Author Topic: EM Drive Developments - related to space flight applications - Thread 3  (Read 1805608 times)

Offline SeeShells

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Just another point of attention.

The resulting electromagnetic force, with the fields confined inside cavity, calculated  from the poyting vector, is F=-(1/c^2).d(integral_vol(S))/dt, where S=ExH
So a constant, at principle, make no difference to resulting force on whole cavity.
...

I think you are overlooking the point I brought up earlier, that the DC magnetic flux can pass through copper. There is no skin effect for the DC offset, so it makes this an open system. Magnetic flux can carry momentum and energy "out" of the frustum and allow it to escape. For DC currents flowing in the copper, the frustum is wide open.
Todd
Hmmm how about perforated copper sheeting? ;)

Offline aero

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Shawyers frustum appeared to have an insulated larger frustum plate, maybe both. In effect, these plates might have been charged capacitors exhibiting the "corona wind".

Interesting. This raises two questions:
- Does the process of resonating microwaves inside a cavity is naturally known to produce an electric potential difference between two conductive plates (both isolated from the frustum and each other by a dielectric gasket) ?
- If so, for EmDrive designs with plates electrically isolated from the frustum, could we just connect the two plates together with a ground wire (outside of the cavity) to prevent any voltage and thus any ion wind around the cavity?
So, I've been SLOWLY working on my own theory at the 40K ft level. Assymetrical (end) plates, insulated from one another are capacitive plates with a dielectric (air) medium. An even better capacitor would be created by putting in a dielectric puck, HDFE, which is what EW did. But don't think thats why they did it.

RF radiation striking the end plates impart Ev and a potential builds up ONLY if the plates are insulated from one another. The EW frustum does not appear to have insulated end-plates and I would not call it a capacitor, nor do I think it is exhibiting the corona/ion effect. Outgassing? Maybe.

Regardless, I've set my design up to be able to insulate the end plates, so I can test the capacitor/corona theory. My first test will be with uninsulated end plates, test #2 will be with insulated end plates...as it stands now.

As I have mentioned before, if anything is escaping the cavity, and if it's somehow real and not quantum, then it can only escape through gaps. Gasket gaps are good in that regard. But if something is escaping, shouldn't the real world direction of the escape be a consideration? It was back when I was looking at forces through gaps, trying to measure evanescent waves. Making the ends like corks in a bottle, instead of like a cap on a pipe made a big difference, percentage wise in the forces.

Retired, working interesting problems

Offline saucyjack

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More info is in the paper, which is in Chinese. Attached.

I don't think the drawing is dimensionally correct.

I hunted throughout the Chinese Physical Society site (http://wulixb.iphy.ac.cn) but didn't find any English translations of the 2014 Chinese paper.  So after going through various machine translation software, I've uploaded to the wiki a more readable English version, with most of the formatting and images intact; may be of help.

http://emdrive.wiki/images/9/9c/2014_Yang_NWPU_Paper.docx

Offline WarpTech

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We are not going to see an EMdrive effect unless some other "teeming streaming background field" is put in that picture of energy_repartition_end minus energy_repartition_start in a closed system.
But equally so, the "teeming streaming background field" will do you no good if you don't have an electromagnetic stress acting on it, so until the "teeming streaming background field" is found we might as well calculate the electromagnetic stress stress energy  tensor components and at least verify whether they are pointing in the right direction.

When there is a DC offset with an AC signal riding on top of it. The copper acts as a low-pass filter. The AC will be reflected. The DC will pass through. When it is charging AND discharging, the DC magnetic flux will be pushed out the Big end due to the gradient force from the geometry acting on it. The polarity of the magnetic flux, Phi does not matter, the force depends on Phi2. You'll have to wait for my paper because I'm still fleshing out the details myself.
Todd

Offline aero

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I went outside for a walk and I found out that actually:


(Group velocity*Electric Permitivity of Free space)*(Phase velocity*Magnetic Permeability of Free space)  = 1

(all these multiplied together give exactly one)

How about that?

great, but not the least bit surprising as vg * vp = c2 and eo * muo = 1/c2
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Offline WarpTech

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[...]
What can I say, except today I have a better handle on it than I did yesterday, or the day before... I'm learning too! Now, I understand that "break even" means that Pin = Pout, or 100% conversion efficiency. It doesn't mean anything more than that.
[...]
Oh, but indeed it does! It means that if the breakeven velocity is physically attainable (which implies at least that it's less than c), then you have free energy forever in a perpetual motion machine of the first kind. That's kinda important  8)

It's a good thing I'm going to bring this debate to a close. The answer is embarrassingly simple. It embarrasses me, you and everyone else who's been gawking at it and writing about this paradox. Not to forget to mention everyone of us who "thinks" we know rocket science! In a way, you are correct. In the Newtonian approximation the break even happens for k=1/c when 2/3 of the initial rest mass has been converted into kinetic energy. That is a significant speed where relativity is no longer negligible.

So then I decided to look at it relativistically rather than solve for an incorrect speed... That's when I said "OUCH!" This is going to hurt.... but you said yourself, you're power source is a battery on board the moving vehicle. :(

Convinced now?
Todd
Sorry, but I've no idea how you got the final line, and I certainly don't accept the expression for Eout without some explanation.

Your in denial! Sit down, have a pint and you'll see how it works. It's just algebra, set them equal and solve for gamma.

Eout is self explanatory. You start with an initial total rest-energy of your vehicle plus charged battery. You expend battery energy Ein, in order to generate thrust, regardless of how it is used to generate force and acceleration, this is the energy from the battery. That energy spent is subtracted from the initial total rest-energy and converted to kinetic energy by whatever means. At t=0, v=0, gamma=1. Plug in the total energy available in your battery for Ein. Then you have the gamma value and the maximum attainable speed. It will never give you more than you started with. Please don't embarrass us further...
Todd


Offline deltaMass

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In a way, you are correct. In the Newtonian approximation the break even happens for k=1/c when 2/3 of the initial rest mass has been converted into kinetic energy. That is a significant speed where relativity is no longer negligible.
You must have been reading someone else's posts, because I never wrote any of that (and nor do I plan to)!  In my simple Newtonian analysis, the value of 'k' is not specified at all. I have no clue where you get this stuff!

Offline WarpTech

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As I have mentioned before, if anything is escaping the cavity, and if it's somehow real and not quantum, then it can only escape through gaps. Gasket gaps are good in that regard. But if something is escaping, shouldn't the real world direction of the escape be a consideration? It was back when I was looking at forces through gaps, trying to measure evanescent waves. Making the ends like corks in a bottle, instead of like a cap on a pipe made a big difference, percentage wise in the forces.

Put a magnet inside the sealed "copper" cavity and see how much gets out. LOL!
Todd

Offline deltaMass

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[...]
What can I say, except today I have a better handle on it than I did yesterday, or the day before... I'm learning too! Now, I understand that "break even" means that Pin = Pout, or 100% conversion efficiency. It doesn't mean anything more than that.
[...]
Oh, but indeed it does! It means that if the breakeven velocity is physically attainable (which implies at least that it's less than c), then you have free energy forever in a perpetual motion machine of the first kind. That's kinda important  8)

It's a good thing I'm going to bring this debate to a close. The answer is embarrassingly simple. It embarrasses me, you and everyone else who's been gawking at it and writing about this paradox. Not to forget to mention everyone of us who "thinks" we know rocket science! In a way, you are correct. In the Newtonian approximation the break even happens for k=1/c when 2/3 of the initial rest mass has been converted into kinetic energy. That is a significant speed where relativity is no longer negligible.

