Author Topic: EM Drive Developments - related to space flight applications - Thread 2  (Read 2101484 times)

Offline TheTraveller

The Traveller: I know, and what I said doesn't contradict you, except that cut-off isn't necessary.

Shawyer always says the small end cutoff wavelength should be just above small end cutoff for best thrust generation. The higher the Df, the higher the resultant thrust from T = 2 Df Po Q / c

So the small end operating at just above cutoff is important to achieve a Df close to 1.
"As for me, I am tormented with an everlasting itch for things remote. I love to sail forbidden seas.
Herman Melville, Moby Dick

Offline deltaMass

  • Full Member
  • ****
  • Posts: 955
  • A Brit in California
  • Liked: 671
  • Likes Given: 275
I'm sorry, but I can't make it any simpler.
N=1,2 is the case when you evaluate something at either end.
But the equation holds for any position in between.

Offline TheTraveller

....
The cutoff wavelength is defined for EACH end, based on it's diameter * 1.71. See attachment 2.

Lambda1 = guide wavelength equation from attachment 1 below using big end cutoff wavelength.

Lambda2 = guide wavelength equation from attachment 1 below using small end cutoff wavelength.

OK

Thanks for the answer

Very clear answer

That's NOT what deltaMass had written  (there is no N=1,2,3,4, in equation 5)

That's exactly the same factor of 1.7    I derived, based on TE110

Now I'm going to sleep.

Did notice you came up with 1.71.
Nice work.
Now we both go to sleep.
"As for me, I am tormented with an everlasting itch for things remote. I love to sail forbidden seas.
Herman Melville, Moby Dick

Offline deltaMass

  • Full Member
  • ****
  • Posts: 955
  • A Brit in California
  • Liked: 671
  • Likes Given: 275
The Traveller: I know, and what I said doesn't contradict you, except that cut-off isn't necessary.

Shawyer always says the small end cutoff wavelength should be just above small end cutoff for best thrust generation. The higher the Df, the higher the resultant thrust from T = 2 Df Po Q / c

So the small end operating at just above cutoff is important to achieve a Df close to 1.

Not if you take that Df at face value, no. Let's do some high school algebra. I use x to mean lambda, and I can't be bothered using suffices throughout.

We start with the relation we know to be true: x2 >= x1 >= x0.
We now quantify this with these definitions:
x2 := a*x1  (a >= 1)
x1 := b*x0 (b >= 1)
The original Df expression:
Df = x0*(x2 - x1) / (x1*x2 - x02)
becomes, using the above definitions:

Df = b*(a - 1) / (a*b2 - 1)

What's Df when x1 = x0?
This is tantamount to setting b=1. Doing that, we get
Df = (a-1) / (a-1) = 1.   QED.
i.e. it doesn't matter what the value of 'a' might be; i.e. it doesn't matter about cutoff.
« Last Edit: 05/14/2015 04:57 AM by deltaMass »

Offline deltaMass

  • Full Member
  • ****
  • Posts: 955
  • A Brit in California
  • Liked: 671
  • Likes Given: 275
I ought to mention that Shawyer does not specify how to compute vg or lambdag. What he does do is specify the functional relation between them - i.e. v(lambda) or lambda(v) - and that's all.

Anyone know how to compute vg without using  lambdag?
Anyone know how to compute lambdag without using  vg?
« Last Edit: 05/14/2015 05:26 AM by deltaMass »

Offline TheTraveller

The Traveller: I know, and what I said doesn't contradict you, except that cut-off isn't necessary.

Shawyer always says the small end cutoff wavelength should be just above small end cutoff for best thrust generation. The higher the Df, the higher the resultant thrust from T = 2 Df Po Q / c

So the small end operating at just above cutoff is important to achieve a Df close to 1.

Not if you take that Df at face value, no. Let's do some high school algebra. I use x to mean lambda, and I can't be bothered using suffices throughout.

We start with the relation we know to be true: x2 >= x1 >= x0.
We now quantify this with these definitions:
x2 := a*x1  (a >= 1)
x1 := b*x0 (b >= 1)
The original Df expression:
Df = x0*(x2 - x1) / (x1*x2 - x02)
becomes, using the above definitions:

Df = b*(a - 1) / (a*b2 - 1)

What's Df when x1 = x0?
This is tantamount to setting b=1. Doing that, we get
Df = (a-1) / (a-1) = 1.   QED.
i.e. it doesn't matter what the value of 'a' might be; i.e. it doesn't matter about cutoff.

Interesting.

I feel what Shawyer is saying is to start a cavity design at the small end. Once you get the small end cut off and guide wavelengths matched to the desired driving Rf wavelength as 1/2x or 1x or 2x, you then switch to the big end and again set the big end diameter to match the driving wavelength. Then adjust length to match the small and big end wavelengths as multiples or subs of each other.

Bit too late for my fuzzy brain. Can see how all this interaction can be folded into an interactive cavity design spreadsheet.
"As for me, I am tormented with an everlasting itch for things remote. I love to sail forbidden seas.
Herman Melville, Moby Dick

Offline TheTraveller

I ought to mention that Shawyer does not specify how to compute vg or lambdag. What he does do is specify the functional relation between them - i.e. v(lambda) or lambda(v) - and that's all.

Anyone know how to compute vg without using  lambdag?
Anyone know how to compute lambdag without using  vg?

Both equations can be found on the net. Standard microwave stuff. Have posted them earlier:
http://www.microwaves101.com/encyclopedias/waveguide-mathematics

Lambda g1 or g2 is guide wavelength for big end (1) or small end (2).

vg1 or vg2 or group velocity 1 or 2 is derived from / determined by guide wavelength for lambda g1/g2.

Shawyer calls lambda g1 and lambda g2 "guide wavelengths" as they are microwave industry definations that I found in less than 5 minutes of Googling.

As far as I can see on microwave industry web sites, vg or group velocity is calculated based on guide wavelength or Shawyers lambda g1 or g2.
« Last Edit: 05/14/2015 05:59 AM by TheTraveller »
"As for me, I am tormented with an everlasting itch for things remote. I love to sail forbidden seas.
Herman Melville, Moby Dick

Offline deltaMass

  • Full Member
  • ****
  • Posts: 955
  • A Brit in California
  • Liked: 671
  • Likes Given: 275
Shawyer's Df says that you want the big end lambda (the smallest lambda ) as close to free space lambda as possible. Of course, he doesn't recommend opening up the big end.

Plays havoc with the Q :)
« Last Edit: 05/14/2015 07:07 AM by deltaMass »

Offline zellerium

  • Full Member
  • *
  • Posts: 171
  • Pittsburgh, PA
  • Liked: 279
  • Likes Given: 400
I've been trying to wrap my brain around why a difference in phase results in a better thrust.  Also, why couldn't EW obtain a thrust without a dielectric?

I think we can all agree that in order for a net thrust, the momentum delivered to the larger end plate is smaller than that delivered to the smaller end plate. So where did the momentum go?

Can momentum be delivered and removed from an orbiting electron?

Take a simple two dimensional case with two atoms, one on the small end, one on the large end, each with their own electron orbiting at a the same angular frequency. If a force is applied to both of them, one in the direction of revolution and the other opposite, one of the forces would slow down the electron and the other would speed it up. The sped up electron requires a larger force to keep it tied to the nucleus and we have a net thrust.

Perhaps this could help explain a couple things:

The dielectric is composed of different elements, thus the electrons are orbiting at a different angular velocity. Using a constant frequency with different elements gives a certain degree of difference in the phase at which momentum is delivered to the electrons.

Shawyer observed more losses with a dielectric because a magnetron outputs a signal at many phases and somehow 'matches' the orbital tendency of the electrons.
I imagine the magnetic component of the wave could be contributing to an alignment of electrons which could amplify this miniscule effect.

Any thoughts?

Offline deltaMass

  • Full Member
  • ****
  • Posts: 955
  • A Brit in California
  • Liked: 671
  • Likes Given: 275
If this is supposed to be a phenomenon based on waves in a can, then surely it applies to any waves. Perhaps someone should look at plain old water waves in a can.

Cum aquis ad sideros
« Last Edit: 05/14/2015 08:22 AM by deltaMass »

Offline deltaMass

  • Full Member
  • ****
  • Posts: 955
  • A Brit in California
  • Liked: 671
  • Likes Given: 275
Quote from: Rodal
Wrong.  (You need AND)
Thanks! - you are quite correct. But no need for Mathematica.
« Last Edit: 05/14/2015 10:40 AM by deltaMass »

Offline deuteragenie

  • Member
  • Posts: 71
  • Germany
  • Liked: 22
  • Likes Given: 0
Any resemblance with existing or future device(s) is purely coincidental.
Made with MEEP.  Does not use the correct frequencies, material, etc. but looks cute.

Is that a 2-Dimensional model with MEEP?
Maxwell's equations in a flat 2-D surface?
modeling the truncated cone as a FLAT trapezium ?

Is the magnetic field (for TM modes)  a point scalar (only able to have + or - sign but the direction is always perpendicular to the surface) instead of being a vector in the azimuthal (circumferential) direction  ?

What are we seeing out of the EM Drive? evanescent wave field?

If the answers are yes, do you have enough memory to run a 3D model instead?

Thanks

I tried my best to warn that this was art and not science.
Anyway, the MEEP ctl file is attached.

Modelling the truncated cone as a FLAT trapezium ? Rather modelling a slice of of the 3D cone on a 2D lattice I guess.
Maxwell's equations in a flat 2-D surface? That is my understanding on how MEEP works.
Is the magnetic field (for TM modes)  a point scalar (only able to have + or - sign but the direction is always perpendicular to the surface) instead of being a vector in the azimuthal (circumferential) direction  ? Ez field is shown.  Producing Hz is straightforward.
What are we seeing out of the EM Drive? evanescent wave field? You'll understand when you look at the attached file :)
It took about 2 minutes to generate in 2D and should be fairly easy to move it to 3D, although I am not too sure which tools are available to visualize the 3D output or to slice it.
« Last Edit: 05/14/2015 11:22 AM by deuteragenie »

Online Mulletron

  • Full Member
  • ****
  • Posts: 1111
  • Liked: 775
  • Likes Given: 1010
Published 12 May 2015
This represents the current state of the art.

http://iopscience.iop.org/0953-8984/27/21/210301/article
http://iopscience.iop.org/0953-8984/27/21 (TOC)

I'm pretty excited about what could come out of applying these materials too...maybe a way forward:
http://www.lap.physik.uni-erlangen.de/lap/?page=research_krstic_chiral&language=en

« Last Edit: 05/14/2015 11:00 AM by Mulletron »
Challenge your preconceptions, or they will challenge you. - Velik

Offline deuteragenie

  • Member
  • Posts: 71
  • Germany
  • Liked: 22
  • Likes Given: 0
Any resemblance with existing or future device(s) is purely coincidental.
Made with MEEP.  Does not use the correct frequencies, material, etc. but looks cute.

You might want to increase your skin thickness or Meep resolution. Looks kind of like a Gaussian source with some of the shorter wavelengths bouncing around and longer wavelengths stepping over the boundary (numerically)?

Yes, I noticed that and tried to improve the resolution with not much success so far.  I'll keep playing.

If you are using perfect metal for the skin, then the thickness won't change the result but will help to avoid the model numerically spanning the skin. If you are modelling copper in the GHz range, please tell me about your model as I have struggled for 6 months trying to find a Drude model for copper at 2 GHz.

