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#2320 by IslandPlaya on 20 Oct, 2014 01:37
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The cat is ok btw.
Just slightly singed.
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#2321 by aero on 20 Oct, 2014 01:49
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The cat is ok btw.
Just slightly singed.
Glad to hear it.
I warned you many pages ago to be careful with those things. They are dangerous when not properly shielded. One Kw of microwave power will boil water quickly, and your cat is mostly water. Didn't you read my warning to your cat? Or just show it to him/her - she/he can probably read it.
Just because a cat mouth won't make word sounds doesn't mean their is anything wrong with their hearing or eyesight. If you want your cat to help with your experiments you should teach it sign language. -
#2322 by Rodal on 20 Oct, 2014 02:12
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Taking into account the frequency drift and bandwidth issues that the researchers have tuning the device under resonance, with concomitant drift in Q (which therefore cannot be a constant during the measurements) these results are quite interesting!
More on that later please, as relation of resonance with Q not clear to me (higher Q => narrower bandwidth not the point, that's perfectly clear). Needs time to think of a clear enough way to express what's not clear, that's difficult.
The Q value depends on the frequency and mode shape of the resonant cavity, as well as the wall material properties and the geometry.
The Q value for the lowest frequency of a cylindrical resonant cavity (TM010) is
Q =( Length / skinDepthForResonantFrequency) / (1 + ( 2*Length / InnerDiameter ))
skinDepthForResonantFrequency = 1/Sqrt[Pi*ResonantFrequency*μ*σ]
so,
Q = Sqrt[Pi*ResonantFrequency*μ*σ]*Length/ (1 + ( 2*Length / InnerDiameter ))
The skinDepth is a function of the frequency and wall material properties: permeability μ and electrical conductivity σ, which can be complex, frequency dependent, and anisotropic functions of field direction. They can also be functions of density, temperature, field strength, and other quantities (they can depend on the electric field and the magnetic field).
Therefore Q is a function of the Length and Diameter of the cylindrical cavity, as well as the frequency, and of the wall material.
The skin depth should be around a couple of micrometers (μm). Therefore the inner wall of the cavity must be carefully plated or polished; otherwise, current flow will be severely perturbed by surface irregularities lowering the cavity Q.
==> so John: there
We do know a lot of things of what's inside these cupric cavities, we know that from the cumulative scientific knowledge that has accumulated since Maxwell's equations and our ability to solve them. When the researchers provide to us S12 plots, and Q measurements, they are providing us information about the inside of the cavities.
For example, given the geometry, frequencies and the measured Q, we know the following:
1) Both inner surface ends of the truncated cone must have been made out of copper. Otherwise the cavity would not have the correct boundary conditions to be a resonant cavity: it would be a waveguide. One wouldn't be able to have high Q and resonance if the inner surface of the ends wouldn't be copper.
2) The inner surface of the copper must have been pretty well polished, in order to get Q~50 000, given the calculated skin depths
3) There cannot be any significant irregularities on the inside surface of the truncated cone.
So there "about hummingbirds". That's the power of science
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#2323 by JohnFornaro on 20 Oct, 2014 03:00
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Frobnicate
Thanks for the learn something new every day moment!
I was starting to think you all were a bunch of Aholeholes, what with all that math you all are throwing around.
I am under a drawing crunch. Will get around to attempting dimensions on those other copper cans later this week.
Maybe someone could illustrate them all sequentially? There are what, three "models"? Four? Highest rez possible, and some scalable diameter. Plus a short name which I can include on the title block.I don't know what is an inch or a foot.
Well, there's an easy memnonic for that:
One inch is the difference between an "Oooh", and an "Ahhh".1) Both inner surface ends of the truncated cone must have been made out of copper. Otherwise the cavity would not have the correct boundary conditions to be a resonant cavity: it would be a waveguide. One wouldn't be able to have high Q and resonance if the inner surface of the ends wouldn't be copper.
2) The inner surface of the copper must have been pretty well polished, in order to get Q~50 000, given the calculated skin depths
3) There cannot be any significant irregularities on the inside surface of the truncated cone.
1) So there's copper on the inside of that PCB board?
2) I'm envisioning polishing on the level of the Hubble's mirrors, not the cursory polish on the devices pictured so far. Plus, the copper is so thin that it would be ...
3) ... easily deformed, particularly by heating, or bumping into something.
Just observin'. -
#2324 by aceshigh on 20 Oct, 2014 04:02
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The cat is ok btw.
Just slightly singed.
good think it was a cat, because gremlins are much more sensitive to microwaves.
I counted 7 seconds
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#2325 by aero on 20 Oct, 2014 04:23
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Ok, that's good. Basically anything sensible we come up with can be computed then for this small data set.
Yes. I looked at some articles on PSP, its just another way of doing beamed propulsion. Since the photons within the resonance cavity (between the mirrors) actually are pushing, all of them count to give thrust. In the EM drive case, it seems that all of the photons are counting to give thrust but they are not leaving the cavity of the engine. I guess that is the trick and the problem. How can the momentum of the photons in the cavity be separated from the photons themselves? Momentum departs, photons remain.
I've been thinking about the double slit experiment which proves that a single electron passes through both slits. The extension to photons is that a single photon passes through both slits. What if the momentum of the photon passing through the left slit was reflected while the momentum of the right photon was passed through the slit? That replaces our problem of separating the momentum from the photon to one of passing a photon through the copper undetectably.
And since the power is now multiplied by Q there is an appreciable amount of power to remain undetected. But momentum is not power. -
#2326 by Mulletron on 20 Oct, 2014 08:01
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Ok I'm aware that I'm the stick in the mud here, but not by virtue of being intentionally obtuse here. I believe I have things in the right perspective. Let me state this another way:
Now you can dump rf energy into a cavity all day long, and that rf cavity is going to eventually absorb (as a function of Q) and re-radiate that energy right back to the universe in which it resides. Given the cavity has a big end and a small end, you have more surface area on the big end in which to radiate heat, giving the illusion of thrust by new science. This isn't new science. I don't need a page long series of equations to characterize this. It is just thermodynamics.
Also it is well known that if an electric current flows through anything, wires, cavity walls, whatever... the result is a perpendicular magnetic field around the conductor. Now pulse that current, you get a pulsating magnetic field.
The NASA test campaign is very telling compared to the other tests, because the NASA tests were low power tests. This allowed them to effectively separate out artifact modes of thrust from the dominant mode of thrust. They concluded, all things considered that the dielectric was important to measured thrust.
If you dump hundreds watts into an empty sealed test article, yep, you're gonna measure some thrust. The thrust you get doesn't need new science to explain.
One can try to get famous by writing page long formulas to explain the obvious, but there is no need.
Empty cavities providing thrust isn't anomalous thrust.
Cavities with dielectric present providing thrust is anomalous thrust. And when you remove the dielectric, the thrust goes away.......that is anomalous. -
#2327 by frobnicat on 20 Oct, 2014 09:31
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Ok I'm aware that I'm the stick in the mud here, but not by virtue of being intentionally obtuse here. I believe I have things in the right perspective. Let me state this another way:
Now you can dump rf energy into a cavity all day long, and that rf cavity is going to eventually absorb (as a function of Q) and re-radiate that energy right back to the universe in which it resides. Given the cavity has a big end and a small end, you have more surface area on the big end in which to radiate heat, giving the illusion of thrust by new science. This isn't new science. I don't need a page long series of equations to characterize this. It is just thermodynamics.
Some thrust from asymmetric heat radiation isn't new science, that's correct. But above that qualitative statement, a single equation is relevant nonetheless : for this classical explanation to hold, the device is just a photon rocket => F < P/c Brady a : power 17W, max classical radiative force 17/3e8 = 56 nN (nano Newtons), much below the claimed thrust of ~90 µN. No need of page long series of equation, but one equation to discard this one classical explanation, in this easy case (at least 2 orders of magnitude off target). And quite a number of equations and computing hours are needed to ascertain an effect as heat radiation thrust when the case is not that clear (Pioneer anomaly...).
Throwing in an equation is better than just words (isn't it dr Rodal ?) but thinking twice about it can't hurt : the F < P/c upper bound has an implied hypothesis that the radiated heat is just lost in the cold of deep space. Now what if the device is inside a chamber and IR photons are bouncing a few times before being absorbed in the cold walls of the chamber ? Then you have an amplification factor and the upper bound becomes F<q P/c where q is the "quality factor" of bouncing IR photons between device and chamber walls (like the "pusher beam" of the photonic laser propulsion scheme)
Do you find likely that not specially prepared surfaces (outer device and inner chamber walls) could bounce IR photons 300 times before absorbing ? I don't => classical radiative thrust discarded, but than needs a minimum of qualitative and quantitative argumentation.Quote
Also it is well known that if an electric current flows through anything, wires, cavity walls, whatever... the result is a perpendicular magnetic field around the conductor. Now pulse that current, you get a pulsating magnetic field.
That can't significantly go any deeper than a few µm if the pulses are GHz.
BTW what is the thickness of copper on the PCBs at the ends in Brady et al apparatus ? Guess it's more than 20µm ? Should be a perfect wall for microwaves, for all practical purpose.QuoteThe NASA test campaign is very telling compared to the other tests, because the NASA tests were low power tests. This allowed them to effectively separate out artifact modes of thrust from the dominant mode of thrust. They concluded, all things considered that the dielectric was important to measured thrust.
If you dump hundreds watts into an empty sealed test article, yep, you're gonna measure some thrust. The thrust you get doesn't need new science to explain.
The relative magnitude of thrust you get in those experiments needs either new science or old science with reasonably detailed quantitative explanations as artifacts. Isn't it dr Rodal ?Quote
One can try to get famous by writing page long formulas to explain the obvious, but there is no need.
Empty cavities providing thrust isn't anomalous thrust.
Cavities with dielectric present providing thrust is anomalous thrust. And when you remove the dielectric, the thrust goes away.......that is anomalous.
Nobody tries to be famous here, we are all already famous enough on some rock scene (otherwise we wouldn't come by here with nicknames). We are here incognito on this family site to advance science, far from our oppressive celebrity
Apart from dr Rodal. But a doctor has and ethic, and can't be motivated by mediocre motives like that.
So doctor : by what I understand you no longer believe the presence of dielectric is significant as
- Brady et al were off resonance target when removing dielectric (dielectric_presence and correct_resonance not tested independently )
- Shawyer no longer uses dielectrics.
Can we reach a consensus whether or not dielectric is significant ?
New physics real effects or old physics artifacts : nothing is obvious, either cases.
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#2328 by Chrochne on 20 Oct, 2014 11:15
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There is a new update about the EmDrive on their official website - emdrive.com
"At the IAC 2014 conference in Toronto, Roger Shawyer stated that 8 sets of test data have now verified EmDrive theory. These data sets resulted from thrust measurements on 7 different thrusters, by 4 independent organisations, in 3 different countries."
You might want to check it. Its from October 2014 there is also one interesting PDF
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#2329 by Star One on 20 Oct, 2014 11:17
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#2330 by Mulletron on 20 Oct, 2014 11:23
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Yes I calculate for between 1-2ghz, the skin depth of copper is 2.1um to 1.46um respectively. That's not the direction I was going. Even with the 4x skin depth recommendation, the copper cavity is way thick enough for skin depth to not be a factor. I don't know how thick it is exactly, but I know it isn't made of copper foil. I acknowledge this and I know I am on shaky ground trying to invoke leaky magnetic influence. IR loss is a given, magnetic induction is not so easy.
Either way, I'm trying to break theories as much as I can. That's science.
I just get bent out of shape when I see new theories for emdrive that try to teach photons new tricks.
I'm trying to explore near field effects in more detail.
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#2331 by Mulletron on 20 Oct, 2014 12:02
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On page 2 of this: http://www.emdrive.com/iac2014presentation.pdf
He makes a point of showing a distinction between thrust (force) and reaction (force). Really tearing into this, and correct me if I got this wrong, the ones that say thrust, are measured from the perspective of the device, and the reaction is where a second object exerts an equal and opposite reaction (force) on the first object providing the thrust (force).
The ones that say thrust all have dielectric in common, and appear reactionless.
The ones that say reaction have no dielectric in common, and are not reactionless.
Designs with dielectric, appear reactionless (postulated to react with the QV).
Designs without dielectric, not reactionless.
This tells me that designs which have no dielectric, would not fly in vacuum, as there is nothing to react against.
The reason I'm saying the above sentence is because photons aren't known to self interact. Set me straight on this sentence, because I am not completely up on 2 photon physics.
Just a side note, a classical rocket in vacuum still works, because the reaction mass is the nozzle itself and its own hot gasses. Hot gas hits the nozzle, the nozzle and gasses provides an equal and opposite reaction.
Edited for grammar/clarity. -
#2332 by frobnicat on 20 Oct, 2014 12:08
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Have we exceeded a million equation search yet?
No but 21769 is already quite a lot. The data set is sparse and with some uncertainties : the risk is overfitting. Scanning on more than 15 bits (32768 combinations) worth of explanation could easily bring up more perfect formula for the specific available data but with less generalisation power : worse at predicting next data points to come. Need more data points before it's worth looking at much more formulas. Not a problem of computing power, not before reaching many billions of formulas.
At this stage with 7 data points, from a "phenomenological theoretically agnostic" point of view, simpler is better, and there is not that much simple equations.
Note : the number of combinations of exponents and added terms were 94 millions but of those only 21769 unique representations (discarding equivalents) made sense in dimensional analysis (kg m s).
Not quite there yet. The theoretical geometrical variable for a resonant cavity box is Sqrt[a^-2+L^-2] and we didn't have square roots of additions allowed, is that right?
Moreover we had a number of formulas very close: [a^-2+L^-2] and [a^-1+L^-1]
So the data was telling us we need to allow Sqrt[a^-2+L^-2]
We had these square roots but they are missing the plus sign:
a^-2 b^-2 L^2 Q^1 P^1 F^-1 c^-1 sqrt(a^-2 L^-2)^-1 1.02 0.58
a^-1 b^-2 L^2 Q^1 P^1 F^-1 c^-1 sqrt(a^-1 L^-1)^-1 1.32 0.58
Before the latest conference "paper" buries us with a fury of posts :
For the previous data, without Brady b, not yet taking into account other parameters ( L is still the wavlength of driving frequency), sieving only through even powers (except for the added term) on with higher thresholds :
6 entries
Thresholds : mean=2.00 stddev=1.35
a b L Q P F c mean stddev
---------------------------------------------------------------------
a b L Q P F c exterm mean stddev
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a^2 b^-2 L^-2 Q^2 P^2 F^-2 c^-2 (a^-2 + b^-2)^-1 1.42 1.34
a^0 b^0 L^2 Q^2 P^2 F^-2 c^-2 (a^-2 + b^-2)^1 1.76 0.93
a^2 b^-2 L^2 Q^2 P^2 F^-2 c^-2 (a^ 2 + b^ 2)^-1 1.42 0.68
a^0 b^-2 L^0 Q^2 P^2 F^-2 c^-2 (a^-2 + L^-2)^-1 1.31 0.77 *
a^-2 b^-2 L^2 Q^2 P^2 F^-2 c^-2 (a^-2 + L^-2)^-1 0.09 1.00 *
a^-2 b^-2 L^2 Q^2 P^2 F^-2 c^-2 (a^ 2 + L^ 2)^1 1.87 0.84 *
a^2 b^-2 L^-2 Q^2 P^2 F^-2 c^-2 (b^-2 + L^-2)^-1 1.23 1.21
a^0 b^0 L^2 Q^2 P^2 F^-2 c^-2 (b^-2 + L^-2)^1 1.95 1.06
a^-2 b^-2 L^2 Q^2 P^2 F^-2 c^-2 (b^ 2 + L^ 2)^1 0.72 1.35
a^2 b^2 L^-2 Q^2 P^2 F^-2 c^-2 |a^-2 - b^-2|^1 1.34 1.25
a^0 b^0 L^2 Q^2 P^2 F^-2 c^-2 |a^-2 - b^-2|^1 1.34 0.60
a^2 b^-2 L^2 Q^2 P^2 F^-2 c^-2 |a^ 2 - b^ 2|^-1 1.84 1.02
a^-2 b^-2 L^2 Q^2 P^2 F^-2 c^-2 |a^-2 - L^-2|^-1 0.79 1.22
a^-2 b^-2 L^2 Q^2 P^2 F^-2 c^-2 |a^ 2 - L^ 2|^1 1.16 0.79
a^2 b^2 L^-2 Q^2 P^2 F^-2 c^-2 |b^-2 - L^-2|^1 0.66 0.62
a^2 b^0 L^0 Q^2 P^2 F^-2 c^-2 |b^-2 - L^-2|^1 1.88 0.88
a^0 b^2 L^0 Q^2 P^2 F^-2 c^-2 |b^-2 - L^-2|^1 -0.57 0.83
a^0 b^0 L^2 Q^2 P^2 F^-2 c^-2 |b^-2 - L^-2|^1 0.65 0.51
a^-2 b^2 L^2 Q^2 P^2 F^-2 c^-2 |b^-2 - L^-2|^1 -1.79 1.19
a^-2 b^-2 L^2 Q^2 P^2 F^-2 c^-2 |b^ 2 - L^ 2|^1 -0.57 0.43
a^0 b^-2 L^0 Q^2 P^2 F^-2 c^-2 sqrt(a^-2 b^-2)^-1 1.59 1.13
a^-2 b^-2 L^2 Q^2 P^2 F^-2 c^-2 sqrt(a^-2 L^-2)^-1 0.98 0.91
a^0 b^-2 L^0 Q^2 P^2 F^-2 c^-2 sqrt(b^-2 L^-2)^-1 0.98 1.26
a^2 b^0 L^2 Q^2 P^2 F^-2 c^-2 (a^-2 * b^-2)^1 1.59 0.80
Checked : 1581201
Validated : 972
There are others (a^ 2 + L^ 2)^1 but that give even worse fits.
Those sqrt(1/x² + 1/y²) were already scanned for all reasonable formula of the form F = QP/c * term. They are not good fit so haven't showed through the sieve. Will explicitly add sqrt(1/x² + 1/y²) to scan with higher powers for the other terms F Q P and c but doubt this will show up more interesting things, as F = QP/c * term seems quite robust. Will also try to integrate the new datas on MiHsC ... please be patient, this is only part time !
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#2333 by frobnicat on 20 Oct, 2014 12:51
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Here is the PDF mentioned above.
http://www.emdrive.com/iac2014presentation.pdf
Is there a paper with more details to be published ?
"4 independent organisations, in 3 different countries" : what organisations ? with what kind of balance ? Someone present at the conference to take notes and give some context ?
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#2334 by Chrochne on 20 Oct, 2014 13:02
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Here is the PDF mentioned above.
http://www.emdrive.com/iac2014presentation.pdf
Is there a paper with more details to be published ?
"4 independent organisations, in 3 different countries" : what organisations ? with what kind of balance ? Someone present at the conference to take notes and give some context ?
...
I would be glad, if there are more details. Thats all there is for the update. Perhaps someone in this debate here have more information. Is there some video from EmDrive conference in Toronto?
By the way guys, do you have any news on the independent testing NASA mentiononed? Like testing at Glenn Research Center and others? Any idea when can se expect some results?
There was some rumor aroud that EmDrive was tested by Boeing and passed. Is that true?
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#2335 by frobnicat on 20 Oct, 2014 13:11
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Striking : interstellar probe page 10, terminal velocity 204429 km/s, thruster efficiency (next page) 0.31. How are we supposed to get this terminal velocity unless there is quite a high wet/dry mass ratio of burned mass ? To get to those speeds with burning less than a few % of the wet mass worth of energy would imply efficiencies much higher than 1. I mean, even if being on an asphalt road between Sun and Proxima and pushing with wheels.
Energy conservation all broken again !
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#2336 by frobnicat on 20 Oct, 2014 13:44
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Still shocked by interstellar probe. (page 10).
Assume around 10 tons wet mass spaceship, starting reactor full of brand new fissile material.
Say we burn instantly m=2 tons out of that and instantly use all that energy to give kinetic energy to the remaining M=8 tons (this is a favorable case as we don't have to accelerate part of the mass that will be burned later).
Ek = Mc²(gamma-1) = Enuclear = mc²
gamma = m/M +1 = 1.25
That is v = c sqrt(1-1/gamma²) = 180000 km/s, bit short of the 204000km/s predicted terminal velocity.
And that is by burning 2 tons of mass from a ship that weighs 10 tons (all included, payload, generators, shields, radiators...). 20% mass to energy conversion would be barely sufficient.
Fusion ? Just blind optimism ? Energy conservation broken ? Where does "efficiency of 0.31" comes from ?
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#2337 by Star One on 20 Oct, 2014 13:51
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Here is the PDF mentioned above.
http://www.emdrive.com/iac2014presentation.pdf
Is there a paper with more details to be published ?
"4 independent organisations, in 3 different countries" : what organisations ? with what kind of balance ? Someone present at the conference to take notes and give some context ?
That's all that is presented on their website at this time and is attached to the latest update. -
#2338 by JohnFornaro on 20 Oct, 2014 14:01
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Woah:
On page 7 of that PDF, they've already got a design for an all electric SSTO! They can't put that in a paper and post it on the intertubes unless it was true!
I didn't read any further, so I may have missed the fine print regarding ticket prices and carry on baggage.
Hey! Mucho thankso for the Hendrix clip!
Do you do yoga? After a weekend of log splitting, I'm looking forward to Hillary's class this PM at 5:30, Downtown. See ya there? -
#2339 by Star One on 20 Oct, 2014 14:23
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Woah:
On page 7 of that PDF, they've already got a design for an all electric SSTO! They can't put that in a paper and post it on the intertubes unless it was true!
I didn't read any further, so I may have missed the fine print regarding ticket prices and carry on baggage.
Hey! Mucho thankso for the Hendrix clip!
Do you do yoga? After a weekend of log splitting, I'm looking forward to Hillary's class this PM at 5:30, Downtown. See ya there?
I've seen similar criticism of LM after their fusion announcement last week but I believe they have committed to publish in peer reviewed journals later on. Maybe he intends to do the same for this?
.