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**Flat Earth Investigations / Re: Ring laser gyros**

« **on:**October 08, 2021, 07:02:19 PM »

Nicely put, thank you.

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Nicely put, thank you.

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I think the important point here is that earth is, at a local level, essentially a rigid body, but in the global scale, it’s far from rigid. You have a series of tectonic plates, floating around on viscous material, with the whole thing rotating away. So if you measure the rate of rotation at a particular spot, you’ll get the aggregate rate of rotation, with any local seismic activity, or oscillatory motion, superimposed on top. Moreover, the entire system is changing slowly - the axis of rotation changes over time, and the whole thing wobbles slightly as well. It’s fascinating stuff, and amazing that we now have devices of such sensitivity.

What we see in all of the graphs, in all of the papers and websites that we’ve discussed so far, is absolutely aligned with that - it all points to the same thing.

What we see in all of the graphs, in all of the papers and websites that we’ve discussed so far, is absolutely aligned with that - it all points to the same thing.

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Actually, the Wiki page accurately says that those types of devices are looking at rotational seismic phenomena right here - https://wiki.tfes.org/Ring_Laser_Gyroscope_-_SeismologyNobody is saying the devices aren't used for seismology. They absolutely are. They are extremely sensitive, and can measure tiny rotations, several orders of magnitude less than the earth's rotation. They are also using them for studying variations in the earth's axis of rotation, amongst other things.

Not sure why you keep reposting that graph. It's not raw data.

I keep posting it for several reasons. Firstly, it's from a site that you linked to. Moreover, it clearly shows earth rate, with far less noise than the apparently unacceptable amount in the examples you have cherry picked for use in the wiki. You kept saying that the amount of noise rendered the various other examples meaningless, but now here we have one with fa less noise, and you've pivoted to some nebulous claim about it measuring seismic activity. You aren't clear though on what, precisely, it is measuring. Seismic activity is, by it's nature, oscillatory. The graph in the example clearly shows a steady state rotation. If it's a steady state rotation, then it must be rotating.

Your request for 'raw data' is also curious, as when you had raw data, such as from the Canadian Honeywell test, you didn't understand what it was, despite it actually demonstrating earth rate, if you knew what to do with the data.

As you pointed out earlier, the angular rate on that one is faster other graphs we saw which are uncorrected for latitude and this data is reprocessed and visualized. Other visualizations we saw in this thread have radically different views.

GINGERino is not a regular seismograph, it is a special tool which studies "rotational seismology" -

...

It's a new field of "rotational seismology" which is "able to observe rotational motion from remote earthquakes".

Yes, it absolutely is. Rotational motion...because it's a highly sensitive rotation detector. Which is why the graph I've shown you, from the site that you linked to, clearly shows oscillatory seismic activity superimposed on the steady state earth rotation.

I suppose the ultimate question really is: 'what RLG data would convince you that the earth was actually rotating?'

You seem to be rejecting every piece of data for a variety of reasons...eg too much noise in a simple device, but then an ultra sensitive device with much less noise must be detecting something else other than earth rate, despite it clearly measuring earth rate.

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It says that the seismic waves are rotating, like a Foucault Pendulum:

https://inside.mines.edu/~rsnieder/Snieder16Fouc.pdfQuoteEarth’s rotation leads to a slow rotation of the transverse polarization of S waves; during the propagation of S waves the particle motion behaves just like a Foucault pendulum.

The device is watching something rotate, and it is assumed to be caused by the rotation of the earth.

You've completely changed your position from the start of our discussion here. I'm curious as to whether you've changed your mind - in which case will you change the wiki to reflect this? - or whether you are just adapting your default 'disagree' position to suit what's in front of you. At the start, your argument was all about the noise levels in the rotation graphs. Then, as you realised that, in fact, a lot of the sources you yourself presented did in fact have very accurate, low noise graphs in them, you've pivoted to an argument based on the fact that the gyros are in fact measuring something other than rotation, even though that is precisely what there are designed to do.

