This one is a 3-axis FOG, for military and space applications.

Innovative Fiber Optic Gyroscope (FOG) technology

https://www.gemrad.com/innovative-fiber-optic-gyroscope-fog-technology/

You can see, clearly, the three FO loops in the photo, which are mounted
in orthogonal directions (xyz arbitrary reference).

Can anyone explain how they work?
Three different outputs provide differential changes, being IRRELEVANT
on which position they are mounted (in particular on missiles and satellites).

This system provides xyz changes 100 times per second.

Which is the frame of reference XYZ they use?

Is it based on relativity (ether or etherless) or anything else?

Try making A DEEP RESEARCH on the web about how this technology has
been developing with increasing accuracy in the last 15 years.

You can try with quantum physics, if you will.

On Wednesday, September 7, 2022 at 7:00:25 PM UTC-3, Richard Hertz wrote:
> This one is a 3-axis FOG, for military and space applications.
>
> Innovative Fiber Optic Gyroscope (FOG) technology
>
> https://www.gemrad.com/innovative-fiber-optic-gyroscope-fog-technology/
>
> You can see, clearly, the three FO loops in the photo, which are mounted
> in orthogonal directions (xyz arbitrary reference).
>
> Can anyone explain how they work?
> Three different outputs provide differential changes, being IRRELEVANT
> on which position they are mounted (in particular on missiles and satellites).
>
> This system provides xyz changes 100 times per second.
>
> Which is the frame of reference XYZ they use?
>
> Is it based on relativity (ether or etherless) or anything else?
>
> Try making A DEEP RESEARCH on the web about how this technology has
> been developing with increasing accuracy in the last 15 years.
>
> You can try with quantum physics, if you will.

Remember that these FOGs, or better MEMS, can be used in ABSOLUTE SPACE. An example is
the antennae stabilization on Webb telescope, almost 1 million Km far away (it's orbiting its Lagrange point).

Also, I'd like VERY MUCH if anyone can explain the output of the photodiode detector (PiN or Avalanche), that
ABSORB TWO ANALOG LIGHT WAVES, transform them into PHOTONS and produce an output current (PER AXIS):

I = I₀ (1 + cos Δθ) ≈ I₀ [1 + 8πN Aω/(λ₀c₀)]

In general, I has a bandwidth lower than 1,000 Hz.

Where happens the conversion to photons?

Some analysis at the input of the photodiode in any FOG:

Some data about the two waves combining at the input of the photodetector in a FOG (using 1550 nm laser diode):

S(t) = A cos (Ω t + θ₁) + A cos (Ω t + θ₂) = A/2 [cos (2Ω t + θ₁+ θ₂) + cos (θ₁ - θ₂)]

Ω₀ = 2π c₀/λ₀ = 2π f₀

S = A/2 [cos (2Ω t + θ₁+ θ₂) + cos Δθ]

After filtering the output for A/2 cos (2Ω t + θ₁+ θ₂), and raising the output of the photodiode to a nominal value of ONE (raised cos),
S is transformed into a current I, which has a limited bandwidth under 1 Khz:

I = I₀ (1 + cos Δθ) ≈ I₀ {1 + cos [8πN Aω/(λ₀c₀)]}

For a I-FOG with N = 200, λ₀ = 1550 nm, R = 0.1 mt and A = 0.0314 m² , it's obtained that

Δθ = 0.004785 ω [unit: 1 radian = 57.2958°]

Δθ = 8.35185E-05 ω [unit: 1° = 0.0174533 radians]

I ≈ I₀ [1 + cos (8.35185E-05 ω)] {ω measured in degrees (°)}

Still remains the question about how come the random signal cos (8.35185E-05 ω) is CONVERTED IN PHOTONS at the entrance
of the photodetector. Or should the equations for photodiodes require a re-writing to work with SLOW CHANGES OF ω with time?

That's a real mystery that requires intelligent answers.
Paul, any ideas?

**********************************
A question emerges: Ω₀ = 2π f₀ is the same within the FO and outside it (in the photodetector input), but the wavelength transitions
from λ= λ₀/n to λ₀ in the segment between the FO end and the entrance of the photodiode, is it TRUE that Δθ is preserved? (maybe
there are 20/30 micrometers between them).

IF Δθ IS PRESERVED IN that small distance, THEN the refraction index n IS NOT RELEVANT AT ALL, and all the explanations based
on relativistic addition of velocities to explain Fizeau ARE PURE BULLSHIT, and always were.

Any opinion?

On Wednesday, September 7, 2022 at 9:51:57 PM UTC-3, Richard Hertz wrote:
> Some analysis at the input of the photodiode in any FOG:
>
> Some data about the two waves combining at the input of the photodetector in a FOG (using 1550 nm laser diode):
>
> S(t) = A cos (Ω t + θ₁) + A cos (Ω t + θ₂) = A/2 [cos (2Ω t + θ₁+ θ₂) + cos (θ₁ - θ₂)]
>
> Ω₀ = 2π c₀/λ₀ = 2π f₀
>
> S = A/2 [cos (2Ω t + θ₁+ θ₂) + cos Δθ]
>
> After filtering the output for A/2 cos (2Ω t + θ₁+ θ₂), and raising the output of the photodiode to a nominal value of ONE (raised cos),
> S is transformed into a current I, which has a limited bandwidth under 1 Khz:
>
> I = I₀ (1 + cos Δθ) ≈ I₀ {1 + cos [8πN Aω/(λ₀c₀)]}
>
> For a I-FOG with N = 200, λ₀ = 1550 nm, R = 0.1 mt and A = 0.0314 m² , it's obtained that
>
> Δθ = 0.004785 ω [unit: 1 radian = 57.2958°]
>
> Δθ = 8.35185E-05 ω [unit: 1° = 0.0174533 radians]
>
> I ≈ I₀ [1 + cos (8.35185E-05 ω)] {ω measured in degrees (°)}
>
> Still remains the question about how come the random signal cos (8.35185E-05 ω) is CONVERTED IN PHOTONS at the entrance
> of the photodetector. Or should the equations for photodiodes require a re-writing to work with SLOW CHANGES OF ω with time?
>
> That's a real mystery that requires intelligent answers.
> Paul, any ideas?
>
> **********************************
> A question emerges: Ω₀ = 2π f₀ is the same within the FO and outside it (in the photodetector input), but the wavelength transitions
> from λ= λ₀/n to λ₀ in the segment between the FO end and the entrance of the photodiode, is it TRUE that Δθ is preserved? (maybe
> there are 20/30 micrometers between them).
>
> IF Δθ IS PRESERVED IN that small distance, THEN the refraction index n IS NOT RELEVANT AT ALL, and all the explanations based
> on relativistic addition of velocities to explain Fizeau ARE PURE BULLSHIT, and always were.
>
> Any opinion?

I wonder why SR, developed for LINEAR MOTION over a single axis x, is used for CIRCULAR MOTION in a plane xy, where there
are TWO COMPONENTS OF VELOCITY. This should be marked as fallacious, as it VIOLATES Lorentz transforms:

For a uni-dimensional inertial motion along x-axis, Lorentz transforms are:

x' = 1/√(1 − v²/c²) (x - v.t)
y' = y
z' = z

But this is not a case with Sagnac effect, that involves two axes xy to write correctly such transforms:

x' = 1/√(1 − v²/c²) (x - v.t)
y' ≠ y (and which is the expression that should be used here? There is not a SR version for rotating, not inertial motion)
z' = z (per definition?)

Now, the biggest MYSTERY: How come SR is applied SIMULTANEOUSLY on a 3-axis FOG?

How come THREE DIFFERENT VARIABLES for x'y'z' are applied SIMULTANEOUSLY and used in real time?

What is the relativity of the relativity that's applied on such instruments?

Just curious about the extended use of the 1905 SR, in particular the addition of THREE VELOCITIES for FOGs rotating on x, y and z
axis SIMULTANEOUSLY.

I hope that SR IS NOT what is applied by industry, so relativity is not accountable for lives of persons that are onboard of a plane
that lose stability on the three axes simultaneously, or they will never manage to stabilize the aircraft.

On Thursday, September 8, 2022 at 5:24:11 PM UTC-7, Richard Hertz wrote:

> I wonder why SR, developed for LINEAR MOTION over a single axis x, is used for CIRCULAR MOTION in a plane xy, where there
> are TWO COMPONENTS OF VELOCITY. This should be marked as fallacious, as it VIOLATES Lorentz transforms:

Dumbestfuck,

SR has been extended to:
1. Uniformly rotating frames
2. Uniformly accelerated frames
3. Uniformly accelerated objects (hyperbolic motion)

Your crass ignorance is showing. Once again.

On 9/8/2022 8:24 PM, Richard Hertz wrote:

> I wonder why SR, developed for LINEAR MOTION over a single axis x,

That is a simplification, so that even someone as dumb as you could
understand it.

Obviously, that failed.

On Thursday, September 8, 2022 at 9:48:11 PM UTC-3, Dono. wrote:
> SR has been extended to:

> 1. Uniformly rotating frames

Cretin. Solve the Ehrenfest’s Paradox first, and then make such "brave" assertion again.

https://link.springer.com/article/10.1140/epjc/s10052-014-3098-6

EXCERPT: Ehrenfest’s Paradox
Ehrenfest’s Paradox is the contradiction that an inertial (laboratory) observer faces in applying special relativistic length contraction to a rotating disk. From an inertial observer’s point of view the rim of a rotating disk undergoes a length contraction due to its transverse motion with velocity v=RΩ and so the circumference of a rotating disk (P′) is shorter than the one non-rotating (P), i.e. P′<P. On the other hand since the radius of the disk is perpendicular to the direction of the rotational motion of the rim, the same observer will not attribute a length contraction to it and so R′=R. Therefore the inertial (laboratory) observer, living in a flat spacetime and thereby using the Euclidean prescription for the circumference of a circle, finds the contradictory result P=2πR=2πR′=P′.

Perhaps it should be left for experiment to decide which relation holds between P and P′, but nevertheless MANY HAVE TRIED HARD TO FIND EITHER A THEORETICAL RESOLUTION to this paradox or otherwise to invalidate it. An apparently favorite resolution in the literature is based on considering the situation from a rotating observer’s point of view and on the idea, introduced by Einstein [11–13], that the spatial geometry in such a frame is non-Euclidean.Footnote2 But, as we will show below, THAT DOES NOT SEEM TO BE LEADING TO ANY KIND OF RESOLUTION OF THE PARADOX but to a somewhat similar paradox from the rotating frame’s point of view.

<snip the shit about accelerated frames or hyperbolic motion>

> Your crass ignorance is showing. Once again.

No, imbecile Dono. What is shown is your religious devotion to support Einstein's relativity at any cost!

You are the fucking Zelensky in this forum (it doesn't matter if he sent to die 200,000 Ukrainians to please his masters).

On Thursday, September 8, 2022 at 8:26:16 PM UTC-7, Richard Hertz wrote:
> On Thursday, September 8, 2022 at 9:48:11 PM UTC-3, Dono. wrote:
>
> > SR has been extended to:
>
> > 1. Uniformly rotating frames
You are frothing at the mouth once again. While you are eating shit.

On Friday, September 9, 2022 at 1:13:37 AM UTC-3, Dono. wrote:
> On Thursday, September 8, 2022 at 8:26:16 PM UTC-7, Richard Hertz wrote:
> > On Thursday, September 8, 2022 at 9:48:11 PM UTC-3, Dono. wrote:
> >
> > > SR has been extended to:
> >
> > > 1. Uniformly rotating frames
> You are frothing at the mouth once again. While you are eating shit.

Imbecile indoctrinated EE gypsy, your butthurt is due to the fact that you can't accept that you wasted two decades
studying relativistic dynamics, after giving up designing graphic cards for Apple II.

Read this analysis, based on NEWTONIAN DYNAMICS, that explain the Sagnac effect without your rotten SR.

Analysis of the “Sagnac’s experiment”
https://physics.bg/home/physics-problems/speed-of-light-constancy/sagnac-experiment/

I quote in advance:

Abstract

This chapter analyses the “Sagnac’s experiment”, carried out by the French physicist Georges Sagnac in the 1912 year. The analysis presented is based on the classical mechanics and relativity of Galileo, which are indisputably valid in our local time-spatial region “on the surface of the Earth”. The experiment demonstrates that in relation to a moving system in the stationary space, the speed of the light differs depending on the speed and the direction of movement of the system in the stationary space. However, the Sagnac experiment is considered as a paradox, because it demonstrates that the speed of light is not the same for all frames of reference – what is not convenient for the modern physics, because the special theory of relativity is created on the base of the claim “the speed of light is the same for all frames of reference”. As further evidence of the authenticity of the presented analysis, the derivation of the equation, which is often used in the rotation analyzes is shown.
........
........
EXCERPT:

"Georges Sagnac’s experiment is unofficially considered mystical, because so far, none of its explanations have been officially accepted. Of course, there are many “modern scientific explanations” which, however, are based on unscientifically proven hypotheses – or on “scientific” references to false theories (see: “What is the Truth and the Proof in the Science?”. Although Sagnac’s experiment proves that the speed of light is not the same in all inertial reference frames, many modern physics journals publish “scientific” explanations based on the special theory of relativity… which is based on the false claim that “the speed of light is the same in all inertial frames” … In other words, this is a classical “circular reference” ! Such an example of a published “scientific” comparison of different explanations is that of Malykin, G.B. “The Sagnac effect: correct and incorrect explanations” (Malykin, 2000). There are other such examples in the scientific literature."

"Despite all this mystification, although there was no valid scientific explanation to this day, nowadays, the result of this experiment has many significant applications in the practice. A wide-ranging application is found in the space navigation, aviation (optical gyroscope), as well as in everyday Earth positioning needs, where no one has observed any “anisotropy” of the meter as a unit of measurement (which is a claim of the special theory of relativity)."

"An additional proof of the credibility of the above-mentioned explanation of the Sagnac experiment is given in the next subsection. This theoretical explanation demonstrates the derivation and origin of the most commonly used equation in the rotational analyses."

*********************************************

The authors are Bulgarian physicists, close to your Romany. The difference in the average IQ of Bulgarian is abysmal, compared with
the "room temperature" IQ of Romanians, like you, relativistic gypsy.

Don't forget to dance in circles with pals, at the meeting of your pagan relativity church.

And suffer your credence on such decadent religion, because IT CAN'T EXPLAIN NOR SAGNAC EFFECT NEITHER DEVELOP RELATIVITY IN
ROTATING FRAMES.

Light speed is neither constant nor isotropic. Deal with it or try 1-way experiments in outer space, far from any massive object, imbecile.

On Thursday, September 8, 2022 at 9:36:57 PM UTC-7, Richard Hertz wrote:
> IT CAN'T EXPLAIN NOR SAGNAC EFFECT NEITHER DEVELOP RELATIVITY IN
> ROTATING FRAMES.
>
> Light speed is neither constant nor isotropic.

Because intellectual midget Ricard Hertz says so? You are digging yourself deeper and deeper , odious kapo piece of shit.

Den 08.09.2022 02:51, skrev Richard Hertz:
> Some analysis at the input of the photodiode in any FOG:
>
> Some data about the two waves combining at the input of the photodetector in a FOG (using 1550 nm laser diode):
>
> S(t) = A cos (Ω t + θ₁) + A cos (Ω t + θ₂) = A/2 [cos (2Ω t + θ₁+ θ₂) + cos (θ₁ - θ₂)]

This is wrong.
When you add two signals with the same frequency and amplitude,
you get a signal with the same frequency, but with an amplitude that
depends on the phase difference between the signals.
You do definitely not get a DC-component.

Since it is the phase difference Δθ = θ₁-θ₂ that matters, we can set:

S(t) = A⋅cos(Ω⋅t + Δθ/2) + A⋅cos(Ω⋅t - Δθ/2) = 2A⋅cos(Ω⋅t)⋅cos(Δθ/2)

When Δθ = 0, the signals are in phase, we get constructive
interference, and the signal is S(t) = 2A⋅cos(Ω⋅t)
When Δθ = π, the signals are π out of phase, we get destructive
interference, and the signal is S(t) = 0.
>
> Ω₀ = 2π c₀/λ₀ = 2π f₀
>
> S = A/2 [cos (2Ω t + θ₁+ θ₂) + cos Δθ]

S(t) = [2A⋅cos(Δθ/2)]⋅cos(Ω⋅t)

>
> After filtering the output for A/2 cos (2Ω t + θ₁+ θ₂), and raising the output of the photodiode to a nominal value of ONE (raised cos),

No filtering is needed. The two light beams shine on
the photodetector, and it is the rms value of the resulting
light signal that matters.

Srms = 2A⋅cos(Δθ/2)/√2
When Δθ = 0, Srms = 2A/√2 = S₀

> S is transformed into a current I, which has a limited bandwidth under 1 Khz:
>
> I = I₀ (1 + cos Δθ) ≈ I₀ {1 + cos [8πN Aω/(λ₀c₀)]}

If the signal S₀ gives the current Iₘ, we can write
the equation for the current:

I = Iₘ⋅cos(Δθ/2) + I₀ where I₀ is the dark-current.

>
> For a I-FOG with N = 200, λ₀ = 1550 nm, R = 0.1 mt and A = 0.0314 m² , it's obtained that
>
> Δθ = 0.004785 ω [unit: 1 radian = 57.2958°]
>
> Δθ = 8.35185E-05 ω [unit: 1° = 0.0174533 radians]
>
> I ≈ I₀ [1 + cos (8.35185E-05 ω)] {ω measured in degrees (°)}

You will have to correct the equation and the numbers,
but the general idea seems OK.

>
> Still remains the question about how come the random signal cos (8.35185E-05 ω) is CONVERTED IN PHOTONS at the entrance
> of the photodetector. Or should the equations for photodiodes require a re-writing to work with SLOW CHANGES OF ω with time?

I don't see what you don't understand.

The signal at the entrance of the photodetector is light.
Are you really asking how light is converted into photons? :-D

The signal Srms is a stream of photons.
If the energy of each photon is higher than the energy gap
between the valence band and the conduction band
an electron may be excited to the conduction band,
and there will be a current in the photodiode.
The quantum efficiency of the photodetector say which
part of the photons will succeed in exciting an electron.

>
> That's a real mystery that requires intelligent answers.
> Paul, any ideas?

I don't see the mystery you see.

>
> **********************************
> A question emerges: Ω₀ = 2π f₀ is the same within the FO and outside it (in the photodetector input), but the wavelength transitions
> from λ= λ₀/n to λ₀ in the segment between the FO end and the entrance of the photodiode, is it TRUE that Δθ is preserved? (maybe
> there are 20/30 micrometers between them).
>
> IF Δθ IS PRESERVED IN that small distance, THEN the refraction index n IS NOT RELEVANT AT ALL, and all the explanations based
> on relativistic addition of velocities to explain Fizeau ARE PURE BULLSHIT, and always were.

Of course the phase difference between the two light beams is preserved
after they come out of the fibre to they reach the photodetector.
How could it measure the phase difference if it wasn't preserved?

How you from this blazingly obvious fact can conclude anything about
the role of the index of refraction in the fibre beats me!

---

A word on the combination of the two light beams at the end.
When you combine two light beams on a screen, you will
always get a diffraction pattern that contains the energy
of the two light beams. There is no way you can make two
light beams cancel each other. The diffraction pattern
will consist of concentric circular 'fringes'. If the two
beams are in phase, the central 'disk' will be light,
and the the inner circular 'fringe' will be dark.
If the two beams are out of phase, it will be opposite;
the central 'disk' will be dark.
It is this central disk that must be sent into the photodetector.
(This is the principle. You will not see the circular fringes anywhere.)

