Author Topic: EM Drive Developments - related to space flight applications - Thread 3  (Read 1804774 times)

Offline deltaMass

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Sicherlich hebt es das Konzept 'Lebensraum' noch hoeher  8)
« Last Edit: 07/15/2015 03:39 AM by deltaMass »

Offline WarpTech

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@WarpTech - I looked at your pdf.
Einstein's ghost will bite you in the neck while you sleep.
Force cannot depend on velocity.

Okay, Okay, Okay, I found the confusing part. I hope everyone is listening. I'm sorry that I've probably made the confusion worse, not better. :(

What is causing the confusion (for me anyway) is the reference frame of the problem. The equations I wrote down were all done from the perspective of an observer in the inertial reference frame from where the vehicle started at rest, with a fully charged battery. I did "everything" from that reference frame.

However, the value of thrust-to-power in this frame is not what is being specified in the paradox, which was my misunderstanding. What should have been specified explicitly, which I failed to understand, is the use of the rocket equation to calculate k. In that equation k is not arbitrary. It depends on the exit velocity of the propellant in the frame of the rocket! When the problem is formulated as such, the thrust-to-power ratio IS constant, in this frame. See attachment 1.

I failed to understand that the thrust depends on the velocity at the exit of the rocket. The c2 in the denominator cancels the 16 orders of magnitude that @deltaMass was saying made the power used have negligible mass. On the contrary, the majority of thrust is due to the mass, because the k value is so small! See attachment 2.

It is the assumption that is made, where k is arbitrarily large, when it is in fact very small because the exit velocity must be less than c, that causes the over-unity paradox.

I got confused by Eout = 0.5*m*v2, because in the frame of the rocket, v=0. So I was thinking that v was measured as the instantaneous velocity in the frame of an inertial observer. It is not. It is measured, as I've been told, as v=a*t, in the frame of the rocket. My mistake.

So, thanks for the education. I apologize for confusing people and I hope my mistakes can be forgiven. Live and learn.
Todd


Offline TheTraveller

@WarpTech - I looked at your pdf.
Einstein's ghost will bite you in the neck while you sleep.
Force cannot depend on velocity.

Inside a microwave waveguide Cullen has shown the bounce Force generated is reletative to the EM waves Guide Wavelength or Group Velocity, which is below c and is determined by Cutoff Wavelength which is determined by waveguide diameter, excitation mode & external frequency.

So no bites in the neck from Einstein's ghost, just microwave physics in action.

Cullen's ghost may bite you in the neck for not understanding and respecting his work.
« Last Edit: 07/15/2015 04:14 AM by TheTraveller »
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Offline TheTraveller

Todd

Todd,

Any EMDrive theory must show the direction of the dielectric generated Force (Small to Big) at 180 deg opposite the non dielectric case (Big to Small) and the dielectric Force to be much weaker than the non dielectric case.

Can your theory do that?
« Last Edit: 07/15/2015 04:25 AM by TheTraveller »
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Online aero

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@leomillert

I didn't look very hard because Bash is not something I use often so am really not good at it. I did notice that the first thing you did was count lines and discard any extra. There is no reason to discard the extra lines or to go farther than discovering that the files don't have the same number of lines. (or columns) If they don't, they are not identical and you're done.

aero
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Offline WarpTech

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@Rodal:
Is it possible with this data to confirm or deny the correctness of 2*P/c versus 2*P/vg?

The group velocity in a waveguide is by definition;  vg = dw/dk, correct?

So, break it apart into dw = vg*dk, then,

dw/dt = P
dk/dt = F

So, F * vg = P

but this is only for waves in a waveguide. For a rocket with propellant, it's the opposite because ue is the group velocity of matter going out the rocket nozzle. In terms of power,

F/P = ue/c2, but this is 1/(phase velocity)!

So, F * c2/ue = P

In other words, apparently EM waves in a waveguide behave differently than matter in a rocket, in this regard.
Or am I botching it again?
Todd


« Last Edit: 07/15/2015 05:22 AM by WarpTech »

Offline deltaMass

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@WarpTech - I looked at your pdf.
Einstein's ghost will bite you in the neck while you sleep.
Force cannot depend on velocity.

Okay, Okay, Okay, I found the confusing part. I hope everyone is listening. I'm sorry that I've probably made the confusion worse, not better. :(

What is causing the confusion (for me anyway) is the reference frame of the problem. The equations I wrote down were all done from the perspective of an observer in the inertial reference frame from where the vehicle started at rest, with a fully charged battery. I did "everything" from that reference frame.

However, the value of thrust-to-power in this frame is not what is being specified in the paradox, which was my misunderstanding. What should have been specified explicitly, which I failed to understand, is the use of the rocket equation to calculate k. In that equation k is not arbitrary. It depends on the exit velocity of the propellant in the frame of the rocket! When the problem is formulated as such, the thrust-to-power ratio IS constant, in this frame. See attachment 1.

I failed to understand that the thrust depends on the velocity at the exit of the rocket. The c2 in the denominator cancels the 16 orders of magnitude that @deltaMass was saying made the power used have negligible mass. On the contrary, the majority of thrust is due to the mass, because the k value is so small! See attachment 2.

It is the assumption that is made, where k is arbitrarily large, when it is in fact very small because the exit velocity must be less than c, that causes the over-unity paradox.

I got confused by Eout = 0.5*m*v2, because in the frame of the rocket, v=0. So I was thinking that v was measured as the instantaneous velocity in the frame of an inertial observer. It is not. It is measured, as I've been told, as v=a*t, in the frame of the rocket. My mistake.

So, thanks for the education. I apologize for confusing people and I hope my mistakes can be forgiven. Live and learn.
Todd
Your great enlightenment (what, number ten is it now in the series of consecutive samadhis?) is unfortunately not shared by me, despite its clearly blinding power over the transformation of your logics. It must be fun in there.

Look: EmDrive is not a rocket, which is why we call it propellantless. Dear Tsiolkovsky had the right of it when he defined exhaust velocity as an invariant when measured with respect to the rocket. Such luxury you do not have because you have no exhaust to manipulate here.

Additional to that misunderstanding (masquerading in your mind, apparently, as a revelation) comes your further misunderstanding that 'v' is relative to the EmDrive. The 'v' used in my proof, clear to most I should have thought, is defined with respect to the original inertial frame in which motion began. In no way should you attempt to interpret it as you now have as relative to the EmDrive itself, for that is not how it is defined in my proof.
« Last Edit: 07/15/2015 05:46 AM by deltaMass »

Online aero

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We continue the program started with posts ____________________________


[img]https://upload.

http://forum.nasaspaceflight.com/index.php?topic=37642.msg1403629#msg1403629

http://forum.nasaspaceflight.com/index.php?topic=37642.msg1404000#msg1404000
http://forum.nasaspaceflight.com/index.php?topic=37642.msg1404004#msg1404004
http://forum.nasaspaceflight.com/index.php?topic=37642.msg1404005#msg1404005
http://forum.nasaspaceflight.com/index.php?topic=37642.msg1404006#msg1404006

showing for the first time (this has not been previously shown anywhere else) what the stress (force/unitArea) on the Big Base looks like.

The stress tensor is obtained using Wolfram Mathematica, post-processed from the transient Finite Difference (using Meep) solution for RF feed ON for an EM Drive.

Notice that the stress at the Big Base is pointed in the direction from the small base towards the big base, as required for a recoil motion to take place and accelerate the EM Drive in the opposite direction, towards the small base. This is in accord with Dr. White and Shawyer.  (It may also work with conventional microwaving of moist air resulting in plasma ions leaking out of the EM Drive or stressed axionic dark matter for example ) However, the stress is not uniformly distributed through the big base (at least for the mode shape TM11 excited in this example) but instead it is distributed mainly in the circumferential outer periphery of the Big Base.

The stress at the Big Base is similar to the one shown at section x=38 near the Big Base, previously shown in this post http://forum.nasaspaceflight.com/index.php?topic=37642.msg1404000#msg1404000, except that the stress is significantly higher and much concentrated at the copper and much less at the center of the Big Base.

   ... snip ...
This seems consistent with results from my attempts to measure force from circular gaps cut in the base of the frustum while investigating evanescent waves. The detected F/P was much higher as the gap approached the outer conic (copper) section. That is, given a circular gap with center radius r, the F/P increased very sharply as r approached the end radius of the cavity. F/P also increased sharply as the width of the gap diminished. Unfortunately both F and P diminished as the gap narrowed, going to zero as the gap closed.

A high F/P is good, but with F in the nano range or smaller, not very practical. That was my evanescent wave conclusion. Of course with slightly larger gaps, higher levels of force were detected with the cavity still resonating strongly.  But F/P never approached values measured experimentally, not repeatably.
« Last Edit: 07/15/2015 05:36 AM by aero »
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Offline WarpTech

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Your great enlightenment (what, number ten is it now in the series of consecutive samadhis?) is unfortunately not shared by me, despite its clearly blinding power over the transformation of your logics. It must be fun in there.

Look: EmDrive is not a rocket, which is why we call it propellantless. Dear Tsiolkovsky had the right of it when he defined exhaust velocity as an invariant when measured with respect to the rocket. Such luxury you do not have because you have no exhaust to manipulate here.

Additional to that misunderstanding (masquerading in your mind, apparently, as a revelation) comes your further misunderstanding that 'v' is relative to the EmDrive. The 'v' used in my proof, clear to most I should have thought, is defined with respect to the original inertial frame in which motion began. In no way should you attempt to interpret it as you now have as relative to the EmDrive itself, for that is not how it is defined in my proof.
You do realize don't you? That you are equating 2 energies, Ein = Eout, that are measured in 2 different reference frames, by two different observers. One that is inertial and one that is not. I thought you or @wallofwolfstreet, said to use an accelerometer and a clock. Such a measurement of velocity is measured in the frame of the rocket.
Todd

Offline deltaMass

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Even with very modest k (say 10-6 N/W) one can achieve excellent mission performance when lots of power is available. Let's go to Pluto (40 AU, 100 Kg). With 1 MW power it takes 1.1 years and maximum speed is 0.1%c.

Offline TheTraveller

The group velocity in a waveguide is by definition;  vg = dw/dk, correct?

vg = c * (lambda0 / lambdag)

where

lambda0 = free wavelength
lambdag = guide wavelength
« Last Edit: 07/15/2015 05:58 AM by TheTraveller »
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Offline tleach

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In other words, apparently EM waves in a waveguide behave differently than matter in a rocket, in this regard.
Or am I botching it again?
Todd

Don't be so hard on yourself!  Leave that to deltaMass ;-)  At least you can do math.  Many of us lurkers have to forcibly uncross our eyes after reading some of the posts on this forum.

I guess what I'm trying to say is, "don't throw the baby out with the bath water."  I don't think you can treat the EM waves in a waveguide exactly like matter in a rocket.  That just nets you a photon rocket, which only generates thrust via emissivity.

I think what your theory is describing (and please forgive me if I'm way off base here) is a frustum shape / EM frequency combo wherein the EM waves propagate down an a tapered waveguide.  As the photons propagate through the tapered waveguide, they repeatedly reflect off of the inside of the walls.  With each bounce, some of the photons are absorbed into the frustum and these photons transfer their momentum into the frustum, exerting Radiation Pressure by Absorption.

Because of the waveguide's taper, the expanding wavefront's frequency drops with each consecutive reflection.  As the frequency shifts, that extra momentum is once again transferred into the frustum in the form of Radiation Pressure By Reflection.

Any photons that make it to the end of the tapered waveguide are emitted and thereby transfer some of their momentum to the frustum via Radiation Pressure by Emission

We are seeing evanescent waves in these images! The power diminishes (exponentially?) as the waves move from front to back. After each reflection, the poynting vectors that hit the wall give up some momentum to the frustum pushing it forward, and redirect themselves more toward the x direction. After each bounce, the vector loses momentum and energy due to heat from copper losses. The less energy and momentum it has when it arrives at the big end plate, and the lowest angle of incidence that can be achieved, the more thrust will be harnessed. Makes me think that "Brass" used by Juan Yang may be better than copper. It's resistivity is 5x higher, and a superconductor may not work as well. Based on this, longer and less taper is better, but I have not calculated an optimum design factor yet.
Todd

What you end up with, is an EM drive wherein Radiation Pressure by Absorption is ADDED to Radiation Pressure by Reflection and then ADDED to Radiation Pressure by Emission.  It's more like a "Perfect Photon Rocket" with all (or some arbitrarily high percentage) of the generated EM photons' momentum being converted directly into momentum of the frustum (i.e. kinetic energy).

Maybe?
T. Thor Leach

Offline TheTraveller

Even with very modest k (say 10-6 N/W) one can achieve excellent mission performance when lots of power is available. Let's go to Pluto (40 AU, 100 Kg). With 1 MW power it takes 1.1 years and maximum speed is 0.1%c.

I'm fairly sure 1N/kW is doable without going superconducting.

With 1MWe that gives us 1,000N pushing 100kg at 1g.

At 40AU and doing a 180 deg flip 1/2 way then decelerating I get

max vel: 0.026c
transit time: 18 days

However 100kg for 1MWe is not a real number.

Using the IXS Clark example: 2MWe at 90t and using 1N/kW EMDrives we get:

2,000N pushing 90t = 0.0023g half way there, then doing a 180 deg flip and decelerating the last half

Max vel: 0.0012c
transit time: 377 days

ref:
http://nathangeffen.webfactional.com/spacetravel/spacetravel.php

Here are Dr. White's Pluto calcs for the IXS Clark with either 0.4N/kW or 4N/kW EMDrives at 2MWe and a 90t crewed ship.
« Last Edit: 07/15/2015 06:24 AM by TheTraveller »
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Offline TheTraveller

Because of the waveguide's taper, the expanding wavefront's frequency drops with each consecutive reflection.  As the frequency shifts, that extra momentum is once again transferred into the frustum in the form of Radiation Pressure By Reflection.

The Group Velocity and Guide wavelength change as the tapered waveguides diameter changes. The EM waves frequency (Guide Wavelength) doesn't change because of the end plate bounce.
« Last Edit: 07/15/2015 06:30 AM by TheTraveller »
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Offline tleach

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The Group Velocity and Guide wavelength change as the tapered waveguides diameter changes. The EM waves frequency (Guide Wavelength) doesn't change because of the end plate bounce.

WarpTech's frustum doesn't have an end plate.
« Last Edit: 07/15/2015 06:31 AM by tleach »
T. Thor Leach

Offline TheTraveller

The Group Velocity and Guide wavelength change as the tapered waveguides diameter changes. The EM waves frequency (Guide Wavelength) doesn't change because of the end plate bounce.

WarpTech's frustum doesn't have an end plate.

Was responding to this statement

Quote
Quote from: tleach on Today at 06:22 AM
Because of the waveguide's taper, the expanding wavefront's frequency drops with each consecutive reflection.  As the frequency shifts, that extra momentum is once again transferred into the frustum in the form of Radiation Pressure By Reflection.
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Offline TheTraveller

Good video on making a DIY VSWR meter. Will work fine at 2.45GHz:


With a variable frequency generator and small Rf amp, a VSWR reading can sweep the frequency range and find the frequency where the Return Loss dB (converted from VSWR) is the max. Then knowing the max Return Loss dB, the side frequencies where the Return Loss is 3dB less than the max dB can be found and the unloaded cavity Q determined as follows:

Unloaded Q = Best Return Loss dB frequency / (high frequency down 3dB - low frequency down 3dB)

If you are not into building the VSWR meter, here is a review of a $90 EBAY VSWR 2.45 GHZ VSWR meter:


If you are interested in obtaining this device, please read the comments on the YouTube page so you buy the correct 2.45GHZ version.

Quote
This Redot model is the 5011 (specifically for 2.4 GHz), not the 1050A (specifically for 100 - 500 MHz).

Here is one source, I'm sure there are others:
http://www.aliexpress.com/item/FREE-SHIPPING-REDDOT-5011-SW-M2400-2-4g-WLAN-WIFI-DIGITAL-SWR-METER-N-Female-Conntctor/1858620818.html

Or you can go fo the Daiwa CN-801SII UHF/VHF RF Wattmeter - GSM/Cellular at $270
http://www.aliexpress.com/item/FREE-SHIPPING-REDDOT-5011-SW-M2400-2-4g-WLAN-WIFI-DIGITAL-SWR-METER-N-Female-Conntctor/1858620818.html

For the extra $200, I would buy the Daiwa even though it doesn't have a USB interface.
« Last Edit: 07/15/2015 08:34 AM by TheTraveller »
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Offline SeeShells

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Would anyone be interested that would model a perforated copper sheet for me in meep?

I'll post the specs of the sheets and angles to run at.

Shell



Excellent point.  There is no fundamental problem whatsoever in modelling this in Meep.  This is what FD and FE programs are for.

It is a pre-processing problem.  It requires for somebody to write a pre-processing mesh routine to locate copper and holes in a grid with holes in them.   The number of nodes required to do this will be much, much larger than the present number of nodes used, and therefore the amount of computer memory and computer time will be much larger.

Another way to handle this (not requiring a different mesh than presently used) would be to use an equivalent constitutive model for the copper with holes (letting wavelengths small enough pass through holes and large wavelengths not pass through, and distributing this wavelength dependence at every node in an average sense).  For example, one may start by modeling a problem with a known exact solution, either a rectangular or cylindrical cross-section cavity of smaller dimensions, modeled with and without the perforated holes to see what difference it makes if any, and if it makes a difference, work out a model for the copper with holes using an equivalent model with copper with no holes for the truncated cone.  Theoretical and/or experimental papers on the effect of perforation on microwaves waveguides would be very helpful for this.  The simplest thing is to refer to papers and ascertain whether the effect of the small wavelength is negligible and henceforth deduce that the effect of the holes is negligible.

The practical problem of holes is lack of stiffness, and this compliance leading to distortion that may affect the Q. 

If these effects are of interest, I would start by modeling the hexagonal cross-section and see what difference that makes.  (The hexagonal model will take much less computer resources than modeling the holes).

A good way to start off the test could be. Let's just take the model Aero has NSF-1701 and model the large end plate, leave the rest of the model solid. We've talked about the end plate having holes or a concentric ring pattern before.  Well tested design in meep and your simulations.

First sheet
Hole pattern 1/16″ on staggered 3/32″ .020" thick

Second sheet
Hole pattern 1/32' on staggered 1/4"  .040 thick

pdf paper of perforated sheets included

Online Rodal

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See this article on a holey waveguide analysis with Meep:

http://ab-initio.mit.edu/wiki/index.php/Meep_Tutorial/Band_diagram,_resonant_modes,_and_transmission_in_a_holey_waveguide

If anybody attempts to model the bases with holes in Meep, please note that this would involve a greater number of FD nodes that at present in the circular cross-section, and therefore the number of FD nodes in the circular cross section would end up  being greater than in the longitudinal direction (unless those are also increased).  If the FD cell defined by a number of FD nodes is too distorted (a long rectangle instead of an ideal square, or worse a geometrical figure with very different inscribed angles) numerical issues ensue, so please look into acceptable Finite Difference mesh distortion before meshing and calculating.
« Last Edit: 07/15/2015 09:45 AM by Rodal »

Online Rodal

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@Rodal:
Is it possible with this data to confirm or deny the correctness of 2*P/c versus 2*P/vg?
IMHO, unfortunately at the moment,  not possible with the limited data available to me from the csv file  .  For group velocity one needs to perform a differentiation (the derivative of frequency with respect to wavevector) .  Here:  they show a routine that calculates the group velocity:  http://ab-initio.mit.edu/wiki/index.php/Meep_Tutorial/Optical_forces using MPB .

Please note that in http://ab-initio.mit.edu/wiki/index.php/Meep_Tutorial/Optical_forces  they state: 


Quote from: Meep ab initio
since we know that the total power transmitted through the waveguide is P = vg U / L (vg is the group velocity, L is the waveguide length and U is defined as before [ U is the total energy of the electromagnetic fields])

MPB  (MIT Photonic-Bands package) can calculate the group velocity:  http://ab-initio.mit.edu/wiki/index.php/MPB_User_Reference, given a set of eigenstates at a given k-point, computing their group velocities (the derivative of frequency with respect to wavevector) using the Hellman-Feynmann theorem  (  https://en.wikipedia.org/wiki/Hellmann%E2%80%93Feynman_theorem ).

With Meep for a constant cross-section waveguide you could compute the band diagram exciting a single mode at a time using a narrow-band source and compute the group velocity as the ratio of flux to energy density (but this assumes this known relationship):

Cycle Averaged (Power/unitArea)/(Energy/unitVolume) = Cycle Averaged (Poynting flux)/(Energy Density)

                                                                                 = group velocity  (see slide attached below)

Interesting reference: Group Velocity and Finite Difference Methods (deals with Group Velocity and Finite Differences in general, not just electromagnetic):  http://math.oregonstate.edu/~bokilv/MTH453-553S07/TrefethenGroup.pdf
« Last Edit: 07/15/2015 01:42 PM by Rodal »

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