.3


Please note the orbital sagnac shows up in these measurements.

Now, because of the vast distance, if you are correct, you should see 1/365 of the rotational sagnac in the measurements and that will show up on this vast distance.

So, if you are correct, then we should see your 1/365 conclusions in the measurements. Guess what. We do not.

That means you are wrong.



jack claims the orbital Sagnac is 1/365 of the rotational Sagnac.

His derivation leads directly to this conclusion.

Dr. Daniel Gezari's calculations prove otherwise: even in the diluted two way averaging form, the orbital Sagnac amounts for a 3/8.4 = 0.357 (35.7%) percentage of the rotational Sagnac.



Both these figures cannot be right at the very same time.


Any sane person would understand that the derivation has been rendered to be useless and worthless, but we are not dealing with such persons here.


Published by the BULLETIN OF THE AMERICAN PHYSICAL SOCIETY, one of the most prestigious journals in the world today.

C.C. Su, "A Local-ether model of propagation of electromagnetic wave," in Bull. Am. Phys. Soc., vol. 45, no. 1, p. 637, Mar. 2000 (Minneapolis, Minnesota).

http://www.ee.nthu.edu.tw/ccsu/


Anyway, the interplanetary Sagnac effect is due to
earth’s orbital motion around the sun as well as earth’s
rotation.

Based on the local-ether model, the propagation is entirely
independent of the earth’s orbital motion around
the sun or whatever and the velocity v for such an earthbound
experiment is referred to an ECI frame and hence
is due to earth’s rotation alone. In the original proposal,
the velocity v was supposed to incorporate earth’s orbital
motion around the sun. Thus, at least, v^2/c^2
=~ 10^-8. Then the amplitude of the phase-difference variation
could be as large as π/3, when the wavelength is
0.6 µm and the path length is 10 m. However, as the velocity
v is the linear velocity due to earth’s rotation alone,
the round-trip Sagnac effect is as small as v^2/c^2∼ 10^-12 which is merely 10^-4 times that due to the orbital motion.


The Sagnac effect is a FIRST ORDER effect in v/c.

Even in the round-trip nature of the Sagnac effect, as it was applied in the Michelson-Morley experiment, thus becoming a second order effect within that context, we can see that the ORBITAL SAGNAC IS 10,000 TIMES GREATER than the rotational Sagnac effect.


Your claims are refuted yet again, and your derivation is rendered to be worthless.


The author PROVED that the orbital Sagnac is 10,000 times greater than the rotational Sagnac, and the peer reviewers agree totally.


It was published in one of the most respected journals in the world.


Both claims cannot be true at the same time, right?


The derivation's conclusion is that the orbital Sagnac is 1/365 of the rotational Sagnac.

The mainstream papers published by these respected journals prove that the orbital Sagnac is 10,000 greater than the rotational Sagnac.


There is no need to bother with the hapless derivation: its conclusions/claims have been shown to be totally erroneous.



Are you mentally insane rabinoz?


Do you understand how to read scientific papers?


You have just been presented with scientific papers which deny the basic conclusion/claim made by jack.


Dr. C.C. Su mailed his analysis to two of the most prestigious journals in the world: BAPS and EPL.

His article was peer reviewed and published.


Here are his own words again:


In the original proposal,
the velocity v was supposed to incorporate earth’s orbital
motion around the sun. Thus, at least, v^2/c^2
=~ 10^-8. Then the amplitude of the phase-difference variation
could be as large as π/3, when the wavelength is
0.6 µm and the path length is 10 m. However, as the velocity
v is the linear velocity due to earth’s rotation alone,
the round-trip Sagnac effect is as small as v^2/c^2∼ 10^-12 which is merely 10^-4 times that due to the orbital motion.


So, jack's own claim, that the orbital Sagnac is 1/365 of the rotational Sagnac has been totally refuted.


Have jack mail his derivation to these same two journals and see if they will publish his thoughts on the matter.

Until then, this thread is over, there nothing to discuss further here.


jack's derivation claims the figure 1/365.

Dr. C.C. Su informs us that the figure is 10,000.

Both these figures cannot be true at the same time.

Please mail the derivation to the same scientific journals which published Dr. Su's paper and see if they will publish it.


jack's derivation is completely wrong.


Here is the correct way to calculate the orbital Sagnac.

Earth's radius = 6357 km; r² = 40411449

Earth's orbital radius = 150,000,000 km r² = 22500000000000000

∆t = 4πR²ω/(c²-v²)
or

I use the linear velocity.

∆t = 4πRv/( c² - v² ), where v is the linear velocity.

For the earth's rotation, it is 0.4638333 km/ sec and the orbit v = 30km/sec.

∆t = 0.62831852628 for the earth's orbit.
Total path of the orbit is 2πr=2π(150,000,000 km) = 942,477,780km

Hence, the sagnac effect for a 1 km path, that means light source in the center and two receivers placed at .5km is:
0.62831852628 / 942,477,780km = 6.6666667 e-10 sec / km

Now, for the earth's rotation.
∆t = 4.1170061 e-7 seconds
Total path of the rotation is 2πr=2π(6357 km) = 39942.21 km


4.1170061 e-7 seconds / 39942.21 km = 1.0307407 e-11 sec / km


The sagnac effect for the earth's orbit is greater than that of the rotation.



The orbital Sagnac, though much larger than the rotational Sagnac, is not being registered by GPS satellites.



The BULLETIN OF THE AMERICAN PHYSICAL SOCIETY, EUROPHYSICS LETTERS JOURNAL, and the JOURNAL OF ELECTROMAGNETIC WAVES AND APPLICATIONS all agree with me.
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sandokhan

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #257 on: June 06, 2017, 06:41:09 AM »
There is nothing else to discuss here until the RE do one basic thing:

Mail the derivation to the SAME journals as did Dr. C.C. Su: the Bulletin of the American Physical Society, Europhysics Letters Journal, and the Journal of Electromagnetic Waves and Applications.

Dr. C.C. Su conclusions were clearly stated and published by these famous journals.

In the original proposal,
the velocity v was supposed to incorporate earth’s orbital
motion around the sun. Thus, at least, v^2/c^2
=~ 10^-8. Then the amplitude of the phase-difference variation
could be as large as π/3, when the wavelength is
0.6 µm and the path length is 10 m. However, as the velocity
v is the linear velocity due to earth’s rotation alone,
the round-trip Sagnac effect is as small as v^2/c^2∼ 10^-12 which is merely 10^-4 times that due to the orbital motion.


The Sagnac effect is a FIRST ORDER effect in v/c.

Even in the round-trip nature of the Sagnac effect, as it was applied in the Michelson-Morley experiment, thus becoming a second order effect within that context, we can see that the ORBITAL SAGNAC IS 10,000 TIMES GREATER than the rotational Sagnac effect.


These conclusions totally agree with my calculations.


The claim/conclusion of the derivation is as follows:

Orbital Sagnac is 1/365 of the rotational Sagnac.


The author PROVED that the orbital Sagnac is 10,000 times greater than the rotational Sagnac, and the peer reviewers agree totally.


It was published in one of the most respected journals in the world.


Both claims cannot be true at the same time, right?


The derivation's conclusion is that the orbital Sagnac is 1/365 of the rotational Sagnac.

The mainstream papers published by these respected journals prove that the orbital Sagnac is 10,000 greater than the rotational Sagnac.


You are going to have to mail your derivation to these same journals if you want anybody to look in your direction.

It is as simple as this.

Until then, there is nothing else to discuss here.
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Badxtoss

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #258 on: June 06, 2017, 07:02:23 AM »
Jesus Sandy, are you incapable of speaking succinctly and in your own words?
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sandokhan

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #259 on: June 06, 2017, 07:10:44 AM »
badxtoss, low content posting is not permitted in the FED section: read the rules.

Have you ever been exposed to the academic world?

Given your whining, you have not.

At the graduate level, you might be required to read some thousands of pages of bibliographical references in order to understand a certain subject.

Yet here you are complaining about two or three paragraphs.

You do not have to worry about the content of my messages: have I ever complained about yours?

Bibliographical references are always used in debates.

As for what I can do on my own, it is much more than you will ever achieve:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1910773#msg1910773
« Last Edit: June 06, 2017, 07:43:26 AM by sandokhan »
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Canadabear

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #260 on: June 06, 2017, 07:48:41 AM »
Quote from: sandokhan on June 06, 2017, 06:41:09 AM
There is nothing else to discuss here until the RE do one basic thing:
...

Until then, there is nothing else to discuss here.

we could discuss all the other "evidence" you have for a flat earth.

you could explain how the foucault Pendulum is working on the flat earth
you could explain how the visual effect of the earth curvature if we look across a sea works on a flat earth
you could explain how the sun set and rise works on a flat earth.

how about you answer these questions.
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Badxtoss

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #261 on: June 06, 2017, 08:52:12 AM »
Quote from: sandokhan on June 06, 2017, 07:10:44 AM
badxtoss, low content posting is not permitted in the FED section: read the rules.

Have you ever been exposed to the academic world?

Given your whining, you have not.

At the graduate level, you might be required to read some thousands of pages of bibliographical references in order to understand a certain subject.

Yet here you are complaining about two or three paragraphs.

You do not have to worry about the content of my messages: have I ever complained about yours?

Bibliographical references are always used in debates.

As for what I can do on my own, it is much more than you will ever achieve:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1910773#msg1910773
If you ever just posted two or three paragraphs that would be amazing.  You don't.  You post pages and pages of stuff pretty much every time.
As for what you've achieved, how about you present these marvelous findings to some of these journals you go on about.
Otherwise you are pretty much just blowing hot air.
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JackBlack

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #262 on: June 06, 2017, 03:23:24 PM »
Quote from: sandokhan on June 06, 2017, 05:47:07 AM
In a relativistic model, the orbital Sagnac effect is MUCH SMALLER than the rotational one.
There is no such thing as the relativistic model.
See how your cognitive dissonance shows up again?
I see you are sticking your head in the sand again.
The relativistic model does exist, and is used quite often.
Even in a non-relativistic model, the Sagnac effect, as we are discussing it, is interested in path length differences around a closed loop, and thus it still shows up the same as what I proved.

It is only other effects which differ.

Quote from: sandokhan on June 06, 2017, 05:47:07 AM
It is very easy to demolish your relativistic model.
No, it is impossible. You have repeatedly tried and failed. Regardless, this is not the place for that.

This is the place to discuss if the A in the formula provided is the area of Earth's orbit or the area of the interferometer loop.
I have provided a derivation, showing that it is the area of the interferomter loop, you have provided a reference (2 actually) which also shows/claims this, Rab provided one as well.

So so far, everything which says anything regarding this discussion agrees with me, except your baseless claim.

So I ask yet again:
CAN YOU SHOW ANYTHING WRONG WITH MY DERIVATION OR PROVIDE YOUR OWN?

You seem to live trying to hide your failures by providing mountains of bullshit, but I'm not going to deal with that bullshit here.


First, we discuss the derivation. Only after you admit you were wrong or prove you were correct by PROVIDING YOUR OWN DERIVATION AND SHOWING WHAT IS WRONG WITH MINE can we move on.

Quote from: sandokhan on June 06, 2017, 05:53:46 AM
Let me remind you, the Sagnac effect is a fringe shift due to 2 counterpropogating light pulses around a rotating ring.
The lunar ranging experiment does not have that and thus is not the Sagnac experiment.
jack, people are beginning to see through your miserable denial of reality.
The lunar laser ranging experiment is an astronomical version of the Sagnac experiment.
No. This isn't my denial.
The lunar laser ranging experiment was bouncing light off the moon. This only had one light path. It did not have 2 counter-propagating paths.

Remember what you said earlier? If not, here, let me remind you:
Quote from: sandokhan on June 03, 2017, 02:26:22 AM
IF SO YOU ARE VIOLATING THE VERY DEFINITION OF THE SAGNAC EFFECT.
"If two pulses of light are sent in opposite directions around a stationary circular loop of radius R" - standard definition

See how even you agree it requires 2 pulses of light that are sent in opposite directions around a loop?

Now then, can you explain what these 2 opposite directions are and what the loop is for the lunar ranging experiment?
If not, then I will continue to dismiss it as NOT THE SAGNAC EFFECT.

Do you know what would happen if they did make it a Sagnac Effect interferometer? They would observe no shift, as instead of a loop, they have a straight line. They are simply bouncing the light off the moon, so the 2 "opposite" directions would be the same and there would be no fringe shift recorded.

So stop bringing up this crap.

Quote from: sandokhan on June 06, 2017, 06:01:44 AM
How about we focus on what the article was about? An aether based model of light, so nothing at all to do with the discussion at hand. Try again.
jack, you are embarrassing yourself.
Once again, that would be you entirely.

You aren't even linking to the article now.
What, are you now afraid of people actually following your links and pointing out your bullshit like I have done repeatedly?

Can you provide a link to that, because that sure seems like the other crap you were linking to before.

Here is a key quote:
Quote
Recently, we have investigated the propagation of electromagnetic waves from a classical approach
i.e. ignoring relativity and instead using aether, the classical model of light.

This is further backed up by their introduction:
Quote
it was generally believed that electromagnetic waves propagate by means of a universal medium called ether.
They then contrast this to relativity.

So they are talking about an aether based model.

They continue with their calculations:
Quote
For a receiver moving at a fixed velocity ve, the classical propagation-range formula given to the first-order normalised speed

They are doing everything in a classical (i.e. aether based) model.

And once again, they aren't actually finding the Sagnac effect.

Quote from: sandokhan on June 06, 2017, 06:01:44 AM
Three papers written by some of the best physicists in the world today, Dr. Daniel Gezari and Dr. C.C. Su, which directly contradict your failed/miserable derivation.
No they don't.
Not in the slightest. They are still not talking about 2 counterpropogating waves around a closed loop and thus still aren't saying anything at all about my CORRECT derivation.

Quote from: sandokhan on June 06, 2017, 06:01:44 AM
Please read:
NO.
I am done reading your bullshit spam.
I have already pointed out the problems with it repeatedly.
Now how about this?
YOU START READING WHAT WE SAY AND ACTUALLY RESPONDING TO IT RATHER THAN IGNORING IT AND REPEATING THE SAME REFUTED CRAP AGAIN AND AGAIN??

Quote from: sandokhan on June 06, 2017, 06:01:44 AM
Until then I win.
No.
Until you can show a problem with my derivation, I win.
Until you can provide your own derivation showing that the A in the formula is the area of the orbital path, you cannot win.

Quote from: sandokhan on June 06, 2017, 06:01:44 AM
Mainstream science agrees with me.
Then why did all your references which actually discuss the issue at hand agree with me?

Quote from: sandokhan on June 06, 2017, 06:01:44 AM
You are done here jack.
You lost again, big time.
One again sticking your head in the sand.
You are the one losing, repeatedly.?

Until you show a problem with my derivation, I win, you lose.

Quote from: sandokhan on June 06, 2017, 06:06:31 AM
Show your derivation of the correct Sagnac time delay!
There is nothing to derive you numskull.
You simply plug in the numbers in the correct formula.
And you think this correct formula just appears by magic?

NO. YOU HAVE TO DERIVE IT.
That is what you need to derive, this "correct" formula.

So go and derive this "correct" formula, for the system I described or one akin to that, showing that the A in that formula is the area of the orbit rather than the interferometer loop.

Quote from: sandokhan on June 06, 2017, 06:06:31 AM
That is what is missing from your derivation: the correct application of the formula.
Nothing is missing from my derivation.
I derived the correct formula and then correctly applied it.


Quote from: sandokhan on June 06, 2017, 06:06:31 AM
jack's derivation is completely wrong.
Then why are you completely unable to show anything actually wrong with it?

Quote from: sandokhan on June 06, 2017, 06:06:31 AM
Here is the correct way to calculate the orbital Sagnac.
Earth's radius = 6357 km; r² = 40411449
Earth's orbital radius = 150,000,000 km r² = 22500000000000000
∆t = 4πR²ω/(c²-v²)
or
I use the linear velocity.
∆t = 4πRv/( c² - v² ), where v is the linear velocity.
For the earth's rotation, it is 0.4638333 km/ sec and the orbit v = 30km/sec.
∆t = 0.62831852628 for the earth's orbit.
Total path of the orbit is 2πr=2π(150,000,000 km) = 942,477,780km
Again, you are full of shit.
Stop repeating the same refuted bullshit.
The R here is the radius of the Sagnac loop. It has nothing to do with the size of Earth or its orbit.

If you wish to claim such bullshit you need to derive this formula showing that the R here is the size of Earth or its orbital radius, rather than the radius of the Sagnac loop.

Quote from: sandokhan on June 06, 2017, 06:41:09 AM
There is nothing else to discuss here until the RE do one basic thing:
No. There is nothing to discuss here until you manage to do one or two basic things:
Show what is wrong with my derivation.
Provide your own derivation showing that the R you use is the radius of Earth's orbit instead of the interferometer.

Until you do that, all the rest just amounts to bullshit spam.

NOW THEN:
STOP ALL THIS BULLSHIT AND SHOW WHAT IS WRONG WITH MY DERIVATION AND PROVIDE YOUR OWN OR ADMIT YOU WERE WRONG!
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sandokhan

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #263 on: June 06, 2017, 08:55:02 PM »
jack, you really need to brush up on your parody skills.

Do you understand the difference between 10,000 and 1/365?

Let's see now.

10,000/(1/365) = 3,650,000

A huge margin of error on your part!

Your derivation is a piece of thrash.



Now then, can you explain what these 2 opposite directions are and what the loop is for the lunar ranging experiment?
If not, then I will continue to dismiss it as NOT THE SAGNAC EFFECT.

Do you know what would happen if they did make it a Sagnac Effect interferometer? They would observe no shift, as instead of a loop, they have a straight line. They are simply bouncing the light off the moon, so the 2 "opposite" directions would be the same and there would be no fringe shift recorded.

The Sagnac effect also applies to TRANSLATIONAL/LINEAR/UNIFORM MOTION.


"For a circular path of radius R, the difference between the different time intervals can also be represented as Δt = 2vl/c^2, where v = ΩR is the speed of the circular motion and l = 2πR is the circumference of the circle.

The travel-time difference of two counterpropagating light beams in moving fiber is proportional to both the total length and the speed of the fiber, regardless of whether the motion is circular or uniform.

In a segment of uniformly moving fiber with a speed of v and a length of Δl, the travel-time difference is 2vΔl/c^2."





The lunar laser ranging experiment is an astronomical version of the Sagnac experiment.

However, G. Sagnac used the fringe-shift method to measure indirectly light travel time;
while Dr. Daniel Gezari uses clocks to measure directly light travel time in both directions.

Shooting light to the moon has to do with the behavior of light like GPS.

The arrival time of light to a receptor is influenced by the motion of
the receptor relative to the earth: this is the basic discovery of G. Sagnac.

This fact has to be incorporated into the lunar laser ranging calculations.

Here is a basic reference which confirms this fact:

Ring-laser tests of fundamental physics and geophysics, G.E. Steadman, 1997, pg 15




Dr. Daniel Gezari emitted a pulse of photons from a point on earth, bounced those photons off a reflector on the moon, and then recorded the photons’ arrival time at that same point on earth.


Please note the theoretical orbital sagnac shows up in these calculations, but is not picked up/registered/recorded by GPS satellites.

Motion of the Earth-Moon system in orbit around the Sun would average out in a two-way measurement, and only appear as a small (∼3 m/s) second-order residual.

Because of the two-way averaging, the orbital Sagnac effect registered is smaller than usual, however it is not 1/365 of the rotational Sagnac effect, in fact even in the diluted form permitted by the two-way averaging calculation, it represents a significant percentage of the rotational Sagnac effect.


THE SMALL (~3M/S) SECOND ORDER RESIDUAL IS THE ORBITAL SAGNAC.


For instance, the Earth’s full 30 km/s orbital velocity along the line-of-sight would produce a second-order residual velocity of only ~3 m/s, so we cannot preclude the possibility that some part of the 8.4 m /s difference between co and c measured here is a real second-order residual due to motion of the Earth-Moon system relative to an absolute frame.

THE 8.4 M/S DIFFERENCE IS THE ROTATIONAL SAGNAC.


Listen to what Dr. Daniel Gezari says:


For instance, the Earth’s full 30 km/s orbital velocity along the line-of-sight would produce a second-order residual velocity of only ~3 m/s, so we cannot preclude the possibility that some part of the 8.4 m /s difference


3/8.4 = 0.357

1/365 = 0.00274

0.357/0.00274 = 130.3


A total refutation of the claims, and thus of the derivation.


Now, because of the vast distance, if you are correct, you should see 1/365 of the rotational sagnac in the measurements and that will show up on this vast distance.

So, if you are correct, then we should see your 1/365 conclusions in the measurements. Guess what. We do not.

That means you are wrong.



jack claims the orbital Sagnac is 1/365 of the rotational Sagnac.

His derivation leads directly to this conclusion.


Dr. Daniel Gezari's calculations prove otherwise: even in the diluted two way averaging form, the orbital Sagnac amounts for a 3/8.4 = 0.357 (35.7%) percentage of the rotational Sagnac.



Both these figures cannot be right at the very same time.



SHOW WHAT IS WRONG WITH MY DERIVATION


Here are your own very words:


Quote from: JackBlack on June 03, 2017, 11:47:37 PM
As such, for a given interferometer (with a fixed area of the light path), treating Earth's orbit as a rotation and Earth's daily rotation as a rotation where one revolution is 24 hours, the orbital Sagnac effect will be approximately 1/365 that of Earth's rotation, as the area is the same, the only thing that changes is the angular velocity which for the orbit is roughly 1/365 that of the day.

Quote from: JackBlack on June 04, 2017, 03:40:32 AM
No. It is easy to calculate that the orbital sagnac is a mere 1/365th that of the rotational.


This is the basic claim made by jack right here in front of everybody, as evidenced/highlighted in red:

Orbital Sagnac is 1/365 of the rotational Sagnac.


Remember well that figure: 1/365.


Let us put your claim/conclusions/the most important part of your derivation to a second real test.


Published by the BULLETIN OF THE AMERICAN PHYSICAL SOCIETY, one of the most prestigious journals in the world today.

C.C. Su, "A Local-ether model of propagation of electromagnetic wave," in Bull. Am. Phys. Soc., vol. 45, no. 1, p. 637, Mar. 2000 (Minneapolis, Minnesota).

http://www.ee.nthu.edu.tw/ccsu/




Anyway, the interplanetary Sagnac effect is due to
earth’s orbital motion around the sun as well as earth’s
rotation.

Based on the local-ether model, the propagation is entirely
independent of the earth’s orbital motion around
the sun or whatever and the velocity v for such an earthbound
experiment is referred to an ECI frame and hence
is due to earth’s rotation alone. In the original proposal,
the velocity v was supposed to incorporate earth’s orbital
motion around the sun. Thus, at least, v^2/c^2
=~ 10^-8. Then the amplitude of the phase-difference variation
could be as large as π/3, when the wavelength is
0.6 µm and the path length is 10 m. However, as the velocity
v is the linear velocity due to earth’s rotation alone,
the round-trip Sagnac effect is as small as v^2/c^2∼ 10^-12 which is merely 10^-4 times that due to the orbital motion.


The Sagnac effect is a FIRST ORDER effect in v/c.

Even in the round-trip nature of the Sagnac effect, as it was applied in the Michelson-Morley experiment, thus becoming a second order effect within that context, we can see that the ORBITAL SAGNAC IS 10,000 TIMES GREATER than the rotational Sagnac effect.


The author PROVED that the orbital Sagnac is 10,000 times greater than the rotational Sagnac, and the peer reviewers agree totally.


It was published in one of the most respected journals in the world.


Both claims cannot be true at the same time, right?


The derivation's conclusion is that the orbital Sagnac is 1/365 of the rotational Sagnac.

The mainstream papers published by these respected journals prove that the orbital Sagnac is 10,000 greater than the rotational Sagnac.



That is what is totally wrong with your derivation.


You are off by a factor of 3,650,000!


Please mail your derivation to the very same journals listed above, just as Dr. C.C. Su did, and see if they will publish it.


Mainstream science agrees with me.

No worthless derivation is needed to properly and correctly apply the Sagnac formula to the orbital case.


You claimed the figure 1/365.


Mainstream science, published by the BAPS, says 10,000.


And that figure, 10,000, is still a very low estimate, as it was obtained in the context of the second order approximation used in the MM experiment.


Do you understand English jack?



In the original proposal,
the velocity v was supposed to incorporate earth’s orbital
motion around the sun. Thus, at least, v^2/c^2
=~ 10^-8. Then the amplitude of the phase-difference variation
could be as large as π/3, when the wavelength is
0.6 µm and the path length is 10 m. However, as the velocity
v is the linear velocity due to earth’s rotation alone,
the round-trip Sagnac effect is as small as v^2/c^2∼ 10^-12 which is merely 10^-4 times that due to the orbital motion.


Here is how you calculate the orbital Sagnac, in agreement with the above referenced paper.

Earth's radius = 6357 km; r² = 40411449

Earth's orbital radius = 150,000,000 km r² = 22500000000000000

∆t = 4πR²ω/(c²-v²)
or

I use the linear velocity.

∆t = 4πRv/( c² - v² ), where v is the linear velocity.

For the earth's rotation, it is 0.4638333 km/ sec and the orbit v = 30km/sec.

∆t = 0.62831852628 for the earth's orbit.
Total path of the orbit is 2πr=2π(150,000,000 km) = 942,477,780km

Hence, the sagnac effect for a 1 km path, that means light source in the center and two receivers placed at .5km is:
0.62831852628 / 942,477,780km = 6.6666667 e-10 sec / km

Now, for the earth's rotation.
∆t = 4.1170061 e-7 seconds
Total path of the rotation is 2πr=2π(6357 km) = 39942.21 km


4.1170061 e-7 seconds / 39942.21 km = 1.0307407 e-11 sec / km


The sagnac effect for the earth's orbit is greater than that of the rotation.



The orbital Sagnac, though much larger than the rotational Sagnac, is not being registered by GPS satellites.


Your derivation is off by a factor of 3,650,000!

My calculation shows the correct figure.

It is as simple as this.



« Last Edit: June 06, 2017, 08:57:54 PM by sandokhan »
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Sam Hill

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #264 on: June 06, 2017, 09:29:11 PM »
Sandy, I thought you were a flat earther?  As such, do you even believe there ARE satellites orbiting?  If the earth is flat, what is an 'orbit' at all?

Please don't reply with yet another copy of your giant pasted wall of text.  just talk to me, write an answer in your own words.
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rabinoz

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #265 on: June 06, 2017, 10:08:35 PM »
Quote from: sandokhan on June 06, 2017, 08:55:02 PM
jack, you really need to brush up on your parody skills.
It is as simple as this.

Just show us where JackBlack''s,  MathsPages', E. J. Post's and Wikipedia's derivations are wrong.

YOU have not shown where JackBlack's derivation might be in error and his results agree with the MathPages reference that YOU gave.

Yes, I know YOU DON'T KNOW HOW TO DO IT! ALL YOU CAN IS COPY, COPY COPY!

YOU only give some references, then drag what you like out of them.
You claim that they are the right answer's!

So stop being a pig-headed idiot an show exactly where that analysis is incorrect.
All we asking is for you to show us how to do the correct derivation.
Surely for a person of your brilliance and mathematical prowess it should as easy as pie - is that π?

Addendum:  Here is a derivation from your favourite source of information, Wikipedia!
Quote from: Wikipedia
Sagnac Effect, Theories
The shift in interference fringes in a ring interferometer can be viewed intuitively as a consequence of the different distances that light travels due to the rotation of the ring.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The time difference is:
         
For , this reduces to 

where A is the area of the ring. The Sagnac area formula applies to any shape of loop.
Although this simple derivation is for a circular ring with an index of refraction of one, the result holds true for any shape of rotating loop with area A.

Wherever we go we seem stuck with and the result holds true for any shape of rotating loop with area, A

Sure, one judges a hypothesis by the results it gives, but look where it takes you!
A sun somewhere up there at an unknown location - at least you can't tell us!
You have to postulate stupidly massive concentrations of your version of aether to make electromagnetic radiation bend like an acrobat and  slow down to walking speed to explain directions of light and the proven distances to the moon and planets.

Come up with an idiotic hypothesis and
end with an idiotic model for your totally unworkable and imaginary flat stationary earth!

A few more should try reading your "Advanced Flat Earth Theory" and really find out what your theories lead to.
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sandokhan

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #266 on: June 06, 2017, 10:09:41 PM »
The Biefeld-Brown technology used for the GPS satellites was hidden from public view.

Satellites orbit above the flat surface of the Earth using electrogravity.


http://users.erols.com/iri/TTBROWN2.htm

In 1985, Dr. Paul LaViolette was in the Library of Congress in Washington, DC and looked up the work "gravity" in the card catalog. Surprisingly, he found the listing for "Electrogravitics Systems," a report that was missing from the stacks. When the librarian tried to locate any other copies through interlibrary loan, she commented, "It must be an exotic document" because she could find only one in the country which was at Wright-Patterson Air Force Base. Thus, LaViolette was successful in obtaining a copy of the formerly classified document. The mystery continued: seven years later when contacting the Wright-Patterson AFB Technical Library, they surprisingly found no reference in the computer-based card catalog. They did locate the document on the shelves, however, after being asked to search for it. To summarize, the report has historic value because:

It validates T.T. Brown's experiments;
It lists the major corporations that were collaborating on electrogravitics;
It includes the requirements for supersonic speed;
It shows the continuity from Project Winterhaven in 1952;
The report includes a list of electrostatic patents;
It had been classified by the Air Force for an undetermined amount of time which underscores its importance.


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sandokhan

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #267 on: June 06, 2017, 10:15:48 PM »
rabinoz, are you mentally insane?

Is that what you are trying by all means to convey to the readers?

To put it more bluntly: are you a f****** idiot?


Here are jack's own claims, based totally on the derivation, they are not mine:

Quote from: JackBlack on June 03, 2017, 11:47:37 PM
As such, for a given interferometer (with a fixed area of the light path), treating Earth's orbit as a rotation and Earth's daily rotation as a rotation where one revolution is 24 hours, the orbital Sagnac effect will be approximately 1/365 that of Earth's rotation, as the area is the same, the only thing that changes is the angular velocity which for the orbit is roughly 1/365 that of the day.

Quote from: JackBlack on June 04, 2017, 03:40:32 AM
No. It is easy to calculate that the orbital sagnac is a mere 1/365th that of the rotational.

Are you such a moron as to not know how to read plain English?


This is the basic claim made by jack right here in front of everybody, as evidenced/highlighted in red:

Orbital Sagnac is 1/365 of the rotational Sagnac.


Remember well that figure: 1/365.





Dr. Daniel Gezari emitted a pulse of photons from a point on earth, bounced those photons off a reflector on the moon, and then recorded the photons’ arrival time at that same point on earth.

Obviously, both the rotational and the orbital Sagnac effects have to be taken into account.

These calculations were done by none other than Dr. Daniel Gezari (CalTech, Goddard Space Flight Center).

For instance, the Earth’s full 30 km/s orbital velocity along the line-of-sight would produce a second-order residual velocity of only ~3 m/s, so we cannot preclude the possibility that some part of the 8.4 m /s difference between co and c measured here is a real second-order residual due to motion of the Earth-Moon system relative to an absolute frame.

3/8.4 = 0.357

1/365 = 0.00274

0.357/0.00274 = 130.3


Please note the orbital sagnac shows up in these measurements.

Now, because of the vast distance, if you are correct, you should see 1/365 of the rotational sagnac in the measurements and that will show up on this vast distance.

So, if you are correct, then we should see your 1/365 conclusions in the measurements. Guess what. We do not.

That means you are wrong.



jack claims the orbital Sagnac is 1/365 of the rotational Sagnac.

His derivation leads directly to this conclusion.


Dr. Daniel Gezari's calculations prove otherwise: even in the diluted two way averaging form, the orbital Sagnac amounts for a 3/8.4 = 0.357 (35.7%) percentage of the rotational Sagnac.



Both these figures cannot be right at the very same time.


Any sane person would understand that the derivation has been rendered to be useless and worthless, but we are not dealing with such persons here.


Published by the BULLETIN OF THE AMERICAN PHYSICAL SOCIETY, one of the most prestigious journals in the world today.

C.C. Su, "A Local-ether model of propagation of electromagnetic wave," in Bull. Am. Phys. Soc., vol. 45, no. 1, p. 637, Mar. 2000 (Minneapolis, Minnesota).

http://www.ee.nthu.edu.tw/ccsu/


Anyway, the interplanetary Sagnac effect is due to
earth’s orbital motion around the sun as well as earth’s
rotation.

Based on the local-ether model, the propagation is entirely
independent of the earth’s orbital motion around
the sun or whatever and the velocity v for such an earthbound
experiment is referred to an ECI frame and hence
is due to earth’s rotation alone. In the original proposal,
the velocity v was supposed to incorporate earth’s orbital
motion around the sun. Thus, at least, v^2/c^2
=~ 10^-8. Then the amplitude of the phase-difference variation
could be as large as π/3, when the wavelength is
0.6 µm and the path length is 10 m. However, as the velocity
v is the linear velocity due to earth’s rotation alone,
the round-trip Sagnac effect is as small as v^2/c^2∼ 10^-12 which is merely 10^-4 times that due to the orbital motion.


The Sagnac effect is a FIRST ORDER effect in v/c.

Even in the round-trip nature of the Sagnac effect, as it was applied in the Michelson-Morley experiment, thus becoming a second order effect within that context, we can see that the ORBITAL SAGNAC IS 10,000 TIMES GREATER than the rotational Sagnac effect.


Your claims are refuted yet again, and your derivation is rendered to be worthless.


The author PROVED that the orbital Sagnac is 10,000 times greater than the rotational Sagnac, and the peer reviewers agree totally.


It was published in one of the most respected journals in the world.


Both claims cannot be true at the same time, right?


The derivation's conclusion is that the orbital Sagnac is 1/365 of the rotational Sagnac.

The mainstream papers published by these respected journals prove that the orbital Sagnac is 10,000 greater than the rotational Sagnac.


There is no need to bother with the hapless derivation: its conclusions/claims have been shown to be totally erroneous.


rabinoz, do you suffer from some kind of a mental disorder which precludes you from understanding these basic things?


Dr. C.C. Su mailed his analysis to two of the most prestigious journals in the world: BAPS and EPL.

His article was peer reviewed and published.


Here are his own words again:


In the original proposal,
the velocity v was supposed to incorporate earth’s orbital
motion around the sun. Thus, at least, v^2/c^2
=~ 10^-8. Then the amplitude of the phase-difference variation
could be as large as π/3, when the wavelength is
0.6 µm and the path length is 10 m. However, as the velocity
v is the linear velocity due to earth’s rotation alone,
the round-trip Sagnac effect is as small as v^2/c^2∼ 10^-12 which is merely 10^-4 times that due to the orbital motion.


So, jack's own claim, that the orbital Sagnac is 1/365 of the rotational Sagnac has been totally refuted.


Have jack mail his derivation to these same two journals and see if they will publish his thoughts on the matter.

Until then, this thread is over, there nothing to discuss further here.


jack's derivation claims the figure 1/365.

Dr. C.C. Su informs us that the figure is 10,000.

Both these figures cannot be true at the same time.

Please mail the derivation to the same scientific journals which published Dr. Su's paper and see if they will publish it.



jack claims the figure 1/365.

The above paper proves, and the peer reviewers agree, that the figure is at least 10,000.


Can you divide 10,000 by 1/365?


You will get 3,650,000: jack's derivation is off by that much.


There is no need for any kind of a derivation, the Sagnac formula speaks for itself.


rabinoz, you are showing off your monumental idiocy with each and every message posted.

You are unable to fathom that jack's claims/conclusions were refuted.


Do you accept the claims made by jack, highlighted in red above?

Then, you have no reason to complain any further.

He claims the figure 1/365, based on his derivation.

Dr. C.C. Su proves that it is 10,000.

It is as simple as this.
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sandokhan

  • Flat Earth Sultan
  • Flat Earth Scientist
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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #268 on: June 06, 2017, 10:26:34 PM »
and his results agree with the MathPages reference that YOU gave

No, they do not.

See how much of a f******* idiot you really are rabinoz?


What is the center of rotation for the orbit of the earth?

Here is the equation.

∆t = 4πRv / ( c² - v²) = 4Aω / ( c² - v²)

Where A = πR² and v = ωR

So, it is easy to calculate the orbital sagnac is more than 60 times that of the rotational.

But, A is based on R and according to mathpages, "circular loop of radius R".

http://www.mathpages.com/rr/s2-07/2-07.htm

Mathpages says you must use the center of rotation which is the sun.


You are an idiot rabinoz, no question about that, you have proven that you are unable to even read the references provided.

Shut the **** up.
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sandokhan

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #269 on: June 06, 2017, 11:15:08 PM »
Just show us where jackblack's derivation is wrong.


HERE ARE TWO MORE DIRECT QUOTES WHICH MAKE A CERTAIN CLAIM, USING A CERTAIN FIGURE:

Quote from: JackBlack on June 04, 2017, 04:37:47 AM
And thus:
dto/dtr=k*wo/k*wr=wo/wo=1/365.

Just like I said.
I also backed up this formula with my own derivation.


jack claims that the orbital Sagnac is 1/365 of the rotational Sagnac, and that, read carefully: "I also backed up this formula with my own derivation."

Using his own very words, jack is telling us that his derivation leads directly to this figure: 1/365.


Here is another quote now:

Quote from: JackBlack on June 04, 2017, 04:01:41 PM
This is the correct calculation:
Δto/Δt r=[4Aiωo/( c² - vo²)] / [4Aiωr/( c² - vr²)]

Note, the area here has nothing to do with the area of Earth's orbit or radius of it or the radius of Earth. It is the area of the interferometer, as my derivation.

No where in any derivation did the area of Earth's orbit come into it.

So to continue:
Δto/Δt r=[4Aiωo/( c² - vo²)] / [4Aiωr/( c² - vr²)]
Obviously, ( c² - vo²) and ( c² - vr²) are very close to the same number, so let's lave them off.
=4Aiωo/ 4Aiωr
Then to simplify:
=ωo/ ωr

And would you look at that? It ends up being just like what we claim.
You have ωo/ ωr.
As Earth rotates roughly 365 times for each orbit, ωr=365*ωo.
Thus we get:
Δto/Δt r=ωo/ ωr
=ωo/ (365*ωo)
=1/365

Just like we claim.


A very clear claim based on jack's derivation.

These are his own very words.


He claims that the derivation, as shown above, leads to the figure 1/365.


And he says that he bases his entire derivation on this: Note, the area here has nothing to do with the area of Earth's orbit or radius of it or the radius of Earth.

This hypothesis, in turn, jack claims leads to this conclusion:

=1/365

Just like we claim.


Let us put jack's claims to a real test.


Published by the BULLETIN OF THE AMERICAN PHYSICAL SOCIETY, one of the most prestigious journals in the world today.

C.C. Su, "A Local-ether model of propagation of electromagnetic wave," in Bull. Am. Phys. Soc., vol. 45, no. 1, p. 637, Mar. 2000 (Minneapolis, Minnesota).

http://www.ee.nthu.edu.tw/ccsu/



Anyway, the interplanetary Sagnac effect is due to
earth’s orbital motion around the sun as well as earth’s
rotation.

Based on the local-ether model, the propagation is entirely
independent of the earth’s orbital motion around
the sun or whatever and the velocity v for such an earthbound
experiment is referred to an ECI frame and hence
is due to earth’s rotation alone. In the original proposal,
the velocity v was supposed to incorporate earth’s orbital
motion around the sun. Thus, at least, v^2/c^2
=~ 10^-8. Then the amplitude of the phase-difference variation
could be as large as π/3, when the wavelength is
0.6 µm and the path length is 10 m. However, as the velocity
v is the linear velocity due to earth’s rotation alone,
the round-trip Sagnac effect is as small as v^2/c^2∼ 10^-12 which is merely 10^-4 times that due to the orbital motion.


The Sagnac effect is a FIRST ORDER effect in v/c.

Even in the round-trip nature of the Sagnac effect, as it was applied in the Michelson-Morley experiment, thus becoming a second order effect within that context, we can see that the ORBITAL SAGNAC IS 10,000 TIMES GREATER than the rotational Sagnac effect.


The author PROVED that the orbital Sagnac is 10,000 times greater than the rotational Sagnac, and the peer reviewers agree totally.


It was published in one of the most respected journals in the world.


Both claims cannot be true at the same time, right?


The derivation's conclusion is that the orbital Sagnac is 1/365 of the rotational Sagnac.

The mainstream papers published by these respected journals prove that the orbital Sagnac is 10,000 greater than the rotational Sagnac.



That is what is totally wrong with your derivation.


You are off by a factor of 3,650,000!


Why are you off by such a huge factor?

Because of the wrong application of the Sagnac formula.


What is the center of rotation for the orbit of the earth?

Here is the equation.

∆t = 4πRv / ( c² - v²) = 4Aω / ( c² - v²)

Where A = πR² and v = ωR

So, it is easy to calculate the orbital sagnac is more than 60 times that of the rotational.

But, A is based on R and according to mathpages, "circular loop of radius R".

http://www.mathpages.com/rr/s2-07/2-07.htm

Mathpages says you must use the center of rotation which is the sun.



Our friend jack brushed aside this requirement to impose his will on everybody else:

Note, the area here has nothing to do with the area of Earth's orbit or radius of it or the radius of Earth.

In this particular case, the orbital path of the Earth, YOU MUST USE THE CORRECT RADIUS IN ORDER TO ACHIEVE A CORRECT RESULT.


jack's dismissal of basic scientific facts led to this monumentally erroneous conclusions, using his own words:

=1/365

Just like we claim.


In the original proposal,
the velocity v was supposed to incorporate earth’s orbital
motion around the sun. Thus, at least, v^2/c^2
=~ 10^-8. Then the amplitude of the phase-difference variation
could be as large as π/3, when the wavelength is
0.6 µm and the path length is 10 m. However, as the velocity
v is the linear velocity due to earth’s rotation alone,
the round-trip Sagnac effect is as small as v^2/c^2∼ 10^-12 which is merely 10^-4 times that due to the orbital motion.



PUBLISHED BY THE BULLETIN OF THE AMERICAN PHYSICAL SOCIETY, one of the most respected journals in the world.


Dr. C.C. Su proves that the correct figure is at least 10,000.


jack's catastrophic interpretation of the Sagnac effect leads to a figure of 1/365.


Let us see by how much jack went wrong.


10,000/(1/365) = 3,650,000

A huge margin of error, just what we would expect of someone like jack.


There is nothing else to discuss here.
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