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.
Quote from: JohnFornaro on 10/25/2014 07:48 pmAlso, 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. ...
Also, can somebody tell me how you get momentum out of group velocity?'Cuz I thought there was no relationship whatsoever between the two.
momentum = groupveloctiy * mass
Quote from: frobnicat on 10/25/2014 11:21 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.
....> Tau=2s (really uncertain, typical rise time, could be much lower)....
... the whole debacle with CERN and the superluminal neutrinos...
people here are ... clueless when it comes to matters of fundamental physics and advanced mathematics from what I've seen.
...nothing you wouldn't see in a first year math class...
...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 ?
Those are just words but really I would be surprised if the thrust excitation giving such response were really instantaneous.
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
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 ?
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?
http://www.falstad.com/mathphysics.htmlNeat...Start at 1:25
Quote from: Supergravity on 10/25/2014 10:18 pmpeople 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.
Quote from: Rodal on 10/25/2014 08:35 pmQuote from: JohnFornaro on 10/25/2014 07:48 pmAlso, 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 askmomentum = groupveloctiy * massCheck 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.
Quote from: Mulletron on 10/26/2014 10:15 amhttp://www.falstad.com/mathphysics.htmlNeat...Start at 1:25Yes 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?
,,,,,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_relationor, 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 1:10The 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).
The value of c, the speed of light, should be taken to be the group velocity of light in whatever material fills the waveguide
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.
QuoteThose 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).
QuoteThis 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 pulsesYes, 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.
QuoteHow 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
... The question is the time behaviour of the thrust output of the device from this rectangular power input
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 ?
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<tcst = target constant amplitude (say 80µN)What is experimental data resembling most ?
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.
Conclusion from the 'experimental results' section of the Cannae drive page.QuoteWhen 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.
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...