Could you very carefully and precisely explain what exactly you think is going on in this graph please?

We have some very obvious seismic activity - clearly oscillatory in nature, superimposed on a flat line that is bang on the generally agreed earth rotation rate. Are you arguing that the flat line is also some kind of seismic activity? And that this activity is somehow a constant, steady state rotation, whose magnitude varies with the sine of the latitude? So it is magically zero on a circle around the monopole North Pole that we call the equator but whose significance is what, precisely, on a flat earth?

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I thought that was a pretty neat shot... It doesn't say much to me about the overall shape of the earth..

If there are sufficient landmarks or other geographical features, and a reasonably cloudless sky, then, as with the Red Bull Space Jump footage, it can be determined whether or not the landmarks are within the viewing scope of a spherical cap, based on the orbital height or altitude of the craft.

What I've seen of the Inspiration footage thus far seems plagued by cloud, though...

Couldn't the same test be done from a Cessna 150 or the top of the Empire State building? The math would work just as well, and landmarks would be easier to see.

In theory, yes, but the visibility from aircraft tends to be limited by haze - usually particulate matter in the lower atmosphere - and not curvature. On a clear day, you might have, say, 20km visibility (met forecasts usually stop discriminating at 10km) at the kind of altitude you might reach in a light aircraft. But if you climb above the haze layer, or indeed the troposphere, such as in the red bull situation, or a rocket, you will be able to see a lot further, as the haze layer is less thick than, say 20km, meaning you can always see through it.

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Recall that the device was described as an 'underground seismic observatory'.

The 'rotation of the earth' affects seismic waves:

https://gfzpublic.gfz-potsdam.de/rest/items/item_5005107_3/component/file_5005141/contentQuoteSUMMARYRotation of the Earth affects the propagation of seismic waves.The global coupling of spheroidal and toroidal modes by the Coriolis force over time is described by normal-mode theory. The local action of the Coriolis force on the propagation of surface waves can be described by coefficients for the coupling between propagating Rayleigh and Love waves as derived by Snieder & Sens-Schonfelder.

....

The rotation of Earth as propagation medium exerts an additional force on moving matter—the Coriolis force. Depending on the angle of the polarization vectors of P and S waves with Earth’s rotation axis, the Coriolis force causes a small transverse component for P waves and a small longitudinal component for S waves. Moreover the Coriolis forcecauses a slow rotationof the shear wave polarization vector akin to the motion of a Foucault pendulum (Snieder et al. 2016b,a).

The Coriolis force 'causes a slow rotation' and the effect is compared to watching a rotating Foucault Pendulum. In the Foucault Pendulum we allegedly watch the pendulum rotate, not the earth. So the device is observing rotating seismic waves and we are assuming that it is the earth rotating and that the movement isn't coming from the seismic waves rotating, like many other things rotate above us diurnally.

I'm not really clear what you're suggesting - are you saying that Coriolis is, or isn't a real effect?

It's not really clear what your point is - seismic observations are a major reason why people keep building ever more precise RLGs. They are also interested in slight variations in the earth's rotational axis, amongst other things. But seismic activity manifests as oscillatory motion, whereas as constant rotation is just that - constant rotation. See that graph again:

The earth rate is the constant line, whereas the seismic activity is the oscillatory signal superimposed on it. If not was just seismic activity, the oscillation would be around zero.

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No.

c is a fixed constant, O(3x10^{5}km/hr).

v is either O(1) or O(30km/hr).

That is, c>>v.

Ah, thank you - I see what you did now.

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I'm curious as to how Sandokhan got to the formula

I have already the provided the link for the derivation of the formula.

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg2117351#msg2117351there's clearly some fairly significant errors along the way

You won't find any. It is a very straightforward derivation.That looks fundamentally wrong to me - the simplification on the right isn't equal to the term on left. Thoughts?

Work out the term on the left, pretty tedious algebra, and you will arrive at the term on the right.

You also have the classic example from the very simple situation where the center of rotation coincides with the geometrical center:

The formula is correct. What you have to deal with now, are the consequences.

From your link:

Keep it simple and just look at the first formula - it's just not correct. l/(c-v) - l/(c+v) does not equal 2lv/c

It's easy to prove by plugging in some numbers - say l=1, c=3 and v=2.

The left hand side would give you 1/(3-2) - 1/(3+2) = 1 - 1/5 = 4/5

The right hand side would give you 2 x 1 x 2 / 3

Always happy to proven wrong, but that doesn't look right to me.

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He already showed you the assumptions and arithmetic required to arrive at Δt=8ωA/c^2. If you disagree with any of the steps, you'll have to pinpoint them

Is my post not pinpoint enough? If the one term doesn't equal the other, then what follows or precedes it can't be right either, can it? Or am I missing something?

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I’m well aware of that equation, but I’m not really clear why you’ve brought it up.Yes, that's rather apparent. Let's help you out.

- You provided a formula for Δϕ
- Sandokhan provided a proposed formula for Δt
- You were confused by this, so I showed you the relation between Δϕ and Δt - all that was required of you was simple algebra
- Since this still eludes you, we can conclude that even though "you are well aware of that equation", you do not understand it in the slightest.
it is missing the Pi and λ terms.It's not missing anything at all. Once again, note that .

Are you with us yet? If we can make it past substitution, we might even be able to discuss physics at some point.

Thank you, and my apologies - I hadn't spotted that he was still talking about 'dt'. Makes sense.

I'm curious as to how Sandokhan got to the formula - there's clearly some fairly significant errors along the way. I'm not talking about minor mistakes - I've no interest in derailing threads when somebody makes a minor mistake (witness this one from TB earlier here: https://forum.tfes.org/index.php?topic=18565.msg246662#msg246662). But, for example, sandokhan says this:

Quote

Δt = (l1 + l2)/(c - v1 - v2) - (l1 + l2)/(c + v1 + v2) = 2[(l1v1 + l2v2)]/c2

That looks fundamentally wrong to me - the simplification on the right isn't equal to the term on left. Thoughts?

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Where have Pi and λ gone?If you had actually read and understood the article you're plucking formulae out of, you would have noticed that . It really would be a good idea to understand what you're discussing before proudly taking a stance on it.

Now, it's still possible that sandy made a small (and largely insignificant for the purpose of this discussion) arithmetic error in his calculations. Can you find it?

I’m well aware of that equation, but I’m not really clear why you’ve brought it up. The reason I quoted the generalised sagnac formula is that it is the one used to get from the interferometer reading to a rotation rate - you can’t measure delta t directly, but you can measure the phase shift.

I’m just pointing out that the equation sandokhan has come up with is not actually the same as the one I showed - it is missing the Pi and λ terms. The latter is particularly important, as without it, the dimensions of the equation change.

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For an interferometer whose center of rotation coincides with its geometrical center, it's even simpler.

Circle

l = 2πr

v1 = v2

My formula: 2(2lv)/c^2 = 4lv/c^2 = 8πωr^{2}/c^2 = 8ωA/c^2

Square

dt = 8rv/c^2 (r = d/2, d = diagonal of the square) = 8ωA/c^2

Everything changes when the center of rotation no longer coincides with the geometrical center of the interferometer.

The ether drift field has a variable speed, latitude dependent. Remember, now I have the formula to PROVE that there is only one possibility for the registered Coriolis effect: it is the ether drift which is rotating above the surface of the Earth.

The formula I showed you was:

You've ended up with 8ωA/c^2

Where have Pi and λ gone?

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If you have a geometrically symmetrical interferometer (square/circle) placed on a rotating platform, then the Coriolis effect formula will COINCIDE with the Sagnac effect formula. It is the only time they will do so.

Ok then. So go back to my previous question - help me understand your formula. You have a symmetrical (let's say square, to keep it simple) RLG on a rotating platform turning at a rate ω. The interferometer of the RLG detects a phase shift Δφ. You are now saying that both formulae would return the same result in this case, but you haven't actually explained how those terms fit into your formula, other than saying v=ωr.

Δt = (l1 + l2)/(c - v1 - v2) - (l1 + l2)/(c + v1 + v2) = 2[(l1v1 + l2v2)]/c2

How would you get from a measured phase shift, Δφ, to a rotation rate, ω, using that formula and v=ωr?

You are also asserting that RLGs are in fact detecting the rotation of the ether. You posted a link in response to my point about latitude variation, but that link doesn't really explain the concept at all - it just says that variations in latitude were observed. Miller may have been an aether proponent, but he still very much subscribed to the round earth model, as the diagrams in that text show. You are proposing a flat earth, with aether / ether rotation that is both detectable by RLGs and also variable according to the sine of the latitude of the device. How do you explain the variation of the measurements with latitude? It makes no sense at all on a flat earth - what is so special about the equator, for example?

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Let us imagine the Earth as a very large scale turntable. To detect the rotation of the turntable itself, you need the Sagnac effect. With the Coriolis effect, you have either of two possibilities: either the turntable is rotating or the ether drift is rotating above its surface.

What Michelson did is to substitute the Coriolis effect formula for the Sagnac effect formula, and then he claimed that the Earth is rotating. Not by a long shot.

The RLGs are detecting the CORIOLIS EFFECT.

A simple yes or no will suffice.

RLGs, as used in navigation systems, or scientific experiments, all use interferometers to measure the phase difference between the two light paths and calculate the rotation rate using the formula I showed above, which is the same formula Sagnac himself came up with.

It's a simple question - are these RLGs measuring rotation correctly or not? Forget about earth rate for a moment, just consider a RLG on a rotating platform turning at a given rate. Would you agree that commercial systems, such as the Honeywell GG1320, actually work correctly?

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So my interferometer detects a phase shift of Δφ radians between the two light beams

Yes, that's the Coriolis effect phase shift. But you won't detect the Sagnac effect.

Each interferometer has two phenomena to deal with: a mechanical effect (Coriolis effect) and an electromagnetic effect (Sagnac effect). Two separate formulas.

So are you suggesting that every RLG that measures rotation via an interferometer is in fact wrong?

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v = ωr

Ok...so v = ωr and

Δt = (l1 + l2)/(c - v1 - v2) - (l1 + l2)/(c + v1 + v2) = 2[(l1v1 + l2v2)]/c2

So my interferometer detects a phase shift of Δφ radians between the two light beams...how do I calculate rotation rate from that? Where does t come into it? And r? And which v are we talking about?

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dt is not the derivative, it is the delta t, difference in time, time shift formula. The notation for the derivative is d/dt (dt is the differential notation).

The ether drift is latitude dependent.

http://www.orgonelab.org/miller.htm

The Coriolis effect is SUBLUMINAL.

The Sagnac effect is SUPERLUMINAL.

That is, if you want the Sagnac effect, the formula must reflect the superluminal velocity. No superluminal velocity, no Sagnac formula.

Coriolis and Sagnac effect formulas for a square ring laser interferometer:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg2153966#msg2153966

Derivation of the Sagnac effect formula:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg2117351#msg2117351

KASSNER EFFECT

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg2234871#msg2234871 (part I)

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg2235136#msg2235136 (part II)

Dr. Gianfranco Spavieri

In both the outward and return paths, the one-way speed is c (in agreement with Einstein’s second postulate) if the length L of the outward path covered by the signal is reduced to L(1 - 2v/c) < L in Eq. (3).

CORIOLIS EFFECT = a path measuring L(1 - 2v/c), a comparison of two separate/different segments

SAGNAC EFFECT = a path measuring L, a comparison of two continuous loops

Therefore, Michelson and Gale, Silberstein, Langevin, Post, Bilger, Anderson, Steadman, Rizzi, Targaglia, Ruggiero, have been measuring ONLY the CORIOLIS EFFECT formula (area and angular velocity), nothing else. The formulas features on the wikipedia and mathpages websites are the CORIOLIS EFFECT equations, not the correct SAGNAC EFFECT formulas.

Here is the crown jewel of all the SAGNAC EFFECT formulas:Δt = (l_{1}+ l_{2})/(c - v_{1}- v_{2}) - (l_{1}+ l_{2})/(c + v_{1}+ v_{2})

The velocity terms are immediately identified: c - v_{1}- v_{2}and c + v_{1}+ v_{2}.Δt = (l_{1}+ l_{2})/(c - v_{1}- v_{2}) - (l_{1}+ l_{2})/(c + v_{1}+ v_{2}) = 2[(l_{1}v_{1}+ l_{2}v_{2})]/c^{2}

So, using your 'crown jewel' formula, if I have a RLG system, with an interferometer, and I measure some phase shift Δφ, how do I calculate the rotation rate, ω? Your formula has neither of those terms in it.

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Show me your formula. Is it by any chance, dt = 4Aω/c^{2}? That's the CORIOLIS EFFECT formula.

Here is the SAGNAC EFFECT formula:

2(V_{1}L_{1}+ V_{2}L_{2})/c^{2}

A huge difference.

Where did you get those formulae from? It would be helpful if you could both explain what the various terms are, and to complete the formulae. The first one doesn't look right at all - the Coriolis effect is simply a function of motion in a non-inertial, rotating frame - the speed of light, c, wouldn't normally come into it. Also that one starts with 'dt = ', but there is no corresponding derivative on the other side of the formula, which means it is essentially meaningless, as dt on its own is zero. I wonder if you've got that from vibrating Coriolis gyro systems? Hard to tell.

Likewise your Sagnac formula bears no resemblance to the Sagnac formula described in several of the papers we've been discussing, Wikipedia, as well as the slide deck I linked to, which is:

where Δ

Your formula for the Sagnac effect has only one side of the equation, so it's not clear what the term actually represents, and it seems to bear no resemblance at all to Sagnac's original formula. It may well be that it is in some way related, but without your source, or some context, it's just meaningless I'm afraid.

You appear to be claiming that RLGs are in fact measuring the rotation of some 'ether', presumably rotating above the FE surface. If that's the case, then you need to explain what the mechanism is by which this mysterious rotation is being detected, and why it is related to the latitude of the sensor. Why would the measured rotation be zero at the equator, and a maximum at the North Pole and southern pole / ice wall / whatever it is ?

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What is your point?

You must be joking, of course.

Here is the point: you are using the wrong formula for the RLGs and the MGX. To detect rotation, you need the SAGNAC EFFECT formula. Your formula (the one that you are endorsing) is the CORIOLIS EFFECT formula. Then, you have two possibilities: either the Earth is rotating, OR, you have a rotational ether drift above the surface of the Earth. The deciding factor is the SAGNAC EFFECT.

Each RLG has TWO FORMULAS: one for the Coriolis effect, and one for the Sagnac effect.

One is a mechanical effect, the slight deflection of the light beams (Coriolis), it is proportional to the area/angular velocity. The other one is an electromagnetic effect (Sagnac), it is proportional to the velocity of the light beams.

All of the graphs have come from papers and briefings made by other people. The specific graph we are discussing came from this slide pack: https://indico.cern.ch/event/736594/contributions/3184374/attachments/1741872/2819336/DiVirgilio_COSMO2018.pdf

It clearly explains the use of the sagnac effect to derive the rotation rate. That graph is derived from the sagnac effect.

[edited to fix link]

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Sure the graph shows rotation. So, based on the results of the RLGs experiments (we might also mention the MGX) you are saying that it is the Earth which is rotating around its own axis? Is that your last word? You still have time to retract your statement.

Not sure it's my last word on the matter, but yes, the graph clearly shows the output from a ring laser gyro measuring the earth's rotation around the polar axis. What is your point?