I tried to change the boundary to copper and the inside to air and then to vacuum, with the definition for copper provided here:
http://www.fzu.cz/~dominecf/meep/data/meep-metals.pdf

Unfortunately, I was not able to produce anything appealing so far with that approach (either I get completely white or dark results). It also did not work when I tried to increase the frequency too much. I need to study physics first then I'll be able to understand what I am supposed to enter as parameters...

Attached is the .ctl file (renamed to .txt because the forum does not allow .ctl to be uploaded).

Offline TheTraveller

Have modified my Shawyer Df calculator and best Df scanner as per the derived Shawyer Df equation, using cutoff wavelength and guide wavelength as per microwave industry supplied equations. I assume Shawyer did not supply these equations in his papers as they are equations that should be known to microwave industry individuals skilled in the art. Anyway they are now in the public record.

The scanner still sweeps the frequency range 0Hz to 10GHz but reports the frequency that generates a Df as close to 1 as possible but not over.

The attached results are very interesting as the frequency needed to get the Df to just below 1 is very close to the Rf driving frequency used to generate Lambda0 or free wavelength in the selected medium.

While I'm still testing the spreadsheet, which meets both of Shawyers boundary conditions, the results for my Flight Thruster design are looking to be very close to what I could build. Bit of dimension tweaking should get the Df 1 frequency to the 3.85GHz Shawyer used.

Will post the spreadsheet after a bit more testing.
« Last Edit: 05/14/2015 11:59 AM by TheTraveller »
"As for me, I am tormented with an everlasting itch for things remote. I love to sail forbidden seas.
Herman Melville, Moby Dick

Online Rodal

  • Senior Member
  • *****
  • Posts: 5835
  • USA
  • Liked: 5913
  • Likes Given: 5253
Have modified my Shawyer Df calculator and best Df scanner as per the derived Shawyer Df equation, using cutoff wavelength and guide wavelength as per microwave industry supplied equations. I assume Shawyer did not supply these equations in his papers as they are equations that should be known to microwave industry individuals skilled in the art. Anyway they are now in the public record.

The scanner still sweeps the frequency range 0Hz to 10GHz but reports the frequency that generates a Df as close to 1 as possible but not over.

The attached results are very interesting as the frequency needed to get the Df to just below 1 is very close to the Rf driving frequency used to generate Lambda0 or free wavelength in the selected medium.

While I'm still testing the spreadsheet, which meets both of Shawyers boundary conditions, the results for my Flight Thruster design are looking to be very close to what I could build. Bit of dimension tweaking should get the Df 1 frequency to the 3.85GHz Shawyer used.

Will post the spreadsheet after a bit more testing.

This is what I get:

DESIGN FACTOR WITH CUT-OFF FREQUENCY CHOSEN AS TE110

Defining the following variables:

bD = big end diameter (m)
sD= small end diameter (m)
f = applied frequency (Hz)
cutOffW1= cutOffWavelength1 (m)
cutOffW2= cutOffWavelength2 (m)
assuming the cutOffWavelengths in the truncated cone to be the lowest possible mode in a cylindrical cavity, TE110, the constant "cst" is given by Pi divided by the 1,1 zero of the derivative of the cylindrical Bessel function:
cst= Pi /X'11
     =Pi/1.84118378134065
     =1.7062895542683174
cM = light speed in selected medium (m/s)
     = 299705000 (m/s) (speed of light in air)
     = 299792458 (m/s) (speed of light in vacuum)



lambda0 = cM /f;

cutOffW1 = bD*cst;

cutOffW2 = sD*cst;

lGuide1 = lambda0/Sqrt[1 - (lambda0/cutOffW1)^2];

lGuide2 = lambda0/Sqrt[1 - (lambda0/cutOffW2)^2];

soCorrectionFactor = 1/(1 - ((lambda0^2)/(lGuide1*lGuide2)));

multiplier = (lambda0/lGuide1) - (lambda0/lGuide2);

designFactor = soCorrectionFactor*multiplier;

designFactor =
           (bD^2 - sD^2)/( (bD^2)*Sqrt[1 - (cM/(bD*cst*f))^2] + (sD ^2)*Sqrt[1 - (cM/(cst*f*sD))^2] )

The Design Factor has a singularity (due to the cut-off frequencies) starting at the cut-off frequency associated with the small diameter

Singularity frequency= cM/(cst*sD)       
                 

When I input the same values you have, I get practically (*) the same output for the Design Factor as you do

EXAMPLE Ref 2

Using speed of light in air

INPUT:

bD = 0.2797; sD = 0.1588; f = 2.45*10^9; cM = 299705000; cst = 1.7062895542683174;

OUTPUT:

Design Factor = 0.540306

singularity frequency = 1.10609*10^9 Hz due to the cut-off frequency associated with the small diameter



(*) small difference is due to the fact that I use 1.7062895542683174 instead of the approximate value of 1.71 you use
« Last Edit: 05/15/2015 11:40 AM by Rodal »

Online SeeShells

  • Senior Member
  • *****
  • Posts: 2326
  • Every action there's a reaction we try to grasp.
  • United States
  • Liked: 2956
  • Likes Given: 2588
Published 12 May 2015
This represents the current state of the art.

http://iopscience.iop.org/0953-8984/27/21/210301/article
http://iopscience.iop.org/0953-8984/27/21 (TOC)

I'm pretty excited about what could come out of applying these materials too...maybe a way forward:
http://www.lap.physik.uni-erlangen.de/lap/?page=research_krstic_chiral&language=en

Excited? I'm more than excited.  I think I've reached my limits of current understanding and applying these effects to the EM device. I need more time to read and digest because every article seems to have a little something that fits in the overall scheme of why. Mullutron I think you have a tiger by the tale here.
Like this one.
Transfer of linear momentum from the quantum vacuum to a magnetochiral molecule
M Donaire1, B A van Tiggelen2 and G L J A Rikken3
Show affiliations

M Donaire et al 2015 J. Phys.: Condens. Matter 27 214002. doi:10.1088/0953-8984/27/21/214002
Received 14 April 2014, accepted for publication 2 July 2014. Published 12 May 2015.
2015 IOP Publishing Ltd

Abstract
In a recent publication [1] we have shown using a QED approach that, in the presence of a magnetic field, the quantum vacuum coupled to a chiral molecule provides a kinetic momentum directed along the magnetic field. Here we explain the physical mechanisms which operate in the transfer of momentum from the vacuum to the molecule. We show that the variation of the molecular kinetic energy originates from the magnetic energy associated with the vacuum correction to the magnetization of the molecule. We carry out a semiclassical calculation of the vacuum momentum and compare the result with the QED calculation.

Online Rodal

  • Senior Member
  • *****
  • Posts: 5835
  • USA
  • Liked: 5913
  • Likes Given: 5253
Have modified my Shawyer Df calculator and best Df scanner as per the derived Shawyer Df equation, using cutoff wavelength and guide wavelength as per microwave industry supplied equations. I assume Shawyer did not supply these equations in his papers as they are equations that should be known to microwave industry individuals skilled in the art. Anyway they are now in the public record.

The scanner still sweeps the frequency range 0Hz to 10GHz but reports the frequency that generates a Df as close to 1 as possible but not over.

The attached results are very interesting as the frequency needed to get the Df to just below 1 is very close to the Rf driving frequency used to generate Lambda0 or free wavelength in the selected medium.

While I'm still testing the spreadsheet, which meets both of Shawyers boundary conditions, the results for my Flight Thruster design are looking to be very close to what I could build. Bit of dimension tweaking should get the Df 1 frequency to the 3.85GHz Shawyer used.

