Author Topic: EM Drive Developments - related to space flight applications - Thread 2  (Read 2106910 times)

Offline aero

  • Senior Member
  • *****
  • Posts: 2744
  • 92129
  • Liked: 705
  • Likes Given: 239
Quote
What relative magnetic permeability did you input into MEEP for copper? Did you input a value less than one?

Unfortunately I don't have a Drude model for copper at 2 GHz. We are cautioned in the meep literature that meep becomes unstable for magnetic permeability less than, or very much less than 1. I'm still using perfect metal and thick sheets at that. It's the computer resolution issue still, not enough memory and not enough CPU.

Thickness of 0.002 * height, or 0.002 * 9 inches = 0.018 inches is the best I can do in under a day of CPU time.

And FYI, I went ahead and calculated the probability of an electron tunnelling through the thin copper (I used Maxima) and found the probability to be 0.0. I guess no one is surprised by that result. That result could be off by a few orders of magnitude, but then what is 0E10? :)
« Last Edit: 02/27/2015 08:17 PM by aero »
Retired, working interesting problems

Online Rodal

  • Senior Member
  • *****
  • Posts: 5838
  • USA
  • Liked: 5919
  • Likes Given: 5267
Quote
What relative magnetic permeability did you input into MEEP for copper? Did you input a value less than one?

Unfortunately I don't have a Drude model for copper at 2 GHz. We are cautioned in the meep literature that meep becomes unstable for magnetic permeability less than, or very much less than 1. I'm still using perfect metal and thick sheets at that. It's the computer resolution issue still, not enough memory and not enough CPU.

Thickness of 0.002 * height, or 0.002 * 9 inches = 0.018 inches is the best I can do in under a day of CPU time.

And FYI, I went ahead and calculated the probability of an electron tunnelling through the thin copper (I used Maxima) and found the probability to be 0.0. I guess no one is surprised by that result. That result could be off by a few orders of magnitude, but then what is 0E10? :)
Is Maxima a descendant of Macsyma?  Are they competing with Mathematica and Maple ?
(I remember using Macsyma at MIT in the late 1970's  :)  )

Offline Mulletron

  • Full Member
  • ****
  • Posts: 1113
  • Liked: 776
  • Likes Given: 1013
It's to bad that we can't find a way that one of the little known or unknown solutions to Maxwell's equations can cause  a momentum.
There is obviously momentum in electromagnetic waves. There are also some little known electromagnetic effects that create torque.

http://arxiv.org/abs/0807.1310

Quoting @Aero
Quote
Unless of course it is surface electrons excited by the high power resonant RF, tunnelling through the 35 micron copper ends.
Quoting @JPLeRouzic
Quote
Yes, and many other things may happen were not tested nor even proposed. I wonder how people can know that a lot of energy is pumped in this device and imagine nothing will get out. At the very least thermal effects should happen. Testing it in (near) vacuum doesn't eliminate the thermal hypothesis. Even Pioneer's acceleration that was due to thermal effects after all: http://spectrum.ieee.org/aerospace/astrophysics/finding-the-source-of-the-pioneer-anomaly
Another thing that strikes me is that people search for a unique cause explaining everything, which is a bit unlikely.

One last thought: If a simulator shows results, build this device and publish results in a mainstream conference. Interesting things may happen ;-)

There are absolutely thermal effects. For every watt pumped in, you get that much heat out as IR. An RF dummy load is an efficient converter of RF to heat. The systemic effects slide with the dummy load attached is the control for heat. Thermal artifacts were a major area of exploration in both threads. Paul March provided lots of data to show that thermal effects were extensively studied over at Eagleworks. There's a thermal analysis on this thread, one of many indications that heat was controlled for: http://forum.nasaspaceflight.com/index.php?topic=36313.msg1326997#msg1326997 Plus the math that has been done numerous times for a photon rocket doesn't add up. So the vacuum did eliminate the thermal hypothesis.
« Last Edit: 02/27/2015 08:44 PM by Mulletron »
Challenge your preconceptions, or they will challenge you. - Velik

Offline Mulletron

  • Full Member
  • ****
  • Posts: 1113
  • Liked: 776
  • Likes Given: 1013
http://www.ebay.com/itm/AERCOM-Microwave-RF-Isolator-Circulator-2-4GHz-20dB-isolation-Low-I-L-TESTED-/281549538390?ssPageName=ADME:L:OU:US:1120
Picked up one of these puppies on Ebay to protect my amp. Another example of broken time reversal symmetry in action.

Got about an oz of very expensive liquid metal from here:
http://www.amazon.com/Gallium-Indium-Eutectic-GaInSn-68-5%25/dp/B00KN92MWW/ref=sr_1_3?ie=UTF8&qid=1425074693&sr=8-3&keywords=galinstan

So back to the copper from way back: http://forum.nasaspaceflight.com/index.php?topic=36313.msg1326742#msg1326742

I've determined that both the 10 and 16 mil are too thin to serve as frustum walls, thus they will become end caps.
https://docs.google.com/file/d/0B4PCfHCM1KYoZUs5dklaRFdoM3M/edit?pli=1

Been working with the supplier with a machine shop I posted about way back:
http://forum.nasaspaceflight.com/index.php?topic=36313.msg1326669#msg1326669
I'm going that route. The quote I got is: price: $120.00 layout + $51.63 for part + freight. So I have to pay the layout, then anyone else who wants one of these:
https://docs.google.com/file/d/0B4PCfHCM1KYoN2VURmltbVlfa3c/edit?pli=1
but built in 16oz copper, with a smooth butt seam inside, and 1/4" flange around edges, can get one for about 50 bucks plus shipping. If all this works out, it'll fulfill my goal of making a replication by DIYers easier. For me, paying the layout plus price about breaks even with buying the sheet myself and fumblefuddeling around trying to solder up a cone at home. So I'm happy. I'll get back with more later, when the items are at home.

Again thanks to @Paul March for providing the dimensions for his DUT over at Eagleworks:
http://forum.nasaspaceflight.com/index.php?topic=36313.msg1327467#msg1327467

Edit:
Links to pics on google drive keep disappearing, hopefully fixed it.
« Last Edit: 03/06/2015 09:40 AM by Mulletron »
Challenge your preconceptions, or they will challenge you. - Velik

Offline aero

  • Senior Member
  • *****
  • Posts: 2744
  • 92129
  • Liked: 705
  • Likes Given: 239
Quote
Is Maxima a descendant of Macsyma?  Are they competing with Mathematica and Maple ?
(I remember using Macsyma at MIT in the late 1970's  :)  )

Yes, Maxima is the direct descendent of Macsyma. No, I wouldn't say "competing" since Maxima is free, hence limited expert support. But yes, Maxima runs Mathematica code. I used Maxima to calculate the masses of the elementary particles using the Heim theory formulas and a .mac file that had been written for Mathematica. Google it. Overcoming the support question was easy once I found a quite good tutorial document for it, here.
http://math.stanford.edu/~paquin/MaximaBook.pdf
Lengthy at 245 pages, but seemingly comprehensive. Actually, I'm using wxmaxima, maxima with a windows like GUI. I found it pretty easy to use once I got past the "shift enter" used to give output; xmaxima installs simultaneously with wxmaxima. Further, it is free and easily installs on either Ubuntu or Windows. I did both just to see which was easier to use. They are the same.

Quote
Stainless Steel 304L (the material of the vacuum chamber) is weakly paramagnetic (the opposite of copper). What relative magnetic permeability did you input into MEEP for the StSt 304L for the chamber? Did you input a value greater than one? Did you report to us that value?  I don't recall.

Same answer as for copper. And I've given up on trying to evaluate evanescent wave coupling to the vacuum chamber. The results I got indicated that, "Yes, there is some coupling, but it is small." Greater than the calculated forces for the as tested "Copper Kettle" but orders of magnitude below the forces calculated for the simulated "pipe" model that I posted yesterday. Therefore, since we can't have one without the other, cavity/cavity coupling via evanescent waves may happen but not significantly.

 The pipe model was simulated in a space environment, vacuum with nothing near. I even moved the PML boundary layer away from the cavity. Bigger lattice, longer runs, but this way I'm confident that the PML is not absorbing or numerically reflecting the RF energy to significantly affect the calculations.

Edit Add: And note that the "pipes" are not really pipes, rather the cone body has been extended on both ends with a cylindrical section and the shaped base plates pressed into the cylindrical section. This is like a cork in a bottle, or more like the old fashion coffee can or peanut can lids that pressed into the end of the can while keeping the length of the resonant cavity as designed.
« Last Edit: 02/28/2015 12:22 AM by aero »
Retired, working interesting problems

Offline aero

  • Senior Member
  • *****
  • Posts: 2744
  • 92129
  • Liked: 705
  • Likes Given: 239
This is what I have for copper:

(define myCu (make dielectric (epsilon 1)
(polarizations
 (make polarizability
(omega 1e-20) (gamma 0.024197) (sigma 4.3873e+41))
(make polarizability
(omega 0.23471) (gamma 0.30488) (sigma 84.489))
(make polarizability
(omega 2.385) (gamma 0.85172) (sigma 1.395))
(make polarizability
(omega 4.2747) (gamma 2.5915) (sigma 3.0189))
(make polarizability
(omega 9.0173) (gamma 3.4722) (sigma 0.59868))
)))
;Additional Information
;Normalization length=1e-06 in meter
;Material_used_is_Cufrom Rakic et al.,Applied Optics (1998)
;Plasma Angular Frequency (and plasma wave vector,kp) in normalized units=6.6236


But this is for much higher frequency, THz and above, not even close to copper behavior at 2 GHz. And it is a Drude-Lorenz model. I don't think I need the extra terms, just the first two lines of data, but normalized at 3E-2, instead of 1E-6. I can normalize the data if I had Gamma and sigma for frequency = 2 GHz. And the first line of data, values for DC, shouldn't change, it's the second line...
Retired, working interesting problems

Online Rodal

  • Senior Member
  • *****
  • Posts: 5838
  • USA
  • Liked: 5919
  • Likes Given: 5267
...
I've determined that both the 10 and 16 mil are too thin to serve as frustum walls, thus they will become end caps.


Been working with the supplier with a machine shop I posted about way back:
http://forum.nasaspaceflight.com/index.php?topic=36313.msg1326669#msg1326669
...
Are you planning to conduct an experiment with a dielectric inside a cylindrical cavity?

Somebody should do so!

Such an experiment would disprove all the theories with formulas advanced so far.  All the formulas (Shawyer's, McCulloch and Notsosureofit...) give zero thrust if both diameters are the same.
« Last Edit: 02/27/2015 09:44 PM by Rodal »

Offline Mulletron

  • Full Member
  • ****
  • Posts: 1113
  • Liked: 776
  • Likes Given: 1013
...
I've determined that both the 10 and 16 mil are too thin to serve as frustum walls, thus they will become end caps.

Been working with the supplier with a machine shop I posted about way back:
http://forum.nasaspaceflight.com/index.php?topic=36313.msg1326669#msg1326669
...
Are you planning to conduct an experiment with a dielectric inside a cylindrical cavity?

Somebody should do so!

Such an experiment would disprove all the theories with formulas advanced so far.  All the formulas (Shawyer's, McCulloch and Notsosureofit...) give zero thrust if both diameters are the same.

Yes, a cylinder experiment absolutely has to be done.
Challenge your preconceptions, or they will challenge you. - Velik

Online Rodal

  • Senior Member
  • *****
  • Posts: 5838
  • USA
  • Liked: 5919
  • Likes Given: 5267
Notsosureofit:

Your equation gives zero thrust for a cylinder (constant diameter along its length). Thus, according to your formula a cylinder will give no thrust, only a geometrical object with decreasing diameter will (ditto for Shawyer's and McCulloch's)

But your formula does not explicitly include a dielectric.

QUESTION: would your line of thinking also give zero thrust for a cylinder with an inserted dielectric at one end of the cavity with constant, homogeneous, isotropic dielectric properties ?  (With no nonlinearities)

Offline Notsosureofit

  • Full Member
  • ****
  • Posts: 656
  • Liked: 704
  • Likes Given: 1364
Notsosureofit:

Your equation gives zero thrust for a cylinder (constant diameter along its length). Thus, according to your formula a cylinder will give no thrust, only a geometrical object with decreasing diameter will (ditto for Shawyer's and McCulloch's)

But your formula does not explicitly include a dielectric.

QUESTION: would your line of thinking also give zero thrust for a cylinder with an inserted dielectric at one end of the cavity with constant, homogeneous, isotropic dielectric properties ?  (With no nonlinearities)

Presumably no, as long as you have frequency dispersion along the axis you should predict a force.

Edit: But it adds complication to the formula in that the second term doesn't cancel out.
« Last Edit: 02/27/2015 10:24 PM by Notsosureofit »

Offline aero

  • Senior Member
  • *****
  • Posts: 2744
  • 92129
  • Liked: 705
  • Likes Given: 239
...
I've determined that both the 10 and 16 mil are too thin to serve as frustum walls, thus they will become end caps.

Been working with the supplier with a machine shop I posted about way back:
http://forum.nasaspaceflight.com/index.php?topic=36313.msg1326669#msg1326669
...
Are you planning to conduct an experiment with a dielectric inside a cylindrical cavity?

Somebody should do so!

Such an experiment would disprove all the theories with formulas advanced so far.  All the formulas (Shawyer's, McCulloch and Notsosureofit...) give zero thrust if both diameters are the same.

Yes, a cylinder experiment absolutely has to be done.

Hey - Good! I've ran some meep numbers with a cylindrical cavity with good results. Let me run some fresh numbers and I'll post what I find. If you have parameters in mind let me know, then I can run your numbers.
Retired, working interesting problems

Online Rodal

  • Senior Member
  • *****
  • Posts: 5838
  • USA
  • Liked: 5919
  • Likes Given: 5267
Notsosureofit:

Your equation gives zero thrust for a cylinder (constant diameter along its length). Thus, according to your formula a cylinder will give no thrust, only a geometrical object with decreasing diameter will (ditto for Shawyer's and McCulloch's)

But your formula does not explicitly include a dielectric.

QUESTION: would your line of thinking also give zero thrust for a cylinder with an inserted dielectric at one end of the cavity with constant, homogeneous, isotropic dielectric properties ?  (With no nonlinearities)

Presumably no, as long as you have frequency dispersion along the axis you should predict a force.

Edit: But it adds complication to the formula in that the second term doesn't cancel out.

Since your existing formula:

NT = P*Q*(1/(4*pi*L*f^3))*(c/(2*pi))^2*X^2*((1/Rs^2)-(1/Rb^2))

gives NT=0 (zero thrust force) for Rs=Rb (a cylinder), just like Shawyer's and McCulloch's also go to zero for that case, then my understanding from your reply is that this formula is a simplification and that there are terms you neglected that do not go to zero for a cylinder.

Do you know the 2nd order term that will still be finite for Rs=Rb (a cylinder) ?   

Or is the issue deriving an equation for the dispersion including the dielectric?

« Last Edit: 02/28/2015 12:30 AM by Rodal »

Offline flux_capacitor

  • Full Member
  • ****
  • Posts: 552
  • France
  • Liked: 654
  • Likes Given: 909
Are you planning to conduct an experiment with a dielectric inside a cylindrical cavity?

Somebody should do so!

Such an experiment would disprove all the theories with formulas advanced so far.  All the formulas (Shawyer's, McCulloch and Notsosureofit...) give zero thrust if both diameters are the same.

Before removing the internal dielectric, Shawyer considered a cylindrical EmDrive, in which a conical dielectric would be mandatory for thrust. See his 1990 UK Patent Application GB2229865 "Electrical Propulsion Unit for Spacecraft", attached below.
« Last Edit: 02/28/2015 12:27 AM by flux_capacitor »

Offline Notsosureofit

  • Full Member
  • ****
  • Posts: 656
  • Liked: 704
  • Likes Given: 1364
Notsosureofit:

Your equation gives zero thrust for a cylinder (constant diameter along its length). Thus, according to your formula a cylinder will give no thrust, only a geometrical object with decreasing diameter will (ditto for Shawyer's and McCulloch's)

But your formula does not explicitly include a dielectric.

QUESTION: would your line of thinking also give zero thrust for a cylinder with an inserted dielectric at one end of the cavity with constant, homogeneous, isotropic dielectric properties ?  (With no nonlinearities)

Presumably no, as long as you have frequency dispersion along the axis you should predict a force.

Edit: But it adds complication to the formula in that the second term doesn't cancel out.

Since your existing formula:

NT = P*Q*(1/(4*pi*L*f^3))*(c/(2*pi))^2*X^2*((1/Rs^2)-(1/Rb^2))

gives NT=0 (zero thrust force) for Rs=Rb (a cylinder), just like Shawyer's and McCulloch's also go to zero for that case, then my understanding from your reply is that this formula is a simplification and that there are terms you neglected that do not go to zero for a cylinder.

Do you know the 2nd order term that will still be finite for Rs=Rb (a cylinder) ?   

Or is the issue deriving an equation for the dispersion including the dielectric?

You need to start back at the dispersion formula and follow the same steps. (sorry, I just got back)

Offline DIYFAN

  • Member
  • Posts: 48
  • Liked: 28
  • Likes Given: 149
Since there is really no proven theory of how would an EM Drive generate thrust in space, thereby apparently violating the law of conservation of momentum, and we are still debating whether the experimental measurements are an artifact, the best way to cut to the chase is to test an EM Drive made of Aluminum, and to also test an iron (or a material coated with an interior thin film of Metglas) for the big flat end.

As much as I applaud any and all experimentalists who are dedicating time, often with little compensation, part of the problem with the Eagleworks experimental results as they compare to those reported elsewhere (Shawyer and the Chinese) is the relatively small response signal.  Not only does this situation threaten the future of the experiment in the U.S. as other labs might refuse to attempt replications, this situation also leads to in-depth discussions about thermal effects, evanescent waves, and other unlikely explanations for the effect.  If the signal response can be amplified and made easily repeatable, more focus could then be placed on improving the effect, and eventually incorporating it into real world applications.  I fully agree, therefore, with Dr. Rodal that an experiment that tests an iron (or a material coated with an interior thin film of Metglas) for the big flat end should be made of highest priority.  Dr. Aquino proposed a theory of why this would amplify the effect.  It ought to be tested as soon as possible, and preferably in the U.S.
« Last Edit: 02/28/2015 01:11 AM by DIYFAN »

Offline ThinkerX

  • Full Member
  • ***
  • Posts: 303
  • Alaska
  • Liked: 113
  • Likes Given: 59
Seems like wherever we go with this project, Shawyer has already been there - and like as not, built a test model.   Given that, is there any way to bring his reasoning/math in line with conventional physics, or do they remain fundamentally irreconcilable?  I keep getting the impression he's on to something major, but consistently fails to communicate that 'something' in a coherent manner. 

Online Rodal

  • Senior Member
  • *****
  • Posts: 5838
  • USA
  • Liked: 5919
  • Likes Given: 5267
Are you planning to conduct an experiment with a dielectric inside a cylindrical cavity?

Somebody should do so!

Such an experiment would disprove all the theories with formulas advanced so far.  All the formulas (Shawyer's, McCulloch and Notsosureofit...) give zero thrust if both diameters are the same.

Before removing the internal dielectric, Shawyer considered a cylindrical EmDrive, in which a conical dielectric would be mandatory for thrust. See his 1990 UK Patent Application GB2229865 "Electrical Propulsion Unit for Spacecraft", attached below.

Great find and thanks  :)  for bringing it to our attention, this is the first time I see this patent.



Very thought provoking, and interesting in its own right.

I note that Shawyer's patent was filed in 1988 !, that's 27 years ago !

This man has been working at this for a long, long time



Now, on a separate matter:

I observe that although the outside metal cavity is indeed cylindrical, it is really not a cylindrical cavity for resonance purposes because the dielectric inside it is conical in shape, so it is a case again of a cavity with non-uniform inner dimensions along its axial length, for resonance purposes.  Shawyer and McCulloch would still calculate "thrust" for this type of cavity.

What I had in mind for people to test is whether

a cylindrical cavity

                               A) empty
                               or
                               B) with a cylindrical dielectric inside at one end (the dielectric OD having the same dimensions as                                     the cavity ID)

will exhibit thrust.

My understanding from Paul March's experiments is that A) will not produce any thrust (as NASA did not get thrust with the empty cavity).

My (rough) reading of the papers brought up by @Mulletron would say B) may produce thrust.  My reading of Shawyer's experience is that B) will not produce significant thrust (because Shawyer has gone through the trouble of using conical dielectrics inside cylindrical cavities, and tapered cones with progressively larger cone angles).   The above patent reinforces this point of view: Shawyer must have surely tested cylindrical cavities with cylindrical dielectrics fitting inside it.  Shawyer must be pursuing the conical shape (with increasing cone angles) for a good reason.
« Last Edit: 02/28/2015 02:02 PM by Rodal »

Offline Notsosureofit

  • Full Member
  • ****
  • Posts: 656
  • Liked: 704
  • Likes Given: 1364
Just catching up.  Yes that dispersion should give thrust (Shawyer !)  I get a formula (scribbled at the moment) that depends on the difference in index of the tapered dielectric (and the equivalent length which doesn't look like it will cancel things out, so far)

Probably can be simplified a bit but not tonight........(mode dependence is still in there)
« Last Edit: 02/28/2015 01:44 AM by Notsosureofit »

Online Rodal

  • Senior Member
  • *****
  • Posts: 5838
  • USA
  • Liked: 5919
  • Likes Given: 5267
Seems like wherever we go with this project, Shawyer has already been there - and like as not, built a test model.   Given that, is there any way to bring his reasoning/math in line with conventional physics, or do they remain fundamentally irreconcilable?  I keep getting the impression he's on to something major, but consistently fails to communicate that 'something' in a coherent manner.
Well, Edison was one of the greatest inventors.  He was completely self-taught (with visits to the Cooper Union). Edison was born only 16 years after Maxwell.  Edison was responsible for some of the greatest electromagnetic inventions ever, yet he was not a theoretical physicist like Maxwell or Hertz. Who was the inventor? Edison or Maxwell?.    Why should Shawyer  be the one expected to explain the physics of why it works (if it does work)?
« Last Edit: 02/28/2015 02:01 AM by Rodal »

Offline zen-in

  • Full Member
  • ****
  • Posts: 531
  • California
  • Liked: 468
  • Likes Given: 365
http://www.ebay.com/itm/AERCOM-Microwave-RF-Isolator-Circulator-2-4GHz-20dB-isolation-Low-I-L-TESTED-/281549538390?ssPageName=ADME:L:OU:US:1120
Picked up one of these puppies on Ebay to protect my amp. Another example of broken time reversal symmetry in action.

Got about an oz of very expensive liquid metal from here:
http://www.amazon.com/Gallium-Indium-Eutectic-GaInSn-68-5%25/dp/B00KN92MWW/ref=sr_1_3?ie=UTF8&qid=1425074693&sr=8-3&keywords=galinstan

So back to the copper from way back: http://forum.nasaspaceflight.com/index.php?topic=36313.msg1326742#msg1326742

I've determined that both the 10 and 16 mil are too thin to serve as frustum walls, thus they will become end caps.


Been working with the supplier with a machine shop I posted about way back:
http://forum.nasaspaceflight.com/index.php?topic=36313.msg1326669#msg1326669
I'm going that route. The quote I got is: price: $120.00 layout + $51.63 for part + freight. So I have to pay the layout, then anyone else who wants one of these:

but built in 16oz copper, with a smooth butt seam inside, and 1/4" flange around edges, can get one for about 50 bucks plus shipping. If all this works out, it'll fulfill my goal of making a replication by DIYers easier. For me, paying the layout plus price about breaks even with buying the sheet myself and fumblefuddeling around trying to solder up a cone at home. So I'm happy. I'll get back with more later, when the items are at home.

Again thanks to @Paul March for providing the dimensions for his DUT over at Eagleworks:
http://forum.nasaspaceflight.com/index.php?topic=36313.msg1327467#msg1327467

You should see if you can find a 2 meter reject cavity.   The can is 3 feet high by over a foot dia and made from 16 or 18 gauge Copper.   There is a seam but you would want to cut an angle on each side of it so you can form it into a cone.   I think the Eagleworks people used a form for bending their cavity; or they had a sheet metal shop do the work.     Getting a sheet of Copper big enough to make the cone part from one piece can be expensive.

Tags: