Author Topic: EM Drive Developments - related to space flight applications - Thread 2  (Read 2164083 times)

Offline hhexo

  • Member
  • Posts: 31
  • An Italian in the UK
  • Liked: 24
  • Likes Given: 1
Thanks. Yes, you are right. It is the next step to fill with numbers and put some plots to see the orders of magnitude. I am not sure yet that I am able to account for a macroscopic effect.
...
Even if the thrust was not macroscopic, ...
...

Ok, I've run some back-of-the-envelope calculations in pure engineer style, i.e. just looking at the orders of magnitude.

I am handwaving constants as follows:

pi^2 = 10^1
G = 10^-11
c^4 = 81 * 10^32 = 10^34
mu0^2 = 16 * pi^2 * 10^-14 = 10^-12
h = 1 (this is the height of the frustum, right? Not Planck!)

and I get that the geometry related effects have to counter-balance a term which is 10^-32 * U0^4.

At this point we are left with a term (1 / ln2(r2/r1)) * (r2^4 * XXX - (r2^6 / r1^2) * YYY), where XXX and YYY are those complicated expressions for the fourth power of the magnetic field that I can't rewrite here. :) But, out of the two terms the second one is clearly dominant if r2 >> r1.

So we're left with 10^-32 * U0^4 * YYY * (r2^6 / r1^2) * (1 / ln2(r2/r1)).

I can't quite follow what the order of magnitude of U0 and YYY would be. Can anybody help me? Are they going to be significant powers of ten?

Anyway, if we just want to balance that 10^-32, we need something like r2/r1 = 10^5.5, for example:
r2 = 10^2.5 = several hundred meters
r1 = 10^-3 = one millimeter
given a frustum height of just one meter... Things get better if h is small, but not too much better.

Not very encouraging in terms of practicality. :( And if U0 or YYY are going to be negative powers of ten, it's going to get even worse.
Damn gravity being such a weak force! :)

Online WarpTech

  • Full Member
  • ****
  • Posts: 1301
  • Do it!
  • Vista, CA
  • Liked: 1350
  • Likes Given: 1813
But what happens when a wave is attenuated in a perfectly conducting circular waveguide? That energy is not lost as "heat" because there is no resistance to dissipate it.

How will the wave attenuate (lose energy) if no wall losses?

My understanding is that, if there are no resistive losses the energy will be stored inductively as circulating current flowing in the frustum. Where else can it go? Once there is current flowing in the copper, it will repel any change in magnetic flux attempting to damp it.

Offline RERT

I've been reading for several days, I have a question regarding the Maxwell Equations simulations of the Frustrum.

I recently read a paper by Tuval and Yahalom (2013). They note that Maxwell's equations are conservative of momentum if the action of forces is taken to be instantaneous. However, if the forces are taken to be propagating at c they are not, in general, for particles. [The momentum of the fields is needed to make conservation of momentum work.]

So, the question is: do the simulations of the frustrum account for the retarded forces and fields on and created by the moving charges in the walls/end caps?

R.

Offline deltaMass

  • Full Member
  • ****
  • Posts: 955
  • A Brit in California
  • Liked: 671
  • Likes Given: 275
It's a smart question but in a closed cavity like this, overall conservation is to be expected irrespective of whether the solution uses retarded or instantaneous wave solutions, I think.
« Last Edit: 05/20/2015 05:26 PM by deltaMass »

Offline Rodal

  • Senior Member
  • *****
  • Posts: 5895
  • USA
  • Liked: 6045
  • Likes Given: 5325
I've been reading for several days, I have a question regarding the Maxwell Equations simulations of the Frustrum.

I recently read a paper by Tuval and Yahalom (2013). They note that Maxwell's equations are conservative of momentum if the action of forces is taken to be instantaneous. However, if the forces are taken to be propagating at c they are not, in general, for particles. [The momentum of the fields is needed to make conservation of momentum work.]

So, the question is: do the simulations of the frustrum account for the retarded forces and fields on and created by the moving charges in the walls/end caps?

R.

No, the simulations (by COMSOL FEA and the exact solution of Greg Egan) does not consider the retarded fields.  It does not consider Jefimenko's equations for example.  The right hand side of Jefimenko's equatons involve a "retarded" time which reflects the "causality" of the expressions. In other words, the left side of each equation is actually "caused" by the right side, unlike the normal differential expressions for Maxwell's equations where both sides take place simultaneously.




Are you referring to this paper ?

Newton's Third Law in the Framework of Special Relativity 
Miron Tuval, Asher Yahalom
(Submitted on 26 Jan 2013) http://arxiv.org/abs/1302.2537
« Last Edit: 05/20/2015 05:42 PM by Rodal »

Offline StrongGR

Thanks. Yes, you are right. It is the next step to fill with numbers and put some plots to see the orders of magnitude. I am not sure yet that I am able to account for a macroscopic effect.
...
Even if the thrust was not macroscopic, ...
...

Ok, I've run some back-of-the-envelope calculations in pure engineer style, i.e. just looking at the orders of magnitude.

I am handwaving constants as follows:

pi^2 = 10^1
G = 10^-11
c^4 = 81 * 10^32 = 10^34
mu0^2 = 16 * pi^2 * 10^-14 = 10^-12
h = 1 (this is the height of the frustum, right? Not Planck!)

and I get that the geometry related effects have to counter-balance a term which is 10^-32 * U0^4.

At this point we are left with a term (1 / ln2(r2/r1)) * (r2^4 * XXX - (r2^6 / r1^2) * YYY), where XXX and YYY are those complicated expressions for the fourth power of the magnetic field that I can't rewrite here. :) But, out of the two terms the second one is clearly dominant if r2 >> r1.

So we're left with 10^-32 * U0^4 * YYY * (r2^6 / r1^2) * (1 / ln2(r2/r1)).

I can't quite follow what the order of magnitude of U0 and YYY would be. Can anybody help me? Are they going to be significant powers of ten?

Anyway, if we just want to balance that 10^-32, we need something like r2/r1 = 10^5.5, for example:
r2 = 10^2.5 = several hundred meters
r1 = 10^-3 = one millimeter
given a frustum height of just one meter... Things get better if h is small, but not too much better.

Not very encouraging in terms of practicality. :( And if U0 or YYY are going to be negative powers of ten, it's going to get even worse.
Damn gravity being such a weak force! :)

You got it. A cone seems a better choice.
« Last Edit: 05/20/2015 05:32 PM by StrongGR »

Offline hhexo

  • Member
  • Posts: 31
  • An Italian in the UK
  • Liked: 24
  • Likes Given: 1


Anyway, if we just want to balance that 10^-32, we need something like r2/r1 = 10^5.5, for example:
r2 = 10^2.5 = several hundred meters
r1 = 10^-3 = one millimeter
given a frustum height of just one meter...

Gah! Sorry, big glaring error there in my last passage. :)
Because it's r2^6 / r1^2, I can't just consider r2/r1 and scale the ratio.

r2=10^5.5
r1=1
is a correct example.

Offline StrongGR



Anyway, if we just want to balance that 10^-32, we need something like r2/r1 = 10^5.5, for example:
r2 = 10^2.5 = several hundred meters
r1 = 10^-3 = one millimeter
given a frustum height of just one meter...

Gah! Sorry, big glaring error there in my last passage. :)
Because it's r2^6 / r1^2, I can't just consider r2/r1 and scale the ratio.

r2=10^5.5
r1=1
is a correct example.

It is my approximation: r2 increasingly large and r1 decreasing toward zero otherwise one should change the final formula. So, in your example it would be better to have r2=10 m and r1=10^-3 m and so on. This is a cone. Another way around, as suggested above, is to fill with some material the cavity. The formula goes like mur^-2 and mur much smaller than 1.

Offline deltaMass

  • Full Member
  • ****
  • Posts: 955
  • A Brit in California
  • Liked: 671
  • Likes Given: 275
Yeah, and this is why some realistic numbers would help get a handle on this.

Offline SeeShells

  • Senior Member
  • *****
  • Posts: 2335
  • Every action there's a reaction we try to grasp.
  • United States
  • Liked: 2982
  • Likes Given: 2601
...
Reading it I had a question (more than one but..). And this is for everyone, why did Eagle Works observe no thrust in a EM device with no HDPE insert? Makes me wonder what effect achiral materials like this that can induce chirality would have with relativity and electromagnetic fields in your equations?

