Author Topic: EM Drive Developments Thread 1  (Read 1473310 times)

Offline Rodal

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Re: EM Drive Developments
« Reply #2640 on: 10/26/2014 12:38 pm »
Notice that Dr. White's proposed "quantum vacuum" force acts perpendicular to both the electric (E) and magnetic (B) fields, so for the transverse electric (TE) modes of resonance (shown in the truncated cone below) that were researched, the force (from plasma source to plasma sink) calculated by Dr. White would act in the radial direction of the truncated cone (because the B field is axial (vertical), and the E field is in the circumferential direction of the cone).

According to Dr. White's "quantum vacuum plasma" formulation, there would be no action in the direction that the thrust forces were measured:  the (vertical) axial direction of the truncated cone.

Electric field (E) in red (circumferential arrows)

Magnetic field (B) in blue (axial arrows)  (Vertical direction)
« Last Edit: 10/26/2014 01:03 pm by Rodal »

Offline Mulletron

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Re: EM Drive Developments
« Reply #2641 on: 10/26/2014 01:24 pm »


According to Dr. White's "quantum vacuum plasma" formulation, there would be no action in the direction that the thrust forces were measured:  the (vertical) axial direction of the truncated cone.


I arrived at the same conclusion. I could not find any preferred direction of thrust using White's formulation, which is why I have been trying to invoke asymmetries (via the cone shape (very slightly modified shape of spacetime from perfectly isotropic, and more toward the shape of the frustum) and chirality). Those were my leaps. My leap was that our isotropic spacetime is a product of all interactions, so if I block some of those interactions via the cone shape, spacetime is no longer perfectly isotropic, giving rise to an opportunity to bias inertia one way more than the other.

« Last Edit: 10/26/2014 01:50 pm by Mulletron »
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Offline JohnFornaro

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Re: EM Drive Developments
« Reply #2642 on: 10/26/2014 01:31 pm »
Also, can somebody tell me how you get momentum out of group velocity?

'Cuz I thought there was no relationship whatsoever between the two.

Yes certainly, that is easy to show. ...

Who does the good doctor think I am?  I use the oracle for all my entertainment needs!

Of course, the doc hisself could use a dictionary.  Note his "careful words":

Quote from: Don't even ask
momentum = groupveloctiy * mass

Check the oracle!  Dey ain't no such thing as "veloctiy".  sheesh.

But on the serious side, the oracle is the only reference tool I have at hand.

The group velocity of a wave is the velocity with which the overall shape of the waves' amplitudes — known as the modulation or envelope of the wave — propagates through space.

My understanding is that you can't multiply this by mass and get any movement of the mass.  Movement of mass implying some kind of velocity with an associated momentum.

Primitive man try for half hour to get it. Me not get it.
Sometimes I just flat out don't get it.

Offline frobnicat

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Re: EM Drive Developments
« Reply #2643 on: 10/26/2014 01:35 pm »
....
> Tau=2s (really uncertain, typical rise time, could be much lower)
....
I modeled the nonlinear coupled equations of motion (obtained from solving the Lagrangian) of the inverted torsional pendulum with Mathematica.

The 2 seconds rise is purely due to the inertial response of the equations of motion to an impulsive rectangular pulse.   There is no doubt about it.  It is due purely to the mass, damping and stiffness matrices of the pendulum.  The 2 sec rise is not due to a time-dependent-loading that takes 2 sec to reach full load.

So what one needs to find is an experimental artifact that acts like an instantaneous  (within the time scale), impulsive pulse.  Not one that takes 2 sec to reach full load.  If the loading function itself would take 2 seconds to reach full load, the response (due to the pendulum equations of motion) would be taking longer than 2 sec.

I don't put in doubt your model but from what I see from "anomalous thrust..." Brady page 15 fig. 19 the underdamped ringing could easily blur the distinction between a step excitation and a more gentle first order rise with time constant about 1s. For what we see with the rectangular pulses of calibration (dips in fig 19, also fig 20 and fig 21 at better scales) the first overshoot is typically bigger (deviation from later averaged drifting baseline response during the pulse) than the second crest 2s later (the other side of the later average). This is far from obvious with the step excitations of the thruster, quite the contrary it appears the first crest  /\ top is closer (to the overall pulse response baseline) than the second crest \/ bottom. The same remarks apply for the fall : first ringing (overshoot) has bigger relative magnitude (to the following level) than the second for calibration pulses, not for thruster pulses.

Those are just words but really I would be surprised if the thrust excitation giving such response were really instantaneous. A first order rate effect with time constant 1 would be at 63% its later level after 1s, 86% after 2s (first ringing overshoot), 98% after 4s (second ringing crest). How would you exclude such a 1s (or even 2s) rise time from those experimental diagrams, have you scraped the data from the images and run that through a deconvolution filter of some sort ? Short of that I remain sceptical of the inevitability to exclude quite not instantaneous, around 1s rise/fall time, candidate effects.

Is it possible to include attached (uploaded) pictures in the body of a post ?
« Last Edit: 10/26/2014 01:44 pm by frobnicat »

Offline JohnFornaro

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Re: EM Drive Developments
« Reply #2644 on: 10/26/2014 01:42 pm »
... the whole debacle with CERN and the superluminal neutrinos...

Well, I didn't read that as a "debacle". 

The impression I got was twofold;  First, the breathless announcement for the purpose of wowing the public and ensuring political interest in the funding stream.  Second, the CERN guys sharing ALL of their results with the community, to find out that they had made a mistake.

The first part is as annoying as considering absurdly benign wormholes, but the second part, I thought, was how science was supposed to work.
Sometimes I just flat out don't get it.

Offline JohnFornaro

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Re: EM Drive Developments
« Reply #2645 on: 10/26/2014 01:51 pm »
people here are ... clueless when it comes to matters of fundamental physics and advanced mathematics from what I've seen.

How about a bit more explanation for us clueless dummies, rather than casual dismissal with blatantly obvious oversimplifications? Like:

Quote from: SuperGenius
...nothing you wouldn't see in a first year math class...

Obviously you're one of those reasonably educated individuals, and those of us who use the oracle aren't?

Besides, I'm a better dancer than you are anyhow.  sheesh.
Sometimes I just flat out don't get it.

Offline Rodal

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Re: EM Drive Developments
« Reply #2646 on: 10/26/2014 02:03 pm »
...
I don't put in doubt your model but from what I see from "anomalous thrust..." Brady page 15 fig. 19 the underdamped ringing could easily blur the distinction between a step excitation and a more gentle first order rise with time constant about 1s. For what we see with the rectangular pulses of calibration (dips in fig 19, also fig 20 and fig 21 at better scales) the first overshoot is typically bigger (deviation from later averaged drifting baseline response during the pulse) than the second crest 2s later (the other side of the later average). This is far from obvious with the step excitations of the thruster, quite the contrary it appears the first crest  /\ top is closer (to the overall pulse response baseline) than the second crest \/ bottom. The same remarks apply for the fall : first ringing (overshoot) has bigger relative magnitude (to the following level) than the second for calibration pulses, not for thruster pulses.

Those are just words but really I would be surprised if the thrust excitation giving such response were really instantaneous. A first order rate effect with time constant 1 would be at 63% its later level after 1s, 86% after 2s (first ringing overshoot), 98% after 4s (second ringing crest). How would you exclude such a 1s (or even 2s) rise time from those experimental diagrams, have you scrapped the data from the images and run that through a deconvolution filter of some sort ? Short of that I remain sceptical of the inevitability to exclude quite not instantaneous, around 1s rise/fall time, candidate effects.

Is it possible to include attached (uploaded) pictures in the body of a post ?

Quote
Those are just words but really I would be surprised if the thrust excitation giving such response were really instantaneous.   

Of course, that's why I wrote <<So what one needs to find is an experimental artifact that acts like an instantaneous  (within the time scale>> resolution.  We can only tell how impulsive the response is from the graph.  From measuring the graph, to me this is about within ~0.2 seconds, so meet me somewhere in-between (I think that 1 sec to 2 sec is too much, as we can discriminate 1 sec to 2 sec  from the graph). 

Quote
This is far from obvious with the step excitations of the thruster, quite the contrary it appears the first crest  /\ top is closer (to the overall pulse response baseline) than the second crest \/ bottom. The same remarks apply for the fall : first ringing (overshoot) has bigger relative magnitude (to the following level) than the second for calibration pulses, not for thruster pulses
Yes, the measured force (whatever its origiin) is not a pure step.  The measured force vs time however, as you know, should not be interpreted as being a replica of the thruster excitation.  Due to the complicated measuring system (an inverted torsional pendulum with nonlinear coupling between swinging and torsion) there are many natural modes of resonance and if the thruster response instead of being a pure step has different frequencies of excitation, the measured response will amplify the frequencies that are near the natural frequencies (some of them parasitic) of the inverted pendulum.

Quote
How would you exclude such a 1s (or even 2s) rise time from those experimental diagrams, have you scrapped the data from the images and run that through a deconvolution filter of some sort ?

Yes, actually, I was planning to do that.  If we can come up with a proposed input excitation (whether an artifact or a real thrust) it would be easy for me to input it in Mathematica and see how it compares with the actual response.

This is an imperfect analogy, but similar argumentation as you very well noticed the InputPower^2 response of the model vs the InputPower^1 actual response
« Last Edit: 10/26/2014 02:07 pm by Rodal »

Offline JohnFornaro

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Re: EM Drive Developments
« Reply #2647 on: 10/26/2014 02:09 pm »
I can't begin to guess how many pages were about symmetries and conservation laws. Make sure you read before you criticize the group. How about you bring some new insight to the table?

Sometimes I just flat out don't get it.

Offline Rodal

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Re: EM Drive Developments
« Reply #2648 on: 10/26/2014 02:12 pm »
http://www.falstad.com/mathphysics.html

Neat

...

Start at 1:25
Yes neat video.  Besides it being an interesting demonstration (we agree on that), do you see here something that could throw a light into the reason for the measured thrust in the EM Drives?
« Last Edit: 10/26/2014 02:13 pm by Rodal »

Offline Rodal

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Re: EM Drive Developments
« Reply #2649 on: 10/26/2014 02:18 pm »
people here are ... clueless when it comes to matters of fundamental physics and advanced mathematics from what I've seen.

How about a bit more explanation for us clueless dummies, rather than casual dismissal with blatantly obvious oversimplifications? Like:

Quote from: SuperGenius
...nothing you wouldn't see in a first year math class...

Obviously you're one of those reasonably educated individuals, and those of us who use the oracle aren't?

Besides, I'm a better dancer than you are anyhow.  sheesh.
Or how about a constructive contribution rather than subjective wordy opinions, a contribution showing what is meant by "higher math" by actually analyzing with "higher math" the EM Drives as an experimental artifact, using what is considered to be non-first year math class?  (the example given of a Fourier transform as being such "higher math" implies an assumption of ergodicity and stationarity, both of which are not met by many physical processes. )
« Last Edit: 10/26/2014 03:24 pm by Rodal »

Offline Rodal

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Re: EM Drive Developments
« Reply #2650 on: 10/26/2014 02:28 pm »
Also, can somebody tell me how you get momentum out of group velocity?

'Cuz I thought there was no relationship whatsoever between the two.

Yes certainly, that is easy to show. ...

Who does the good doctor think I am?  I use the oracle for all my entertainment needs!

Of course, the doc hisself could use a dictionary.  Note his "careful words":

Quote from: Don't even ask
momentum = groupveloctiy * mass

Check the oracle!  Dey ain't no such thing as "veloctiy".  sheesh.

But on the serious side, the oracle is the only reference tool I have at hand.

The group velocity of a wave is the velocity with which the overall shape of the waves' amplitudes — known as the modulation or envelope of the wave — propagates through space.

My understanding is that you can't multiply this by mass and get any movement of the mass.  Movement of mass implying some kind of velocity with an associated momentum.

Primitive man try for half hour to get it. Me not get it.
You asked for an example. I gave you an example where momentum = groupvelocity *mass, exactly. 

The Wikipedia article defines the group velocity in exactly the same way that I defined it: as the partial derivative of the angular frequency with respect to the wavenumber.  See everything that the wikipedia article states under definition.

The relationship of group velocity to phase velocity depends on the dispersion relation.

Due to dispersion, wave velocity is not uniquely defined, giving rise to the distinction of phase velocity and group velocity when the dispersion relation is not linear.

Don't worry about not understanding this right away.  (Nonlinear) Dispersion is a nonlinear phenomenom, therefore it is non-intuitive:  human minds are good at thinking linearly, and not about thinking nonlinearly.  Nonetheless we live in a world where radiowaves, waterwaves, electromagnetic waves, encounter nonlinear dispersion.  In a perfect vacuum the dispersion relation is linear, but we humans don't live in a vacuum.

I don't like the fact that people that don't identify themselves with their real name, but instead use monickers, can constantly write and modify Wikipedia articles, at their will,  but if you like to use use it, here is the article on Wikipedia on dispersion that applies:  http://en.wikipedia.org/wiki/Dispersion_relation

or, better, this lecture (video):  http://ocw.mit.edu/courses/physics/8-03-physics-iii-vibrations-and-waves-fall-2004/video-lectures/lecture-12/

physical demonstration starting approximately at 1:10

The MIT Prof. (Lewin) discusses cut-off frequency, nonlinear dispersion, group velocity and a lot of the stuff we are discussing.  The lecture is at an introductory level (the third Physics course for undergraduates).

« Last Edit: 10/26/2014 04:37 pm by Rodal »

Offline Mulletron

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Re: EM Drive Developments
« Reply #2651 on: 10/26/2014 02:52 pm »
http://www.falstad.com/mathphysics.html

Neat

...

Start at 1:25
Yes neat video.  Besides it being an interesting demonstration (we agree on that), do you see here something that could throw a light into the reason for the measured thrust in the EM Drives?

The video is a neat demonstration of the link between optics and rf. No dichotomy is required. Optics gets all the attention due to its familiarity in daily life. Hence why the refractive index we commonly encounter is actually based off of an optical value of 589nanometers. Foam and wax are to rf prisms as glass is to optical prisms. Just a neat thing I found while researching evanescent wave coupling/quantum tunneling. I have to force myself to think a different/better way so I can apply this to copper cans.

« Last Edit: 10/26/2014 02:53 pm by Mulletron »
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Offline Rodal

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Re: EM Drive Developments
« Reply #2652 on: 10/26/2014 03:53 pm »
,,,,,
You asked for an example. I gave you an example where momentum = groupvelocity *mass, exactly. 

The Wikipedia article defines the group velocity in exactly the same way that I defined it: as the partial derivative of the angular frequency with respect to the wavenumber.  See everything that the wikipedia article states under definition.

The relationship of group velocity to phase velocity depends on the dispersion relation.

Due to dispersion, wave velocity is not uniquely defined, giving rise to the distinction of phase velocity and group velocity when the dispersion relation is not linear.

I don't like the fact that people that don't identify themselves with their real name, but instead use monickers, can constantly write and modify Wikipedia articles, at their will,  but if you like to use use it, here is the article on Wikipedia on dispersion that applies:  http://en.wikipedia.org/wiki/Dispersion_relation

or, better, this lecture (video):  http://ocw.mit.edu/courses/physics/8-03-physics-iii-vibrations-and-waves-fall-2004/video-lectures/lecture-12/

physical demonstration starting approximately at 1:10

The MIT Prof. (Lewin) discusses cut-off frequency, nonlinear dispersion, group velocity and a lot of the stuff we are discussing.  The lecture is at an introductory level (the third Physics course for undergraduates).



If you like Wikipedia, also see the article on cut off frequency http://en.wikipedia.org/wiki/Cutoff_frequency

the cut off frequency for waveguides:
Quote
The value of c, the speed of light, should be taken to be the group velocity of light in whatever material fills the waveguide

and the relationship to dispersion (Prof. Lewin in his lecture -video above-  shows the nonlinear dispersion [omega vs kz] occurring near the cut off frequency for his example).

Quote
The wave equations are also valid below the cutoff frequency, where the longitudinal wave number is imaginary. In this case, the field decays exponentially along the waveguide axis and the wave is thus evanescent.
 
« Last Edit: 10/26/2014 03:58 pm by Rodal »

Offline frobnicat

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Re: EM Drive Developments
« Reply #2653 on: 10/26/2014 04:12 pm »
 @John,
Concerning group velocity and momentum, maybe you should start from this phrase of the oracle :
"The group velocity is often thought of as the velocity at which energy or information is conveyed along a wave."

Group velocity is velocity of energy, read mass_energy as the two are equivalent. Speed of energy_mass times amount of energy_mass yields momentum.

"However, if the wave is travelling through an absorptive medium, this does not always hold."
From this point onward I give up, dr Rodal's erudite answers are your only recourse.

Offline Rodal

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Re: EM Drive Developments
« Reply #2654 on: 10/26/2014 06:05 pm »
« Last Edit: 10/26/2014 06:07 pm by Rodal »

Offline frobnicat

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Re: EM Drive Developments
« Reply #2655 on: 10/26/2014 06:10 pm »
Quote
Those are just words but really I would be surprised if the thrust excitation giving such response were really instantaneous.   

Of course, that's why I wrote <<So what one needs to find is an experimental artifact that acts like an instantaneous  (within the time scale>> resolution.  We can only tell how impulsive the response is from the graph.  From measuring the graph, to me this is about within ~0.2 seconds, so meet me somewhere in-between (I think that 1 sec to 2 sec is too much, as we can discriminate 1 sec to 2 sec  from the graph). 

I won't bargain on that, I stand on 1s to 2s rise (and fall) times are compatible with what we see of fig. 19 and fig. 21 Brady's anomalous thrusts. This is not a matter of time scale resolution, this is a matter of settling time of the "filter" constituted by the balance when step like signals are applied. This is an important matter as actual thrust vs time (when power is rectangular pulse) is a strong constraint on possible explanations. I'm considering the RF power generator has a very short time constant, and the standing resonant microwave reaches stationary amplitudes in µs, so from the point of view of RF energy or power we have instantaneous input excitation of the drive at nominal values for all practical purpose. The question is the time behaviour of the thrust output of the device from this rectangular power input. I don't see how real effects (momentum conservation with ghostly particles or local cosmic horizon) could have "inertia" in the thrust/power ratio for longer than a few µs (Q times length_of_cavity/c)

Quote
Quote
This is far from obvious with the step excitations of the thruster, quite the contrary it appears the first crest  /\ top is closer (to the overall pulse response baseline) than the second crest \/ bottom. The same remarks apply for the fall : first ringing (overshoot) has bigger relative magnitude (to the following level) than the second for calibration pulses, not for thruster pulses
Yes, the measured force (whatever its origiin) is not a pure step.  The measured force vs time however, as you know, should not be interpreted as being a replica of the thruster excitation.  Due to the complicated measuring system (an inverted torsional pendulum with nonlinear coupling between swinging and torsion) there are many natural modes of resonance and if the thruster response instead of being a pure step has different frequencies of excitation, the measured response will amplify the frequencies that are near the natural frequencies (some of them parasitic) of the inverted pendulum.

Not clear to me, can we clarify terminology ?

P(t) : Microwave power input to the thruster (instantaneous compared to time scales, rectangular pulse)
F(t) : thrust output of the thruster (the interesting magnitude vs time shape, perfect steps or not ?)
Fb(t) : excitation input of the balance (almost the same as F, but with added DC power parasitic term )
Obs(t) : Measure output of the balance (as displacement, with natural frequencies, underdamped...)

"if the thruster response instead of being a pure step..." you are referring to F ?

Quote
Quote
How would you exclude such a 1s (or even 2s) rise time from those experimental diagrams, have you scrapped the data from the images and run that through a deconvolution filter of some sort ?

Yes, actually, I was planning to do that.  If we can come up with a proposed input excitation (whether an artifact or a real thrust) it would be easy for me to input it in Mathematica and see how it compares with the actual response.

Input excitation of balance, that is Fb ?

Ok, could we try with 3 easy ones :

- Pure rectangle, 30 seconds
    Fb(t) = cst for 0<t<30
    Fb(t) = 0 otherwise

- "charge/discharge" rectangle : first order tau=1s rise and decay 
    Fb(t)=0 for t<0
    Fb(t)=cst*(1-exp(-t/1)) for 0<t<30
    Fb(t)=cst*exp(-(t-30)/1) for 30<t

- "charge/discharge" rectangle : first order tau=2s rise and decay 
    Fb(t)=0 for t<0
    Fb(t)=cst*(1-exp(-t/2)) for 0<t<30
    Fb(t)=cst*exp(-(t-30)/2) for 30<t

cst = target constant amplitude (say 80µN)

What is experimental data resembling most ?

Quote
This is an imperfect analogy, but similar argumentation as you very well noticed the InputPower^2 response of the model vs the InputPower^1 actual response

This was for one Shawyer's thrust vs power graph. I prefer to concentrate on Brady, as someone put it (you ? Aero ? Mulletron ?) it is better documented, and maybe of more reliable value. I don't completely give up on this "warm jet effect". Even if it seems unlikely it would have been gone unnoticed and gave results somehow in agreements between different labs, the alternatives are either as much hazardous (magnetic couplings) or more hard to swallow (axions anyone ?)

We don't even know from the experimenters the most basic geometric characteristics of their device. Who know (save them) how much they are tinkering with things before "it works", what kind of "secret recipe" procedures are followed and in which we could maybe see some confirmation or invalidation of such or such possible explanation. One thing stands out : the effect is hard to get at, and that's not because of ultra faint magnitude (µN when nN are "routinely" investigated with short range gravitation studies). So what makes it so hard to make it reproducible and reliable ? Driving a high Q at resonant frequency ? Mmm. Well. Maybe. Only maybe.

My latest derivation for the warm air jet hypothesis : F in function of microwave volumetric heating Pow and tau time constant of the rise/fall

F = Cf/rho * ((2*p*Tau)/V)^0.5 * (Pow/(C*T))^1.5

That is by using numerical values for almost constants ( 20°C ambient ... )
F = 1.41e-6 * 1/sqrt(V) * sqrt(Tau) * Pow^1.5

V=0.027m^3  Tau=2s  Pow=4W  yields F=97µN

But that needs tinkering until the leak area A is small enough (but not stoppered) to give high thrust, but big enough so that Tau is below some level (otherwise you see a ramp up). How much tinkering is going on ?
F = 9.7e-12 * Pow^2/A
Tau = 4.74e-11 * V*Pow/A^2

Intuitively I'm in favor of some electromagnetic explanation, but think this is still worth investigating.
« Last Edit: 10/26/2014 06:23 pm by frobnicat »

Offline Rodal

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Re: EM Drive Developments
« Reply #2656 on: 10/26/2014 06:47 pm »
... The question is the time behaviour of the thrust output of the device from this rectangular power input
Here it looks to me that we were referring to different things.  The inverted torsional pendulum has a time response to an (instantaneous) impulsive load of 2s, not 1s
Quote
Not clear to me, can we clarify terminology ?

P(t) : Microwave power input to the thruster (instantaneous compared to time scales, rectangular pulse)
F(t) : thrust output of the thruster (the interesting magnitude vs time shape, perfect steps or not ?)
Fb(t) : excitation input of the balance (almost the same as F, but with added DC power parasitic term )
Obs(t) : Measure output of the balance (as displacement, with natural frequencies, underdamped...)

"if the thruster response instead of being a pure step..." you are referring to F ?
It's great that you realize that we are referring to different quantities and the best way to resolve this is to define what we mean.  Here I was saying (admittedly in less than perfect language) that if Fb(t) instead of being an impulse has a significant delay-to-reach-steady-state (time-dependent rise), then Obs(t) will have a response longer than 2 sec.  From my side I'm a black box that if given Fb(t) as an input, I can give you Obs(t) as an output.

I make no comments whatsoever on P(t) and F(t).

Quote
Input excitation of balance, that is Fb ?

Ok, could we try with 3 easy ones :

- Pure rectangle, 30 seconds
    Fb(t) = cst for 0<t<30
    Fb(t) = 0 otherwise

- "charge/discharge" rectangle : first order tau=1s rise and decay 
    Fb(t)=0 for t<0
    Fb(t)=cst*(1-exp(-t/1)) for 0<t<30
    Fb(t)=cst*exp(-(t-30)/1) for 30<t

- "charge/discharge" rectangle : first order tau=2s rise and decay 
    Fb(t)=0 for t<0
    Fb(t)=cst*(1-exp(-t/2)) for 0<t<30
    Fb(t)=cst*exp(-(t-30)/2) for 30<t

cst = target constant amplitude (say 80µN)

What is experimental data resembling most ?

It's great working with you. That's exactly what I needed.

However, now I'm going to enjoy the Sun.  I'll work on that when the Sun goes away  :)
« Last Edit: 10/26/2014 06:52 pm by Rodal »

Offline ThinkerX

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Re: EM Drive Developments
« Reply #2657 on: 10/26/2014 07:27 pm »
Conclusion from the 'experimental results' section of the Cannae drive page.


Quote
When operated with a TM010 EM wave, the QDrive POC cavity generates a linear net imbalance of Lorentz forces exerted on the cavity. FEM numerical predictions for net unbalanced Lorentz force on the POC cavity is within 40% of experimental results. Variations between experimental results and numerical method predictions are within the margin of error based on limitations of the experimental measurements of Cavity Q. EM field energy in the POC cavity during experimental runs was less than 3% of design maximum. Field energy levels in the experiment limit correlations between numerical method predictions and experimental results.

'...linear net imbalance of Lorentz forces?'  Does this mean 'thrust?'

'...less than 3% of design maximum?'  Does this mean he could have created an imbalance on the order of 33 times greater than what was measured? 

Apart from that, I note a lot of pages under construction on that site...including some very interesting ones. 

Offline Rodal

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Re: EM Drive Developments
« Reply #2658 on: 10/26/2014 07:44 pm »
Conclusion from the 'experimental results' section of the Cannae drive page.


Quote
When operated with a TM010 EM wave, the QDrive POC cavity generates a linear net imbalance of Lorentz forces exerted on the cavity. FEM numerical predictions for net unbalanced Lorentz force on the POC cavity is within 40% of experimental results. Variations between experimental results and numerical method predictions are within the margin of error based on limitations of the experimental measurements of Cavity Q. EM field energy in the POC cavity during experimental runs was less than 3% of design maximum. Field energy levels in the experiment limit correlations between numerical method predictions and experimental results.

'...linear net imbalance of Lorentz forces?'  Does this mean 'thrust?'

'...less than 3% of design maximum?'  Does this mean he could have created an imbalance on the order of 33 times greater than what was measured? 

Apart from that, I note a lot of pages under construction on that site...including some very interesting ones.

If they get a "net imbalance of Lorentz forces" it means that they get thrust, yes, but it also means to me that the Finite Element problem was not well posed if they are dealing with classical physics (Maxwell's equations and equations or equations of motion from a Lagrangian or Hamiltonian).

To obtain a finite element solution one of the things one has to be careful of are the boundary conditions. 

If one deals with the free-free body floating in space, the finite-element matrix is singular (the determinant of the matrix is zero) because a free-free body has rigid-body modes.  In order to obtain a proper solution one has to identify the rigid-body modes and extract them from the finite-element solution in order to obtain a non-singular matrix and hence a solution.  I don't know what Finite Element code they used or how they went about eliminating the rigid-body modes but I do know that if you pose the problem correctly using classical physics (as done by Greg Egan) there should be conservation of momentum and hence hence no acceleration of the center of mass.  Therefore if they obtain unbalanced forces in the FEM solution the following has to be addressed:

a) properly taking into account hidden momentum
b) correct treatment of boundary conditions
« Last Edit: 10/26/2014 07:48 pm by Rodal »

Offline JohnFornaro

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Re: EM Drive Developments
« Reply #2659 on: 10/26/2014 08:20 pm »
You asked for an example. I gave you an example where momentum = groupvelocity *mass, exactly.

The Wikipedia article defines the group velocity in exactly the same way that I defined it...

I was noticing the similarity between your explanation and that of the oracle, which caused a bit of head scratching at your seeming dismissal of the oracle.  I need to study a bit more, my good troglodite.

As always, appreciate your pedagology.  That chart is most excellent!
Sometimes I just flat out don't get it.

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