--
Paul

https://paulba.no/

On Friday, September 9, 2022 at 5:04:10 AM UTC-7, Paul B. Andersen wrote:
> Den 08.09.2022 02:51, skrev Richard Hertz:
> > Some analysis at the input of the photodiode in any FOG:
> >
> > Some data about the two waves combining at the input of the photodetector in a FOG (using 1550 nm laser diode):
> >
> > S(t) = A cos (Ω t + θ₁) + A cos (Ω t + θ₂) = A/2 [cos (2Ω t + θ₁+ θ₂) + cos (θ₁ - θ₂)]
> This is wrong.
> When you add two signals with the same frequency and amplitude,
> you get a signal with the same frequency, but with an amplitude that
> depends on the phase difference between the signals.
> You do definitely not get a DC-component.
>
> Since it is the phase difference Δθ = θ₁-θ₂ that matters, we can set:
>
> S(t) = A⋅cos(Ω⋅t + Δθ/2) + A⋅cos(Ω⋅t - Δθ/2) = 2A⋅cos(Ω⋅t)⋅cos(Δθ/2)
>
> When Δθ = 0, the signals are in phase, we get constructive
> interference, and the signal is S(t) = 2A⋅cos(Ω⋅t)
> When Δθ = π, the signals are π out of phase, we get destructive
> interference, and the signal is S(t) = 0.
> >
> > Ω₀ = 2π c₀/λ₀ = 2π f₀
> >
> > S = A/2 [cos (2Ω t + θ₁+ θ₂) + cos Δθ]
> S(t) = [2A⋅cos(Δθ/2)]⋅cos(Ω⋅t)
> >
> > After filtering the output for A/2 cos (2Ω t + θ₁+ θ₂), and raising the output of the photodiode to a nominal value of ONE (raised cos),
> No filtering is needed. The two light beams shine on
> the photodetector, and it is the rms value of the resulting
> light signal that matters.
>
> Srms = 2A⋅cos(Δθ/2)/√2
> When Δθ = 0, Srms = 2A/√2 = S₀
> > S is transformed into a current I, which has a limited bandwidth under 1 Khz:
> >
> > I = I₀ (1 + cos Δθ) ≈ I₀ {1 + cos [8πN Aω/(λ₀c₀)]}
> If the signal S₀ gives the current Iₘ, we can write
> the equation for the current:
>
> I = Iₘ⋅cos(Δθ/2) + I₀ where I₀ is the dark-current.
> >
> > For a I-FOG with N = 200, λ₀ = 1550 nm, R = 0.1 mt and A = 0.0314 m² , it's obtained that
> >
> > Δθ = 0.004785 ω [unit: 1 radian = 57.2958°]
> >
> > Δθ = 8.35185E-05 ω [unit: 1° = 0.0174533 radians]
> >
> > I ≈ I₀ [1 + cos (8.35185E-05 ω)] {ω measured in degrees (°)}
> You will have to correct the equation and the numbers,
> but the general idea seems OK.
> >
> > Still remains the question about how come the random signal cos (8.35185E-05 ω) is CONVERTED IN PHOTONS at the entrance
> > of the photodetector. Or should the equations for photodiodes require a re-writing to work with SLOW CHANGES OF ω with time?
> I don't see what you don't understand.
>
> The signal at the entrance of the photodetector is light.
> Are you really asking how light is converted into photons? :-D
>
> The signal Srms is a stream of photons.
> If the energy of each photon is higher than the energy gap
> between the valence band and the conduction band
> an electron may be excited to the conduction band,
> and there will be a current in the photodiode.
> The quantum efficiency of the photodetector say which
> part of the photons will succeed in exciting an electron.
> >
> > That's a real mystery that requires intelligent answers.
> > Paul, any ideas?
> I don't see the mystery you see.
> >
> > **********************************
> > A question emerges: Ω₀ = 2π f₀ is the same within the FO and outside it (in the photodetector input), but the wavelength transitions
> > from λ= λ₀/n to λ₀ in the segment between the FO end and the entrance of the photodiode, is it TRUE that Δθ is preserved? (maybe
> > there are 20/30 micrometers between them).
> >
> > IF Δθ IS PRESERVED IN that small distance, THEN the refraction index n IS NOT RELEVANT AT ALL, and all the explanations based
> > on relativistic addition of velocities to explain Fizeau ARE PURE BULLSHIT, and always were.
> Of course the phase difference between the two light beams is preserved
> after they come out of the fibre to they reach the photodetector.
> How could it measure the phase difference if it wasn't preserved?
>
> How you from this blazingly obvious fact can conclude anything about
> the role of the index of refraction in the fibre beats me!
>
> ---
>
> A word on the combination of the two light beams at the end.
> When you combine two light beams on a screen, you will
> always get a diffraction pattern that contains the energy
> of the two light beams. There is no way you can make two
> light beams cancel each other. The diffraction pattern
> will consist of concentric circular 'fringes'. If the two
> beams are in phase, the central 'disk' will be light,
> and the the inner circular 'fringe' will be dark.
> If the two beams are out of phase, it will be opposite;
> the central 'disk' will be dark.
> It is this central disk that must be sent into the photodetector.
> (This is the principle. You will not see the circular fringes anywhere.)
>
>
> --
> Paul
>
> https://paulba.no/
Pearls before the nazi kapo. Note that I no longer call Richard Hertz a PIG.. Comparing him with the pigs is an insult. For the pigs.

On Friday, September 9, 2022 at 9:04:10 AM UTC-3, Paul B. Andersen wrote:
> Den 08.09.2022 02:51, skrev Richard Hertz:
> > Some analysis at the input of the photodiode in any FOG:
> >
> > Some data about the two waves combining at the input of the photodetector in a FOG (using 1550 nm laser diode):
> >
> > S(t) = A cos (Ω t + θ₁) + A cos (Ω t + θ₂) = A/2 [cos (2Ω t + θ₁+ θ₂) + cos (θ₁ - θ₂)]
> This is wrong.
> When you add two signals with the same frequency and amplitude,
> you get a signal with the same frequency, but with an amplitude that
> depends on the phase difference between the signals.
> You do definitely not get a DC-component.
>
> Since it is the phase difference Δθ = θ₁-θ₂ that matters, we can set:
>
> S(t) = A⋅cos(Ω⋅t + Δθ/2) + A⋅cos(Ω⋅t - Δθ/2) = 2A⋅cos(Ω⋅t)⋅cos(Δθ/2)
>
> When Δθ = 0, the signals are in phase, we get constructive
> interference, and the signal is S(t) = 2A⋅cos(Ω⋅t)
> When Δθ = π, the signals are π out of phase, we get destructive
> interference, and the signal is S(t) = 0.
> >
> > Ω₀ = 2π c₀/λ₀ = 2π f₀
> >
> > S = A/2 [cos (2Ω t + θ₁+ θ₂) + cos Δθ]
> S(t) = [2A⋅cos(Δθ/2)]⋅cos(Ω⋅t)
> >
> > After filtering the output for A/2 cos (2Ω t + θ₁+ θ₂), and raising the output of the photodiode to a nominal value of ONE (raised cos),
> No filtering is needed. The two light beams shine on
> the photodetector, and it is the rms value of the resulting
> light signal that matters.
>
> Srms = 2A⋅cos(Δθ/2)/√2
> When Δθ = 0, Srms = 2A/√2 = S₀
> > S is transformed into a current I, which has a limited bandwidth under 1 Khz:
> >
> > I = I₀ (1 + cos Δθ) ≈ I₀ {1 + cos [8πN Aω/(λ₀c₀)]}
> If the signal S₀ gives the current Iₘ, we can write
> the equation for the current:
>
> I = Iₘ⋅cos(Δθ/2) + I₀ where I₀ is the dark-current.
> >
> > For a I-FOG with N = 200, λ₀ = 1550 nm, R = 0.1 mt and A = 0.0314 m² , it's obtained that
> >
> > Δθ = 0.004785 ω [unit: 1 radian = 57.2958°]
> >
> > Δθ = 8.35185E-05 ω [unit: 1° = 0.0174533 radians]
> >
> > I ≈ I₀ [1 + cos (8.35185E-05 ω)] {ω measured in degrees (°)}
> You will have to correct the equation and the numbers,
> but the general idea seems OK.
> >
> > Still remains the question about how come the random signal cos (8.35185E-05 ω) is CONVERTED IN PHOTONS at the entrance
> > of the photodetector. Or should the equations for photodiodes require a re-writing to work with SLOW CHANGES OF ω with time?
> I don't see what you don't understand.
>
> The signal at the entrance of the photodetector is light.
> Are you really asking how light is converted into photons? :-D
>
> The signal Srms is a stream of photons.
> If the energy of each photon is higher than the energy gap
> between the valence band and the conduction band
> an electron may be excited to the conduction band,
> and there will be a current in the photodiode.
> The quantum efficiency of the photodetector say which
> part of the photons will succeed in exciting an electron.
> >
> > That's a real mystery that requires intelligent answers.
> > Paul, any ideas?
> I don't see the mystery you see.
> >
> > **********************************
> > A question emerges: Ω₀ = 2π f₀ is the same within the FO and outside it (in the photodetector input), but the wavelength transitions
> > from λ= λ₀/n to λ₀ in the segment between the FO end and the entrance of the photodiode, is it TRUE that Δθ is preserved? (maybe
> > there are 20/30 micrometers between them).
> >
> > IF Δθ IS PRESERVED IN that small distance, THEN the refraction index n IS NOT RELEVANT AT ALL, and all the explanations based
> > on relativistic addition of velocities to explain Fizeau ARE PURE BULLSHIT, and always were.
> Of course the phase difference between the two light beams is preserved
> after they come out of the fibre to they reach the photodetector.
> How could it measure the phase difference if it wasn't preserved?
>
> How you from this blazingly obvious fact can conclude anything about
> the role of the index of refraction in the fibre beats me!
>
> ---
>
> A word on the combination of the two light beams at the end.
> When you combine two light beams on a screen, you will
> always get a diffraction pattern that contains the energy
> of the two light beams. There is no way you can make two
> light beams cancel each other. The diffraction pattern
> will consist of concentric circular 'fringes'. If the two
> beams are in phase, the central 'disk' will be light,
> and the the inner circular 'fringe' will be dark.
> If the two beams are out of phase, it will be opposite;
> the central 'disk' will be dark.
> It is this central disk that must be sent into the photodetector.
> (This is the principle. You will not see the circular fringes anywhere.)
>
>
> --
> Paul
>
> https://paulba.no/

You are completely right, except with the formula I = I₀ (1 + cos Δθ) ≈ I₀ {1 + cos [8πN Aω/(λ₀c₀)]}, which I copied correctly.

I made a mistake when googling the solution for (cos x + cos y). I, without realizing it, used (cos x . cos y), which was next.

cos x + cos y = 2 cos [(x+y)/2] . cos [(x-y)/2] , the CORRECT FORMULA

cos x . cos y = 1/2 [cos (x+y) + cos (x-y)] , the WRONG FORMULA (I didn't realize my mistake, and kept writing)

Even with this HUGE MISTAKE, the formula for the output of the photodiode is correct (I extracted it from a book).

I = I₀ (1 + cos Δθ) ≈ I₀ {1 + cos [8πN Aω/(λ₀c₀)]} IS CORRECT.

The value "1" is due to the conversion of the signal A⋅cos(Ω⋅t) into a constant stream of photons with energy hΩ/2π.

But the simplest open loop I-FOG is prepared to provide an analog output I = I₀ (1 + cos Δθ), which is plotted as a raised cosine.
When ω = 0 (no lateral motion), I =2 I₀ (maximum value).

In a real application, ω changes temporarily and very quickly (like a sudden turn of an aircraft), so the multiplying signal
cos (θ₁ - θ₂)/2 is a DUMPED cosine which results of the form of ω(t).

The signal cos [8πN Aω(t)/(λ₀c₀)] is the ENVELOPE of the signal A⋅cos(Ω⋅t), so I assume that photonic conversion takes place
centered at Ω/2π, with SLOW VARIATIONS that give the analog output slow change.

As far as I read, a typical analog output of a CHEAP IFOG is about 24 mV/°/sec (the photodiode current is converted to voltage).
This signal is digitalized and post-processed with additional circuitry in the FOG assembly.

Den 09.09.2022 18:45, skrev Richard Hertz:
> On Friday, September 9, 2022 at 9:04:10 AM UTC-3, Paul B. Andersen wrote:
>> Den 08.09.2022 02:51, skrev Richard Hertz:
>>> Some analysis at the input of the photodiode in any FOG:
>>>
>>> Some data about the two waves combining at the input of the photodetector in a FOG (using 1550 nm laser diode):
>>>
>>> S(t) = A cos (Ω t + θ₁) + A cos (Ω t + θ₂) = A/2 [cos (2Ω t + θ₁+ θ₂) + cos (θ₁ - θ₂)]
>> This is wrong.
>> When you add two signals with the same frequency and amplitude,
>> you get a signal with the same frequency, but with an amplitude that
>> depends on the phase difference between the signals.
>> You do definitely not get a DC-component.
>>
>> Since it is the phase difference Δθ = θ₁-θ₂ that matters, we can set:
>>
>> S(t) = A⋅cos(Ω⋅t + Δθ/2) + A⋅cos(Ω⋅t - Δθ/2) = 2A⋅cos(Ω⋅t)⋅cos(Δθ/2)
>>
>> When Δθ = 0, the signals are in phase, we get constructive
>> interference, and the signal is S(t) = 2A⋅cos(Ω⋅t)
>> When Δθ = π, the signals are π out of phase, we get destructive
>> interference, and the signal is S(t) = 0.
>>>
>>> Ω₀ = 2π c₀/λ₀ = 2π f₀
>>>
>>> S = A/2 [cos (2Ω t + θ₁+ θ₂) + cos Δθ]
>> S(t) = [2A⋅cos(Δθ/2)]⋅cos(Ω⋅t)
>>>
>>> After filtering the output for A/2 cos (2Ω t + θ₁+ θ₂), and raising the output of the photodiode to a nominal value of ONE (raised cos),
>> No filtering is needed. The two light beams shine on
>> the photodetector, and it is the rms value of the resulting
>> light signal that matters.
>>
>> Srms = 2A⋅cos(Δθ/2)/√2
>> When Δθ = 0, Srms = 2A/√2 = S₀
>>> S is transformed into a current I, which has a limited bandwidth under 1 Khz:
>>>
>>> I = I₀ (1 + cos Δθ) ≈ I₀ {1 + cos [8πN Aω/(λ₀c₀)]}
>> If the signal S₀ gives the current Iₘ, we can write
>> the equation for the current:
>>
>> I = Iₘ⋅cos(Δθ/2) + I₀ where I₀ is the dark-current.
>>>
>>> For a I-FOG with N = 200, λ₀ = 1550 nm, R = 0.1 mt and A = 0.0314 m² , it's obtained that
>>>
>>> Δθ = 0.004785 ω [unit: 1 radian = 57.2958°]
>>>
>>> Δθ = 8.35185E-05 ω [unit: 1° = 0.0174533 radians]
>>>
>>> I ≈ I₀ [1 + cos (8.35185E-05 ω)] {ω measured in degrees (°)}
>> You will have to correct the equation and the numbers,
>> but the general idea seems OK.
>>>
>>> Still remains the question about how come the random signal cos (8.35185E-05 ω) is CONVERTED IN PHOTONS at the entrance
>>> of the photodetector. Or should the equations for photodiodes require a re-writing to work with SLOW CHANGES OF ω with time?
>> I don't see what you don't understand.
>>
>> The signal at the entrance of the photodetector is light.
>> Are you really asking how light is converted into photons? :-D
>>
>> The signal Srms is a stream of photons.
>> If the energy of each photon is higher than the energy gap
>> between the valence band and the conduction band
>> an electron may be excited to the conduction band,
>> and there will be a current in the photodiode.
>> The quantum efficiency of the photodetector say which
>> part of the photons will succeed in exciting an electron.
>>>
>>> That's a real mystery that requires intelligent answers.
>>> Paul, any ideas?
>> I don't see the mystery you see.
>>>
>>> **********************************
>>> A question emerges: Ω₀ = 2π f₀ is the same within the FO and outside it (in the photodetector input), but the wavelength transitions
>>> from λ= λ₀/n to λ₀ in the segment between the FO end and the entrance of the photodiode, is it TRUE that Δθ is preserved? (maybe
>>> there are 20/30 micrometers between them).
>>>
>>> IF Δθ IS PRESERVED IN that small distance, THEN the refraction index n IS NOT RELEVANT AT ALL, and all the explanations based
>>> on relativistic addition of velocities to explain Fizeau ARE PURE BULLSHIT, and always were.
>> Of course the phase difference between the two light beams is preserved
>> after they come out of the fibre to they reach the photodetector.
>> How could it measure the phase difference if it wasn't preserved?
>>
>> How you from this blazingly obvious fact can conclude anything about
>> the role of the index of refraction in the fibre beats me!
>>
>> ---
>>
>> A word on the combination of the two light beams at the end.
>> When you combine two light beams on a screen, you will
>> always get a diffraction pattern that contains the energy
>> of the two light beams. There is no way you can make two
>> light beams cancel each other. The diffraction pattern
>> will consist of concentric circular 'fringes'. If the two
>> beams are in phase, the central 'disk' will be light,
>> and the the inner circular 'fringe' will be dark.
>> If the two beams are out of phase, it will be opposite;
>> the central 'disk' will be dark.
>> It is this central disk that must be sent into the photodetector.
>> (This is the principle. You will not see the circular fringes anywhere.)
>>
>>
>> --
>> Paul
>>
>> https://paulba.no/

I thought I had posted a response to this, but now I can't find it.

I will try again.

>
> You are completely right, except with the formula I = I₀ (1 + cos Δθ) ≈ I₀ {1 + cos [8πN Aω/(λ₀c₀)]}, which I copied correctly.
>
> I made a mistake when googling the solution for (cos x + cos y). I, without realizing it, used (cos x . cos y), which was next.
>
> cos x + cos y = 2 cos [(x+y)/2] . cos [(x-y)/2] , the CORRECT FORMULA
>
> cos x . cos y = 1/2 [cos (x+y) + cos (x-y)] , the WRONG FORMULA (I didn't realize my mistake, and kept writing)
>
> Even with this HUGE MISTAKE, the formula for the output of the photodiode is correct (I extracted it from a book).
>
> I = I₀ (1 + cos Δθ) ≈ I₀ {1 + cos [8πN Aω/(λ₀c₀)]} IS CORRECT.

Yes, I was wrong.

The following is still correct:
S(t) = [2A⋅cos(Δθ/2)]⋅cos(Ω⋅t)
Srms = 2A⋅cos(Δθ/2)/√2
When Δθ = 0, Srms = 2A/√2 = S₀

This is what I got wrong:
The current in the photodetector is proportional
to the power density [W/m²] of the signal.
The power density is proportional to the _square_
of the amplitude of the signal.

If the signal with power density proportional to S₀² gives
the current Iₘ, we can write the equation for the current:

I = Iₘ⋅cos²(Δθ/2) = Iₘ⋅(1 + cos(Δθ))/2 (dark current ignored)

Your I₀ = Iₘ/2.

>
> The value "1" is due to the conversion of the signal A⋅cos(Ω⋅t) into a constant stream of photons with energy hΩ/2π.

Nonsense.
The "1" comes from the equality: cos²(α) = (1 + cos(2α))/2

It is no 'conversion' of a signal to photons.
The light signal S(t) = [2A⋅cos(Δθ/2)]⋅cos(Ω⋅t) IS a stream of photons.

>
> But the simplest open loop I-FOG is prepared to provide an analog output I = I₀ (1 + cos Δθ), which is plotted as a raised cosine.
> When ω = 0 (no lateral motion), I =2 I₀ (maximum value).
>
> In a real application, ω changes temporarily and very quickly (like a sudden turn of an aircraft), so the multiplying signal
> cos (θ₁ - θ₂)/2 is a DUMPED cosine which results of the form of ω(t).

I am not sure what you are trying to say.
But note that even if Δθ may be negative, cos²(Δθ/2) is always positive.
So ω and -ω will give the same result.
The current in the photodetector is always positive.

>
> The signal cos [8πN Aω(t)/(λ₀c₀)] is the ENVELOPE of the signal A⋅cos(Ω⋅t), so I assume that photonic conversion takes place
> centered at Ω/2π, with SLOW VARIATIONS that give the analog output slow change.
>
> As far as I read, a typical analog output of a CHEAP IFOG is about 24 mV/°/sec (the photodiode current is converted to voltage).
> This signal is digitalized and post-processed with additional circuitry in the FOG assembly.
>
>

--
Paul

https://paulba.no/