sandokhan lies regarding the Sagnac effect

What you have done on your own jack amounts to a piece of shit.

Nothing more.

I, on the other hand, was able to come up with the global formula for the natural logarithm, something unheard of.

My bibliographical references are the very best, and totally refute your failed ideas about science.


 One beam moves forward in the big arc and backwards in the small arc. The other goes the other way.
Both beams of light travel through the both arcs.

What you did is to actually construct an interferometer using the angular velocity of the Earth.

Now you admit that the interferometer has a light source built in.

Take a look at how the correct calculation actually is performed, as exemplified in my previous message.

In the rectangular shaped interferometer we get for the CCW path:

ct₁ = vt₁ + 4a + 4b

One beam.

For the CW path:

ct₂ = 4a + 4b - ct₂

The second beam.

This is the correct calculation.

Now, we get the shift:

dt = 2Lv/c²

L = 4a + 4b

No loop at all, just the path of the light.


But a circular path isn't simple.
A circular path is only simple when you have the circle centred on the centre of rotation. As soon as you move it away it is no longer that simple.

A nice simple question to highlight this is to ask what is v meant to be?

V can be the rotational speed or the orbital speed.


So now lets let alpha be the angle subtended by the arcs, and omega be the angular velocity.
So, one beam of light will propagate in the same direction as the big arc and the opposite direction as the little arc.
What this means is its trip along the big arc will be longer than if it was stationary and shorter along the little arc.

That is not how you analyze the Sagnac in the lab.

You have to use the source of light as a reference point.

What you are doing, again, is to build up an interferometer using the angular velocity/angle subtended of the Earth, and then go back and forth with the two beams.

You cannot do that.

This is where you went wrong.

You cannot use the sun as the source of light for the interferometer.

You need to perform your analysis of the Sagnac right there in the lab.

By using the Sun as a reference for your source of light, and R, you simply eliminated the Earth-Sun distance from your calculations and then claimed victory.

Here is how to do the Sagnac properly:

Here is how to correctly calculate the orbital Sagnac effect:

Earth's radius = 6357 km; r² = 40411449

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

∆t = 4πR²ω/(c²-v²)
or

I use the linear velocity.

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

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

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

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

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


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


That is exactly the calculation used by both Dr. Su and by Dr. Gezari to arrive at the conclusion that the orbital Sagnac was greater than the rotational Sagnac, and that a local-aether model was needed to explain the situation.

Their papers were peer reviewed.

You have nothing at all going for you.

You practically eliminated the crucial distance Earth-Sun from your calculations.

But that is not what the noted authors Tartaglia and Ruggiero did for the same paralleogram-shaped interferometer.

They obtained the correct value for the shift, and only at the end apply v.


Take a look at how the rotational Sagnac is calculated.

You have a center of rotation and an angular velocity, you multiply both and you get v.

That is how GPS technology works.

What you did is to eliminate the radius from the calculations by actually building up an interferometer with arcs, using the Sun-Earth distance as a pivot.

But you cannot have the Earth go back and forth to accomodate for your interferometer.

If your interferometer already has a light source, then you apply the Sagnac shift formula as it is being done in all the referenced papers I provided.



The simple fact is the aether has been debunked long ago.

The Dayton Miller and the Galaev experiments prove otherwise.

See how little you know about science?

The most significant development since Miller has been the
experiments of Yuri Galaev of the Institute of Radiophysics and
Electronics in the Ukraine. Galaev made independent measure-
ments of ether-drift using radiofrequency and optical wave
bands. His research "confirmed Miller's results down
to the details".

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1722791#msg1722791

Yuri Galaev, Ph.D.; Senior research officer of the Institute for Radiophysics & Electronics National Academy of Sciences of Ukraine, and corresponding member of the Russian Academy of Natural Sciences (RANS)

THE MEASURING OF ETHER-DRIFT VELOCITY AND KINEMATIC ETHER VISCOSITY WITHIN OPTICAL WAVES BAND Yu.M. Galaev The Institute of Radiophysics and Electronics of NSA in Ukraine


Dr. Galaev remarks:

Orbital component of the ether drift velocity, stipulated by the Earth movement around the Sun with the velocity 30 km/sec, was not detected [during the Dayton Miller experiments].


Dr. Galaev also concludes:

The method action is based on the development regularities of viscous liquid or gas streams in the directing systems. The significant measurement results have been obtained statistically. The development of the ether drift required effects has been shown. The measured value of the ether kinematic viscosity on the value order has coincided with its calculated value.


The most precise experiments ever undertaken in ether-drift detection thus prove that the Earth does not orbit the Sun at a speed of 30km/s.

Again, this does not deal with the OP at all.

Here is what you wrote earlier.

Do you have anything other than pathetic dismissal?

The precise calculation provided in the bibliographical reference showed that you are wrong.


Can you show in this graphic WHERE THE LOOP OF THE ROTATIONAL AND/OR ORBITAL SAGNAC IS LOCATED?

IF YOU CANNOT YOU LOSE.



Where is the loop jack?

No loop at all.

Just a straight path to the receiver.

Then, you need the theory I provided here for the uniform/translational Sagnac.


APPLY YOUR DERIVATION TO THE REAL LIFE SITUATION WHERE YOU DO NOT HAVE A LOOP AT ALL.

What you have done on your own jack amounts to a piece of shit.
Nothing more.

No. Vastly more.
I have provided a derivation for the Sagnac effect due to Earth's orbit on an interferometer on Earth.
A derivation you are yet to show any problem with (you lying about it is not a problem with it).
And a derivation you are yet to be able to come up with yourself.

I, on the other hand, was able to come up with the global formula for the natural logarithm, something unheard of.

With regards to this topic, all you have done is provide shit.
You have tried to distract people with garbage, you have blatantly lied and tried to misrepresent things.

Now deal with the OP (yourself, not just linking to crap) or SHUT UP!!!

What you did is to actually construct an interferometer using the angular velocity of the Earth.

STOP LYING!!!
I used a small angle (alpha) which the 2 arc's subtended. Both moved with Earth's orbit.

Take a look at how the correct calculation actually is performed, as exemplified in my previous message.

Your previous message had nothing to do with the discussion at hand.
Deal with the OP and provide your own derivation.

In the rectangular shaped interferometer we get for the CCW path:
ct₁ = vt₁ + 4a + 4b

There is no v and there is no a or b.
Try again.

No loop at all, just the path of the light.

I take it you don't understand what a loop is?

But a circular path isn't simple.
A circular path is only simple when you have the circle centred on the centre of rotation. As soon as you move it away it is no longer that simple.
A nice simple question to highlight this is to ask what is v meant to be?
V can be the rotational speed or the orbital speed.

The orbital speed of what?
There are many different orbital speeds corresponding to the different parts of the circle.

This is why your analysis fails.

That is not how you analyze the Sagnac in the lab.

No, this is how you can do it in the lab. It is travelling around the loop. It doesn't matter where it starts in the loop or where it ends.

What you are doing, again, is to build up an interferometer using the angular velocity/angle subtended of the Earth, and then go back and forth with the two beams.

Again, stop lying. I am doing no such thing.

You cannot use the sun as the source of light for the interferometer.

I didn't.
I take it you are running out of ideas?

Here is how to do the Sagnac properly:

No, you don't do it by spouting a bunch of numbers which don't apply to the situation.
Provide a picture, do the derivation.
If you can't, then you are full of shit.

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

You have 2 separate radii to consider here. One for each section of the arc.

∆t = 4πR²ω/(c²-v²)

Again, this applies for a circular interferometer, where R is the radius of the interferometer.
As such, putting the orbital radius only works if the interferometer is the size of the orbit.

So no, this is a completely incorrect way to determine the Sagnac effect for the OP.

You have nothing at all going for you.

I have a derivation you are yet to refute.
So I have everything going for me.
Meanwhile, you are grasping at straws for whatever excuse you can use to avoid the derivation I provided because you know you can't refute it or provide your own without showing you are wrong.

You practically eliminated the crucial distance Earth-Sun from your calculations.

No, the pretty much irrelevant Earth-sun distance.
This is because the shift is proportional to the area of the loop, not how far away the loop is from the centre of rotation.

If your interferometer already has a light source, then you apply the Sagnac shift formula as it is being done in all the referenced papers I provided.

You mean like the one which clearly stated it is:
dt=4*A*w/c^2, where A is the AREA OF THE LOOP?

I did, but you rejected that because it didn't fit your delusional garbage.

And skipping your irrelevant spam.

Again, this does not deal with the OP at all.
Here is what you wrote earlier.
Do you have anything other than pathetic dismissal?
The precise calculation provided in the bibliographical reference showed that you are wrong.

Once again, they don't as they do not deal with what I have shown.
DEAL WITH THE OP OR SHUT UP!!

And once again, your irrelevant crap will be ignored.

Deal with the OP or admit you were wrong before moving on.

What you have done on your own jack amounts to a piece of shit.

Nothing more.

Well, what have you done relevant to the Sagnac delay? Nothing, nada, zilch!
Yes, you know very well that even if you were capable of that,

I, on the other hand, was able to come up with the global formula for the natural logarithm, something unheard of.

Are you a total idiot?
Have you totally lost your mind?
I seriously suspect that you have gone senile and can do nothing more than copy-n-paste anything you think might be vaguely relevant.

But look at

By the way, you forgot to mention that I discovered the global natural logarithm formula:
<<  even back there you were trying to bignote yourself  >>
https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1910773#msg1910773

LN V =  2ⁿ x ((-2 + {2 + [2 + (2 + 1/V + V)^1/2]^1/2...}^1/2))^1/2   (n+1 evaluations)

By summing the nested continued square root function, we finally obtain:

LN V = 2ⁿ x (V^1/2n+1 - 1/V^1/2n+1)

This is the first explicit global formula for the natural logarithm, which can be used immediately to find LN V without resorting to logarithm tables, or calculators which feature the logarithm key: all we need is a calculator which has the four basic operations and the square root key. It links algebraic functions with elementary and higher transcendental functions.

Your "famous global formula for the natural logarithm" is not a "formula for the natural logarithm" but just a good approximation. - effectively just a series approximation
And it does not link "algebraic functions with elementary and higher transcendental functions"! It is just another, albeit very good, approximation to ln(x)!

So cut the crap about how smart you are, you're just a big bag of hot-air, trying to be king frog in this little pond.

Your "famous global formula for the natural logarithm" is not a "formula for the natural logarithm" but just a good approximation. - effectively just a series approximation

I disagree.
It isn't a good approximation at all.

It is just replacing one function with another, much worse one.
It means a computer, instead of using the nice optimised algorithms for ln, instead needs to use the formula for powers. It needs to do that for 2^n (which they can do easily), and then if it is good it can just carry that through to 2^(n+1), and even 1/2^(n+1)=2^(-n-1).
But the v^(1/2^(n+1)) it is no longer a power of 2 so it needs to compute it.
And how do computers compute powers? well they use the exponentail function (technically an approximation to it).
So if you take a^b, it will compute it as exp(c*b), where exp(c)=a.
i.e. they compute exp(ln(a)*b).

So how great is that?
He made a formula for the natural log, which uses the natural log to calculate the answer......
What a genius.....

jack, you are not very good at math.

Do not kid yourself.

You are in the same category as rabinoz, an imbecile with no scientific background who enjoys being humiliated on a daily basis.

And how do computers compute powers? well they use the exponentail function (technically an approximation to it).

Can I call you an idiot?

I think I can.

You have multiple SQUARE ROOTS, easily computed by using Newton's iteration formula (quadratic convergence) or continued fractions.

I have reduced the task of computing a logarithm, without using log tables, to that of simple additions/divisions using any iterative method.

If you do not know enough about math, then you should not be posting at all.

The parallelogram in the following graphic is virtually the same as the loop in your graphic.

Here is a list of the scientific publications signed A. Tartaglia and A. M. Ruggiero:

http://porto.polito.it/view/creators/Ruggiero=3AMatteo_Luca=3A004059=3A.scopus_impact.html

Two of the best experts in the world regarding the Sagnac effect.

Their paper was published in a peer reviewed journal.

This then is my derivation for the parallelogram loop (which does have a perimeter as well):

EXACTLY THE SAME SITUATION.

THE OBSERVER IS MOVING WITH A SPEED V.

Sagnac effect and pure geomatry, Tartaglia/Ruggiero







Published in the  AMERICAN JOURNAL OF PHYSICS, vol. 83, pp. 427-432. - ISSN 0002-9505

In the rectangular shaped interferometer we get for the CCW path:

ct1 = vt1 + 4a + 4b

One beam.

For the CW path:

ct2 = 4a + 4b - vt2

The second beam.

This is the correct calculation.

Now, we get the shift:

dt = 2Lv/c2

L = 4a + 4b

No loop at all, just the path of the light.


You have obtained as a final result, based on your piece of shit derivation, 1/365.

There is no v and there is no a or b.

What?!

Their calculation is extremely precise.

It runs contrary to yours.

dt = 2Lv/c2

v = angular velocity x radius

Earth rotational angular velocity = 7.27 × 10−5 rad/s

Radius = 6,378 km

v_r = 0.463 km/s

Earth orbital angular velocity = 2 x 10-7 rad/s

Radius = 150,000,000 km

v_o = 30 km/s


If we substitute these values in the Sagnac formula, we get

v_o/v_r = 30km/s/0.463 km/s =~ 60

The orbital Sagnac is larger than the rotational Sagnac at least by a factor of 60.


Take a look at the two times obtained by these authors.

In the rectangular shaped interferometer we get for the CCW path:

ct1 = vt1 + 4a + 4b

One beam.

For the CW path:

ct2 = 4a + 4b - vt2

The second beam.

This is the correct calculation.


I will next deal with the sector of a circle area in your derivation.

A quick calculation.

Let tr = 4(Ar)(wr)/c² be the sagnac for the rotation and
to = 4(Ao)(wo)/c² be the sagnac for the orbit.

Let's choose distance for the experiment say 30km.
Light takes 0.0001 seconds to travel for that distance.
In that time period, it is your contention that tr/to > 1. That is your whole point.

tr/to = 4(Ar)(wr)/c² / 4(Ao)(wo)/c² = (Ar)(wr)/(Ao)(wo)

The distance to the sun is 150,000,000 km
The linear speed of the earth's orbit is 30km/s
The earth's radius is 6360 km
The linear speed of the earth's rotation is 0.462 km/sec

Let's calculate the angular speed for each.
w = v/r
wr = 0.462/6360 = 7.264e-5
wo = 30/150,000,000 = 2.0e-7

A = (θ/360)πR²
To calculate θ/360 for the earth's rotation, it is θ/360 = 30km/ 2 π Rr
Hence, Ar = 30Rr/2 = 15 Rr

The same is true for Ao
Ao = 15 Ro

So,
Ar = 15 Rr = 15 * 6360 = 93900
Ao = 15 Ro = 15 * ( 150,000,000 ) = 2250000000

Now, lets do the ratio of tr/to
tr/to = (Ar wr ) / (Ao wo ) = (93900 * 7.264e-5) / (2250000000 * 2.0e-7) = 6.82 / 450 = 0.015

As we can see, the rotational sagnac in any time period is smaller than the orbit according to the calculations on your link.

In fact, the orbital is 450/6.82 = 66.
This is consistent with my logic.

So according to your link, the orbital sagnac is 66 times greater.

Check this out, the earth's orbital speed is 30km/s and the rotational is 0.462.
30/0.462 = 66.

As we can see, the ratio of the sagnacs is based only on the linear speeds of each and the orbital sagnac is 66 times greater for two way. One way is half that. It is not at all based on the ratios of the angular speeds. We can see with math that is false.


A = π r ²
That is where you are making your error.

It is A = π r ² only if the light makes a complete circle back to the receiver.

That is why they used A so that a subset of the circle can be used.
Since we are dealing with 30 km, we must use the area of a sector of the circle representing the 30 km. That is where you are confused.

I will try to teach you though judging from that past, I will not get anywhere.

The normal derivation is
2πr ( 1/ (c-v) - 1/(c+v))

You see, the 2πr is the distance between the emitter and the receiver for the light path.

Now, we do not have the full 2πr. We only have a portion of it. So, we have 2πrK where 0 <= K <= 1. Remember, we must know the distance betweern the emitter and the receiver. Mathpages assumes light traversed the entire circle ± vt. That is not our experiment.

Now, let's continue.

∆t = 2πrK ( 1/ (c-v) - 1/(c+v)) = 2πrK ( 2v / ( c² - v²) ) = 4πrKv / (c² - v²).

Let's use angular velocity ω = v/r.

Hence,

∆t = 4πrKv / (c² - v²) = 4πr²Kω / (c² - v²)

Now, let's use the area created by the distance between the the emitter and the receiver.

A = Kπr²

∆t = 4πr²Kω / (c² - v²) = 4Aω / (c² - v²).

Since v is so small, it is usually eliminated from the denominator.

Hence,

∆t = 4Aω /c².

Now, let's try to understand A.

A = Kπr²

Since K is only a subset of the circumference, then K can be expressed as θ/360 since the full circumference is 360 degrees and we are only using a portion of it.

Hence, A = (θ/360) πr²

which is the familiar area of a sector of a circle.

Now, this is two way light travel and the total correction.


∆t = 4Aω /c²

Assume 30 km of a distance between the emitter and the receiver




Rotation of the earth

Earth radius 6378
ω - 7.292 e-5 radians/s

A = (θ/360) πr²

Calculate the number of degrees 30 km represents of the earth's circumference.

θ = (30) (360 ) / ( 2 π r ) = (30) (360 ) / ( 2 π 6378 ) = 0.27

A = (θ/360) πr² = (0.27/360) π (6378)² = 95798.77

Now for the calculation of the sagnac correction for 30 km
∆t = 4Aω /c² = 4 (95798.77)(7.292 e-5 )/c² = 3.105 e-10

So, the rotational sagnac is
∆t = 3.105 e-10




Orbit of the earth

Earth distance to sun 150,000,000
ω - 2.0 e-7

A = (θ/360) πr²

Calculate the number of degrees 30 km represents of the earth's orbit.

θ = (30) (360 ) / ( 2 π r ) = (30) (360 ) / ( 2 π 150,000,000 ) = 1.146 e-5

A = (θ/360) πr² = (1.146 e-5/360) π (150000000)² = 2249025000

Now for the calculation of the sagnac correction for 30 km
∆t = 4Aω /c² = 4 (2249025000)(2.0 e-7 )/c² = 2.0e-8

So, the orbital sagnac is
∆t = 2.0e-8


Now, let's go a little further, if you can, to understand what this means.

Take the ratios of the two sagnacs

2.0e-8 / 3.105 e-10 = 64.4

The linear velocity of the rotation of the earth is 0.465 km/s
The linear velocity of the orbit of the earth is 30 km/s

Take the linear velocity ratios
30/ 0.465 = 64.5

Ths difference above is rounding.

Hence, this proves the sagnac correction is a function purely of the linear velocity NOT THE ANGULAR VELOCITY of the receiver on the circular path.

If this statement is false, then this ratio would have other factors so that sagnac ratio would not equal the linear velocity ratio.

http://www.anti-relativity.com/forum/viewtopic.php?f=3&t=3912&start=60


What I am looking at right now is this:

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

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

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

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

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

What I like best about sandokhan is how every single scientist he references and quotes believes the earth is a globe.

It's true.  He repeatedly says the greatest physicists of all time agree with him, but when asked to produce a quote of them agreeing that the world is motionless, flat, etc, he grows oddly quite.

If the subtended angle gets smaller and smaller, the shape of the annular sector will approach that of a parallelogram.

In one case the formula leads to a figure of 1/365.

In the other one, accepted by mainstream science, peer reviewed leads to a figure of 60 (minimum value, in the case investigated by Dr. Su, the figure is 10,000).

Then, the derivation being discussed here must be very wrong.


How is the Sagnac actually calculated?

As each beam travels a common path but in opposite directions, they will be detected after different times of travel.

The time the beam travelling in the direction of rotation takes to travel before being detected, t₁

t₁= (2πr + Δl₁)/c

Δl₁ = rωt₁

t₂ = (2πr + Δl₂)/c

The main point: the difference between t₁ and t₂ gives an equation for the time elapsed between when the beam travelling in opposition to the rotation is detected, and when the beam travelling in the direction of rotation is detected.

The beams have to reach the point to where they were split, they recombine and are detected.


The same goes for the interferometer in the shape of a parallelogram.

The two beams of light are sent from the origin, they make full circuit (counterclockwise and clockwise) and then return to be detected at the same origin.

In this case

t₁ = (4a + 4b)/(c - v)

t₂ = (4a + 4b)/(c + v)

L = 4a + 4b


In both cases, the counter-propagating beams make a full circuit before returning to the point of origin to be detected.


In the case of the derivation being discussed here, we are faced with this:

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

Now as I said, the total time for each one is given by:
t1=t1a+t1b.
t2=t2a+t2b.


For starters, what is missing is not necessarily the contribution to the overall equation by the actual rotation of the Earth, but the nowhere to be found (R2 - R1) sides, which are taken into consideration (as they should) in the interferometer in the shape of parallelogram. The interferometer is moving at an angle, thus the sides do contribute to the shift, just as in the case of the parallelogram shaped interferometer.

Case in point

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

v = ωR
v₂ = ωR₂
v₁ = ωR₁

s = Rφ (s = arclength)

dt = 2φωR₂²/c² - 2φωR₁²/c²

dt = 2v₂s₂/c² - 2v₁s₁/c²

Therefore the derivation leads to a bizarre situation where actually two speeds are needed to complete the formula. And the second term is negative.

But this is not Sagnac.

Moreover, the derivation has to be started all over again, if the rotational Sagnac is needed: that is, we will have r₂ and r₁ starting from the center of the Earth to the interferometer. Certainly the same graphic could not be used for both the derivation of the orbital and the rotational Sagnac.


dt=4*A*w/c^2

Thus 4*A/c^2 is constant, thus you get the relation:
dt=k*w, where k=4*A/c^2.
As such, the shift is directly proportional to the angular velocity.

Thus comparing 2:
dto=k*wo
dtr=k*wr
And thus:
dto/dtr=k*wo/k*wr=wo/wo=1/365.

Completely false.

Viewed from the Earth the arcs of the graphic will be concave, and the s=Rφ formula cannot be used.
A new derivation will be needed for the rotational Sagnac, using r₂ and r₁ and the rotational speed of the Earth around its own axis.

The loop will have to resemble very closely the shape of the first interferometer, to approximate the length of the two arcs.

dt_r = 2φω_rr₂²/c² - 2φω_rr₁²/c²

dt_r = 2v_2rs₁/c² - 2v_1rs₂/c²


dt_o/dt_r =  [2v_2os₂ - 2v_1os₁]/[2v_2rs₁ - 2v_1rs₂]

Case in point

Love it!

Quote from: JackBlack

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

v = ωR
v₂ = ωR₂
v₁ = ωR₁

s = Rφ (s = arclength)

dt = 2φωR₂²/c² - 2φωR₁²/c²

dt = 2v₂s₂/c² - 2v₁s₁/c²

Therefore the derivation leads to a bizarre situation where actually two speeds are needed to complete the formula. And the second term is negative.

But this is not Sagnac.

Moreover, the derivation has to be started all over again, if the rotational Sagnac is needed: that is, we will have r₂ and r₁ starting from the center of the Earth to the interferometer. Certainly the same graphic could not be used for both the derivation of the orbital and the rotational Sagnac.

Quote from: JackBlack

dt=4*A*w/c^2

Thus 4*A/c^2 is constant, thus you get the relation:
dt=k*w, where k=4*A/c^2.
As such, the shift is directly proportional to the angular velocity.

Thus comparing 2:
dto=k*wo
dtr=k*wr
And thus:
dto/dtr=k*wo/k*wr=wo/wo=1/365.

Completely false.

Viewed from the Earth the arcs of the graphic will be concave, and the s=Rφ formula cannot be used.

Get real! Why can't the s=Rφ be used? Of course arclength = radius x angle!

A new derivation will be needed for the rotational Sagnac, using r₂ and r₁ and the rotational speed of the Earth around its own axis.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Failed again!

Read:
          Sagnac Effect, E. J. POST, Rev. Mod. Phys. 39, 475 (1967) – Published 1 April 1967

In Section III. General Aspects of the Theory, near end p. 478 it unambiguously states:

Summarizing, the experiments of Sagnac, Pogany and Michelson-Gale and the results of Harress, as re-interpreted by Harzer, demonstrate beyond doubt the following features of the Sagnac effect. The observed fringe shift
a) obeys formula (1);
b) does not depend on the shape of the surface A;
c) does not depend on the location of the centre of rotation;
d) does not depend on the presence of a comoving refracting medium in the path of the beam.

And if it made the slightest bit of difference the segment lengths are so small compared to the earth's orbit that those segments are effectively straight.

The whole problem here, Mr Sandokhan, is you trying to tenaciously hold onto an orbital Sagnac delay that is roughly 365 time as large as everybody else gets.

Whether you like the idea or not, from what I can see all the the papers that tou quote from were written but authors who firmly believed:

• The earth is a Globe about 12,700 km in diameter,
• The earth orbits the sun with at average distance of about 150,000 km,
• The stars are a few to many millions of light-years away.

They might differ on many aspects of physics, such as the ether, its existence and its propers, the cause of gravitation and the validity of relativity, but I have seen none that doubt the heliocentric system.

Even your idol, Nikola Tesla was a staunch believer in the Heliocentric Globe. Not only that, but one of the reasons for his dislike of Einstein and his relativity is that he thought Einstein was trying to disprove Newton's theories.

You should read, HOW COSMIC FORCES SHAPE OUR DESTINIES, ("Did the War Cause the Italian Earthquake") by Nikola Tesla
also at — How Cosmic Forces Shape Our Destinies — ("Did the War Cause the Italian Earthquake"), New York American, February 7, 1915  in which he states:

NATURAL FORCES INFLUENCE US

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Accepting all this as true let us consider some of the forces and influences which act on such a wonderfully complex automatic engine with organs inconceivably sensitive and delicate, as it is carried by the spinning terrestrial globe in lightning flight through space. For the sake of simplicity we may assume that the earth's axis is perpendicular to the ecliptic and that the human automaton is at the equator. Let his weight be one hundred and sixty pounds then, at the rotational velocity of about 1,520 feet per second with which he is whirled around, the mechanical energy stored in his body will be nearly 5,780,000 foot pounds, which is about the energy of a hundred-pound cannon ball.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The sun, having a mass 332,000 times that of the earth, but being 23,000 times farther, will attract the automaton with a force of about one-tenth of one pound, alternately increasing and diminishing his normal weight by that amount

Though not conscious of these periodic changes, he is surely affected by them.

The earth in its rotation around the sun carries him with the prodigious speed of nineteen miles per second . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
From the above address.

Sure, Nicola Tesla had a lot of "different ideas", but he most certainly did not believe in a flat stationary earth.

Get real! Why can't the s=Rφ be used? Of course arclength = radius x angle!

Viewed from the Earth the graphic will have concave arcs.

s = R x @ is applied to a circle, convex arcs.

r2 and r1 will have to be drawn to the same interferometer which has convex arcs as viewed from the Earth.

Even if it has the same area, you will reach the same situation encountered in the piece of shit derivation posted by jack.


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

v = ωR
v₂ = ωR₂
v₁ = ωR₁

s = Rφ (s = arclength)

dt = 2φωR₂²/c² - 2φωR₁²/c²

dt = 2v₂s₂/c² - 2v₁s₁/c²

Therefore the derivation leads to a bizarre situation where actually two speeds are needed to complete the formula. And the second term is negative.

But this is not Sagnac.


Let us now use the SAME INTERFEROMETER, ignoring the convex/concave arcs issue, using r1 and r2 (radius from the center of the Earth to the arcs of the interferometer).

dt_r = 2φω_rr₂²/c² - 2φω_rr₁²/c²

dt_r = 2v_2rs₁/c² - 2v_1rs₂/c²


dt_o/dt_r =  [2v_2os₂ - 2v_1os₁]/[2v_2rs₁ - 2v_1rs₂]

This is what you will get, not

dto=k*wo
dtr=k*wr
And thus:
dto/dtr=k*wo/k*wr=wo/wo=1/365.


You cannot have Sagnac with TWO DIFFERENT SPEEDS APPLIED TO A SINGLE INTERFEROMETER, and one of them being negative.

Something else is going on.

You are in the same category as rabinoz, an imbecile with no scientific background who enjoys being humiliated on a daily basis.

Nope, that would just be you.
You are yet to humiliate anyone.

You have multiple SQUARE ROOTS, easily computed by using Newton's iteration formula (quadratic convergence) or continued fractions.

This is irrelevant the topic at hand.

You are unable to focus on multiple things at once, so deal with the OP.

The parallelogram in the following graphic is virtually the same as the loop in your graphic.

I don't particularly care. It isn't your derivation.
PROVIDE YOUR OWN DERIVATION!!!
Show what is wrong with mine explicitly, the exact step that is wrong and what it should be.
If you can't, shut up.

Do it yourself. If you are unable to it shows you have no idea what you are talking about and thus are not in any position to judge.
I will ignore any outside links.

You have obtained as a final result, based on your piece of shit derivation, 1/365.

No, based upon my correct derivation which you are yet to refute.

v = angular velocity x radius

Again, there isn't just a single v.
The entire point is that there is a slight difference between the 2 (or a slight difference in length or both) which results in the shift.
If you had a rectangle which as just translating, there would be no shift.

I will next deal with the sector of a circle area in your derivation.

Really? You are going to finally deal with it?

Let tr = 4(Ar)(wr)/c² be the sagnac for the rotation and
to = 4(Ao)(wo)/c² be the sagnac for the orbit.

The area of the orbit and the radius of Earth is irrelevant.
We are discussing the effect of both on a single interferometer and thus they use the same area.

Thus the correct formulas would be:
tr = 4(Al)(wr)/c² be the sagnac for the rotation and
to = 4(Al)(wo)/c² be the sagnac for the orbit.

Where al is the area of the loop.

Thus the ratio simplifies to:
to/tr=wo/wr

Nothing more needed.

tr/to = 4(Ar)(wr)/c² / 4(Ao)(wo)/c² = (Ar)(wr)/(Ao)(wo)

Again, it is using a single loop. Thus the area is the same. Thus this reduces to tr/to=wr/wo.

Ar = 15 Rr = 15 * 6360 = 93900
Ao = 15 Ro = 15 * ( 150,000,000 ) = 2250000000

No it is not.
It is a single interferometer.
It is not a large ring the size of Earth's orbit compared with a large ring the size of Earth.

It is a small ring on Earth.

So once again, you have failed to address the problem and instead are just making up own horribly dishonest comparison.

Remember, I already did another one which showed the rotational Sagnac effect is much much much greater, by simply having a massive loop.

In order to have an honest comparison you need the same loop or you need to normalise the effect to the area.

So according to your link, the orbital sagnac is 66 times greater.

Not when you do an honest calculation.
When you do an honest calculation, using the same loop with the same area you get:
tr/to=wr/wo=365.

Thus the rotational Sagnac is 365 times greater.

A = π r ²
That is where you are making your error.

No. This is where you are making your error.
You are trying to use 2 completley different interferometers.
Remember, in the interferometer is a small one composed of 2 arcs.
There is not a single r, there are 2 separate Rs, and the area is alpha*(R2^2-R1^2)/2

It is A = π r ² only if the light makes a complete circle back to the receiver.

Yes, a circle of radius r.
But that isn't what is happening in this case.
The light isn't circling the sun.
It is going around a small loop away from the sun.
So you can't use that formula.

Since we are dealing with 30 km, we must use the area of a sector of the circle representing the 30 km. That is where you are confused.

No.
You would use that if the light went around the sector, that is, go along the circular arc, then go all the way into the sun, then outwards to the other end of the arc.
That isn't what is happening.
Instead, the light travels along the arc, then goes in a small bit, then travels along another arc, then goes back out.

As such, we need to use the area of the annular sector. Not a circular sector.
It is the difference between the 2 circular sectors.
That is alpha*(R2^2-R1^2)/2.

Again, if the sector has a width of b, and the arc is ~ l long, this becomes:
alpha*((R1+b)^2-R1^2)/2
=alpha*(R1^2+2*R1*b+b^2-R1^2)/2
=alpha*(2*R1*b+b^2)/2
=2*l*b/2 - alpha*b^2/2
=l*b+alpha*b^2/2

So the orbital radius is irrelevant.

I will try to teach you though judging from that past, I will not get anywhere.

And that's your problem.
You refuse to even entertain the possibility that you might be wrong. You refuse to address any issues that are raised.
Instead you act like an extremely arrogant teacher that is lecturing to ignorant children that should just accept whatever BS you say.

That is not how it works.
I am not an ignorant child.
I can think for myself.
I realise what you are saying is garbage.
Stop trying to teach people.
Try to debate. DEFEND YOUR CLAIMS!!
Show what is actually wrong with my derivation.
Show what the correct derivation is.
If you can't, SHUT UP!

The normal derivation is
2πr ( 1/ (c-v) - 1/(c+v))

Which only applies to a circle.
It does not apply in our case.
The axial components do not contribute at all.
Instead only 2 arc sections do, and their contributions partially cancel each other.

Now, we do not have the full 2πr. We only have a portion of it. So, we have 2πrK

Or you can do what I did and make it alpha*r, where alpha is the angle subtended, noting that it is in radians.
Or are you afraid that shows your problem too early?

That is not our experiment.

That is correct. So you can't do something simple like this without extensive modification.

The light has to travel a bit extra (or less) along one arc, but then has the opposite in the other arc.

∆t = 2πrK ( 1/ (c-v) - 1/(c+v)) = 2πrK ( 2v / ( c² - v²) ) = 4πrKv / (c² - v²).

Which is effectively what I did before, but this only applies to ONE of the arcs.
You need to use both arcs.

So the first part of dt (along the big arc)
dtp1=2*alpha*R2*v2/(c^2-v2^2), where v2=w*R2.
Thus we get:
dt=2*alpha*R2*R2*w/(c^2-v2^2)=2*alpha*R2^2*w/(c^2-v2^2)

But then we have the other arc as well, where in order to correctly calculate it we need to note their relation, and not that this arc is effectively backwards compared to the first (that is light travelling forwards along the large arc travels backwards along this arc and vice versa), thus this will be negative.
dtp2=-2*alpha*R1^2*w/(c^2-v1^2).

Then we add these 2 parts together:
dt=dtp1+dtp2=2*alpha*R2^2*w/(c^2-v2^2)-2*alpha*R1^2*w/(c^2-v1^2)
=2*alpha*w*(R2^2/(c^2-v2^2)-R1^2/(c^2-v1^2))

And by noting that c is much larger than either v, this simplifies to:
=2*alpha*w*(R2^2/c^2-R1^2/c^2)
=2*alpha*(R2^2-R1^2)*w/c^2.

You can't just ignore one arc, nor can you ignore the fact that the light is travelling in opposite in directions along the arc (relative to the rotation direction).

Before you bitch and moan and say you can't break it into parts, YOU CAN!!
For example, consider it for a cricle. You should get a result of:
dt=4*pi*R^2*w/(c^2-v^2).