So then I decided to look at it relativistically rather than solve for an incorrect speed... That's when I said "OUCH!" This is going to hurt.... but you said yourself, you're power source is a battery on board the moving vehicle. :(

Convinced now?
Todd
Sorry, but I've no idea how you got the final line, and I certainly don't accept the expression for Eout without some explanation.

Your in denial! Sit down, have a pint and you'll see how it works. It's just algebra, set them equal and solve for gamma.

Eout is self explanatory. You start with an initial total rest-energy of your vehicle plus charged battery. You expend battery energy Ein, in order to generate thrust, regardless of how it is used to generate force and acceleration, this is the energy from the battery. That energy spent is subtracted from the initial total rest-energy and converted to kinetic energy by whatever means. At t=0, v=0, gamma=1. Plug in the total energy available in your battery for Ein. Then you have the gamma value and the maximum attainable speed. It will never give you more than you started with. Please don't embarrass us further...
Todd
I'm sorry to say that any embarrassment belongs to you. What you have done here is set up equations that guarantee that over-unity cannot occur, then performed some algebraic manipulation to show that...presto!...over-unity cannot occur.

Sorry - profoundly underwhelmed.
« Last Edit: 07/10/2015 04:38 AM by deltaMass »

Offline WarpTech

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In a way, you are correct. In the Newtonian approximation the break even happens for k=1/c when 2/3 of the initial rest mass has been converted into kinetic energy. That is a significant speed where relativity is no longer negligible.
You must have been reading someone else's posts, because I never wrote any of that (and nor do I plan to)!  In my simple Newtonian analysis, the value of 'k' is not specified at all. I have no clue where you get this stuff!

You said,

...
1. Writing 'u' for the phase velocity, you get
k = u/c2 Newton/Watt = 1/c when u=c,
or in other words a pure photon rocket. But experimental evidence suggests a much higher value for k, and so if your formula is correct, it is predicting a superluminal phase velocity.
Is that your intent? Do you think that this observation is important?
...

I used 1/c as an example to solve for a particular case where break even was < c. But then I had an "AH HA Moment". LOL! Here is the Newtonian version too, in this case, the velocity goes to infinity rather than c, when 100% of the initial rest-energy has been spent. I hope you realize what this is saying. That for whatever energy is available in the battery to use for thrust, there will be a limiting velocity because the battery will go dead. It will not suddenly start to recharge when the speed exceeds some limit.
Todd

Offline TheTraveller

More info is in the paper, which is in Chinese. Attached.

I don't think the drawing is dimensionally correct.

I hunted throughout the Chinese Physical Society site (http://wulixb.iphy.ac.cn) but didn't find any English translations of the 2014 Chinese paper.  So after going through various machine translation software, I've uploaded to the wiki a more readable English version, with most of the formatting and images intact; may be of help.

http://emdrive.wiki/images/9/9c/2014_Yang_NWPU_Paper.docx

Thanks for doing that.

As I'm restricted to bed for the next few weeks, will try to straighten out some of the formatting issues.
« Last Edit: 07/10/2015 05:15 AM by TheTraveller »
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Offline zen-in

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...

Yes clever Chinese, getting Q of 117,500 without needing to go to spherical end plates.

Anybody translate the tags pointing to the 2 flat to/from spherical converter sections? I think I know how to calc the min length but knowing what those tags say may help.

Earlier in this thread there was a discussion on how the Chinese calculated Q.  The conclusion was they are not calculating Q correctly.   The actual Q of their cavity, based on network analyzer sweeps was closer to 3,000.  I don't recall the exact number but it was  1000 < Q < 5000.

Offline TheTraveller


...

Yes clever Chinese, getting Q of 117,500 without needing to go to spherical end plates.

Anybody translate the tags pointing to the 2 flat to/from spherical converter sections? I think I know how to calc the min length but knowing what those tags say may help.

Earlier in this thread there was a discussion on how the Chinese calculated Q.  The conclusion was they are not calculating Q correctly.   The actual Q of their cavity, based on network analyzer sweeps was closer to 3,000.  I don't recall the exact number but it was  1000 < Q < 5000.

