sandokhan lies regarding the Sagnac effect

You were assigned a task, remember?

And I said go fuck yourself. Remember?
You were assigned a task first.
You complete you task before giving me any.

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

No. That would require you to actually debate which you seem incapable of.
Instead it is just you repeating the same (or similar) bulslhit and me handing your ass to you.

Do you think you have a chance with me in a direct debate, where you will be required to debate, addressing the points I raise and not just trying to bury me with your BS?

Because you are yet to address the derivation I provided and you are yet to provide your own.

You have failed to meet the OP in any way.

You can't even answer simple questions.
Instead just you spam the same refuted crap.

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

No, as you are yet to show anything wrong with the equation.
Instead my final result agrees with mainstream science with what the Sagnac effect is:
dt=4*A*w/c^2.

So it sure looks like mine is correct.

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

No. Your manipulation of my derivation led to that garbage.
My derivation led to the common form expressed by many:
dt=4*A*w/c^2.

As I have pointed out many times, you can't bitch about the conclusion not being what you like. You need to show a problem with the derivation itself.

Those two are: C + V AND C - V.
Not ωR2 AND ωR1.

No. Those 2 are typically c+wr and c-wr.
But that only holds for a circle centred on the centre of rotation.

Like I said, if you think I am wrong, explain why. Explain why the shift I got is wrong by showing an error in the derivation, and provide your own derivation.

WRONG!!! Let's put your word to the test.
This is what you get instead:

The above is what I got. Stop lying.
If you want to put my word to the test you need to show an error in the derivation.

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

Again, you are yet to justify that.

In what way is the system I constructed not an interferometer?
It is a closed loop with 2 counterpropagating beams of light undergoing rotation.

Once again, if you cannot use an interferometer like this to measure the orbital Sagnac, that means that you cannot measure the orbital Sagnac here on Earth, instead you would need a circular interferometer centred on the sun.

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

No. That is the final one based on your piece of shit, not mine.

My one ended up with a simple ratio of wo/wr, based upon dt=4*A*w/c^2.

You are yet to show an error with that. Instead you just go off on a tangent.

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

Yes, as the loop is a different distance from the 2.
But you have the same loop (or equivalent loops) and the same area.

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

You don't stand a chance with me here, jack.
The sooner you understand this fact, the better for you.

No, I have been wiping the floor with you, repeatedly.
You are the one that doesn't stand a chance with me or basically anyone who honestly debates you and forces you to stay on topic.

The only way you can stand a chance if you are allowed to run free to repeatedly bury your opponents in mountains pure bullshit, never staying on the same topic for long enough to be refuted and accept defeat.

Here is the correct derivation for a parallelogram shaped interferometer:

No, that is a link to a link.
Do it here, yourself.
If you can't, it shows you don't understand it.

As such, everything that follows from that is null and void.
Your "correct" calculation is also completely wrong, as my derivation shows.
As such, everything following from it is garbage.

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

No, I don't.
You are yet to debunk anything.

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

I would never do something like that. You have a massive natural advantage over me.

Your entire derivation is based on faulty premises.

A premise you are yet to show any actual problem with.
Instead you just bitch and moan about if it is the Sagnac effect, or you bitch about the conclusion.

You are yet to show anything at all that is wrong with my derivation.

Here are the best mainstream treatises on the Sagnac effect:

No, They are not mainstream.
Arguably the first one is, but the others are not. The others are fringe groups.

But they agree. They have 2 terms, one positive, one negative. And they all boil down to the same result, dt=4*A*w/c^2.

But none deal directly with the interferometer in question. Instead using either a circle or an arbitrary polygon.

So once again, you are just refuting yourself.

You don't understand.

Right now I am about to deflower the roof of your mouth and blow the globe out of your brains.

Maybe then you'll understand what you did.

Instead my final result agrees with mainstream science with what the Sagnac effect is:
dt=4*A*w/c^2.

So it sure looks like mine is correct.

But it does not.

This is what you put forth before your readers.

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

v = ωR
v₂ = ωR₂
v₁ = ωR₁

s = Rφ (s = arclength)

dt = 2φωR₂²/c² - 2φωR₁²/c²

dt = 2v₂s₂/c² - 2v₁s₁/c²


BUT THIS IS WRONG!

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

You cannot have TWO SPEEDS, only one.

There is no negative term in the Sagnac.


Your manipulation of my derivation led to that garbage.

No manipulation at all.

A straightforward calculation.

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

v = ωR
v₂ = ωR₂
v₁ = ωR₁

s = Rφ (s = arclength)

dt = 2φωR₂²/c² - 2φωR₁²/c²

dt = 2v₂s₂/c² - 2v₁s₁/c²

This is a piece of shit.

Just like your result.

Two pieces of shit created by your piece of shit derivation.


That is the final one based on your piece of shit, not mine.

My one ended up with a simple ratio of wo/wr, based upon dt=4*A*w/c^2.

Only a simpleton like yourself would end up with that.

Using the equation you provided you end up with this:

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]


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

But you cannot eliminate the radii.

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

v = ωR
v₂ = ωR₂
v₁ = ωR₁

s = Rφ (s = arclength)

dt = 2φωR₂²/c² - 2φωR₁²/c²

dt = 2v₂s₂/c² - 2v₁s₁/c²


BUT THIS IS WRONG!

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

You cannot have TWO SPEEDS, only one.

There is no negative term in the Sagnac.


Your "correct" calculation is also completely wrong, as my derivation shows.

The calculation was peer reviewed.

It is correct.

Your piece of shit derivation is being ripped into shreds right here right now.

A big difference.

But they agree. They have 2 terms, one positive, one negative. And they all boil down to the same result, dt=4*A*w/c^2.

But none deal directly with the interferometer in question. Instead using either a circle or an arbitrary polygon.

NOT THE SAME RESULT.

WHAT YOU GET IS THIS:

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

v = ωR
v₂ = ωR₂
v₁ = ωR₁

s = Rφ (s = arclength)

dt = 2φωR₂²/c² - 2φωR₁²/c²

dt = 2v₂s₂/c² - 2v₁s₁/c²


BUT THIS IS WRONG!

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

You cannot have TWO SPEEDS, only one.

There is no negative term in the Sagnac.


THEY GET THE CORRECT RESULT, NO MATTER THE GEOMETRICAL SHAPE.

DUFOUR AND PRUNIER USED A VERY COMPLEX GEOMETRICAL SHAPE, MADE UP OF BOTH STATIONARY AND ROTATING INTERFEROMETERS.

THEY OBTAINED THE CORRECT RESULT.

YOU GOT A SAGNAC WHICH USES TWO SPEEDS, AND HAS TWO FINAL TERMS, ONE OF WHICH IS NEGATIVE.


Now, you can go **** yourself, perhaps then you'll understand what you did you wrong.

You don't understand.
Right now I am about to deflower the roof of your mouth and blow the globe out of your brains.
Maybe then you'll understand what you did.

Wow, so now you are going to rape me because I kept handing your ass to you?

Not a very rational response.

And yet again, you are just repeating the same refuted BS.

Looks like I will need to try baby steps with you.
So lets take it one step at a time.

You either agree on that step or you explain the problem and what it should be.
No trying to run ahead or anything like that.

So to start with, do you agree that this pictures shows an interferometer, that is a loop with 2 counterpropagating beams of light:

The main loop is in red, with a small extension in pink to go to the light source and detector.
So one beam goes around clockwise and one goes counter clockwise.

In this thread you got fucked up really badly.

And it can get worse.

There are no baby steps, you imbecile.

This is what you put forth in front of your readers.

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

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

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

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

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

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

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

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

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

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

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.


There is nothing else to discuss here.

Your derivation has nothing to do with the Sagnac.

Sagnac = one term, one speed, nothing else

Your derivation = two terms, one of them is negative, two speeds.


An intelligent person will understand these facts pretty fast.

But not you.


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





Yet, they obtained the correct result.

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

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

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


But not in the case of your piece of shit derivation.

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

v = ωR
v₂ = ωR₂
v₁ = ωR₁

s = Rφ (s = arclength)

dt = 2φωR₂²/c² - 2φωR₁²/c²

dt = 2v₂s₂/c² - 2v₁s₁/c²


BUT THIS IS WRONG!

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

You cannot have TWO SPEEDS, only one.

There is no negative term in the Sagnac.


Let me spell out for you.

Your equations, derived by you, posted by you, cherished by you, finally lead to a formula which has nothing to do with Sagnac.

A very straightforward calculation.

dt = 2v₂s₂/c² - 2v₁s₁/c²

TWO SPEEDS. ONE TERM IS NEGATIVE. TWO TERMS.

THIS IS NOT SAGNAC.


Need I remind you that you can go and **** yourself now?

In this thread you got fucked up really badly.
And it can get worse.
There are no baby steps, you imbecile.

Well if you weren't such an imbecile, we could do it all at once, but you seem incapable of handling multiple bits of information at once.
So we need to take baby steps to step you through it.

Now answer the question:
Do you agree that this pictures shows an interferometer, that is a loop with 2 counterpropagating beams of light?

The main loop is in red, with a small extension in pink to go to the light source and detector.
So one beam goes around clockwise and one goes counter clockwise.
Yes or no.

No other bullshit to try and distract from your inevitable defeat.
We will take this one step at a time until we reach the conclusion.

Your derivation leads to this:

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


Your own words.

Based on the graphic you provided.

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.


Can you understand this much, or are you fucking mad?


Your equation.

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


This is what you posted.

It immediately leads to this:

v = ωR
v₂ = ωR₂
v₁ = ωR₁

s = Rφ (s = arclength)

dt = 2φωR₂²/c² - 2φωR₁²/c²

dt = 2v₂s₂/c² - 2v₁s₁/c²

If you didn't know the Sagnac formula looks like this:

dt = 2vs/c²

s=@R (usually @=2π)


A SINGLE TERM. ONE SPEED. NOTHING ELSE.

Your derivation leads to this:

dt = 2v₂s₂/c² - 2v₁s₁/c²

TWO TERMS, ONE OF THEM IS NEGATIVE. TWO SPEEDS.

Conclusion: your piece of shit derivation is useless and worthless.



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





Yet, they obtained the correct result.

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

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

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


But not in the case of your piece of shit derivation.


We will take this one step at a time until we reach the conclusion.

Go **** yourself jack.

We already have a conclusion.

I do not have to explain anything else, or to investigate any of the steps you took to arrive at a final result.

If the result is wrong, then the entire derivation is wrong.

Here is your conclusion 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

If you didn't know the Sagnac formula looks like this:

dt = 2vs/c²

s=@R (usually @=2π)


A SINGLE TERM. ONE SPEED. NOTHING ELSE.

Your derivation leads to this:

dt = 2v₂s₂/c² - 2v₁s₁/c²

TWO TERMS, ONE OF THEM IS NEGATIVE. TWO SPEEDS.

Conclusion: your piece of shit derivation is useless and worthless.

You don't understand.

No, you do not understand! You are wasting your time posting the same old stuff over and over.

Until you show us your analysis of both the rotational and orbital Sagnac delays for the system represented below you have nothing!

Look at:

Quote from: JackBlack on October 30, 2017, 03:08:32 PM

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.

So, what is it to be? Your analysis to counter JackBlack's or admit what everybody knows,
Your choice!

rabinoz, you are just as ****ing stupid (and possibly ****ing mad) as your friend.

You cannot calculate the Sagnac from a 150,000,000 km distance where the arcs could measure 1 meter, 100 meters, or a few kilometers in size.

The angle subtended will be on the order of 0.0000000001 radians.

The Sagnac will not be picked up.


Your mad friend dismisses established authors which have calculated the Sagnac correctly.

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 on the Sagnac effect in the world today.

They have calculated the Sagnac for a parallelogram shaped interferometer the correct way.







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 cannot pick up the Sagnac in a lab, where s might measure 1 meter, from a distance of 150,000,000 km.


Moreover, the interferometer exemplified in this graphic, COULD NEVER BE BUILT.

https://i.imgur.com/BrbOgFL.png

You just will not be able to match the R2-R1 and r1-r2 differences and the sizes of the s1 and s2 distances.

Take a look at a very simple calculation.

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 = 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

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

0.11996/-0.0013362 = -89.777


HOW ARE YOU GOING TO BUILT THIS INTERFEROMETER WHEN THE R1-R2 AND r1-r2 not to mention the s1-s2 DISTANCES JUST CANNOT BE MATCHED?

The Sagnac cannot be picked up from a distance of 150,000,000 km, as your mad friend has done.


Obviously, this is the wrong approach.


Your mad friend has posted the equations he thinks directly represent the graphic proposed by him.


They lead to this equation:

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

Let me repeat.

Your mad friend has tried to pick up the Sagnac from a distance of 150,000,000 km, using a subtended angle of the size of some 0.000000001 radians.


According to him, this is the final product of his warped logic:

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


Then, it is very simple to see where this madman went wrong.


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 the correct formula for the Sagnac.

dt = 2vs/c²

s=@R (usually @=2π)


A SINGLE TERM. ONE SPEED. NOTHING ELSE.


Until you show us your analysis of both the rotational and orbital Sagnac delays for the system represented below you have nothing!

Go **** yourself rabinoz.

You have already been provided with a classic analysis done the right/correct way for a 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.


There is no error in this calculation, you moron.


Here is where your mad friend's approach has taken you, by contrast:

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


Then, it is very simple to see where this madman went wrong.


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.


His final equation has TWO SPEEDS, TWO TERMS (ONE OF WHICH IS NEGATIVE).

But this is not Sagnac.

These are HIS EQUATIONS.

Yet here you are like the imbecile that you are, proclaiming that they are correct.


This is the Sagnac formula:


dt = 2vs/c²

s=@R (usually @=2π)


A SINGLE TERM. ONE SPEED. NOTHING ELSE.


Here is what you mad friend has achieved:

dt = 2v₂s₂/c² - 2v₁s₁/c²

TWO SPEEDS, TWO TERMS, ONE OF THEM NEGATIVE.

This is what you get when you try to pick the Sagnac from a distance of 150,000,000 km, where s might measure 1 meter or a few kilometers in size.


You know what you have to do rabinoz.

Find a single reference where the Sagnac includes TWO TERMS, TWO SPEEDS.

You won't find any.

Then, you cannot come here and complain.

The final equation posted by your mad colleague is totally wrong.

It is as simple as this.

I was asked to debunk his derivation, which I did.

His final equation leads to this:

dt = 2v₂s₂/c² - 2v₁s₁/c²

TWO SPEEDS, TWO TERMS.

But this is not the Sagnac formula.


Can you understand this much, or are you fucking mad as well?

rabinoz, you are just as fucking stupid (and possibly fucking mad) as your friend.

Quit stalling a post your analysis of the Sagnac loop given in the diagram.  Do it symbolically with the symbo's given.

 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!

Yes, I know that you can't do it, but simply copying your previous rubbish does nothing to help your case..

Just remember that the earth really is a Globe orbiting a Sun about 150,000 km away. From what I have seen:
Nicola Tesla said so and the authors of all your papers seem to say the same thing.

They might differ on the existence and/or the properties of ether, the correctness of SR or GR or the ultimatecause of gravitation, but
they all certainly believe that the earth really is a Globe orbiting a Sun about 150,000 km away.

Care to comment?

Have a nice day!

Here, let me exemplify how to CORRECTLY calculate the Sagnac for a triangular shaped interferometer.

C will be the center of a circle, where the triangle will be inscribed.

Sides BC and AC are equal (isosceles triangle).

Points A and B do not have to touch the circle.

Side AB is closest to the side of the circle.


dt = (tAC + tCB + tBA) - (tAB + tBC + tCA)

The travel times along AC and BC will not differ.

Then, tAB - tBA = 4ωA/c², where A the area of the triangle.

4ωA/c² = (4ωhL/2)/c² where h is the height of the triangle to point D (midpoint between A and B), and L is the distance from A to B.

Then the final answer:

dt = 2vL/c² where v = the speed at point D, the midpoint between A and B


ONE TERM. ONE SPEED.


The correct way.


Now the calculation for square shaped interferometer.


Square, sides ABCD

Rotated clockwise.

Sides = 2^1/2R

R = half the diagonal of the square

v = ωR

Travel time counterclockwise:

8R/(2^1/2c + ωR)

Travel time clockwise:

8R/(2^1/2c - ωR)


dt = 8R²ω/c²


FINAL ANSWER:

dt = 4Aω/c²


One term. One speed.


This is the correct way to calculate the Sagnac.

One does not use a 150,000,000 radius to the interferometer with a subtended angle of some 0.000000001 radians.

That would be madness.

dt = 2vL/c² where v = the speed at point D, the midpoint between A and B

That's funny, everybody else gets an expression involving only angular velocity and hence satisfies:

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

I guess you're smarter than E. J. POST and all the others. I suppose we knew all along that you we've smartest person on earth.
You think you know far better that the authors of all these papers.

Must go, got vital things to do, Li, e shopping!

That's funny, everybody else gets an expression involving only angular velocity

See how little you know about the Sagnac effect?

As I said, A. Tartaglia and A. Ruggiero are two of the best experts on the Sagnac effect in the world. They even attempted to explain the Sagnac using GR.

Their papers are being quoted and referenced all over the scientific community.

Sagnac effect and pure geometry, published in a scientific journal, peer reviewed:




Compare this with your hare brained statement.

Professor Ruyong Wang is another expert on the Sagnac effect, with worldwide recognition.


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

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

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

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

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

Compare this with your stupid statement and draw the necessary conclusions.

Your derivation leads to this:

I know what my derivation leads to and I know it shows you are wrong.
But you seem incapable of handling lots of information at once, so we need to step through this.

Unless you are willing to accept that there is nothing wrong with my derivation up until that point.

So I ask again:
Do you agree that this pictures shows an interferometer, that is a loop with 2 counterpropagating beams of light?

The main loop is in red, with a small extension in pink to go to the light source and detector.
So one beam goes around clockwise and one goes counter clockwise.
Yes or no.

No other bullshit to try and distract from your inevitable defeat.
We will take this one step at a time until we reach the conclusion.

CONFORMAL MAPPING OF ANNULAR SECTORS/TRAPEZOIDS ONTO A RECTANGLE



Conformal mapping, an important area of complex variables analysis.

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

Numerical conformal mapping of an circular sector onto a rectangle:

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

Numerical conformal mapping of a trapezoid onto a rectangle:

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

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


Any annular section/trapezoidal shaped interferometer then can be analyzed in terms of a rectangular shaped interferometer.

But this has already been done.

Here is the formula:

dt = 2Lv/c²


Next, I am going to attempt a mathematical demonstration that the interferometer represented by an annular section in the following graphic CANNOT be actually physically built.

https://i.imgur.com/BrbOgFL.png

CONFORMAL MAPPING OF ANNULAR SECTORS/TRAPEZOIDS ONTO A RECTANGLE

Why? When it is totally unnecessary.

CONFORMAL MAPPING OF ANNULAR SECTORS/TRAPEZOIDS ONTO A RECTANGLE

So you can't answer simple questions and instead resort to even more pathetic distractions.

Next, I am going to attempt a mathematical demonstration that the interferometer represented by an annular section in the following graphic CANNOT be actually physically built.

Considering it can be constructed using fibre optics, I would say it definitely can be built.

But does that mean you are saying the interferometer is not possible?

If so, you will need to explain why it isn't possible.

So you can't answer simple questions and instead resort to even more pathetic distractions.

Conformal mapping is no distraction. It is one of the most useful tools used in engieering and in physics today, in dealing with irregularly shaped interferometers.

I do not have to debate anything anymore: the algorithms for mapping an annular section onto a rectangle (or a trapezoid onto a rectangle) are well established (see the links provided).

As such the Sagnac for an annular section can be reduced to a Sagnac for a rectangle.


But I always strive to give my readers the very best information, that is why I will analyze the geometry of the interferometer as it relates to R2, R1, r1, r2 and s2 and s1.

The data is as follows:

R2 - R1 = r2 - r1

s2o2 = s2r2

s1o1 = s1r1

I could even remind everyone here that the interferometer is impossible to build given the fact that the s2/s1 arcs are convex when looking from the Sun to the Earth, and concave when looking the opposite way. As such the calculations cannot be applied looking from the viewpoint of the Earth (concave arcs). However, I will be reminded that the area is equal, so at least a theoretical analysis is owed to the readers.

https://i.imgur.com/BrbOgFL.png

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.

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.


The interferometer is impossible to build given the fact that the s2/s1 arcs are convex when looking from the Sun to the Earth, and concave when looking the opposite way. As such the calculations cannot be applied looking from the viewpoint of the Earth (concave arcs). However, I will be reminded that the area is equal, so at least a theoretical analysis is owed to the readers.


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

v_2o = R2 x (2 x 10^-7)  o stands for orbital

v_1o = R1 x (2 x 10^-7)  o stands for orbital

s₂ = R2 x θ_o

s₁ = R1 x θ_o


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

v_2r = r2 x (7.27 x 10^-5)  r stands for rotational

v_1r = r1 x (7.27 x 10^-5)  r stands for rotational

s₂ = r2 x θ_r

s₁ = r1 x θ_r


R2 - R1 = r2 - r1

r2 x θ_r = R2 x θ_o

r1 x θ_r = R1 x θ_o

r2/r1 = R2/R1

(r2 x R1) = (r1 x R2)

Since the two areas must be equal,

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



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

R2 = r2 - r1 + R1

(r2 x R1) = r1r2 - r1² + (R1 x r1)

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

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


The entire set up doesn't make any sense.


A sample calculation.

r2 = 6401 km

Let us choose s₂ = 1 km

Then, θ_r = 1/6401

r1 = 6400 km

s₁ = 6400 x 1/6401


R2 = 150,000,000 km

θ_o = 1/150,000,000

s₂ = 1 km

R1 = s₁/θ_o  = 6400/6401 x 150,000,000 = 149,976,566.2


But this is impossible.

150,000,000 - 149,976,566.2 = 23433.8 km = R2 - R1

But R2 - R1 = r2 - r1, where r2 - r1 = 1 km


Let us now require that R2 - R1 = 1km

R1 = 149,999,999 km

s₁ = R1 x θ_o = 149,999,999/150,000,000 = 0.999999993

But now r1 = s₁/θ_r = 6400.999955


Evidently, a huge error was made along the way, during the derivation.

I believe it was committed right here:

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

Thus the area between them is:
A=alpha*R22/2-alpha*R12/2
=alpha*(R22-R12)/2
Thus 2*A=alpha*(R22-R12).

Notice that was in the formula above?
That means we can sub it in.


An area was substituted for velocity x Radius, which is totally wrong.

This is where the crucial error was made.


That is why nothing seems to work out properly, we get a Sagnac with two terms, one of them being negative, and two speeds.


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


dt_o/dt_r =  [2v_2os₂ - 2v_1os₁]/[2v_2rs₂ - 2v_1rs₁]

If we substitute the data obtained above, for both cases investigated, into this formula we obtain, 62.767 respectively 64.34, thus the orbital Sagnac is greater than the rotational Sagnac.


The interferometer described in the graphic requires that,

R2 - R1 = r2 - r1

r2 x θ_r = R2 x θ_o

r1 x θ_r = R1 x θ_o

r2/r1 = R2/R1

(r2 x R1) = (r1 x R2)


From this we reach an immediate contradiction:

R2 = r2 - r1 + R1

(r2 x R1) = r1r2 - r1² + (R1 x r1)

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

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



Now it becomes even easier to see right away where the error was committed.

dt_o/dt_r =  [2v_2os₂ - 2v_1os₁]/[2v_2rs₂ - 2v_1rs₁]

If we substitute the data obtained above, for both cases investigated, into this formula we obtain, 62.767 respectively 64.34, thus the orbital Sagnac is greater than the rotational Sagnac.

The formula works.

The orbital Sagnac is greater than the rotational Sagnac.

A total debunking of the failed derivation.

Conformal mapping is no distraction. It is one of the most useful tools used in engieering and in physics today, in dealing with irregularly shaped interferometers.

Sure sounds like a distraction.

I do not have to debate anything anymore: the algorithms for mapping an annular section onto a rectangle (or a trapezoid onto a rectangle) are well established (see the links provided).

Yes you do. You are just making baseless claims without backing them up, and you are yet to refute my claims.

I could even remind everyone here that the interferometer is impossible to build given the fact that the s2/s1 arcs are convex when looking from the Sun to the Earth, and concave when looking the opposite way.

Baby steps.
We are just dealing with one interferometer first, and how it applies to the orbit.
We can get to comparing the orbit with the rotation later

Either accept that the interferometer in question is possible, or explain why it isn't.

Evidently, a huge error was made along the way, during the derivation.
I believe it was committed right here:
Thus dt=2*omega*alpha*R22/c2-2*omega*alpha*R1^2/c2
=2*omega*alpha*(R22-R12)/c2

Does that mean you accept the derivation up to that point?

If not, answer the question from before and we take it in baby steps.
Remember, we are just dealing with the orbital Sagnac to begin with. No running ahead.

I could even remind everyone here that the interferometer is impossible to build given the fact that the s2/s1 arcs are convex when looking from the Sun to the Earth, and concave when looking the opposite way. As such the calculations cannot be applied looking from the viewpoint of the Earth (concave arcs).

Big deal!
A 1 km arc subtends about 0.00899° at on the centre of the earth and the chord of that arc is about 0.99999999076 km.
A 1 km arc subtends about 3.81971863420548805845316 x 10^-10° at on the centre of the sun and the chord of that arc is about 0.999999999999999999999983 km.
Get real and stop making excuses for your inability to do a little albebra.

The orbital Sagnac will ALWAYS be larger than the rotational Sagnac (unless the interferometer is in the shape of rectangle where s₂ = s₁, and then we apply the usual Sagnac formula dt = 2vL/c²)

Proof:

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


The final ratio will be:

dt_o/dt_r =  [v_2os₂ - v_1os₁]/[v_2rs₂ - v_1rs₁]

=

dt_o/dt_r =  [v_2os₂ - v_1os₁]/[v_2rs₂ - v_1rs₁]


r1 >= 6,400 km
r2 < 12,800 km
s₂ > s₁
R2 <= 150,000,000 km
R1 > 149,987,200 km


Let us divide both the numerator and the denominator of the fraction by s₂:

(v_2o - v_1os₁/s₂)/(v_2r - v_1rs₁/s₂)

s₁/s₂ <1

v_2o, always larger than 25 (2 x 10^-7 x a number ranging from 150,000,000 to 149,987,200)
v_2o > v_1o

v_2r, always less than 1 (7.27 x 10^-5 x a number ranging from 12,800 to 6,400)
 
v_1o = v_2o -  ε

The numerator becomes:

v_2o - v_2os₁/s₂ + εs₁/s₂


Let us analyze the term

v_2o(1 - s₁/s₂)

This term can NEVER be less than 0

v_2o<s₂/(s₂ - s₁)

By definition s₂ > s₁


The denominator will ALWAYS be less than 1.

v_2r, always less than 1 (7.27 x 10^-5 x a number ranging from 12,800 to 6,400)

v_2r - v_2rs₁/s₂ + εs₁/s₂

εs₁/s₂ always less than 1

v_2r(1 - s₁/s₂)

Always less than 1


Thus the numerator will ALWAYS be greater than the denominator (unless we have the case where, of course, s₂ = s₁, where the shape of the interferometer becomes a rectangle):

dt_o/dt_r =  [v_2os₂ - v_1os₁]/[v_2rs₂ - v_1rs₁]

The orbital Sagnac will ALWAYS be larger than the rotational Sagnac for this type of interferometer and for the graphic posted earlier.

The orbital Sagnac will ALWAYS be larger than the rotational Sagnac (unless the interferometer is in the shape of rectangle where s₂ = s₁, and then we apply the usual Sagnac formula dt = 2vL/c²)
Proof:
Thus dt=2*omega*alpha*R22/c2-2*omega*alpha*R1^2/c2
=2*omega*alpha*(R22-R12)/c2

Stop just posting massive walls of crap which have already been refuted.

Like I said, we are doing this in baby steps as you are unable to handle lots of information at once.

So do you accept the derivation up to this point (with the correct version with powers instead of just 2)? YES OR NO???
If not, either tell me what part you disagree with or we start at the beginning.

Let us see, again, what this derivation entails.

https://i.imgur.com/BrbOgFL.png

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.

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.

https://www.theflatearthsociety.org/forum/index.php?topic=72601.msg1980338#msg1980338 (complete debunking of the derivation, full details)

s₂ = r2 x θ_r

s₁ = r1 x θ_r


R2 - R1 = r2 - r1

r2 x θ_r = R2 x θ_o

r1 x θ_r = R1 x θ_o

r2/r1 = R2/R1

(r2 x R1) = (r1 x R2)

Since the two areas must be equal,

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



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

R2 = r2 - r1 + R1

(r2 x R1) = r1r2 - r1² + (R1 x r1)

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

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


The entire set up doesn't make any sense.


which have already been refuted.

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

R2 = r2 - r1 + R1

(r2 x R1) = r1r2 - r1² + (R1 x r1)

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

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


To see at a glance the impossible derivation that our mad friend, jack, has proposed, here is a simple example:

r2 = 6401 km

Let us choose s₂ = 1 km

Then, θ_r = 1/6401

r1 = 6400 km

s₁ = 6400 x 1/6401


R2 = 150,000,000 km

θ_o = 1/150,000,000

s₂ = 1 km

R1 = s₁/θ_o  = 6400/6401 x 150,000,000 = 149,976,566.2


But this is impossible.

150,000,000 - 149,976,566.2 = 23433.8 km = R2 - R1

But R2 - R1 = r2 - r1, where r2 - r1 = 1 km


Let us now require that R2 - R1 = 1km

R1 = 149,999,999 km

s₁ = R1 x θ_o = 149,999,999/150,000,000 = 0.999999993

But now r1 = s₁/θ_r = 6400.999955


The interferometer described in the graphic requires that,

R2 - R1 = r2 - r1

r2 x θ_r = R2 x θ_o

r1 x θ_r = R1 x θ_o

r2/r1 = R2/R1

(r2 x R1) = (r1 x R2)


From this we reach an immediate contradiction:

R2 = r2 - r1 + R1

(r2 x R1) = r1r2 - r1² + (R1 x r1)

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

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


dt_o/dt_r =  [2v_2os₂ - 2v_1os₁]/[2v_2rs₂ - 2v_1rs₁]

If we substitute the data obtained above, for both cases investigated, into this formula we obtain, 62.767 respectively 64.34, thus the orbital Sagnac is greater than the rotational Sagnac.

The formula works.

The orbital Sagnac is greater than the rotational Sagnac.


Earlier today I posted the proof that the orbital Sagnac is larger than the rotational Sagnac, using the very equations posted by our friend.

https://www.theflatearthsociety.org/forum/index.php?topic=72601.msg1980644#msg1980644

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


An analysis of this ratio reveals that the numerator will ALWAYS be greater than the denominator (unless we have the case where, of course, s₂ = s₁.


A total victory for FE.

Let us see, again, what this derivation entails.

Then lets see.
You have shown you are unable to understand it in full, so we are going to go through it step by step.

Why do you seem so afraid of it?
Is it because you know it will show you to be wrong as you are unable to refute any of it and being forced to step through it and accept each bit will confirm that?

Answer the question:
Do you agree that this pictures shows an interferometer, that is a loop with 2 counterpropagating beams of light?

The main loop is in red, with a small extension in pink to go to the light source and detector.
So one beam goes around clockwise and one goes counter clockwise.
Yes or no.

Until we step through it bit by bit I will just keep bringing it up, as you are yet to refute anything.

jack, take a look at yourself.

Totally defeated.

Begging for mercy.

Unable to accept the very simple proofs which destroy your fucked up derivation.

Pleading with me to take a look at your "baby steps".

Beaten to a pulp.


Do you agree that this pictures shows an interferometer, that is a loop with 2 counterpropagating beams of light?

Are you telling your readers that you are totally mad?


https://i.imgur.com/BrbOgFL.png

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.

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.

https://www.theflatearthsociety.org/forum/index.php?topic=72601.msg1980338#msg1980338 (complete debunking of the derivation, full details)

s₂ = r2 x θ_r

s₁ = r1 x θ_r


R2 - R1 = r2 - r1

r2 x θ_r = R2 x θ_o

r1 x θ_r = R1 x θ_o

r2/r1 = R2/R1

(r2 x R1) = (r1 x R2)

Since the two areas must be equal,

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



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

R2 = r2 - r1 + R1

(r2 x R1) = r1r2 - r1² + (R1 x r1)

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

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


The entire set up doesn't make any sense.


You don't have anything, just a fucked up derivation, which is worthless and useless.

jack, take a look at yourself.
Totally defeated.
Begging for mercy.

Nope. Not defeated at all.
Beating you to a pulp with a simple question you cannot answer, because you know that answering it will just lead down the path of your defeat.

I have already dealt with your failed refutations.

Do you agree that this pictures shows an interferometer, that is a loop with 2 counterpropagating beams of light?
Are you telling your readers that you are totally mad?

So still unable to answer a simple question?

Stop trying to jump ahead, deal with it 1 question at a time, as that seems to be all you are capable of (if you are even capable of that).

So I ask again:
Do you agree that this pictures shows an interferometer, that is a loop with 2 counterpropagating beams of light?

It is a simple yes or no question to begin with. No jumping ahead to trying to use the same interferometer for 2 things.

A total victory for the FE has to be celebrated in a very special way.

([v_2o - v_1os₁/s₂)/(v_2r - v_1rs₁/s₂)

This is the ratio of the orbital Sagnac/rotational Sagnac.

v_2o = ω_o x R2
v_1o = ω_o x R1

ω_o = 2 x 10^-7

range of R2 (149,987,200 to 150,000,000)

v_2r = ω_r x r2
v_1r = ω_r x r1

ω_r = 7.27 x 10^-5

range of r2 (6,400 to 12,800)

ω_o(R2 - R1s₁/s₂)/ω_r(r2 - r1s₁/s₂)

let now s₁ approach the value of s₂

example s₂ x 0.9999 = s₁

let now r1 approach the value of r2 (since by hypothesis, R2 - R1 = r2 - r1, R1 will have to approach the value of R2 by similar amount)

example r2 x 0.999 = r1

ω_oR2(1 - 0.999x0.9999)/ω_rr2(1 - 0.999x0.9999)

We finally get the ratio we were seeking, after all this time:

ω_oR2/ω_rr2

Since R2 will always be much larger than r2, THE ORBITAL SAGNAC WILL ALWAYS BE LARGER THAN THE ROTATIONAL SAGNAC.

We can also observe where the derivation went totally wrong: ONLY IN THE CASE WHERE R2 = r2, would we be left with a ratio of the angular velocities.

A total victory for the FE.

Earlier I posted the correct calculation for the ratio, for a interferometer in the shape of a rectangle:

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.


ω_oR2/ω_rr2


largest value of R2 = 150,000,000

largest value of r2 = 12,800 (an interferometer in outer space)

The ratio will measure 32.23

smallest value for R2 = 149,987,200

smallest value for r2 = 6,400

The ratio will measure 64.47


So the values for the orbital Sagnac vs. the rotational Sagnac are practically the same, if the interferometer is placed very close to the surface of the Earth, which will happen in virtually all cases considered by modern science.


https://www.theflatearthsociety.org/forum/index.php?topic=72601.msg1980338#msg1980338 (complete debunking of the derivation, full details)

https://www.theflatearthsociety.org/forum/index.php?topic=72601.msg1980644#msg1980644 (an introduction to the analysis of the orbital Sagnac/rotational Sagnac ratio)

The interferometer described in the graphic requires that,

R2 - R1 = r2 - r1

r2 x θr = R2 x θo

r1 x θr = R1 x θo

r2/r1 = R2/R1

(r2 x R1) = (r1 x R2)


From this we reach an immediate contradiction:

R2 = r2 - r1 + R1

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

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

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

Let us remember that these were the requirements issued forth by the author of this thread:

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.

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.

A total victory for the FE has to be celebrated in a very special way.

But you are yet to acheive victory.
All you have done is try to avoid defeat. But my derivation unrefuted.

You being unwilling to step through bit by bit or say up until what point you agree with shows that you know my derivation is correct and that you are full of shit and are unwilling to admit it. Either because you are deluding yourself, you are too arrogant to admit you are wrong or because you're just another pathetic troll.

Which is it?

If you don't think it is any of them, then either tell me what is the first part of the derivation you disagree with and we can take it from there, or we go through from the start, step by step.

As you are yet to refute the derivation at all and instead just repeatedly bitch and moan about the conclusion, you are yet to refute anything at all.
So until we do this you will remain defeated.

Do you agree that this pictures shows an interferometer, that is a loop with 2 counterpropagating beams of light?
https://i.imgur.com/BrbOgFL.png
The main loop is in red, with a small extension in pink to go to the light source and detector.
So one beam goes around clockwise and one goes counter clockwise.
Yes or no.

For those interested in a square interferometer, with side length l, I have worked out part of the math:
The total time around the loop, for the beam going forward on the outside is the sum of the "t" values from these three equations, with the last one use twice (once for the path in, once for the path out):
t*c=l*cos(w*t/2)+2*R2*sin(w*t/2)
t*c=l*cos(w*t/2)-2*R1*sin(w*t/2)
(t*c)^2=Ri^2+Ro^2-2*Ri*Ro*cos(ai-ao+wt)

And the loop going the other wise is likewise from:
t*c=l*cos(w*t/2)-2*R2*sin(w*t/2)
t*c=l*cos(w*t/2)+2*R1*sin(w*t/2)
(t*c)^2=Ri^2+Ro^2-2*Ri*Ro*cos(ai-ao-wt)

Where R1 is the distance from the centre of rotation to the middle of the near leg of the loop. R2 is likewise but for the far leg.
l is the length of each leg.
w is the angular velocity.
Ri=sqrt(l^2/4+R1)^2 (similar to R1 but to the end of a leg instead of the middle)
Ro=sqrt(l^2/4+R2)^2

I'll try doing some more math later to see if it can be simplified.

Sandy, I see you are posting a bunch more lies in the liars only section, yet you have been unable to rationally respond here.
Why is that?

I also see you are now fully accepting of the derivation up to this point:
dt = t₁ - t₂ = 2θ_oω_o(R₂² - R₁²)/c²

But as I showed you before, the area of an annular sector is given by:
A=θ(R₂² - R₁²)/2

This is because the area of a circular sector is given by:
A=θR²/2
where in the case of θ=2π, this becomes the area of a circle:
A=2πR²/2=πR²

And an annular sector is just the difference between 2 circular sectors.
So rearranging gives:
2A=θ(R₂² - R₁²)
And then subbing that back in we get:
dt=2ω_oθ_o(R₂² - R₁²)/c²
dt=2ω_o(2A)/c²
dt=4Aω_o/c²

This shows that it is clearly proportional to the area and the angular velocity, not the tangential velocities, and thus the ratio will reduce to ωo/ωr, with no velocity term present.

Your latest pathetic attempts involve dividing or multiply by 0.

You try and have what amounts to this:
[vo*(1-y)]/[vr*(1-x)], and look at what happens when both x and y approach 0, treating it as if you just cancel the 0.

However there are many things of that form.
As an example, let y=0.1*x.
This becomes [vo*(1-0.1*x)]/[vr*(1-x)]=[0.1*vo*(1-x)]/[vr*(1-y)]=0.1*vo/vr.
Notice how you can get a completely different result?
You can't just go, (1-x)=0, (1-y)=0, thus 1-x=1-y. That is a classic division by 0 error.

So instead, you actually need to analyse what they are and eliminate them properly as variables.
If you can't do that, then you can't eliminate them like that.
The only time you can eliminate them without matching the variables is when they approach a specific non-0 value, for example if you had 5-y, where y approaches 1, 5-y approaches 4, and thus can be treated as 4.

So good job failing yet again.

If you start with:

Another source of your dishonesty stems from continually changing the association between r1, r2, s1, s2 and so on. But that can be fixed by simply pushing it to the other side.
But again, you can't use the same annular section for both, as it wont be an annular section for both. Instead you use 2 slightly different annular sections which approximate the rectangular loop, ensuring they have the same area.

dto/dtr=[vo(1 - (R1/R2)*(s1/s2))]/[vr(1 - (r1/r2)*(s1/s2))] (assuming s2 are the same for both).
The s1 values will be different for each, and due to the ratio of r1/r2 and s1/s2 being equal, this can be simplified:
dto/dtr=vo(1 - (R1/R2)^2)/vr(1 - (r1/r2)^2)
Then, just like you did in the liars only section, we let R1=R2-e, and r1=r2-e, but as it is about to get a bit more complex, I will just look at the orbital one for now:
vo(1 - (R1/R2)^2)=v0[1-((R2-e)/R2)^2]
=vo*[1-(R2^2-2*e*R2+e^2)/R2^2]
=vo*[R2^2-R2^2+2*e*R2-e^2]/R2^2
=vo*[2*e*R2-e^2]/R2^2
=vo*e*[2*R2-e]/R2^2

Now, e is much smaller than R2 (or r2), so 2*R2-e will be basically 2*R2 (as the e term does not contribute significantly, note this means R2 must be nowhere near 0 to compare the 2).

So continuing from the above:
=vo*e*[2*R2-e]/R2^2
=vo*e*2*R2/R2^2
=2*v0*e/R2

Now we can stick this (and the equivalent for the rotation) back into the ratio:
dto/dtr=[vo(1 - (R1/R2)^2)]/[vr(1 - (r1/r2)^2) ]
=[2*v0*e/R2]/[2*vr*e/r2]
=[v0/R2]/[vr/r2]

And of course, v=w*r, so v/r=w, so this further simplifies to:
dto/dtr=wo/wr.

Just like it always has (when done honestly and correctly).

So you latest piece of shit has also been torn to shreds.
Going to admit you were wrong yet, or are you going to keep harping on with your a*0/b*0=a/b?