So lets break this circle into 2 sections, each has the radius (and thus the velocity the same). alpha=pi., v1=v2=v, R1=R2=R
So dtp1=2*alpha*R2^2*w/(c^2-v2^2)
=2*pi*R/(c^2-v^2)

Now, because the light is still travelling in the same direction relative to the rotation, they have the same sign (both positive).
dtp2=2*alpha*R1^2*w/(c^2-v1^2)
=2*pi*R^2*w/(c^2-v^2)

Add the 2 together:
dt=dtp1+dtp2=2*pi*R^2*w/(c^2-v^2)+2*pi*R^2*w/(c^2-v^2)
=4*pi*R^2*w/(c^2-v^2)
Just like we needed.

So it sure seems like you can break the loop into parts and add up the contribution from each part.

Assume 30 km of a distance between the emitter and the receiver

Why?
We are discussing a loop. There is 0 distance between the emitter and the receiver.

So lets sub that in.
dt=0. So looks like Sagnac doesn't exist. That sure makes me think you did something wrong.
Perhaps it is because you tried to model the system as a section of a circle than what it is? 2 circular arcs joined together.

So, the rotational sagnac is

Not the one for the system we are analysing.
Try again.

Take the ratios of the two sagnacs

These are the ratio's for 2 completely different things.

Hence, this proves the sagnac correction is a function purely of the linear velocity NOT THE ANGULAR VELOCITY of the receiver on the circular path.

No it doesn't, as you are not focusing on the system in question.

If this statement is false, then this ratio would have other factors so that sagnac ratio would not equal the linear velocity ratio.

That is because you didn't actually make the ratio, instead you made up crap which is not the Sagnac effect.

What I am looking at right now is this:

Something you are yet to refute.
Let me know when you manage to find any error and can provide a correct derivation for the system in question.

In one case the formula leads to a figure of 1/365.

Yes, in the case you are unable to show any error with.

In the other one

which you are completely unable to do yourself or explain in any way and instead just repeatedly link to crap.
As such, it is irrelevant.
You are clearly unable to judge if it is correct or if it even applies.

Then, the derivation being discussed here must be very wrong.

Or the derivation you are linking to is wrong or has nothing at all to do with what we are discussing?

How is the Sagnac actually calculated?
As each beam travels a common path but in opposite directions, they will be detected after different times of travel.
The time the beam travelling in the direction of rotation takes to travel before being detected, t₁

And this already shows your problem.
The beam is entirely on one side of the centre of rotation.
How is it capable of completing a loop while continuing to move in the direction of rotation?
It is physically impossible.
In this case each beam will move in the direction of rotation for one part of the loop, and move back against the direction of rotation for another part.

You need to include that in your derivation. If you don't you will not get the correct result because you do not have a correct analysis of the system.

The same goes for the interferometer in the shape of a parallelogram.

Yes, the same goes, where you need to consider both the section where the light is moving with the motion, and the section where the light moves against the motion.
You can't just pretend they are the same.

The only time that works is when you have a fibre optic conveyor where you have a beam of light continually moving against the motion of the conveyor or continually moving with the motion of the conveyor.

If the entire loop moves together, that does not work.

In the case of the derivation being discussed here, we are faced with this:

A derivation you are yet to show any error in.

For starters, what is missing is ... the nowhere to be found (R2 - R1) sides

Because they do not contribute at all.
As was already explained.
All it does it needless complicates the math.

As an example, beam A, moving from the outside loop into the inside loop.
You now have 2 dimensions (3 including time) it is travelling in and need to include both.
For simplicity of the analysis, I will just focus on this one part, which will start at an angle of 0 and a time of 0.
Thus, in polar+temporal coordinates it starts at (R2,0,0), which match the cartesian+time coordinates.
But at the end, it has moved to R1, after some time tin, which means it has now rotated.
So we now get, in polar coordinates:
(R1,w*tin,tin), and in cartesian coordinates (R1*cos(w*tin),R1*sin(w*tin),tin).

And of course, this distance is c*tin.
So c*tin=(R1*cos(w*tin)-R2)^2+R1^2*sin(w*tin)^2

So this is already quite complex and ugly.
It also has a section where it goes out from R1 to R2, and it works out basically the same, but I will do it slightly differently by choosing to have it finish at an angle of 0 (as the starting and finishing angle is irrelevant.
So now it finishes at (R2,0,tout), and it starts at (R1,-w*tout,0), or in cartesian coordinates, (R1*cos(-w*tout),R1*sin(-w*tout),0) and through some trig identities (R1*cos(w*tout),-R1*sin(w*tout),0).
So now we have:
c*tout=(R2-R1*cos(w*tout))^2+R1^2*sin(w*tout)^2.

And, it isn't just beam A that does this.
Beam B does it as well, just at different sections. But it doesn't matter where it starts or ends, you obtain the same result.
So what this means, the total time taken for beam A will be:
ta=alpha*R2/(c-omega*R2)+alpha*R1/(c+omega*R1)+solve(c*tin=(R1*cos(w*tin)-R2)^2+R1^2*sin(w*tin)^2,tin)+solve(c*tout=(R2-R1*cos(w*tout))^2+R1^2*sin(w*tout)^2,tout)

Meanwhile for tb it will be:
tb=alpha*R2/(c+omega*R2)+alpha*R1/(c-omega*R1)+solve(c*tin=(R1*cos(w*tin)-R2)^2+R1^2*sin(w*tin)^2,tin)+solve(c*tout=(R2-R1*cos(w*tout))^2+R1^2*sin(w*tout)^2,tout)

So now to find the difference we have:
dt=ta-tb
=alpha*R2/(c-omega*R2)+alpha*R1/(c+omega*R1)+solve(c*tin=(R1*cos(w*tin)-R2)^2+R1^2*sin(w*tin)^2,tin)+solve(c*tout=(R2-R1*cos(w*tout))^2+R1^2*sin(w*tout)^2,tout)-(alpha*R2/(c+omega*R2)+alpha*R1/(c-omega*R1)+solve(c*tin=(R1*cos(w*tin)-R2)^2+R1^2*sin(w*tin)^2,tin)+solve(c*tout=(R2-R1*cos(w*tout))^2+R1^2*sin(w*tout)^2,tout))
=alpha*R2/(c-omega*R2)+alpha*R1/(c+omega*R1)+solve(c*tin=(R1*cos(w*tin)-R2)^2+R1^2*sin(w*tin)^2,tin)+solve(c*tout=(R2-R1*cos(w*tout))^2+R1^2*sin(w*tout)^2,tout)-alpha*R2/(c+omega*R2)-alpha*R1/(c-omega*R1)-solve(c*tin=(R1*cos(w*tin)-R2)^2+R1^2*sin(w*tin)^2,tin)-solve(c*tout=(R2-R1*cos(w*tout))^2+R1^2*sin(w*tout)^2,tout)
=alpha*R2/(c-omega*R2)-alpha*R2/(c+omega*R2)+alpha*R1/(c+omega*R1)-alpha*R1/(c-omega*R1)+solve(c*tin=(R1*cos(w*tin)-R2)^2+R1^2*sin(w*tin)^2,tin)-solve(c*tin=(R1*cos(w*tin)-R2)^2+R1^2*sin(w*tin)^2,tin)+solve(c*tout=(R2-R1*cos(w*tout))^2+R1^2*sin(w*tout)^2,tout)-solve(c*tout=(R2-R1*cos(w*tout))^2+R1^2*sin(w*tout)^2,tout)
=alpha*R2/(c-omega*R2)-alpha*R2/(c+omega*R2)+alpha*R1/(c+omega*R1)-alpha*R1/(c-omega*R1)

just like before.

So yet again, all you have done is appeal to a pointless pathetic distraction.
You are grasping at whatever straws you can to pretend you aren't wrong.

which are taken into consideration (as they should) in the interferometer in the shape of parallelogram.

For a parallelogram where these sides are not axial segments they need to be.
When these are axial segments, they don't need to be as they cancel, as shown above.

Therefore the derivation leads to a bizarre situation where actually two speeds are needed to complete the formula. And the second term is negative.

There is nothing bizarre about it at all.
I have already explained why this is the case.
Of course there are 2 speeds.
You have 2 circular arcs.
These 2 circular arcs, which are rotating about the point which is the centre of the circle they are an arc of, at the same angular velocity, with different radii, MUST have 2 different velocities.

So nothing bizarre about that. You need it.
if you don't, and instead have a formula with a single velocity, you have done something wrong.

And yes, the second term is negative, just as you would expect it to be, because the light travelling along the small arc is moving in opposite directions to the light travelling along the large arc (for each individual beam), just like has been explained to you many times.

But this is not Sagnac.

Yes it is. You have 2 counterpropagating beams travelling around a loop with a phase shift produced by it.

Moreover, the derivation has to be started all over again, if the rotational Sagnac is needed: that is, we will have r₂ and r₁ starting from the center of the Earth to the interferometer. Certainly the same graphic could not be used for both the derivation of the orbital and the rotational Sagnac.

Technically it could, it would just be much more complex. The purpose of this graphic was to show it was dependent upon the area of the loop, not how far away the loop is from the centre of rotation. It served its purpose. But that is why i produced the image in the OP, which can be used for both by approximating it to a loop like this one for Earth's rotation or Earth's orbit.

Dealing with non-circular arcs is a lot more complex.

Completely false.

If it was false you would be able to show an error. But you can't.

Viewed from the Earth the arcs of the graphic will be concave, and the s=Rφ formula cannot be used.

That's right. Instead you need to use a much more complex formula which will still produce the same result, that it is proportional to the area.

I also see that you are happy with approximating it is a parallelogram, but not happy to just flip it to approximate it as its mirror image. Why is that? Is it because you know it will show you are wrong?

A new derivation will be needed for the rotational Sagnac, using r₂ and r₁ and the rotational speed of the Earth around its own axis.
The loop will have to resemble very closely the shape of the first interferometer, to approximate the length of the two arcs.

And you will end up with the same result, the shift is proportional to the area of the loop.

Also, the loop will actually be a significantly different shape, with φ being vastly different.
if l is the length of the small arc, then φ*R1=l.
Thus φ=l/R1.
So, assuming a length of 1 m, then for the rotational Sagnac, where R1=3 678 100 m, φ_r=2.7e-7.
For the orbital Sagnac, where R1=~150 000 000 000 m, φ_o=6.7e-12.

The section of the circle they take up will be vastly different.

So your dishonesty now stems from you converting it to an linear form and not simplifying further, to try and make it far too difficult to analyse. An extra issue is that for the approximation used, we actually have slightly different s values.
But don't worry, I shall continue from a prior point and go through (noting the φ will have subscripts, and the area of the 2 loops are the same):

dt_o = 2φ_oω_oR₂²/c² - 2φ_oω_oR₁²/c²
=2φ_oω_o*(R₂² - R₁²)/c²
=4*w_o*A/c²

dt_r = 2φ_rω_rr₂²/c² - 2φ_rω_rr₁²/c²
=2φ_rω_*(rr₂²- r₁²)/c²
=4*w_r*A/c²

Now the comparison:
dt_o/dt_r=[4*w_o*A_o/c²]/[4*w_r*A_r/c²]
=w_o/w_r

So yet again, we are reduced to the ratio of the angular velocities.

But if you don't like that, perhaps we can try your formula:
dt_o/dt_r =  [2v_2os₂ - 2v_1os₁]/[2v_2rs₁ - 2v_1rs₂]

For simplicity, lets let R₁=150 000 000 km, and r₂=6400 km. (I know, not perfect, but I don't care).
And the difference between the 2 "R" or "r" values will be 1 km.
So R₂=150000001 km and r₁=6379 km.

We note w_o=1.99238E-07, and w_r=7.27221E-05
Thus, v_1o=29.88577486 km/s and v_2o=29.88577506 km/s
Likewise v_1r=0.465348412 km/s and v_2r=0.465421134 km/s

Now we arbitrarily define s₁=1 km.
s₂=R₂s₁/R₁=1.000000007 km. (To do it properly, that would mean that for Earth, you need s₂=0.99984375, so that may cause some issues).

Sticking those numbers straight in, we get dt_o/dt_r=0.005479686, ~=1/182
Putting in the more correct version for the more correct approximation, we instead get 0.00273994, or roughly 1/365

So even using that formula, the orbital Sagnac is much smaller than the rotational one.

