### Author Topic: EM Drive Developments - related to space flight applications - Thread 3  (Read 1881373 times)

#### WarpTech

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##### Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #1220 on: 06/08/2015 12:09 AM »
...Because you assert that P is frame-dependent. For a long time now we have used P to denote input power Pin.  This can be thought of as being supplied by a battery.

1) You are addressing Todd's paper

2)  When Todd uses P in his equations 2 and 3, P depends on velocity
Then I suggest to Todd that he explicitly differentiate between Pin and Pout in his paper.

How do you define which is which? The equation works both ways. Hydroelectric power for instance.

#### deltaMass

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##### Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #1221 on: 06/08/2015 12:12 AM »
I'm fascinated in a way by WarpTech's insistence of including elements of metric tensors, and of the Lorentz boost gamma, in his calculations.  It is surely clear that space is "almost flat" and that the relative velocities involved in foreseeable lab tests, or even in space in the mid-term, will have a gamma factor as close to unity as makes "no difference".

So Todd, can you quantify please by what kind of percentage your calculations would be affected if you dropped these SR and GR references? 0.000001%? 0.00000000001%? less?

And if you agree that they are so small, then why do you bother with them when they don't change the outcome in a materially significant way? That's a serious question. I just don't get why you would bother with them.

Because, when mass is accelerated, it's inertia content increases. That inertia is stored as length contraction and time dilation of of every sub-atomic particle in the object. In other words, inertial mass curves space-time. Without that, the inertia seems to disappear! Leading to preposterous conclusions like "free energy" and perpetual motion machines. At the human scale of a meter stick moving at v << c, these factors may be imperceivable small, but for millions of sub-atomic particles whose tiny wavelength depends on its momentum,

lambda(t) = h/p(t), i.e., it's inertia content....

ANY warping of space-time is significant.

In the case of your battery scenario. In that situation, you are starting with a total inertial mass energy of;

E = ship(m) + charged battery(M(t)) = (m + M(t))*c^2. The limiting velocity will be determined by the equation;

E2 = ((m + M(t))c2)2 + (p(t)c)2

The thrust-to-power ratio will just tell you how fast you can get there. What happens is, the energy stored in the battery is discharged into accelerating all the particles to a new momentum. If no mass was ejected in the process, then once the ultimate momentum state is reached the battery is discharged and all the sub-atomic particles of mass that were accelerated will have been contracted. Work was done to curve space-time for the matter that was transported, relative to the rest-frame where it started. Without that, there is "hidden inertia", not accounted for, even at low speed.

Todd
So Todd, can you quantify please by what kind of percentage your calculations would be affected if you dropped these SR and GR references? 0.000001%? 0.00000000001%? less?

Take the case, for example of a 100 Kg EmDrive above GSO accelerating for a few minutes with an initial thrust of say 100 uN. Or whatever floats your boat really, so long as  it is not totally beyond our expectations.
« Last Edit: 06/08/2015 12:22 AM by deltaMass »

#### deltaMass

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##### Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #1222 on: 06/08/2015 12:17 AM »
...Because you assert that P is frame-dependent. For a long time now we have used P to denote input power Pin.  This can be thought of as being supplied by a battery.

1) You are addressing Todd's paper

2)  When Todd uses P in his equations 2 and 3, P depends on velocity
Then I suggest to Todd that he explicitly differentiate between Pin and Pout in his paper.

How do you define which is which? The equation works both ways. Hydroelectric power for instance.
I am surprised you would ask that, given the water under the bridge. It is self-evident what the answer is, so I can only assume there's some agenda behind that question. But to be neutral, clear and informative, here is the definition for something like EmDrive
Pin = battery power supplied
Pout = kinetic power of EmDrive

When I use P, you can take it to mean Pin because I rarely if ever talk about Pout explicitly.

#### Rodal

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##### Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #1223 on: 06/08/2015 12:30 AM »
...
I am surprised you would ask that, given the water under the bridge. It is self-evident what the answer is, so I can only assume there's some agenda behind that question. But to be neutral, clear and informative, here is the definition for something like EmDrive
Pin = battery power supplied
Pout = kinetic power of EmDrive

When I use P, you can take it to mean Pin because I rarely if ever talk about Pout explicitly.