Thanks nice work!
Excellent question that has been puzzling me for a long time as well.  To examine this, first let's recapitulate the statement in Brady et.al.' report:

Quote
There appears to be a clear dependency between thrust magnitude and the presence of some sort of dielectric RF resonator in the thrust chamber. The geometry, location, and material properties of this resonator must be evaluated using numerous COMSOLŪ iterations to arrive at a viable thruster solution. We performed some very early evaluations without the dielectric resonator (TE012 mode at 2168 MHz, with power levels up to ~30 watts) and measured no significant net thrust

My exact solution for the truncated cone, given the dimensions reported by Paul March and also used in NASA's COMSOL FEA analysis, for mode shape TE012, gives me a natural frequency of:

2.2024 GHz

So they operated at frequency of 2.168 GHz which is 1.59% away from the frequency given by the exact solution. Perhaps the reason was that that was the frequency given by the Finite Element analysis for mode TE012 (let's recall that the Finite Element solution converges from below to the correct eigensolution, and only for an infinite Finite Element mesh one can theoretically converge to the eigensolution).  So, they were looking in the right frequency range for TE012.

My recollection is that Paul March thought that the reason maybe that to generate thrust without a dielectric one needs to provide Amplitude, Frequency and Phase Modulation, and that at the time of the experiments detailed in the NASA report ( http://www.libertariannews.org/wp-content/uploads/2014/07/AnomalousThrustProductionFromanRFTestDevice-BradyEtAl.pdf ), they were not providing the needed AM, FM and PM modulation.  I was not able to find the exact quotation (I wonder why, given the fantastic search function we have at this site  ;) ).  I was only able to find this quotation from Paul March, in answer to this question:

Quote from: Star-Drive
(Shawyer and the Chinese used the magnetron excited TE012 mode in their frustum cavities without dielectrics being present.)
Thanks for digging it out, I know you're more savvy with the search function than I am. I keep on coming back to this paper and this statement. Yes and I know the effects are very very small.
http://iopscience.iop.org/0295-5075/93/4/41002/fulltext/epl_93_4_41002.html
<An important point to appreciate here is that the permeability has two distinct contributions; a reciprocal part (μab(x, ω) = μba(x, ω)) which is just the standard vacuum contribution and a non-reciprocal part (μab(x, ω) ≠ μba(x, ω)) due to the external magnetic field interacting with the magnetic polarisability of the particles. The non-reciprocal piece breaks time-reversal invariance and results from the Faraday effect in magnetic media. Although the non-reciprocal parts arise from the point interaction where the external magnetic field couples to the scattering object, this is enough to require the correlation functions and thereby the Green tensors, to take a different form. Note here that we are choosing to work with the momentum density associated with the canonical energy momentum tensor rather than the Poynting vector; the latter is expected to integrate to zero [9] so a permeability is required to distinguish the two vectors. Using the expectation values of the noise currents at zero temperature, we find for the two-point function of the physical.

and

one usually calculates an energy or a force starting from the energy-momentum tensor of electromagnetism. Of course the measurable observable, that is the energy obtained by integration over all space, is but one component of a full tensor of observables. In particular, the time-space components are the associated momentum density of the fields which upon integration over all space gives the total momentum contained in the field [7]. In the case of a single parallel plate, it is precisely because the incident vacuum photons from the left and right tend to cancel that the plate does not develop a net momentum. This preserves the translational invariance of the vacuum. A natural question then to ask is how to find a system whereby a momentum can be displayed.>

I honestly think we have a "physical" system in the EM Drive if we use magneto-chiral matter (HDPE is a weak one in the way the carbon atoms are layered). REF: http://www.nature.com/nmat/journal/v7/n9/full/nmat2256.html

Offline Rodal

  • Senior Member
  • *****
  • Posts: 5895
  • USA
  • Liked: 6045
  • Likes Given: 5325
...
I honestly think we have a "physical" system in the EM Drive if we use magneto-chiral matter (HDPE is a weak one in the way the carbon atoms are layered). REF: http://www.nature.com/nmat/journal/v7/n9/full/nmat2256.html
Having worked with polymers for decades, I agree that the extruded HDPE and PTFE (from McMaster Carr) that NASA Eagleworks used as dielectrics would be quite a stretch to be considered "strongly magneto-chiral", and be responsible for the measured thrust forces.  Also, (I have not read those papers in a long time, so please point out whether I'm wrong in my assessment, which I'll appreciate to know) my recollection is that they had to resort to a 4th order term in a perturbation series of the nonlinear problem to get the term that they rely on for the claimed effect.  My recollection is that they do not support what is the range of validity for that 4th order term to be significant, and it is also my recollection that the authors do not prove that the series in question (is it an asymptotic series ? ) is convergent and therefore whether the neglected higher order terms sum do not overwhelm the 4th order term they consider...

It looks like the magneto-chiral effect is very interesting for R&D but I have a difficulty embracing it as an explanation when the EM Drive's with the highest claimed thrust forces (Cannae superconducting. Shawyer and Prof Yang) did not use a polymer dielectric.
« Last Edit: 05/20/2015 06:19 PM by Rodal »

Offline deltaMass

  • Full Member
  • ****
  • Posts: 955
  • A Brit in California
  • Liked: 671
  • Likes Given: 275
Quote
Note here that we are choosing to work with the momentum density associated with the canonical energy momentum tensor rather than the Poynting vector; the latter is expected to integrate to zero
Doesn't that imply zero net thrust?

Offline aero

  • Senior Member
  • *****
  • Posts: 2784
  • 92129
  • Liked: 724
  • Likes Given: 249
As promised I am posting the latest draft about general relativity and electromagnetic field. The relevant conclusion is that there is thrust. Thanks to the comments by Jose Rodal, it can be shown that this can be meaningful and the best geometry is that of the frustum tending to a cone. There is no violation of conservation law due to the presence of the gravity that can escape the device producing a reaction.

I post it here for your comments that are very welcome as usual. You can find the final equation at page 12 for your evaluations. Later on, I will post a version with a somewhat different presentation to arxiv.

I like you paper but have a request for a point of clarification. I got really confused when you went from equation 55 to equation 56. Why did you invert the c4/G term? (Actually I got confused way before that but I won't go there. :)
Retired, working interesting problems

Offline Rodal

  • Senior Member
  • *****
  • Posts: 5895
  • USA
  • Liked: 6045
  • Likes Given: 5325
Quote
Note here that we are choosing to work with the momentum density associated with the canonical energy momentum tensor rather than the Poynting vector; the latter is expected to integrate to zero
Doesn't that imply zero net thrust?
Certainly yes for a linear isotropic system, but apparently not for an anisotropic nonlinear system as the one they are considering.  They are considering a 4th order term.
« Last Edit: 05/20/2015 06:31 PM by Rodal »

Offline hhexo

  • Member
  • Posts: 31
  • An Italian in the UK
  • Liked: 24
  • Likes Given: 1
r2=10^5.5
r1=1
is a correct example.

It is my approximation: r2 increasingly large and r1 decreasing toward zero otherwise one should change the final formula. So, in your example it would be better to have r2=10 m and r1=10^-3 m and so on. This is a cone.

Yes. But the dependence on r2 is stronger than the one on r1, because r2 is to the sixth power whereas r1 is just squared.
So: if you keep r1 constant and vary r2, you don't need many orders of magnitude to have a reasonable effect; if you keep r2 constant and vary r1, you need to go down a lot of orders of magnitude.

If we have an r2 of one meter, r1 must be 10^-16, which is a tenth of a femtometer, which is less than the size of a proton. We cannot possibly manufacture such a cone and I think it would not work anyway because the resulting "small end plate" doesn't have enough electrons to behave like an ideal EM-reflecting plate. Although maybe it doesn't need to and it's just the cavity shape that matters.

What if we considered other shapes? We started with the truncated cone because of the EmDrive, but what if we considered a pillbox cavity (which is basically a rounded cone...)? All the equations would be different, for sure, but it might be worth exploring. Maybe in another thread. :)

Another way around, as suggested above, is to fill with some material the cavity. The formula goes like mur^-2 and mur much smaller than 1.

Yes, varying mu may be profitable too.

Offline SeeShells

  • Senior Member
  • *****
  • Posts: 2335
  • Every action there's a reaction we try to grasp.
  • United States
  • Liked: 2982
  • Likes Given: 2601
Quote
Note here that we are choosing to work with the momentum density associated with the canonical energy momentum tensor rather than the Poynting vector; the latter is expected to integrate to zero
Doesn't that imply zero net thrust?
Certainly yes for a linear isotropic system, but apparently not for an anisotropic nonlinear system as the one they are considering.  They are considering a 4th order term.
I see it!
Thank you to taking the time to answer and helping me by answering some questions that have a hard time going away. I guess I'm like a old dog with a bone, a good and bad thing. If you do any tests in the future on why that anomaly reared its head with the insertion of the HTPE I'd love to know what caused it. You would hate to see another anomaly pop up. You're dealing with a pretty big one right now ... thrust in a Can.

Offline Rodal

  • Senior Member
  • *****
  • Posts: 5895
  • USA
  • Liked: 6045
  • Likes Given: 5325
r2=10^5.5
r1=1
is a correct example.