Good job disproving yourself yet again.

Even if it has the same area, you will reach the same situation encountered in the piece of shit derivation posted by jack.

Do you mean you will reach the same situation encountered with my correct analysis, which reaches the correct conclusion that the orbital Sagnac is much smaller?

dt_o/dt_r =  [2v_2os₂ - 2v_1os₁]/[2v_2rs₁ - 2v_1rs₂]
This is what you will get, not
dto=k*wo
dtr=k*wr
And thus:
dto/dtr=k*wo/k*wr=wo/wo=1/365.

Actually, the 2 are equivalent, at least when done correctly.

You cannot have Sagnac with TWO DIFFERENT SPEEDS APPLIED TO A SINGLE INTERFEROMETER, and one of them being negative.

But that isn't happening.
We are analysing the 2 components (Earth's rotation and Earth's orbit) separately.

You could do it differently by treating the orbit as purely translational and then using Earth's sidereal rotation instead, and that gives you the same result, with the orbit not contributing significantly.

jack, you are not very good at math.

But still orders of magnitude above you!

Do not kid yourself.
You are in the same category as rabinoz, an imbecile with no scientific background who enjoys being humiliated on a daily basis.

Please don't insult JackBlack like that! He's certainly no imbecile and I he doesn't get "humiliated on a daily basis".

Quote from: JackBlack

And how do computers compute powers? well, they use the exponential function (technically an approximation to it).

Can I call you an idiot? I think I can.

I guess being called an idiot by someobe like you is high praise indeed!

You have multiple SQUARE ROOTS, easily computed by using Newton's iteration formula (quadratic convergence) or continued fractions.

Oh, really! Sure "Newton's iteration formula" works, but is quite inefficient in computation time.
Anyone that had done a little computer programming and numerical analysis would know that!

I have reduced the task of computing a logarithm, without using log tables, to that of simple additions/divisions using any iterative method.

If you do not know enough about math, then you should not be posting at all.

If your approximate method of computing ln(x) is so efficient, why isn't everyone using it?
Your method might be fine if your are only playing around with simply calculators that have nothing above sqrt(x).

But anyone seriously programming stuff like this on a real computer would know that the standard IEEE 754-2008 floating point format already contains the integer part of log₂(x) as the exponent.
All that remains is finding the log₂(mantissa), where the 0 < mantissa < 1 and most of the work is done.

Sandokhan, you can keep on playing around with your toy calculators and toy flat earth, while real people, idiots like JackBlack and even imbeciles like myself do the real work on the real Globe earth.

By the way it is totally untrue that I have "no scientific background". I am/was, not a physicist nor scientist, but an engineer.
And I have not been "humiliated on a daily basis" - in fact, I find your squirming and ramblings slightly amusing.

Some might find this amusing:

Now we know that Pythagoras never existed actually, as there were no ancient Greece/Rome/Egypt in our radical new chronology, and that the conspirators invented the irrational number concept in order to deceive the public regarding the Pythagorean comma (instead of a circle of fifths, we would have a spiral of fifths); they also invented, through J.S. Bach, the equal temperament scale in order to hide the real scale they used to produce levitation of large blocks of stone.

If you have a week to spare, read the rest and the link in:
          Flat Earth Believers / Re: Alternative Flat Earth Theory « Message by sandokhan on June 09, 2010, 07:35:44 PM »

And

But it is correct.

You simply have not read yet my Irrational Numbers do not Exist thread in the .net website (look for it there).

You have not done the research needed to understand this very important issue: the work done for the past 150 years which demonstrates that irrational numbers (algebraic or transcendental) do not exist, they are a mathematical pipe dream.

i would like to know just who cares about an "Irrational Numbers do not Exist thread" posted by a clearly irrational person.
Especially when we read:

LN V = 2ⁿ x (V^1/2n+1 - 1/V^1/2n+1)

This is the first explicit global formula for the natural logarithm, which can be used immediately to find LN V without resorting to logarithm tables, or calculators which feature the logarithm key: all we need is a calculator which has the four basic operations and the square root key. It links algebraic functions with elementary and higher transcendental functions.

Get real! Why can't the s=Rφ be used? Of course arclength = radius x angle!
Viewed from the Earth the graphic will have concave arcs.

So what! You always ignore this bit! Read:
          Sagnac Effect, E. J. POST, Rev. Mod. Phys. 39, 475 (1967) – Published 1 April 1967

In Section III. General Aspects of the Theory, near end p. 478 it unambiguously states:

Summarizing, the experiments of Sagnac, Pogany and Michelson-Gale and the results of Harress, as re-interpreted by Harzer, demonstrate beyond doubt the following features of the Sagnac effect. The observed fringe shift
a) obeys formula (1);
b) does not depend on the shape of the surface, A;
c) does not depend on the location of the centre of rotation;
d) does not depend on the presence of a comoving refracting medium in the path of the beam.

And if it made the slightest bit of difference the segment lengths are so small compared to the earth's orbit that those segments are effectively straight.

I can't shout much louder,  I'm getting a hoarse throat! 

Again, there isn't just a single v.
The entire point is that there is a slight difference between the 2 (or a slight difference in length or both) which results in the shift.
If you had a rectangle which as just translating, there would be no shift.

Your charade is over jackblack.

YOU CANNOT HAVE TWO Vs.

THE SHIFT IS BETWEEN C + V AND C - V, AND NOT BETWEEN ωR2 AND ωR1.

Your derivation DOES NOT INCLUDE SAGNAC.

Name a single bibliographical reference which uses TWO SPEEDS FOR THE SAGNAC.

Name a single one.

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

Here are the best mainstream treatises on the Sagnac effect:

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

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

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

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

Your derivation IS NOT THE SAGNAC EFFECT.


It is as simple as this.

Of course there are 2 speeds.
You have 2 circular arcs.
These 2 circular arcs, which are rotating about the point which is the centre of the circle they are an arc of, at the same angular velocity, with different radii, MUST have 2 different velocities.

YOU CANNOT!

This is not Sagnac.

Sagnac = one velocity and one shift.

Your entire derivation is wrong.

It leads to this:

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

v = ωR
v₂ = ωR₂
v₁ = ωR₁

s = Rφ (s = arclength)

dt = 2φωR₂²/c² - 2φωR₁²/c²

dt = 2v₂s₂/c² - 2v₁s₁/c²


BUT THIS IS WRONG!

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

You cannot have TWO SPEEDS, only one.

There is no negative term in the Sagnac.


Thus the correct formulas would be:
tr = 4(Al)(wr)/c² be the sagnac for the rotation and
to = 4(Al)(wo)/c² be the sagnac for the orbit.

It cannot be.

Your entire derivation is wrong.

It leads to this:

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

v = ωR
v₂ = ωR₂
v₁ = ωR₁

s = Rφ (s = arclength)

dt = 2φωR₂²/c² - 2φωR₁²/c²

dt = 2v₂s₂/c² - 2v₁s₁/c²


BUT THIS IS WRONG!

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

You cannot have TWO SPEEDS, only one.

There is no negative term in the Sagnac.


Again, it is using a single loop. Thus the area is the same. Thus this reduces to tr/to=wr/wo.

IT DOES NOT!

dt=4*A*w/c^2

Thus 4*A/c^2 is constant, thus you get the relation:
dt=k*w, where k=4*A/c^2.
As such, the shift is directly proportional to the angular velocity.

Thus comparing 2:
dto=k*wo
dtr=k*wr
And thus:
dto/dtr=k*wo/k*wr=wo/wo=1/365.

Completely false.

Viewed from the Earth the arcs of the graphic will be concave, and the s=Rφ formula cannot be used.
A new derivation will be needed for the rotational Sagnac, using r₂ and r₁ and the rotational speed of the Earth around its own axis.

The loop will have to resemble very closely the shape of the first interferometer, to approximate the length of the two arcs.

dt_r = 2φω_rr₂²/c² - 2φω_rr₁²/c²

dt_r = 2v_2rs₁/c² - 2v_1rs₂/c²


dt_o/dt_r =  [2v_2os₂ - 2v_1os₁]/[2v_2rs₁ - 2v_1rs₂]


As such, we need to use the area of the annular sector. Not a circular sector.
It is the difference between the 2 circular sectors.
That is alpha*(R2^2-R1^2)/2.

You cannot derive the Sagnac by using R1 or R2 to construct an interferometer.

Your derivation leads to this disaster, WHICH IS NOT SAGNAC:

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

v = ωR
v₂ = ωR₂
v₁ = ωR₁

s = Rφ (s = arclength)

dt = 2φωR₂²/c² - 2φωR₁²/c²

dt = 2v₂s₂/c² - 2v₁s₁/c²


BUT THIS IS WRONG!

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

You cannot have TWO SPEEDS, only one.

There is no negative term in the Sagnac.



dto = 2φoωoR22/c2 - 2φoωoR12/c2
=2φoωo*(R22 - R12)/c2
=4*wo*A/c2

dtr = 2φrωrr22/c2 - 2φrωrr12/c2
=2φrω*(rr22- r12)/c2
=4*wr*A/c2

Now the comparison:
dto/dtr=[4*wo*Ao/c2]/[4*wr*Ar/c2]
=wo/wr

WRONG!!!

This is what you get instead:

As such, we need to use the area of the annular sector. Not a circular sector.
It is the difference between the 2 circular sectors.
That is alpha*(R2^2-R1^2)/2.

You cannot derive the Sagnac by using R1 or R2 to construct an interferometer.

Your derivation leads to this disaster, WHICH IS NOT SAGNAC:

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

v = ωR
v₂ = ωR₂
v₁ = ωR₁

s = Rφ (s = arclength)

dt = 2φωR₂²/c² - 2φωR₁²/c²

dt = 2v₂s₂/c² - 2v₁s₁/c²


BUT THIS IS WRONG!

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

You cannot have TWO SPEEDS, only one.

There is no negative term in the Sagnac.


Viewed from the Earth the arcs of the graphic will be concave, and the s=Rφ formula cannot be used.
A new derivation will be needed for the rotational Sagnac, using r₂ and r₁ and the rotational speed of the Earth around its own axis.

The loop will have to resemble very closely the shape of the first interferometer, to approximate the length of the two arcs.

dt_r = 2φω_rr₂²/c² - 2φω_rr₁²/c²

dt_r = 2v_2rs₁/c² - 2v_1rs₂/c²


dt_o/dt_r =  [2v_2os₂ - 2v_1os₁]/[2v_2rs₁ - 2v_1rs₂]

This is the final answer for your piece of shit derivation.

You are done here jack.

Your derivation is useless and worthless.

So even using that formula, the orbital Sagnac is much smaller than the rotational one.

REALLY?

Let us use different figures to start with.

R1 = 150,000,000 km

r1 = 6400 km

Let R2 = 140,000,000 km

r2 = 6200 km

v2o = 30km/s

v1o = 28

v2r = 0.46528

v1r = 0.45074


s2 = 15 km

s1 = 5 km

dt_o/dt_r =  [2v_2os₂ - 2v_1os₁]/[2v_2rs₁ - 2v_1rs₂]

620/-8.87 = -70


YOUR DERIVATION IS VARIABLE AND AS SUCH IS A PIECE OF SHIT! Moreover, your value of 150,000,001 cannot be true! You can only have v<=150,000,000km.

I can always provide figures which will lead TO ANYTHING ELSE BUT 1/365.

Your derivation led to a situation where you can never have 1/365.


Here is the correct derivation:

dt = 2Lv/c2

v = angular velocity x radius

Earth rotational angular velocity = 7.27 × 10−5 rad/s

Radius = 6,378 km

vr = 0.463 km/s

Earth orbital angular velocity = 2 x 10-7 rad/s

Radius = 150,000,000 km

vo = 30 km/s


If we substitute these values in the Sagnac formula, we get

vo/vr = 30km/s/0.463 km/s =~ 60

The orbital Sagnac is larger than the rotational Sagnac at least by a factor of 60.


rabinoz...make yourself useful.

http://apps.nrbook.com/bateman/Vol1.pdf

One of the most famous treatises of the 20th century: HIGHER TRANSCENDENTAL FUNCTIONS, 3 VOLS.

Of course there are 2 speeds.
You have 2 circular arcs.
These 2 circular arcs, which are rotating about the point which is the centre of the circle they are an arc of, at the same angular velocity, with different radii, MUST have 2 different velocities.

If you analyse the Sagnac loop
      from an inertial (or near-inertial) frame of reference you have one velocity, c (if in a vacuum), and two distances
but if you analyse it
      on the rotating frame (non-inertial frame) you get two velocities and one distance.
Same Sagnac loop, same angular velocity.

rabinoz...make yourself useful.

http://apps.nrbook.com/bateman/Vol1.pdf
One of the most famous treatises of the 20th century: HIGHER TRANSCENDENTAL FUNCTIONS, 3 VOLS.

So? What's the point of that book, since transcendental functions all give transcendental answers and all transcendental numbers are irrational.

And you assert that:

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The odd thing about the discovery of the irrational numbers is the fact that the most celebrated "proofs" (geometrical/algebraic) were offered by Pythagoras himself to the public through his disciples. What Hippasus had uncovered was something much more interesting; that is, that THERE ARE NO IRRATIONAL NUMBERS (there exist only natural and rational numbers [with a finite decimal part]), and that Pythagoras was planning to inject the false concept of the irrational number to the public (scientific/philosophical). The two proofs offered by Pythagoras do not demonstrate ANYTHING regarding the existence of irrational numbers; Hippasus was assasinated by his colleagues so as not to reveal to the world what Pythagoras was actually trying to do: to mislead the coming generations of mathematicians.

http://www-groups.dcs.st-and.ac.uk/~history/Biographies/Kronecker.html

The only mathematician who realized that there were no irrational numbers in the real/physical world, and who continuously attacked R. Dedekind and G. Cantor for their mathematical pipe dreams, was Leopold Kronecker.

Kronecker is well known for his remark:-

God created the integers, all else is the work of man.
Irrational numbers are totally man-invented.

So your giving us a treatise on "HIGHER TRANSCENDENTAL FUNCTIONS" is, by your own definition a waste of time, since you claim that transcendental numbers being a sub-set of irrational numbers cannot exist.

But I, unlike you, believe in both irrational numbers, like 2^1/2, and transcendental numbers like e and π.

So, run away and don't come back until you have an exact value for 2^1/2 that is rational.

Let us now use a different set of values.

R1 = 150,000,000 km

r1 = 6400 km

Let R2 = 149,999,000 km

r2 = 6200 km

v2o = 30km/s

v1o = 29.9998

v2r = 0.46528

v1r = 0.45074


s2 = 15 km

s1 = 5 km


dt_o/dt_r =  [2v_2os₂ - 2v_1os₁]/[2v_2rs₁ - 2v_1rs₂]


600.002/-8,8694=-67.648


YOUR DERIVATION IS A PIECE OF SHIT JACK!


YOUR DERIVATION IS VARIABLE AND AS SUCH IS A PIECE OF SHIT! Moreover, your value of 150,000,001 cannot be true! You can only have v<=150,000,000km.

I can always provide figures which will lead TO ANYTHING ELSE BUT 1/365.

Your derivation led to a situation where you can never have 1/365.


Here is the correct derivation:

dt = 2Lv/c2

v = angular velocity x radius

Earth rotational angular velocity = 7.27 × 10−5 rad/s

Radius = 6,378 km

vr = 0.463 km/s

Earth orbital angular velocity = 2 x 10-7 rad/s

Radius = 150,000,000 km

vo = 30 km/s


If we substitute these values in the Sagnac formula, we get

vo/vr = 30km/s/0.463 km/s =~ 60

The orbital Sagnac is larger than the rotational Sagnac at least by a factor of 60.



rabinoz, I told you to make yourself useful, not a fool of yourself.

If you analyse the Sagnac loop
      from an inertial (or near-inertial) frame of reference you have one velocity, c (if in a vacuum), and two distances
but if you analyse it
      on the rotating frame (non-inertial frame) you get two velocities and one distance.

THE TWO SPEEDS ARE AS FOLLOWS: C + V AND C - V.

NOT ωR2 AND ωR1.


You cannot have two distances.



There is no such thing as STR:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1978311#msg1978311

Your charade is over jackblack.
YOU CANNOT HAVE TWO Vs.

Again, by definition you MUST have two "v"s.

Both arcs are moving at the same angular velocity about the same point, but with a different distance from that point.
As such, there must be two "v"s.
In the case of a circle interferometer on Earth, then due to Earth's orbit around the sun, only 2 points on the share the same velocity and there are an infinite number of velocities on the circle.

Your derivation DOES NOT INCLUDE SAGNAC.

My derivation is Sagnac.

Sagnac = one velocity and one shift.

No. One angular velocity, which is what I had.

Your entire derivation is wrong.
It leads to this:

And you are yet to show why it is wrong.

Discussing what it leads to is just saying you don't like the conclusion.
It doesn't show anything wrong with the derivation.

BUT THIS IS WRONG!

Prove it.

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

It can be made up of as many terms as you like.

You cannot have TWO SPEEDS, only one.

You can only have 1 ANGULAR speed, which is what I have.

There is no negative term in the Sagnac.

Yes there is.
The shift, dt=ta-tb.
It is entirely based upon the difference between times taken for 2 beams of light.
As such, you MUST have a negative term, which can be removed after simplificaiton.

It cannot be.

No. It MUST be.
If you use 2 different areas you are not making an honest comparison.

Your entire derivation is wrong.

Again, if it is wrong you would have been able to show exactly where it is wrong instead of repeatedly bitching about the conclusion.

Again, it is using a single loop. Thus the area is the same. Thus this reduces to tr/to=wr/wo.
IT DOES NOT!

It does.
If you wish to disagree, show why.

Completely false.

No, completely true.
You have been unable to refute it.

The loop will have to resemble very closely the shape of the first interferometer, to approximate the length of the two arcs.

And in the end it will boil down to the same formula, with dt=4*a*omega/c^2.

You cannot derive the Sagnac by using R1 or R2 to construct an interferometer.

Why not?
It is a loop with 2 counterpropagating beams of light.
Why can't I use it?
Is it because it shows you are wrong?

Your derivation leads to this disaster

The only "disaster" is that you have been shown to be full of shit.
That isn't good enough to claim it is wrong. You need to show an error in the derivation which you are yet to do.

I have shown that it does lead to Sagnac.
You end up with dt=4*A*w/c^2.
The classic formula for Sagnac, where A is the area of the loop.

This is the final answer for your piece of shit derivation.

I haven't provided a piece of shit derivation.
I have provided a sound derivation you are yet to refute.
If you were talking about that, does that mean you will shut up about it and stop lying and saying it is wrong?

Your derivation is useless and worthless.

Nope. It is working quite well to show you are full of shit.

So even using that formula, the orbital Sagnac is much smaller than the rotational one.
REALLY?

Yes, really.
I explained what each of the values were then did the math.
The orbital Sagnac was much smaller.

Let us use different figures to start with.

You mean completely change the comparison to make it a horribly dishonest comparison?

R1 = 150,000,000 km
r1 = 6400 km
Let R2 = 140,000,000 km
r2 = 6200 km

And this is your problem. You are now using 2 completely different interferometers.
The one for measuring the orbital Sagnac now has a distance of 10 000 000 km between the 2 arcs.
The one for the rotational Sagnac has a distance of 2000 km between the arcs.
So you have made the orbital one larger by a factor of roughly 5000. That will exaggerate the orbital Sagnac.

Do it properly such that the difference is the same for both.

s2 = 15 km
s1 = 5 km

Now you seem to just be making up figures as you go.
Remember, the 2 arcs need to subtend the same angle. If they don't, then the sections connecting them are no longer axial and thus contribute.
So let's see what angle it subtends?
R1 is 150 000 000 km. I assume s2 corresponds to that?
Then the angle subtended is 15/150 000 000=0.0000001=1e-7.
What about R2 and s1?
That is 5/140000000=3.6e-8
Not even close.

So you are just making up numbers to stick in.

As such, your numerical is meaningless.

Try again, within the restrictions of the problem.

You can't just make up all the numbers.
You can make up three of R1, R2, r1 and r2.
The 4th one will be a direct result of the other three. The difference between the 2 MUST be the same. In fact, you can only really choose 2 values, as the distance from the centre of Earth to the sun is already set as well, but as it doesn't matter where the loops are (as long as they are entirely on one side of the centre of rotation), you can choose 3.

All 4 "v"s will be based upon the angular velocity and thus you can't make up them either and instead must use the correct values.
You can only choose 1 s value.
The other value is then based upon this s value and the R values.
To do the approximation correctly you will need 3 s values, where the final one is based on the r values.

YOUR DERIVATION IS VARIABLE AND AS SUCH IS A PIECE OF SHIT!

No. You ignored the constraints relating the variables and thus ended up with a piece of shit.
Do it properly.

Here is the correct derivation:

I have already provided the correct derivation, until you refute it or DERIVE YOUR OWN (not starting at the end of one), mine remains correct.

dt = 2Lv/c2

No. That is for a FOC, or a circular interferometer centred on the axis of rotation.
It does not apply in this case.

If you wish to claim it applies you need to derive it, as I showed dt=4*A*w/c^2 applies.

Let us now use a different set of values.
R1 = 150,000,000 km
r1 = 6400 km
Let R2 = 149,999,000 km
r2 = 6200 km

Just like before, you fail.
The difference between R1 and R2 is 1000 km.
The difference between r1 and r2 is 2000 km.
So of course it is going to fail because you are using 2 completely different loops.
Try it with r2=6300 km.

s2 = 15 km
s1 = 5 km

And here you fail yet again.
They are still subtending different angles.
Try it with s1=14.9999 km.

Then try it with the third s value for the rotational Sagnac to complete the approximation, otherwise you still get the wrong result.

Better still, use the actual values for v1 and v2, not ones you have as rounded.

You cannot have two distances.

The interferometer is constructed from 2 circular arcs. as 2 arms and 2 axial segments as others.
You need to have two different distances.
If you don't, you just end up with a single arc and no shift at all (because the area is 0).

Again, quit with the external BS. Explain yourself, or shut up.

YOUR DERIVATION IS VARIABLE AND AS SUCH IS A PIECE OF SHIT! Moreover, your value of 150,000,001 cannot be true! You can only have v<=150,000,000km.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
v = angular velocity x radius

If v is a velocity, presumably in km/s and 150,000,000 km" is a distance,
Please explain just what "v <=150,000,000 km" is supposed to mean!

Again, by definition you MUST have two "v"s.

YOU CANNOT HAVE TWO SPEEDS.

You are making things up as you go along jack.

Here is a challenge for you.

Name a single bibliographical reference which uses TWO SPEEDS FOR THE SAGNAC.

Name a single one.

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

Here are the best mainstream treatises on the Sagnac effect:

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

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

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

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

Your derivation IS NOT THE SAGNAC EFFECT.


My derivation is Sagnac.

It could be anything else BUT Sagnac.

YOU CANNOT HAVE TWO Vs.

THE SHIFT IS BETWEEN C + V AND C - V, AND NOT BETWEEN ωR2 AND ωR1.

Your derivation DOES NOT INCLUDE SAGNAC.

Sagnac = one velocity and one shift.

Your entire derivation is wrong.

It leads to this:


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

v = ωR
v₂ = ωR₂
v₁ = ωR₁

s = Rφ (s = arclength)

dt = 2φωR₂²/c² - 2φωR₁²/c²

dt = 2v₂s₂/c² - 2v₁s₁/c²


BUT THIS IS WRONG!

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

You cannot have TWO SPEEDS, only one.

There is no negative term in the Sagnac.


One angular velocity, which is what I had.

YOU HAVE TWO SPEEDS.

THIS IS NOT SAGNAC.

THE SHIFT IS BETWEEN C + V AND C - V, AND NOT BETWEEN ωR2 AND ωR1.


You are history here jack.


Prove it.

Sure thing.

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

v = ωR
v₂ = ωR₂
v₁ = ωR₁

s = Rφ (s = arclength)

dt = 2φωR₂²/c² - 2φωR₁²/c²

dt = 2v₂s₂/c² - 2v₁s₁/c²


BUT THIS IS WRONG!

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

You cannot have TWO SPEEDS, only one.

There is no negative term in the Sagnac.


SAGNAC = ONE SPEED, ONE SHIFT

YOU HAVE TWO SPEEDS AND A NEGATIVE TERM.

THAT IS NOT SAGNAC AT ALL.


The shift, dt=ta-tb.
It is entirely based upon the difference between times taken for 2 beams of light.
As such, you MUST have a negative term, which can be removed after simplificaiton.


NOT IN THE FINAL FORMULA.

THERE IS NO SUCH THING AS SAGNAC SHIFT WITH TWO SPEEDS, TWO TERMS ONE OF WHICH IS NEGATIVE.

NO SUCH THING.

You can only have 1 ANGULAR speed, which is what I have.

You have two speeds.

Again, if it is wrong you would have been able to show exactly where it is wrong

It is catastrophically wrong.

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

v = ωR
v₂ = ωR₂
v₁ = ωR₁

s = Rφ (s = arclength)

dt = 2φωR₂²/c² - 2φωR₁²/c²

dt = 2v₂s₂/c² - 2v₁s₁/c²


BUT THIS IS WRONG!

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

You cannot have TWO SPEEDS, only one.

There is no negative term in the Sagnac.



THIS IS WHAT YOUR DERIVATION FINALLY LEADS TO:

dt = 2v₂s₂/c² - 2v₁s₁/c²

TWO DIFFERENT SPEEDS, TWO TERMS ONE OF WHICH IS NEGATIVE.

THIS IS NOT SAGNAC.

Here are the best references on Sagnac:

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

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

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

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

Your derivation IS NOT THE SAGNAC EFFECT.


You end up with dt=4*A*w/c^2.
The classic formula for Sagnac, where A is the area of the loop.

You don't.

You end up with this piece of shit.

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

v = ωR
v₂ = ωR₂
v₁ = ωR₁

s = Rφ (s = arclength)

dt = 2φωR₂²/c² - 2φωR₁²/c²

dt = 2v₂s₂/c² - 2v₁s₁/c²


BUT THIS IS WRONG!

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

You cannot have TWO SPEEDS, only one.

There is no negative term in the Sagnac.


Subtend the same angle you say.

R2 = 150,000,000

s = r x @

s2 = 15 km

15 = 150,000,000 x @

@ = 1x10^-7

s1 = 149,990,000 x @ = 14.999


r2 = 6450 km

s2r = 15km

@ = 0.0023256

r1 = 6500 km

s1r = 15.1164

v2r = 0.468915

v1r = 0.47255


dt_o/dt_r =  [2v_2os₂ - 2v_1os₁]/[2v_2rs₁ - 2v_1rs₂]

0.11996/0.0001342 = 1057.74

A long way off from 1/365.

Let us different figures.

r2 = 16,400 km
r1 = 6,400 (10,000 difference just like the orbital calculation)
s2 = 15 km
@ = 0.000915
s1r = 5.853 km
v2r = 1.1923
v1r = 0.46528

0.11996/-0.0013362 = -89.777


REMEMBER: YOU CANNOT A VARIABLE RATIO.

Here is the correct calculation for a interferometer in the shape of parallelogram:

dt = 2Lv/c2

v = angular velocity x radius

Earth rotational angular velocity = 7.27 × 10−5 rad/s

Radius = 6,378 km

vr = 0.463 km/s

Earth orbital angular velocity = 2 x 10-7 rad/s

Radius = 150,000,000 km

vo = 30 km/s


If we substitute these values in the Sagnac formula, we get

vo/vr = 30km/s/0.463 km/s =~ 60

The orbital Sagnac is larger than the rotational Sagnac at least by a factor of 60.



I have debunked your entire derivation.

It turned out to be both useless and worthless.


Here is what you need to do jack.

Name a single bibliographical reference which uses TWO SPEEDS FOR THE SAGNAC.

Name a single one.

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

Here are the best mainstream treatises on the Sagnac effect:

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

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

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

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

Your derivation IS NOT THE SAGNAC EFFECT.

Again, by definition you MUST have two "v"s.
YOU CANNOT HAVE TWO SPEEDS.

STOP JUST REPEATING THE SAME REFUTED BULLSHIT!!!
I have already explained why you need the 2 "v"s.
You have 2 circular arcs which are rotating around the same point.
These arcs have different radii and thus MUST have different "v"s.
If you think you can have the same, draw a diagram showing how that is possible.

You are making things up as you go along jack.

No. That would be you. Making up whatever BS excuses you can to avoid admitting you were wrong.

Here is a challenge for you.

No. This entire thread is a challenge for you, to defend your BS claims or accept that you were wrong.
Deal with that by showing what is wrong with my derivation and providing your own, or admit defeat. Then we can move on.

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

As I said EVERY ONE IN EXISTENCE!!!
It is a time difference between 2 beams of light. As such it must use a negative term.

Here are the best mainstream treatises on the Sagnac effect:

Again, stop just linking to things you don't understand.
These either agree with me that it is the area of the loop, i.e. dt=4*A*w/c^2, and thus the orbital sagnac is much smaller, or do not address this.

ONE TERM FOR THE SAGNAC. NOT TWO TERMS. ONE SPEED. NOT TWO SPEEDS. NO NEGATIVE TERMS INCLUDED.
Your derivation IS NOT THE SAGNAC EFFECT.

No. 2 terms (or as many as you want) in the derivation, including negative terms as it is the time difference for 2 beams of light.
In the final result it can often be simplified to 1 term.
Again, it is only ANGULAR velocity.
When you deal with linear or tangential velocities, it is only one speed for a circular loop.

My derivation is the derivation of the shift (measured as a time difference) between 2 counterpropagating beams of light moving around a closed loop that is rotatin.
That is Sagnac.

Your entire derivation is wrong.
It leads to this:

Again, bitching about what it leads to doesn't show it is wrong.
You need to show an error in the derivation which you are yet to do.

Prove it.
Sure thing.

I didn't say just copy and paste what I said or repeat the same refuted BS. I said prove it.

Repeating the same baseless claims isn't proving anything, except that lack any sense of honesty or rationality.

The shift, dt=ta-tb.
It is entirely based upon the difference between times taken for 2 beams of light.
As such, you MUST have a negative term, which can be removed after simplificaiton.

NOT IN THE FINAL FORMULA.

And I didn't have a negative in my final formula.
Remember what it was?
dt=4*A*w/c^2.
Where was the extra term? Where was the negative term?

Grow up.

You can only have 1 ANGULAR speed, which is what I have.
You have two speeds.

I have a single angular speed, which is what is required.

Again, if it is wrong you would have been able to show exactly where it is wrong
It is catastrophically wrong.

WHAT IS WRONG?

STOP JUST BITCHING ABOUT THE CONCLUSION!!!
Show a problem with the derivation.

If you want to complain that it has 2 terms, use the final form I had, dt=4*A*w/c^2.
It doesn't have 2 terms. It doesn't have a negative term.

If you want to ignore that and use some other form, such as one at the start, then derive Sagnac completely without using 2 terms or any negative terms.
This will be impossible, as by definition, Sagnac is a difference and that requires 2 terms, one of which is negative.

You end up with dt=4*A*w/c^2.
The classic formula for Sagnac, where A is the area of the loop.
You don't.
You end up with this piece of shit.

No, you end up with that piece of shit by manipulating what I had.

Subtend the same angle you say.
R2 = 150,000,000
15 = 150,000,000 x @
s1 = 149,990,000 x @

From this I assume you have R1=149990000
Thus a difference of 10 000 km.

r2 = 6450 km
r1 = 6500 km

Yet here it is only a difference of 50 km.
So you still have the loop for the orbit MUCH larger.

Try again.

I have debunked your entire derivation.

No you haven't. You have continually bitched about it not giving you the conclusion you wanted with you repeatedly lying about it to try and make up whatever excuse you can to dismiss it.

You are yet to show a single error with it, and you have been completely unable to provide your own derivation.

I have already explained why you need the 2 "v"s.
You have 2 circular arcs which are rotating around the same point.
These arcs have different radii and thus MUST have different "v"s.

YOU CANNOT HAVE TWO Vs, TWO SPEEDS. PERIOD!

Here is what you need to do jack.

Name a single bibliographical reference which uses TWO SPEEDS FOR THE SAGNAC.

Name a single one.

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

Here are the best mainstream treatises on the Sagnac effect:

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

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

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

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

Your derivation IS NOT THE SAGNAC EFFECT.


As I said EVERY ONE IN EXISTENCE!!!
It is a time difference between 2 beams of light. As such it must use a negative term.

They do not.

Each reference provided USED A SINGLE TERM FOR THE SAGNAC SHIFT.

AND IT IS POSITIVE.

NOT TWO SPEEDS.

NOT TWO TERMS.

NO NEGATIVE TERM.

You have just been disproven.


In the final result it can often be simplified to 1 term.
Again, it is only ANGULAR velocity.

Your entire derivation is wrong.

It leads to this:


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

v = ωR
v₂ = ωR₂
v₁ = ωR₁

s = Rφ (s = arclength)

dt = 2φωR₂²/c² - 2φωR₁²/c²

dt = 2v₂s₂/c² - 2v₁s₁/c²


BUT THIS IS WRONG!

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

You cannot have TWO SPEEDS, only one.

There is no negative term in the Sagnac.


You can no longer fool anybody here jack.


dt=4*A*w/c^2.
Where was the extra term? Where was the negative term?

YOUR FINAL RESULT IS BASED ON A FALSE EQUATION. REMEMBER?

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

v = ωR
v₂ = ωR₂
v₁ = ωR₁

s = Rφ (s = arclength)

dt = 2φωR₂²/c² - 2φωR₁²/c²

dt = 2v₂s₂/c² - 2v₁s₁/c²


BUT THIS IS WRONG!

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

You cannot have TWO SPEEDS, only one.

There is no negative term in the Sagnac.


I have a single angular speed, which is what is required.

YOU DON'T!

YOU HAVE TWO SPEEDS.

dt = 2v₂s₂/c² - 2v₁s₁/c²


if you want to complain that it has 2 terms, use the final form I had, dt=4*A*w/c^2.
It doesn't have 2 terms. It doesn't have a negative term.

Your final form was derived from this:

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

v = ωR
v₂ = ωR₂
v₁ = ωR₁

s = Rφ (s = arclength)

dt = 2φωR₂²/c² - 2φωR₁²/c²

dt = 2v₂s₂/c² - 2v₁s₁/c²



WHAT YOU GET IS TWO SPEEDS.

TWO FINAL TERMS.

ONE OF WHICH IS NEGATIVE.

Here is what you need to do jack.

Name a single bibliographical reference which uses TWO SPEEDS FOR THE SAGNAC.

Name a single one.

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

Here are the best mainstream treatises on the Sagnac effect:

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

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

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

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

Your derivation IS NOT THE SAGNAC EFFECT.


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


Yet here it is only a difference of 50 km.
So you still have the loop for the orbit MUCH larger.

Let us different figures.

r2 = 16,400 km
r1 = 6,400 (10,000 difference just like the orbital calculation)
s2 = 15 km
@ = 0.000915
s1r = 5.853 km
v2r = 1.1923
v1r = 0.46528

0.11996/-0.0013362 = -89.777


REMEMBER: YOU CANNOT A VARIABLE RATIO.

Here is the correct calculation for a interferometer in the shape of parallelogram:

dt = 2Lv/c2

v = angular velocity x radius

Earth rotational angular velocity = 7.27 × 10−5 rad/s

Radius = 6,378 km

vr = 0.463 km/s

Earth orbital angular velocity = 2 x 10-7 rad/s

Radius = 150,000,000 km

vo = 30 km/s


If we substitute these values in the Sagnac formula, we get

vo/vr = 30km/s/0.463 km/s =~ 60

The orbital Sagnac is larger than the rotational Sagnac at least by a factor of 60.



I have debunked your entire derivation.

It turned out to be both useless and worthless.

I have already explained why you need the 2 "v"s.
You have 2 circular arcs which are rotating around the same point.
These arcs have different radii and thus MUST have different "v"s.
YOU CANNOT HAVE TWO Vs, TWO SPEEDS. PERIOD!

STOP JUST ASSERTING THE SAME CRAP!!!

Tell me how the 2 arcs of different radii can be circling the centre point with the same angular velocity yet magically have 2 different speeds?
If you can't THEN SHUT UP!!!

No stupid pathetic appeals to ignorance. Explain how or shut up.

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

Yes one term:
dt=4*A*w/c^2.

They do not.
Each reference provided USED A SINGLE TERM FOR THE SAGNAC SHIFT.

No. They ended up with one. But the derivation requires 2.

You have just been disproven.

Nope. You are still yet to show any error in the derivation, so I remain unrefuted.

Your entire derivation is wrong.
It leads to this:

Again, bitching about what it leads to just means you don't like it.
It doesn't show anything wrong with it.
It leads to dt=4*A*w/c^2.
Thus the orbital sagnac is 1/365 times the rotational one.

If you wish to claim otherwise you need to show an error in my derivation and provide your own derivation.
If you can't, then you have no basis to say it is wrong.
Saying you don't like the conclusion is not showing a problem with the derivation.

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

Then stop manipulating it into 2.
Leave it as the one term that I had it as:
dt=4*A*w/c^2.

You can no longer fool anybody here jack.

You are the only one trying to fool people here, and I suspect you are failing miserably.

dt=4*A*w/c^2.
Where was the extra term? Where was the negative term?
YOUR FINAL RESULT IS BASED ON A FALSE EQUATION. REMEMBER?

No. This was the final result:
dt=4*A*w/c^2.

Where is the extra term? Where is the negative term?

dt = 2φωR₂²/c² - 2φωR₁²/c²

This is your garbage, which you have intentionally made complex to pretend it isn't the Saganc.

I have a single angular speed, which is what is required.
YOU DON'T!
YOU HAVE TWO SPEEDS.

I have 1 angular speed. There are 2 tangential speeds, as is required for this system.

if you want to complain that it has 2 terms, use the final form I had, dt=4*A*w/c^2.
It doesn't have 2 terms. It doesn't have a negative term.
Your final form was derived from this:

I don't care.
Use the final form. Not your bastardisation of another form.

It is the final form which has a single term, just like the others.

If you go back, you can then bring in more terms, just like the others.

TWO FINAL TERMS.

No. ONE FINAL TERM!!!
dt=4*A*w/c^2.

Here is what you need to do jack.

No. Here is what you need to do:
Stop your pathetic distractions.
Deal with the derivation at hand.
Stop bitching about multiple speeds or multiple terms or any crap like that.
Deal with the derivation itself, with how I determined what the time difference is.
You need to show an error in that, a mathematical error, and provide your own derivation.

If you can't, you have no basis to claim my result is wrong.

Yet here it is only a difference of 50 km.
So you still have the loop for the orbit MUCH larger.
Let us different figures.

STOP JUST USING DIFFERENT FIGURES!!!
Just go back and use the simplified formula.
Every time you use different figures you are just making shit up and violating the conditions required for the loop.

s2 = 15 km
@ = 0.000915
s1r = 5.853 km

And now you break the equivalence of the 2 loops.

You need to have a small difference between the 2 arcs if you want to treat them as the same loop, and even then, this method requires treating the 2 loops separately.
You can pretend the 2 arcs are the same.
As you pointed out, a convex arc is different to a concave arc.

Here is the correct calculation for a interferometer in the shape of parallelogram:

Don't just jump to a calculation.
DO THE DERIVATION!!!
If you can't, it is useless.

Here is the correct calculation for any interferometer, as the vast majority of your sources agree:
dt=4*A*w/c^2, where A is the area of the loop.

Thus dto=4*A*wo/c^2
dtr=4*A*wr/c^2.
Thus dto/drt=(4*A*wo/c^2)/(4*A*wr/c^2)
=wo/wr.

wr=365 rev/year
ro=1 rev/year.
If we substitute these values in the Sagnac formula, we get
dto/dtr=1/365.

I have debunked your entire derivation.

You have debunked nothing.
You spouting baseless crap and bitching and moaning about the conclusion doesn't debunk anything.
You need to show a problem with the math.
If you can't, you fail.

It turned out to be both useless and worthless.

Nope. That would be you.

jack, get some fresh air and try to gain some congruence into your thought.

You were assigned a task, remember?

Find a single source which uses TWO SPEEDS FOR THE SAGNAC and TWO DIFFERENT TERMS, ONE OF WHICH IS NEGATIVE, FOR THE FINAL RESULT.

You didn't find any, did you?


You are history here jack.

You failed miserably.

Did you really think you have chance with me in a direct debate?


Tell me how the 2 arcs of different radii can be circling the centre point with the same angular velocity yet magically have 2 different speeds?

Very simple, it's all your fault.

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

v = ωR
v₂ = ωR₂
v₁ = ωR₁

s = Rφ (s = arclength)

dt = 2φωR₂²/c² - 2φωR₁²/c²

dt = 2v₂s₂/c² - 2v₁s₁/c²


BUT THIS IS WRONG!

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

You cannot have TWO SPEEDS, only one.

There is no negative term in the Sagnac.


Yes one term:
dt=4*A*w/c^2.

YOUR FINAL RESULT IS BASED ON A FALSE EQUATION. REMEMBER?

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

v = ωR
v₂ = ωR₂
v₁ = ωR₁

s = Rφ (s = arclength)

dt = 2φωR₂²/c² - 2φωR₁²/c²

dt = 2v₂s₂/c² - 2v₁s₁/c²

Your derivation led to this piece of shit. The same derivation, likewise, led to your piece of shit.

Which makes the entire derivation a huge piece of shit.


They ended up with one. But the derivation requires 2.

Those two are: C + V AND C - V.

Not ωR2 AND ωR1.

One final term. NOT TWO FINAL TERMS, ONE OF WHICH IS NEGATIVE.

It leads to dt=4*A*w/c^2.
Thus the orbital sagnac is 1/365 times the rotational one.
Here is the correct calculation for any interferometer, as the vast majority of your sources agree:
dt=4*A*w/c^2, where A is the area of the loop.

Thus dto=4*A*wo/c^2
dtr=4*A*wr/c^2.
Thus dto/drt=(4*A*wo/c^2)/(4*A*wr/c^2)
=wo/wr.

wr=365 rev/year
ro=1 rev/year.
If we substitute these values in the Sagnac formula, we get
dto/dtr=1/365.

WRONG!!! Let's put your word to the test.

This is what you get instead:

As such, we need to use the area of the annular sector. Not a circular sector.
It is the difference between the 2 circular sectors.
That is alpha*(R2^2-R1^2)/2.

You cannot derive the Sagnac by using R1 or R2 to construct an interferometer.

Your derivation leads to this disaster, WHICH IS NOT SAGNAC:

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

v = ωR
v₂ = ωR₂
v₁ = ωR₁

s = Rφ (s = arclength)

dt = 2φωR₂²/c² - 2φωR₁²/c²

dt = 2v₂s₂/c² - 2v₁s₁/c²


BUT THIS IS WRONG!

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

You cannot have TWO SPEEDS, only one.

There is no negative term in the Sagnac.


dt_o/dt_r =  [2v_2os₂ - 2v_1os₁]/[2v_2rs₁ - 2v_1rs₂]

This is the final correct ratio of your piece of shit derivation.

YOU NEED TWO DIFFERENT SETS OF RADII, FOR BOTH THE ROTATIONAL AND FOR THE ORBITAL CALCULATIONS.

YOUR HARE BRAINED FORMULA IS USELESS:

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

YOU NEED TWO SETS OF RADII.

HERE IS THE CORRECT FORMULA, DIRECTLY DERIVED FROM YOUR PIECE OF SHIT DERIVATION:

dt_o/dt_r =  [2v_2os₂ - 2v_1os₁]/[2v_2rs₁ - 2v_1rs₂]


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

The sooner you understand this fact, the better for you.


Here is the correct derivation for a parallelogram shaped interferometer:

https://www.theflatearthsociety.org/forum/index.php?topic=72601.msg1979685#msg1979685

Here is the correct calculation for a interferometer in the shape of parallelogram:

dt = 2Lv/c2

v = angular velocity x radius

Earth rotational angular velocity = 7.27 × 10−5 rad/s

Radius = 6,378 km

vr = 0.463 km/s

Earth orbital angular velocity = 2 x 10-7 rad/s

Radius = 150,000,000 km

vo = 30 km/s


If we substitute these values in the Sagnac formula, we get

vo/vr = 30km/s/0.463 km/s =~ 60

The orbital Sagnac is larger than the rotational Sagnac at least by a factor of 60.


See how easy it is to debunk your piece of shit derivation?


You need to show a problem with the math.

Don't try and play stupid tricks with me jack.

Your entire derivation is based on faulty premises.

As such, it leads to this piece of shit: your equations.

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

v = ωR
v₂ = ωR₂
v₁ = ωR₁

s = Rφ (s = arclength)

dt = 2φωR₂²/c² - 2φωR₁²/c²

dt = 2v₂s₂/c² - 2v₁s₁/c²


BUT THIS IS WRONG!

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

You cannot have TWO SPEEDS, only one.

There is no negative term in the Sagnac.



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


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

Here are the best mainstream treatises on the Sagnac effect:

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

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

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

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

Your derivation IS NOT THE SAGNAC EFFECT.

jack, get some fresh air and try to gain some congruence into your thought.

You were assigned a task, remember?
Find a single source which uses TWO SPEEDS FOR THE SAGNAC and TWO DIFFERENT TERMS, ONE OF WHICH IS NEGATIVE, FOR THE FINAL RESULT.

You, Mr Sandokhan don't get to assign anyone tasks around here especially when you take no notice of what is written anyway.
You take no notice even of your own papers!

You didn't find any, did you?

RUBBISH!
Look at:

Explicitly derive the Sagnac effect due to Earth's orbit and Earth's rotation on a ring interferometer constructed as in the following (very much not to scale) diagram:

The Earth is shown in blue.
The sun is shown in red.
This has a square loop interferometer, of side length l, which is sitting on the surface of Earth (which has a radius of r, and is following a circular orbit of radius r), at the equator.
This is taken on the equinox, simplified to a perfectly circular orbit of 365 days for simplicity (so it will just be an approximation).
The lengths of the interferometer are straight, but are sufficiently short to be approximated as axial spokes or as sections of a circular arc centred on either the centre of Earth or the centre of Earth's orbit.
Also note that the interferometer stands vertically, that is one arm is on the ground, and another is l above the ground.

The path segments, one on the surface of earth and one standing l above the ground are at different radii from the centres of rotation.

Hence in analysing the rotational Sagnac the linear velocities are:

• For the rotational Sagnac (about the Earth's centre): (ω_E × r) and (ω_E × (r + l)), where ω_E is the earth's rotational angular velocity.
  
  
• For the orbital Sagnac (about the Sun's centre): (ω_O × R) and (ω_O × (R - l)), where ω_O is the earth's orbital angular velocity.

And please, Mr Sandokhan, remember that

The observed fringe shift
b) does not depend on the shape of the surface A;
c) does not depend on the location of the centre of rotation;
d) does not depend on the presence of a comoving refracting medium in the path of the beam.

From:  Sagnac Effect, E. J. POST, Rev. Mod. Phys. 39, 475 (1967) – Published 1 April 1967

You do not get the option of leaving this bit out if it suits you!