sandokhan lies regarding the Sagnac effect

How does the jackbot plead, guilty or not guilty?

Invariably, the chatbot's response is not guilty, even though it has committed some of the most atrocious errors in the history of scientific discussions on the internet.

YES!

Then, there is nothing to debate here any longer.

That's right. For a given loop, A=A.
For the rotation we have wr, for the orbit we have wo.

∆t=4*A*w/c^2

Thus ∆to=4A*wo/c^2
and ∆tr=4A*wr/c^2.
Comparing the 2:
∆to/∆tr=(4A*wo/c^2)/(4A*wr/c^2)=wo/wr.
Thus it is quite simple to see that the orbital Sagnac is much less than the rotational one due to the much lower angular velocity.

Once again, if you use different loops, your comparison is meaningless.


That comparison IS INVALID.

It is complete BS.


The bot has THE SAME AREA for BOTH the orbital and rotational Sagnac evaluations.


The orbital Sagnac has radii, R1 and R2.

The rotational Sagnac has radii, r1 and r2.


But for the chatbot this does not represent a problem at all.


Let us see what happens if indeed A = A for the two Sagnac effects.


θo = angle subtended by the two radii, R2 and R1 = orbital angle

s2 = R2 x θo

s1 = R1 x θo


θr = angle subtended by the two radii, r2 and r1 = rotational angle

s2 = r2 x θr

s1 = r1 x θr


R2 - R1 = r2 - r1

r2 x θr = R2 x θo

r1 x θr = R1 x θo

r2/r1 = R2/R1

(r2 x R1) = (r1 x R2)

Since the two areas must be equal,

r1/R1 = (r2 + r1)/(R2 + R1)



Right away, one runs into huge problems with this scenario.

R2 = r2 - r1 + R1

(r2 x R1) = r1r2 - r12 + (R1 x r1)

r2(R1 - r1) = r1(R1 - r1)

So we end up with: r2 = r1, which is impossible.


Therefore, one cannot use the TWO SETS OF RADII FOR THE SAME AREA to evaluate two different Sagnac effects, which use different radii.


REMEMBER, the original message allowed for NO MINUTE DIFFERENCES AT ALL.

The jackbot's original derivation meant: A = A and R2 - R1 = r2 - r1.

That's all.

From now on, I will not allow any departure from the data set in the original piece of analysis, where no minute differences were implied at all, or even conjectured: jackbot's data simply stated that A = A and  R2 - R1 = r2 - r1.


With this data, we reach a direct contradiction in no time at all: r2 = r1.


That is, the following comparison is invalid:

Thus ∆to=4A*wo/c^2
and ∆tr=4A*wr/c^2.
Comparing the 2:
∆to/∆tr=(4A*wo/c^2)/(4A*wr/c^2)=wo/wr.
Thus it is quite simple to see that the orbital Sagnac is much less than the rotational one due to the much lower angular velocity.


Can the jackbot understand where it went wrong?

Nobody is complaining about the area of a sector of a circle (that is why your trolling question is meaningless).

The problem is with the invalid comparison.


The proper radii have to be used for each Sagnac effect, then and only then, a comparison can be made.


∆to=4A*wo/c^2, where the area is (alpha/2)*(R2^2-R1^2)

By contrast,

∆tr=4A*wr/c^2, where the area now features the term: (r2^2-r1^2)


The ratio will involve TWO LINEAR VELOCITIES AND TWO SETS OF DIFFERENT RADII, and not two angular velocities.


This is exactly how JPL/CalTech and ESA have calculated the orbital and the rotational Sagnac.


http://tycho.usno.navy.mil/ptti/2003papers/paper34.pdf

Dr. Massimo Tinto, Jet Propulsion Laboratory, Principal Scientist


Time Delay Interferometry with Moving Spacecraft Arrays

Massimo Tinto, F. B. Estabrook, and J.W. Armstrong
Jet Propulsion Laboratory, California Institute of Technology, Pasadena, CA 91109

Aberration due to LISA’s orbit about the Sun dominates its instantaneous rotation.

The formula is 2VL/c.

V = RΩ


The formula is 2VL/c.


The linear velocity is being used, and not the angular velocity.

The orbital radius is being used (R = 150,000,000 km) in the equation for the linear velocity, and not the rotational radius (R = 5,000,000 km for LISA).


Algebraic approach to time-delay data analysis: orbiting case
K Rajesh Nayak and J-Y Vinet

https://www.cosmos.esa.int/documents/946106/1027345/TDI_FOR_.PDF/2bb32fba-1b8a-438d-9e95-bc40c32debbe

This is an IOP article, published by the prestigious journal Classic and Quantum Gravity:

http://iopscience.iop.org/article/10.1088/0264-9381/22/10/040/meta

The conclusion of the paper:

For the LISA geometry, R⊙/L is of the order 30 and the orbital contribution to the Sagnac phase is larger by this factor.

And as Earth's rotational motion (measured as angular velocity or ω) is 365 times as fast as Earth's orbital motion, that means the rotational sagnac effect will be 365 times that of the orbital sagnac effect.

Thus it is quite simple to see that the orbital Sagnac is much less than the rotational one due to the much lower angular velocity.

Quote from: sandokhan on November 13, 2017, 11:59:49 AM

Please explain why Dr. Kelly measures the Sagnac according to the linear velocity and NOT the angular velocity of the loop.

Because they are a charlatan trying to claim special relativity is broken.

The jackbot is insisting that the orbital Sagnac is much lower than the rotational Sagnac due to a lower angular velocity.

But only the jackbot could have left the TWO DIFFERENT RADII OUT OF THE EQUATION.


The orbital Sagnac uses THE ORBITAL RADIUS: R = 150,000,000 KM.

The rotational Sagnac uses the ROTATIONAL RADIUS: r = 6,400 km.


How do the distinguished scientists at JPL/CalTech calculate the ORBITAL SAGNAC?


THEY USE THE LINEAR VELOCITY AND THE ORBITAL RADIUS (R = 150,000,000 KM).


There is nothing else to debate here.


The best scientists in the world, who have been asked to perform the crucial calculations for the LISA project, are using the LINEAR VELOCITY AND THE ORBITAL RADIUS to calculate the orbital Sagnac.


A total refutation of the jackbot's failed analysis.

How does the jackbot plead, guilty or not guilty?

Not guilty.
Now how about you answer the question:

Do you accept that the area of an annular sector is given by:
A=θ_o(R₂² - R₁²)/2
or equally that the following equation holds:
2A=θ_o(R₂² - R₁²)

Yes or no?

Until you do you are just making yourself look more and more pathetic.

The question has been answered in my previous message.

The jackbot HAS NEVER PROVIDED ANY FORMULAS FOR THE dto and dtr comparison, and indeed it cannot, for then the entire edifice of its failed analysis would come crashing down.

The jackbot has TRICKED its readers with promises which can never be delivered, not in this the real world.

It is the comparison of the dto and dtr which is INVALID.



Now, for the coup de grace.

It is well known that the jackbot has no knowledge of higher mathematics, resorting to simple algebra to advance its failed hypotheses.

CONFORMAL MAPPING OF ANNULAR SECTORS/TRAPEZOIDS ONTO A RECTANGLE



Conformal mapping, an important area of complex variables analysis.

Conformal mappings are invaluable for solving problems in engineering and physics that can be expressed in terms of functions of a complex variable but that exhibit inconvenient geometries. By choosing an appropriate mapping, the analyst can transform the inconvenient geometry into a much more convenient one.

Numerical conformal mapping of an circular sector onto a rectangle:

https://books.google.ro/books?id=VMKcK0CYfmMC&pg=PA92&lpg=PA92&dq=conformal+mapping+annular+sector+onto+rectangle&source=bl&ots=j76uwWjuT2&sig=8H11fmFmAw1_AJpP60YrJCZndOY&hl=ro&sa=X&ved=0ahUKEwin8qfGzKnXAhWI56QKHX3VCaIQ6AEIgQEwDg#v=onepage&q=conformal%20mapping%20annular%20sector%20onto%20rectangle&f=false

Numerical conformal mapping of a trapezoid onto a rectangle:

https://www.researchgate.net/publication/288738306_An_approximate_conformal_mapping_of_a_trapezoid_onto_a_rectangle_and_its_inversion_obtained_by_the_block_method

https://ac.els-cdn.com/037704278790152X/1-s2.0-037704278790152X-main.pdf?_tid=c8c3b184-c2d2-11e7-9f43-00000aab0f6c&acdnat=1509959802_4e7a2f1be3f5865a22c13c120ac6f022


Conformal mapping is one of the most useful tools used in engineering and in physics today, in dealing with irregularly shaped interferometers.

The Sagnac for an annular section can be reduced to a Sagnac for a rectangle.





The formula the rectangular Sagnac is:


dt = 2Lv/c^2

v = linear velocity of the entire interferometer (the motion can be translational or rotational)

L = 4a + 4b


All that matters in the Sagnac is THE LINEAR VELOCITY.


Why is the bot avoiding like the plague the following results from JPL/CalTech and ESA?

The ratio will involve TWO LINEAR VELOCITIES AND TWO SETS OF DIFFERENT RADII, and not two angular velocities.


This is exactly how JPL/CalTech and ESA have calculated the orbital and the rotational Sagnac.


http://tycho.usno.navy.mil/ptti/2003papers/paper34.pdf

Dr. Massimo Tinto, Jet Propulsion Laboratory, Principal Scientist


Time Delay Interferometry with Moving Spacecraft Arrays

Massimo Tinto, F. B. Estabrook, and J.W. Armstrong
Jet Propulsion Laboratory, California Institute of Technology, Pasadena, CA 91109

Aberration due to LISA’s orbit about the Sun dominates its instantaneous rotation.

The formula is 2VL/c.

V = RΩ


The formula is 2VL/c.


The linear velocity is being used, and not the angular velocity.

The orbital radius is being used (R = 150,000,000 km) in the equation for the linear velocity, and not the rotational radius (R = 5,000,000 km for LISA).


Algebraic approach to time-delay data analysis: orbiting case
K Rajesh Nayak and J-Y Vinet

https://www.cosmos.esa.int/documents/946106/1027345/TDI_FOR_.PDF/2bb32fba-1b8a-438d-9e95-bc40c32debbe

This is an IOP article, published by the prestigious journal Classic and Quantum Gravity:

http://iopscience.iop.org/article/10.1088/0264-9381/22/10/040/meta

The conclusion of the paper:

For the LISA geometry, R⊙/L is of the order 30 and the orbital contribution to the Sagnac phase is larger by this factor.


Thus it is quite simple to see that the orbital Sagnac is much less than the rotational one due to the much lower angular velocity.

The jackbot is insisting that the orbital Sagnac is much lower than the rotational Sagnac due to a lower angular velocity.

But only the jackbot could have left the TWO DIFFERENT RADII OUT OF THE EQUATION.


The orbital Sagnac uses THE ORBITAL RADIUS: R = 150,000,000 KM.

The rotational Sagnac uses the ROTATIONAL RADIUS: r = 6,400 km.


How do the distinguished scientists at JPL/CalTech calculate the ORBITAL SAGNAC?


THEY USE THE LINEAR VELOCITY AND THE ORBITAL RADIUS (R = 150,000,000 KM).


There is nothing else to debate here.


The best scientists in the world, who have been asked to perform the crucial calculations for the LISA project, are using the LINEAR VELOCITY AND THE ORBITAL RADIUS to calculate the orbital Sagnac.


A total refutation of the jackbot's failed analysis.

The question has been answered in my previous message.

Try answering it without covering it with mountains of crap.

All it takes to answer is a simple yes or no. You do not need the mountains of crap you have provided.
Skimming through your previous post I don't see any indication of it being answered. So try again:

Do you accept that the area of an annular sector is given by:
A=θ_o(R₂² - R₁²)/2
or equally that the following equation holds:
2A=θ_o(R₂² - R₁²)

Yes or no?

That is all you need to type out, yes or no. No mountains of BS.

What a pathetic chatbot.

It is avoiding the conclusions of its own failed analysis.

Let us review what the jackbot is claiming here.

That's right. For a given loop, A=A.
For the rotation we have wr, for the orbit we have wo.

∆t=4*A*w/c^2

Thus ∆to=4A*wo/c^2
and ∆tr=4A*wr/c^2.
Comparing the 2:
∆to/∆tr=(4A*wo/c^2)/(4A*wr/c^2)=wo/wr.
Thus it is quite simple to see that the orbital Sagnac is much less than the rotational one due to the much lower angular velocity.

Once again, if you use different loops, your comparison is meaningless.


That comparison IS INVALID.

It is complete BS.


The bot has THE SAME AREA for BOTH the orbital and rotational Sagnac evaluations.


The orbital Sagnac has radii, R1 and R2.

The rotational Sagnac has radii, r1 and r2.


But for the chatbot this does not represent a problem at all.


Let us see what happens if indeed A = A for the two Sagnac effects.


θo = angle subtended by the two radii, R2 and R1 = orbital angle

s2 = R2 x θo

s1 = R1 x θo


θr = angle subtended by the two radii, r2 and r1 = rotational angle

s2 = r2 x θr

s1 = r1 x θr


R2 - R1 = r2 - r1

r2 x θr = R2 x θo

r1 x θr = R1 x θo

r2/r1 = R2/R1

(r2 x R1) = (r1 x R2)

Since the two areas must be equal,

r1/R1 = (r2 + r1)/(R2 + R1)



Right away, one runs into huge problems with this scenario.

R2 = r2 - r1 + R1

(r2 x R1) = r1r2 - r12 + (R1 x r1)

r2(R1 - r1) = r1(R1 - r1)

So we end up with: r2 = r1, which is impossible.


Therefore, one cannot use the TWO SETS OF RADII FOR THE SAME AREA to evaluate two different Sagnac effects, which use different radii.


REMEMBER, the original message allowed for NO MINUTE DIFFERENCES AT ALL.

The jackbot's original derivation meant: A = A and R2 - R1 = r2 - r1.

That's all.

From now on, I will not allow any departure from the data set in the original piece of analysis, where no minute differences were implied at all, or even conjectured: jackbot's data simply stated that A = A and  R2 - R1 = r2 - r1.


With this data, we reach a direct contradiction in no time at all: r2 = r1.


WHY IS THE JACKBOT AVOIDING THIS VERY SIMPLE PROOF OF ITS FAILURE?


It is demanding a response to question which has already been answered (right on this page), but AVOIDING to account for its conclusions.


Are these not your claims, jackbot?

And as Earth's rotational motion (measured as angular velocity or ω) is 365 times as fast as Earth's orbital motion, that means the rotational sagnac effect will be 365 times that of the orbital sagnac effect.

Thus it is quite simple to see that the orbital Sagnac is much less than the rotational one due to the much lower angular velocity.

Quote from: sandokhan on November 13, 2017, 11:59:49 AM

Please explain why Dr. Kelly measures the Sagnac according to the linear velocity and NOT the angular velocity of the loop.

Because they are a charlatan trying to claim special relativity is broken.

The jackbot is insisting that the orbital Sagnac is much lower than the rotational Sagnac due to a lower angular velocity.

But only the jackbot could have left the TWO DIFFERENT RADII OUT OF THE EQUATION.


The orbital Sagnac uses THE ORBITAL RADIUS: R = 150,000,000 KM.

The rotational Sagnac uses the ROTATIONAL RADIUS: r = 6,400 km.


How do the distinguished scientists at JPL/CalTech calculate the ORBITAL SAGNAC?


THEY USE THE LINEAR VELOCITY AND THE ORBITAL RADIUS (R = 150,000,000 KM).


There is nothing else to debate here.


The best scientists in the world, who have been asked to perform the crucial calculations for the LISA project, are using the LINEAR VELOCITY AND THE ORBITAL RADIUS to calculate the orbital Sagnac.


A total refutation of the jackbot's failed analysis.


There is nothing else to debate here.

The discussion is over.


Here is the claim made by the jackbot:

Thus it is quite simple to see that the orbital Sagnac is much less than the rotational one due to the much lower angular velocity.


Algebraic approach to time-delay data analysis: orbiting case
K Rajesh Nayak and J-Y Vinet

https://www.cosmos.esa.int/documents/946106/1027345/TDI_FOR_.PDF/2bb32fba-1b8a-438d-9e95-bc40c32debbe

This is an IOP article, published by the prestigious journal Classic and Quantum Gravity:

http://iopscience.iop.org/article/10.1088/0264-9381/22/10/040/meta

The conclusion of the paper:

For the LISA geometry, R⊙/L is of the order 30 and the orbital contribution to the Sagnac phase is larger by this factor.

Time Delay Interferometry with Moving Spacecraft Arrays

Massimo Tinto, F. B. Estabrook, and J.W. Armstrong
Jet Propulsion Laboratory, California Institute of Technology, Pasadena, CA 91109

Aberration due to LISA’s orbit about the Sun dominates its instantaneous rotation.

The formula is 2VL/c.

V = RΩ


The formula is 2VL/c.


Why is the jackbot avoiding like the plague these very simple proofs which contradict its failed analysis?

SAGNAC EFFECT WITHOUT AN AREA OR A LOOP

I always keep the best for the last.

The biggest breakthrough in the physics of the Sagnac effect has been the use of PHASE-CONJUGATE MIRRORS.

In fact, had this technology been available to G. Sagnac and A. Michelson it would have saved them a lot of time and energy in constructing experiments and deriving formulas.


https://arxiv.org/ftp/physics/papers/0609/0609202.pdf

Test of the one-way speed of light and the first-order experiment of
Special Relativity using phase-conjugate interferometers

Professor Ruyong Wang



The equation which expresses the relationship between interference fringes and time differences is F=dt[c/λ].

This experiment shows us two important points. First, it confirms the phase reversal of a PCM and demonstrates the Sagnac effect in an arc segment AB, not a closed path. Second, it gives us important implications: The result, φ = 4πRΩL/cλ, can be re-written as φ = 4πvL/cλ where v is the speed of the moving arc segment AB (where R is the radius of the circular motion, Ω is the rotational rate).

If we increase the radius of the circular motion as shown in Fig. 6, the arc segment AB will approach a linear segment AB, the circular motion will approach the linear motion, the phase-conjugate Sagnac experiment will approach the phase-conjugate first-order experiment as shown in Fig. 4, and the phase shift is always φ = 4πvL/cλ.

The Sagnac formula for a straight line path which is rotating becomes:

φ = 4πvL/c²

where v = RΩ


That is, the formula for a rotating linear segment is:

φ = 4πvL/c²

where v = RΩ


Using the values for v for both the orbital Sagnac and the rotational Sagnac we can see immediately that v_o/v_r = 60.

Using the values for v for both the orbital Sagnac and the rotational Sagnac we can see immediately that v_o/v_r = 60.

Not 30? I was told everyone agreed it was 30. Is it 30, or 60?

Why is the orbital Sagnac for LISA 30 times greater than the rotational Sagnac?

I included the calculations as well.

https://arxiv.org/pdf/gr-qc/0306125.pdf (pg 5, because the plane of the interferometer is 60 deg. from the ecliptic, the angular orbital velocity has to be divided by 2; otherwise we would get a value of 60)

What a pathetic chatbot.

Not sure if you are a chatbot, but you are most certainly pathetic.

Answer the question:
Do you accept that the area of an annular sector is given by:
A=θ_o(R₂² - R₁²)/2
or equally that the following equation holds:
2A=θ_o(R₂² - R₁²)

Yes or no?

Answer this before moving on to anything else.

I have already answered the question (in fact, the first message on this page): the trolling features of the jackbot apparently disable its ability to read properly.

There is no need to address any further questions: the use of nonlinear optics, mainly the phase-conjugate mirror has been the most extraordinary breakthrough for the Sagnac effect.

Here is the Sagnac formula for A ROTATING LINEAR SEGMENT.

SAGNAC EFFECT WITHOUT AN AREA OR A LOOP

The biggest breakthrough in the physics of the Sagnac effect has been the use of PHASE-CONJUGATE MIRRORS.

In fact, had this technology been available to G. Sagnac and A. Michelson it would have saved them a lot of time and energy in constructing experiments and deriving formulas.


https://arxiv.org/ftp/physics/papers/0609/0609202.pdf

Test of the one-way speed of light and the first-order experiment of
Special Relativity using phase-conjugate interferometers

Professor Ruyong Wang



The equation which expresses the relationship between interference fringes and time differences is F=dt[c/λ].

This experiment shows us two important points. First, it confirms the phase reversal of a PCM and demonstrates the Sagnac effect in an arc segment AB, not a closed path. Second, it gives us important implications: The result, φ = 4πRΩL/cλ, can be re-written as φ = 4πvL/cλ where v is the speed of the moving arc segment AB (where R is the radius of the circular motion, Ω is the rotational rate).

If we increase the radius of the circular motion as shown in Fig. 6, the arc segment AB will approach a linear segment AB, the circular motion will approach the linear motion, the phase-conjugate Sagnac experiment will approach the phase-conjugate first-order experiment as shown in Fig. 4, and the phase shift is always φ = 4πvL/cλ.

The Sagnac formula for a straight line path which is rotating becomes:

φ = 4πvL/c²

where v = RΩ


That is, the formula for a rotating linear segment is:

φ = 4πvL/c²

where v = RΩ


Using the values for v for both the orbital Sagnac and the rotational Sagnac we can see immediately that v_o/v_r = 60.

A total refutation of the "analysis" signed the jackbot.

I have already answered the question (in fact, the first message on this page):

Provide a link, or better yet, provide a quote of where you have answered it.
Saying the first message on this page is rather meaningless as different people can have a different number of posts per page.

If you are unable to do so, then ANSWER THE QUESTION!
No mountains surrounding your answer either.
Here it is yet again:
Do you accept that the area of an annular sector is given by:
A=θ_o(R₂² - R₁²)/2
or equally that the following equation holds:
2A=θ_o(R₂² - R₁²)

Yes or no?

The jackbot has not read the message at all (a simple search for that first message using "question").

As I have already said, the use the phase-conjugate mirrors changes everything.

Now, we can derive the exact Sagnac formula for a rotating linear segment.

It makes this entire thread superfluous.

There is no need to address any further questions: the use of nonlinear optics, mainly the phase-conjugate mirror has been the most extraordinary breakthrough for the Sagnac effect.

Here is the Sagnac formula for A ROTATING LINEAR SEGMENT.

SAGNAC EFFECT WITHOUT AN AREA OR A LOOP

The biggest breakthrough in the physics of the Sagnac effect has been the use of PHASE-CONJUGATE MIRRORS.

In fact, had this technology been available to G. Sagnac and A. Michelson it would have saved them a lot of time and energy in constructing experiments and deriving formulas.


https://arxiv.org/ftp/physics/papers/0609/0609202.pdf

Test of the one-way speed of light and the first-order experiment of
Special Relativity using phase-conjugate interferometers

Professor Ruyong Wang



The equation which expresses the relationship between interference fringes and time differences is F=dt[c/λ].

This experiment shows us two important points. First, it confirms the phase reversal of a PCM and demonstrates the Sagnac effect in an arc segment AB, not a closed path. Second, it gives us important implications: The result, φ = 4πRΩL/cλ, can be re-written as φ = 4πvL/cλ where v is the speed of the moving arc segment AB (where R is the radius of the circular motion, Ω is the rotational rate).

If we increase the radius of the circular motion as shown in Fig. 6, the arc segment AB will approach a linear segment AB, the circular motion will approach the linear motion, the phase-conjugate Sagnac experiment will approach the phase-conjugate first-order experiment as shown in Fig. 4, and the phase shift is always φ = 4πvL/cλ.

The Sagnac formula for a straight line path which is rotating becomes:

φ = 4πvL/c²

where v = RΩ


That is, the formula for a rotating linear segment is:

φ = 4πvL/c²

where v = RΩ


Using the values for v for both the orbital Sagnac and the rotational Sagnac we can see immediately that v_o/v_r = 60.

A total refutation of the "analysis" signed the jackbot.

The jackbot has not read the message at all (a simple search for that first message using "question").

And there you go lying yet again.
A simple search for question on the last three pages reveals:
Me repeatedly telling you to answer the question.
You asking the mods how much longer your spamming can continue.
You claiming to have answered the question.
You asking other questions.
You deflecting the question.

There is not a single time you have answered the question.
If there was, and it was easy to find, you would have provided a link and a quote.

But you haven't.
So this is just a further indicator of your dishonest and how pathetic your position is.
Not only are you unable to answer a simple question you need to pretend you already have to avoid dealing with your complete failure.

NOW ANSWER THE QUESTION OR GET LOST!
Do you accept that the area of an annular sector is given by:
A=θ_o(R₂² - R₁²)/2
or equally that the following equation holds:
2A=θ_o(R₂² - R₁²)

Yes or no?

Nobody is complaining about the area of a sector of a circle (that is why your trolling question is meaningless).

The problem is with the invalid comparison.

Why is the physics illiterate jackbot avoiding the exact Sagnac formula for the rotating linear segment?

Truly pathetic.

As I have already said, the use the phase-conjugate mirrors changes everything.

Now, we can derive the exact Sagnac formula for a rotating linear segment.

It makes this entire thread superfluous.

There is no need to address any further questions: the use of nonlinear optics, mainly the phase-conjugate mirror has been the most extraordinary breakthrough for the Sagnac effect.

Here is the Sagnac formula for A ROTATING LINEAR SEGMENT.

SAGNAC EFFECT WITHOUT AN AREA OR A LOOP

The biggest breakthrough in the physics of the Sagnac effect has been the use of PHASE-CONJUGATE MIRRORS.

In fact, had this technology been available to G. Sagnac and A. Michelson it would have saved them a lot of time and energy in constructing experiments and deriving formulas.


https://arxiv.org/ftp/physics/papers/0609/0609202.pdf

Test of the one-way speed of light and the first-order experiment of
Special Relativity using phase-conjugate interferometers

Professor Ruyong Wang



The equation which expresses the relationship between interference fringes and time differences is F=dt[c/λ].

This experiment shows us two important points. First, it confirms the phase reversal of a PCM and demonstrates the Sagnac effect in an arc segment AB, not a closed path. Second, it gives us important implications: The result, φ = 4πRΩL/cλ, can be re-written as φ = 4πvL/cλ where v is the speed of the moving arc segment AB (where R is the radius of the circular motion, Ω is the rotational rate).

If we increase the radius of the circular motion as shown in Fig. 6, the arc segment AB will approach a linear segment AB, the circular motion will approach the linear motion, the phase-conjugate Sagnac experiment will approach the phase-conjugate first-order experiment as shown in Fig. 4, and the phase shift is always φ = 4πvL/cλ.

The Sagnac formula for a straight line path which is rotating becomes:

φ = 4πvL/c²

where v = RΩ


That is, the formula for a rotating linear segment is:

φ = 4πvL/c²

where v = RΩ


Using the values for v for both the orbital Sagnac and the rotational Sagnac we can see immediately that v_o/v_r = 60.

A total refutation of the "analysis" signed the jackbot.

Why is the physics illiterate jackbot avoiding the exact Sagnac formula for the rotating linear segment?

Again, no idea about your delusions. I don't know who this Jackbot is.
However the physically literate me is avoiding your BS to make you focus on one point at a time rather than trying to bury the thread with mountains of BS.

Especially as you seem unable to even distinguish between different things, like LISA and an interferometer on Earth and seem unable to realise that they will have vastly different Sagnac effects.

Quote from: sandokhan on February 08, 2018, 12:16:53 AM

Nobody is complaining about the area of a sector of a circle (that is why your trolling question is meaningless).

Notice how even this isn't a direct answer?
The reason I had to demand you answer it is because you continually rejected the derivation without cause.
You were unable to say what part you disagreed with so we stepped through it, bit by bit.

But I will take this as an admission that the formula is correct.

So, continuing on from that:
You have already accepted the derivation up to this point:
dt=t₁ - t₂ = 2θ_oω_o(R₂² - R₁²)/c²

You have now accepted that the following relation holds:
2A=θ_o(R₂² - R₁²)

Note that the RHS of this appears in the Sagnac formula and thus can be substituted in.

Thus we end up with this:
dt=2(2A)ω_o/c²
dt=4Aω_o/c²

Where A is the area of the loop.

This is consistent with the formula previously provided by me, showing my derivation is correct.
Importantly, this "A" is the area of the loop, not the area of the orbit as you repeatedly claimed, and repeatedly tried to use to lie and claim the orbital Sagnac should be much greater.
This conclusively shows your claims are wrong.
So are you going to recant and admit that it is not the area of the orbit which is important but the area of the loop?

And now this question will continue to appear until you either admit you were wrong or show how that is wrong, as you have effectively just accepted that it is the area of the loop rather than the orbit which determines the Sagnac effect.

Only after we have dealt with this can we move on.

The jackbot has a very short memory.

dt=t1 - t2 = 2θoωo(R22 - R12)/c2

It cannot have two speeds.

And a negative term.

This is what the jackbot put forth before its readers.

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

v = ωR
v₂ = ωR₂
v₁ = ωR₁

s = Rφ (s = arclength)

dt = 2φωR₂²/c² - 2φωR₁²/c²

dt = 2v₂s₂/c² - 2v₁s₁/c²


BUT THIS IS WRONG!

The Sagnac shift is made up of one term, not two.

You cannot have TWO SPEEDS, only one.

There is no negative term in the Sagnac.

dto/dtr=[4*wo*Ao/c2]/[4*wr*Ar/c2]
=wo/wr

ONE NEEDS TWO SETS OF RADII.

The radii (along with angle subtended) eliminated to produce the area.

But you cannot eliminate the radii.

 dt=2*omega*alpha*R22/c2-2*omega*alpha*R1^2/c2
=2*omega*alpha*(R22-R12)/c2[/i]

v = ωR
v₂ = ωR₂
v₁ = ωR₁

s = Rφ (s = arclength)

dt = 2φωR₂²/c² - 2φωR₁²/c²

dt = 2v₂s₂/c² - 2v₁s₁/c²


BUT THIS IS WRONG!

The Sagnac shift is made up of one term, not two.

You cannot have TWO SPEEDS, only one.

There is no negative term in the Sagnac.


Let us proceed along your line of thought, and see where it will get you.


θo = angle subtended by the two radii, R2 and R1 = orbital angle

s2 = R2 x θo

s1 = R1 x θo


θr = angle subtended by the two radii, r2 and r1 = rotational angle

s2 = r2 x θr

s1 = r1 x θr


R2 - R1 = r2 - r1

r2 x θr = R2 x θo

r1 x θr = R1 x θo

r2/r1 = R2/R1

(r2 x R1) = (r1 x R2)

Since the two areas must be equal,

r1/R1 = (r2 + r1)/(R2 + R1)



Right away, one runs into huge problems with this scenario.

R2 = r2 - r1 + R1

(r2 x R1) = r1r2 - r12 + (R1 x r1)

r2(R1 - r1) = r1(R1 - r1)

So we end up with: r2 = r1, which is impossible


The entire set up doesn't make any sense.


