Author Topic: EM Drive Developments - related to space flight applications - Thread 3  (Read 1876215 times)

Offline Star One

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Well, I'm in the following group:

1: Shawyer's paper is completely confused. Right at the start he attributes the force to a greater radiation pressure upon the wide end, yet it pushes itself small end forward; this is based upon a completely confused discussion of reaction forces and thrust. This notion that there would be no force on the side walls "according to Maxwell's equations" is simply flat-out wrong. Maxwell's equations, as applied, yield zero thrust; the force on the side walls precisely balances out the pressure difference between the ends. (They're also Lorentz invariant so there's no special relativity corrections to be made)

2: All explanations where the measured force is impacted upon the cavity walls by incident electromagnetic radiation are likewise wrong, whenever they involve speculations about the quantum vacuum or not. The measured force corresponds to the incident electromagnetic radiation deviating from conventional predictions by >50% (Shawyer, Chinese results), or >2.5% (EW results), which is in gross contradiction to experiments that measure electromagnetic radiation directly (many are precise to parts per billion or better).

3: Regarding EW's experiments, their readings contradict each other (when flipped 180 degrees). Other experiments are substantially worse still, with high voltage wires, stiff waveguides being heated, etc. pushing the cavity mechanically.

What results do you expect to get if there's no thrust but you got a bunch of high voltage wires, substantial heat, electrical current in the wires, and vibration? You can't seriously expect to get a literal zero.

And you think the professional experimenters, not talking about the DIY versions here, haven't already considered this. You put forward an argument as if this was the first time anyone had thought of these issues. A greater part of the last thread was examining such issues amongst other things.
I don't think many competent physicists have any interest in Shawyer's theories. What he's writing is so wrong it is painful to read. Experimental physics requires, at least, good knowledge of mechanics, and the glaring bit about the pressure puts anyone with such knowledge off.

I'm less interested in the theories than the practical results at this stage in time. Attempts at explanations should not hinder the experimental investigation. Just because no one has put a widely accepted theory together yet should dissuade scientific investigation.

Offline kdhilliard

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Brand new. Interview with the inventor of EmDrive. Good info in there.



That was done by the same guy who posted a 50 minute recording of an impromptu presentation by Shawyer last year.





These offer a good introduction to Shawyer's ideas, but without any mathematical formulas.  They don't have any live video, but instead show slides.  It was recorded in a cafe and there is a lot of background noise, but it is sill easy to make out all that Shawyer says.

I listened to these last night hoping to glean some hints that would explain Shawyer's reasoning for why he expects the drive to accelerate small end first.

Quote from: Part 1, 0:33
What the EmDrive thruster does is to produce a force, which we call the thrust, in one direction.  This is a force that you can measure.  If you put your hand against the end plate that's producing the thrust you'll feel it pushing against you.  And, as with all machines that follow Newton's principles, it will therefore accelerate in the opposite direction.  So this is not a reactionless thruster, because those things just don't exist outside of science fiction, but it is a propellantless thruster.

That leaves me bewildered as to his thought process.  I hear what he says about feeling the force of the large end plate against your hand, but how does that make the drive accelerate in the other direction when it should just balance forces to hold the drive in place until you pull your hand out of the way and let it accelerate large end first?

By analogy, consider a ping pong ball being held underwater in a pool.  The force exerted by the water pressure onto the ball's surface increases with depth, so the upward forces on the lower portions of the ball are greater than the downward forces on its upper portions, resulting in a net upward force.  Your stationary  hand on top of the ping pong ball feels this force, and acts to keep the ball from accelerating upward, but it doesn't cause the ball to accelerate downward, and I'm sure that Shawyer wouldn't claim that it would.

I guess this bothers me so much because Shawyer sounds like a smart guy, and this appears to be such a simple and obvious contradiction of logic that there must be more to his argument.  (Unless he's just pulling our legs.)


Traveller, you seem to be the one here most familiar with Shawyer's works.  Do you follow his line of reasoning?

~Kirk

Offline deltaMass

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This is what happens when a smart person is faced with data that is irreconcilable with known physics. They know that they are expected to explain it, and they also know that they cannot. In this case Shawyer simply babbles nonsense (and I'm putting that as kindly as I can without resorting to insult). It's cognitive dissonance in the flesh.
« Last Edit: 05/23/2015 08:05 PM by deltaMass »

Offline Star One

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This is what happens when a smart person is faced with data that is irreconcilable with known physics. They know that they are expected to explain it, and they also know that they cannot. In this case Shawyer simply babbles nonsense (and I'm putting that as kindly as I can without resorting to insult). It's cognitive dissonance in the flesh.

If there's anything this I wouldn't bet on it being outside known physics, rather known physics but in a different way if that makes sense.

Offline deltaMass

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Not really :). If it were known physics, you could tell us all about it!

Offline Rodal

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I just opened my box of Mechanics books.  I found Maxwell's book

Matter and motion
by James Clark Maxwell

Notes and appendices by Sir Joseph Larmor
Cambridge University, 1920

page 40

The Third Law of Motion

Law III. Reaction is always equal and opposite to action, that is to say, the action of two bodies upon each other are always equal and in opposite directions.

When the bodies between which the action takes place are not acted on by any other force, the changes in their respective momenta produced by the action are equal and opposite directions.

The changes in the velocities of the two bodies are also in opposite directions, but not equal, except in the case of equal masses.  In other cases the changes of velocity are in the inverse ratio of the masses.



Following D'Alembert's convention of fictional inertial forces.  See  http://forum.nasaspaceflight.com/index.php?topic=37642.msg1378839#msg1378839  for the more usual convention

So, Shawyer has two forces that he says follow Newton's 3rd law, and Shawyer says that he follows Maxwell

then, from the image above, you must have

Summation of forces = 0

Reaction force vector is in opposite direction to Thrust force vector.  Hence they have opposite sign.

Reaction - Thrust =0 (essentially Shawyer shows a D'Alembert's Free-Body-Diagram  https://en.wikipedia.org/wiki/D%27Alembert%27s_principle where the forces sum up to zero)

Reaction = Thrust

assign a portion of the mass of the truncated cone to the Reaction force and the other portion to the Thrust force

massReaction + massThrust =total mass

then

massReaction*accelerationReaction = massThrust*accelerationThrust

imagine that the truncated cone is split apart as a gun and a bullet, essentially when you turn on the power to the EM Drive there is an explosive force inside the sends the Small End and the Big End in opposite directions,

In that case the Reaction is the force on the bullet (the Small End) and the Thrust (force on the Big End) is the recoil force on the gun, then acceleration of the bullet is

accelerationBullet = (massGun/massBullet)*recoilAccelerationGun

(the recoil acceleration is in opposite direction to the bullet acceleration)

no problem with understanding that.  However, the EM Drive remains as one EM Drive (it does not separate into two wagequides), therefore we must have :

accelerationThrust = - accelerationReaction = accelerationEMDrive

(both the acceleration of the Big End and the Small End are in the same direction)

massReaction*accelerationEMDrive= massThrust*(-accelerationEMDrive)

therefore:

massReaction = -  massThrust

in other words, for what Shawyer claims that happens to happen, one must have the mass associated with the Thrust force to be negative mass

According to his theory, separating the EM Drive into two distinct waveguides (instead of one closed cavity), one waveguide is associated with the Big End and the other waveguide is associated with the Small End.  Then for the small end to accelerate with the Reaction Force, that means that the portion of the total mass associated with the Thrust force, the mass of the Big End waveguide, must have negative mass.

and the total mass of the EM Drive must be zero:

massReaction + massThrust = total mass

                                           = 0



Again:

1) Mass of waveguide associated with the Small End is positive, normal mass

2) Mass of waveguide associated with the Big End is negative, exotic mass

3) Total mass of EM Drive cavity is zero.



Conclusion: unless the total mass of the EM Drives being experimented by Shawyer is zero, and a portion of their mass (associated with the Big End) is negative, exotic mass, his theory cannot explain what is being claimed


Quote from: Wolfgang Pauli
not even wrong
« Last Edit: 05/23/2015 11:20 PM by Rodal »

Offline deltaMass

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Interesting. Maybe he should have a chat with Professor Woodward  8)

Offline CW

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(...)

Reaction - Thrust =0 (essentially Shawyer shows a Free-Body-Diagram where the forces sum up to zero)

(...)

Dear Dr. Rodal,

If, as you write, the following is supposed to be true:

   Reaction - Thrust = 0        | with Thrust, according to Newton, being equal to (-Reaction), then it follows

   Reaction - (-Reaction) = 0        or
   Reaction + Reaction = 0

That cannot be right. Once the arrow or vector convention is set, all vectors must be treated equally. Vectors are simply added. Vector subtraction a-b is also just vector addition a+(-1)*b .

The equation should be    Reaction + Thrust = 0 , under the premise that anyone even gives a darn about conventions anymore.
« Last Edit: 05/23/2015 09:26 PM by CW »
Reality is weirder than fiction

Online WarpTech

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(...)
therefore:

massReaction = -  massThrust

in other words, for what Shawyer claims that happens to happen, one must have the mass associated with the Thrust force to be negative mass

According to his theory, separating the EM Drive into two distinct waveguides (instead of one closed cavity), one waveguide is associated with the Big End and the other waveguide is associated with the Small End.  Then for the small end to accelerate with the Reaction Force, that means that the portion of the total mass associated with the Thrust force, the mass of the Big End waveguide, must have negative mass.

and the total mass of the EM Drive must be zero:

massReaction + massThrust = total mass

                                           = 0



Again:

1) Mass of waveguide associated with the Small End is positive, normal mass

2) Mass of waveguide associated with the Big End is negative, exotic mass

3) Total mass of EM Drive cavity is zero.



Conclusion: unless the total mass of the EM Drives being experimented by Shawyer is zero, and a portion of their mass (associated with the Big End) is negative, exotic mass, his theory cannot explain what is being claimed


Quote from: Wolfgang Pauli
not even wrong

The "effective mass" term is proportional to 1/cut-off-wavelengths at each end. If we define the baseline at the small end, the "relative" effective mass at the big end is negative.

« Last Edit: 05/23/2015 09:09 PM by WarpTech »

Offline txdrive

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The other way to put it - we can actually set up a situation where radiation pressure upon the big end is greater than that upon the little end. In space, put a flashlight inside the cavity, pointed upon the big end (which is not perfectly reflective). The radiation pressure upon the big end will be greater than that upon the small end, and the cavity will accelerate big end forward.

(The flashlight, if unsupported, will move in the opposite direction, like a photon rocket, but it could in principle be held in place, e.g. using magnetic levitation)
« Last Edit: 05/23/2015 09:12 PM by txdrive »

Online dustinthewind

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1) Mass of waveguide associated with the Small End is positive, normal mass

2) Mass of waveguide associated with the Big End is negative, exotic mass
...


Quote from: Wolfgang Pauli
not even wrong

I think that is effectively the concept behind phased based propulsion.  The current working with the electric field of light, say at the top of the cavity, provides normal repulsion due to the changing magnetic field (or the currents resistance to encountering a changing magnetic field makes a counter current).  Counter currents repel.  If at the bottom the current may be working against the electric field of light and we don't get repulsion as the small end.  Rather the bottom experiences attraction which is opposite of the normal repulsion due to changing magnetic fields (the light appears to have negative mass and attracts the bottom plate).  I am not saying this is what is happening but I suspect it is a possibility. 

I think if it was happening the bottom plate might lose its Q and heat up more so than the top part.  Is the ratio of heat on the bottom plate in ratio to the top as it should be? 

The drive still has positive mass and resists being accelerated but the effective mass of the radiation and near field should be imbalanced in effective mass from front to back. 
« Last Edit: 05/24/2015 12:47 AM by dustinthewind »

Offline txdrive

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Quote from: txdrive
I don't think many competent physicists have any interest in Shawyer's theories. What he's writing is so wrong it is painful to read. Experimental physics requires, at least, good knowledge of mechanics, and the glaring bit about the pressure puts anyone with such knowledge off.

Yet both Shawyer and the Chinese claim their theories closely calculate the value of their measured thrust? Surely that must open the possibility of their unconventional application of classic theory being correct and that no new physics is involved nor needed?
It's like you come across some "make money online" ad, and they're claiming that their earnings match their calculations, which have glaring arithmetical errors. Anyone can claim anything.
« Last Edit: 05/23/2015 09:21 PM by txdrive »

Offline Rodal

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...
The equation should be    Reaction + Thrust = 0 , under the premise that anyone even gives a darn about conventions anymore.
I'm trying to interpret Shawyer's diagram using D'Alembert's principle.  I cannot make sense of his "convention" for his force construction to work.  Maybe something is lost in translation or my imagination is not good enough to understand what he is showing.

His force convention does not follow any of the books I have in Mechanics (the fact that he has these two equal an opposite forces which should result in a body in equilibrium, hence having no acceleration).
It leads to a contradiction whatever way I adopt for a consistent convention.


Let's say that we instead interpret Shawyer as you suggest.

Then work out the bullet/gun split: one comes up with the accelerations having different signs which I agree is a more conventional view.  If one consistently follows the same convention all the way through, (since the bullet and the gun both have real positive masses), then one ends up with the same result I have above that the mass of the Big End is the negative of the mass of the Small End and that the Total Mass of the EM Drive must be zero, for  Shawyer's construction to hold in an EM Drive that does not split apart and accelerates as unit in one direction.
« Last Edit: 05/23/2015 09:48 PM by Rodal »

Offline phaseshift

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The other way to put it - we can actually set up a situation where radiation pressure upon the big end is greater than that upon the little end. In space, put a flashlight inside the cavity, pointed upon the big end (which is not perfectly reflective). The radiation pressure upon the big end will be greater than that upon the small end, and the cavity will accelerate big end forward.

(The flashlight, if unsupported, will move in the opposite direction, like a photon rocket, but it could in principle be held in place, e.g. using magnetic levitation)

Is what I'm hearing is that Shawyer is saying: In space I can literally push against the windshield with a flashlight, but not with my hand, and get thrust?
"It doesn't have to be a brain storm, a drizzle will often do" - phaseshift

Offline CW

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...
The equation should be    Reaction + Thrust = 0 , under the premise that anyone even gives a darn about conventions anymore.
I'm using D'Alembert's principle looking at Shawyer's diagram.  His force convention does not follow any of the books I have in Mechanics (the fact that he has these two equal an opposite forces which should result in a body in equilibrium, hence having no acceleration).


Let's say that we instead interpret Shawyer as you suggest.

Then work out the bullet/gun split: one comes up with the accelerations having different signs which I agree is a more conventional view.  If one consistently follows the same convention all the way through, for the bullet and the gun to both have real positive masses, then one ends up with the same result I have above that the mass of the Big End is the negative of the mass of the Small End and that the Total Mass of the EM Drive must be zero, according to Shawyer.

I fear that the available documents from Mr. Shawyer are unusable for any reasonable discussion. Judging by the available reports of a number of groups telling that something seems to or is going on, I feel that Mr. Shawyer might have found something by sheer coincidence. It reminds me of the logical implication that tells us that starting from a wrong premise, any conclusion is possible - even the right one.
;)
« Last Edit: 05/23/2015 09:51 PM by CW »
Reality is weirder than fiction

Offline Star One

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...
The equation should be    Reaction + Thrust = 0 , under the premise that anyone even gives a darn about conventions anymore.
I'm using D'Alembert's principle looking at Shawyer's diagram.  His force convention does not follow any of the books I have in Mechanics (the fact that he has these two equal an opposite forces which should result in a body in equilibrium, hence having no acceleration).


Let's say that we instead interpret Shawyer as you suggest.

Then work out the bullet/gun split: one comes up with the accelerations having different signs which I agree is a more conventional view.  If one consistently follows the same convention all the way through, for the bullet and the gun to both have real positive masses, then one ends up with the same result I have above that the mass of the Big End is the negative of the mass of the Small End and that the Total Mass of the EM Drive must be zero, according to Shawyer.

I fear that the available documents from Mr. Shawyer are unusable for any reasonable discussion. Judging by the available reports of a number of groups telling that something seems to or is going on, I feel that Mr. Shawyer might have found something by sheer coincidence. It reminds me of the logical implication that tells us that starting from a wrong premise, any conclusion is possible - even the right one.
;)

I agree. Even with my limited understanding I've been left scratching my head.:)

Offline Rodal

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Following Newton's 2nd Law convention this time

Matter and motion
by James Clark Maxwell

Notes and appendices by Sir Joseph Larmor
Cambridge University, 1920

page 40

The Third Law of Motion

Law III. Reaction is always equal and opposite to action, that is to say, the action of two bodies upon each other are always equal and in opposite directions.

When the bodies between which the action takes place are not acted on by any other force, the changes in their respective momenta produced by the action are equal and opposite directions.

The changes in the velocities of the two bodies are also in opposite directions, but not equal, except in the case of equal masses.  In other cases the changes of velocity are in the inverse ratio of the masses.



So, Shawyer has two forces that he says follow Newton's 3rd law, and Shawyer says that he follows Maxwell

Reaction = - Thrust

assign a portion of the mass of the truncated cone to the Reaction force and the other portion to the Thrust force

massReaction + massThrust =total mass

then

massReaction*accelerationReaction = - massThrust*accelerationThrust

imagine that the truncated cone is split apart as a gun and a bullet, essentially when you turn on the power to the EM Drive there is an explosive force inside the sends the Small End and the Big End in opposite directions,

In that case the Reaction is the force on the bullet (the Small End) and the Thrust (force on the Big End) is the recoil force on the gun, then acceleration of the bullet is

accelerationBullet = (massGun/massBullet)*(-  recoilAccelerationGun)

(the recoil acceleration is in opposite direction to the bullet acceleration)

no problem with understanding that.  However, the EM Drive remains as one EM Drive (it does not separate into two wagequides), therefore we must have :

accelerationThrust = accelerationReaction = accelerationEMDrive

(both the acceleration of the Big End and the Small End are in the same direction)

massReaction*accelerationEMDrive= - massThrust*accelerationEMDrive

therefore:

massReaction = -  massThrust

in other words, for what Shawyer claims that happens to happen, one must have the mass associated with the Thrust force to be negative mass

According to his theory, separating the EM Drive into two distinct waveguides (instead of one closed cavity), one waveguide is associated with the Big End and the other waveguide is associated with the Small End.  Then for the small end to accelerate with the Reaction Force, that means that the portion of the total mass associated with the Thrust force, the mass of the Big End waveguide, must have negative mass.

and the total mass of the EM Drive must be zero:

massReaction + massThrust = total mass

                                           = 0



Again:

1) Mass of waveguide associated with the Small End is positive, normal mass

2) Mass of waveguide associated with the Big End is negative, exotic mass

3) Total mass of EM Drive cavity is zero.



Conclusion: unless the total mass of the EM Drives being experimented by Shawyer is zero, and a portion of their mass (associated with the Big End) is negative, exotic mass, his theory cannot explain what is being claimed


Quote from: Wolfgang Pauli
not even wrong
« Last Edit: 05/23/2015 11:01 PM by Rodal »

Offline phaseshift

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Finally completed a simple UI.

One thing to add is the ability to enter the small plate diameter - and switch between small plate diameter or design factor and have the other parameter computed and displayed.

Also a toggle for a small end cylinder. :)

Now to start messing with modes.

I prefer this better than a spreadsheet.  I wrote an Excel like spreadsheet years ago and even with the knowledge I have about them find them constricting.
« Last Edit: 05/23/2015 10:07 PM by phaseshift »
"It doesn't have to be a brain storm, a drizzle will often do" - phaseshift

Offline CW

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(...)

So, Shawyer has two forces that he says follow Newton's 3rd law, and Shawyer says that he follows Maxwell

Reaction = - Thrust

assign a portion of the mass of the truncated cone to the Reaction force and the other portion to the Thrust force

massReaction + massThrust =total mass

then

massReaction*accelerationReaction = massThrust*accelerationThrust

(...)

Dear Dr. Rodal,

I believe your're introducing a sign error in your considerations. It looks to me as if you only take the absolute values or norm of 'accelerationReaction' and 'accelerationThrust' and equate them. If Newton's 3rd law is correctly applied, then it is IMHO written

massReaction*accelerationReaction + massThrust*accelerationThrust = 0,   or
massReaction*accelerationReaction = -massThrust*accelerationThrust

Or is there a reason to ignore the vector directions that eludes me? Otherwise I can easily see, why negative mass seemingly arises. If the EM-drive accelerates, it can't have anything to do with Newton's 3rd law. At least not in our measly 3+1 space, IMHO.
;)
« Last Edit: 05/23/2015 10:35 PM by CW »
Reality is weirder than fiction

Offline Rodal

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(...)

So, Shawyer has two forces that he says follow Newton's 3rd law, and Shawyer says that he follows Maxwell

Reaction = - Thrust

assign a portion of the mass of the truncated cone to the Reaction force and the other portion to the Thrust force

massReaction + massThrust =total mass

then

massReaction*accelerationReaction = massThrust*accelerationThrust

(...)

Dear Dr. Rodal,

I believe your're introducing a sign error in your considerations. It looks to me as if you only take the absolute values or norm of 'accelerationReaction' and 'accelerationThrust' and equate them. If Newton's 3rd law is correctly applied, then it is IMHO written

massReaction*accelerationReaction + massThrust*accelerationThrust = 0,   or
massReaction*accelerationReaction = -massThrust*accelerationThrust

Or is there a reason to ignore the vector directions that eludes me? Otherwise I can easily see, why negative mass seemingly arises. If the EM-drive accelerates, it can't have anything to do with Newton's 3rd law. At least not in our measly 3+1 space, IMHO.
;)

Yes there was a (-) sign missing in one of the equations in second message:  http://forum.nasaspaceflight.com/index.php?topic=37642.msg1378839#msg1378839

(which I had rapidly copied, and I thought that I had put the minus signs everywhere, but I forgot to put one in that equation)

Thanks. Please let me know whether you find other missing signs.

But in the end, it is not an error in convention, it is an error in Shawyer having two forces for an accelerating body in different directions.  For the body to accelerate, the force has to be in the same direction as the acceleration. 

 A rigid body (that does not break apart or elongates like rubber) cannot have inertial forces in opposite directions. There is where the problem lies

I worked it out both ways, and using consistent conventions (no matter what convention) Shawyer's construction implies zero total mass.

If you can make Shawyer's construction to work, with any convention, I would be delighted to see it, as then we could end the theoretical part of the thread saying , AHA ! we got it, there is no violation of CoM, and the EM Drive can be analyzed with classical physics.
« Last Edit: 05/23/2015 11:22 PM by Rodal »

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