The calcs in the latest paper are correct. Check it out.

Suggest the earlier low Q was related to cavity bandwidth needed to accept their magnetron output bandwidth.

Quote
Abstract
A microwave resonator system is made, which has a tapered resonant cavity, a microwave source, and a transmission device.

Because of the electromagnetic pressure gradient on the tapered resonant cavity, a net electromagnetic force along the axis of the cavity may be observed, which is needed to verify experimentally the use of the independent microwave resonator system.

It is also needed to keep the independent microwave resonator system in resonating state, which is the important procedure to demonstrate the possibility of net electromagnetic force.

Thus, a low-signal resonating experiment on the tapered resonant cavity combined with resonating parts is completed to accurately find out the resonant frequency of 2.45 GHz and to analyze the influence of temperature on the resonant state.

Experimental result shows that the resonant frequency and quality factor of the independent microwave resonator system are 2.44895 GHz and 117495.08 respectively.

When the temperature of the tapered resonant cavity wall rises, the resonant frequency will be decreased and the quality factor changed separately.
"As for me, I am tormented with an everlasting itch for things remote. I love to sail forbidden seas.
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Offline TheTraveller

What I read into the latest Chinese paper is:

1) They have worked out how to build a +100K Q cavity with flat end plates.

2) They have worked out how to drive this cavity with a narrow band Rf coax source.

3) They have worked out how to build a Rf control system that will automatically adjust the frequency to always stay in the middle of a dynamically changing cavity resonance bandwidth.

To me this suggests they are probably not that far away from going commercial with their propellantless thruster.
"As for me, I am tormented with an everlasting itch for things remote. I love to sail forbidden seas.
Herman Melville, Moby Dick

Offline WBY1984

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The layman EMDrive thread has disappeared down the list, so could anyone give a brief (and simple) description of where you guys are at on this issue?

Thanks.

Offline zen-in

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...

Yes clever Chinese, getting Q of 117,500 without needing to go to spherical end plates.

Anybody translate the tags pointing to the 2 flat to/from spherical converter sections? I think I know how to calc the min length but knowing what those tags say may help.

Earlier in this thread there was a discussion on how the Chinese calculated Q.  The conclusion was they are not calculating Q correctly.   The actual Q of their cavity, based on network analyzer sweeps was closer to 3,000.  I don't recall the exact number but it was  1000 < Q < 5000.

The calcs in the latest paper are correct. Check it out.

Suggest the earlier low Q was related to cavity bandwidth needed to accept their magnetron output bandwidth.

...

Experimental result shows that the resonant frequency and quality factor of the independent microwave resonator system are 2.44895 GHz and 117495.08 respectively.

When the temperature of the tapered resonant cavity wall rises, the resonant frequency will be decreased and the quality factor changed separately.
[/quote]

No, this was discussed before.  The method the Chinese use to calculate Q is wrong.   The top illustration you show is a linear/linear plot that shows the correct way to measure Q.

    Q = (Center Freq)/(Half power Bandwidth)

The second figure you showed is a different graph than what was discussed earlier and it is a linear-log plot.  The half power bandwith is the width measured at 3 dB below the reference.  So (Half Power Bandwith) = .1 MHz.  This makes the Q = 2.5/.0001  = 25000.   I think even this is unrealistic.   Here is the Wikipedia page on Q:
https://en.wikipedia.org/wiki/Q_factor
I have read a lot of well intentioned but misinformed ideas in this forum on RF, waveguides, filters, Q,  etc, etc.   The idea that the higher the Q the more thrust you get is wrong.   It is just another red flag.   If you get thrust then you have to be transferring power out of the cavity.  But if the cavity is losing power the Q can't be high.   So this idea that a high Q is needed for thrust is just another way the em-drive supplies free energy and therefore is the stuff of someones' dreams.
http://urgentcomm.com/mag/make-vhf-cavities-over-counter-hardware
The extremely high Q claimed for the Chinese and other em-drive cavities is completely wrong.  I have already showed they are not calculating Q correctly.   The graph shown is not a photo taken from a network analyzer and so I believe it is just made up data.   These cavities are similar in most respects to cavity filters used for VHF and UHF repeaters.   A 145 MHz cavity typically has a Q = 350.   Scaling this up to 2.5 GHz and the Q may be as high as 1,000 - 2,000.  The skin effect and other factors increase the losses at higher frequencies.
« Last Edit: 07/10/2015 09:12 AM by zen-in »

Offline deltaMass

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In a way, you are correct. In the Newtonian approximation the break even happens for k=1/c when 2/3 of the initial rest mass has been converted into kinetic energy. That is a significant speed where relativity is no longer negligible.
You must have been reading someone else's posts, because I never wrote any of that (and nor do I plan to)!  In my simple Newtonian analysis, the value of 'k' is not specified at all. I have no clue where you get this stuff!

You said,

...
1. Writing 'u' for the phase velocity, you get
k = u/c2 Newton/Watt = 1/c when u=c,
or in other words a pure photon rocket. But experimental evidence suggests a much higher value for k, and so if your formula is correct, it is predicting a superluminal phase velocity.
Is that your intent? Do you think that this observation is important?
...

I used 1/c as an example to solve for a particular case where break even was < c. But then I had an "AH HA Moment". LOL! Here is the Newtonian version too, in this case, the velocity goes to infinity rather than c, when 100% of the initial rest-energy has been spent. I hope you realize what this is saying. That for whatever energy is available in the battery to use for thrust, there will be a limiting velocity because the battery will go dead. It will not suddenly start to recharge when the speed exceeds some limit.
Todd
Dude. Your Newtonian expression for kinetic energy is not 0.5*m*v2, so how can I take this seriously?

Offline TheTraveller

The extremely high Q claimed for the Chinese and other em-drive cavities is completely wrong.  I have already showed they are not calculating Q correctly.   The graph shown is not a photo taken from a network analyzer and so I believe it is just made up data.   These cavities are similar in most respects to cavity filters used for VHF and UHF repeaters.   A 145 MHz cavity typically has a Q = 350.   Scaling this up to 2.5 GHz and the Q may be as high as 1,000 - 2,000.  The skin effect and other factors increase the losses at higher frequencies.

So easy to claim the Chinese data has been made up, despite having no proof. Along with silly claims that would also say Eagleworks doesn't know how to measure Q either as per attached measured Q of 50,995.

I'll not bother to send you my data as you will claim it is also made up.
« Last Edit: 07/10/2015 09:56 AM by TheTraveller »
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Offline flux_capacitor

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The layman EMDrive thread has disappeared down the list, so could anyone give a brief (and simple) description of where you guys are at on this issue?

Thanks.

I suggest you read the posts on the EmDrive subreddit entitled "As the Frustum Turns" which are written by @bitofaknowitall on a regular basis and consist of a summary of the most important stories discussed here in this NSF thread. The last one to date is from 7th July 2015 and covers the week of 28 June - 4 July 2015 (Episode 5).

Offline Rodal

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I went outside for a walk and I found out that actually:


(Group velocity*Electric Permitivity of Free space)*(Phase velocity*Magnetic Permeability of Free space)  = 1

(all these multiplied together give exactly one)

How about that?

great, but not the least bit surprising as vg * vp = c2 and eo * muo = 1/c2

Of course, it was a contrived attempt at humor concerning several posts remarking the not least beat surprising fact that 2c is not the same thing as c^2 (unless c=2): :)

Have you considered going outside and walking around a bit? LOL
Phase velocity + group velocity = 2c.
Wrong

Group velocity X phase velocity = 2c.
Also wrong

Phase velocity + group velocity = 2c.
Wrong
« Last Edit: 07/10/2015 12:05 PM by Rodal »

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