Meaning, you should use  Pin,  Pout explicitly from now on, and set the example , that you are telling others to follow.   It would certainly have helped me in reading and interpreting your long post, using P, when addressing Todd's paper.
« Last Edit: 06/08/2015 12:33 AM by Rodal »

#### WarpTech

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##### Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #1224 on: 06/08/2015 12:38 AM »
How do you define which is which? The equation works both ways. Hydroelectric power for instance.
I am surprised you would ask that, given the water under the bridge. It is self-evident what the answer is, so I can only assume there's some agenda behind that question. But to be neutral, clear and informative, here is the definition for something like EmDrive
Pin = battery power supplied
Pout = kinetic power of EmDrive

When I use P, you can take it to mean Pin because I rarely if ever talk about Pout explicitly.

It was a joke. I have a VERY sarcastic sense of humor. LOL! I do solar power, so Pout, Pin, Pout/Pin, Performance Ratio, Incentive factors, all that mumbo jumbo gets so ingrained into my common language that I take it for granted people will interpret it the way I do. I should know better, but hey, I'm faster than I am thorough. Character flaw? "I" know what I meant.

#### not_a_physicist

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##### Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #1225 on: 06/08/2015 12:41 AM »
Every particle of matter in both devices still absorbed the increased inertia when it was accelerated by the rocket engine to the new potential, (v - v0)^2. Note, this is the "change" in velocity, relative to where it started from, relative to its rest frame. The rest frame it started in is a preferred frame for that object, but it was not at rest in any "absolute" sense of the word.

If you and the EM drive were aboard the spaceship after the conventional rocket finished accelerating it, would you be able to take apart and inspect the EM drive to tell that it was "spent" (as in already at its maximum velocity)? Presumably (if I am understanding your predictions correctly), you could tell it was spent by running it and seeing if it accelerated your ship further, but could you tell by looking at its particles?

Thanks for all your patience with us!

No problem. If you are in the same inertial frame as the EM Drive, then you, and all your tools would be scaled as well, so no. On board the rocket however, you would've felt the acceleration when the rocket was running, so you "know" work was done to your body and to the EM Drive to get it there.

Taking it apart in this inertial frame, it would appear totally normal but if you compared your EM Drive to the one you left behind, using Lorentz Transformations between the two frames, you will see yours is different than that one left behind. This is because "you and your EM Drive" are the ones that were accelerated to a higher velocity potential. Work was done to get you there, and you could measure it happening by the physical forces acting on every sub-atomic particle. It is not symmetrical. Frame invariance implies symmetry, this is not symmetrical because only 1 EM Drive was accelerated by the rocket, to this new inertial frame.

...(quote clipped)...

Does that about sum it up? I really do hope I'm helping to facilitate understanding of all these things.

Todd

Thanks for the detailed response! I think I follow your explanation. It is reassuring that the on-board EM drive would appear totally normal to the observer on the spaceship with it after being accelerated by the conventional rocket.

I have a question that this brings up, though: suppose this observer is intent on determining if the EM drive is in its starting rest frame. Taking it apart will do no good, as you say, because his particles and its particles have been equally affected by the acceleration they've felt so far. But, if he turns the EM drive on and tries to accelerate his spaceship with it, it will not cause any acceleration, because it is already at its maximum velocity relative to its rest frame. If he were to have turned it on before firing his conventional rocket, he would have gotten some acceleration out of the EM drive. In this way, he has a experiment to run to tell whether or not the drive is in its starting rest frame.

So, I have a hard time reconciling these two things:
- He can't inspect his EM drive to determine if it is in its rest frame (since it appears normal).
- He can activate his EM drive to determine if it is in its rest frame.

Can you explain that? How does the EM drive "remember" that it is out of its rest frame in a way that is immune to inspection, but accessible by running it? Or is my premise wrong?

Thanks again for all your patience! I have certainly learned a lot from yours and others' explanations in this thread. (I was personally confused about the twin paradox in the way you describe a few weeks ago, for example.)
« Last Edit: 06/08/2015 12:46 AM by not_a_physicist »

#### wallofwolfstreet

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##### Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #1226 on: 06/08/2015 12:45 AM »
Okay.... but how does that solve the problem?  Instead of referring to the gradient of (F/P), you are referring to the gradient of (P/F)2, correct?

Yes, the gradient derivative of (P/F)^2 is an acceleration vector that opposes the thrust.

Quote
So in my example, wtih a ball just sitting in a gravitational field, the F/P ratio is just given by 1/v.  There is no spatial dependence.  It is not uniquely defined.

There is no such thing as the gradient of (P/F)^2 for this case.

So is there something special in my hypothetical case that causes you theory to be inapplicable?

(P/F)^2 = v2^2 - v1^2, the difference between two potentials, better?

Or if I define v1 at infinity to be 0, does that make it clearer? It differs only by a limit of integration and I am showing the indefinite integral solution. I guess it could be expressed clearer, but the meaning is there.

In a gravitational field this would be;

(P/F)^2 = g(h2 - h1), same thing.

(P/F)^2 is not a constant wrt the center of mass in a gravitational field, it depends on the relative distance from it and varies like 1/r, like a potential.

Todd

I really admire you for having the courage to get your ideas out there, especially when those ideas are under your real name.

That said, I have to be honest with you.  I have tried my best to understand where you are coming from using what I was taught, and there is no having it.  I will go on record as saying that I think your paper is very confused, incredibly hand-wavy and outright wrong.  Harsh, but no more harsh than the comments I received when I was first subjected to peer review.

It is trivial to think of situations where the thrust to power ratio is infinite.  A rock in a gravitational field (my original example given here: http://forum.nasaspaceflight.com/index.php?topic=37642.msg1385991#msg1385991).  Two charges spaced apart.  Two magnets spaced apart.  Capacitor plates, solenoids, etc.  Basic household stuff.  By the logic of you paper, none of these objects should ever be able to move, because they have a limiting velocity of 0 (as per equation (5)).  Perhaps foreseeing this issue, you use "constant" as a qualifier to the F/P ratio, yet nothing in your math indicates that a constant F/P should behave any differently than an instantaneous F/P ratio.

I have asked you about the CoE issue, when the drive is expending power while not gaining velocity (as would occur once it has achieved it's max delta-v for it's given F/P) , and I received this answer:

Quote
First of all, although you want there to be "only a single rock", matter is composed of particles which are constituents of fields. The EM field and Gravitational field span the entire universe. So it's not only a rock, it is a rock + its fields. The two are in constant equilibrium, Power-in = Power-out. The power an atom absorbs from the ZPF is equal to the power it radiates back into the ZPF. When this symmetry is broken the rock will accelerate and then you have an external gravitational field. Where as, if you drop in a 2nd object as a "test particle" it will also be in equilibrium with its surrounding ZPF, but when it comes near your "rock" it the symmetry of the interaction between the field and the objects will be broken, and they will attract each other. Just like to boats on a rough sea!

I let this go at the time, but this statement is not grounded in any physics I ever learned.  It honestly reads like a physics word ad lib.

Sorry to come at you like this, but that is peer review.  People reading this forum might read your posts and think they are uncontroversial statements, generally accepted in the mainstream.  That really isn't the case.  Maybe you and I just share different mainstreams.

Quote
(P/F)^2 = v2^2 - v1^2, the difference between two potentials, better?

No, that's not at all better.  For the gradient of a function to exist in any meaningful it must, at the bare minimum, be uniquely defined over the domain of interest.  (P/F)2 is not uniquely defined over any domain.  Maybe this is the place where your "constant" F/P caveat falls into place?  Well, no, because the gradient of a constant is zero, and if the gradient is zero, what in the world is the point of the paper given:
Quote
The gradient of this potential is an acceleration vector that opposes the thrust

That alone would be enough for a a reviewer to send you back to the drawing board.  You say  F/P is constant.  You say the gradient of (P/F)2 is an acceleration vector that opposes thrust.  The gradient is obviously zero however.
« Last Edit: 06/08/2015 12:51 AM by wallofwolfstreet »

#### WarpTech

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##### Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #1227 on: 06/08/2015 01:02 AM »
...
So Todd, can you quantify please by what kind of percentage your calculations would be affected if you dropped these SR and GR references? 0.000001%? 0.00000000001%? less?

Take the case, for example of a 100 Kg EmDrive above GSO accelerating for a few minutes with an initial thrust of say 100 uN. Or whatever floats your boat really, so long as  it is not totally beyond our expectations.

In the paper, I said, "Where, for a constant thrust-to-power ratio of 1 N/W, the limiting velocity is 1 m/s.", I did not use gamma there, because as you imply, it is negligible. Later, I use Newtonian gravity and Newtonian kinetic energy to explain hovering, I didn't use GR metric terms there either. However, if I leave out the gamma's and g_uv in equations (5, 6, 7),  then taking the gradient derivative will be incorrect and it will make a HUGE difference in understanding the model and the mechanism of how it works. Without those factors, what I'm saying will make no sense.

Todd

#### deltaMass

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##### Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #1228 on: 06/08/2015 01:06 AM »
...
So Todd, can you quantify please by what kind of percentage your calculations would be affected if you dropped these SR and GR references? 0.000001%? 0.00000000001%? less?

Take the case, for example of a 100 Kg EmDrive above GSO accelerating for a few minutes with an initial thrust of say 100 uN. Or whatever floats your boat really, so long as  it is not totally beyond our expectations.

In the paper, I said, "Where, for a constant thrust-to-power ratio of 1 N/W, the limiting velocity is 1 m/s.", I did not use gamma there, because as you imply, it is negligible. Later, I use Newtonian gravity and Newtonian kinetic energy to explain hovering, I didn't use GR metric terms there either. However, if I leave out the gamma's and g_uv in equations (5, 6, 7),  then taking the gradient derivative will be incorrect and it will make a HUGE difference in understanding the model and the mechanism of how it works. Without those factors, what I'm saying will make no sense.

Todd
It would be churlish of me to ask you for a percentage estimate for a third time, since it appears clear that I will not receive one.

#### Rodal

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##### Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #1229 on: 06/08/2015 01:14 AM »
...
...
I really admire you for having the courage to get your ideas out there, especially when those ideas are under your real name.

That said, I have to be honest with you.  I have tried my best to understand where you are coming from using what I was taught, and there is no having it.  I will go on record as saying that I think your paper is very confused, incredibly hand-wavy and outright wrong.  Harsh, but no more harsh than the comments I received when I was first subjected to peer review.

It is trivial to think of situations where the thrust to power ratio is infinite.  A rock in a gravitational field (my original example given here: http://forum.nasaspaceflight.com/index.php?topic=37642.msg1385991#msg1385991).  Two charges spaced apart.  Two magnets spaced apart.  Capacitor plates, solenoids, etc.  Basic household stuff.  By the logic of you paper, none of these objects should ever be able to move, because they have a limiting velocity of 0 (as per equation (5)).  Perhaps foreseeing this issue, you use "constant" as a qualifier to the F/P ratio, yet nothing in your math indicates that a constant F/P should behave any differently than an instantaneous F/P ratio.

I have asked you about the CoE issue, when the drive is expending power while not gaining velocity (as would occur once it has achieved it's max delta-v for it's given F/P) , and I received this answer:

...
That alone would be enough for a a reviewer to send you back to the drawing board.  You say  F/P is constant.  You say the gradient of (P/F)2 is an acceleration vector that opposes thrust.  The gradient is obviously zero however.

You are not performing any peer review here, this is not a peer-review journal, and you are not one of the editors.

You are stating your own subjective opinions, anonymously,  using a monicker, in an Internet forum.  That's all.

And concerning peer-review, I have reviewed papers for peer-reviewed journals and I (and the Professors that directed the Lab at  the University) never addressed errors in the papers I reviewed by making subjective general statements like "your paper is very confused, incredibly hand-wavy and outright wrong" .  I always addressed what I thought were the technical errors specifically one by one, and concentrated on the mathematical aspects.
« Last Edit: 06/08/2015 01:30 AM by Rodal »

#### WarpTech

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##### Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #1230 on: 06/08/2015 01:30 AM »
...
I let this go at the time, but this statement is not grounded in any physics I ever learned.  It honestly reads like a physics word ad lib.

Sorry to come at you like this, but that is peer review.  People reading this forum might read your posts and think they are uncontroversial statements, generally accepted in the mainstream.  That really isn't the case.  Maybe you and I just share different mainstreams.

Quote
(P/F)^2 = v2^2 - v1^2, the difference between two potentials, better?

No, that's not at all better.  For the gradient of a function to exist in any meaningful it must, at the bare minimum, be uniquely defined over the domain of interest.  (P/F)2 is not uniquely defined over any domain.  Maybe this is the place where your "constant" F/P caveat falls into place?  Well, no, because the gradient of a constant is zero, and if the gradient is zero, what in the world is the point of the paper given:
Quote
The gradient of this potential is an acceleration vector that opposes the thrust

That alone would be enough for a a reviewer to send you back to the drawing board.  You say  F/P is constant.  You say the gradient of (P/F)2 is an acceleration vector that opposes thrust.  The gradient is obviously zero however.

Where to start? Please read a few things first please, so I don't have to explain it all. Thank you!

https://www.researchgate.net/publication/222942820_Quantum_Ground_States_as_Equilibrium_Particle-Vacuum_Interaction_States
https://www.researchgate.net/publication/24268956_Electromagnetic_Potentials_
Basis_for_Energy_Density_and_Power_Flux
_Based_on_Vacuum_%28Spacetime_Metric%29_Engineering

This is an excerpt from a paper I can't post here, regarding the interaction between a charged particle harmonic oscillator and the EM ZPF, that should explain it more clearly than I can...
Todd
« Last Edit: 06/08/2015 04:29 PM by Chris Bergin »

#### rfmwguy

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##### Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #1231 on: 06/08/2015 01:40 AM »
For those of us working @ 2.4 ghz, here is an academic paper on low cost mw leakage gear. Safety first, friends...

#### Devilstower

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##### Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #1232 on: 06/08/2015 01:44 AM »
While my skills are too rusty to be of much use in elaborating on the theory or design of the EM Drive, I've been following this thread and the previous threads closely. While unable to solve mysteries central to this discussion, it seems we may have inadvertently shed light on two other well-known puzzles—the limiting factor on the Drake Equation and the solution to the Fermi Paradox. To wit: at some point, technological civilizations learn it's possible to construct an over unity device from a common kitchen appliance.

(Apologies, I couldn't resist.)

#### wallofwolfstreet

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##### Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #1233 on: 06/08/2015 01:52 AM »
...
...
I really admire you for having the courage to get your ideas out there, especially when those ideas are under your real name.

That said, I have to be honest with you.  I have tried my best to understand where you are coming from using what I was taught, and there is no having it.  I will go on record as saying that I think your paper is very confused, incredibly hand-wavy and outright wrong.  Harsh, but no more harsh than the comments I received when I was first subjected to peer review.

It is trivial to think of situations where the thrust to power ratio is infinite.  A rock in a gravitational field (my original example given here: http://forum.nasaspaceflight.com/index.php?topic=37642.msg1385991#msg1385991).  Two charges spaced apart.  Two magnets spaced apart.  Capacitor plates, solenoids, etc.  Basic household stuff.  By the logic of you paper, none of these objects should ever be able to move, because they have a limiting velocity of 0 (as per equation (5)).  Perhaps foreseeing this issue, you use "constant" as a qualifier to the F/P ratio, yet nothing in your math indicates that a constant F/P should behave any differently than an instantaneous F/P ratio.

I have asked you about the CoE issue, when the drive is expending power while not gaining velocity (as would occur once it has achieved it's max delta-v for it's given F/P) , and I received this answer:

...
That alone would be enough for a a reviewer to send you back to the drawing board.  You say  F/P is constant.  You say the gradient of (P/F)2 is an acceleration vector that opposes thrust.  The gradient is obviously zero however.

You are not performing any peer review here, this is not a peer-review journal, and you are not one of the editors.

You are stating your own subjective opinions, anonymously,  using a monicker, in an Internet forum.  That's all.

And concerning peer-review, I have reviewed papers for peer-reviewed journals and I (and the Professors that directed the Lab at  the University) never addressed errors in the papers I reviewed by making subjective general statements like "your paper is very confused, incredibly hand-wavy and outright wrong" .  I always addressed what I thought were the technical errors specifically one by one, and concentrated on the mathematical aspects.

There is a point where addressing every technical error, one by one, is pointless.  If you have a problem with the technical issues I brought up, address them.  I gave subjective analysis, but I brought up specific issues as well to validate that subjective opinion in the post.  If you think those were made incorrectly, state why.  Surely I can hold you to the same standard of criticism as you hold me?  What are the technical issues with my critique of Todd's paper that make it incorrect?

Of course this isn't peer review. If I was asked to legitimately peer review Todd's paper, I would have responded back to the editor that I could make neither heads nor tails of it.  I commend you on your peer review etiquette.  I suppose you are a better person than the reviewers I have had.

As to:
Quote
You are stating your own subjective opinions, anonymously, using a monicker, in an Internet forum.
What in the world does that matter?  If I doxxed myself would my criticism somehow become more meaningful?
« Last Edit: 06/08/2015 02:04 AM by wallofwolfstreet »

#### wallofwolfstreet

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##### Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #1234 on: 06/08/2015 02:00 AM »
...
I let this go at the time, but this statement is not grounded in any physics I ever learned.  It honestly reads like a physics word ad lib.

Sorry to come at you like this, but that is peer review.  People reading this forum might read your posts and think they are uncontroversial statements, generally accepted in the mainstream.  That really isn't the case.  Maybe you and I just share different mainstreams.

Quote
(P/F)^2 = v2^2 - v1^2, the difference between two potentials, better?

No, that's not at all better.  For the gradient of a function to exist in any meaningful it must, at the bare minimum, be uniquely defined over the domain of interest.  (P/F)2 is not uniquely defined over any domain.  Maybe this is the place where your "constant" F/P caveat falls into place?  Well, no, because the gradient of a constant is zero, and if the gradient is zero, what in the world is the point of the paper given:
Quote
The gradient of this potential is an acceleration vector that opposes the thrust

That alone would be enough for a a reviewer to send you back to the drawing board.  You say  F/P is constant.  You say the gradient of (P/F)2 is an acceleration vector that opposes thrust.  The gradient is obviously zero however.

Where to start? Please read a few things first please, so I don't have to explain it all. Thank you!

This is an excerpt from a paper I can't post here, regarding the interaction between a charged particle harmonic oscillator and the EM ZPF, that should explain it more clearly than I can...
Todd

Todd, if you address one issue in my post, please let it be this, because vector calc is not something I need shy away from:  Immediately after introducing equation (5), you write:

Quote
The gradient of this potential is an acceleration vector that opposes the thrust.

If we assume F/P is constant, this gradient is zero (ignoring all the issues with the fact F/P is not unique in the first place).  If the gradient is zero, there is no acceleration that opposes thrust.  How am I to resolve this?
« Last Edit: 06/08/2015 04:29 PM by Chris Bergin »

#### Rodal

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##### Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #1235 on: 06/08/2015 02:13 AM »
...
There is a point where addressing every technical error, one by one, is pointless.  If you have a problem with the technical issues I brought up, address them.  I gave subjective analysis, but I brought up specific issues as well to validate that subjective opinion in the post.  If you think those were made incorrectly, state why.  Surely I can hold you to the same standard of criticism as you hold me?  Address the technical.

Of course this isn't peer review. If I was asked to legitimately peer review Todd's paper, I would have responded back to the editor that I could make neither heads nor tails of it.  I commend you on your peer review etiquette.  I suppose you are a better person than the reviewers I have had.

As to:
Quote
You are stating your own subjective opinions, anonymously, using a monicker, in an Internet forum.
What in the world does that matter?  If I doxxed myself would my criticism somehow become more meaningful?

If you think it is pointless for you to address what you perceive as a technical error in a technical manner, then please refrain from writing.  Don't use this thread to express your feelings.  Don't write ""your paper is very confused, incredibly hand-wavy and outright wrong" in response to a busy engineering executive who took his precious time to explain his viewpoint in a more formal way.  Just stick to the technical aspects.  Please refer to the guidelines on page one of this thread.

Thank you.

#### wallofwolfstreet

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##### Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #1236 on: 06/08/2015 02:23 AM »
...
There is a point where addressing every technical error, one by one, is pointless.  If you have a problem with the technical issues I brought up, address them.  I gave subjective analysis, but I brought up specific issues as well to validate that subjective opinion in the post.  If you think those were made incorrectly, state why.  Surely I can hold you to the same standard of criticism as you hold me?  Address the technical.

Of course this isn't peer review. If I was asked to legitimately peer review Todd's paper, I would have responded back to the editor that I could make neither heads nor tails of it.  I commend you on your peer review etiquette.  I suppose you are a better person than the reviewers I have had.

As to:
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You are stating your own subjective opinions, anonymously, using a monicker, in an Internet forum.
What in the world does that matter?  If I doxxed myself would my criticism somehow become more meaningful?

If you think it is pointless for you to address what you perceive as a technical error in a technical manner, then please refrain from writing.  Don't use this thread to express your feelings.  Don't write ""your paper is very confused, incredibly hand-wavy and outright wrong" in response to a busy engineering executive who took his precious time to explain his viewpoint in a more formal way.  Just stick to the technical aspects.  Please refer to the guidelines on page one of this thread.

Thank you.

I did address technical errors.  Read my post again if you don't believe me.  If those issues I raised aren't actually issues, then I have no problem admitting I was in the wrong.  Once again, where are the technical issues in my post?

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Don't write ""your paper is very confused, incredibly hand-wavy and outright wrong" in response to a busy engineering executive who took his precious time to explain his viewpoint in a more formal way.

You must be playing a joke on me.  If someone posts an opinion online, claiming a new theoretical basis, nobody who raises real objections to that post can be accused of taking up "precious time".  That is the price of wanting to have other people listen to your ideas and try to understand them in earnest.

#### ThinkerX

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##### Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #1237 on: 06/08/2015 03:12 AM »
'Warp Drive' - I am noticing a pattern with your posts.  You explain things, but fail to clearly define key elements - most recently the whole P in / P out bit, leave other elements of your equations out.  Other posters here seize on these missing bits, and the result is pages of back and forth bickering.

I strongly suggest you go back through your papers and

1) make certain all the elements of your formula are defined beforehand (no more shortcuts with P, for example), and

2) incorporate detailed step by step examples in those papers that account for the objections you KNOW will be raised - things like 'pushing on a wall' and the multistage spacecraft example.

I do find your idea interesting, but by NOT doing those things you are hurting your argument very badly.

#### TheTraveller

##### Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #1238 on: 06/08/2015 03:17 AM »
For those of us working @ 2.4 ghz, here is an academic paper on low cost mw leakage gear. Safety first, friends...

Thanks. Most appropriate for the magnetron builders.

Working with the EMDrive Calculator, I working to design the smallest TM011 flat end plate unit possible, that uses a simple Rf excitation in the middle of the small end and that is resonate at 2.45 GHz so can use the low cost 5-20W WiFi amps available.

Plan to keep total mass, including battery and USB IR (wireless data link) USB link to max 2.5kg and ALL to fit inside a clear sealable perspex cylinder that can easily be oriented as desired. Will then sit directly on a 5kg scale that has 0.001g resolution. Can pump down the perspex cylinder, using a small lab pump I have to very significantly reduce hot air effects. Electro caps will be replaced with solid tants.

Work In Progress.
« Last Edit: 06/08/2015 03:20 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

#### WarpTech

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##### Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #1239 on: 06/08/2015 03:19 AM »
...
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That alone would be enough for a a reviewer to send you back to the drawing board.  You say  F/P is constant.  You say the gradient of (P/F)2 is an acceleration vector that opposes thrust.  The gradient is obviously zero however.

You are not performing any peer review here, this is not a peer-review journal, and you are not one of the editors.

You are stating your own subjective opinions, anonymously,  using a monicker, in an Internet forum.  That's all.

And concerning peer-review, I have reviewed papers for peer-reviewed journals and I (and the Professors that directed the Lab at  the University) never addressed errors in the papers I reviewed by making subjective general statements like "your paper is very confused, incredibly hand-wavy and outright wrong" .  I always addressed what I thought were the technical errors specifically one by one, and concentrated on the mathematical aspects.

I'm no newbie to forums. His comments are worthy of discussion and I am giving them due consideration. Honestly, I appreciate that someone is actually giving me criticism that makes me think! However, there are some gaps in our common knowledge base that confound a simple explanation. I hope @wallofwolfstreet will read up on the work I base my model on, and I will sort out where I went wrong in defining the potential.

I must agree and admit my mistake, the way equation (5) is written in my paper is informal, and it really does only mean something as an equilibrium condition at a relative change in velocity, (delta-v)2. When this potential is in equilibrium with the gravitational potential relative to the center of the increased inertial mass, there is no force. So he is correct when he's saying thrust-to-power is zero! Sorry, I didn't see that. At that point, it would be pressure-to-power rather than thrust, since no work is being done and therefore, the force is zero.

There is also a point confounding the issue. In this paper, like the energy paradox problem, the question of "where the power is coming from" was ignored. There is no battery, no fuel and no propellant. The total energy of the system is increasing as it goes, not decreasing like a rocket. Therefore, it must represent only what happens when an object gains inertia from an outside-source, such as radiation pressure. As soon as you put the battery on-board, the velocity is limited by the equation;

E2=(mc2)2 + (pc)2

I'm sure there will be a revision 5 coming soon.

Todd

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