IF YOU BLINDLY SUBSTITUTE THE AREA FOR THE ANGULAR VELOCITY X RADIUS FACTOR, YOU END UP WITH R2=R1 AND r2=r1.


REMEMBER, the original message allowed for NO MINUTE DIFFERENCES AT ALL.

The jackbot's original derivation meant: A = A and R2 - R1 = r2 - r1.

That's all.

From now on, I will not allow any departure from the data set in the original piece of analysis, where no minute differences were implied at all, or even conjectured: jackbot's data simply stated that A = A and  R2 - R1 = r2 - r1.


With this data, we reach a direct contradiction in no time at all: r2 = r1.


Here are the jackbot's own words:

That's right. For a given loop, A=A.
For the rotation we have wr, for the orbit we have wo.

∆t=4*A*w/c^2

Thus ∆to=4A*wo/c^2
and ∆tr=4A*wr/c^2.
Comparing the 2:
∆to/∆tr=(4A*wo/c^2)/(4A*wr/c^2)=wo/wr.
Thus it is quite simple to see that the orbital Sagnac is much less than the rotational one due to the much lower angular velocity.

Once again, if you use different loops, your comparison is meaningless.


He is telling his readers that THE SAME INTERFEROMETER HAS TO BE USED TO COMPARE THE TWO SAGNAC EFFECTS.

Obviously, it had no idea what the entire contraption entailed at that time.

But now, it is too late to change things.

The jackbot's derivation leads to the situation where r2 = r1, a total nonsense.


Now, it takes A SINGLE COUNTEREXAMPLE TO DEFEAT A CERTAIN HYPOTHESIS.


You can no longer avoid this very simple proof.


Why is the physics illiterate jackbot avoiding the exact Sagnac formula for the rotating linear segment?

Truly pathetic.

As I have already said, the use the phase-conjugate mirrors changes everything.

Now, we can derive the exact Sagnac formula for a rotating linear segment.

It makes this entire thread superfluous.

There is no need to address any further questions: the use of nonlinear optics, mainly the phase-conjugate mirror has been the most extraordinary breakthrough for the Sagnac effect.

Here is the Sagnac formula for A ROTATING LINEAR SEGMENT.

SAGNAC EFFECT WITHOUT AN AREA OR A LOOP

The biggest breakthrough in the physics of the Sagnac effect has been the use of PHASE-CONJUGATE MIRRORS.

In fact, had this technology been available to G. Sagnac and A. Michelson it would have saved them a lot of time and energy in constructing experiments and deriving formulas.


https://arxiv.org/ftp/physics/papers/0609/0609202.pdf

Test of the one-way speed of light and the first-order experiment of
Special Relativity using phase-conjugate interferometers

Professor Ruyong Wang



The equation which expresses the relationship between interference fringes and time differences is F=dt[c/λ].

This experiment shows us two important points. First, it confirms the phase reversal of a PCM and demonstrates the Sagnac effect in an arc segment AB, not a closed path. Second, it gives us important implications: The result, φ = 4πRΩL/cλ, can be re-written as φ = 4πvL/cλ where v is the speed of the moving arc segment AB (where R is the radius of the circular motion, Ω is the rotational rate).

If we increase the radius of the circular motion as shown in Fig. 6, the arc segment AB will approach a linear segment AB, the circular motion will approach the linear motion, the phase-conjugate Sagnac experiment will approach the phase-conjugate first-order experiment as shown in Fig. 4, and the phase shift is always φ = 4πvL/cλ.

The Sagnac formula for a straight line path which is rotating becomes:

φ = 4πvL/c²

where v = RΩ


That is, the formula for a rotating linear segment is:

φ = 4πvL/c²

where v = RΩ


Using the values for v for both the orbital Sagnac and the rotational Sagnac we can see immediately that v_o/v_r = 60.

A total refutation of the "analysis" signed the jackbot.

Thus we end up with this:
dt=2(2A)ωo/c2
dt=4Aωo/c2

Where A is the area of the loop.

No.

You end up with this:

v = ωR
v₂ = ωR₂
v₁ = ωR₁

s = Rφ (s = arclength)

dt = 2φωR₂²/c² - 2φωR₁²/c²

dt = 2v₂s₂/c² - 2v₁s₁/c²


BUT THIS IS WRONG!

The Sagnac shift is made up of one term, not two.

You cannot have TWO SPEEDS, only one.

There is no negative term in the Sagnac.



The Sagnac formula has only one term, that's it.

Not two terms, one of which would be negative.

The jackbot's entire setup is faulty.

Moreover we end up with r2 = r1, as proven in my previous message.


The correct formula for the rotating linear segment is:

φ = 4πvL/c²

where v = RΩ


Using the values for v for both the orbital Sagnac and the rotational Sagnac we can see immediately that v_o/v_r = 60.

A total refutation of the "analysis" signed the jackbot.

This is consistent with the formula previously provided by me, showing my derivation is correct.
Importantly, this "A" is the area of the loop, not the area of the orbit as you repeatedly claimed, and repeatedly tried to use to lie and claim the orbital Sagnac should be much greater.

The entire derivation is wrong.

The jackbot ends up with TWO TERMS FOR THE SAGNAC, ONE OF WHICH IS NEGATIVE.

The entire setup is faulty.


The formula for a rotating linear segment is:

φ = 4πvL/c²

where v = RΩ


Using the values for v for both the orbital Sagnac and the rotational Sagnac we can see immediately that v_o/v_r = 60.

A total refutation of the "analysis" signed the jackbot.

You don't stand a chance with me here, jackbot.

Looks like sandokhan you can legitimately have a defeater of JackBlack in your signature too.

Looks like sandokhan you can legitimately have a defeater of JackBlack in your signature too.

This has always been the case.

Now, there is no way out for the jackbot.

THE JACKBOT HAS TO NAME A SINGLE SOURCE WHICH USES TWO TERMS FOR THE SAGNAC, TWO DIFFERENT SPEEDS AND A NEGATIVE TERM.


Name a single bibliographical reference which uses TWO TERMS FOR THE SAGNAC, one of which is a negative term.

Here are the best mainstream treatises on the Sagnac effect:

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

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

http://www.kritik-relativitaetstheorie.de/Anhaenge/Wolfgang-Engelhardt-Sagnac.pdf

ONE TERM FOR THE SAGNAC. NOT TWO TERMS. ONE SPEED. NOT TWO SPEEDS. NO NEGATIVE TERMS INCLUDED.

The jackbot's derivation IS NOT THE SAGNAC EFFECT.

Looks like sandokhan you can legitimately have a defeater of JackBlack in your signature too.

As he is still yet to refute me in the slightest, he cannot legitimately have one, nor can you as you are yet to defeat me.

Instead you both just spout crap and ignore refutations.

The jackbot has a very short memory.

Again, no idea who this imaginary fiend of yours is, but I have a fairly good memory.

You seem completely unable to address the question.
You have agreed the derivation is correct.
No going back on that now.

So skipping your mountains of bullshit where you might try and bury an answer in, answer the question without all the BS:
Are you going to recant and admit that it is not the area of the orbit which is important but the area of the loop?

You have just accepted that that is the case, that is the area of the loop.
So will you show a shred of integrity by admitting you were wrong and that you do not use the area of the orbit as you repeatedly tried?

Here are the best mainstream treatises on the Sagnac effect:
http://www.mathpages.com/rr/s2-07/2-07.htm
http://www.naturalphilosophy.org/pdf/ebooks/Kelly-TimeandtheSpeedofLight.pdf
http://www.kritik-relativitaetstheorie.de/Anhaenge/Wolfgang-Engelhardt-Sagnac.pdf

Which all agree with me.
dt=4Aw/c^2.
Where A is the area of the loop.
Regardless of the geometry of the loop and where the centre of rotation is.

So thanks for disproving yourself yet again.

Quote from: Shifter on February 08, 2018, 02:31:37 PM

Looks like sandokhan you can legitimately have a defeater of JackBlack in your signature too.

As he is still yet to refute me in the slightest, he cannot legitimately have one, nor can you as you are yet to defeat me.

Instead you both just spout crap and ignore refutations.

Dude, your arguments were utterly annihilated in your debate with me in my class. In fact, I will change my sig to reflect this. You were more than defeated. You were annihilated. I would be a more gracious 'winner' if you weren't so rude and smug to everyone. You need to be humble and learn humility

slightest

No, but with the full and mighty force of REAL science.

When it comes to the complex shapes of an interferometer, nothing beats the interferometer designed by Dufour and Prunier.




Yet, they obtained the correct result.

http://www.conspiracyoflight.com/pdf/Dufour_and_Prunier-On_the_Fringe_Movement_Registered_on_a_Platform_in_Uniform_Motion_%281942%29.pdf

A. Dufour and F. Prunier created Sagnac interferometers that were composites of moving and stationary paths, including stationary sources and stationary detectors. This was essentially to test if the relativistic approach could be distinguished from the classical approach.

In all cases of this experimental test, the Sagnac effect was the same.


THE JACKBOT'S DERIVATION HAS NOTHING TO DO WITH THE SAGNAC.

This is what it ends up with:

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

v = ωR
v₂ = ωR₂
v₁ = ωR₁

s = Rφ (s = arclength)

dt = 2φωR₂²/c² - 2φωR₁²/c²

dt = 2v₂s₂/c² - 2v₁s₁/c²


BUT THIS IS WRONG!

The Sagnac shift is made up of one term, not two.

You cannot have TWO SPEEDS, only one.

There is no negative term in the Sagnac.


They agree with me


The formulas for the Sagnac in all three references feature ONE TERM.

NOT TWO TERMS.

NOT A NEGATIVE TERM.

The jackbot's derivation has nothing to do with the Sagnac.


Here is the CORRECT DERIVATION.

Now, it takes A SINGLE COUNTEREXAMPLE TO DEFEAT A CERTAIN HYPOTHESIS.


You can no longer avoid this very simple proof.


Why is the physics illiterate jackbot avoiding the exact Sagnac formula for the rotating linear segment?

Truly pathetic.

As I have already said, the use the phase-conjugate mirrors changes everything.

Now, we can derive the exact Sagnac formula for a rotating linear segment.

It makes this entire thread superfluous.

There is no need to address any further questions: the use of nonlinear optics, mainly the phase-conjugate mirror has been the most extraordinary breakthrough for the Sagnac effect.

Here is the Sagnac formula for A ROTATING LINEAR SEGMENT.

SAGNAC EFFECT WITHOUT AN AREA OR A LOOP

The biggest breakthrough in the physics of the Sagnac effect has been the use of PHASE-CONJUGATE MIRRORS.

In fact, had this technology been available to G. Sagnac and A. Michelson it would have saved them a lot of time and energy in constructing experiments and deriving formulas.


https://arxiv.org/ftp/physics/papers/0609/0609202.pdf

Test of the one-way speed of light and the first-order experiment of
Special Relativity using phase-conjugate interferometers

Professor Ruyong Wang



The equation which expresses the relationship between interference fringes and time differences is F=dt[c/λ].

This experiment shows us two important points. First, it confirms the phase reversal of a PCM and demonstrates the Sagnac effect in an arc segment AB, not a closed path. Second, it gives us important implications: The result, φ = 4πRΩL/cλ, can be re-written as φ = 4πvL/cλ where v is the speed of the moving arc segment AB (where R is the radius of the circular motion, Ω is the rotational rate).

If we increase the radius of the circular motion as shown in Fig. 6, the arc segment AB will approach a linear segment AB, the circular motion will approach the linear motion, the phase-conjugate Sagnac experiment will approach the phase-conjugate first-order experiment as shown in Fig. 4, and the phase shift is always φ = 4πvL/cλ.

The Sagnac formula for a straight line path which is rotating becomes:

φ = 4πvL/c²

where v = RΩ


That is, the formula for a rotating linear segment is:

φ = 4πvL/c²

where v = RΩ


Using the values for v for both the orbital Sagnac and the rotational Sagnac we can see immediately that v_o/v_r = 60.

A total refutation of the "analysis" signed the jackbot.

Quote from: Shifter on February 08, 2018, 02:31:37 PM

Looks like sandokhan you can legitimately have a defeater of JackBlack in your signature too.

This has always been the case.

Now, there is no way out for the jackbot.

THE JACKBOT HAS TO NAME A SINGLE SOURCE WHICH USES TWO TERMS FOR THE SAGNAC, TWO DIFFERENT SPEEDS AND A NEGATIVE TERM.


Name a single bibliographical reference which uses TWO TERMS FOR THE SAGNAC, one of which is a negative term.

Here are the best mainstream treatises on the Sagnac effect:

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

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

http://www.kritik-relativitaetstheorie.de/Anhaenge/Wolfgang-Engelhardt-Sagnac.pdf

ONE TERM FOR THE SAGNAC. NOT TWO TERMS. ONE SPEED. NOT TWO SPEEDS. NO NEGATIVE TERMS INCLUDED.

The jackbot's derivation IS NOT THE SAGNAC EFFECT.

Can you name a single source, from your many great physicists that agree with your conclusions about the shape of the earth and its movement?

Quote from: JackBlack on February 08, 2018, 02:48:16 PM

Quote from: Shifter on February 08, 2018, 02:31:37 PM

Looks like sandokhan you can legitimately have a defeater of JackBlack in your signature too.

As he is still yet to refute me in the slightest, he cannot legitimately have one, nor can you as you are yet to defeat me.

Instead you both just spout crap and ignore refutations.

Dude, your arguments were utterly annihilated in your debate with me in my class. In fact, I will change my sig to reflect this. You were more than defeated. You were annihilated. I would be a more gracious 'winner' if you weren't so rude and smug to everyone. You need to be humble and learn humility

Exactly.

The most amazing thing is that the jackbot refuses to understand that reality is starring it right in the face.


Why is the physics illiterate jackbot avoiding the exact Sagnac formula for the rotating linear segment?

Truly pathetic.

As I have already said, the use the phase-conjugate mirrors changes everything.

Now, we can derive the exact Sagnac formula for a rotating linear segment.

It makes this entire thread superfluous.

There is no need to address any further questions: the use of nonlinear optics, mainly the phase-conjugate mirror has been the most extraordinary breakthrough for the Sagnac effect.

Here is the Sagnac formula for A ROTATING LINEAR SEGMENT.

SAGNAC EFFECT WITHOUT AN AREA OR A LOOP

The biggest breakthrough in the physics of the Sagnac effect has been the use of PHASE-CONJUGATE MIRRORS.

In fact, had this technology been available to G. Sagnac and A. Michelson it would have saved them a lot of time and energy in constructing experiments and deriving formulas.


https://arxiv.org/ftp/physics/papers/0609/0609202.pdf

Test of the one-way speed of light and the first-order experiment of
Special Relativity using phase-conjugate interferometers

Professor Ruyong Wang



The equation which expresses the relationship between interference fringes and time differences is F=dt[c/λ].

This experiment shows us two important points. First, it confirms the phase reversal of a PCM and demonstrates the Sagnac effect in an arc segment AB, not a closed path. Second, it gives us important implications: The result, φ = 4πRΩL/cλ, can be re-written as φ = 4πvL/cλ where v is the speed of the moving arc segment AB (where R is the radius of the circular motion, Ω is the rotational rate).

If we increase the radius of the circular motion as shown in Fig. 6, the arc segment AB will approach a linear segment AB, the circular motion will approach the linear motion, the phase-conjugate Sagnac experiment will approach the phase-conjugate first-order experiment as shown in Fig. 4, and the phase shift is always φ = 4πvL/cλ.

The Sagnac formula for a straight line path which is rotating becomes:

φ = 4πvL/c²

where v = RΩ


That is, the formula for a rotating linear segment is:

φ = 4πvL/c²

where v = RΩ


Using the values for v for both the orbital Sagnac and the rotational Sagnac we can see immediately that v_o/v_r = 60.

A total refutation of the "analysis" signed the jackbot.

Dude, your arguments were utterly annihilated in your debate with me in my class.

There you go with your delusions again.
Your ass was repeatedly handed to you. You failed to defeat anyone.

I would be a more gracious 'winner' if you weren't so rude and smug to everyone. You need to be humble and learn humility

Perhaps you and Sandy should follow your advice.

You and Sandy are both being smug arrogant assholes with delusions of grandeur.

Sandy, yet again you seem to have failed to answer a simple question.
Perhaps you can answer it this time:
Do you accept that you were wrong, and it is the area of the loop that the Sagnac shift is based upon as this derivation which you have agreed is correct shows, not the area of the orbit as you had been repeatedly claiming?

Yes or no?

The jackbot has to learn to derive a proper Sagnac formula.

Its derivation has nothing to do with the Sagnac.

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

v = ωR
v2 = ωR2
v1 = ωR1

s = Rφ (s = arclength)

dt = 2φωR22/c2 - 2φωR12/c2

dt = 2v2s2/c2 - 2v1s1/c2


BUT THIS IS WRONG!

The Sagnac shift is made up of one term, not two.

You cannot have TWO SPEEDS, only one.

There is no negative term in the Sagnac.


One has to have an asshole for a brain to end up with TWO TERMS FOR THE SAGNAC, ONE OF WHICH IS NEGATIVE.


Here is the correct formula for the rotating linear segment.

SAGNAC EFFECT WITHOUT AN AREA OR A LOOP

The biggest breakthrough in the physics of the Sagnac effect has been the use of PHASE-CONJUGATE MIRRORS.


https://arxiv.org/ftp/physics/papers/0609/0609202.pdf

Test of the one-way speed of light and the first-order experiment of
Special Relativity using phase-conjugate interferometers

Professor Ruyong Wang



This experiment shows us two important points. First, it confirms the phase reversal of a PCM and demonstrates the Sagnac effect in an arc segment AB, not a closed path. Second, it gives us important implications: The result, φ = 4πRΩL/cλ, can be re-written as φ = 4πvL/cλ where v is the speed of the moving arc segment AB (where R is the radius of the circular motion, Ω is the rotational rate).

If we increase the radius of the circular motion as shown in Fig. 6, the arc segment AB will approach a linear segment AB, the circular motion will approach the linear motion, the phase-conjugate Sagnac experiment will approach the phase-conjugate first-order experiment as shown in Fig. 4, and the phase shift is always φ = 4πvL/cλ.

The Sagnac formula for a straight line path which is rotating becomes:

φ = 4πvL/c2

where v = RΩ


That is, the formula for a rotating linear segment is:

φ = 4πvL/c2

where v = RΩ


Using the values for v for both the orbital Sagnac and the rotational Sagnac we can see immediately that vo/vr = 60.

A total refutation of the "analysis" signed the jackbot.