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 #120 on: June 01, 2017, 09:49:10 PM »
No, they weren't.
They were catapulted side by side with unequal momentum and unequal projection angle.

He likely didn't adequately control for the spinning mechanism to not effect the launch and failed to hold his arm straight as he launched it.


A sure sign of a loser.

He is now projecting his cognitive dissonance all the way to Harvard University.


For those who don't know, here is the biography of Dr. Edward Purcell:

In 1952, he shared the Nobel Prize in Physics for discovering a way to detect the extremely weak magnetism of the atomic nucleus. The method, measuring nuclear magnetic resonance, is widely used to study the structure of molecules and to measure magnetic fields.


Another Nobel prize winner.

But for a ----- like jackblack this is no problem at all.


He was a past president of the American Physical Society and a member of the National Academy of Sciences, American Philosophical Society and American Academy of Arts and Sciences. In 1967, he won the Oersted Medal of the American Association of Physics Teachers, and in 1979, he received the National Medal of Science.


Do you really think that the British Scientific Research Association Journal would have published Dr. DePalma's paper in 1976 if they thought that the spinning and nonspinning balls were launched with a different momentum/angle?

This experiment was also outlined personally by DePalma to Dr. Edward Purcell, one of the most eminent experimental physicists from Harvard at that time. According to DePalma, Purcell, after contemplating the experiment for several minutes, remarked "This will change everything."


The experiments were even carried out in vacuum.

A ball spinning at 27,000 RPM and a non-spinning ball were catapulted side-by-side with equal momentum and projection angle. In defiance of all who reject the ether as unrealistic, the spinning ball actually weighed less, and traveled higher than its non-spinning counterpart. Those who attribute this to an aerodynamic or atmospheric effect, please note that it works just as well in a vacuum. Also note, this effect has since been verified by other researchers. The decrease in weight of the spinning ball - anti-gravity - can explain why the spinning object goes higher and falls faster than the identical non-rotating control. Current thinking is that there is no special interaction between rotation and gravity. The behavior of rotating objects is simply the addition of ether energy to whatever motion the rotating object is making.


The law of universal gravitation totally violated: FOR THE SAME MASS OF THE STEEL BALLS, AND THE SAME SUPPOSED LAW OF ATTRACTIVE GRAVITY, THE ROTATING BALL WEIGHED LESS AND TRAVELED HIGHER THAN THE NON-ROTATING BALL.


The very hypothesis requires an equal momentum and an equal projection angle: this would have been the very first thing checked out by the peer reviewers of the prestigious British journal of physics.


Within a complete vacuum, DePalma took two steel balls and catapulted them at equal angles, with an equal amount of force.

The only difference was that one ball was rotating 27,000 times per minute and the other was stationary. The rotating ball traveled higher into the air and then descended faster than its counterpart, which violated all known laws of physics.

The only explanation for this effect is that both balls are drawing energy into themselves from an unseen source, and the rotating ball is thus “soaking up” more of this energy than its counterpart – energy that would normally exist as gravity, moving down into the earth.

With the addition of torsion-field research we can see that the spinning ball was able to harness naturally spiraling torsion waves in its environment, which gave it an additional supply of energy.


The precise application of Newton’s laws … have to be restricted to non-rotating mechanical objects in field-free space. In a gravitational field, the possibility of extraction of greater energy by a new mechanical dimension [rotation] opens up the possibility of an anti-gravitational interaction—Bruce DePalma, March, 1977



The Earth is totally stationary, no doubt about that: the orbital Sagnac effect which is larger than the rotational Sagnac, IS NOT BEING REGISTERED BY THE GPS SATELLITES.


https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1911899#msg1911899


THE ORBITAL SUN'S GRAVITATIONAL POTENTIAL IS NOT BEING RECORDED BY THE GPS SATELLITES EITHER.


https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1846706#msg1846706


Thus, the hypotheses of the Ruderfer experiment are totally fulfilled:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1846721#msg1846721

Why is there no requirement for a Sagnac correction due to the earth’s orbital motion? Like the transit time in the spinning Mossbauer experiments, any such effect would be completely canceled by the orbital-velocity effect on the satellite clocks.



BOTH THE ORBITAL SAGNAC AND THE SOLAR GRAVITATIONAL POTENTIAL ARE MISSING FROM THE RECORDINGS OF THE GPS SATELLITES.

THUS THE EARTH DOES NOT ORBIT THE SUN AT ALL: IT IS COMPLETELY STATIONARY.


Any movement of the Earth, be it a mere few meters would be recorded immediately by the GPS satellites.

As a final proof that it is movement of the receiver which is significant--not whether that movement is in a curved or straight line path--a test was run using the highly precise differential carrier phase solution. The reference site was stationary on the earth and assumed to properly apply the Sagnac effect. However, at the remote site the antenna was moved up and down 32 centimeters (at Los Angeles) over an eight second interval. The result of the height movement was that the remote receiver followed a straight line path with respect to the center of the earth.

The Sagnac effect was still applied at the remote receiver. The result was solved for position that simply moved up and down in height the 32 centimeters with rms residuals
which were unchanged (i.e. a few millimeters). If a straight line path did not need the Sagnac adjustment to the ranges the rms residuals should have increased to multiple meters. This shows again that it is any motion--not just circular motion which causes the Sagnac effect.

http://web.stcloudstate.edu/ruwang/ION58PROCEEDINGS.pdf

(Conducting a Crucial Experiment of the Constancy of the Speed of Light Using GPS, R. Wang/R. Hatch)
« Last Edit: June 03, 2017, 09:15:34 AM by sandokhan »

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Tessa Yuri

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #121 on: June 01, 2017, 10:46:57 PM »
sandokhan, the lack of certain largely ignored alternative scientific theories from GPS satellite data has absolutely nothing to do with whether or not the Earth is stationary. That entire argument is a non sequitur.
Tessa believes in the scientific method.
Yuri believes the Earth is a flat disk.
     _________              _________         _________
.<`X######I---I|    |I[][][][][][][][]I|     |I[][][][][][][][]I|
-=o--o====o--o=-=o-o====o-o=-=o-o====o-o=

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sandokhan

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #122 on: June 01, 2017, 11:53:07 PM »
The orbital Sagnac effect is totally MAINSTREAM SCIENCE.

It has to be.

We are being told that the Earth is orbiting the Sun.

Then this movement MUST BE/HAS TO BE RECORDED BY THE GPS SATELLITES.

But it is not.

The orbital Sagnac effect is part of the basic calculations done at Nasa:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1911899#msg1911899


The GPS satellites ONLY record/register the rotational Sagnac effect.


The larger orbital Sagnac effect is completely missing.

Which means that the Earth is stationary.


Also, the solar orbital gravitational potential IS ALSO MISSING.


Totally mainstream, basic fact of science.



https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1846706#msg1846706


Thus, the hypotheses of the Ruderfer experiment are totally fulfilled:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1846721#msg1846721

Why is there no requirement for a Sagnac correction due to the earth’s orbital motion? Like the transit time in the spinning Mossbauer experiments, any such effect would be completely canceled by the orbital-velocity effect on the satellite clocks.

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JackBlack

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #123 on: June 02, 2017, 12:04:32 AM »
A sure sign of a loser.
Nope. That would be you. Repeatedly asserting the same refuted nonsense and failing to address the comments made.
If you wish to discuss it, go start your own thread and I will happily refute you again.

The same applies to all your other spam.
Go make your own thread, or address the topic.

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Tessa Yuri

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  • The shortest distance between two points is a lie.
Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #124 on: June 02, 2017, 12:27:36 AM »
The GPS satellites ONLY record/register the rotational Sagnac effect.


The larger orbital Sagnac effect is completely missing.

Which means that the Earth is stationary.


Also, the solar orbital gravitational potential IS ALSO MISSING.


Totally mainstream, basic fact of science.

That's not how GPS works. It has no way of tracking its own progression through space, because that isn't what it is designed for. If you're saying GPS is fake, fine then adopt that theory, but come up with better proofs than 'it hasn't found what it isn't looking for'.
Tessa believes in the scientific method.
Yuri believes the Earth is a flat disk.
     _________              _________         _________
.<`X######I---I|    |I[][][][][][][][]I|     |I[][][][][][][][]I|
-=o--o====o--o=-=o-o====o-o=-=o-o====o-o=

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tomato

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #125 on: June 02, 2017, 02:31:48 AM »
Again pajama boy, you spam the same crap twice.
This time I will just link to the refutation:
https://www.theflatearthsociety.org/forum/index.php?topic=70566.msg1915381#msg1915381

Refute this :

STATIONARY EARTH SCENARIO :

Atmosphere = the canal with perfectly still water

The wind = boat propeller

The balloon = passenger in a boat (or a boat or a passenger in a boat & a boat)

The boat sails 30 knots per hour towards west

After one hour the boat is 30 nm westward from it's starting position (within earth's frame of reference and with respect to the frame of reference of absolute space, also).

As soon as we turn off the engine which propels the propeller of the boat, there will be no need for restoration of anything (non-pre-existing initial inertia).

The consequence /  the effect = the boat will simply rest at the calm water of the canal with no kind of perturbation/disturbance/commotion.

cikljamas, man, I love your enthusiasm, but I'm not sure if you're sure what you're saying here!

A boat is moving through water at 30 knots (1 knot = 1 nm per hour).
You suddenly shut the engine off.
I imagine that the boat would immediately start to slow down. Why do you say a boat wouldn't want to slow down suddenly when its engine is shut off suddenly? The boat would keep moving and not slow down? This seems as ridiculous to me as saying the floor is a spinning ball!
« Last Edit: June 02, 2017, 02:35:40 AM by tomato »
Tomato puree

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rabinoz

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #126 on: June 02, 2017, 04:53:24 AM »
Why is there no requirement for a Sagnac correction due to the earth’s orbital motion? Like the transit time in the spinning Mossbauer experiments, any such effect would be completely canceled by the orbital-velocity effect on the satellite clocks.
You have claimed that:
Quote
SAGNAC EFFECT
Let Δto = the Sagnac correction for the earth's orbital path
And Δtr = the Sagnac correction for the earth's rotational path

RE claim: Δto/Δtr = 1/365

Now, Δto = [4Aoωo/( c2 - vo2)] / [4Arωr/( c2 - vr2)]
and, Δtr = [Aoωo/(c2 - vo2)] / [Arωr/( c2 - vr2)]

Obviously, ( c2 - vo2) and ( c2 - vr2) are very close to the same number, so let's leave them off.

Hence, Δto/Δtr = Aoωo/Arωr
Not too bad so far!
But, I do believe that both Ao and Ar should be the area of the Sagnac loop, not the areas of the orbits!
That puts a whole new complexion on your infamous orbital Sagnac effect.
And note, if you use the proper angular velocity of the earth, calculated from the sidereal day, the orbital Sagnac effect is "automatically" included.

Don't take my amateurish word for it, maybe your favourite scientist, JackBlack, can check this out.

Quote from: sandokhan
Ao = πRo2 and Ar = πRr2
Incorrect and irellevant!
Quote from: sandokhan
So, Δto/Δtr = πRo2ωo / πRr2ωr or, Δto/Δtr = Ro2 ωo / Rr2ωr.[/size]
Incorrect and irrelevant!

Try again! Yes, you'll call me an ignorant idiot,
but when I see your "wonderful new radical chronology", I know who might be best called an ignorant idiot.

Have a nice day!
PS: 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 #127 on: June 02, 2017, 05:27:38 AM »
rabinoz, you seem to be drunk again.

You are not making any sense at all.


But, I do believe that both Ao and Ar should be the area of the Sagnac loop, not the areas of the orbits!

The equation for the sagnac is:

4Aω/( c² - v²)

One must calculate the area swept out by the path and that is A = πR², where R is measured from the Sun to the center of the Earth (radius of the orbital path loop).


This is how the orbital Sagnac is measured correctly.



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

This is an IOP article.

The author recognizes the earth's orbital Sagnac is missing whereas the earth's rotational Sagnac is not.

He uses GPS and a link between Japan and the US to prove this.

In GPS the actual magnitude of the Sagnac correction
due to earth’s rotation depends on the positions of
satellites and receiver and a typical value is 30 m, as the
propagation time is about 0.1s and the linear speed due
to earth’s rotation is about 464 m/s at the equator. The
GPS provides an accuracy of about 10 m or better in positioning.
Thus the precision of GPS will be degraded significantly,
if the Sagnac correction due to earth’s rotation
is not taken into account. 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.


In an intercontinental microwave link between Japan and
the USA via a geostationary satellite as relay, the influence
of earth’s rotation is also demonstrated in a high-precision
time comparison between the atomic clocks at two remote
ground stations.
In this transpacific-link experiment, a synchronization
error of as large as about 0.3 µs was observed unexpectedly.


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.
Thereby, it is evident
that the wave propagation in GPS or the intercontinental
microwave link depends on the earth’s rotation, but
is entirely independent of earth’s orbital motion around
the sun or whatever. As a consequence, the propagation
mechanism in GPS or intercontinental link can be viewed
as classical in conjunction with an ECI frame, rather than
the ECEF or any other frame, being selected as the unique
propagation frame. In other words, the wave in GPS or the
intercontinental microwave link can be viewed as propagating
via a classical medium stationary in a geocentric
inertial frame.


The author actually present a local-ether model (MLET, Modified Lorentz Ether Theory) in order to account for the MISSING ORBITAL SAGNAC EFFECT.


And note, if you use the proper angular velocity of the earth, calculated from the sidereal day, the orbital Sagnac effect is "automatically" included.

That is not how the orbital Sagnac is calculated.

Calculations performed at the NASA Goddard Space Flight Center.

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.


Even in the official version of heliocentricity, the Earth's orbit around the Sun is assumed to be nearly circular.

The radius of the earth very nearly circular orbit around the sun is 1.50⋅10^11 m.

THE GPS SATELLITES' ORBIT AROUND THE EARTH IS ALSO NEARLY CIRCULAR, YET THE SAGNAC EFFECT IS CALCULATED PRECISELY USING THE KNOWN FORMULA:

4Aω/( c² - v²)

https://www.faa.gov/about/office_org/headquarters_offices/ato/service_units/techops/navservices/gnss/faq/gps/

The orbits are nearly circular.

http://www.navipedia.net/index.php/GPS_Space_Segment

Orbits are nearly circular.


Eccentricity of Earth's orbit around the Sun (official science information):

Earth's orbit has an eccentricity of 0.0167.

Eccentricity of GPS satellites orbit around the Earth:

Orbits are nearly circular, with eccentricity less than 0.02.

The orbital eccentricity of an astronomical object is a parameter that determines the amount by which its orbit around another body deviates from a perfect circle. A value of 0 is a circular orbit, values between 0 and 1 form an elliptical orbit.



In addition, the orbital solar gravitational potential is not being registered either by the GPS satellites' clocks.

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1846706#msg1846706


Thus, the hypotheses of the Ruderfer experiment are totally fulfilled:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1846721#msg1846721

Why is there no requirement for a Sagnac correction due to the earth’s orbital motion? Like the transit time in the spinning Mossbauer experiments, any such effect would be completely canceled by the orbital-velocity effect on the satellite clocks.


The Ruderfer experiment puts an end to all of the RE illusions: it is the highest achievement of the 20th century in terms of ether drift theory, its conclusions cannot be disputed.

Since BOTH the orbital Sagnac and the orbital solar gravitational potential effect upon the GPS clocks are missing, the hypotheses of the Ruderfer experiment are totally fulfilled.

This means that the ether does exist.


This is the reason why rabinoz, each and every relativist who encounters the orbital Sagnac effect and the Ruderfer experiment simply gives up on the STR/GTR.

Their last stand is the MLET (modified Lorentz ether theory).

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rabinoz

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #128 on: June 02, 2017, 03:45:26 PM »
rabinoz
Well said! You are still not making any sense at all, you seem to be still drunk on your own omniscience.

Quote from: sandokhan
But, I do believe that both Ao and Ar should be the area of the Sagnac loop, not the areas of the orbits!
The equation for the sagnac is:
4Aω/( c² - v²)
One must calculate the area swept out by the path and that is A = πR², where R is measured from the Sun to the center of the Earth (radius of the orbital path loop).
No! swept out by the path taken by the propagating signals. And the propagating signals do not travel to the sun and back.
Sure 4Aω/( c2 - v2), where "A is the area enclosed by the light path".

Quote from: sandokhan
This is how the orbital Sagnac is measured correctly.
Sure if you have a Sagnac loop with propagation paths going in each direction all the way around the earth's orbit.

Quote from: sandokhan

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

This is an IOP article.

The author recognizes the earth's orbital Sagnac is missing whereas the earth's rotational Sagnac is not.
True, then says
Quote

The actual value of the Sagnac range correction due to the Earth’s rotation depends on the positions of satellites and receiver and a typical value is 30 m. The GPS provides an accuracy of about 10 m or better in positioning. Thus the precision of GPS will be degraded significantly, if the Sagnac correction due to the Earth’s rotation is not taken into account. On the other hand,
the orbital motion of the Earth around the Sun has a linear speed about 100 times that of the Earth’s rotation. Thus the present high-precision GPS would be entirely impossible if the ignored correction due to orbital motion is really necessary. Thereby, it is concluded that the wave propagation in GPS is actually in accord with the classical propagation in an ECI frame.

The "linear speed about 100 times that of the Earth’s rotation" is totally irrelevant - the Sagnac effect does not measure "linear velocity".

I don't know that I agree with the large value he gives for the "orbital Sagnac".
The signals are not propagated around the earth's orbit.

Maybe he should have written:
Either
    "Thereby, it is concluded that the wave propagation in GPS is actually in accord with the classical propagation in an ECI frame.[/font]"
or
    This high value of "orbital Sagnac" is incorrectly calculated.

The Sagnac effect does not depend on uniform linear velocity but on acceleration.

In the following paragraph the paper says:
Quote
      GPS is not the only experiment in which the wave propagation is referred uniquely to an ECI frame. In an intercontinental microwave link between Japan and USA via a geostationary satellite as relay, the influence of the Earth’s rotation is also demonstrated in a high-precision time comparison between the atomic clocks at two remote stations. In this transpacific-link experiment, a synchronization error as large as about 0.3 µs was observed unexpectedly and then is attributed to the Sagnac effect due to the Earth’s rotation after a detailed analysis,
while no effects of the Earth’s orbital motion are reported, although they would be easier to observe if they do exist.
In the intercontinental microwave link, the propagation path is around the surface of the rotating earth.

I note also that you include
Quote from: sandokhan
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.
Thereby, it is evident that the wave propagation in GPS or the intercontinental microwave link depends on the earth’s rotation, but is entirely independent of earth’s orbital motion around the sun or whatever.
This bit "Thereby, it is evident that the wave propagation in GPS or the intercontinental microwave link depends on
             the earth’s rotation, but is entirely independent of earth’s orbital motion around the sun."
Is it really so "evident"?
It could also mean that the Sagnac effect due to the "earth’s orbital motion around the sun" is so much smaller (like 1/365) that is was not separately detected.

All the "earth’s orbital motion around the sun" does is add one extra revolution per year and had the earth's rotational period been taken as one sidereal day (which it is) it that extra revolution would be there anyway!


But, I'm not going to waste time over it. You pick and chose what to accept out of all these papers.

Michelson himself had not the slightest doubt that the earth rotated about it's axis and orbited the sun.
This paper also shows not the slightest doubt that the author accepts the same things. He's just suggesting a "local aether".

You totally reject these things, so why should I not doubt some points raised.

Every time I look more deeply into your claims, I find something that you "conveniently" interpret in your own weird way.
Not only that, but you quote all these sources where the authors clearly have no doubts about the heliocentric globe,
and then you interpret it in some weird way to support you own ridiculous model.

The paper you quoted here is a case in point. The author shows not the slightest in the heliocentric globe, nor in orbiting GPS satellites, yet you use one little point out of it to support you radical "cosmology".
I am still waiting!
PS: 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".

*

JackBlack

  • 23451
Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #129 on: June 02, 2017, 05:09:32 PM »
cikljamas, man, I love your enthusiasm, but I'm not sure if you're sure what you're saying here!

A boat is moving through water at 30 knots (1 knot = 1 nm per hour).
You suddenly shut the engine off.
I imagine that the boat would immediately start to slow down. Why do you say a boat wouldn't want to slow down suddenly when its engine is shut off suddenly? The boat would keep moving and not slow down? This seems as ridiculous to me as saying the floor is a spinning ball!
He is actually saying the opposite, that the boat will stop instantly, that if you shut off the engine the boat will no longer move.

In reality, it starts to slow down once you cut the engine, but it doesn't come to a stop instantly.

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JackBlack

  • 23451
Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #130 on: June 02, 2017, 06:06:27 PM »
4Aω/( c² - v²)

One must calculate the area swept out by the path and that is A = πR², where R is measured from the Sun to the center of the Earth (radius of the orbital path loop).
I pointed out this bullshit before, NO, THAT IS NOT HOW YOU DO IT.

A nice simple example, consider a single arc, where the light merely propagates along it, back and forth, so the area of the "loop" would be 0.

This means going around the loop is the same in each direction.
This means it will produce NO sagnac effect.
And this remains true regardless of where this "loop" is placed.

Now lets try 2 arcs, one at R1 and one at R2, where the sections connecting the 2 arcs are along the radii.
What this means is that going between them is the same for both directions.
In each one you have it going to the inner arc along a radii, and going to the outer arc along a radii, so the path length and time taken will be equal for those sections.
The only thing giving rise to a difference will be the sections along the arc.

So now lets let alpha be the angle subtended by the arcs, and omega be the angular velocity.
So, one beam of light will propagate in the same direction as the big arc and the opposite direction as the little arc.
What this means is its trip along the big arc will be longer than if it was stationary and shorter along the little arc.
The distance it has to travel is alpha*R2+omega*R2*t1a+alpha*R1-omega*R1*t1b, where t1a is the time taken to go along the big arc and t1b is the time taken to go along the little arc.
This is because in t1a, the arc will have moved along a bit, and the light needs to travel the length of the arc and that bit it has moved along, while for t1b (from the perspective of the light) the arc has travelled back a bit, shrinking the distance.

Meanwhile, the light going the other way has to travel a distance of alpha*R1+omega*R1*t2b+alpha*R2-omega*R2*t2a, for similar reasons.

(Note: the a is the big arc, the b is the little arc, this is to make it simpler later on).

The total time for each will be the sum of the a and b part, so now we need to figure out what they are.
Well, in t1a, the light (travelling at speed c) travels a distance alpha*R2+omega*t1a.
Thus t1a*c=alpha*R2+omega*R2*t1a
Thus t1a=alpha*R2/(c-omega*R2).
Similarly, t2b=alpha*R1/(c-omega*R1).
And t1b*c=alpha*R1-omega*t1b
Thus t1b=alpha*R1/(c+omega*R1)
and t2a=alpha*R2/(c+omega*R2).

Now as I said, the total time for each one is given by:
t1=t1a+t1b.
t2=t2a+t2b.
And then we find the difference as (note: may give minus sign, I haven't checked, but the important part is the magnitude)
dt=t1-t2
=(t1a+t1b)-(t2a+t2b)
=t1a+t1b-t2a-t2b
=t1a-t2a+t1b-t2b
=(t1a-t2a)+(t1b-t2b)

Rather than try to solve it all at once, for simplicity we break it into 2 parts:
dta=t1a-t2a
And dtb=t1b-t2b.
And thus dt=dta+dtb

Now then:
dta=t1a-t2a
=alpha*R2/(c-omega*R2)-alpha*R2/(c+omega*R2)
=alpha*R2*(1/(c-omega*R2)-1/(c+omega*R2))
=alpha*R2*((c+omega*R2)/(c-omega*R2)*(c+omega*R2)-(c-omega*R2)/(c+omega*R2)*(c-omega*R2))
=alpha*R2*((c+omega*R2)-(c-omega*R2))/(c+omega*R2)*(c-omega*R2))
=alpha*R2*(c+omega*R2-c+omega*R2)/(c2-(omega*R2)2)
=alpha*R2*2*omega*R2/(c2-(omega*R2)2)
=2*omega*alpha*R22/(c2-(omega*R2)2)
And then if we assume omega*R2=v<<c (i.e. our system is moving much slower than the speed of light, and 30 km/s is still much slower than the speed of light at roughly 300 000 km/s, then we can simplify (c2-(omega*R2)2) to c2
And thus we end up with dta=2*omega*alpha*R2^2/c2

Now we do the same for dtb.
dtb=t1b-t2b
=alpha*R1/(c+omega*R1)-alpha*R1/(c-omega*R1)
=alpha*R1*(1/(c+omega*R1)-1/(c-omega*R1))
(combining some steps from before for brevity, and noting that omega*R1 will be tiny compared to c just like omega*R2)
=alpha*R1*(c-omega*R1-c-omega*R1)/c2
=-2*omega*alpha*R12/c2

Thus dt=2*omega*alpha*R22/c2-2*omega*alpha*R1^2/c2
=2*omega*alpha*(R22-R12)/c2

Now, can we simplify this any more?
I know, lets work out the area.
Note that alpha has always been in radians.
A circle has an area pi*r2.
This circle is a circular sector which subtends an angle of 2*pi.
If it only subtends an angle of alpha, then it will only have an area of alpha/(2*pi) of the circle.
Thus A (for a circular sector) is (alpha/(2*pi))*pi*r2
=alpha*r2/2
This means from the center to the outer arc you have an area of:
A2=alpha*R22/2.
And for the inner one you have an area:
A1=alpha*R12/2.
Thus the area between them is:
A=alpha*R22/2-alpha*R12/2
=alpha*(R22-R12)/2
Thus 2*A=alpha*(R22-R12).

Notice that was in the formula above?
That means we can sub it in.
As a reminder we had:
dt=2*omega*alpha*(R22-R12)/c2
By subbing in the above we get:
dt=2*omega*2*A/c2
And thus:
dt=4*A*omega/c2

Now that certainly looks like the formula you had (with one simplification).
But notice a key fact?
It uses the area of the loop, not the area of the orbit.
So no, you are full of shit and have repeatedly been using the wrong radius/area to try and pretend that the orbital Sagnac effect should be much greater.

You can do the same without the simplification, but it gets more complicated, and it still isn't based upon the area of the orbit.


This is how the orbital Sagnac is measured correctly.
No. What I did was how to measure it correctly. What you did is pure bullshit which I have pointed out numerous times.
Now quite spamming your ignorant bullshit, especially as it has already been refuted many times.

If you wish to assert your bullshit then derive the formula and show what is wrong with my derivation.
Until then, stop spamming the same refuted ignorant bullshit.
« Last Edit: June 02, 2017, 06:20:03 PM by JackBlack »

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JackBlack

  • 23451
Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #131 on: June 02, 2017, 06:18:57 PM »
Also, don't worry if that is too long and hard for you to understand, I can try a much simpler way:
By having a small loop on Earth (akin to what I had, 2 arcs along Earth's orbit, with a different radius, R1 and R2), you effectively have 2 pieces of 2 much larger interferometers.

You have a small piece (alpha/2*pi) of each a small one and a big one, where each one is the massive one going all the way around Earth's orbit.

(I will again use the simplification such that you have (c2-v2)=c2 rather than dealing with much more complicated math which doesn't make much of a difference.

So for the big one, we have it at R2, all the way around Earth's orbit, and thus it will register a shift of:
dt2=4*A2*omega/c2

But we don't have the full one. Instead we just have a small piece of it. As such, we won't get the full effect, instead we will get a smaller effect, based upon the proportion we have. (akin to removing some loops for an optic fibre one)
We only have a small section, which subtends an angle of alpha instead of the full 2*pi.
This means we will only record a fraction, alpha/(2*pi) of this one.
But remember, from before, this is the ratio of the area of a circular sector to the area of the circle. Thus we can just replace this A with the area of our arc (Aa).
Thus our part of the big one will record a shift of:
dt2=4*Aa2*omega/c2
The same would apply for the little one, but remember, the light travels around this one in the opposite direction. That means the shift will be in the opposite direction.
That means we will get dt1=-4*Aa1*omega/c2.

Then just like before we can add them up (akin to adding more loops for an optic fibre one).
This means dt=dt1+dt2
=4*Aa2*omega/c2-4*Aa1*omega/c2

That means what we are left with is:
dt=4*(Aa2-Aa1)*omega/c2
And that difference, Aa2-Aa1 is the area of the physical loop we have, the section of an annuls.

So what that means is we have:
dt=4*A*omega/c2

Again, the area is the area of the detector, not the area of an orbit.

So once again, YOU ARE FULL OF SHIT.
Now quit spouting the same refuted ignorant bullshit.

If you wish to claim that you use the area of the orbit, PROVE IT. DERIVE THE FORMULA SHOWING THAT IS THE AREA YOU USE!
Also show what was wrong with my derivations.

If you are unable to do that, then shut up.

*

sandokhan

  • Flat Earth Sultan
  • Flat Earth Scientist
  • 7262
Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #132 on: June 02, 2017, 09:34:44 PM »
jack and rabinoz, go ahead and write to your local university, even Nasa, and inform them that you do not agree and do not accept how the Sagnac formula/effect is currently being calculated: see what kind of an answer you will receive.

This is basic and classic cognitive dissonance: you are trying to change the laws of physics in order to feel safe.


Remember this?

According to me, it is purely a force based upon the tension in the rope.

So no contradiction there (or anywhere else between my analysis and reality).

According to ME...

Here you are at it again, this time around modifying even the Sagnac formula/effect to suit your purposes.

Also show what was wrong with my derivations.

Go ahead, and send your derivation to any university, the very same IOP journal which published the referenced article, and see how they will answer you.

Nobody is going to even look into your direction, since they will infer clearly that you have lost touch with reality.

DO YOU HAVE ANY PAPERS THAT AGREE WITH YOUR CALCULATIONS? You do not: nobody pays any attention to the mathematical equivalent of a severe episode of cognitive dissonance.


HERE IS THE STANDARD USED BY THE US NAVY TO CALCULATE THE SAGNAC:

https://web.archive.org/web/20120915150646/http://tycho.usno.navy.mil/ptti/ptti2006/paper28.pdf

The term “Sagnac effect” is part of the vocabulary of only the observer in the rotating reference frame. The corresponding correction applied by the inertial observer might be called a “velocity correction.” While the interpretation of the correction is different in the two frames, the numerical value is the same in either frame.

Note, that the observer on the rotating path AND the observer at the center of rotation calculate exactly the same sagnac. There is no disagreement.



Here is how the orbital Sagnac is to be calculated correctly.

Calculations performed at the NASA Goddard Space Flight Center.

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.


Here are two references which examine in great detail the errors of the 1/365 approximation desired by the relativists:

http://webcache.googleusercontent.com/search?q=cache:v60nWuQi1JQJ:www.anti-relativity.com/forum/viewtopic.php%3Ff%3D3%26t%3D6176%26start%3D30%26view%3Dprint+&cd=1&hl=ro&ct=clnk&gl=ro

http://webcache.googleusercontent.com/search?q=cache:o4iphVdmMSAJ:www.anti-relativity.com/forum/viewtopic.php%3Ff%3D3%26t%3D6212%26start%3D45%26view%3Dprint+&cd=2&hl=ro&ct=clnk&gl=ro


Another relativist tried to use the 1/365 silly argument.

But it doesn't work like that in the real world.

Here is the correct answer to this question:

"You used the same radius for the orbital path and the rotational path. This violates mathpages (the standard used by the relativists to explain the Sagnac effect).

You then claimed it is 1/365 of the amount. Well, that does not even show up.

In addition, by your logic, it is the rotational speed. That would imply lab sagnacs, which rotate mych faster, should show huge sagnacs. They do not, they show c+v and c-v just like the earth's rotation.

You never could understand this. It is purely a function of linear speed.

Finally, this IOP paper makes it clear, the orbital sagnac is consistent with all other sagnacs and hence it should exist. Since it does not, this author invented a local-aether which IOP thought enough of to publish it.

You do not seem to understand what I am saying and I doubt you can.

The orbital sagnac is missing and that is why GPS works. I have not disagreed it is missing.

Also, you do not know what you are doing when you calculate sagnac since you do not understand what the radius of the path means.

There is something else going on with light and this is not within your grasp. You make a faulty calculation that is not supported by any mainstream math but since it agrees somewhat with the evidence, you proclaim it is correct.

You see, they are not seeing 1/365 the rotational sagnac as you would proclaim should exist in shooting a laser to the mirror on the moon and back.

So, get over it, your calculation is wrong and proven wrong by the mirror on the moon. Also, your calculation violates mathpages.


Here are the calculations for the moon.

http://arxiv.org/abs/0912.3934?context=physics.gen-ph

You will 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."



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

This is an IOP article.

The author recognizes the earth's orbital Sagnac is missing whereas the earth's rotational Sagnac is not.

The calculations performed for both the rotational and the orbital Sagnac use exactly the correct formula.

If the author had sent your catastrophic piece of bullshit to the journal, nobody would have paid any attention to him.

Mainstream science accepts and understands that the orbital Sagnac is missing.


Here is how to correctly calculate the Sagnac, exactly as done in the above referenced papers.

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1911899#msg1911899

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 equation for the sagnac is:

4Aω/( c² - v²)

One must calculatate the area swept out by the path and that is A = πR², where R is measured from the Sun to the center of the Earth (radius of the orbital path loop).



Please note that the orbital Sagnac is NOT being registered/recorded by the GPS satellites.


In addition, the orbital solar gravitational potential is not being registered either by the GPS satellites' clocks.

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1846706#msg1846706


Thus, the hypotheses of the Ruderfer experiment are totally fulfilled:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1846721#msg1846721

Why is there no requirement for a Sagnac correction due to the earth’s orbital motion? Like the transit time in the spinning Mossbauer experiments, any such effect would be completely canceled by the orbital-velocity effect on the satellite clocks.


The Ruderfer experiment puts an end to all of the RE illusions: it is the highest achievement of the 20th century in terms of ether drift theory, its conclusions cannot be disputed.

Since BOTH the orbital Sagnac and the orbital solar gravitational potential effect upon the GPS clocks are missing, the hypotheses of the Ruderfer experiment are totally fulfilled.

This means that the ether does exist.
« Last Edit: June 02, 2017, 09:53:10 PM by sandokhan »

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JackBlack

  • 23451
Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #133 on: June 03, 2017, 12:09:39 AM »
jack and rabinoz, go ahead and write to your local university, even Nasa, and inform them that you do not agree and do not accept how the Sagnac formula/effect is currently being calculated: see what kind of an answer you will receive.
No, I agree with NASA/the unis. I don't agree with your pathetic bullshit.

This is basic and classic cognitive dissonance: you are trying to change the laws of physics in order to feel safe.
No. That would be you. You are trying to change them to pretend all the bullshit you spout is perfectly fine.


Remember this?
Yes. I do.
Remember how I repeatedly refuted your bullshit, showing how you were wrong every time?
Remember how you failed to address any of the objections I raised and instead kept on repeating the same refuted bullshit?
Remember how you then ran away like a scared, defeated kid that didn't want to admit they were wrong?

Here you are at it again, this time around modifying even the Sagnac formula/effect to suit your purposes.
No. Here I am at it again, showing the world the truth, which includes how you are full of shit.

Go ahead, and send your derivation to any university, the very same IOP journal which published the referenced article, and see how they will answer you.
They will say that it was already known and not worth publishing.

How about you send your delusional bullshit to them and get it rejected for being pure bullshit?

I told you to show what is wrong with my derivation. Can you do that?
If not, my derivation stands correct.

So, are you going to refute it and do your own?
If not, then like I said, shut up.

So either refute my derivation and provide your own, or stop repeating the same ignorant refuted bullshit.

*

sandokhan

  • Flat Earth Sultan
  • Flat Earth Scientist
  • 7262
Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #134 on: June 03, 2017, 01:00:30 AM »
jack, you are truly delusional.

You have been utterly defeated, in a very embarrassing manner, each and every time we met.

You have absolutely nothing going for you, scientifically speaking.

It takes less than 30 seconds to debunk your poorly written messages.


You have to learn to properly present your case, using subscripts/superscripts, some diagrams, which will convey your point to your readers. Ask rabinoz to draw these diagrams for you, go back and forth, until you have a clear idea of what it is you want to communicate.



They will say that it was already known and not worth publishing.

You are way to generous with your assessment.

They will wipe their asses with your "analysis" and flush the toilet with it right into the sewer system.

Nobody can make any sense out of what you wrote.

You need to use subscripts/superscripts and diagrams to convey the point you are trying to make in an understandable manner.


I told you to show what is wrong with my derivation.

Nobody can understand what you are trying to say: you owe to yourself and to your readers to please mail your analysis, as it is, to the same IOP journal referenced above, and see what kind of a response you will receive.

Do you have any references to support your view? You do not.


On the other hand, I have references which show exactly how the Sagnac is to be calculated.

There is only one correct way to calculate the Sagnac (be it rotational or orbital).


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

This is an IOP article.

The author recognizes the earth's orbital Sagnac is missing whereas the earth's rotational Sagnac is not.

He uses GPS and a link between Japan and the US to prove this.

In GPS the actual magnitude of the Sagnac correction
due to earth’s rotation depends on the positions of
satellites and receiver and a typical value is 30 m, as the
propagation time is about 0.1s and the linear speed due
to earth’s rotation is about 464 m/s at the equator. The
GPS provides an accuracy of about 10 m or better in positioning.
Thus the precision of GPS will be degraded significantly,
if the Sagnac correction due to earth’s rotation
is not taken into account. 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.


In an intercontinental microwave link between Japan and
the USA via a geostationary satellite as relay, the influence
of earth’s rotation is also demonstrated in a high-precision
time comparison between the atomic clocks at two remote
ground stations.
In this transpacific-link experiment, a synchronization
error of as large as about 0.3 µs was observed unexpectedly.


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.
Thereby, it is evident
that the wave propagation in GPS or the intercontinental
microwave link depends on the earth’s rotation, but
is entirely independent of earth’s orbital motion around
the sun or whatever. As a consequence, the propagation
mechanism in GPS or intercontinental link can be viewed
as classical in conjunction with an ECI frame, rather than
the ECEF or any other frame, being selected as the unique
propagation frame. In other words, the wave in GPS or the
intercontinental microwave link can be viewed as propagating
via a classical medium stationary in a geocentric
inertial frame.


The author actually present a local-ether model (MLET, Modified Lorentz Ether Theory) in order to account for the MISSING ORBITAL SAGNAC EFFECT.


I repeat: this is an IOP article, the highest standard of mainstream science.

The paper was peer reviewed and published.

Each and every scientist working at that journal understood the meaning of these words:

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.



My advice: go ahead and mail your analysis to the same IOP scientific journal and ask them to take a look at it.

Do you really think that anybody in his/her right mind will pay any kind of attention to this kind of ramblings? Rest assured: nobody will.
« Last Edit: June 03, 2017, 09:17:09 AM by sandokhan »

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JackBlack

  • 23451
Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #135 on: June 03, 2017, 02:01:35 AM »
jack, you are truly delusional.

You have been utterly defeated, in a very embarrassing manner, each and every time we met.

You have absolutely nothing going for you, scientifically speaking.

It takes less than 30 seconds to debunk your poorly written messages.
And there you go almost describing yourself.
Except the 30 seconds part, you have described yourself.
Due to the sheer amount of bullshit you normally post, it typically takes a lot longer.

You want to discuss the Saganac effect, fine.
Either show what is wrong with my derivation and derive it yourself, admit you were wrong, or shut up.
They are your options.
I have refuted your bullshit and you lying about it pretending it never happened wont magically make it disappear.

Here it is again:
Now lets try 2 arcs, one at R1 and one at R2, where the sections connecting the 2 arcs are along the radii.
What this means is that going between them is the same for both directions.
In each one you have it going to the inner arc along a radii, and going to the outer arc along a radii, so the path length and time taken will be equal for those sections.
The only thing giving rise to a difference will be the sections along the arc.

So now lets let alpha be the angle subtended by the arcs, and omega be the angular velocity.
So, one beam of light will propagate in the same direction as the big arc and the opposite direction as the little arc.
What this means is its trip along the big arc will be longer than if it was stationary and shorter along the little arc.
The distance it has to travel is alpha*R2+omega*R2*t1a+alpha*R1-omega*R1*t1b, where t1a is the time taken to go along the big arc and t1b is the time taken to go along the little arc.
This is because in t1a, the arc will have moved along a bit, and the light needs to travel the length of the arc and that bit it has moved along, while for t1b (from the perspective of the light) the arc has travelled back a bit, shrinking the distance.

Meanwhile, the light going the other way has to travel a distance of alpha*R1+omega*R1*t2b+alpha*R2-omega*R2*t2a, for similar reasons.

(Note: the a is the big arc, the b is the little arc, this is to make it simpler later on).

The total time for each will be the sum of the a and b part, so now we need to figure out what they are.
Well, in t1a, the light (travelling at speed c) travels a distance alpha*R2+omega*t1a.
Thus t1a*c=alpha*R2+omega*R2*t1a
Thus t1a=alpha*R2/(c-omega*R2).
Similarly, t2b=alpha*R1/(c-omega*R1).
And t1b*c=alpha*R1-omega*t1b
Thus t1b=alpha*R1/(c+omega*R1)
and t2a=alpha*R2/(c+omega*R2).

Now as I said, the total time for each one is given by:
t1=t1a+t1b.
t2=t2a+t2b.
And then we find the difference as (note: may give minus sign, I haven't checked, but the important part is the magnitude)
dt=t1-t2
=(t1a+t1b)-(t2a+t2b)
=t1a+t1b-t2a-t2b
=t1a-t2a+t1b-t2b
=(t1a-t2a)+(t1b-t2b)

Rather than try to solve it all at once, for simplicity we break it into 2 parts:
dta=t1a-t2a
And dtb=t1b-t2b.
And thus dt=dta+dtb

Now then:
dta=t1a-t2a
=alpha*R2/(c-omega*R2)-alpha*R2/(c+omega*R2)
=alpha*R2*(1/(c-omega*R2)-1/(c+omega*R2))
=alpha*R2*((c+omega*R2)/(c-omega*R2)*(c+omega*R2)-(c-omega*R2)/(c+omega*R2)*(c-omega*R2))
=alpha*R2*((c+omega*R2)-(c-omega*R2))/(c+omega*R2)*(c-omega*R2))
=alpha*R2*(c+omega*R2-c+omega*R2)/(c2-(omega*R2)2)
=alpha*R2*2*omega*R2/(c2-(omega*R2)2)
=2*omega*alpha*R22/(c2-(omega*R2)2)
And then if we assume omega*R2=v<<c (i.e. our system is moving much slower than the speed of light, and 30 km/s is still much slower than the speed of light at roughly 300 000 km/s, then we can simplify (c2-(omega*R2)2) to c2
And thus we end up with dta=2*omega*alpha*R2^2/c2

Now we do the same for dtb.
dtb=t1b-t2b
=alpha*R1/(c+omega*R1)-alpha*R1/(c-omega*R1)
=alpha*R1*(1/(c+omega*R1)-1/(c-omega*R1))
(combining some steps from before for brevity, and noting that omega*R1 will be tiny compared to c just like omega*R2)
=alpha*R1*(c-omega*R1-c-omega*R1)/c2
=-2*omega*alpha*R12/c2

Thus dt=2*omega*alpha*R22/c2-2*omega*alpha*R1^2/c2
=2*omega*alpha*(R22-R12)/c2

Now, can we simplify this any more?
I know, lets work out the area.
Note that alpha has always been in radians.
A circle has an area pi*r2.
This circle is a circular sector which subtends an angle of 2*pi.
If it only subtends an angle of alpha, then it will only have an area of alpha/(2*pi) of the circle.
Thus A (for a circular sector) is (alpha/(2*pi))*pi*r2
=alpha*r2/2
This means from the center to the outer arc you have an area of:
A2=alpha*R22/2.
And for the inner one you have an area:
A1=alpha*R12/2.
Thus the area between them is:
A=alpha*R22/2-alpha*R12/2
=alpha*(R22-R12)/2
Thus 2*A=alpha*(R22-R12).

Notice that was in the formula above?
That means we can sub it in.
As a reminder we had:
dt=2*omega*alpha*(R22-R12)/c2
By subbing in the above we get:
dt=2*omega*2*A/c2
And thus:
dt=4*A*omega/c2

Now can you refute it?
Can you derive your own.

You have to learn to properly present your case, using subscripts/superscripts, some diagrams, which will convey your point to your readers.
Any intelligent person can easily understand what I am saying.

there is a detector made of 4 pieces, one is an arc along Earth's orbit with radius R1 another is a similar arc with radius R2, both subtend and angle of alpha, and both ends are joined by a line that would be colinear with the centre of the orbit.

Are you saying you are too much of a moron to understand something that simple?
Do you not understand things like R1 and isntead need R1?
Are you incapable of making a picture yourself from the words?

Here, I'll be nice and make you a picture. What will your next BS excuse be?
Your dog ate your derivation?
Don't know how to follow hyperlinks?
http://i.imgur.com/NEacN3P.png
The red bit is the interferometer with the 2 counter propogating light rays.

Regardless, that in no way stops you from deriving it yourself.

They will say that it was already known and not worth publishing.
You are way to generous with your assessment.
They will wipe their asses with your "analysis" and flush the toilet with it right into the sewer system.
Nobody can make any sense out of what you wrote.
No. Any honest, rational person, who is fluent in English and has a basic grasp of what the Sagnac effect is would easily understand what I am saying.

All I did was derive what was already known, that the Sagnac effect is proportional to the area of the detector, and thus has nothing to do with how large the rotating system is. You can have a tiny detector in a massive rotating system and it will produce the exact same effect as the same tiny detector in a much smaller rotating system which is rotating at the same rate.

Now, like I said, show what was wrong with my derivation and derive it yourself, or shut up/admit you were wrong.

Continuing in this way is only making yourself more of an embarrassment.

*

sandokhan

  • Flat Earth Sultan
  • Flat Earth Scientist
  • 7262
Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #136 on: June 03, 2017, 02:10:14 AM »
The Sagnac effect is devastating to RET.

Not only can it not be explained in terms of STR/GTR, but the orbital Sagnac effect is not being registered by GPS satellites.

Combining this with the Ruderfer experiment, we have a most direct proof of the existence of ether.


All I did was derive what was already known, that the Sagnac effect is proportional to the area of the detector, and thus has nothing to do with how large the rotating system is.

Your derivation is not supported by any referenced papers: you are all on your own.

That is why I told you to please mail this piece of thrash to your local university.

Here is how mainstream science deals with the orbital Sagnac using the correct formula, not your whimsical fantasies.


CORRECT ORBITAL SAGNAC CALCULATION


Biography of Dr. Daniel Gezari:

https://science.gsfc.nasa.gov/sed/bio/daniel.y.gezari

Dr. Gezari joined the Goddard Space Flight Center in Greenbelt, Maryland in 1978.  He has been a pioneer in the the development of several new fields of observational astronomy, including:

1. First application of speckle interferometry (measuring diffraction-limited stellar diameters
        in images blurred by the Earth's atmosphere).
2. First 350 µm observations, and the study of cool star forming clouds.
3. Pioneered the use of array detectors for infrared astronomical imaging, especially at thermal
        infrared wavelengths (5-25 µm).
4. Produced the only infrared astronomical catalog in the field of observational astronomy.
5. In addition, Dr. Gezari has developed several useful compendia of astronomical research resources.

Research Areas

infrared astrophysics
star formation/interstellar medium
Galactic Center
infrared array camera development
optical spatial interferometry
astronomical database management
propagation of light in space
invariance of the speed of light


Positions and Appointments

Post Doctoral Research Fellow - California Institute of Technology (1974 - 1976)
Resident Research Associate - National Research Council (1976 - 1978)
Astrophysicist - NASA/Goddard (1976 - 2008)
Astrophysicist (Emeritus) - NASA/Goddard (2008 - present)
Research Associate - American Museum of Natural History, Astrophysics Dept. (2008 - present)

Education

A.B. (Physics), Cornell University, 1966
M.S. (Physics), New York University, 1969
Ph.D. (Astronomy), S.U.N.Y. at Stony Brook, 1973
Robert A. Millikan Research Fellow in Astriohysics, California Institute of Technology, 1974-1976


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.


There is only one correct way to calculate the Sagnac (be it rotational or orbital).


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

This is an IOP article.

The author recognizes the earth's orbital Sagnac is missing whereas the earth's rotational Sagnac is not.

He uses GPS and a link between Japan and the US to prove this.

In GPS the actual magnitude of the Sagnac correction
due to earth’s rotation depends on the positions of
satellites and receiver and a typical value is 30 m, as the
propagation time is about 0.1s and the linear speed due
to earth’s rotation is about 464 m/s at the equator. The
GPS provides an accuracy of about 10 m or better in positioning.
Thus the precision of GPS will be degraded significantly,
if the Sagnac correction due to earth’s rotation
is not taken into account. 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.


In an intercontinental microwave link between Japan and
the USA via a geostationary satellite as relay, the influence
of earth’s rotation is also demonstrated in a high-precision
time comparison between the atomic clocks at two remote
ground stations.
In this transpacific-link experiment, a synchronization
error of as large as about 0.3 µs was observed unexpectedly.


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. Thereby, it is evident
that the wave propagation in GPS or the intercontinental
microwave link depends on the earth’s rotation, but
is entirely independent of earth’s orbital motion around
the sun or whatever. As a consequence, the propagation
mechanism in GPS or intercontinental link can be viewed
as classical in conjunction with an ECI frame, rather than
the ECEF or any other frame, being selected as the unique
propagation frame. In other words, the wave in GPS or the
intercontinental microwave link can be viewed as propagating
via a classical medium stationary in a geocentric
inertial frame.


The author actually present a local-ether model (MLET, Modified Lorentz Ether Theory) in order to account for the MISSING ORBITAL SAGNAC EFFECT.


I repeat: this is an IOP article, the highest standard of mainstream science.

The paper was peer reviewed and published.

Each and every scientist working at that journal understood the meaning of these words:

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.



Nobody, other than yourself, can make any kind of a sense out of the ramblings you presented in your previous message.

There are no diagrams, no properly labeled equations: certainly you are not applying the Sagnac properly, you need to use the correct RADIUS in the formula.

Go ahead and mail your analysis to any peer review, just as the above authors had to do: everybody will be laughing at your poor understanding of the Sagnac effect.

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sandokhan

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #137 on: June 03, 2017, 02:12:47 AM »
You want to discuss the Saganac effect, fine.

The Sagnac effect, by itself, is a sure proof of the existence of ether.

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 #138 on: June 03, 2017, 02:26:22 AM »
What the frell is this?

http://i.imgur.com/NEacN3P.png

Is this supposed to be a bloody joke?

Have you ever published any kind of a scientific paper?

How is that diagram related to what you wrote earlier?

there is a detector made of 4 pieces, one is an arc along Earth's orbit with radius R1 another is a similar arc with radius R2, both subtend and angle of alpha, and both ends are joined by a line that would be colinear with the centre of the orbit.

What detector? Are you such a moron as to not be able to clearly convey your ideas to your readers?

What four pieces? Is R1 the distance to a satellite? R2 the distance to the surface of the Earth? Nobody knows what you are saying.

Note: the a is the big arc, the b is the little arc, this is to make it simpler later on
So, one beam of light will propagate in the same direction as the big arc and the opposite direction as the little arc.
What this means is its trip along the big arc will be longer than if it was stationary and shorter along the little arc.
The distance it has to travel is alpha*R2+omega*R2*t1a+alpha*R1-omega*R1*t1b, where t1a is the time taken to go along the big arc and t1b is the time taken to go along the little arc.


You are not making any sense. No FE/RE/neutral has any idea what you mean.  What big arc/little arc? Are you kidding us?

ARE YOU USING TWO DIFFERENT RADII TO DESCRIBE THE SAGNAC EFFECT? 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

Please note the radius R is the same for both pulses of light (one cannot use two different radii: R has to be the same).


Nobody understands what you are trying to say my friend.

Clear up your misunderstandings and come up with a much better version of your ramblings.


Until then please note that I have offered you not one but two MAINSTREAM PAPERS which directly refute your analysis.
« Last Edit: June 03, 2017, 06:18:49 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 #139 on: June 03, 2017, 02:41:20 AM »
CONCLUSIONS OF THE TWO MAINSTREAM PAPERS CITED ABOVE:

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.


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.

In the second paper, authored by Dr. Daniel Gezari, THE ORBITAL SAGNAC SHOWS UP IN THE CALCULATIONS, BUT IT IS NOT BEING PICKED UP BY THE 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.


Both these papers, authored by respected and well-known authors, calculate the orbital Sagnac exactly as I have described, the correct way.

A direct refutation of any other whimsical analysis.


But now we have a huge problem.



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.



Since the orbital Sagnac is not being picked up, and also the solar gravitational potential is not being recorded either, the hypotheses of the Ruderfer ether drift experiment are fulfilled.



http://file.scirp.org/pdf/JMP20120800006_80885197.pdf (pg. 718-720, 744)

Another observation that also clearly conflicts with the
constancy and isotropy of the velocity of light was discovered
during the implementation and calibration of
set-ups for Very Long Baseline Interferometry (VLBI)
radio astronomy observations. The resolution of optical
and radio astronomy observations can be improved by
orders of magnitude by analyzing the data recorded in
different observatories over the earth surface using interferometric
methods. The condition is that these data be
synchronous. The method consists in superposing coherently
the data recorded in different observatories with the
help of computers taking into account the instantaneous
position of the antennas etc. For the (VLBI) radio astronomy
observations clock synchronization at intercontinental
distances via the GPS achieve 0.1 ns. Nevertheless,
on testing the so synchronized clocks by confronting
them with the arrival of the wave fronts from distant
pulsars, which according to the TR may be synchronous,
it was observed that the pulsar signal reaches the foregoing
side of Earth 4.2 μs before the rear side along the
orbital motion of Earth. This discrepancy exceeds
the time resolution by more than four orders of magnitude.
Nevertheless along the transverse direction the arrival
of the pulsar signal was synchronous. This apparent
discrepancy in the GPS clock synchronization is again
raising very hot debates about the nature of space. Some
people speak of scandalous clocks that are biased
along the Earth’s orbital motion, others see in these
facts definitive prove that the velocity of light along different
directions within the solar system is not the same.


Many people believe that GR accounts for all the observed
effects caused by gravitational fields. However, in
reality GR is unable to explain an increasing number of
clear observational facts, several of them discovered recently
with the help of the GPS. For instance, GR
predicts the gravitational time dilation and the slowing of
the rate of clocks by the gravitational potential of Earth,
of the Sun, of the galaxy etc. Due to the gravitational
time dilation of the solar gravitational potential, clocks in
the GPS satellites having their orbital plane nearly parallel
to the Earth-Sun axis should undergo a 12 hour period
harmonic variation in their rate so that the difference
between the delay accumulated along the half of the orbit
closest to the Sun amounts up to about 24 ns in the time
display, which would be recovered along the half of the
orbit farthest from the Sun. Such an oscillation exceeds
the resolution of the measurements by more than two
orders of magnitude and, if present, would be very easily
observed. Nevertheless, contradicting the predictions of
GR, no sign of such oscillation is observed. This is the
well known and so long unsolved non-midnight problem.
In fact observations show that the rate of the
atomic clocks on Earth and in the 24 GPS satellites is
ruled by only and exclusively the Earth’s gravitational
field and that effects of the solar gravitational potential
are completely absent. Surprisingly and happily the GPS
works better than expected from the TR.


Obviously the gravitational
slowing of the atomic clocks on Earth cannot be due to
relative velocity because these clocks rest with respect to
the laboratory observer. What is immediately disturbing
here is that two completely distinct physical causes produce
identical effects, which by it alone is highly suspicious.
GR gives only a geometrical interpretation to the
gravitational time dilation. However, if motions cause
time dilation, why then does the orbital motion of Earth
suppress the time dilation caused by the solar gravitational
potential on the earthbased and GPS clocks?
Absurdly
in one case motion causes time dilation and in the
other case it suppresses it. This contradiction lets evident
that what causes the gravitational time dilation is not the
gravitational potential and that moreover this time dilation
cannot be caused by a scalar quantity. If the time dilation
shown by the atomic clocks within the earthbased
laboratories is not due to the gravitational potential and
cannot be due to relative velocity too then it is necessarily
due to some other cause. This impasse once more
puts in check the central idea of the TR, according to
which the relative velocity with respect to the observer is
the physical parameter that rules the effects of motions.
The above facts show that the parameter that rules the
effects of motions is not relative velocity but a velocity
of a more fundamental nature.


See also http://www.hrpub.org/download/20150510/UJPA2-18403649.pdf (pg. 147)


On the other hand, the time dilation effect of the solar
gravitational field on the atomic clocks orbiting with
Earth round the Sun, which is predicted by GR but not
observed, is a highly precise observation. It exceeds by
orders of magnitude the experimental precision and
hence is infinitely more reliable.
If the orbital motion of
Earth round the Sun suppresses the time dilation due to
the solar gravitational field and moreover does not show
the predicted relativistic time dilation due to this orbital
motion, then it seems reasonable that a clock in a satellite
orbiting round the Earth in a direct equatorial orbit or in a
jet flying round the Earth too should give no evidence of
such a relativistic time dilation. The relativistic time dilation
alleged in both these round the world Sagnac experiments
is in clear and frontal contradiction with the
absence of such a relativistic time dilation effect in the
case of the orbiting Earth round the Sun.




THE SOLAR ORBITAL GRAVITATIONAL POTENTIAL DOES NOT SHOW UP, IT IS NOT BEING REGISTERED EITHER BY THE GPS SATELLITES.


Ruderfer, Martin (1960) “First-Order Ether Drift
Experiment Using the Mössbauer Radiation,”
Physical Review Letters, Vol. 5, No. 3, Sept. 1, pp
191-192

Ruderfer, Martin (1961) “Errata—First-Order Ether
Drift Experiment Using the Mössbauer Radiation,”
Physical Review Letters, Vol. 7, No. 9, Nov. 1, p 361


in 1961, M. Ruderfer proved mathematically and experimentally, using the spinning Mossbauer effect, the FIRST NULL RESULT in ether drift theory.


Analysis of the spinning Mossbauer experiments is a natural step toward analysis of the
slightly more complex and much larger-scale Global Positioning System (GPS). This
system constitutes a large scale near-equivalent to the spinning Mossbauer experiments.
The transit time between the satellite and ground-based receivers is routinely measured.
In addition, the atomic clocks on the satellite are carefully monitored; and high precision
corrections are provided as part of the information transmitted from the satellites.
Because the satellites and the receivers rotate at different rates (unlike the Mossbauer
experiments), a correction for the motion of the receiver during the transit time is
required. This correction is generally referred to as a Sagnac correction, since it adjusts
for anisotropy of the speed of light as far as the receiver is concerned. Why is there no
requirement for a Sagnac correction due to the earth’s orbital motion? Like the transit
time in the spinning Mossbauer experiments, any such effect would be completely
canceled by the orbital-velocity effect on the satellite clocks.


Specifically, there is substantial independent experimental evidence that clock speed always affects the clock frequency and, as the GPS system shows, the spin velocity of the earth clearly affects the clock rate. This being the case, the null result of the rotating Mössbauer experiments actually implies that an ether drift must exist or else the clock effect would not be canceled and a null result would not be present.

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sandokhan

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #140 on: June 03, 2017, 04:55:42 AM »
Now, the standard/correct derivation of the Sagnac effect formula.



http://www.mathpages.com/rr/s2-07/2-07.htm (most referenced paper used by the relativists themselves)

If two pulses of light are sent in opposite directions around a stationary circular loop of radius R

Please note the radius R is the same for both pulses of light (one cannot use two different radii: R has to be the same).

Light source and interferometer are at S, both fixed on a rotating disk.

Let to be the time taken for a light signal to traverse the circumference of the circle and to return to the source/interferometer, when both the disc and the observer are stationary. Thus, to is the path length 2πr divided by the speed of light:

to = 2πr/c

Let ds' be the distance SS' and ds" the distance SS". Let t' be the time measured by an observer situated in the stationary laboratory for the light to go from S to S' in the anticlockwise direction.

The time measured by that observer is:

t' = (2πr - ds')/c

But, t' is also the time taken for the disc to move a distance ds' in the clockwise direction. Therefore t' = ds'/v (v =rω), ds' = t'v and,

ds' = (2πr - ds')v/c

ds'/v = 2πr/(c + v)

or

t' = 2πr/(c + v)


Following similar calculations one gets for t", the time measured by a stationary observer for the light to go from S to S" in a clockwise direction,

t" = 2πr/(c - v)

The difference between the times for the light to go clockwise and anticlockwise is:

dt = 2πr/(c - v)  - 2πr/(c + v) = 4πrv/(c2 - v2)

Since v = rω and A is the area of the circle,

dt = 4Aω/(c2 - v2)

The term v2 is negligible for practical tests and may be ignored, giving the final Sagnac formula:

dt = 4Aω/c2 (I)


COMPARISON OF THE SAGNAC EFFECT WITH STR

STR stipulates that the time t' recorded by an observer moving at velocity v is slower than the time to recorded by a stationary observer, according to:

to = t'γ

where γ = (1 - v2/c2)-1/2 = 1 + v2/2c2 + O(v/c)4...

to = t'(1 + v2/2c2)


dtR = (to - t')/to = v2/(v2 + 2c2)

dtR = relativity time ratio



Now, to - t' = 2πr/c - 2πr/(c + v) = 2πrv/(c + v)c

dt' = to - t' = tov/(c + v)


dtS = (to - t')/to = v/(v + c)


dtS = Sagnac ratio


dtS/dtR = (2c2 + v2)/v(v + c)

When v is small as compared to c, as is the case in all practical experiments, this ratio
reduces to 2c/v.

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.
« Last Edit: June 03, 2017, 05:35:15 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 #141 on: June 03, 2017, 06:07:36 AM »
Now, the standard/correct derivation of the Sagnac effect formula.



http://www.mathpages.com/rr/s2-07/2-07.htm (most referenced paper used by the relativists themselves)

If two pulses of light are sent in opposite directions around a stationary circular loop of radius R

Please note the radius R is the same for both pulses of light (one cannot use two different radii: R has to be the same).

Light source and interferometer are at S, both fixed on a rotating disk.

Let to be the time taken for a light signal to traverse the circumference of the circle and to return to the source/interferometer, when both the disc and the observer are stationary. Thus, to is the path length 2πr divided by the speed of light:
to = 2πr/c
Let ds' be the distance SS' and ds" the distance SS". Let t' be the time measured by an observer situated in the stationary laboratory for the light to go from S to S' in the anticlockwise direction.
The time measured by that observer is:
t' = (2πr - ds')/c
But, t' is also the time taken for the disc to move a distance ds' in the clockwise direction. Therefore t' = ds'/v (v =rω), ds' = t'v and,
ds' = (2πr - ds')v/c
ds'/v = 2πr/(c + v)
or
t' = 2πr/(c + v)
Following similar calculations one gets for t", the time measured by a stationary observer for the light to go from S to S" in a clockwise direction,
t" = 2πr/(c - v)
The difference between the times for the light to go clockwise and anticlockwise is:
dt = 2πr/(c - v)  - 2πr/(c + v) = 4πrv/(c2 - v2)
Since v = rω and A is the area of the circle,
dt = 4Aω/(c2 - v2)
The term v2 is negligible for practical tests and may be ignored, giving the final Sagnac formula:
dt = 4Aω/c2 (I)
Agreed, now analyse the case when the centre of rotation is far outside the Sagnac ring, as it is with a comparatively small ring of the Michelson–Gale–Pearson experiment and the centre of rotation the axis of the earth.

And I believe you will still find that it is the area of the Sagnac loop and not the area swept by the radius of the earth that determines the time difference.
That was the case analysed in IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND? « Reply #130 on: Today at 11:06:27 AM »

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sandokhan

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #142 on: June 03, 2017, 06:24:36 AM »
rabinoz, you know better than to try to use tricks like these with me.

There is no "far outside the Sagnac ring".

The equation for the sagnac is:

4Aω/( c² - v²)

One must calculatate the area swept out by the path and that is A = πR², where R is measured from the Sun to the center of the Earth (radius of the orbital path loop), for the orbital Sagnac.


Is that so? Would you like to bring the Michelson-Gale-Pearson experiment into our discussion?


And I believe you will still find that it is the area of the Sagnac loop and not the area swept by the radius of the earth that determines the time difference.

I invite you to send your beliefs to your local university and see what kind of a response you will receive.

Until then, this is how modern science calculates the orbital Sagnac:

https://www.theflatearthsociety.org/forum/index.php?topic=70614.msg1915700#msg1915700


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tomato

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #143 on: June 03, 2017, 06:53:03 AM »
I'm curious what all this discussion about Sagnac interference is for? Its results aren't evidence for aether anyway.
Tomato puree

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sandokhan

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #144 on: June 03, 2017, 07:02:35 AM »
Subject matter of this thread: stationary/geocentric Earth vs. orbiting Earth

Its results aren't evidence for aether anyway.

But they are.

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1886058#msg1886058

The optical whirlwind effect of an artificial rotation of an overall system really shows itself, without unexpected compensation, as an effect of the first order of the movement in comparison with the ether.  The experience directly reveals […] the linear delay […] that the overall rotation of the optical system produces in the ether between the two systems of inverse waves T and R during their propagation around the circuit.

G. Sagnac


Have you forgotten about the Ruderfer experiment?

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1846721#msg1846721

No orbital Sagnac + no orbital solar gravitational potential = existence of ether

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tomato

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #145 on: June 03, 2017, 07:24:59 AM »
Subject matter of this thread: stationary/geocentric Earth vs. orbiting Earth

Its results aren't evidence for aether anyway.

But they are.

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1886058#msg1886058

The optical whirlwind effect of an artificial rotation of an overall system really shows itself, without unexpected compensation, as an effect of the first order of the movement in comparison with the ether.  The experience directly reveals […] the linear delay […] that the overall rotation of the optical system produces in the ether between the two systems of inverse waves T and R during their propagation around the circuit.

G. Sagnac

I wasn't here before - I don't know about the Ruderfer experiment.

Per my reading from this pdf, the "optical whirlwind effect" is saying that, with aether stationary, the light in the spinning ring itself feels a "wind" since the ring is moving against stationary aether. No? Am I wrong? Are there any additional effects here?
Tomato puree

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cikljamas

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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #146 on: June 03, 2017, 08:34:27 AM »
Sandokhan, i have a few questions for you, if you don't mind :

1. According to you what causes Sagnac rotational effect?

2. Do you really believe that there is a mirror on the moon? If so, who put it there? Neil Armstrong?

3. What is your stance on 1986 Challenger disaster? Aren't you aware that nobody died in that event, all these guys are still alive?

4. Do you really believe that satellites (as a solid physical objects) orbit the earth? Do they orbit around round earth or above the flat earth? If satellites orbit the flat earth what is their trajectory?

5. Do you believe that gravity is a product of earth's upward motion or do you believe that the earth is PERFECTLY STILL?

6. If the earth is flat how do you explain southern summer path of the sun? How do you explain southern circumpolar stars? How do you explain sun's direct path along the equator during the equinoxes?

7. How come you have never referred to my ZIGZAG argument? Do you consider it wrong? If yes, then would you be so kind to point out what is wrong with this argument?

8. Do you believe in geocentric universe with much smaller (how much smaller) distances (than modern science claims) between celestial objects or do you believe in geocentric universe with large (which is in accordance with current scientific paradigm) distances between celestial objects?

9. What is the nature of celestial objects according to you (are they solid physical bodies, or do you think that their nature is electromagnetically determined (lights in the sky))?

10. What is "gravity"?
"I can't breathe" George Floyd RIP

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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 #147 on: June 03, 2017, 08:57:39 AM »
1. The rotational Sagnac effect is caused by the rotation of the ether field above the flat surface of the Earth. This same field causes terrestrial gravity and the antigravitational effects listed here (Allais, DePalma, Kozyrev, Lamoreaux, Biefeld-Brown).

Ring laser gyroscopes are being activated/acted upon by these ether strings/telluric currents.

2. No one has managed to go beyond the first Dome, not even the Nazis with their mercury gyro implosion of the atom UFOs.

I believe that there is a small satellite which orbits right in front the Dome, matching the exact orbit of the Moon, and which acts as a mirror in order to account for the lunar laser ranging experiment.

3. It is very possible that the version posted by you is true; Dr. Peter Beter wrote extensively on the secret space shuttle program.

4. Each GPS satellite uses the Biefeld-Brown effect to orbit above the flat surface of the Earth. The same goes for the ISS/Atlantis spacecrafts.

5. The Earth is perfectly still: the UA was introduced to the FES by people who had no knowledge of ether theory, by the time I came here in 2007 it was too late, nobody seems to care even now. The orbital Sagnac effect disproves that the Earth has any kind of an upward motion.

6. I have written extensively on the true path of the Sun: each and every topic raised was addressed many times. In fact, I was the only FE who has done this, I introduced by bipolar map which features both northern and southern circumpolar orbits for the stars.

https://www.theflatearthsociety.org/forum/index.php?topic=64997.0

7. I have never commented on the zigzag argument, given the responses from the RE you are on the right path to demonstrate a major fallacy of the heliocentrical theory.

8. https://www.theflatearthsociety.org/forum/index.php?topic=30499.270

9. Solid disks, covered by plasma clouds, using the double torsion/Schauberger implosion to orbit above the first Dome.

10. Terrestrial gravity: pressure exerted by the dextrorotatory subquark strings upon matter. Planetary/stellar gravity: rotational pressure exerted by a very dense field of ether/aether which keeps the planets/stars in orbit.

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sandokhan

  • Flat Earth Sultan
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Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #148 on: June 03, 2017, 09:13:43 AM »
VIKTOR SCHAUBERGER: GENIUS OF DOUBLE TORSION PHYSICS

http://discaircraft.greyfalcon.us/Viktor%20Schauberger.htm

"If water or air is rotated into a twisting form of oscillation known as ‘colloidal’, a build up of energy results, which, with immense power, can cause levitation. This form of movement is able to carry with it its own means of power generation. This principle leads logically to its application in the design of the ideal airplane or submarine... requiring almost no motive power."  V. Schauberger

WHO WAS VIKTOR SCHAUBERGER?

http://free-energy.xf.cz/SCHAUBERGER/Living_Energies.pdf (best work on double torsion/implosion)

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tomato

  • 175
  • Shine on you crazy diamonds.
Re: IF THE EARTH IS AT REST, CAN SHE STILL BE ROUND?
« Reply #149 on: June 03, 2017, 09:16:56 AM »
1. The rotational Sagnac effect is caused by the rotation of the ether field above the flat surface of the Earth. This same field causes terrestrial gravity and the antigravitational effects listed here (Allais, DePalma, Kozyrev, Lamoreaux, Biefeld-Brown).

sandokhan, are we in agreement that the "rotation of the ether field above the flat surface of the Earth," in other words the aether winds acting on the stationary Earth, are just the same thing as the effect of the motion of the Earth through stationary aether?

(If not, I give these lines from the pdf:

"According to Fresnel, in massive bodies an excess of ether is dragging ... While, in Sagnac’s theory, the ether is the same everywhere and the light has the same velocity in every direction. Starting with this theory of motionless ether, Sagnac, in 1910 ... built the interferometer for testing this and called it “effet tourbillonnaire optique”; this translates as “optical whirlwind effect”. The optical whirlwind effect started from the hypothesis that the time of light propagation between two points of a system must be distorted like when the system is immobile and submitted to the action of the ether’s wind."

page 2, paragraph 3)
« Last Edit: June 03, 2017, 09:22:57 AM by tomato »
Tomato puree