EM Drive Developments - related to space flight applications - Thread 2

[phaseshift]

I suggest that we wait for the person who has read most of Shawyer's papers (TheTraveller).  Perhaps TheTraveller can find another paper on the Demonstrator Engine by Shawyer besides the one I quoted, and check whether Shaywer quotes the same DesignFactor 0.844 or the more sensible number 0.484

This is my source: http://www.emdrive.com/IAC-08-C4-4-7.pdf

Meanwhile, if you have the time and find it worthwhile to do so, perhaps you could ascertain whether the following dimensions make sense

(* Shawyer Experimental *)
rfFrequency=2.45*10^9;
cavityLength=0.156;(estimated from photographs)
bigDiameter=0.16; (given by Shawyer)
smallDiameter=0.1025; (obtained from the Design Factor, bigDiameter and frequency provided by Shawyer)
Design Factor = 0.497;

Almost an exact match!  small diameter was .100m and cavityLength was .155m.  2.5 millimeters off on the small diameter is easily within my margin of error.

This is also supports the correctness of the DF equation.

[aero]

Regarding the Demo cavity dimensions - It would be very helpful if someone with knowledge and tools for image interpretation were to look at the Demo cavity as was done for the Flight thruster. In looking at the image again, it appears to me like there might be a cylinder section on both ends. The one on the small end is obvious but is there a short cylinder between the two flanges at the big end? And if so, why? Did Shawyer make his Demo thruster as two cylinders joined by a conic section?

[phaseshift]

Regarding the Demo cavity dimensions - It would be very helpful if someone with knowledge and tools for image interpretation were to look at the Demo cavity as was done for the Flight thruster. In looking at the image again, it appears to me like there might be a cylinder section on both ends. The one on the small end is obvious but is there a short cylinder between the two flanges at the big end? And if so, why? Did Shawyer make his Demo thruster as two cylinders joined by a conic section?

aero,

I considered that when analyzing the dimensions - you don't even need image processing software - look at the reflection of the "workbench" on the cylinder, then on the cone, then on the next 'cylinder'.  It's rather obvious that it is a cylinder rather than more of the cone.  I believe given the dimensions (thickness of this area), and the possibility of a concave surface on the large plate that the larger cylinder contains the large plate.  For some reason, perhaps just because he was still figuring out construction methods, Shawyer did it this way rather than the fully external plate on the higher fidelity devices.

[aero]

Regarding the Demo cavity dimensions - It would be very helpful if someone with knowledge and tools for image interpretation were to look at the Demo cavity as was done for the Flight thruster. In looking at the image again, it appears to me like there might be a cylinder section on both ends. The one on the small end is obvious but is there a short cylinder between the two flanges at the big end? And if so, why? Did Shawyer make his Demo thruster as two cylinders joined by a conic section?

aero,

I considered that when analyzing the dimensions - you don't even need image processing software - look at the reflection of the "workbench" on the cylinder, then on the cone, then on the next 'cylinder'.  It's rather obvious that it is a cylinder rather than more of the cone.  I believe given the dimensions (thickness of this area), and the possibility of a concave surface on the large plate that the larger cylinder contains the large plate.  For some reason, perhaps just because he was still figuring out construction methods, Shawyer did it this way rather than the fully external plate on the higher fidelity devices.

I think in the past we were enthralled by the geometry of the Eagleworks thruster cavity. Are you confirming that you think there are cylinders on both ends of the Demo frustum section? If so, then how long is, (what is the height of), the big end cylinder (estimate). Then what is the height of the frustum section? Lastly, what is your estimate of the length of the small cylinder? That is, what is the overall electrical length of the EM thruster "Demo"?

Now, can these pieces be related to the terms used in Shawyer's design factor equation?

[phaseshift]

Regarding the Demo cavity dimensions - It would be very helpful if someone with knowledge and tools for image interpretation were to look at the Demo cavity as was done for the Flight thruster. In looking at the image again, it appears to me like there might be a cylinder section on both ends. The one on the small end is obvious but is there a short cylinder between the two flanges at the big end? And if so, why? Did Shawyer make his Demo thruster as two cylinders joined by a conic section?

aero,

I considered that when analyzing the dimensions - you don't even need image processing software - look at the reflection of the "workbench" on the cylinder, then on the cone, then on the next 'cylinder'.  It's rather obvious that it is a cylinder rather than more of the cone.  I believe given the dimensions (thickness of this area), and the possibility of a concave surface on the large plate that the larger cylinder contains the large plate.  For some reason, perhaps just because he was still figuring out construction methods, Shawyer did it this way rather than the fully external plate on the higher fidelity devices.

Are you confirming that you think there are cylinders on both ends of the frustum section? If so, then how long is, (what is the height of), the big end cylinder (estimate). Then what is the height of the frustum section? Lastly, what is your estimate of the length of the small cylinder? That is, what is the overall electrical length of the EM thruster "Demo"?

Now, can these pieces be related to the terms used in Shawyer's design factor equation?

Yes, I am saying that there are two cylinders.

See above for dimensions.   Those are the dimensions I got from the model - the thickness of the large plate is not terribly important - thick enough to contain a spherically concave surface if it is there.  The small plate is at the end of the cone.  Given the possibility of Shawyer mistyping his DF of .844 when it should have been .484 the numbers all work.

[aero]

Yes, I am saying that there are two cylinders.

See above for dimensions.   Those are the dimensions I got from the model - the thickness of the large plate is not terribly important - thick enough to contain a spherically concave surface if it is there.  The small plate is at the end of the cone.  Given the possibility of Shawyer mistyping his DF of .844 when it should have been .484 the numbers all work.

Could you provide the link to the dimensions that you refer to? "See above" leaves a lot of room for me to choose the wrong post. And the length of the cylinder on the big end is kind of important if I were to model it with FDTD software (Meep), as is the length and radius of the small end cylinder and length of the frustum section. Not that I have any immediate plans to do so, but I might model it in 2D just for grins. It could be interesting to see what the wave forms might look like.

[phaseshift]

Yes, I am saying that there are two cylinders.

See above for dimensions.   Those are the dimensions I got from the model - the thickness of the large plate is not terribly important - thick enough to contain a spherically concave surface if it is there.  The small plate is at the end of the cone.  Given the possibility of Shawyer mistyping his DF of .844 when it should have been .484 the numbers all work.

Could you provide the link to the dimensions that you refer to? "See above" leaves a lot of room for me to choose the wrong post. And the length of the cylinder on the big end is kind of important if I were to model it with FDTD software (Meep), as is the length and radius of the small end cylinder and length of the frustum section. Not that I have any immediate plans to do so, but I might model it in 2D just for grins. It could be interesting to see what the wave forms might look like.

My apologies, I thought the post was only a few before this one but it's actually spread out over the last couple pages.

Large diameter: .28m - given by Shawyer
Small diameter: .17m - calculated using .484 DF - measured to be .169m
Cone Length (surface to surface): .187m (measured)

This is why the lengths of the cylinders don't matter - the plate surfaces are at the ends of the cone section.  Though in the case of the small plate it probably starts about 5mm inside the cylinder and is moved forward until there is phase lock - it "should" be at the cone end. Either way the cylinders can be any size as long as they are large enough to contain whatever it is they contain - the large one by the way is "about" an inch thick which is also what I would calculate for the depression on a spherical surface. - I can determine that more precisely if you want but it will have to wait till tomorrow

Cheers

[Rodal]

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

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

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

I have taken out a l0^-1 that so becomes l0^-2.

Sorry, but I still don't understand how that converts c⁴/G to (c⁴/G)^-1.

I suspect that there is a transcription error somewhere, and that (c⁴/G)^-1 was intended all along.

Thanks for pointing out this. I will check all tomorrow. Please, consider that a term c^4/G is coming out from the construction of the tensor in eq.(46). I have checked physical dimensions step by step by I'll redo it for sure.

I don't see any problem with the units the way they are in Eq. 55 and 56.

I see a units problem if you do what @aero is proposing.

Units of (G/c^4) from Eq. 23 is 1/Force,
hence (c^4)/G has units of Force
Units of L from Eq. 55 is  ((c^4)/G )/(length^2)
hence L has units of Force/(Length^2)
L has units of stress (force per surface area)
(U_o)²/µ_o has units of Energy per unit volume (Energy density)
(Marco made this clear in Eq. 22 for example)
Eq. 56 has units of (G/c^4) *(Length^2)*((Energy/Volume)^2)
hence (1/Force)*(Length^2)*(Force*Length/(Length^3))^2
hence (1/Force)*(Length^2)*(Force^2)/(Length^4)
hence Force/(Length^2) , which are the units of stress, which are the proper units of L from Eq. 55

Therefore the factor of G/(c^4) in Eq. 56 has the correct units, and it is the proper factor to use. 
Ditto for Eq. 55 you have to use (c^4)/G in order for L to have the units of stress.
There is no problem with the units in Eq. 55 or in Eq. 56.

On the other hand if you make the change proposed by @aero, the units get screwed up.


The answer to @aero's question,(Why did you invert the c⁴/G term?) is:
that term is due to the equality in Eq. 28 which you used to substitute for l_o, which introduces the inverse square of  c⁴/G .  Also notice that that's the reason that the term (Pi)^2 appeared in Eq. 56. 

This makes perfect sense because that term is the proper term needed to have an equality hold, otherwise one cannot have an equal sign in Eq. 55 and Eq. 56

@aero: Look at Eq. 28 and Eq. 23.  You need to account for the equalities in Eq. 28 and Eq. 23 to follow what happened. You need to substitute the value of lo (from Eq. 28) into Eq. 55, so that Eq. 55 is no longer in terms of lo.

Then set r= r₁ in Eq. 55 and take the limit of Eq. 55 for r₁/r₂ -> 0

[aero]

I guess the step between Eqn. 55 and Eqn 56 is just to large for me.

[WarpTech]

I guess the step between Eqn. 55 and Eqn 56 is just to large for me.

Marco loses me at the end there too. Difficult to follow it all the way through, damn index gymnastics makes my eyes glaze over. One thing to note however, equation 3;

alpha*w = sqrt(4pi*eps0*G) has units of Coulombs/kg.

alpha*w*E has units of acceleration, in the sense that this is a Lorentz force q*E/m, divided by mass. Some people believe this is a gravitational acceleration derived from the EM field, but I do not see it as such. It is many orders of magnitude larger than gravity would be. This is essentially the Planck charge / Planck mass

Todd

[WarpTech]

Warptech

<snip>

Ok
If Im following your thought train correctly then, from my limited perspective on handling microwaves.

The generation side of the circuit requires cyclic refreshing to achieve resonance. {Unknown time-element at the moment}

Once resonance is achieved, {and not before} the energy is dumped into the "Load".

The operational functionality requires simplicity of operation.

The functionality requires the ability to alter running characteristics in realtime.

The length of the initial resonance chamber {ideally} needs to be automatically configurable {hence shawyers use of piezoelectric actuators inside his system}

The mechanical nature of the beam chopper requires simple operation but also an ability to tune in relation to time taken to achieve resonance compared with port-opening cycles. {potentially a software function to find the maximum thrust by adjusting resonance/port operation timing automatically, Microcontrollers are good for this, I use Arduino}

Thinking about the end product it may be easier engineering wise to have multiple attenuation chambers. This allows simplicity in design and beam chopper operation.

2 attenuators at 180, single beam splitter opening with weight adjusted disk to account for mass removed on one side.{less efficient model}

4 attenuation chambers equally spaced so any two are logically 180 degrees from each other, Both can be fed from the resonant chamber simultaneously via 2 opposed holes in the beam splitter. This also allows for slower rotation rate of the splitter as it has 2 holes not 1.

A variable speed rotary port opening  mechanism. The shape of the beam choppers pass-through-port determines the efficiency of the opening process, square, circle, ellipse, triangular

For shorter port opening times use 2 disks counter rotating with respect to each other. {or just smaller port openings}.

Personally I tend to favour 4 attenuators because 2 will be simultaneously active while the resonator recharges to pump the other 2 attenuators. {this is also because I have no idea how long it will take to attenuate the signals in relation to achieving resonance}

? any use or.. just junk?
Terrible drawing but you get the basic concept...

...

This is great arc! You understood what I meant exactly. I had this idea too (without the jockey shorts in the middle there), but I think it's better to use a single resonant cavity for each attenuator and an Iris diaphragm shutter. The advantage being, less distortion of the cylindrical waves as they enter the attenuator because the iris remains circular, and also the ability to use the iris as a high-pass filter for testing purposes.

Regarding the duty cycle, it will depend on the input power, the ability to store energy and the desired thrust. Just keep in mind, if you had the same amplifier setup without the frustum attenuator, it would be more efficient as a microwave photon rocket. I'm afraid this is why I'm not taking the EM drive so seriously anymore. Once I realized I can gain maximum efficiency by just letting the stored energy out as thrust, from a pulsed microwave cavity source of power Q*P, there's not much point in bottling it up in a frustum. The only advantage I see is that it can be used "safely", as opposed to cooking everything down-wind of the exhaust. In space, who cares?

Todd

[deltaMass]

The task of the theorist, surely, is to figure out why the time-averaged thrust is about 1000x that of an equivalently-powered photon rocket in continuous mode. Isn't it?

[deltaMass]

IMO the fastest way to put this to bed or elevate it to the status of a major new propulsion paradigm (let alone new physics) is to stuff it onto a CubeSat and test it in space. And the only way that's going to happen is using laser frequencies, else it won't physically fit.

[ThinkerX]

Regarding the duty cycle, it will depend on the input power, the ability to store energy and the desired thrust. Just keep in mind, if you had the same amplifier setup without the frustum attenuator, it would be more efficient as a microwave photon rocket. I'm afraid this is why I'm not taking the EM drive so seriously anymore. Once I realized I can gain maximum efficiency by just letting the stored energy out as thrust, from a pulsed microwave cavity source of power Q*P, there's not much point in bottling it up in a frustum. The only advantage I see is that it can be used "safely", as opposed to cooking everything down-wind of the exhaust. In space, who cares?

One other possible 'sideways' advantage to the frustum route is this device will generate a lot of heat, which could be converted back into electrical energy with a bit of clever engineering.  Won't get anywhere near unity, but should cut down on the power bill.  Might not be near as much waste heat without the frustum.
 

[The Amazing Catstronaut]

If or when power levels ramp up to "considerable" watts.. we might see unusual things emerge in the environment immediately around the drive. 

I appreciate that you don't have any hard numbers yet to support your conjecture, but what, roughly, do you consider to be "considerable watts"?

[StrongGR]

But then you have that U_0^4 that depends on Q^2 and P^2. As said before, I am not certain that I get a really macroscopic effect even if an interferometric device, sensible enough, can grant observation of the effect. I have to work out some numerics to see really what is going on. There is also the contribution coming from the square of the mode that can take the effect down.

Let's keep working this out.

Hm... from equation (22) I get that a volume integral of the square of the mode times U0/2mu0 is equal to QP/omega. Omega is the resonant angular frequency (which I think is the linear frequency divided by 2pi).
Q is about 10^4 (for the devices we've seen).
Let's say P is about 10^3 (1KW).
omega is about 10^9 (because gigahertz).
mu0 is still about 10^-6.

I don't know about that volume integral, but mu0*QP/omega is about 10^-8. Unless the volume integral is a substantially negative power of ten, this isn't helping much. Also, it looks like increasing the resonance frequency actually makes things worse (is that right?!).
U0^4 becomes about 10^-32 divided by the fourth power of that volume integral.

This effect is starting to become too small.
Still, these are just back-of-the-envelope ballpark calculations, and I can't figure out all those integrals as my maths isn't good enough. So, as you say, we need proper numerics.
I'm just not too hopeful.

Just working!

[Rodal]

This thread has gone over 1 million views !!!!

[Notsosureofit]

I guess the step between Eqn. 55 and Eqn 56 is just to large for me.

Marco loses me at the end there too. Difficult to follow it all the way through, damn index gymnastics makes my eyes glaze over. One thing to note however, equation 3;

alpha*w = sqrt(4pi*eps0*G) has units of Coulombs/kg.

alpha*w*E has units of acceleration, in the sense that this is a Lorentz force q*E/m, divided by mass. Some people believe this is a gravitational acceleration derived from the EM field, but I do not see it as such. It is many orders of magnitude larger than gravity would be. This is essentially the Planck charge / Planck mass

Todd

Yes the apparent acceleration (of the photons in the chamber) is of the order of 10^15 m/s^2 in my calculation.  very large.

[Rodal]

I guess the step between Eqn. 55 and Eqn 56 is just to large for me.

Marco loses me at the end there too. Difficult to follow it all the way through, damn index gymnastics makes my eyes glaze over. One thing to note however, equation 3;

alpha*w = sqrt(4pi*eps0*G) has units of Coulombs/kg.

alpha*w*E has units of acceleration, in the sense that this is a Lorentz force q*E/m, divided by mass. Some people believe this is a gravitational acceleration derived from the EM field, but I do not see it as such. It is many orders of magnitude larger than gravity would be. This is essentially the Planck charge / Planck mass

Todd

Yes the apparent acceleration (of the photons in the chamber) is of the order of 10^15 m/s^2 in my calculation.  very large.

and let's recall that this apparent acceleration of the photons (of the order of 10^15 m/s^2)  in the cavity is invoked by Dr. McCulloch for his theory of Unruh radiation being responsible for the EM Drive's change in momentum  http://www.ptep-online.com/index_files/2015/PP-40-15.PDF

[TheTraveller]

Following Roger Shawyers kindly laid bread crumb trail, my EM Drive spreadsheet now can calc the effective internal guide wavelength and external Rf wavelength that will give resonance from end plate to end plate.

As an example for the Flight Thruster big and small end diameters as below, the required end plate to end plate spacing to achieve resonance with an external Rf of 3.85GHz is as below.

Alteration of either the big, small end or Rf frequency will now automatically generate a new Df and from that the end plate spacing needed to achieve resonance with the external Rf.

big diameter      m   0.2440000
small diameter   m   0.1450000
cavity length      m   0.1603484
rf frequency       Hz   3,850,000,000
Calculated Df      Df   0.49094
slant angle        Deg   28.8

1,000 point numerically integrated guide wavelength of the above example is: 0.0801741816

I will publish the spreadsheet but would 1st like to run / verify it against other frustum dimensions and what the calculated / measured resultant resonance was and in what mode.

With this spreadsheet if we know either end diameter, Rf frequency and Df, the other diameter and spacing can now be determined as all 3 dimensions and external Rf wavelength (4 variables) affect each other.

NEXT STEPS:

1) Determine the best way to inject the coax Rf into the Flight Thruster? Loop or Stub?

2) Determine the best location to inject the Rf into the Flight Thruster?

3) Determine the best way to impedance match the Flight Thruster to the impedance of the Rf generator so as to get optimal VSWR and energy delivery to inside the Flight Thruster?