Will post the spreadsheet after a bit more testing.

designFactor =
           (bD^2 - sD^2)/( (bD^2)*Sqrt[1 - (cM/(bD*cst*f))^2)] + (sD ^2)*Sqrt[1 - (cM/(cst*f*sD))^2)] )

The Design Factor has a singularity (due to the cut-off frequencies) starting at the cut-off frequency associated with the small diameter

Singularity frequency= cM/(cst*sD)     

LIMITS

1) Limit[DesignFactor, sD -> bD] = 0

2) Limit[DesignFactor, bD -> Infinity] =1

3) Limit[DesignFactor, sD -> 0] = 1/Sqrt[1 - (cM/(bD * cst * f ))^2]

4) Limit[DesignFactor, f -> Infinity]  = (bD^2 - sD^2)/(bD^2 + sD^2)

5) Limit[DesignFactor, f -> 0] =0
« Last Edit: 05/14/2015 05:16 PM by Rodal »

Online SeeShells

  • Senior Member
  • *****
  • Posts: 2326
  • Every action there's a reaction we try to grasp.
  • United States
  • Liked: 2956
  • Likes Given: 2588
Published 12 May 2015
This represents the current state of the art.

http://iopscience.iop.org/0953-8984/27/21/210301/article
http://iopscience.iop.org/0953-8984/27/21 (TOC)

I'm pretty excited about what could come out of applying these materials too...maybe a way forward:
http://www.lap.physik.uni-erlangen.de/lap/?page=research_krstic_chiral&language=en

Excited? I'm more than excited.  I think I've reached my limits of current understanding and applying these effects to the EM device. I need more time to read and digest because every article seems to have a little something that fits in the overall scheme of why. Mullutron I think you have a tiger by the tale here.
Like this one.
Transfer of linear momentum from the quantum vacuum to a magnetochiral molecule
M Donaire1, B A van Tiggelen2 and G L J A Rikken3
Show affiliations

M Donaire et al 2015 J. Phys.: Condens. Matter 27 214002. doi:10.1088/0953-8984/27/21/214002
Received 14 April 2014, accepted for publication 2 July 2014. Published 12 May 2015.
2015 IOP Publishing Ltd

Abstract
In a recent publication [1] we have shown using a QED approach that, in the presence of a magnetic field, the quantum vacuum coupled to a chiral molecule provides a kinetic momentum directed along the magnetic field. Here we explain the physical mechanisms which operate in the transfer of momentum from the vacuum to the molecule. We show that the variation of the molecular kinetic energy originates from the magnetic energy associated with the vacuum correction to the magnetization of the molecule. We carry out a semiclassical calculation of the vacuum momentum and compare the result with the QED calculation.
another good publication.
http://arxiv.org/pdf/1404.5990v2.pdf

Offline Notsosureofit

  • Full Member
  • ****
  • Posts: 656
  • Liked: 704
  • Likes Given: 1361
Post-sleep (?) rambling thoughts on entropy:

It has always bothered me that (to me) it sounds like Shawyer's theory should make the frustrum go in the other direction, not that I spend any time thinking about it, but anyway;

Very roughly, it seemed to me like entropy change of a photon gas would give an isotropic force from the surface integral of the entropy change times the temperature divided by the work that was done (by the pressure expanding the surface)

But what if the temperature was inside the integral.  A symmetrical cavity would still have no net force as long as the surface temperature change was symmetrical.

Of course, EW and @RODAL's calculations show the big end hotter.

But, if the radiation temperature of the photons is much LESS than the ambient, raising the big end temperature gives a NEGATIVE force in that direction when the entropy increases.   Sort of in keeping w/ the 1p on the big end 2p on the small end thoughts.

These are just sleep (or non-sleep) induced images so use a 10lb salt bag.  (havn't found my old thermo text)  Including the surface integral of entropy change of photons opens a BIG can of worms, but it might explain EW not getting thrust w/o dielectric if the cavity surface has uneven characteristics (ie circuit boards on the ends, ferinstance)

As stated above, 10lb bag of salt.


Tags: