Bob Knodel and the laser ring gyroscope

By contrast the formula derived by me and proven to be correct by Professor Yeh is this:

2(V₁L₁ + V₂L₂)/c²

My analysis is correct, as also proven by Professor Yeh.

The nearest result from Professor Yeh seems to be:

and that is only for his Phase Conjugate Mirror fibre optic gyroscope so is totally irrelevant here!

So before wasting more time going over you derivation again and again,
let's see just what values your "beautiful and striking generalization of the Sagnac effect formula" gives in some practical situations.

Here is what you claimed:

A beautiful and striking generalization of the Sagnac effect formula:



Everything you want to know about the universe in just one formula: the most important formula in physics, by far.

Obviously your diagram came from the Conspiracy of light, Michelson-Gale Interferometer Simulator.
Now surely you would trust the "Conspiracy of Light" site to give the correct answers or is everybody but you wrong?

That simulator let's the delay for various latitudes be calculated so why not check your "most important formula in physics" against those values.
For example for a loop of the dimensions used by Michelson, Gale and Pearson the delays are:
Latitude     Delay sec      Fringes
      0°         0.00                 0.000
    45°         4.77 x 10^-16    0.251
    90°         6.74 x 10^-16    0.355 

Now you calculate the Sagnac delay for those dimensions and at latitudes 0°, 45° and 90°. If yours differ markedly from the above please explain why.

And I notice in your latest post that you again show that you don't know the difference between a normal Sagnac Loop and Dr. P. Yeh's Phase Conjugate Mirror fibre optic gyroscope!

My formula is totally up to date.

"It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong."
Richard P. Feynman

Here is the experiment performed by Professor Yeh, one of the top experts in the world in laser optics, which confirms the CORRECT SAGNAC EFFECT formula:
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2(V₁L₁ + V₂L₂)/c²

Dr. P. Yeh
PhD, Caltech, Nonlinear Optics
Principal Scientist of the Optics Department at Rockwell International Science Center
Professor, UCSB
"Engineer of the Year," at Rockwell Science Center
Leonardo da Vinci Award in 1985
Fellow of the Optical Society of America, the Institute of Electrical and Electronics Engineers

Did you seriously think that we didn't know the difference between the two?

Now please show your "CORRECT SAGNAC EFFECT formula" as used in Sagnac's experiment and the Michelson-Gale-Pearson experiment!

It is nigh time to welcome both rabinoz and jackblack as flat earth believers.

Sagnac is for a rotating interferometer, but yours works with any motion, even linear motion without any rotation at all.

The SAGNAC EFFECT can detect LINEAR/UNIFORM/TRANSLATIONAL MOTION as well.

You didn't know that?

The translational/linear Sagnac effect IS A FACT OF SCIENCE.

Professor Ruyong Wang, in two well-designed experiments showed unambiguously that an identical Sagnac effect appearing in uniform radial motion occurs in linear inertial motion.

He tested the travel-time difference between two counter-propagating light beams in uniformly moving fiber.

The travel-time difference was found to be:

Δt = 2vΔL/c^2

where ΔL is the length of the fiber segment moving with the source and detector at a v, whether the segment was moving uniformly or circularly.



https://arxiv.org/ftp/physics/papers/0609/0609222.pdf (first experiment conducted by R. Wang)

https://arxiv.org/ftp/physics/papers/0609/0609202.pdf (second experiment carried out by R. Wang)

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

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

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

Here is another demonstration.





If you can't do the derivation correctly for a stationary loop

But you just stated the opposite:

I know. (that for a stationary interferometer there are no transit times).

Then, if you know that the SAGNAC EFFECT applies to a rotational interferometer (or to an interferometer in linear motion), your requests are meaningless, since you are well outside the scope of our debate here. A stationary interferometer has no transit times. We need to understand the SAGNAC EFFECT, which in our case here, involves rotation.

Your errors have already been pointed out and you just ignore them.

Let's put your word to the test.



Point A is located at the detector
Point B is in the bottom right corner
Point C is in the upper right corner
Point D is in the upper left corner

l1 is the upper arm.
l2 is the lower arm.


so the total delay for the A > D > C > B > A path (clockwise) should be: l1/(c - v1) + l2/(c + v2)

It can't be.

Now, you have TWO OPPOSING DIRECTIONS, while you assign the SAME SIGN, a catastrophic mistake if you are seeking the SAGNAC EFFECT formula.

Moreover, where are the loops ??

What in effect you are saying is this:

A > D > C combined with A > B.

NO LOOP AT ALL.

A tremendous error committed by Michelson.

The SAGNAC EFFECT requires TWO LOOPS. Now we are dealing with the clockwise loop.

What you want is this:

A > D > C > B > A path (clockwise path)

Now we have a loop.

This loop includes two opposite directions, two opposite terms/components.

That is why one must a positive sign, while the other must have a negative sign.

Sagnac phase components for the A > D > C > B > A path (clockwise path):

l₁/(c - v₁)

-l₂/(c + v₂)

Now, we do the same thing for the counterclockwise path, the A > B > C > D > A path:

l₂/(c - v₂)

-l₁/(c + v₁)


For the single continuous clockwise path we now have the total time difference:

l₁/(c - v₁) - l₂/(c + v₂)


For the single continuous counterclockwise path we have the total difference:

l₂/(c - v₂) - l₁/(c + v₁)


TWO LOOPS as required by the definition of the Sagnac effect.

If we change the sign of the second term/phase component to +, that is:

l₁/(c - v₁)

l₂/(c + v₂)

then, we no longer have a LOOP, and moreover we are using the wrong sign for the direction of the second transit time; each transit time has a different direction, hence we must use opposite signs to correctly designate them in our analysis.

Let us remember the very defintion of the Sagnac effect: two loops are required to properly derive the formula.

I have the loops, you have nothing.

I use the signs correctly, you use the same sign for beams traveling in opposite directions.

You have ONE interferometer, with two loops: a counterclockwise loop and a clockwise loop.

A > B > C > D > A is a continuous counterclockwise path, a negative sign -

A > D > C > B > A is a continuous clockwise path, a positive sign +

The Sagnac phase difference for the clockwise path has a positive sign.

The Sagnac phase difference for the counterclockwise has a negative sign.


Make sure you understand the definition of the term LOOP.

Loop = a structure, series, or process, the end of which is connected to the beginning.

Connecting the two sides through a single mathematical description closes the loop; treating each side separately does not. The Sagnac effect requires, by definition, a structure, the end of which is connected to the beginning.

Each time transit has a DEFINITE DIRECTION.

Opposite directions have opposite signs.

Let us go back to the original derivation of the Sagnac effect.



To get the time transits IN OPPOSITE DIRECTIONS, you must assign a NEGATIVE SIGN to one of them.



The CORIOLIS EFFECT and the GLOBAL SAGNAC EFFECT are two different phenomena: they require two different formulas.

Professor Yeh's experiment, peer-reviewed at the highest possible level, in the Journal of Optics Letters proves I am correct.


The use of the phase conjugate mirror permits the revealing of the final formula, the total phase difference:

φ = -2(φ₂ - φ₁) = 4π(R₁L₁ + R₂L₂)Ω/λc

2(v₁l₁ + v₂l₂)/c²

Exactly the formula obtained by Professor Yeh:

φ = -2(φ₂ - φ₁) = 4π(R₁L₁ + R₂L₂)Ω/λc = 4π(V₁L₁ + V₂L₂)/λc

Since Δφ = 2πc/λ x Δt, Δt = 2(R₁L₁ + R₂L₂)Ω/c² = 2(V₁L₁ + V₂L₂)/c²

CORRECT SAGNAC FORMULA:

2(V₁L₁ + V₂L₂)/c²


You... were saying what?

No errors whatsoever in my derivation which is backed up by Professor Yeh's fantastic experiment.

SAME FORMULA.

By contrast, you derived the CORIOLIS EFFECT.

YOU HAVE NO LOOPS.

YOU USED BEAMS IN OPPOSITE DIRECTIONS WITH THE SAME SIGN.

Your derivation has nothing to do with the SAGNAC EFFECT.

You have been proven wrong, yet again.

Let us know take a look at jackblack's errors, which lead to the CORIOLIS EFFECT formula.

Let us take a look at the humongous error perpetrated by Michelson and also by the RE.

Now instead of adding and subtracting based upon direction, we will add the terms of the same colour, corresponding to the one beam rotating around the interferometer and then find the difference.
dt=l₁/(c - v₁)+l₂/(c + v₂)-l₁/(c + v₁)-l₂/(c - v₂)
=l₁/(c - v₁)-l₁/(c + v₁)+l₂/(c + v₂)-l₂/(c - v₂)
=l₁(c + v₁-c + v₁)/(c² - v₁²)+l₂(c - v₂-c - v₂)/(c² - v₂²)
=2*l₁v₁/(c² - v₁²)-2*l₂v₂/(c² - v₂²)

Now, what the frell is this?

The author of this unscientific piece of garbage cannot distinguish between two opposite directions.

We no longer have a Sagnac interferometer whose center of rotation coincides with its geometrical center: the interferometer is located away from the center of rotation, as such each and every direction MUST HAVE THE CORRECT SIGN.

This guy has the same sign for opposite directions:

l₁/(c - v₁)+l₂/(c + v₂)

and

-l₁/(c + v₁)-l₂/(c - v₂)

Catastrophically wrong!!!

Here is the correct analysis:

Sagnac phase components for the A > D > C > B > A path (clockwise path):

l₁/(c - v₁)

-l₂/(c + v₂)

Sagnac phase components for the A > B > C > D > A path (counterclockwise path):

l₂/(c - v₂)

-l₁/(c + v₁)


For the single continuous clockwise path we add the components:

l₁/(c - v₁) - l₂/(c + v₂)

For the single continuous counterclockwise path we add the components:

l₂/(c - v₂) - l₁/(c + v₁)


The proper signs, in accordance with the direction, are in place.

What jackblack did is to substract the phase differences for TWO SEPARATE OPEN SEGMENTS, and not for the TWO LOOPS (as required by the defintion of the Sagnac effect).

He assigned the wrong signs, moreover, he did not complete the counterclockwise and the clockwise addition of the components of the phase differences.

We have the following terms, both have the same direction, that means one of them corresponds to the red in the inner segment and one to orange on the outer segment. I will colour code them for clarity:
l₁/(c - v₁)
l₂/(c - v₂)

Then, we have the remaining terms, in the opposite direction, likewise meaning one is for orange one is for red, noting that red travelled along l1 in the previous one so now it must travel along l2 in this one:
l₁/(c + v₁)
l₂/(c + v₂)

Then, if they ARE in opposite direction, they must have the OPPOSITE SIGN.

Here is the correct analysis:

Sagnac phase components for the A > D > C > B > A path (clockwise path):

l₁/(c - v₁)

-l₂/(c + v₂)

Sagnac phase components for the A > B > C > D > A path (counterclockwise path):

l₂/(c - v₂)

-l₁/(c + v₁)


For the single continuous clockwise path we add the components:

l₁/(c - v₁) - l₂/(c + v₂)

For the single continuous counterclockwise path we add the components:

l₂/(c - v₂) - l₁/(c + v₁)


jackblack assigned the SAME SIGN, even though he just said a few lines earlier, that they are in fact in opposite direction.


Where are your loops???

You are still comparing two OPEN SEGMENTS: defying the very definition of the Sagnac effect.

Path 1 - A>B, D>C.
Path 2 - C>D, B>A

Completely wrong!

The paths are very clear:

A > B > C > D > A is a continuous counterclockwise path, a negative sign -
A > D > C > B > A is a continuous clockwise path, a positive sign +

Yes, ignoring the sign which I don't particular care about at this time

You CANNOT ignore the sign, since by your own admission you have light beams travelling in opposite directions.


My analysis is correct, as also proven by Professor Yeh.

and that is only for his Phase Conjugate Mirror fibre optic gyroscope so is totally irrelevant here!

And I notice in your latest post that you again show that you don't know the difference between a normal Sagnac Loop and Dr. P. Yeh's Phase Conjugate Mirror fibre optic gyroscope!

You mean I also have to explain the physics of the phase-conjugate mirror to you?

Phase Conjugate Mirror

"Let us begin with the properties of a phase conjugate mirror. A phase conjugate mirror is like a mirror, in that it reflects incident light back towards where it came from, but it does so in a different way than a regular mirror.

In a regular mirror, light that strikes the mirror normal to its surface, is reflected straight back the way it came (A). This is also true of a phase conjugate mirror (B). When the light strikes a normal mirror at an angle, it reflects back in the opposite direction, such that the angle of incidence is equal to the angle of reflection. (C)



In a phase conjugate mirror, on the other hand, light is always reflected straight back the way it came from, no matter what the angle of incidence. (D)

This difference in the manner of reflection has significant consequences. For example if we place an irregular distorting glass in the path of a beam of light, the parallel rays get bent in random directions, and after reflection from a normal mirror, each ray of light is bent even farther, and the beam is scattered.



With a phase conjugate mirror, on the other hand, each ray is reflected back in the direction it came from. This reflected conjugate wave therefore propagates backwards through the distorting medium, and essentially "un-does" the distortion, and returns to a coherent beam of parallel rays travelling in the opposite direction."

BOTH THE PHASE-CONJUGATE MIRROR AND THE REGULAR MIRRORS LEAD TO THE VERY SAME RESULTS; ONE, THE PHASE CONJUGATE MIRROR, LEADS TO THE USE OF A MUCH SMALLER INTERFEROMETER, A BETTER TECHNOLOGY PUT TO GOOD USE.



Steven Lehar


http://www.physics.iitm.ac.in/~cvijayan/opc-review.pdf

Optical phase conjugation: principles, techniques, and applications

Over the last three decades, optical phase conjugation (OPC) has been one of the major
research subjects in the field of nonlinear optics and quantum electronics.

Optical phase conjugation (OPC)1 is a new laser-based technique developed since
1970s. As this technique is feasible for use in many significant applications, the study
of OPC has become one of the most active research subjects in the areas of nonlinear
optics and quantum electronics.

In the area of OPC-related studies, a huge number (more than thousands) of
research papers and conference presentations have been published since 1970s.

The use of the phase-conjugate mirror in optical interferometry has produced huge breakthroughs.



Exactly.



Self-pumped phase-conjugate fiber-optic gyro, I. McMichael, P. Yeh, Optics Letters 11(10):686-8 · November 1986 

http://www.dtic.mil/dtic/tr/fulltext/u2/a170203.pdf (appendix 5.1)


Exactly the formula obtained by Professor Yeh:

φ = -2(φ₂ - φ₁) = 4π(R₁L₁ + R₂L₂)Ω/λc = 4π(V₁L₁ + V₂L₂)/λc

Since Δφ = 2πc/λ x Δt, Δt = 2(R₁L₁ + R₂L₂)Ω/c² = 2(V₁L₁ + V₂L₂)/c²

CORRECT SAGNAC FORMULA:

2(V₁L₁ + V₂L₂)/c²


THE VERY SAME FORMULA, OBTAINED FOR A SAGNAC INTERFEROMETER WHICH FEATURES TWO DIFFERENCE LENGTHS AND TWO DIFFERENT VELOCITIES.


That simulator let's the delay for various latitudes be calculated so why not check your "most important formula in physics" against those values.
For example for a loop of the dimensions used by Michelson, Gale and Pearson the delays are:
Latitude     Delay sec      Fringes
      0°         0.00                 0.000
    45°         4.77 x 10-16    0.251
    90°         6.74 x 10-16    0.355 

Now you calculate the Sagnac delay for those dimensions and at latitudes 0°, 45° and 90°. If yours differ markedly from the above please explain why.

These are the data FOR THE CORIOLIS EFFECT formula, the one employed for the simulator, the one derived by Michelson.

The simulator measures the CORIOLIS EFFECT for the MGX, at various latitudes.

http://www.conspiracyoflight.com/Michelson-Gale_webapp/Michelson-Gale.htm

You seem to be a loss recognizing the CORIOLIS EFFECT formula, no need for you to feel lost.

Spinning Earth and its Coriolis effect on the circuital light beams

http://www.ias.ac.in/article/fulltext/pram/087/05/0071

The final formula is this:

4AΩsinΦ/c^2

This is the CORIOLIS EFFECT formula.

That is why the corresponding values for the GLOBAL SAGNAC EFFECT must be much greater than these for the 45° value latitude: the very debate we are having here. At 0° and at 90° latitude, the lengths and velocities are equal again, so the LOCAL SAGNAC EFFECT formula is being applied.

These much higher fringe values are NOT recorded.

All the MGX and RLGs are recording is the CORIOLIS EFFECT.


Let us compare the results for the MGX experiment (1925), same formula featured in the simulator, the same latitude as for Clearing, Illinois.

LATITUDE: 41°46' = 41.76667°

FULL CORIOLIS EFFECT FOR THE MGX:

4AΩsinΦ/c²

FULL SAGNAC EFFECT FOR THE MGX:

4Lv(cos²Φ₁ + cos²Φ₂)/c²


Sagnac effect/Coriolis effect ratio:

R((cos²Φ₁ + cos²Φ₂)/hsinΦ

R = 4,250 km

h = 0.33924 km

The rotational Sagnac effect is much greater than the Coriolis effect for the MGX.

Φ₁ = Φ = 41°46' = 41.76667°

Φ₂ = 41°45' = 41.75°

R((cos²Φ₁ + cos²Φ₂) = 4729.885

hsinΦ = 0.225967

4729.885/0.225967 = 20,931.72

THE ROTATIONAL SAGNAC EFFECT IS 21,000 TIMES GREATER THAN THE CORIOLIS EFFECT.

Michelson and Gale recorded ONLY the Coriolis effect, and not the rotational Sagnac effect.

If the ring laser gyroscope measures/detects the CORIOLIS EFFECT, then the Earth is stationary (existence of the ether drift).

If the ring laser gyroscope measures/detects the SAGNAC EFFECT, then the Earth is rotating (no ether drift).

Since 1913, all interferometers, especially the ring laser gyroscopes have detected ONLY the CORIOLIS EFFECT, nothing else.

It is nigh time to welcome both rabinoz and jackblack as flat earth believers.

No. It is nigh time for you to start addressing the issues raised, rather than repeatedly ignoring them and spouting the same refuted nonsense.

The SAGNAC EFFECT can detect LINEAR/UNIFORM/TRANSLATIONAL MOTION as well.

No it can't.
Don't you recall what you have been repeatedly saying:

Sagnac effect = the rotation of the interferometer

HERE IS THE DEFINITION OF THE SAGNAC EFFECT:
Two pulses of light sent in opposite direction around a closed loop (either circular or a single uniform path), while the interferometer is being rotated'.

Or were you completely wrong with those statements before?

No dealing with more complex setups until you have the basics done.
There is no Sagnac effect for an interferometer undergoing uniform translation motion, just relative translational motion, where different parts of the interferometer are moving relative to each other, which contains key parts which are rotating.

If you can't do the derivation correctly for a stationary loop
But you just stated the opposite:

No I didn't.
You have completely avoided the derivation.
It not producing a shift doesn't mean you can do the derivation, nor does it mean it is pointless.

It provides a baseless, a simple starting point.
If you can't figure it out, anything more is far too complex.

so the total delay for the A > D > C > B > A path (clockwise) should be: l1/(c - v1) + l2/(c + v2)
It can't be.

Technically correct. That is because it completely ignoring 2 of the arms. This would only be the case when arm 2 and arm 4 have a length of 0.
This is because you are assuming that arm 2 and 4 do not contribute to the shift at all, and instead only arm 1 and arm 3 do.
This is the time taken to transit arm 1 and arm 3.

Now, you have TWO OPPOSING DIRECTIONS, while you assign the SAME SIGN

Yes, because they are times taken.
To assign a negative time you would be saying that the light is travelling backwards in time as it propagates along the loop.
According to your "reasoning" if the loop is stationary, it takes no time for the light to travel arm 1 and arm 3 as the light magically moves backwards in arm 3.
This makes absolutely no sense and is the massive error in your analysis.

What in effect you are saying is this:
A > D > C combined with A > B.

NO LOOP AT ALL.

No, that is what you have.
With my derivation you have 2 light paths, one path goes around A>D>C>B>A, with each component adding time taken (but with the vertical arms ignored).
The other path goes A>B>C>D>A.
This produces the 2 times provided:
l₁/(c - v₁)+l₂/(c + v₂)
l₁/(c + v₁)+l₂/(c - v₂)

Meanwhile, you have what amounts to this nonsense:
{l₁/(c - v₁) + l₂/(c - v₂)} -  {l₁/(c + v₁) + l₂/(c + v₂)}

This is meant to be a time difference, so it amounts to one path of light, with a time of:
l₁/(c - v₁) + l₂/(c - v₂)
and a second path with a time of:
l₁/(c + v₁) + l₂/(c + v₂)

But notice how they have the same c+v or c-v?
This means the light is propogating in the same direction relative to the velocity of the loop.
i.e. you have a setup like this:

You have your light travel along B>C, but then magically go A>D.
This makes no sense. You are not having your light travel along the loop, and you are completely ignoring the effects of the crossing paths.

So no, I'm not the one with the complete nonsensical paths. That would be you entirely.
Either you have paths which do not correspond to the interferometer, or you have your light magically travelling backwards in time so it arrives before it leaves.

This is why I am demanding you start with the basics, a simple stationary loop.
Until you can get your head around a stationary loop, with only positive times, anything more complex is well beyond you.

So again:
Can you figure out how long it takes for the light to propagate around a stationary loop?
If so, provide the derivation here. If not, give up as you have no chance of understanding the Sagnac effect.

Sure, for the case being discussed here we have rotation (MGX).

Sagnac effect while in uniform/translational/linear motion is a proven fact.

Professor Ruyong Wang, in two well-designed experiments showed unambiguously that an identical Sagnac effect appearing in uniform radial motion occurs in linear inertial motion.

He tested the travel-time difference between two counter-propagating light beams in uniformly moving fiber.

The travel-time difference was found to be:

Δt = 2vΔL/c^2

where ΔL is the length of the fiber segment moving with the source and detector at a v, whether the segment was moving uniformly or circularly.



https://arxiv.org/ftp/physics/papers/0609/0609222.pdf (first experiment conducted by R. Wang)

https://arxiv.org/ftp/physics/papers/0609/0609202.pdf (second experiment carried out by R. Wang)

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

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

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

Here is another demonstration.




To assign a negative time you would be saying that the light is travelling backwards in time as it propagates along the loop.

You don't have a loop and there is no negative time.

You simply do not understand the SAGNAC EFFECT.

The RE standard for the Sagnac effect:

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

If two pulses of light are sent in opposite directions around a stationary circular loop of radius R, they will travel the same inertial distance at the same speed, so they will arrive at the end point simultaneously. This is illustrated in the left-hand figure below.



If the interferometer is being rotated, both pulses begin with an initial separation of 2piR from the end point, so the difference between the travel times is:



Opposite directions, therefore WE SUBSTRACT THE DIFFERENCE IN TIME TRAVEL.

Moreover, we are dealing with TWO LOOPS.

Can everyone understand that the differences in time travel have to be substracted?

This is the correct way to derive the Sagnac formula:

Sagnac phase component for the clockwise path:

2πR(1/(c - v))

Sagnac phase component for the counterclockwise path:

-2πR(1/(c + v))

The continuous clockwise loop has a positive sign +

The continuous counterclockwise loop has a negative sign -

Good.

That is, if we want to find out the difference in travel times (opposite directions) we must substract them.


For an interferometer which is now located AWAY FROM THE CENTER OF ROTATION, the situation is a bit more complicated, but the same principle applies.



Point A is located at the detector
Point B is in the bottom right corner
Point C is in the upper right corner
Point D is in the upper left corner

l₁ is the upper arm.
l₂ is the lower arm.

Let us remember that now we are dealing with DIFFERENT VELOCITIES for each arm, and DIFFERENT LENGTHS of each arm, a situation a bit more complex than the previous case analyzed here.


We need to designate the TWO LOOPS, as required by the definition of the Sagnac effect.

HERE IS THE DEFINITION OF THE SAGNAC EFFECT:

Two pulses of light sent in opposite direction around a closed loop (either circular or a single uniform path), while the interferometer is being rotated.

Loop = a structure, series, or process, the end of which is connected to the beginning.

A single continuous pulse A > B > C > D > A, while the other one, A > D > C > B > A is in the opposite direction, and has the negative sign.


So, for the first loop, the clockwise path, the A > D > C > B > A path, we have to deal with beams which are traveling IN OPPOSITE DIRECTIONS, that is, in order to find out the total time travel we need to substract the time differences, just like we did the first time: in effect we are adding two transit times, one of which is traveling in a opposite direction to the first, hence the opposite signs.

We substracted the time differences the first time around for the interferometer whose center of rotation coincides with its geometric center.

Now, we have a loop consisting of two different paths, which travel in opposite directions.

Therefore, to get the TOTAL TIME DIFFERENCE FOR THE CLOCKWISE PATH, we substract the time differences: again, in effect we are adding the transit times, but since one of them has an opposite direction, it will have a different sign than the first transit time, just like in the first example of the Sagnac interferometer.

Very simple, and at the same time we are dealing with a LOOP, as required by the defintion of the Sagnac effect.

Sagnac phase components for the A > D > C > B > A path (clockwise path):

l₁/(c - v₁)

-l₂/(c + v₂)

Now, we do the same thing for the counterclockwise path, the A > B > C > D > A path:

l₂/(c - v₂)

-l₁/(c + v₁)


For the single continuous clockwise path we now have the total time difference:

l₁/(c - v₁) - l₂/(c + v₂)


For the single continuous counterclockwise path we have the total difference:

l₂/(c - v₂) - l₁/(c + v₁)


TWO LOOPS as required by the definition of the Sagnac effect.

If we change the sign of the second term/phase component to +, that is:

l₁/(c - v₁)

l₂/(c + v₂)

then, we no longer have a LOOP, and moreover we are using the wrong sign for the direction of the second transit time; each transit time has a different direction, hence we must use opposite signs to correctly designate them in our analysis.

Let us remember the very defintion of the Sagnac effect: two loops are required to properly derive the formula.


Now, to obtain the final answer, WE SUBSTRACT THE TOTAL TIME DIFFERENCES FOR EACH PATH, since we are dealing with a counterclockwise path and a clockwise path, if we want the time phase, we need to substract the total time differences for each LOOP. Each loop has a different direction, as such it must have a different sign assigned to it.

The net phase difference will be (let us remember that the counterclockwise phase difference has a negative sign attached to it, that is why the substraction of the phase differences becomes an addition):

{l₁/(c - v₁) - l₂/(c + v₂)} - (-){l₂/(c - v₂) - l₁/(c + v₁)} = {l₁/(c - v₁) - l₂/(c + v₂)} + {l₂/(c - v₂) - l₁/(c + v₁)}

Rearranging terms:

l₁/(c - v₁) - l₁/(c + v₁) + {l₂/(c - v₂) - l₂/(c + v₂)} =

2(v₁l₁ + v₂l₂)/c²

Exactly the formula obtained by Professor Yeh:

φ = -2(φ₂ - φ₁) = 4π(R₁L₁ + R₂L₂)Ω/λc = 4π(V₁L₁ + V₂L₂)/λc

Since Δφ = 2πc/λ x Δt, Δt = 2(R₁L₁ + R₂L₂)Ω/c² = 2(V₁L₁ + V₂L₂)/c²

CORRECT SAGNAC FORMULA:

2(V₁L₁ + V₂L₂)/c²


A stationary loop does not involve the Sagnac effect. For the Sagnac effect only arms l1 and l2 come into play, as simple as this. Only the arms involved in the rotation analysis are used, and the Sagnac effect has nothing to do with a stationary interferometer. You are trolling the upper forums, desperately trying to avoid the final conclusion: I am right, you are wrong.

The analysis for the Sagnac effect is completely different than for a stationary interferometer, yet this very simple fact seems to escape your attention.

According to your "reasoning" if the loop is stationary, it takes no time for the light to travel arm 1 and arm 3 as the light magically moves backwards in arm 3.

We are not concerned here AT ALL with the stationary case: my analysis only applies to the rotational case. In the SAGNAC EFFECT, only arms l1 and l2 come into play to provide the time differences: we are now rotating the interferometer. What you are stating to your readers is that you do not understand the workings of the Sagnac effect.




With my derivation you have 2 light paths, one path goes around A>D>C>B>A, with each component adding time taken (but with the vertical arms ignored).
The other path goes A>B>C>D>A.
This produces the 2 times provided:
l1/(c - v1)+l2/(c + v2)
l1/(c + v1)+l2/(c - v2)

You are lying through your teeth jack, and that is not nice.

What is you have is this:

Path 1 - A>B, D>C.
Path 2 - C>D, B>A

YOU ARE DISREGARDING THE CORRECT DEFINITION OF THE SAGNAC EFFECT IN FULL VIEW.

l1/(c - v1)+l2/(c + v2)

You are comparing two sides, NOT TWO LOOPS.

You admitted that those beams have opposite directions.

Then, we have the remaining terms, in the opposite direction, likewise meaning one is for orange one is for red, noting that red travelled along l1 in the previous one so now it must travel along l2 in this one:
l1/(c + v1)
l2/(c + v2)

Then, if you have opposite directions, you must use different signs.

A humongous error on your part.

Here is the correct analysis:

Sagnac phase components for the A > D > C > B > A path (clockwise path):

l1/(c - v1)

-l2/(c + v2)

Sagnac phase components for the A > B > C > D > A path (counterclockwise path):

l2/(c - v2)

-l1/(c + v1)


For the single continuous clockwise path we add the components:

l1/(c - v1) - l2/(c + v2)

For the single continuous counterclockwise path we add the components:

l2/(c - v2) - l1/(c + v1)


jackblack assigned the SAME SIGN, even though he just said a few lines earlier, that they are in fact in opposite direction.


This is meant to be a time difference, so it amounts to one path of light, with a time of:
l1/(c - v1) + l2/(c - v2)
and a second path with a time of:
l1/(c + v1) + l2/(c + v2)

THOSE ARE NOT TWO LOOPS, BUT TWO SEGMENTS.

I am not comparing two segments at all, you are.

The correct way is to compare two loops.

The definition of the Sagnac effect involves two loops.

The paths are very clear:

A > B > C > D > A is a continuous counterclockwise path, a negative sign -
A > D > C > B > A is a continuous clockwise path, a positive sign +

The net phase difference will be (let us remember that the counterclockwise phase difference has a negative sign attached to it, that is why the substraction of the phase differences becomes an addition):

{l₁/(c - v₁) - l₂/(c + v₂)} - (-){l₂/(c - v₂) - l₁/(c + v₁)} = {l₁/(c - v₁) - l₂/(c + v₂)} + {l₂/(c - v₂) - l₁/(c + v₁)}

Rearranging terms:

l₁/(c - v₁) - l₁/(c + v₁) + {l₂/(c - v₂) - l₂/(c + v₂)} =

2(v₁l₁ + v₂l₂)/c²

Exactly the formula obtained by Professor Yeh:

φ = -2(φ₂ - φ₁) = 4π(R₁L₁ + R₂L₂)Ω/λc = 4π(V₁L₁ + V₂L₂)/λc

Since Δφ = 2πc/λ x Δt, Δt = 2(R₁L₁ + R₂L₂)Ω/c² = 2(V₁L₁ + V₂L₂)/c²

CORRECT SAGNAC FORMULA:

2(V₁L₁ + V₂L₂)/c²


My formula is in complete agreement with the formula published by Professor Yeh in the Journal of Optics Letters.

Your formula is in complete agreement with the published CORIOLIS EFFECT formula:

Spinning Earth and its Coriolis effect on the circuital light beams

http://www.ias.ac.in/article/fulltext/pram/087/05/0071

The final formula is this:

4AΩsinΦ/c^2

This is the CORIOLIS EFFECT formula, not the SAGNAC EFFECT formula.


You are about to become a flat earth believer, no doubt about that.

Sagnac effect while in uniform/translational/linear motion is a proven fact.

No it isn't.
It has never been observed in a system that is moving with uniform translation motion.
But we are not discussing that.
Like I said, start with the basics.

Can everyone understand that the differences in time travel have to be substracted?

No, we need to find the difference in travel time.
We aren't dealing with multiple time differences (not with how you are trying to derive it, we would if you wanted to focus on the shift from each leg and add them up).

So, for the first loop, the clockwise path, the A > D > C > B > A path, we have to deal with beams which are traveling IN OPPOSITE DIRECTIONS, that is, in order to find out the total time travel we need to substract the time differences

No we don't.
No one in the right mind does that.
To find the total travel time we need to add up the individual travel times. There are no time differences here.
A time difference is the final result.

This is why I am saying to start with a stationary loop.
You cannot even understand something as simple as adding times taken for multiple steps to find the total time.
According to your insanity, the time taken to travel around a stationary loop is 0.
Do you really think that is true?

So now, like I said, start with the basics.
How long does it take light to travel around a stationary loop?

Let us see just how easy it is to catch jackblack lying:

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, you are admitting that in the case of rotation the only thing giving rise to a difference will be the arms l1 and l2.

Exactly what I said repeatedly.

Yet, you are asking for a stationary loop which is not the SAGNAC EFFECT.

You are trolling the upper forums.

To find the total travel time we need to add up the individual travel times. There are no time differences here.

Certainly there are time differences, different transit times, since we are dealing with two arms which feature different velocities and different lengths.

YOU STILL DO NOT HAVE TWO LOOPS, which is the SAGNAC EFFECT DEFINITION.

I win.

To assign a negative time you would be saying that the light is travelling backwards in time as it propagates along the loop.

You don't have a loop and there is no negative time.

You simply do not understand the SAGNAC EFFECT.

The RE standard for the Sagnac effect:

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

If two pulses of light are sent in opposite directions around a stationary circular loop of radius R, they will travel the same inertial distance at the same speed, so they will arrive at the end point simultaneously. This is illustrated in the left-hand figure below.



If the interferometer is being rotated, both pulses begin with an initial separation of 2piR from the end point, so the difference between the travel times is:



Opposite directions, therefore WE SUBSTRACT THE DIFFERENCE IN TIME TRAVEL.

Moreover, we are dealing with TWO LOOPS.

Can everyone understand that the differences in time travel have to be substracted?

This is the correct way to derive the Sagnac formula:

Sagnac phase component for the clockwise path:

2πR(1/(c - v))

Sagnac phase component for the counterclockwise path:

-2πR(1/(c + v))

The continuous clockwise loop has a positive sign +

The continuous counterclockwise loop has a negative sign -

Good.

That is, if we want to find out the difference in travel times (opposite directions) we must substract them.


For an interferometer which is now located AWAY FROM THE CENTER OF ROTATION, the situation is a bit more complicated, but the same principle applies.



Point A is located at the detector
Point B is in the bottom right corner
Point C is in the upper right corner
Point D is in the upper left corner

l₁ is the upper arm.
l₂ is the lower arm.

Let us remember that now we are dealing with DIFFERENT VELOCITIES for each arm, and DIFFERENT LENGTHS of each arm, a situation a bit more complex than the previous case analyzed here.


We need to designate the TWO LOOPS, as required by the definition of the Sagnac effect.

HERE IS THE DEFINITION OF THE SAGNAC EFFECT:

Two pulses of light sent in opposite direction around a closed loop (either circular or a single uniform path), while the interferometer is being rotated.

Loop = a structure, series, or process, the end of which is connected to the beginning.

A single continuous pulse A > B > C > D > A, while the other one, A > D > C > B > A is in the opposite direction, and has the negative sign.


So, for the first loop, the clockwise path, the A > D > C > B > A path, we have to deal with beams which are traveling IN OPPOSITE DIRECTIONS, that is, in order to find out the total time travel we need to substract the time differences, just like we did the first time: in effect we are adding two transit times, one of which is traveling in a opposite direction to the first, hence the opposite signs.

We substracted the time differences the first time around for the interferometer whose center of rotation coincides with its geometric center.

Now, we have a loop consisting of two different paths, which travel in opposite directions.

Therefore, to get the TOTAL TIME DIFFERENCE FOR THE CLOCKWISE PATH, we substract the time differences: again, in effect we are adding the transit times, but since one of them has an opposite direction, it will have a different sign than the first transit time, just like in the first example of the Sagnac interferometer.

Very simple, and at the same time we are dealing with a LOOP, as required by the defintion of the Sagnac effect.

Sagnac phase components for the A > D > C > B > A path (clockwise path):

l₁/(c - v₁)

-l₂/(c + v₂)

Now, we do the same thing for the counterclockwise path, the A > B > C > D > A path:

l₂/(c - v₂)

-l₁/(c + v₁)


For the single continuous clockwise path we now have the total time difference:

l₁/(c - v₁) - l₂/(c + v₂)


For the single continuous counterclockwise path we have the total difference:

l₂/(c - v₂) - l₁/(c + v₁)


TWO LOOPS as required by the definition of the Sagnac effect.

If we change the sign of the second term/phase component to +, that is:

l₁/(c - v₁)

l₂/(c + v₂)

then, we no longer have a LOOP, and moreover we are using the wrong sign for the direction of the second transit time; each transit time has a different direction, hence we must use opposite signs to correctly designate them in our analysis.

Let us remember the very defintion of the Sagnac effect: two loops are required to properly derive the formula.


Now, to obtain the final answer, WE SUBSTRACT THE TOTAL TIME DIFFERENCES FOR EACH PATH, since we are dealing with a counterclockwise path and a clockwise path, if we want the time phase, we need to substract the total time differences for each LOOP. Each loop has a different direction, as such it must have a different sign assigned to it.

The net phase difference will be (let us remember that the counterclockwise phase difference has a negative sign attached to it, that is why the substraction of the phase differences becomes an addition):

{l₁/(c - v₁) - l₂/(c + v₂)} - (-){l₂/(c - v₂) - l₁/(c + v₁)} = {l₁/(c - v₁) - l₂/(c + v₂)} + {l₂/(c - v₂) - l₁/(c + v₁)}

Rearranging terms:

l₁/(c - v₁) - l₁/(c + v₁) + {l₂/(c - v₂) - l₂/(c + v₂)} =

2(v₁l₁ + v₂l₂)/c²

Exactly the formula obtained by Professor Yeh:

φ = -2(φ₂ - φ₁) = 4π(R₁L₁ + R₂L₂)Ω/λc = 4π(V₁L₁ + V₂L₂)/λc

Since Δφ = 2πc/λ x Δt, Δt = 2(R₁L₁ + R₂L₂)Ω/c² = 2(V₁L₁ + V₂L₂)/c²

CORRECT SAGNAC FORMULA:

2(V₁L₁ + V₂L₂)/c²


A stationary loop does not involve the Sagnac effect. For the Sagnac effect only arms l1 and l2 come into play, as simple as this. Only the arms involved in the rotation analysis are used, and the Sagnac effect has nothing to do with a stationary interferometer. You are trolling the upper forums, desperately trying to avoid the final conclusion: I am right, you are wrong.

The analysis for the Sagnac effect is completely different than for a stationary interferometer, yet this very simple fact seems to escape your attention.

According to your "reasoning" if the loop is stationary, it takes no time for the light to travel arm 1 and arm 3 as the light magically moves backwards in arm 3.

We are not concerned here AT ALL with the stationary case: my analysis only applies to the rotational case. In the SAGNAC EFFECT, only arms l1 and l2 come into play to provide the time differences: we are now rotating the interferometer. What you are stating to your readers is that you do not understand the workings of the Sagnac effect.




With my derivation you have 2 light paths, one path goes around A>D>C>B>A, with each component adding time taken (but with the vertical arms ignored).
The other path goes A>B>C>D>A.
This produces the 2 times provided:
l1/(c - v1)+l2/(c + v2)
l1/(c + v1)+l2/(c - v2)

You are lying through your teeth jack, and that is not nice.

What is you have is this:

Path 1 - A>B, D>C.
Path 2 - C>D, B>A

YOU ARE DISREGARDING THE CORRECT DEFINITION OF THE SAGNAC EFFECT IN FULL VIEW.

l1/(c - v1)+l2/(c + v2)

You are comparing two sides, NOT TWO LOOPS.

You admitted that those beams have opposite directions.

Then, we have the remaining terms, in the opposite direction, likewise meaning one is for orange one is for red, noting that red travelled along l1 in the previous one so now it must travel along l2 in this one:
l1/(c + v1)
l2/(c + v2)

Then, if you have opposite directions, you must use different signs.

A humongous error on your part.

Here is the correct analysis:

Sagnac phase components for the A > D > C > B > A path (clockwise path):

l1/(c - v1)

-l2/(c + v2)

Sagnac phase components for the A > B > C > D > A path (counterclockwise path):

l2/(c - v2)

-l1/(c + v1)


For the single continuous clockwise path we add the components:

l1/(c - v1) - l2/(c + v2)

For the single continuous counterclockwise path we add the components:

l2/(c - v2) - l1/(c + v1)


jackblack assigned the SAME SIGN, even though he just said a few lines earlier, that they are in fact in opposite direction.


This is meant to be a time difference, so it amounts to one path of light, with a time of:
l1/(c - v1) + l2/(c - v2)
and a second path with a time of:
l1/(c + v1) + l2/(c + v2)

THOSE ARE NOT TWO LOOPS, BUT TWO SEGMENTS.

I am not comparing two segments at all, you are.

The correct way is to compare two loops.

The definition of the Sagnac effect involves two loops.

The paths are very clear:

A > B > C > D > A is a continuous counterclockwise path, a negative sign -
A > D > C > B > A is a continuous clockwise path, a positive sign +

The net phase difference will be (let us remember that the counterclockwise phase difference has a negative sign attached to it, that is why the substraction of the phase differences becomes an addition):

{l₁/(c - v₁) - l₂/(c + v₂)} - (-){l₂/(c - v₂) - l₁/(c + v₁)} = {l₁/(c - v₁) - l₂/(c + v₂)} + {l₂/(c - v₂) - l₁/(c + v₁)}

Rearranging terms:

l₁/(c - v₁) - l₁/(c + v₁) + {l₂/(c - v₂) - l₂/(c + v₂)} =

2(v₁l₁ + v₂l₂)/c²

Exactly the formula obtained by Professor Yeh:

φ = -2(φ₂ - φ₁) = 4π(R₁L₁ + R₂L₂)Ω/λc = 4π(V₁L₁ + V₂L₂)/λc

Since Δφ = 2πc/λ x Δt, Δt = 2(R₁L₁ + R₂L₂)Ω/c² = 2(V₁L₁ + V₂L₂)/c²

CORRECT SAGNAC FORMULA:

2(V₁L₁ + V₂L₂)/c²


My formula is in complete agreement with the formula published by Professor Yeh in the Journal of Optics Letters.

Your formula is in complete agreement with the published CORIOLIS EFFECT formula:

Spinning Earth and its Coriolis effect on the circuital light beams

http://www.ias.ac.in/article/fulltext/pram/087/05/0071

The final formula is this:

4AΩsinΦ/c^2

This is the CORIOLIS EFFECT formula, not the SAGNAC EFFECT formula.

and that is only for his Phase Conjugate Mirror fibre optic gyroscope so is totally irrelevant here!
And I notice in your latest post that you again show that you don't know the difference between a normal Sagnac Loop and Dr. P. Yeh's Phase Conjugate Mirror fibre optic gyroscope!

YOU claim that:

BOTH THE PHASE-CONJUGATE MIRROR AND THE REGULAR MIRRORS LEAD TO THE VERY SAME RESULTS; ONE,.

Optical phase conjugation: principles, techniques, and applications Guang S. He

Sure the PCM Fibre OPtic Gyroscope both measure rotation but only YOU seem to claim that the expressions for delay are the same!

So please post evidence for your claim that "BOTH THE PHASE-CONJUGATE MIRROR AND THE REGULAR MIRRORS LEAD TO THE VERY SAME" expression for the Sagnac delay.

And don't you dare just repeat your own incorrect "analysis" - that is not evidence.

You mean I also have to explain the physics of the phase-conjugate mirror to you?

No you do not! You have to answer the above.

I choose not believe YOU no matter how many times you post the same old rubbish!

Let us see just how easy it is to catch jackblack lying:

You mean lets see how easy it is to lie and pretend I am lying.

So, you are admitting that in the case of rotation the only thing giving rise to a difference will be the arms l1 and l2.

In that particular case, where it is an annular sector, not a rectangular loop.

Yet, you are asking for a stationary loop which is not the SAGNAC EFFECT.

Again, I am asking for it as a baseline.
The Sagnac effect deals with time differences in beams of light propagating around loops.
As such, even though a stationary loop does not produce the Sagnac effect, it is still very relevant.
According to you, it takes no time for the light to propagate around the loop.
That is pure insanity.

Again, your error has been pointed out repeatedly and you are yet to rationally address it.
As you seem to be completely unwilling to address the issues and instead insist upon repeating the same refuted nonsense, I will keep asking for the basics.

Now can you provide the derivation for how long it take light to propogate around a stationary loop?
If not, then you shouldn't be trying anything more complex.
You continually avoiding this very simple task if you trolling the fora.

Sure, for the case being discussed here we have rotation (MGX).

Sagnac effect while in uniform/translational/linear motion is a proven fact.

Sure but only relative "uniform/translational/linear motion" never yet any measured absolute "uniform/translational/linear motion".

Professor Ruyong Wang, in two well-designed experiments showed unambiguously that an identical Sagnac effect appearing in uniform radial motion occurs in linear inertial motion.
He tested the travel-time difference between two counter-propagating light beams in uniformly moving fiber.
The travel-time difference was found to be:
Δt = 2vΔL/c^2

where ΔL is the length of the fiber segment moving with the source and detector at a v, whether the segment was moving uniformly or circularly.

Yes, but all those are just segments of the fibre and a Wang himself says.

https://arxiv.org/ftp/physics/papers/0609/0609222.pdf (first experiment conducted by R. Wang)
https://arxiv.org/ftp/physics/papers/0609/0609202.pdf (second experiment carried out by R. Wang)

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

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

As I said, the "difference of two counterpropagating light beams" in a loop of fibre.

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

<< There's no point in repeating the same old.  >>

None of this is measuring absolute "uniform/translational/linear motion".

Now read the conclusion to:
Test of the one-way speed of light and the first-order experiment of Special Relativity using phase-conjugate interferometers by Ruyong Wang, Yi Zheng and Aiping Yao

7. CONCLUSIONS

The phase-conjugate Sagnac experiment can be repeated and the one-way Sagnac experiment can be conducted using the PCM. We can expect that the phase shift is φ = 4πvL/cλ in the one-way Sagnac experiment with path length L and speed v, even with an increasingly larger radius of the rotation.

Based on these and the experimental fact of the generalized Sagnac effect, it is very important to examine whether there is the same phase shift for the test of the one-way speed of light and the phase-conjugate first-order experiment in a system moving uniformly in a straight line. The sensitivities of these experiments are very high.

He does describe a test in the conclusion but says it "can be repeated and the one-way Sagnac experiment can be conducted using the PCM".

Up until the date of that paper there seemed to be no "Test of the one-way speed of light and the first-order experiment of Special Relativity" by Ruyong Wang, Yi Zheng and Aiping and their phase-conjugate interferometers.

So what is your point in claiming "Sagnac effect while in uniform/translational/linear motion is a proven fact."
And a Sagnac device must have a fixed geometry that is rotated or move about an off centre axis.
Ruyong Wang's devices measuring movement have a varying geometry and are not designed to measure rotation.

Either Sandy knows what he has is bunk and so spends all his time spamming this forum trying to convince others he isn't as dumb as he writes.

Or he has tried to publish all his bunk and got shot down and so spends all his time spamming this forum trying to convince others he isn't as dumb as he writes.

My formula is totally up to date.

 This is not about formula. You are out of date with most of your sources. And if you quote some contemporary ones they don't support your ideas and sometimes contradict with your earlier and outdated sources. You just quote them for... I don't know for what reasons.

Quote from: sandokhan on February 21, 2019, 03:13:02 AM

My formula is totally up to date.

 This is not about formula. You are out of date with most of your sources. And if you quote some contemporary ones they don't support your ideas and sometimes contradict with your earlier and outdated sources. You just quote them for... I don't know for what reasons.

Obfuscation.

Let us see just how easy it is to catch jackblack lying:

Quote from: JackBlack on June 02, 2017, 06:06:27 PM

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, you are admitting that in the case of rotation the only thing giving rise to a difference will be the arms l1 and l2.

Exactly what I said repeatedly.

Yet, you are asking for a stationary loop which is not the SAGNAC EFFECT.

You are trolling the upper forums.

To find the total travel time we need to add up the individual travel times. There are no time differences here.

Certainly there are time differences, different transit times, since we are dealing with two arms which feature different velocities and different lengths.

YOU STILL DO NOT HAVE TWO LOOPS, which is the SAGNAC EFFECT DEFINITION.

I win.

To assign a negative time you would be saying that the light is travelling backwards in time as it propagates along the loop.

You don't have a loop and there is no negative time.

You simply do not understand the SAGNAC EFFECT.

The RE standard for the Sagnac effect:

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

If two pulses of light are sent in opposite directions around a stationary circular loop of radius R, they will travel the same inertial distance at the same speed, so they will arrive at the end point simultaneously. This is illustrated in the left-hand figure below.



If the interferometer is being rotated, both pulses begin with an initial separation of 2piR from the end point, so the difference between the travel times is:



Opposite directions, therefore WE SUBSTRACT THE DIFFERENCE IN TIME TRAVEL.

Moreover, we are dealing with TWO LOOPS.

Can everyone understand that the differences in time travel have to be substracted?

This is the correct way to derive the Sagnac formula:

Sagnac phase component for the clockwise path:

2πR(1/(c - v))

Sagnac phase component for the counterclockwise path:

-2πR(1/(c + v))

The continuous clockwise loop has a positive sign +

The continuous counterclockwise loop has a negative sign -

Good.

That is, if we want to find out the difference in travel times (opposite directions) we must substract them.


For an interferometer which is now located AWAY FROM THE CENTER OF ROTATION, the situation is a bit more complicated, but the same principle applies.



Point A is located at the detector
Point B is in the bottom right corner
Point C is in the upper right corner
Point D is in the upper left corner

l₁ is the upper arm.
l₂ is the lower arm.

Let us remember that now we are dealing with DIFFERENT VELOCITIES for each arm, and DIFFERENT LENGTHS of each arm, a situation a bit more complex than the previous case analyzed here.


We need to designate the TWO LOOPS, as required by the definition of the Sagnac effect.

HERE IS THE DEFINITION OF THE SAGNAC EFFECT:

Two pulses of light sent in opposite direction around a closed loop (either circular or a single uniform path), while the interferometer is being rotated.

Loop = a structure, series, or process, the end of which is connected to the beginning.

A single continuous pulse A > B > C > D > A, while the other one, A > D > C > B > A is in the opposite direction, and has the negative sign.


So, for the first loop, the clockwise path, the A > D > C > B > A path, we have to deal with beams which are traveling IN OPPOSITE DIRECTIONS, that is, in order to find out the total time travel we need to substract the time differences, just like we did the first time: in effect we are adding two transit times, one of which is traveling in a opposite direction to the first, hence the opposite signs.

We substracted the time differences the first time around for the interferometer whose center of rotation coincides with its geometric center.

Now, we have a loop consisting of two different paths, which travel in opposite directions.

Therefore, to get the TOTAL TIME DIFFERENCE FOR THE CLOCKWISE PATH, we substract the time differences: again, in effect we are adding the transit times, but since one of them has an opposite direction, it will have a different sign than the first transit time, just like in the first example of the Sagnac interferometer.

Very simple, and at the same time we are dealing with a LOOP, as required by the defintion of the Sagnac effect.

Sagnac phase components for the A > D > C > B > A path (clockwise path):

l₁/(c - v₁)

-l₂/(c + v₂)

Now, we do the same thing for the counterclockwise path, the A > B > C > D > A path:

l₂/(c - v₂)

-l₁/(c + v₁)


For the single continuous clockwise path we now have the total time difference:

l₁/(c - v₁) - l₂/(c + v₂)


For the single continuous counterclockwise path we have the total difference:

l₂/(c - v₂) - l₁/(c + v₁)


TWO LOOPS as required by the definition of the Sagnac effect.

If we change the sign of the second term/phase component to +, that is:

l₁/(c - v₁)

l₂/(c + v₂)

then, we no longer have a LOOP, and moreover we are using the wrong sign for the direction of the second transit time; each transit time has a different direction, hence we must use opposite signs to correctly designate them in our analysis.

Let us remember the very defintion of the Sagnac effect: two loops are required to properly derive the formula.


Now, to obtain the final answer, WE SUBSTRACT THE TOTAL TIME DIFFERENCES FOR EACH PATH, since we are dealing with a counterclockwise path and a clockwise path, if we want the time phase, we need to substract the total time differences for each LOOP. Each loop has a different direction, as such it must have a different sign assigned to it.

The net phase difference will be (let us remember that the counterclockwise phase difference has a negative sign attached to it, that is why the substraction of the phase differences becomes an addition):

{l₁/(c - v₁) - l₂/(c + v₂)} - (-){l₂/(c - v₂) - l₁/(c + v₁)} = {l₁/(c - v₁) - l₂/(c + v₂)} + {l₂/(c - v₂) - l₁/(c + v₁)}

Rearranging terms:

l₁/(c - v₁) - l₁/(c + v₁) + {l₂/(c - v₂) - l₂/(c + v₂)} =

2(v₁l₁ + v₂l₂)/c²

Exactly the formula obtained by Professor Yeh:

φ = -2(φ₂ - φ₁) = 4π(R₁L₁ + R₂L₂)Ω/λc = 4π(V₁L₁ + V₂L₂)/λc

Since Δφ = 2πc/λ x Δt, Δt = 2(R₁L₁ + R₂L₂)Ω/c² = 2(V₁L₁ + V₂L₂)/c²

CORRECT SAGNAC FORMULA:

2(V₁L₁ + V₂L₂)/c²


A stationary loop does not involve the Sagnac effect. For the Sagnac effect only arms l1 and l2 come into play, as simple as this. Only the arms involved in the rotation analysis are used, and the Sagnac effect has nothing to do with a stationary interferometer. You are trolling the upper forums, desperately trying to avoid the final conclusion: I am right, you are wrong.

The analysis for the Sagnac effect is completely different than for a stationary interferometer, yet this very simple fact seems to escape your attention.

According to your "reasoning" if the loop is stationary, it takes no time for the light to travel arm 1 and arm 3 as the light magically moves backwards in arm 3.

We are not concerned here AT ALL with the stationary case: my analysis only applies to the rotational case. In the SAGNAC EFFECT, only arms l1 and l2 come into play to provide the time differences: we are now rotating the interferometer. What you are stating to your readers is that you do not understand the workings of the Sagnac effect.




With my derivation you have 2 light paths, one path goes around A>D>C>B>A, with each component adding time taken (but with the vertical arms ignored).
The other path goes A>B>C>D>A.
This produces the 2 times provided:
l1/(c - v1)+l2/(c + v2)
l1/(c + v1)+l2/(c - v2)

You are lying through your teeth jack, and that is not nice.

What is you have is this:

Path 1 - A>B, D>C.
Path 2 - C>D, B>A

YOU ARE DISREGARDING THE CORRECT DEFINITION OF THE SAGNAC EFFECT IN FULL VIEW.

l1/(c - v1)+l2/(c + v2)

You are comparing two sides, NOT TWO LOOPS.

You admitted that those beams have opposite directions.

Then, we have the remaining terms, in the opposite direction, likewise meaning one is for orange one is for red, noting that red travelled along l1 in the previous one so now it must travel along l2 in this one:
l1/(c + v1)
l2/(c + v2)

Then, if you have opposite directions, you must use different signs.

A humongous error on your part.

Here is the correct analysis:

Sagnac phase components for the A > D > C > B > A path (clockwise path):

l1/(c - v1)

-l2/(c + v2)

Sagnac phase components for the A > B > C > D > A path (counterclockwise path):

l2/(c - v2)

-l1/(c + v1)


For the single continuous clockwise path we add the components:

l1/(c - v1) - l2/(c + v2)

For the single continuous counterclockwise path we add the components:

l2/(c - v2) - l1/(c + v1)


jackblack assigned the SAME SIGN, even though he just said a few lines earlier, that they are in fact in opposite direction.


This is meant to be a time difference, so it amounts to one path of light, with a time of:
l1/(c - v1) + l2/(c - v2)
and a second path with a time of:
l1/(c + v1) + l2/(c + v2)

THOSE ARE NOT TWO LOOPS, BUT TWO SEGMENTS.

I am not comparing two segments at all, you are.

The correct way is to compare two loops.

The definition of the Sagnac effect involves two loops.

The paths are very clear:

A > B > C > D > A is a continuous counterclockwise path, a negative sign -
A > D > C > B > A is a continuous clockwise path, a positive sign +

The net phase difference will be (let us remember that the counterclockwise phase difference has a negative sign attached to it, that is why the substraction of the phase differences becomes an addition):

{l₁/(c - v₁) - l₂/(c + v₂)} - (-){l₂/(c - v₂) - l₁/(c + v₁)} = {l₁/(c - v₁) - l₂/(c + v₂)} + {l₂/(c - v₂) - l₁/(c + v₁)}

Rearranging terms:

l₁/(c - v₁) - l₁/(c + v₁) + {l₂/(c - v₂) - l₂/(c + v₂)} =

2(v₁l₁ + v₂l₂)/c²

Exactly the formula obtained by Professor Yeh:

φ = -2(φ₂ - φ₁) = 4π(R₁L₁ + R₂L₂)Ω/λc = 4π(V₁L₁ + V₂L₂)/λc

Since Δφ = 2πc/λ x Δt, Δt = 2(R₁L₁ + R₂L₂)Ω/c² = 2(V₁L₁ + V₂L₂)/c²

CORRECT SAGNAC FORMULA:

2(V₁L₁ + V₂L₂)/c²


My formula is in complete agreement with the formula published by Professor Yeh in the Journal of Optics Letters.

Your formula is in complete agreement with the published CORIOLIS EFFECT formula:

Spinning Earth and its Coriolis effect on the circuital light beams

http://www.ias.ac.in/article/fulltext/pram/087/05/0071

The final formula is this:

4AΩsinΦ/c^2

This is the CORIOLIS EFFECT formula, not the SAGNAC EFFECT formula.

I’m not sure if Jack Black lies, but I know for a fact that you do. Not only do you lie constantly but you also twist and distort.
How many of the scientists you regularly misquote were or are flat earth believers?......let me answer thatfor you......none.

You tell lies about about science. You tell lies about maths and you tell lies about history.
It looks like you are both a serial and compulsive liar.

Let's put your word to the test.

You are here after a SIX MONTH BAN, FOR HAVING LIED TO EVERYONE: you had multiple accounts/alts, one of which was nightsky.

You were lying all the time to your readers.

A sure sign of compulsive lying.

Yet, here you are having the audacity to accuse someone else.


Is Gauss' Easter formula a lie?

If not, then I am not lying about history.

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg758652#msg758652

Are the paleomagnetic datings of the artefacts found in southern Italy, including Pompeii, a lie?

If not, then I am not lying about history.

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1683846#msg1683846


So far, no lies on my part.


The math part.

Here is my global natural logarithm formula, proven to be true by myself, also the RE checked it and found it to be true:



Here is my global cosine/exponential formula:



Here is my global arctangent formula:



My formula/algorithm for the zeros of the Riemann zeta function works out perfectly, a feat unmatched and unheard of, without using mathematical analysis, only basic arithmetic.


No lies on my part so far.


My global SAGNAC FORMULA is totally proven by the fact that Professor Yeh's formula was published in one of the best journals in the world.


The fact that yet these scientists have to come to their senses and realize that the fact that terrestrial gravity cannot be a force of attraction, is something else, it is their responsibility as true scientists to explore the consequences and direct conclusions of their findings which obviously contradict Newton's law of gravitation.

Is Gauss' Easter formula a lie?

If not, then I am not lying about history.

I wouldn't go so far as saying a lie. Perhaps more of a misinterpretation. From a 'Scientific American' article titled, 'Algorithms Calendar Calculations: Easter Is a Quasicrystal'

"In 1800 German mathematician Carl Friedrich Gauss invented a simple algorithm that incorporated the church’s rules for calculating Easter’s date. Unfortunately, Gauss’s work contained a minor oversight: it gives April 13 for the year 4200 when the correct date should be April 20. He corrected this error by hand in his own copy of the published paper. The first flawless algorithm was presented in 1876 in the journal Nature by an anonymous American. In 1965 Thomas H. O’Beirne of Glasgow University published two such procedures in his book Puzzles and Paradoxes (Oxford University Press). O’Beirne’s method puts the various cycles and adjustments into an arithmetical scheme."

You were given a link which directly contradicts your drivel.

Yet, here you are merrily and happily wondering aloud about "misinterpretations".

No misinterpretations at all.

GAUSS EASTER FORMULA APPLIED TO THE CHRONOLOGY OF HISTORY

According to the official RE equations of orbital mechanics, the ones in question here, the vernal equinox fell on March 21, in the year 325 AD.

I am going to prove to you that no such thing ever happened, thus showing the utter fallacy of the differential equation approach to understanding orbital mechanics.

You also seem to forget that just as Einstein fudged his Mercury equation to fit the results, so the conspirators who invented the RE differential equations of motion had to modify drastically not only the masses of the planets and the Sun, but also their corresponding distances from Earth, in order to, at least, offer accurate results for a time span not extending beyond some three hundred years.


Now, Gauss' Easter formula is the most accurate astronomical dating tool at our disposal.

A brief summary of the dating of the First Council of Nicaea and the startling conclusions following the fact that the Gregorian calendar reform never occurred in 1582 AD.


Let us turn to the canonical mediaeval ecclesial tractate - Matthew Vlastar’s Collection of Rules Devised by Holy Fathers, or The Alphabet Syntagma. This rather voluminous book represents the rendition of the rules formulated by the Ecclesial and local Councils of the Orthodox Church.

Matthew Vlastar is considered to have been a Holy Hierarch from Thessalonica, and written his tractate in the XIV century. Today’s copies are of a much later date, of course. A large part of Vlastar’s Collection of Rules Devised by Holy Fathers contains the rules for celebrating Easter. Among other things, it says the following:


“The Easter Rules makes the two following restrictions: it should not be celebrated together with the Judaists, and it can only be celebrated after the spring equinox. Two more had to be added later, namely: celebrate after the first full moon after the equinox, but not any day – it should be celebrated on the first Sunday after the equinox. All of these restrictions, except for the last one, are still valid (in times of Matthew Vlastar – the XIV century – Auth.), although nowadays we often celebrate on the Sunday that comes later. Namely, we always count two days after the Lawful Easter (that is, the Passover, or the full moon – Auth.) and end up with the subsequent Sunday. This didn’t happen out of ignorance or lack of skill on the part of the Elders, but due to lunar motion”

Let us emphasize that the quoted Collection of Rules Devised by Holy Fathers is a canonical mediaeval clerical volume, which gives it all the more authority, since we know that up until the XVII century, the Orthodox Church was very meticulous about the immutability of canonical literature and kept the texts exactly the way they were; with any alteration a complicated and widely discussed issue that would not have passed unnoticed.

So, by approximately 1330 AD, when Vlastar wrote his account, the last condition of Easter was violated: if the first Sunday happened to be within two days after the full moon, the celebration of Easter was postponed until the next weekend. This change was necessary because of the difference between the real full moon and the one computed in the Easter Book. The error, of which Vlastar was aware, is twenty-four hours in 304 years.

Therefore the Easter Book must have been written around AD 722 (722 = 1330 - 2 x 304). Had Vlastar known of the Easter Book’s 325 AD canonization, he would have noticed the three-day gap that had accumulated between the dates of the computed and the real full moon in more than a thousand years. So he either was unaware of the Easter Book or knew the correct date when it was written, which could not be near 325 AD.

G. Nosovsky: So, why the astronomical context of the Paschalia contradicts Scaliger’s dating (alleged 325 AD) of the Nicaean Council where the Paschalia was canonized?

This contradiction can easily be seen from the roughest of calculations.

1) The difference between the Paschalian full moons and the real ones grows at the rate of one day in 300 years.

2) A two-day difference had accumulated by the time of Vlastar, which is roughly dated 1330 AD.

3) Ergo, the Paschalia was compiled somewhere around 730 AD, since

1330 – (300 x 2) = 730.

It is understood that the Paschalia could only be canonized by the Council sometime later. But this fails to correspond to Scaliger’s dating of its canonization as 325 AD in any way at all!

Let us emphasize, that Matthew Vlastar himself, doesn’t see any contradiction here, since he is apparently unaware of the Nicaean Council’s dating as the alleged year 325 AD. A natural hypothesis: this traditional dating was introduced much later than Vlastar’s age. Most probably, it was first calculated in Scaliger’s time.

With the Easter formula derived by C.F. Gauss in 1800, Nosovsky calculated the Julian dates of all spring full moons from the first century AD up to his own time and compared them with the Easter dates obtained from the Easter Book. He reached a surprising conclusion: three of the four conditions imposed by the First Council of Nicaea were violated until 784, whereas Vlastar had noted that “all the restrictions except the last one have been kept firmly until now.” When proposing the year 325, Scaliger had no way of detecting this fault, because in the sixteenth century the full-moon calculations for the distant past couldn’t be performed with precision.

Another reason to doubt the validity of 325 AD is that the Easter dates repeat themselves every 532 years. The last cycle started in 1941, and previous ones were 1409 to 1940, 877 to 1408 and 345 to 876. But a periodic process is similar to drawing a circle—you can choose any starting point. Therefore, it seems peculiar for the council to have met in 325 AD and yet not to have begun the Easter cycle until 345.

Nosovsky thought it more reasonable that the First Council of Nicaea had taken place in 876 or 877 AD, the latter being the starting year of the first Easter cycle after 784 AD, which is when the Easter Book must have been compiled. This conclusion about the date of the First Council of Nicaea agreed with his full-moon calculations, which showed that the real and the computed full moons occurred on the same day only between 700 and 1000 AD. From 1000 on, the real full moons occurred more than twenty-four hours after the computed ones, whereas before 700 the order was reversed. The years 784 and 877 also match the traditional opinion that about a century had passed between the compilation and the subsequent canonization of the Easter Book.

G. Nosovky:

The Council that introduced the Paschalia – according to the modern tradition as well as the mediaeval one, was the Nicaean Council – could not have taken place before 784 AD, since this was the first year when the calendar date for the Christian Easter stopped coinciding with the Passover full moon due to slow astronomical shifts of lunar phases.

The last such coincidence occurred in 784 AD, and after that year, the dates of Easter and Passover drifted apart forever. This means the Nicaean Council could not have possibly canonized the Paschalia in IV AD, when the calendar Easter Sunday would coincide with the Passover eight (!) times – in 316, 319, 323, 343, 347, 367, 374, and 394 AD, and would even precede it by two days five (!) times, which is directly forbidden by the fourth Easter rule, that is, in 306 and 326 (allegedly already a year after the Nicaean Council), as well as the years 346, 350, and 370.

Thus, if we’re to follow the consensual chronological version, we’ll have to consider the first Easter celebrations after the Nicaean Council to blatantly contradict three of the four rules that the Council decreed specifically for this feast! The rules allegedly become broken the very next year after the Council decrees them, yet start to be followed zealously and in full detail five centuries (!) after that.

Let us note that J.J. Scaliger could not have noticed this obvious nonsense during his compilation of the consensual ancient chronology, since computing true full moon dates for the distant past had not been a solved problem in his epoch.

The above mentioned absurdity was noticed much later, when the state of astronomical science became satisfactory for said purpose, but it was too late already, since Scaliger’s version of chronology had already been canonized, rigidified, and baptized “scientific”, with all major corrections forbidden.


Now, the ecclesiastical vernal equinox was set on March 21st because the Church of Alexandria, whose staff were reputed to have astronomical expertise, reckoned that March 21st was the date of the equinox in 325 AD, the year of the First Council of Nicaea.

The Council of Laodicea was a regional synod of approximately thirty clerics from Asia Minor that assembled about 363–364 AD in Laodicea, Phrygia Pacatiana, in the official chronology.

The major concerns of the Council involved regulating the conduct of church members. The Council expressed its decrees in the form of written rules or canons.

However, the most pressing issue, the fact that the calendar Easter Sunday would coincide with the Passover eight (!) times – in 316, 319, 323, 343, 347, 367, 374, and 394 AD, and would even precede it by two days five (!) times, which is directly forbidden by the fourth Easter rule, that is, in 306 and 326 (allegedly already a year after the Nicaean Council), as well as the years 346, 350, and 370 was NOT presented during this alleged Council of Laodicea.


We are told that the motivation for the Gregorian reform was that the Julian calendar assumes that the time between vernal equinoxes is 365.25 days, when in fact it is about 11 minutes less. The accumulated error between these values was about 10 days (starting from the Council of Nicaea) when the reform was made, resulting in the equinox occurring on March 11 and moving steadily earlier in the calendar, also by the 16th century AD the winter solstice fell around December 11.


But, in fact, as we see from the information presented in the preceeding paragraphs, the Council of Nicaea could not have taken place any earlier than the year 876-877 e.n., which means that in the year 1582, the winter solstice would have arrived on December 16, not at all on December 11.

Papal Bull, Gregory XIII, 1582:

Therefore we took care not only that the vernal equinox returns on its former date, of which it has already deviated approximately ten days since the Nicene Council, and so that the fourteenth day of the Paschal moon is given its rightful place, from which it is now distant four days and more, but also that there is founded a methodical and rational system which ensures, in the future, that the equinox and the fourteenth day of the moon do not move from their appropriate positions.


Given the fact that in the year 1582, the winter solstice would have arrived on December 16, not at all on December 11, this discrepancy could not have been missed by T. Brahe, or G. Galilei, or J. Kepler - thus we can understand the fiction at work in the official chronology.

Newton agrees with the date of December 11, 1582 as well; moreover, Britain and the British Empire adopted the Gregorian calendar in 1752 (official chronology); again, more fiction at work: no European country could have possibly adopted the Gregorian calendar reformation in the period 1582-1800, given the absolute fact that the winter solstice must have falled on December 16 in the year 1582 AD, and not at all on December 11 (official chronology).


The conclusions are as follows:

No historical or astronomical proof exists that before 1700 AD any gradual shift in the orientation of Earth's axis of rotation (axial precession) ever took place. The 10 day cumulative error in the Vernal Equinox date since the Council of Nicaea until the year 1582 AD is due just to the reform of the Julian calendar: if we add the axial precession argument, then  the cumulative errors would have added to even more than 10 days, because of the reverse precessional movement. No axial precession means that the Earth did not ever orbit around the Sun, as we have been led to believe. And it means that the entire chronology of the official history has been forged at least after 1750 AD.

In the FE theory, the approximately 50 seconds of arc per year (1 degree/71.6 years) change of longitude of the Pole Star is due to the movement of the Pole Star itself and NOT due to any axial precession of the Earth.


EXPLICIT DATING GIVEN BY MATTHEW VLASTAR



It is indeed amazing that Matthew Vlastar’s Collection of Rules Devised by Holy Fathers – the book that every Paschalia researcher refers to – contains an explicit dating of the time the Easter Book was compiled. It is even more amazing that none of the numerous researchers of Vlastar’s text appeared to have noticed it (?!), despite the fact that the date is given directly after the oft-quoted place of Vlastar’s book, about the rules of calculating the Easter date. Moreover, all quoting stops abruptly immediately before the point where Vlastar gives this explicit date.



What could possibly be the matter? Why don’t modern commentators find themselves capable of quoting the rest of Vlastar’s text? We are of the opinion that they attempt to conceal from the reader the fragments of ancient texts that explode the entire edifice of Scaliger’s chronology. We shall quote this part completely:



Matthew Vlastar:



“There are four rules concerning the Easter. The first two are the apostolic rules, and the other two are known from tradition. The first rule is that the Easter should be celebrated after the spring equinox. The second is that is should not be celebrated together with the Judeans. The third: not just after the equinox, but also after the first full moon following the equinox. And the fourth: not just after the full moon, but the first Sunday following the full moon… The current Paschalia was compiled and given to the church by our fathers in full faith that it does not contradict any of the quoted postulates. (This is the place the quoting usually stops, as we have already mentioned – Auth.). They created it the following way: 19 consecutive years were taken starting with the year 6233 since Genesis (= 725 AD – Auth.) and up until the year 6251 (= 743 AD – Auth.), and the date of the first full moon after the spring equinox was looked up for each one of them. The Paschalia makes it obvious that when the Elders were doing it; the equinox fell on the 21st of March” ([518]).



Thus, the Circle for Moon – the foundation of the Paschalia – was devised according to the observations from the years 725-743 AD; hence, the Paschalia couldn’t possibly have been compiled, let alone canonized, before that.


I have just proven to you that the spring equinox could not, and did not, fall on March 21, in the year 325 AD, CONTRARY to the figures implied by the RE equations of orbital mechanics.

According to the official RE equations of orbital mechanics, the ones in question here, the vernal equinox fell on March 21, in the year 325 AD.

That is incorrect and has nothing to do with, "According to the official RE equations of orbital mechanics." It was a a church thing to lay out the Christian holidays, standardize them across the religion. That's what the big meeting of bishops in Nicaea in 325 AD was all about. Pretty much everyone knows this except for you. Has nothing to with RE equations nor with orbital mechanics. Quit posting stuff that is totally irrelevant, then misinterpreting it, and plastering all over here.

"According to the Nicaean Ecumenical Council of 325 AD, Easter should be celebrated on the first Sunday after the first full moon after the Spring (Vernal) Equinox. All Eastern and Western churches follow this Ecumenical Council ruling. The differences come from the different interpretation of these rules.
The Eastern Christian Churches (Orthodox) follow the Julian calendar in calculating Easter dates, while the Western Christian Churches (Catholic, Protestant) follow the Gregorian calendar.
The Eastern Churches have fixed the above “Vernal Equinox” to be March 21 of the Julian calendar, which currently is April 3 in the Gregorian calendar. So, in Eastern churches, Easter falls between April 4 and May 8, while in Western churches Easter falls between March 22 and April 25."

From:
'On the Julian, Gregorian and Lunar Calendars, the First Ecumenical Council of Nicaea, the Vernal Equinox and the Paschal Full Moon. - Why is Easter Celebrated on Different Sundays in the Eastern Orthodox and Western Catholic and Protestant Churches?'

Panos Antsaklis
Notre Dame, Indiana

So far, no lies on my part.

Except basically every statement you have made int he thread.

Now stop with the pathetic distractions (This topic is laser ring gyroscopes, not Easter or logarithms). Your formula is pure garbage, not proven by anyone and just baselessly asserted by you, however it has been refuted by countless others.

Now can you show the derivation for how long it takes light to propagate around a stationary loop?

You derived a formula which has NO LOOPS, and is a comparison of TWO SEGMENTS.

You used the same sign for beams traveling in opposite directions.

Your final formula is this:

4AΩ/c^2

However, THIS IS THE CORIOLIS EFFECT FORMULA.

Here is a reference which uses the CORIOLIS FORCE to derive the very same formula, which is proportional to the area of the interferometer:

Spinning Earth and its Coriolis effect on the circuital light beams

http://www.ias.ac.in/article/fulltext/pram/087/05/0071

The final formula is this:

4AΩ/c^2

This is the CORIOLIS EFFECT formula, not the SAGNAC EFFECT formula.

A direct disproval of your failed claims that this would be the Sagnac effect formula.


My formula has the LOOPS, and the CORRECT SIGNS.

And there is nothing that you can do about it.

The experimental proof, using the VERY SAME FORMULA, was published in one of the two best journals on OPTICS in the world: the Journal of Optics Letters (the other one is Applied Optics).

It was peer-reviewed at the highest level possible.

It was published by the top expert in the world on LASER OPTICS: Professor Yeh.

Dr. P. Yeh
PhD, Caltech, Nonlinear Optics
Principal Scientist of the Optics Department at Rockwell International Science Center
Professor, UCSB
"Engineer of the Year," at Rockwell Science Center
Leonardo da Vinci Award in 1985
Fellow of the Optical Society of America, the Institute of Electrical and Electronics Engineers



Self-pumped phase-conjugate fiber-optic gyro, I. McMichael, P. Yeh, Optics Letters 11(10):686-8 · November 1986 

http://www.dtic.mil/dtic/tr/fulltext/u2/a170203.pdf (appendix 5.1)


Exactly the formula obtained by Professor Yeh:

φ = -2(φ₂ - φ₁) = 4π(R₁L₁ + R₂L₂)Ω/λc = 4π(V₁L₁ + V₂L₂)/λc

Since Δφ = 2πc/λ x Δt, Δt = 2(R₁L₁ + R₂L₂)Ω/c² = 2(V₁L₁ + V₂L₂)/c²

CORRECT SAGNAC FORMULA:

2(V₁L₁ + V₂L₂)/c²


THE VERY SAME FORMULA, OBTAINED FOR A SAGNAC INTERFEROMETER WHICH FEATURES TWO DIFFERENCE LENGTHS AND TWO DIFFERENT VELOCITIES.


Please read:

http://www.dtic.mil/dtic/tr/fulltext/u2/a170203.pdf

ANNUAL TECHNICAL REPORT PREPARED FOR THE US OF NAVAL RESEARCH.

THE FORMULA WAS PUBLISHED IN THE JOURNAL OF OPTICS LETTERS.

PAGE 18 OF THE PDF DOCUMENT, SECTION 3.0 PROGRESS:

Our first objective was to demonstrate that the phase-conjugate fiberoptic gyro (PCFOG) described in Section 2.3 is sensitive to rotation. This phase shift plays an important role in the detection of the Sagnac phase shift due to rotation.


SELF-PUMPED PHASE CONJUGATE FIBER OPTIC GYRO, PUBLISHED BY OPTICS LETTERS

PUBLISHED BY THE US OF NAVAL RESEARCH, PHYSICS DIVISION

HIGHEST POSSIBLE SCIENTIFIC LEVEL


PAGE 38 OF THE PDF DOCUMENT, PAGE 6 OF APPENDIX 3.1


it does demonstrate the measurement of the Sagnac phase shift Eq. (3)


HERE IS EQUATION (3) OF THE PAPER, PAGE 3 OF APPENDIX 3.1:


φ = -2(φ₂ - φ₁) = 4π(R₁L₁ + R₂L₂)Ω/λc = 4π(V₁L₁ + V₂L₂)/λc

Since Δφ = 2πc/λ x Δt, Δt = 2(R₁L₁ + R₂L₂)Ω/c² = 2(V₁L₁ + V₂L₂)/c²

CORRECT SAGNAC FORMULA:

2(V₁L₁ + V₂L₂)/c²


THE VERY SAME FORMULA, OBTAINED FOR A SAGNAC INTERFEROMETER WHICH FEATURES TWO DIFFERENCE LENGTHS AND TWO DIFFERENT VELOCITIES.


The US OF NAVAL RESEARCH, Physics Division, thus confirms my formula.

There is nothing else to discuss here.

My global SAGNAC FORMULA is totally proven by the fact that Professor Yeh's formula was published in one of the best journals in the world.

Except for one minor detail. Prof Pochi Yeh's Sagnac delay is for a device with a Phase Conjugate Mirror and yours isn't!
And please don't claim that it makes no difference.

So now would you show your analysis of the passive loop as used in the Michelson-Gale-Pearson  measurement.

By the way, you do remember the book you referred to in your vain attempt to prove that Silberstein derived the Coriolis effect and not the Sagnac effect?
This book: Ring Interferometry, De Grigorii B. Malykin, Vera I. Pozdnyakova

Well, De Grigorii B. Malykin also wrote a paper in Russian, The Sagnac effect: correct and incorrect explanations, G. B. Malykin, and look what the abstract says:

The Sagnac effect: correct and incorrect explanations

Abstract: Different explanations for the Sagnac effect are discussed. It is shown that this effect is a consequence of the relativistic law of velocity composition and that it can also be explained adequately within the framework of general relativity. When certain restrictions on the rotational velocity are imposed, the Sagnac effect can be attributed to the difference in the time dilation (or phase change) of material particle wave functions in the scalar (or correspondingly vector) gravitational potential of the inertial forces in a rotating reference system for counterpropagating waves. It is also shown that all the nonrelativistic interpretations of the Sagnac effect, which are unfortunately sometimes found in scientific papers, monographs and textbooks, are wrong in principle, even though the results they yield are accurate up to relativistic corrections in some special cases.   

Luckily the abstract is in English.
Yes, YOUR expert states quite clearly that the"Sagnac effect" "can also be explained adequately within the framework of general relativity" and that "the nonrelativistic interpretations of the Sagnac effect" "are wrong".

The fact that yet these scientists have to come to their senses and realize that the fact that terrestrial gravity cannot be a force of attraction, is something else, it is their responsibility as true scientists to explore the consequences and direct conclusions of their findings which obviously contradict Newton's law of gravitation.

Those "scientists have . . come to their senses"! It's you that needs to wake u.

But do "these scientists" really claim "that terrestrial gravity" is "a force of attraction"?
I'm no scientist but even I do not claim claim "that terrestrial gravity" is "a force of attraction" just that it behaves as one.

So what "findings . . . . obviously contradict Newton's law of gravitation"?

You do not even understand yet the implications of Newton's law of universal gravitation. It is a Law and not a theory.
And as such it simply states that gravitation behaves as a force of attraction, not that gravitation is a force of attraction.

Now according to Einstein's Theory of General Relativity, gravitation is not "a force of attraction" but an "inertial force".

That is incorrect and has nothing to do with, "According to the official RE equations of orbital mechanics."

Are you actually saying that the RE equations of orbital mechanics DO NOT predict that the vernal equinox for the year 325 AD fell on March 21?

You have 23 hours to modify your statement.

Otherwise, you are claiming that you are a flat earth believer.

If you made an error, and now you realize that actually those equations MUST PREDICT that the vernal equinox fell on March 21, in the year 325 AD, then you have a huge problem.

Not only does Gauss' Easter formula directly contradict this statement, but also you have to deal with another fact.

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1935048#msg1935048

Except for one minor detail. Prof Pochi Yeh's Sagnac delay is for a device with a Phase Conjugate Mirror and yours isn't!

My formula has the LOOPS, and the CORRECT SIGNS.

And there is nothing that you can do about it.

The experimental proof, using the VERY SAME FORMULA, was published in one of the two best journals on OPTICS in the world: the Journal of Optics Letters (the other one is Applied Optics).

It was peer-reviewed at the highest level possible.

It was published by the top expert in the world on LASER OPTICS: Professor Yeh.

Dr. P. Yeh
PhD, Caltech, Nonlinear Optics
Principal Scientist of the Optics Department at Rockwell International Science Center
Professor, UCSB
"Engineer of the Year," at Rockwell Science Center
Leonardo da Vinci Award in 1985
Fellow of the Optical Society of America, the Institute of Electrical and Electronics Engineers



Self-pumped phase-conjugate fiber-optic gyro, I. McMichael, P. Yeh, Optics Letters 11(10):686-8 · November 1986 

http://www.dtic.mil/dtic/tr/fulltext/u2/a170203.pdf (appendix 5.1)


Exactly the formula obtained by Professor Yeh:

φ = -2(φ₂ - φ₁) = 4π(R₁L₁ + R₂L₂)Ω/λc = 4π(V₁L₁ + V₂L₂)/λc

Since Δφ = 2πc/λ x Δt, Δt = 2(R₁L₁ + R₂L₂)Ω/c² = 2(V₁L₁ + V₂L₂)/c²

CORRECT SAGNAC FORMULA:

2(V₁L₁ + V₂L₂)/c²


THE VERY SAME FORMULA, OBTAINED FOR A SAGNAC INTERFEROMETER WHICH FEATURES TWO DIFFERENT LENGTHS AND TWO DIFFERENT VELOCITIES.


Please read:

http://www.dtic.mil/dtic/tr/fulltext/u2/a170203.pdf

ANNUAL TECHNICAL REPORT PREPARED FOR THE US OF NAVAL RESEARCH.

THE FORMULA WAS PUBLISHED IN THE JOURNAL OF OPTICS LETTERS.

PAGE 18 OF THE PDF DOCUMENT, SECTION 3.0 PROGRESS:

Our first objective was to demonstrate that the phase-conjugate fiberoptic gyro (PCFOG) described in Section 2.3 is sensitive to rotation. This phase shift plays an important role in the detection of the Sagnac phase shift due to rotation.


SELF-PUMPED PHASE CONJUGATE FIBER OPTIC GYRO, PUBLISHED BY OPTICS LETTERS

PUBLISHED BY THE US OF NAVAL RESEARCH, PHYSICS DIVISION

HIGHEST POSSIBLE SCIENTIFIC LEVEL


PAGE 38 OF THE PDF DOCUMENT, PAGE 6 OF APPENDIX 3.1


it does demonstrate the measurement of the Sagnac phase shift Eq. (3)


HERE IS EQUATION (3) OF THE PAPER, PAGE 3 OF APPENDIX 3.1:


φ = -2(φ₂ - φ₁) = 4π(R₁L₁ + R₂L₂)Ω/λc = 4π(V₁L₁ + V₂L₂)/λc

Since Δφ = 2πc/λ x Δt, Δt = 2(R₁L₁ + R₂L₂)Ω/c² = 2(V₁L₁ + V₂L₂)/c²

CORRECT SAGNAC FORMULA:

2(V₁L₁ + V₂L₂)/c²


THE VERY SAME FORMULA, OBTAINED FOR A SAGNAC INTERFEROMETER WHICH FEATURES TWO DIFFERENCE LENGTHS AND TWO DIFFERENT VELOCITIES.


The US OF NAVAL RESEARCH, Physics Division, thus confirms my formula.

There is nothing else to discuss here.

The US OF NAVAL RESEARCH, Physics Division, thus confirms my formula.

No it does not confirm your formula! Your's is simply an incorrect formula for the Sagnac delay in a passive loop.
And Prof Pochi Yeh's delay is for a device with a Phase Conjugate Mirror and not for your passive loop - quite different.

There is nothing else to discuss here.

Oh yes there is! Read the OP again.

Bob Knodel is a famous (?) Youtube flat earther who tried an experiment to prove the earth was fixed - non rotating. He obtained a UD$20,000 laser ring gyroscope, an extremely accurate device, to show that the earth was not spinning. The gyroscope registered a 15 degree per hour drift.

Does this prove a rotating earth? Or like Bob Knodel said, did it measure the "heavenly energies"?

If the gyroscope did measure a 15 degree per hour rotating earth, can this be possible still with a flat earth and local sun-moon circling system?

There is no mention of the Sagnac effect!
It simply says "Bob Knodel . . . . obtained a UD$20,000 laser ring gyroscope, an extremely accurate device, to show that the earth was not spinning".
The device was part of a device used for inertial navigation systems. Here is a complete INS based on fibre optic gyroscopes plus accelerometers.

EMCORE-OrionTM EN-1000 Micro Inertial Navigation System (MINAV)
The EMCORE-OrionTM EN-1000 high-precision Micro Inertial Navigation System (MINAV) is developed primarily for applications where navigation aids such as GPS are unavailable or denied.

The EMCORE-OrionTM MINAV is a state-of-the-art, fiber optic gyro-based Inertial Navigation System incorporating EMCORE’s proprietary integrated optics devices to enhance performance, providing standalone aircraft grade navigator performance in 1/3 the size of competing systems. . . . . In a GPS denied environment the EN-1000 MINAV will gyrocompass to approximately 1 milliradian.

Who cares whether these devices use the Sagnac effect or the Fred effect?
Whatever YOU call it, they work and depend on the earth's rotation to find true North (" to approximately 1 milliradian").
Try explanating what sort of aether rotation would  allow a "gyrocompass" to find true North on any flat earth map.

You now have to answer the OP and all your pages of copy-pasta will not do!

You derived a formula which has NO LOOPS, and is a comparison of TWO SEGMENTS.

No, I derived a formula which has 2 counter-propagating beams of light travelling around a loop, focusing on the two segments you have been focusing on, deriving the difference in time required for the beams to travel.
Meanwhile, instead of focusing on the loops, you decided to calculate the time difference between the segments, for a single beam, which has nothing to do with the Sagnac effect.

Now again, can you show a derivation for the very simple problem of light propagating around a stationary loop?
If not, it isn't surprising that you don't understand what you are doing wrong and cannot correctly derive the Sagnac effect for a rotating loop.

So can you do this very simple task? If not, don't bother commenting on the Sagnac effect ever again because you clearly don't understand.

And repeating your same refuted nonsense again and again doesn't help your case.

I sometimes wonder if sandokhan even reads what he posts. Seems more that he has some keywords and then takes some text from somewhere and copies/pastes it here.

Except for one minor detail. Prof Pochi Yeh's Sagnac delay is for a device with a Phase Conjugate Mirror and yours isn't!
My formula has the LOOPS, and the CORRECT SIGNS.
And there is nothing that you can do about it.

Oh, yes there is! I can inform everyone of your errors!

You say your formula for the Sagnac delay is the same as Prof Pochi Yeh's. Here's what you clsim:

CORRECT SAGNAC FORMULA:
2(V₁L₁ + V₂L₂)/c²

For Professor Yeh's phase-conjugate fiber-optic gyroscope:

They would be the same except that:
          yours has (V₁L₁ + V₂L₂) where Prof Yeh's has 2π(R₁L₁ + R₂L₂)Ω.
In your analysis L₁ and L₂ are two SIDES of the one loop but in Prof Yeh's L₁ and L₂ are the total lengths of two loops.

Those differences make yours a whole different loop structure and yours hides the necessity for rotation which Prof Yeh's explicitly includes.
So, whatever YOU claim, while your formula might look similar to Prof Yeh's it is simply an incorrect analysis of a different structure.

Repeatedly claiming yours is the same as Prof Pochi Yeh's fibre optic gyroscope with a phase conjugate mirror will get you nowhere!

A repeated error is simply a compounded error.

Why is it that you are the only person who seems get an obviously incorrect expression like that?
Look at this from what appears to be the source of your diagram:

Conspiracy of Light, The Michelson-Gale Experiment
In refining his argument, he proposed that it was not necessary for the light to go all the way around the globe - since there should be a velocity difference for any closed path rotating on the surface of the earth. He presented the following equation to calculate the time difference expected, using the shift in the interference fringes when the two beams overlap at the detector as a measure of the time difference:

Fig.1:

where:  Vo = the tangential velocity of the earth's rotation at the equator (465m/s)
              A = the area of the circular path
              R = the radius of the earth (6371000 m)
              c = speed of light (3E8 m/s)
              f = the latitude in degrees where the experiment is conducted.
              l = wavelength of the light

And those 2's should be 4's because even Michelson didn't initially get it quite right and it was corrected by Silberstein:

   The experiment remained in abeyance for several years, until Silberstein published a paper in 1921 on the theory of light propagation in rotating systems [2]. In this article, Silberstein discusses Michelson's proposed experiment and through calculations of his own demonstrated that the time difference expected in such an experiment would be double what Michelson suggested.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
After taking all these factors into account, the expected fringe shift becomes:

But what I find so telling is that you claim everyone else, including those that deny relativity, is wrong and that only you are right.

what gyro did Bob use, because small laser gyro (up to a perimeter length of 60 cm) used for navigation are unable to detect Earth’s rotation and need to be dithered to avoid lock-in.

That is incorrect and has nothing to do with, "According to the official RE equations of orbital mechanics."

Are you actually saying that the RE equations of orbital mechanics DO NOT predict that the vernal equinox for the year 325 AD fell on March 21?

You have 23 hours to modify your statement.

Otherwise, you are claiming that you are a flat earth believer.

If you made an error, and now you realize that actually those equations MUST PREDICT that the vernal equinox fell on March 21, in the year 325 AD, then you have a huge problem.

Not only does Gauss' Easter formula directly contradict this statement, but also you have to deal with another fact.

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1935048#msg1935048

You are not entirely incorrect. But you are using things to run with completely out of context and to fabricate some sort of foundation for all of your claims that land squarely in obfuscation.

I too can copy/paste a wall of text:

"THE REFORM OF THE JULIAN CALENDAR
ROSCOE LAMONT.
The average length of the Julian year is 3654 days, which is about 11 ¼ minutes greater than the tropical year. Since there A 1440 minutes in a day, the number of years that must elapse before the excess amounts to one day will be found by dividing 1440 by 11%, which gives 128. Therefore at the end of 128 years the time of the equinox would be one day earlier in the calendar. When the Julian calendar was established in the year 45 B. C. the equinox occurred about March 24, and in 325 A. D. it came about March 21. At this rate, in a little over 10,00o years the equinox would fall on the first of January, and in 22,000 years the fourth of July would come at the time of the winter solstice. This would not do at all, and in order to prevent any such occurrence the calendar was reformed again.

But the idea of the reformers was to restore the equinox to March 21, the day on which they supposed it came in the year 325, when a Council of the Church was held at Nicaea, in Asia Minor, which made a decision as to the time of celebrating Easter. One sect of the Chris-tians, following the Jewish practice, observed the fourteenth day of the moon, on whatever day of the week it came, which fell on or next followed the date of the equinox; but the greater number condemned this practice, wishing to have nothing in common with the Jews, who boasted that without instruction from them the Christians wouldn’t know when to celebrate Easter. When the Nicene Council was held the
Romans placed the date of the equinox at March 18, and the Alexandrians at March 21, but modern astronomers who have written on this
subject say that the equinox in the year 325 came on March 20. The Alexandrian determination appears to have been accepted, and the rule laid down that Easter was to be celebrated on the first Sunday after the fourteenth day of the moon which falls on or next after March 21, the date of the equinox, the computation to be made by the Church of Alexandria, the most skilled in the science of astronomy, and the Church of Rome to make it known. The Romans, however, continued to make their own computation, for Hefele, in his History of the Church Councils, says that in the very next year, 326, the Romans celebrated Easter on a different day from the Alexandrians, and that the same thing happened in the years 330, 333, 340, 341 and 343. But though councils might by decree fix the equinox at March 21, it none the less continued to come one day earlier in the month every 128 years, falling on March 11 in 1582 when the reform of the calendar was carried into effect, and to restore the date to March 21 the omission of ten days, or, better, the dropping of ten monthly dates, was necessary."