It is my approximation: r2 increasingly large and r1 decreasing toward zero otherwise one should change the final formula. So, in your example it would be better to have r2=10 m and r1=10^-3 m and so on. This is a cone.

...

What if we considered other shapes? We started with the truncated cone because of the EmDrive, but what if we considered a pillbox cavity (which is basically a rounded cone...)? All the equations would be different, for sure, but it might be worth exploring. Maybe in another thread. :)

...
Marco's present equations in his present papers are general enough to already include (in the limit for cone angle approaching zero) a cylindrical pillbox cavity as the one used by NASA in their actual experiments.  We (in a previous post) actually considered such a shape, and it is clearly inferior to a pointy cone.  The action takes place towards the apex of the cone.  The pointy conical geometry is important for this effect.  Notice that there is a singularity at a cone's vertex, we need some help like that from geometry, that I don't see in nice rounded shapes.  Considering a material with different permeability I think that a material discontinuity interface (providing a jump rather than a gradual variation) is also helpful in this regards :)
« Last Edit: 05/20/2015 07:02 PM by Rodal »

Offline squid

  • Member
  • Posts: 15
  • USA
  • Liked: 37
  • Likes Given: 2
Consider the typical SRF cavities described in this paper: http://arxiv.org/pdf/physics/0003011.pdf

Each cell in the stack of cavities could be thought of as being two cones arranged back to back: <><><><> (see Figure 5). We would then expect any forces to try and pull the two half-cavities apart.

These were operated at 1.3 GHz, 200 kW CW RF, with a measured Q of ~ 5e9.

Applying the quoted formula of F ~ Po*Q/c, we get a force of 3e6 N.

The cavities have a wall thickness of about 2 mm, and a major radius of ~100 mm, giving a strain of > 2000 MPa in the niobium. The yield strength of niobium is somewhere in the range of 80-150 MPa, depending on temper and annealing. And yet the cavities did not fly apart.

Anyone care to poke holes in this?
« Last Edit: 05/20/2015 06:57 PM by squid »

Offline phaseshift

  • Full Member
  • *
  • Posts: 104
  • Seattle, WA
  • Liked: 84
  • Likes Given: 97
I'm incredibly appreciate of all the work that went into the dimensional estimates of the various drives but I'm having some difficulty understanding where the dimensional values came from regarding the attached image:

Shawyer reported a large plate diameter of .28m - so using that as a given.

Going straight off the image (not taking into account perspective) at the junction of the cone and the cylinder I get about .169m for the small plate diameter. This is a difference of 40mm with what was originally estimated and about 73mm from Dr. Rodal's calculation (96.13) of the small plate diameter.  This is an enormous difference and I can't believe it. I'm going to model this and lay it on top of the photo in perspective so I can find out what the numbers are closer to.

Also the cone length was estimated to be .345m - which is greater than the large plate diameter. Between which two points is this measurement for?





« Last Edit: 05/20/2015 07:17 PM by phaseshift »
"It doesn't have to be a brain storm, a drizzle will often do" - phaseshift

Offline Rodal

  • Senior Member
  • *****
  • Posts: 5895
  • USA
  • Liked: 6045
  • Likes Given: 5325
I'm incredibly appreciate of all the work that went into the dimensional estimates of the various drives but I'm having some difficulty understanding where the dimensional values came from regarding the attached image:

Shawyer reported a large plate diameter of .28m - so using that as a given.

Going straight off the image (not taking into account perspective) at the junction of the cone and the cylinder I get about .169m for the small plate diameter. This is a difference of 40mm with what was originally estimated and about 73mm from Dr. Rodal's calculation (96.13) of the small plate diameter.  This is an enormous difference and I can't believe it. I'm going to model this and lay it on top of the photo in perspective so I can find out what the numbers are closer to.

Also the cone length was estimated to be .345m - which is larger than the plate diameter. Between which two points is this measurement for?
I can't answer that beyond hypotheticals like 1) Shawyer's paper may have a typo in the numbers (a small difference in the DesignFactor makes a big difference in the dimensions for example), 2) TheTraveller's latest interpretation of the DesignFactor may still not be Shawyer's DesignFactor (which I repeat was not explicitly defined by Shawyer in his papers), 3) the picture you are using may not represent what Shawyer meant by the Demonstration engine in his paper, 4) the "length" is arbitrary for this design because the location of the internal movable wall at the small end is unknown (I don't know whether the guesstimator used the maximum length for example or the average length or the minimum length) etc. etc. 
« Last Edit: 05/20/2015 07:28 PM by Rodal »

Tags: