sandokhan lies regarding the Sagnac effect

JackBlack, I interject just for a moment as I see FE victory has been levied with potential for a special celebration. In the past, important victory was followed by declaration of a FE parade.  I'm not sure if you have ever attended or been part of a flat Earth sponsored parade but they are more than memorable.

Upon your surrender in this thread, consideration for a victory parade can commence. If finalized, you will be invited to attend as a special guest with a promise not to belittle you in the course of the parade.   

You have been defeated so you may want to strongly consider this offer.       

JackBlack, I interject just for a moment as I see FE victory has been levied with potential for a special celebration. In the past, important victory was followed by declaration of a FE parade.  I'm not sure if you have ever attended or been part of a flat Earth sponsored parade but they are more than memorable.

Upon your surrender in this thread, consideration for a victory parade can commence. If finalized, you will be invited to attend as a special guest with a promise not to belittle you in the course of the parade.   

You have been defeated so you may want to strongly consider this offer.     

And how do manage to come to the delusion conclusion that "JackBlack has been defeated"?

It could only happen on somewhere like this:
that you must inhabit with sceptimatic and the like. Why do you even bother with these occasional visitations to the real world?

In the past, important victory was followed by declaration of a FE parade.  I'm not sure if you have ever attended or been part of a flat Earth sponsored parade but they are more than memorable   

I'd love to go to a Flat Earth sponsored parade. I doubt I'd take part, but that would indeed be something to see! Has there ever been one?

Please let us know if (another?) one ever happens!

JackBlack, I interject just for a moment as I see FE victory has been levied with potential for a special celebration. In the past, important victory was followed by declaration of a FE parade.  I'm not sure if you have ever attended or been part of a flat Earth sponsored parade but they are more than memorable.

Upon your surrender in this thread, consideration for a victory parade can commence. If finalized, you will be invited to attend as a special guest with a promise not to belittle you in the course of the parade.   

You have been defeated so you may want to strongly consider this offer.     

You must be new here.

Quote from: gotham on November 12, 2017, 04:40:49 PM

In the past, important victory was followed by declaration of a FE parade.  I'm not sure if you have ever attended or been part of a flat Earth sponsored parade but they are more than memorable   

I'd love to go to a Flat Earth sponsored parade. I doubt I'd take part, but that would indeed be something to see! Has there ever been one?

Please let us know if (another?) one ever happens!

I would put on my FES shirt and march along! lol It would be fun. And I generally don't do parades or protests.

You have already been beaten to a pulp jack.

What more do you want?

By now, you are down to plain lying and cheating.

Do you really think that your readers cannot see these things?


The two interferometers have the same area, but different radii and subtended angles.

In order to compare the two Sagnac shifts, YOU CANNOT SUBSTITUTE THE AREA of one interferometer for the other, because then what you are saying is that R2=r2 and R1=r1.

Here is the correct ratio:

(ω_oR2)/(ω_rr2)

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

And this formula is correct even if the figures of R1 and r1 approach the values of R2, respectively r2, at different rates.

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


ω_o(R2 - R1s₁/s₂)/ω_r(r2 - r1s₁/s₂)

(R2 - R1s₁/s₂) will always be greater than (r2 - r1s₁/s₂)

example: R2=150,000,000; R1=149,999,999; r2=6401; r1=6400; s₂x0.999=s₁

(R2 - R1s₁/s₂) = 150,001
(r2 - r1s₁/s₂) = 7.4


R2 = R1 + ε = R1 + r2 - r1

r2 = r1 + ε = r1 + R2 - R1

(by hypothesis, R2 - R1 = r2 - r1)

a = s₁/s₂


[ω_o/ω_r]{(R1 + r2 - r1 - aR1)/(r1 + R2 - R1 - ar1)}

(R1 + r2 - r1 - aR1)>(r1 + R2 - R1 - ar1)

R1(1 - a)>r1(1 - a)

R1>r1

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

(ω_oR2)/(ω_rr2)



What YOU have done is a terrible mistake, one which I was very careful to avoid from the very start. This alone shows that you have no expertise when dealing with very simple ratios. And you have a very short memory:

https://www.theflatearthsociety.org/forum/index.php?topic=72601.msg1980674#msg1980674


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


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

s₂ = R2 x θ_o

s₁ = R1 x θ_o


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

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.


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

Let us now mistankenly substitute the AREA for the VELOCITY X RADIUS, just like jack did:

θ_o/θ_r = (r2 + r1)/(R2 + R1)

Substituting this into our correct formula, we get:

ω_oθ_o(R2 + R1)/ω_rθ_r(r2 + r1)

As was proved above, using the interferometer presented in the graphic posted by jack, we end up with r1 = r2, this meaning that R2 = R1, and everything cancels out in the ratio, with the exception of the angular velocities.

So this is how jack got his final formula, fraudulently and erroneously.

He substituted blindly an area for a velocity x radius term, taking full advantage of the fact that r1 = r2 given the impossible setup, just like proven above, and everything canceled out.


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.

Let us first take a look at your terrible computational errors.

vo(1-y)/vr(1-x)

You said y=0.1x

vo(1-0.1x)/vr(1-x)=

vo(0.1)[10-x]/vr(1-x)

What you wrote is:

[0.1*vo*(1-x)]/[vr*(1-y)]

If we bring back the 0.1 factor within the parenthesis, we get, according to your faulty logic:

vo*(0.1 - 0.1x)

This is incorrect.

The correct answer is:

vo(0.1)[10-x]/vr(1-x)


[0.1*vo*(1-x)]/[vr*(1-y)]

You made another terrible computational error.

vo(0.1)(10-x)/vr(1-x)=

vo(0.1)(10-x)/10vr(1/10 - 0.1x)


Using your line of thought, we get:

[0.1*vo*(1-x)]/[vr*(1-y)]

0.1vo(1-x)/vr(1-0.1x)


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.

No dishonesty was ever included in my messages.

YOU CANNOT HAVE TWO SLIGHTLY DIFFERENT ANNULAR SECTIONS WHICH APPROXIMATE A RECTANGULAR LOOP.

The formula for a rectangular loop is DIFFERENT.

YOU NEED THE SAME AREA FOR THE TWO INTERFEROMETERS FROM THE VERY START.

Slightly different annular sections will mean THAT R2-R1 DOES NOT EQUAL ANYMORE r2-r1, which was the essential hypothesis for your setup, from the very start.

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)

Notwithstanding the fact that now you have s1o DIFFERENT FROM s1r, and as such you cannot compare the two Sagnac shifts, since you are dealing with DIFFERENT geometrical figures, you made a terrible blunder.

One which I was very careful to avoid from the very start.

YOU BLINDLY ASSUMED THAT s1/s2 = r1/r2 = R1/R2, and proceeded to get this:

dto/dtr=vo(1 - (R1/R2)^2)/vr(1 - (r1/r2)^2)


Remember this?

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


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

s₂ = R2 x θ_o

s₁ = R1 x θ_o


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

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.


Remember this?

Thus the numerator will ALWAYS be greater than the denominator (unless we have  s2 = s1, and r2 = r1, which is impossible)

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1981067#msg1981067


Let us now mistankenly substitute the AREA for the VELOCITY X RADIUS, just like jack did:

θo/θr = (r2 + r1)/(R2 + R1)

Substituting this into our correct formula, we get:

ωoθo(R2 + R1)/ωrθr(r2 + r1)

As was proved above, using the interferometer presented in the graphic posted by jack, we end up with r1 = r2, this meaning that R2 = R1, and everything cancels out in the ratio, with the exception of the angular velocities.


THE ONLY WAY YOU ARE GOING TO END UP WITH wo/wr IS TO ASSUME THAT R2=R1 AND r2=r1, THEN EVERYTHING WILL CANCEL OUT.


But this amounts to plain cheating and lying.

Your readers can see through your thin veils of deception jack.

You are done here.

You cannot substitute s1/s2=R1/R2=r1/r2 blindly into the formula, since then you are assuming that R2=R1 and r2=r1, given the hypothesis that R2 - R1 = r2 - r1. I was very careful to avoid this error, as I have said, from the very start.


Here is how to correctly compare the two ratios:

(ω_oR2)/(ω_rr2)

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

And this formula is correct even if the figures of R1 and r1 approach the values of R2, respectively r2, at different rates.

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


ω_o(R2 - R1s₁/s₂)/ω_r(r2 - r1s₁/s₂)

(R2 - R1s₁/s₂) will always be greater than (r2 - r1s₁/s₂)

example: R2=150,000,000; R1=149,999,999; r2=6401; r1=6400; s₂x0.999=s₁

(R2 - R1s₁/s₂) = 150,001
(r2 - r1s₁/s₂) = 7.4


R2 = R1 + ε = R1 + r2 - r1

r2 = r1 + ε = r1 + R2 - R1

(by hypothesis, R2 - R1 = r2 - r1)

a = s₁/s₂


[ω_o/ω_r]{(R1 + r2 - r1 - aR1)/(r1 + R2 - R1 - ar1)}

(R1 + r2 - r1 - aR1)>(r1 + R2 - R1 - ar1)

R1(1 - a)>r1(1 - a)

R1>r1

JackBlack, I interject just for a moment as I see FE victory has been levied with potential for a special celebration.

Yes it has been "levied", without just cause.
He baseless asserted victory even when his ass was handed to him.

Upon your surrender in this thread
You have been defeated so you may want to strongly consider this offer.     

Why would I surrender after he has been so horribly defeated?
You might want to go back and read the thread, perhaps with your eyes open this time. I wasn't the one that was defeated.

I told you that I was going to make a flat earth believer out of you.

You are totally defeated here (and elsewhere, and always) jack.

Your cognitive dissonance condition detaches you from reality.

Your terrible blunders have been enumerated here in plain sight, so that everyone can see your cheating.

You have already been beaten to a pulp jack.

Nope. The only one that has been beaten to a pulp is you, with you pathetically flailing as you try and pretend you haven't been.

What more do you want?

How about for you to rationally address the arguments raised.
Perhaps starting with answering simple yes or no questions that have previously been asked of you.

Also, it would be great if you could be consistent rather than repeatedly contradict yourself.

By now, you are down to plain lying and cheating.

Once again you are describing yourself.
Do you really think that your readers cannot see these things?

The two interferometers have the same area, but different radii and subtended angles.
In order to compare the two Sagnac shifts, YOU CANNOT SUBSTITUTE THE AREA of one interferometer for the other, because then what you are saying is that R2=r2 and R1=r1.

Yes, the have the same area, and as my derivation showed, which you seem to accept up until the point of substituting in the areas, the Seebeck is proportional to the area.
It is not saying R1=r1 and R2=r2.
If I were pretending they were identical it would require θo*R1=θr*r1 and θo*R2=θr*r2.

Try again, perhaps without so much dishonesty.

Here is the correct ratio:
(ω_oR2)/(ω_rr2)

No, that is the correct ratio when you multiple both sides by 0.
It is more accurately expressed in this form as:
(0ω_oR2)/(0ω_rr2)

Or if you don't want to dishonestly divide by 0 to blatantly misrepresent the ratio it would be:
(ω_o)/(ω_r)

As ω_r is much greater than ω_o, the THE ROTATIONAL SAGNAC WILL ALWAYS BE LARGER THAN THE ORBITAL SAGNAC for the same interferometer.

If you disagree, DEAL WITH THE ARGUMENT PRESENTED RATHER THAN IGNORING AND REPEATING THE SAME BULLSHIT!!!!!

Okay?

What YOU have done is a terrible mistake

Yet you are completley unable to show what that mistake it.
Instead you can only lie about what I have done to pretend I made some mistake.

Meanwhile you have repeatedly made mistakes which I have pointed out and you have been unable to defend those mistakes you made.

Once again, both interferometers must be annular sector. We are approximating an interferometer as an annular sector to make the math simpler, using slightly different approximations for each, but keeping the area constant.
So you bitching and moaning, claiming R1=R2 and r1=r2 is just another pathetic distraction from you, it is a straw man so you can pretend to win by refuting an argument I never made.

If you would like to do it otherwise, such as using a square or a rectangular interferometer (or even a circular one), in the place of the annular sectors, you are free to do so, but the math is quite complex and I will demand you actually derive it rather than just baselessly assert crap.

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

Yet another lie.
I substituted the area, for the area.
I do not convert to velocity at all in my derivation.

Let us now mistakenly substitute the AREA for the VELOCITY X RADIUS, just like jack did:

No, I didn't.
How about you correctly substitute θ_o(R₂²-R₁²)=2*A, like I actually did.
If you disagree with this substitution, feel free to point out why that is not the case. Explain what your area magically is.

Because substituting that CORRECT substitution into the CORRECT formula, you end up with:
dt=4*A*ω/c², just like all mainstreams sources indicate is the Sagnac shift for a rotating interferometer (not a FOC).

As you didn't do this correct substitution and instead did a completely different one which you have already admitted is mistaken, further lies stemming from that substitution are garbage.

Let us first take a look at your terrible computational errors.

My bad, I fucked up, but it wasn't actually a computational error.
It was an error in the setup of the problem.
You needed both x and y to reach 1 at the same point (which corresponds to the 2 arcs overlapping and you getting your ratio to be 0/0.

What I should have said is y=0.9+0.1x
Substituting this in we have:
vo(1-y)/vr(1-x)
=vo(1-(0.9+0.1x)/vr(1-x)
=0.1vo(10-9+x)/vr(1-x)
=0.1vo(1-x)/vr(1-x)
=0.1vo/vr.

That is the correct form.

You made another terrible computational error.
vo(0.1)(10-x)/vr(1-x)=
vo(0.1)(10-x)/10vr(1/10 - 0.1x)

No, I never did anything like this.
You are trying to use the one error as if it is multiple different errors, once again showing your dishonesty.

No dishonesty was ever included in my messages.

Almost your entire messages have been filled with dishonesty.

YOU CANNOT HAVE TWO SLIGHTLY DIFFERENT ANNULAR SECTIONS WHICH APPROXIMATE A RECTANGULAR LOOP.

Yes you can.
A rectangular roop would by an annular section of infinite radius.
As you reduce the radius to some finite value it will be an approximation of the original rectangle.

The formula for a rectangular loop is DIFFERENT.

I know. It has horrible math to try and solve which I can't be bothered solving, especially as you will just ignore it like you repeatedly ignored the math for the annular sections until you figure out a way to lie about it, using it to pretend to get a different result.
That is why we are approximating them as annular sections.

If you want, you can do the math.
You can even start from the formulas I provided which need to be solved.
If you make some simplifications, including that the in and out sections do not contribute significantly, that sin(x)=x and cos(x)=1, for small x, you get half the expected value.
Or if you can figure out the

YOU NEED THE SAME AREA FOR THE TWO INTERFEROMETERS FROM THE VERY START.

Yes, which is why instead of leaving it in terms of the angle subtended and the 2 radii, I showed that the area can be substituted in.

Slightly different annular sections will mean THAT R2-R1 DOES NOT EQUAL ANYMORE r2-r1, which was the essential hypothesis for your setup, from the very start.

No. You can still have R2-R1=r2-r1

Notwithstanding the fact that now you have s1o DIFFERENT FROM s1r, and as such you cannot compare the two Sagnac shifts, since you are dealing with DIFFERENT geometrical figures, you made a terrible blunder.

From the start of this thread, I have been discussing a rectangle, which is approximated as an annular section to determine the shift, as the math for the rectangle is too complex.
Try to keep up.

One which I was very careful to avoid from the very start.

Nope. If you have the 2 identical then only one is an annular section centered on the centre of rotation.
That means only one can use that simple math and the other needs to use much more complicated math.
Or, like I said before, you can approximate them as 2 slightly different annular sectors.

So no, not only did you make that mistake, your entire argument currently rests upon that mistake.

You cannot treat them as identical shape, yet still treat them as annular sectors rotating about the centre of the circles that make them.

YOU BLINDLY ASSUMED THAT s1/s2 = r1/r2 = R1/R2, and proceeded to get this:

And there you go lying again.
I assumed s1r/s2r=r1/r2, as by definition, that holds for an annular sector.
Similarly, I assumed s1o/s2o=R1/R2, as again, by definition that holds for an annular sector.
Try again, and quit with your strawman of myposition.

Remember this?
https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1981067#msg1981067

You mean where you worked out the ratio to be 0/0, yet presented it as somethign completely different?

Here is how to correctly compare the two ratios:
(ω_oR2)/(ω_rr2)

No. You cannot times both sides by 0.
If you wish to do that, present it honestly, the ratio there is 0/0.

Now deal with what has been actually been said, or fuck off.
Quit with your pathetic strawmen.
Quit with your pathetic copy paste BS.

I told you that I was going to make a flat earth believer out of you.

And I told you that like so many other things you have said, you are full of shit.

You are totally defeated here (and elsewhere, and always) jack.
Your cognitive dissonance condition detaches you from reality.
Your terrible blunders have been enumerated here in plain sight, so that everyone can see your cheating.

Projecting again I see.
Everywhere I have seen you post, except the liars only section, you have repeatedly had your ass handed to you.
You are yet to defeat me, and I can't recall you defeating anyone.

Instead you resort to tactics where you repeatedly post the same refuted BS, ignoring the refutations; or you repeatedly jump between topics so you can pretend you were never shown to be wrong, only to go back to the original topic completely ignoring the refutations of your BS.
You are also (for the most part) completely unable to rationally and honestly respond to arguments which have been made; instead you make up your own argument, lie while claiming the other people have made them, and then refute them.

As for those watching wanting to know why I have said this:

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 1, treating it as if you just cancel the 0.

However there are many things of that form.

[For example let] y=0.9+0.1x
Substituting this in we have:
vo(1-y)/vr(1-x)
=vo(1-(0.9+0.1x)/vr(1-x)
=0.1vo(10-9+x)/vr(1-x)
=0.1vo(1-x)/vr(1-x)
=0.1vo/vr.

This is because of the dishonest way he is presenting the ratio.
He has:

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

Ignoring the fact that you can't actually have the same s1/s2 value while keeping the math that simple, he has this:
ω_o(R2 - R1s₁/s₂)/ω_r(r2 - r1s₁/s₂)
Letting s₁/s₂=x for simplicity this can be simplified to:
ω_oR2(1 - (R1/R2)*x)/ω_rr2(1 - (r1/r2)*x)

Notice that this is similar to what I presented.
You have the top and bottom both have a term akin to (1-y).
in the top case, y=(R1/R2)*x, in the bottom case it is (r1/r2)*x
In order for his method to hold, these 2 ratios need to be the same. They are not.
So instead, you need to do more math to eliminate the variables correctly.
Yes, they will both look quite close to 1 (and in fact can tend to 1 as x does, making it more akin to 1-x^2, but lets ignore that for now as it doesn't change the argument. They tend to 1 at a different rate and thus the (1-y) tends to 0 at a different rate)), but due to the small numbers used, that can make a world of difference, especially as x is quite close to 1 making the entire thing quite close to 0.

In fact, we can compare them.
So lets set up some hypothetical interferometers. r1=6400 km, R2=150 000 000 km
We will make it really small, only 1 m difference. So r2=6400.001 km and R1=149 999 999.999 km.
So R1/R2=0.9999999999933
r1/r2=0.999999844

Not the same number.
So ignoring the fact that x isn't actually 1 (instead it would have a ratio quite similar to the above and the result will merely be twice the value below), and instead pretending it is, we get:
1-(R1/R2)*x=6.67E-12
1-(r1/r2)*x=1.56E-07

That means the slope between y and x (which was 0.1 before, with 0.9 to offset it correctly)), is 23437, or the ratio (1-y)/(1-x) (which was 10 before), which he has hidden with 0/0 is actually 4.27E-05 (in this case, it varies slightly as you change the size of the interferomter, it is also suspicially close to r1/R1, almost as if that magically invents the R1/r1 term he has).

So sticking that in his garbage you end up with the ratio of shifts being:
4.27E-05 * v_o/v_r
Sticking in the often used values of v_o=30 km/s and v_o=0.460 km/s (from google)
You end up with a final ratio of 2.78E-03, or roughly 1/359.38. (It would be more accurate if I used the actual radius of Earth and actual speeds).

My bad, I fucked up, but it wasn't actually a computational error.

You still don't get it jack.

You totally fucked up everything here in sight.


It is not saying R1=r1 and R2=r2.
If I were pretending they were identical it would require θo*R1=θr*r1 and θo*R2=θr*r2.

But that is exactly what you require when you SUBSTITUTE AN AREA WHICH INVOLVES r1 and r2 INTO A FORMULA WHICH FEATURES R1 AND R2.

Please learn.

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

s₂ = R2 x θ_o

s₁ = R1 x θ_o


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

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.


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


(0ωoR2)/(0ωrr2)

You have totally lost it jack.

Here is the correct derivation in plain sight:

Here is the correct ratio:

(ω_oR2)/(ω_rr2)

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

And this formula is correct even if the figures of R1 and r1 approach the values of R2, respectively r2, at different rates.

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


ω_o(R2 - R1s₁/s₂)/ω_r(r2 - r1s₁/s₂)

(R2 - R1s₁/s₂) will always be greater than (r2 - r1s₁/s₂)

example: R2=150,000,000; R1=149,999,999; r2=6401; r1=6400; s₂x0.999=s₁

(R2 - R1s₁/s₂) = 150,001
(r2 - r1s₁/s₂) = 7.4


R2 = R1 + ε = R1 + r2 - r1

r2 = r1 + ε = r1 + R2 - R1

(by hypothesis, R2 - R1 = r2 - r1)

a = s₁/s₂


[ω_o/ω_r]{(R1 + r2 - r1 - aR1)/(r1 + R2 - R1 - ar1)}

(R1 + r2 - r1 - aR1)>(r1 + R2 - R1 - ar1)

R1(1 - a)>r1(1 - a)

R1>r1



A rectangular roop would by an annular section of infinite radius.
As you reduce the radius to some finite value it will be an approximation of the original rectangle.

Do I have to teach you Geometry 101 as well?

The formula for a rectangular interferometer is totally different than the formula for an annular sector.

What you are saying is that s1o DOES NOT EQUAL s1r. But then you can no longer compare two Sagnac shifts involving different areas.

Moreoever a rectangle means that the Qo (the subtended angle for R2 and R1) IS NO LONGER THE SAME FOR R2 AND R1. The same goes for Qr.


I assumed s1r/s2r=r1/r2, as by definition, that holds for an annular sector.
Similarly, I assumed s1o/s2o=R1/R2, as again, by definition that holds for an annular sector.


That is another fuck up out of your twisted mind.

YOU CANNOT ASSUME  s1r/s2r=r1/r2 OR s1o/s2o=R1/R2 and at the same time have R2 - R1 = r2 - r1.

That is a beginner's mistake, exactly your line of thought.


s1r/s2r EQUALS s1o/s2o by definition (same area of the interferometer).

So you cannot substitute blindly r1/r2 = R1/R2 since you have DIFFERENT RADII.

It is the same area but DIFFERENT RADII.


If one proceeds along your faulty logic, you get this:

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

s₂ = R2 x θ_o

s₁ = R1 x θ_o


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

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


Can you understand this much jack?


Let us now mistankenly substitute the AREA for the VELOCITY X RADIUS, just like jack did:

θo/θr = (r2 + r1)/(R2 + R1)

Substituting this into our correct formula, we get:

ωoθo(R2 + R1)/ωrθr(r2 + r1)

As was proved above, using the interferometer presented in the graphic posted by jack, we end up with r1 = r2, this meaning that R2 = R1, and everything cancels out in the ratio, with the exception of the angular velocities.


THE ONLY WAY YOU ARE GOING TO END UP WITH wo/wr IS TO ASSUME THAT R2=R1 AND r2=r1, THEN EVERYTHING WILL CANCEL OUT.


But this amounts to plain cheating and lying.

Your readers can see through your thin veils of deception jack.

You are done here.




Ignoring the fact that you can't actually have the same s1/s2 value while keeping the math that simple

But they are same jack.

s1o = s1r and s2o = s2r by DEFINITION, same area for the interferometer.


in the top case, y=(R1/R2)*x, in the bottom case it is (r1/r2)*x
In order for his method to hold, these 2 ratios need to be the same. They are not.

Exactly.

I proceeded to present the case where R1 = R2 x b and r1 = r2 x b (same b), and THEN PRESENTED THE GENERAL CASE WHERE b0 does not equal br.

Remember?

(ω_oR2)/(ω_rr2)

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

And this formula is correct even if the figures of R1 and r1 approach the values of R2, respectively r2, at different rates.

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


ω_o(R2 - R1s₁/s₂)/ω_r(r2 - r1s₁/s₂)

(R2 - R1s₁/s₂) will always be greater than (r2 - r1s₁/s₂)

example: R2=150,000,000; R1=149,999,999; r2=6401; r1=6400; s₂x0.999=s₁

(R2 - R1s₁/s₂) = 150,001
(r2 - r1s₁/s₂) = 7.4


R2 = R1 + ε = R1 + r2 - r1

r2 = r1 + ε = r1 + R2 - R1

(by hypothesis, R2 - R1 = r2 - r1)

a = s₁/s₂


[ω_o/ω_r]{(R1 + r2 - r1 - aR1)/(r1 + R2 - R1 - ar1)}

(R1 + r2 - r1 - aR1)>(r1 + R2 - R1 - ar1)

R1(1 - a)>r1(1 - a)

R1>r1


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


There are no errors whatsoever in my derivation. I am very good with formulas, please understand this fact.


This, then, is the correct final formula:

(ω_oR2)/(ω_rr2) = v_o/v_r

My bad, I fucked up, but it wasn't actually a computational error.
You still don't get it jack.
You totally fucked up everything here in sight.

Nope. I fucked up 1 thing, which doesn't help your case, and when corrected still shows you to be completely wrong.

You couldn't refute the main argument with that, and instead focused on the error. It is almost like you know you are wrong so are nitpicking at whatever you can.
Once again, you have just repeated the same refuted bullshit and failed to address the arguments raised, so I will ignore the vast majority of it.
You aren't even addressing the straw men you make up.

Last chance before I go back to asking simple questions, one at a time, which you seem unable to deal with because you know they lead to your defeat.

Here is the correct derivation in plain sight:

Here is the correct ratio:

(ω_oR2)/(ω_rr2)

You might want to learn what a derivation is
It isn't just saying a formula. You need to show how you got it.

example: R2=150,000,000; R1=149,999,999; r2=6401; r1=6400; s₂x0.999=s₁
(R2 - R1s₁/s₂) = 150,001
(r2 - r1s₁/s₂) = 7.4

Once again, you are just making numbers up.
Using the numbers for R1 and R2, you would end up with s2*0.999999993=s1, not 0.999 like you claim.
If instead you use r1 and r2 you get 0.999843774. So good job lying, yet again.

In order to get 0.999 (keeping r1 and R2), you would need r2=6406.406406 and R1=149850000.000

Do you understand that?
You need a section of the interferometer which goes 150000 km away from Earth. That is almost to the distance of the moon, while you have another one which only stretched 6.4 km away. You have a massively different area, so no wonder you get a completely different result.
This just goes to highlight that you can't treat them both as the exact same shape. Instead they must have a different s1/s1 ratio to keep them both being annular sectors.
If you want to keep them as the exact same shape, rather than approximating one or both as an annular sector which is different to the physical loop you CAN'T use the simple math. That simple math only holds for annular sectors.

Do I have to teach you Geometry 101 as well?

Get this through your thick skull:
YOU ARE NOT HERE TO TEACH!!!
I am not here as an ingorant child to just accept whatever bullshit you spout.

This is a place for debate.
That means you need address the arguments raised rather than dismissing them and preaching the same refuted BS.

The formula for a rectangular interferometer is totally different than the formula for an annular sector.

Yes, I know. But they both work out the be 4Aw/c^2.
But the math is a hell of a lot more complicated, so you can approximate it as an annular sector.

What you are saying is that s1o DOES NOT EQUAL s1r. But then you can no longer compare two Sagnac shifts involving different areas.

Again, I am approximating the same shape as 2 separate annular sectors. These have the same area.

Moreoever a rectangle means that the Qo (the subtended angle for R2 and R1) IS NO LONGER THE SAME FOR R2 AND R1

Yes it is. Both are 0.

That is another fuck up out of your twisted mind.
YOU CANNOT ASSUME  s1r/s2r=r1/r2 OR s1o/s2o=R1/R2 and at the same time have R2 - R1 = r2 - r1.

Do I need to teach you geometry?
By definition, that holds for an annular sector.
Unless this holds, you CANNOT be dealing with annular sectors and thus CANNOT use the math for a simple annular sector.
Instead you will need to use much more complex math.
Like I said, the are 2 slightly different annular sectors.
If you would like to prove that 2 slightly different annular sectors, such that they have s1r, s2r, s1o and s2o, cannot have the above relations with R2-R1=r2-r1, PROVE IT!!

s1r/s2r EQUALS s1o/s2o by definition (same area of the interferometer).

No. It doesn't.
The interferometers are annular sectors.
The area is given by (using your Q because I can't be bothered to copy theta):
A=Q*(R2^2-R1^2)/2
Notice that there are 4 variables in this equation, yes R1 and R2 linked, but that is by another variable so it just pushes the problem, but it can make the point easier:
R1=R2-e.
Thus:
A=Q*(R2^2-(R2-e)^2)/2
A=Q*(R2^2-R2^2+2*R2*e+e^2)/2
A=Q*(2*R2*e+e^2)/2
So yet again, 4 variables.
All we were constraining to be equal were A and e. This still leaves 2 variables.

Q=2*A/(2*R2*e+e^4)

I will make it nice and simple for you. I will provide an annular sector.
I will tell you its its r1, r2, Q, s1, s2, s1/s2 and A.
I will then make another one, using just R2, A and setting R2-R1=r2-r1, giving you the same values.

So, first, lets let r1=6400, r2=6401, and s1=1.
Thus the values (in the order above) are:
6400, 6401, 0.00015625, 1, 1.00015625, 0.999843774410248, 1.000078125

Now, lets use R2=150 000 000.
Then we get:
149999999, 150000000, 0.0000000066671875, 1.00007811833281, 1.000078125, 0.999999993333333, 1.000078125

Notice the area is identical, R2-R1=1=r2-r1, but s1/s2=0.999843774410248 in the first case and 0.999999993333333 in the latter.

So no, you can.
In fact, YOU MUST.
s1/s2=R1/R2
=(R2-e)/R2
=1-e/R2.
So if you take an annular sector, and try to keep s1/s2 constant, and try to keep e constant, you will not be able to vary the radius.

So you cannot substitute blindly r1/r2 = R1/R2 since you have DIFFERENT RADII.

You're not very good at this are you?
This is what I mean about contradicting yourself.
Above you said that s1/s2 must be the same.
But by definition, s1/s2=r1/r2.

But now you are saying the can't be the same.
Guess what?
THAT IS WHAT I HAVE BEEN SAYING ALL ALONG!!!
So thanks for telling everyone that you are wrong and I am right.
Good job objecting to your own claim.

THE ONLY WAY YOU ARE GOING TO END UP WITH wo/wr IS TO ASSUME THAT R2=R1 AND r2=r1, THEN EVERYTHING WILL CANCEL OUT.
But this amounts to plain cheating and lying.

Nope. I subbed in the area.
Meanwhile, the only way you get v0/vr is if you assume(1-(R1/R2)^2)=(1-(r1/r2)^2), or by letting R1=R2 and thus getting 0vo/0vr and then dividing both sides by 0.

I proceeded to present the case where R1 = R2 x b and r1 = r2 x b (same b)

Where you showed the area is 0, or where you ended up with 0/0, as that is the only way to have the same b?
Remember, where I tore that to shreds for that reason?

Repeating it is pointless.

THEN PRESENTED THE GENERAL CASE WHERE b0 does not equal br.

No, as you have not done that.

There are no errors whatsoever in my derivation. I am very good with formulas, please understand this fact.

There are many errors, as I have shown.
You are good with deceit, including dishonestly manupulating formulas, like turning:
wo/wr
into:
0/0
and that into:
vo/vr.

As you are yet to refute my derivation, the correct formula remains as:
dt=4*A*w/c^2, and thus the rotational Sagnac remains as 365 times that of the orbit.


Now remember, last chance for you to rationally respond before I resort back to simple questions.
Deal with what I have said, without just spamming the same refuted BS, or I go back to the simple questions.

Using the numbers for R1 and R2, you would end up with s2*0.999999993=s1, not 0.999 like you claim.
If instead you use r1 and r2 you get 0.999843774. So good job lying, yet again.

In order to get 0.999 (keeping r1 and R2), you would need r2=6406.406406 and R1=149850000.000


R1 = 149,999,997
R2 = 150,000,000

s2 = 1
s1 = 0.99999998
r2 = 6401
r1 = 6400.999872

(R2 - R1s₁/s₂) = 6
(r2 - r1s₁/s₂) = 0.0002561

R2 - R1s₁/s₂ will ALWAYS BE GREATER THAN r2 - r1s₁/s₂.


I am not here as an ingorant child

Take a look at your earlier calculation, posted today:

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.

Let us first take a look at your terrible computational errors.

vo(1-y)/vr(1-x)

You said y=0.1x

vo(1-0.1x)/vr(1-x)=

vo(0.1)[10-x]/vr(1-x)

What you wrote is:

[0.1*vo*(1-x)]/[vr*(1-y)]

If we bring back the 0.1 factor within the parenthesis, we get, according to your faulty logic:

vo*(0.1 - 0.1x)

This is incorrect.

The correct answer is:

vo(0.1)[10-x]/vr(1-x)


[0.1*vo*(1-x)]/[vr*(1-y)]

You made another terrible computational error.

vo(0.1)(10-x)/vr(1-x)=

vo(0.1)(10-x)/10vr(1/10 - 0.1x)


Using your line of thought, we get:

[0.1*vo*(1-x)]/[vr*(1-y)]

0.1vo(1-x)/vr(1-0.1x)


You have a total ignorance of basic math.

You cannot factor a value of 0.1 out of a parenthesis, and you have the audacity to claim you are not an ignorant child?

Now, lets use R2=150 000 000.
Then we get:
149999999, 150000000, 0.0000000066671875, 1.00007811833281, 1.000078125, 0.999999993333333, 1.000078125

Notice the area is identical, R2-R1=1=r2-r1, but s1/s2=0.999843774410248 in the first case and 0.999999993333333 in the latter.

So no, you can.
In fact, YOU MUST.
s1/s2=R1/R2

BUT YOU CANNOT!

The range of R2 is 150,000,000 to 149,987,200.

The range of r2 is 12,800 to 6,400.


R2 - R1s₁/s₂ will ALWAYS BE GREATER THAN r2 - r1s₁/s₂.



Let us use your numbers.

s1/s2 = 0.999843774

R1/R2 = 0.9999999933

You are an ignorant child.

If you are off by a couple of digits, say 0.999999666, then you get this:

R1 = 149,999,949.9 km

Your R1 value was 149,999,999

A difference of 49.1 km.

You see where your twisted calculations take you?

The difference was supposed to be a single kilometer. Now you have 49.1 km.

So s1/s2 can NEVER equal R1/R2.


TAKE A LOOK AT WHAT IS ACTUALLY BEING REQUIRED.

1.00007811833281, 1.000078125 (s1 and s2), your data for s1o and s2o.

An interferometer which has to be built to a precision of 0.00000000666719km.

That is, down to the level of actual MICRONS. It is not feasible at all.

If you are off by a couple of digits, say 0.999999666, then you get this:

R1 = 149,999,949.9 km

Your R1 value was 149,999,999

A difference of 49.1 km.

So if you take an annular sector, and try to keep s1/s2 constant, and try to keep e constant, you will not be able to vary the radius.

Then, such a pair of interferometers CANNOT BE BUILT, exactly what I told from the very start.

Just take a look at your data:

I will tell you its its r1, r2, Q, s1, s2, s1/s2 and A.
I will then make another one, using just R2, A and setting R2-R1=r2-r1, giving you the same values.

So, first, lets let r1=6400, r2=6401, and s1=1.
Thus the values (in the order above) are:
6400, 6401, 0.00015625, 1, 1.00015625, 0.999843774410248, 1.000078125

Now, lets use R2=150 000 000.
Then we get:
149999999, 150000000, 0.0000000066671875, 1.00007811833281, 1.000078125, 0.999999993333333, 1.000078125

Notice the area is identical, R2-R1=1=r2-r1, but s1/s2=0.999843774410248 in the first case and 0.999999993333333 in the latter.


You have DIFFERENT s1o and s2o as opposed to s1r and s2r. Given the huge difference in magnitude between R2 and r2, that will not be the same interferometer, that is, no comparison can be made.

You need one single interferometer whose orbital and rotational Sagnac could be compared: no such interferometer in the shape of an annular sector which can satisfy all the requirements could be built.

Then we are back to square one.


Here is the formula for a rectangular interferometer:


http://image.ibb.co/m9YATG/ta1.jpg
http://image.ibb.co/k3Mx8G/ta2.jpg
http://image.ibb.co/me5w2b/ta3.jpg
http://image.ibb.co/iXXLTG/ta4.jpg
http://image.ibb.co/mKHyFw/ta5.jpg



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


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.


dt=4*A*w/c^2, and thus the rotational Sagnac remains as 365 times that of the orbit.

The interferometer which could be used to compare the orbital and rotational Sagnac CANNOT BE BUILT AT ALL. The figures won't match.

Moreover, dt can expressed in terms of single speed, for the R2 R1 interferometer, the only thing is it cannot be compared to a r2 r1 interferometer since the data won't match:

dt_o = 2v_2os₂(1 - s₁/s₂ + ε([1-ε/R2]/R2))/c²


But we cannot build an interferometer which will satisfy the requirements of the combined orbital and of the rotational Sagnac.

But the rectangular interferometer CAN BE USED IMMEDIATELY TO COMPARE THE DATA.

AND THE DATA IS TOTALLY AGAINST YOU:

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.



You are here to learn, and I am here to teach you.

THE ONLY ASSUMPTION YOU CAN MAKE IS R2 - R1 = r2 - r1.

That's it.


You cannot substitute an area involving r2 and r1 into a formula which features R2 and R1.

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

s2 = R2 x θo

s1 = R1 x θo


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

s2 = r2 x θr

s1 = r1 x θr


R2 - R1 = r2 - r1

r2 x θr = R2 x θo

r1 x θr = R1 x θo

r2/r1 = R2/R1

(r2 x R1) = (r1 x R2)

Since the two areas must be equal,

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



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

R2 = r2 - r1 + R1

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

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

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


Here is what you posted:

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)

You blindly substituted s1/s2 = R1/R2 into the formula.

However, as we have seen above, there is no mathematical compatibility between the two values.

I have observed this error from the very start, and was very careful not to fall into the trap of just substituting s1/s2 = R1/R2.


Meanwhile, the only way you get v0/vr is if you assume(1-(R1/R2)^2)=(1-(r1/r2)^2), or by letting R1=R2 and thus getting 0vo/0vr and then dividing both sides by 0.

THAT WAS YOUR CALCULATION AND YOUR ASSUMPTION NOT MINE.

You are accusing me of the very errors committed by you?

Here is the correct derivation.

(ω_oR2)/(ω_rr2)

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

And this formula is correct even if the figures of R1 and r1 approach the values of R2, respectively r2, at different rates.

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


ω_o(R2 - R1s₁/s₂)/ω_r(r2 - r1s₁/s₂)

(R2 - R1s₁/s₂) will always be greater than (r2 - r1s₁/s₂)

example: R2=150,000,000; R1=149,999,999; r2=6401; r1=6400; s₂x0.999=s₁

(R2 - R1s₁/s₂) = 150,001
(r2 - r1s₁/s₂) = 7.4


R2 = R1 + ε = R1 + r2 - r1

r2 = r1 + ε = r1 + R2 - R1

(by hypothesis, R2 - R1 = r2 - r1)

a = s₁/s₂


[ω_o/ω_r]{(R1 + r2 - r1 - aR1)/(r1 + R2 - R1 - ar1)}

(R1 + r2 - r1 - aR1)>(r1 + R2 - R1 - ar1)

R1(1 - a)>r1(1 - a)

R1>r1


See how nice and easy everything works out? No s1/s2 comparison to R1/R2.

No 0/0, no comparison of the areas.

Just a perfect logic using very simple calculations, which even an ignorant child like you can comprehend.



R1 = 149,999,997
R2 = 150,000,000

s2 = 1
s1 = 0.99999998
r2 = 6401
r1 = 6400.999872

(R2 - R1s₁/s₂) = 6
(r2 - r1s₁/s₂) = 0.0002561

R2 - R1s₁/s₂ will ALWAYS BE GREATER THAN r2 - r1s₁/s₂.


AGAIN, A SINGLE INTERFEROMETER (OR EVEN A PAIR OF SUCH INTERFEROMETERS) CANNOT BE BUILT AT ALL TO SATISFY THE DATA. But that is what I said from the very start of this thread.

However, we do have the formula for a rectangular interferometer (see above), which can be used immediately to see that the orbital Sagnac is at least 60 times greater than the rotational Sagnac.

<< I fail to see your derivation of the Sagnac delay anywhere in your massive walls of text -- Where ist it? >>

And just remember that your result be independent of both the shape of the loop and the centre of rotation.

Just remember that all derivations, including that of Sagnac himself, get an expression involving only the projected loop area of the signal path and the angular velocity.

Read:

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

So we have a rather good check on the validity of yor results.

rabinoz, you are just a barnacle on jack's ass. A most unenviable position to be in, given the torrents of shit unloaded by jack on this thread.

R2 - R1s₁/s₂ will ALWAYS BE GREATER THAN r2 - r1s₁/s₂.

That is all you have to know.

rabinoz, you are just a barnacle on jack's ass. A most unenviable position to be in, given the torrents of shit unloaded by jack on this thread.

R2 - R1s₁/s₂ will ALWAYS BE GREATER THAN r2 - r1s₁/s₂.

That is all you have to know.

No, it's not. I want to either see your derivation of the Sagnac Delay or your admission that you cannot derive it.

Every reference I have see claims 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

So the orbital Sagnac is roughly 1/365 of the rotational Sagnac.

If you can show us your derivation of the Sagnac delay, not just the supposed ratio, then we might understand.

Oscar Wilde wrote, "I am not young enough to know everything."
And Einstein wrote,  "The more I learn, the more I realize how much I don't know."

On that basis the  I-am-the-greatest know-it-all, sandokhan, must be all of 5 years old and young enough "young enough to know everything"!

So the orbital Sagnac is roughly 1/365 of the rotational Sagnac.

Let's put your statement to the test.

(ω_oR2)/(ω_rr2)

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

And this formula is correct even if the figures of R1 and r1 approach the values of R2, respectively r2, at different rates.

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


ω_o(R2 - R1s₁/s₂)/ω_r(r2 - r1s₁/s₂)

(R2 - R1s₁/s₂) will always be greater than (r2 - r1s₁/s₂)

R2 = R1 + ε = R1 + r2 - r1

r2 = r1 + ε = r1 + R2 - R1

(by hypothesis, R2 - R1 = r2 - r1)

a = s₁/s₂


[ω_o/ω_r]{(R1 + r2 - r1 - aR1)/(r1 + R2 - R1 - ar1)}

(R1 + r2 - r1 - aR1)>(r1 + R2 - R1 - ar1)

R1(1 - a)>r1(1 - a)

R1>r1

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

(ω_oR2)/(ω_rr2)

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.

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

Let us now proceed as jack requires.

dto = 2w_ohL/c2

dtr = 2w_rhL/c2

Then, obviously we would be left with a ratio of the angular velocities of the loops, and NOT the circumferential speeds at the end point.

Let us now proceed correctly.

dto = 2v_oL/c2

where ωR = 30km/s (center of loop located at the Sun, since we are calculating the correct orbital Sagnac effect)

dtr = 2v_rL/c2

where ωr = 0.465km/s (calculation for the rotational Sagnac)

Then the final ratio will be: v_o/v_r = ~60.


How do the best physicists in the world actually compare the Sagnac effects?

Dr. A.G. Kelly, one of the best experts of the 20th century regarding the Sagnac effect (Dr. Kelly discovered the huge errors in the Hafele-Keating paper).

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





The linear velocities (the circumferential tangent speeds at the end point) are being compared and NOT the angular velocities.

It is the circumferential speeds at the end point which are being compared, and not the angular velocities of the loop itself.

COMPARISON OF THE SAGNAC EFFECT WITH STR

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

t_o = t'γ

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

t_o = t'(1 + v²/2c²)


dt_R = (t_o - t')/t_o = v²/(v² + 2c²)

dt_R = relativity time ratio



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

dt' = t_o - t' = t_ov/(c + v)


dt_S = (t_o - t')/t_o = v/(v + c)


dt_S = Sagnac ratio


dt_S/dt_R = (2c² + v²)/v(v + c)

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

Well don't say I didn't give you a chance.
You just keep on repeating the same old crap, without addressing what has been said.
I made a mistake, acknowledged that mistake, and corrected it, yet you continue to focus on that mistake instead of the corrected version.
What point are you trying to make?
That you are so pathetic and hopeless that you can't make any rational argument at all to defend your position so now need to continually focus on this mistake to pretend you have achieved something for once in your pathetic existence?

Making a mistake doesn't make one an ignorant child. However continually focusing on that mistake and ignoring the correction does make one a pathetic child.

Oh well, I gave you a chance, now it is time for the simple version.

In your linked post, you have this:

dt = t₁ - t₂ = {2θ_oω_o(R₂² - R₁²}/c²

Does that mean you agree to the derivation up to this point?
YES OR NO!!
No other bullshit, just answer the question.

Your bravado does not impress anyone.

You lose all the way.

My linked post?

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1981067#msg1981067

dt = t1 - t2 = {2θoωo(R22 - R12}/c2

v2o = ωo x R2
v1o = ωo x R1

dt = 2v2s2/c2 - 2v1s1/c2

Sure.


Now, let me ask you a question.

How is the Sagnac effect measured, according to the angular velocity of the loop, OR the linear velocity (circumferential tangent speed at the end point) which has to involve a radius and an angular velocity?

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





As for the chances being offered...

This forum has offered you jack more than enough chances to come to your senses, to realize that there is no actual curvature at the surface of the Earth.

I told you that I will eventually make a flat earth believer out of you.

Also, I will say one thing (and I wont bother responding to anything regarding it until you have answered my above question), have you actually bothered looking at your claimed derivation for the rectangular interferometer?

It is effectively a fibre optic conveyer. The 4 mirrors need to remain stationary while the observer moves.
This is not the Sagnac effect.
For the Sagnac effect you need the mirrors to move with the observer.

So in order to record the orbital shift for this interferometer you need a loop which remains stationary w.r.t. Earth's orbit. We have no such loop and as such cannot measure it.
It also makes it impossible to measure the orbital and rotational shift with the same interferometer, as it would require the mirrors to be stationary w.r.t Earth's orbit, while remaining stationary in a reference frame moving with Earth along its orbit but not spinning.
The mirrors can't do both.

Here is the correct derivation for a rectangular interferometer moving with the person at some velocity v.
I will even use their setup (at least for the most part).
First the simple version using relativity:
The loop is stationary w.r.t. the person in an inertial reference frame. As such, there is no shift.
This is similar to their derivation, except the final point is at (4a+4b,0,0) for both.
Thus dt=(4a+4b)-(4a+4b)=0.

Now in a different reference frame or in an aether model, where the loop and person move at velocity v, which is no longer as simple.
The CW path (light initially moving opposite observer):
We start at (0,0,0), where I am having this as (x, y, t) and when I only have 2, it is (x,y).
We move towards the point (-a,0), but in this time, the mirror moves towards us, resulting in the reflection being at:
(vt₁-a,0,t₁)

That is to say:
ct₁=a-vt₁
Thus t₁(c+v)=a.
Thus t₁=a/(c+v).

Similarly, as we go around the loop we have the points:
(vt₂-a,2b,t₂)
(vt₃+a,2b,t₃)
(vt₄+a,0,t₄)
(vt₅,0,t₅)

I will also let dt_i=t_i-t_i-1, where t₀=0.
That is, dt corresponds to the time increment required to get to that time point from the prior one.
So for the second part:
dt₂c=sqrt(dt₂²v²+4b²).

dt₃c=2a+dt₃v
dt₃=2a/(c-v)

dt₄c=sqrt(dt₄²v²+4b²).
Note that this is the exact same relation as dt₂
Thus dt₄=dt₂

dt₅c=a-dt₅v
dt₅=a/(c+v)

t_CW=t₁+dt₂+dt₃+dt₄+dt₅
=a/(c+v)+dt₂+dt₂+a/(c+v)
=2a/(c+v)+2a/(c-v)+2dt₂

Now for the CCW path we would have the same, except the v is now negative.
So first we just need to look at dt₂
Originally it was:
dt₂c=sqrt(dt₂²v²+4b²).

But now v is negative so it becomes:
dt₂c=sqrt(dt₂²(-v)²+4b²).
dt₂c=sqrt(dt₂²v²+4b²).
Which is the same as before.
So this means our total time for this path is:
t_CCW=2a/(c+(-v))+2a/(c-(-v))+2dt₂
=2a/(c-v)+2a/(c+v)+2dt₂

And thus our shift is:
dt=t_CW-t_CCW
=[2a/(c+v)+2a/(c-v)+2dt₂]-[2a/(c-v)+2a/(c+v)+2dt₂]
=2a/(c+v)+2a/(c-v)+2dt₂-2a/(c-v)-2a/(c+v)-2dt₂
=2a/(c+v)-2a/(c+v)+2a/(c-v)-2a/(c-v)+2dt₂-2dt₂
=0+0+0
=0.

Oh would you look at that, translation produces NO SHIFT AT ALL!!

The only part I didn't do completely was this:
dt₂c=sqrt(dt₂²v²+4b²)
But as that is the same all 4 times, IT DOESN'T MATTER!!!

So your "derivation" amounts to a pile of shit.

See, this is why I say you can't just link to crap.
You need to do it yourself, explaining each step.

You either didn't understand what was being said and wanted it to be like that;
Or you knew that it didn't apply but wanted to try and hide it.

Your bravado does not impress anyone.
You lose all the way.

Except I still win.
You are yet to defeat me.

My linked post?
https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1981067#msg1981067
dt = t1 - t2 = {2θoωo(R22 - R12}/c2
v2o = ωo x R2
v1o = ωo x R1
dt = 2v2s2/c2 - 2v1s1/c2
Sure.

No jumping ahead or any insane substitutions. Leave it as it is (especially when you remove indications of which is which, if you want to do it like this, indicate the the s is for the orbit).
You have said yes.
That means you agree the shift is given by:
dt=2θ_oω_o(R₂² - R₁²)/c²

Which can equally be written as:
dt=2θ_o(R₂² - R₁²)ω_o/c²

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

Yes or no?

How is the Sagnac effect measured, according to the angular velocity of the loop, OR the linear velocity (circumferential tangent speed at the end point) which has to involve a radius and an angular velocity?

It is measured with angular velocity, given dt=4*A*w/c^2.

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

Because they are a charlatan trying to claim special relativity is broken.
I don't give a damn as they are not you, and you are the one here debating.
If they would like to come here and get their ass handed to them, they are welcome to.

As for the chances being offered...
This forum has offered you jack more than enough chances to come to your senses, to realize that there is no actual curvature at the surface of the Earth.
I told you that I will eventually make a flat earth believer out of you.

Except you haven't come close to making me reject all of the evidence for the curve and all the evidence refuting a flat Earth.

So what you should have said is that the forum has offered more than enough chances for me to abandon all sense of reason, rejecting all evidence and falsely think Earth is flat.

I made a mistake, acknowledged that mistake, and corrected it, yet you continue to focus on that mistake instead of the corrected version.

But that was an intentional mistake, designed to fool your readers.

Here are your accompanying words:

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 good job failing yet again.

Another source of your dishonesty stems from continually changing the association between r1, r2, s1, s2 and so on.

Let us take a look at your honesty then.

vo(1-y)/vr(1-x)

You said y=0.1x

vo(1-0.1x)/vr(1-x)=

vo(0.1)[10-x]/vr(1-x)

What you wrote is:

[0.1*vo*(1-x)]/[vr*(1-y)]

If we bring back the 0.1 factor within the parenthesis, we get, according to your faulty logic:

vo*(0.1 - 0.1x)

This is incorrect.

The correct answer is:

vo(0.1)[10-x]/vr(1-x)


[0.1*vo*(1-x)]/[vr*(1-y)]

You made another terrible computational error.

vo(0.1)(10-x)/vr(1-x)=

vo(0.1)(10-x)/10vr(1/10 - 0.1x)


Using your line of thought, we get:

[0.1*vo*(1-x)]/[vr*(1-y)]

0.1vo(1-x)/vr(1-0.1x)



NOT ONLY YOU FAILED TO PROPERLY FACTOR OUT THE 0.1 TERM, BUT YOU ACTUALLY USED THE WRONG DIVISOR, 1-Y INSTEAD OF 1-X.

You are not fooling anybody here jack.

What I should have said is y=0.9+0.1x
Substituting this in we have:
vo(1-y)/vr(1-x)
=vo(1-(0.9+0.1x)/vr(1-x)
=0.1vo(10-9+x)/vr(1-x)
=0.1vo(1-x)/vr(1-x)
=0.1vo/vr.

You are trying to use the one error as if it is multiple different errors, once again showing your dishonesty.

But you did. You used the wrong divisor in your original message.

[0.1*vo*(1-x)]/[vr*(1-y)]




Your next response has another huge error in it.

0.1vo(10-9+x)/vr(1-x)
=0.1vo(1-x)/vr(1-x)
=0.1vo/vr.

What you mean is this:

0.1vo(1 + x)/vr(1 - x)

10-9 = 1.

How did the 1+x term become 1-x?

So to use your own words, another fuck up signed jack.

I made a mistake, acknowledged that mistake, and corrected it, yet you continue to focus on that mistake instead of the corrected version.
But that was an intentional mistake, designed to fool your readers.

Seriously?
I asked a simple question and you can't answer it?
It was an accidental mistake, (and there was another there that you missed, and which I missed the second time as well).
And apparently another typographical error as well.
I never said I was perfect.

Your next response has another huge error in it.
0.1vo(10-9+x)/vr(1-x)
=0.1vo(1-x)/vr(1-x)
=0.1vo/vr.

I notice you ignored the first 2 lines of that.
here let me put it in as well:

vo(1-y)/vr(1-x)
=vo(1-(0.9+0.1x)/vr(1-x)
0.1vo(10-9+x)/vr(1-x)
=0.1vo(1-x)/vr(1-x)
=0.1vo/vr.

Now tell me what I mean.

It is quite clear where the error is. It appears in a single line:
=vo(1-(0.9+0.1x)/vr(1-x)
0.1vo(10-9+x)/vr(1-x)
=0.1vo(1-x)/vr(1-x)

That plus, should be a minus.
I wouldn't call a typo that does not affect the conclusion (as the typo was just for that one line) a huge error.

But i notice you didn't correct that one. Instead you dishonestly left it there and pretended that was the correct solution.
Was that because you knew correcting this properly would show you to be full of shit?

How did the 1+x term become 1-x?

It didn't.
The (1-(0.9+0.1x) became
0.1(10-(9+x)) which became
0.1(10-9-x) which became
0.1(1-x).

There was just a typo where the (10-9-x) was shown as (10-9+x).

But don't worry, you dont need to address that. You have a simpler question to address:
Do you accept that the area of an annular sector is given by:
A=θ_o(R₂² - R₁²)/2
or equally that the following equation holds:
2A=θ_o(R₂² - R₁²)

Yes or no?

It is effectively a fibre optic conveyer. The 4 mirrors need to remain stationary while the observer moves.
This is not the Sagnac effect.

But it is the Sagnac effect.

The entire interferometer moves with speed v.

It can be a rotational movement or a translational movement.


So in order to record the orbital shift for this interferometer you need a loop which remains stationary w.r.t. Earth's orbit. We have no such loop and as such cannot measure it.

But we do have a loop, the very orbital path of the Earth around the Sun.

Please learn.

"The motion of the earth's orbit is also a sagnac effect. We should see light path distance differentials caused by the orbit just like we see if for earth's rotation.

The orbital path is simply longer and nothing else.

The earth - sun orbital frame is a sagnac rotating frame.

The Sagnac correction for the earth's rotation is applied because as the light moves toward the receiver, the receiver rotates with the earth changing the distance the signal travels.

In the same light, if the unit had been at the equator at noon, then it should see the full effect of the Sagnac effect of the earth's revolution around the sun.
In other words, assume a satellite is low on the horizon in the east at the equator.

We should measure a sagnac correction for the earth's rotation on its axis and a sagnac correction of the earth's rotation/revolution around the sun.
If sagnac is true for the earth's rotation, then light travels at one speed c. the speed of light cannot be increased by circular motion and presumably not by linear motion either.

If light travels at one speed c, then as the earth moves in it's revolution loop at 30k/s, while light moves c through space, the unit at the equator at noon would move with the earth' rotation and the earth's revolution cutting the distance the signal must travel to meet the unit.
The earth is rotating at 1000 mph. This shows up in GPS as c+v and c-v as you would expect with Sagnac.

All that is fine.

When the satellite emits at c, the earth rotates the receiver at v and so a correction is needed.

This is all OK.

Now, the earth is revolving around the sun at 67000 mph, as we are told by the heliocentrists.

Let's say the unit is at the equator and the satellite is low on the horizon in the east at noon.

That means the unit is traveling at the orbital speed of the earth at 67,000 MPH.

The satellite emits at one speed c in space. While the light travels through space toward the unit at c, the unit moves with the earth at 67,000 MPH. The unit cuts the distance that the light must travel.

This is not being seen by any experiements nor GPS."

Try to imagine the center of the earth as the sun and the earth's orbit as the surface. So, the earth is in a rotational/circular pattern in its orbit. Just like the rotational sagnac, the earth rotates toward where the satellite emitted the signal, and with the orbit, the earth, the unit is orbited toward where the signal was emitted.

Then we simply apply the well-known formula to this situation, where R = 150,000,000 km, v = wR (=30km/s).

The orbital Sagnac is much greater than the rotational Sagnac and that it is missing.


First the simple version using relativity:
The loop is stationary w.r.t. the person in an inertial reference frame. As such, there is no shift.
This is similar to their derivation, except the final point is at (4a+4b,0,0) for both.
Thus dt=(4a+4b)-(4a+4b)=0.

What an ignorant child.

The Sagnac effect is far larger than the effect forecast by relativity theory.

STR has no possible function in explaining the Sagnac effect.

The Sagnac effect is a non-relativistic effect.

COMPARISON OF THE SAGNAC EFFECT WITH SPECIAL RELATIVITY, starts on page 7, calculations/formulas on page 8

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

page 8

Because many investigators claim that the
Sagnac effect is made explicable by using the
Theory of Special Relativity, a comparison of
that theory with the actual test results is given
below. It will be shown that the effects
calculated under these two theories are of very
different orders of magnitude, and that
therefore the Special Theory is of no value in
trying to explain the effect.

COMPARISON OF THE SAGNAC EFFECT WITH STR

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

t_o = t'γ

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

t_o = t'(1 + v²/2c²)


dt_R = (t_o - t')/t_o = v²/(v² + 2c²)

dt_R = relativity time ratio



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

dt' = t_o - t' = t_ov/(c + v)


dt_S = (t_o - t')/t_o = v/(v + c)


dt_S = Sagnac ratio


dt_S/dt_R = (2c² + v²)/v(v + c)

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


Thus the Sagnac effect is far larger than any
purely Relativistic effect. For example,
considering the data in the Pogany test (8 ),
where the rim of the disc was moving with a
velocity of 25 m/s, the ratio dtS/dtR is about
1.5 x 10^7. Any attempt to explain the Sagnac
as a Relativistic effect is thus useless, as it is
smaller by a factor of 10^7.


Referring back to equation (I), consider a disc
of radius one kilometre. In this case a fringe
shift of one fringe is achieved with a velocity
at the perimeter of the disc of 0.013m/s. This
is an extremely low velocity, being less than
lm per minute. In this case the Sagnac effect
would be 50 billion times larger than the
calculated effect under the Relativity Theory.


Post (1967) shows that the two (Sagnac and STR) are of very different orders of magnitude. He says that the dilation factor to be applied under SR is “indistinguishable with presently available equipment” and “is still one order smaller than the Doppler correction, which occurs when observing fringe shifts” in the Sagnac tests. He also points out that the Doppler effect “is v/c times smaller than the effect one wants to observe." Here Post states that the effect forecast by SR, for the time dilation aboard a moving object, is far smaller than the effect to be observed in a Sagnac test.


That is to say:
ct1=a-vt1
Thus t1(c+v)=a.
Thus t1=a/(c+v).


Another total fuck up signed jack.

YOU FORGOT THE LIGHT SIGNAL ALONG THE 2B PATH. THESE ARE NOT THE SIDES OF A ISOSCELES TRIANGLE, OR OF AN ANGULAR SECTION.

THE TWO SIDES, 2B, DO RECORD A SAGNAC SHIFT.

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


You are an ignorant child.

That means you agree the shift is given by:
dt=2θoωo(R22 - R12)/c2

Which can equally be written as:
dt=2θo(R22 - R12)ωo/c2

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

See, this is where you go wrong.

If you want to compare the two Sagnac effect, the whole point and subject of this thread, by substituting the area instead of writing the equation likeso:

[2v2os2 - 2v1os1]/c2

you are going to get a total fuck up.

Please learn.

A very simple proof.

s2 = r2 x θr

s1 = r1 x θr


R2 - R1 = r2 - r1

r2 x θr = R2 x θo

r1 x θr = R1 x θo

r2/r1 = R2/R1

(r2 x R1) = (r1 x R2)

Since the two areas must be equal,

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



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

R2 = r2 - r1 + R1

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

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

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


It is measured with angular velocity, given dt=4*A*w/c^2.

NOT IF YOU WANT TO COMPARE THE TWO SAGNAC EFFECT.

YOU COMPARE SAGNAC EFFECT USING LINEAR VELOCITY (CIRCUMFERENTIAL TANGENT SPEED AT THE END POINT) AND NOT THE ANGULAR VELOCITY.

A BIG MISTAKE ON YOUR PART.

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

Dr. A.G. Kelly is one of the top foremost experts on the Sagnac effect in the world.

And he did break the special relativity.

Please learn.

The Sagnac effect is far larger than the effect forecast by relativity theory.

STR has no possible function in explaining the Sagnac effect.

The Sagnac effect is a non-relativistic effect.

COMPARISON OF THE SAGNAC EFFECT WITH SPECIAL RELATIVITY, starts on page 7, calculations/formulas on page 8

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

page 8

Because many investigators claim that the
Sagnac effect is made explicable by using the
Theory of Special Relativity, a comparison of
that theory with the actual test results is given
below. It will be shown that the effects
calculated under these two theories are of very
different orders of magnitude, and that
therefore the Special Theory is of no value in
trying to explain the effect.

COMPARISON OF THE SAGNAC EFFECT WITH STR

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

t_o = t'γ

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

t_o = t'(1 + v²/2c²)


dt_R = (t_o - t')/t_o = v²/(v² + 2c²)

dt_R = relativity time ratio



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

dt' = t_o - t' = t_ov/(c + v)


dt_S = (t_o - t')/t_o = v/(v + c)


dt_S = Sagnac ratio


dt_S/dt_R = (2c² + v²)/v(v + c)

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


Thus the Sagnac effect is far larger than any
purely Relativistic effect. For example,
considering the data in the Pogany test (8 ),
where the rim of the disc was moving with a
velocity of 25 m/s, the ratio dtS/dtR is about
1.5 x 10^7. Any attempt to explain the Sagnac
as a Relativistic effect is thus useless, as it is
smaller by a factor of 10^7.


Referring back to equation (I), consider a disc
of radius one kilometre. In this case a fringe
shift of one fringe is achieved with a velocity
at the perimeter of the disc of 0.013m/s. This
is an extremely low velocity, being less than
lm per minute. In this case the Sagnac effect
would be 50 billion times larger than the
calculated effect under the Relativity Theory.


Post (1967) shows that the two (Sagnac and STR) are of very different orders of magnitude. He says that the dilation factor to be applied under SR is “indistinguishable with presently available equipment” and “is still one order smaller than the Doppler correction, which occurs when observing fringe shifts” in the Sagnac tests. He also points out that the Doppler effect “is v/c times smaller than the effect one wants to observe." Here Post states that the effect forecast by SR, for the time dilation aboard a moving object, is far smaller than the effect to be observed in a Sagnac test.

Is Dr. Post a charlatan?

Both Dr. Kelly and Dr. Post are two of the very best experts on the Sagnac effect of the 20th century.


So the question remains the same.

How is the Sagnac effect measured, according to the angular velocity of the loop, OR the linear velocity (circumferential tangent speed at the end point) which has to involve a radius and an angular velocity?

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





The only charlatan around here is you jack.

[bitching and moaning]

You aren't reading and responding to what I have said.
You are just spouting the same refuted crap.
Grow up.
Answer the question asked of you.

Once we deal with the annular sectors then perhaps we can move on to this magic rectangle of yours:

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

Yes or no?

See, this is where you go wrong.

That is not answering the question.

ANSWER IT!!
Is that the correct formula, YES OR NO?

If you want to compare the two Sagnac effect, the whole point and subject of this thread, by substituting the area instead of writing the equation likeso:
[2v2os2 - 2v1os1]/c2

Again, if you want to do it like this, you need to clearly indicate that the v and s belong to the orbital one, not the rotational one, as the rotational one will have different values.
s1/s2 must be different for the 2 or you can't have an annular sector and thus can't have the simple math.

NOW ANSWER THE QUESTION!!!

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

Yes or no?

Answer the question asked of you.

And I can tell you to fuck off.

This is what YOU tell people who ask questions of you, remember?

You avoided answering my question.

No problem.

Answer the question asked of you.
And I can tell you to fuck off.
This is what YOU tell people who ask questions of you, remember?
You avoided answering my question.
No problem.

No. That isn't what I tell people who ask questions of me.
That is what I tell people that refuse to address what I have said.
I did answer your question.

Here, let me remind you:

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

How is the Sagnac effect measured, according to the angular velocity of the loop, OR the linear velocity (circumferential tangent speed at the end point) which has to involve a radius and an angular velocity?

It is measured with angular velocity, given dt=4*A*w/c^2.

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

YES OR NO?