Author Topic: EM Drive Developments - related to space flight applications - Thread 3  (Read 1798591 times)

Offline TheTraveller


...

Have bought a few to measure up and using the EMDrive Calculator to see what resonate frequencies and excitation modes it can work with, versus predicted Df and thrust.

It may be a Copper colored glaze on glass so the conductivity would not be very high.   On thread 2 someone mentioned they had a machinist spin a cone for them.   That's a good way to go, especially if they can put a straight section in at the thin end (closed) end.   The only downside is the surface after metal spinning is not very smooth, because of the way the metal gets moved.

Seller said solid copper. Will know for sure when they arrive.

Spreadsheet is being modified to auto calc Df, thrust & resonance in many modes at once. Then easy to see where the sweet spots are.
« Last Edit: 06/06/2015 10:53 PM by TheTraveller »
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Offline wallofwolfstreet

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It goes into maintaining a state of compressed, time dilated matter, just as in a gravitational field. In order to do so, it is pushing up-hill against a gravity well that is resisting the acceleration. There is constant force on your feet while standing on the ground. Power is being applied by the gravitational field, yet you're not moving or gaining energy. Where is the power going? It is going into maintaining your state of compression.

Therefore, the EM Drive, if it is creating and can maintain a gradient in (P/F)^2, then it it will accelerate until it reaches a gradient in the potential energy per unit mass that exerts an equal and opposite force. At which point, it is the same as standing on the ground just to maintain the relative compression. If you turn off the engine, it will no longer "feel" like an accelerated reference frame. It will become a weightless inertial reference frame, and so the feeling of compression to the floor would disappear.

I guess it would be correct to say, the power goes into maintaining a gravitational field, aka an accelerated reference frame, even though it's not gaining any relative velocity.

Todd

Power is being applied by a gravitational field when I am standing on the ground?  Power is necessary to maintain a state of "compression" or a gravitational field?  This is news to me, and certainly goes against any classical model. 

Thought experiment:  Imagine a universe in which there is only a single rock, one single rock, floating in emptiness.  The laws of physics apply in this world just as they do in ours.  You say there is power expended in maintaining this rock's gravitational field.  Is the total energy thus constantly increasing/decreasing in this theoretical universe?  How is this reconciled with CoE?

I see I am asking you a lot of questions, but what you are saying is completely foreign to me.   

Offline TheTraveller

It goes into maintaining a state of compressed, time dilated matter, just as in a gravitational field. In order to do so, it is pushing up-hill against a gravity well that is resisting the acceleration. There is constant force on your feet while standing on the ground. Power is being applied by the gravitational field, yet you're not moving or gaining energy. Where is the power going? It is going into maintaining your state of compression.

Therefore, the EM Drive, if it is creating and can maintain a gradient in (P/F)^2, then it it will accelerate until it reaches a gradient in the potential energy per unit mass that exerts an equal and opposite force. At which point, it is the same as standing on the ground just to maintain the relative compression. If you turn off the engine, it will no longer "feel" like an accelerated reference frame. It will become a weightless inertial reference frame, and so the feeling of compression to the floor would disappear.

I guess it would be correct to say, the power goes into maintaining a gravitational field, aka an accelerated reference frame, even though it's not gaining any relative velocity.

Todd

Power is being applied by a gravitational field when I am standing on the ground?  Power is necessary to maintain a state of "compression" or a gravitational field?  This is news to me, and certainly goes against any classical model. 

Thought experiment:  Imagine a universe in which there is only a single rock, one single rock, floating in emptiness.  The laws of physics apply in this world just as they do in ours.  You say there is power expended in maintaining this rock's gravitational field.  Is the total energy thus constantly increasing/decreasing in this theoretical universe?  How is this reconciled with CoE?

I see I am asking you a lot of questions, but what you are saying is completely foreign to me.   

From one engineer to another, you may enjoy reading this dumbed down version of the EMDrive Basic Theory. This presentation is not on the EMDrive site. Was sent by Roger Shawyer to Mulletron, who made it public.
« Last Edit: 06/06/2015 11:13 PM by TheTraveller »
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Offline Prunesquallor

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....

I am sorry to keep dragging this out. My feeble engineer's brain keeps dropping into thought experiments.
Given the conclusions reached in your "Resolution" paper, what would be the result of this scenario?

Assume two identical self-powered emdrive "stages" operating in series. Turn on Stage 1. According to your Resolution, it will attain some delta-v (with respect to the initial power-up?) that depends on its thrust-to-power ratio. Now physically detach Stage 2 and turn on the power.

What will happen?

Assuming identical input power is available, when you detach Stage 2 from Stage 1, it now has half the mass, so it can accelerate up to ~ 2*delta-v.

Let's change the question up a bit.  After Stage 1 has been powered for long enough that it's maximum delta-v has been achieved, turn it off.  Don't detach it this time.  Your spaceship, made of the two identical emdrive stages, is now floating through space at delta-v plus whatever initially velocity it had.  Let's say that the spaceship is in deep space, negligible gravity.  In this inertial frame, the spaceship has no way of "knowing" whether it is moving or standing still.  How does Stage 2 "know" it is currently moving at it's limiting velocity, without invoking a preferred frame? ...

Because, when the engine was running, every sub-atomic particle of matter was accelerated and in dong so, their momentum increased. Relative to where it started from, the wavelength of every matter-wave has been reduced in size and this represents the real stored energy of inertia. Therefore, the matter that was accelerated "knows" it was accelerated because it possesses more inertia than when it started. In this regard, when the engine is turned off, it is equivalent to orbiting at a constant gravitational potential (v^2), in free-fall at a constant velocity, as opposed to hovering at this potential when the engine was running.

Einstein's Equivalence Principle still rocks!  ::)

Todd

But...
Stage 2 has gone through the same acceleration and momentum change and should have experience the same subatomic effects as Stage 1. So why would it operate?  In fact after reaching the first delta-v and turning off Stage 1, the two stages should be indistguishable. So why could Stage 2 accelerate and Stage 1 be "dead"?
Retired, yet... not

Offline TheTraveller

....

I am sorry to keep dragging this out. My feeble engineer's brain keeps dropping into thought experiments.
Given the conclusions reached in your "Resolution" paper, what would be the result of this scenario?

Assume two identical self-powered emdrive "stages" operating in series. Turn on Stage 1. According to your Resolution, it will attain some delta-v (with respect to the initial power-up?) that depends on its thrust-to-power ratio. Now physically detach Stage 2 and turn on the power.

What will happen?

Assuming identical input power is available, when you detach Stage 2 from Stage 1, it now has half the mass, so it can accelerate up to ~ 2*delta-v.

Let's change the question up a bit.  After Stage 1 has been powered for long enough that it's maximum delta-v has been achieved, turn it off.  Don't detach it this time.  Your spaceship, made of the two identical emdrive stages, is now floating through space at delta-v plus whatever initially velocity it had.  Let's say that the spaceship is in deep space, negligible gravity.  In this inertial frame, the spaceship has no way of "knowing" whether it is moving or standing still.  How does Stage 2 "know" it is currently moving at it's limiting velocity, without invoking a preferred frame? ...

Because, when the engine was running, every sub-atomic particle of matter was accelerated and in dong so, their momentum increased. Relative to where it started from, the wavelength of every matter-wave has been reduced in size and this represents the real stored energy of inertia. Therefore, the matter that was accelerated "knows" it was accelerated because it possesses more inertia than when it started. In this regard, when the engine is turned off, it is equivalent to orbiting at a constant gravitational potential (v^2), in free-fall at a constant velocity, as opposed to hovering at this potential when the engine was running.

Einstein's Equivalence Principle still rocks!  ::)

Todd

But...
Stage 2 has gone through the same acceleration and momentum change and should have experience the same subatomic effects as Stage 1. So why would it operate?  In fact after reaching the first delta-v and turning off Stage 1, the two stages should be indistguishable. So why could Stage 2 accelerate and Stage 1 be "dead"?

Inside the EMDrive frustum all it knows of the world outside is when acceleration causes internal Doppler shift, dropping Q and stored energy, which converts into Kinetic. Reverse happens if something tries to move it small end to big end. Acts like a Kinetic energy sink.

Push big end toward small end and it is a Kinetic source, push it the other and it is a Kinetic sink.
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Offline ThinkerX

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Let's change the question up a bit.  After Stage 1 has been powered for long enough that it's maximum delta-v has been achieved, turn it off.  Don't detach it this time.  Your spaceship, made of the two identical emdrive stages, is now floating through space at delta-v plus whatever initially velocity it had.  Let's say that the spaceship is in deep space, negligible gravity.  In this inertial frame, the spaceship has no way of "knowing" whether it is moving or standing still.  How does Stage 2 "know" it is currently moving at it's limiting velocity, without invoking a preferred frame? ...

For some reason I am reminded here of the examples given for relativistic spacecraft - the ones that go 'you are traveling at 99.99999% the speed of light, then you hit the boosters, so you should be going faster than light.' 

To me, this...debate...appears to be of the same nature.


Offline TheTraveller

Quote
Let's change the question up a bit.  After Stage 1 has been powered for long enough that it's maximum delta-v has been achieved, turn it off.  Don't detach it this time.  Your spaceship, made of the two identical emdrive stages, is now floating through space at delta-v plus whatever initially velocity it had.  Let's say that the spaceship is in deep space, negligible gravity.  In this inertial frame, the spaceship has no way of "knowing" whether it is moving or standing still.  How does Stage 2 "know" it is currently moving at it's limiting velocity, without invoking a preferred frame? ...

For some reason I am reminded here of the examples given for relativistic spacecraft - the ones that go 'you are traveling at 99.99999% the speed of light, then you hit the boosters, so you should be going faster than light.' 

To me, this...debate...appears to be of the same nature.

Inside the EMDrive is an antenna, which is attached to the side wall and moves as the EMDrive moves. The internal wave propagation is more complex that there just being a standing wave. The small vertical line in the centre of the attached frustum drawing is a 1/4 wave stub antenna radiating Rf energy toward both end plates.
« Last Edit: 06/06/2015 11:50 PM by TheTraveller »
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Offline wallofwolfstreet

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Inside the EMDrive frustum all it knows of the world outside is when acceleration causes internal Doppler shift, dropping Q and stored energy, which converts into Kinetic. Reverse happens if something tries to move it small end to big end. Acts like a Kinetic energy sink.

Push big end toward small end and it is a Kinetic source, push it the other and it is a Kinetic sink.

That doesn't resolve the energy paradox though.  The EMdrive doesn't have to be accelerating at all to act as a free energy machine, so converting EM energy to kinetic energy when an acceleration occurs solves nothing.  This has been fleshed out a few times now, so I'm just going to link to my previous post on the matter: http://forum.nasaspaceflight.com/index.php?topic=37642.msg1385685#msg1385685.

Free energy doesn't require acceleration, so CoE during acceleration just isn't enough.
« Last Edit: 06/06/2015 11:54 PM by wallofwolfstreet »

Offline Prunesquallor

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Let's change the question up a bit.  After Stage 1 has been powered for long enough that it's maximum delta-v has been achieved, turn it off.  Don't detach it this time.  Your spaceship, made of the two identical emdrive stages, is now floating through space at delta-v plus whatever initially velocity it had.  Let's say that the spaceship is in deep space, negligible gravity.  In this inertial frame, the spaceship has no way of "knowing" whether it is moving or standing still.  How does Stage 2 "know" it is currently moving at it's limiting velocity, without invoking a preferred frame? ...

For some reason I am reminded here of the examples given for relativistic spacecraft - the ones that go 'you are traveling at 99.99999% the speed of light, then you hit the boosters, so you should be going faster than light.' 

To me, this...debate...appears to be of the same nature.

I really wasn't trying to start a debate with my thought experiment, I was testing to see if I understood the implications of Todd's Resolution. My thought experiment results in what appears to me to be a paradox (two identical vehicles experiencing identical momentum changes subsequently behave differently). It is almost surely a misunderstanding on my part, and I'm just trying to figure out where I am going astray.
Retired, yet... not

Offline TheTraveller

Inside the EMDrive frustum all it knows of the world outside is when acceleration causes internal Doppler shift, dropping Q and stored energy, which converts into Kinetic. Reverse happens if something tries to move it small end to big end. Acts like a Kinetic energy sink.

Push big end toward small end and it is a Kinetic source, push it the other and it is a Kinetic sink.

That doesn't resolve the energy paradox though.  The EMdrive doesn't have to be accelerating at all to act as a free energy machine, so converting EM energy to kinetic energy when an acceleration occurs solves nothing.  This has been fleshed out a few times now, so I'm just going to link to my previous post on the matter: http://forum.nasaspaceflight.com/index.php?topic=37642.msg1385685#msg1385685.

Free energy doesn't require acceleration, so CoE during acceleration just isn't enough.

If the EMDrive is not acceleration (in motor mode) nor opposing acceleration (in generator mode), some small fraction of cavity energy does turn into thermal energy, assuming constant input, due to wall losses.

How is this a Free Energy Source?
« Last Edit: 06/07/2015 12:20 AM by TheTraveller »
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Online SeeShells

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Because, when the engine was running, every sub-atomic particle of matter was accelerated and in dong so, their momentum increased. Relative to where it started from, the wavelength of every matter-wave has been reduced in size and this represents the real stored energy of inertia. Therefore, the matter that was accelerated "knows" it was accelerated because it possesses more inertia than when it started. In this regard, when the engine is turned off, it is equivalent to orbiting at a constant gravitational potential (v^2), in free-fall at a constant velocity, as opposed to hovering at this potential when the engine was running.

Einstein's Equivalence Principle still rocks!  ::)

Todd
[/quote]
So let me get this right. The real reference frame isn't the gravitational field but spacetime? Since that is how an object "knows" it's speed from the stored compression of the matter it's made of? So if we backwards think this and look at the Emdrive removing that spacetime compression while it's running...

Offline frobnicat

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Inside the EMDrive frustum all it knows of the world outside is when acceleration causes internal Doppler shift, dropping Q and stored energy, which converts into Kinetic. Reverse happens if something tries to move it small end to big end. Acts like a Kinetic energy sink.

Push big end toward small end and it is a Kinetic source, push it the other and it is a Kinetic sink.

That doesn't resolve the energy paradox though.  The EMdrive doesn't have to be accelerating at all to act as a free energy machine, so converting EM energy to kinetic energy when an acceleration occurs solves nothing.  This has been fleshed out a few times now, so I'm just going to link to my previous post on the matter: http://forum.nasaspaceflight.com/index.php?topic=37642.msg1385685#msg1385685.

Free energy doesn't require acceleration, so CoE during acceleration just isn't enough.

Yes, and actually accelerating a spacecraft with an EM drive doesn't even require an EM drive accelerating when it's operated. It can be operated intermittently : when it is on it pushes on an actuator that lengthen in such a way as to cancel any acceleration of it, but the other side of the actuator communicate the gained momentum to rest of spacecraft that do accelerate. Switch off EM drive, pull it back to its original position, that will pull back the spacecraft, but not to the point of cancelling the gained thrust momentum. The EM drive is accelerated when it is off, it is not accelerated when it is on : the thrust is "buffered" through the intermediate actuator to mass of rest of spacecraft.

So, does that mean that an EM drive kind of records accelerations even when it is passively accelerated (by outside agent) while being off ? You must be kidding.

If EM drives like to be free to accelerate to thrust, so be it, if they don't like to be accelerated to thrust, so be it, either way, a buffering intermediate link can accommodate for those whims, at a very, very, very modest energetic cost. So, a 40s long thrust of 50ľN for 50W at constant velocity, that is exactly the same conditions as claimed results at Eagleworks, if it is reproducible many times, can be used intermittently to reach break even velocities, in principle (if not in practice). What could possibly make it, in principle, not reproducible many times if it can genuinely work one time ? Growing distance with the lab, or cumulative disturbance of the Force ?

Online WarpTech

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....

I guess it would be correct to say, the power goes into maintaining a gravitational field, aka an accelerated reference frame, even though it's not gaining any relative velocity.
... (EDIT: This previous post could've been worded better..)


Power is being applied by a gravitational field when I am standing on the ground?  Power is necessary to maintain a state of "compression" or a gravitational field?  This is news to me, and certainly goes against any classical model. 

Thought experiment:  Imagine a universe in which there is only a single rock, one single rock, floating in emptiness.  The laws of physics apply in this world just as they do in ours.  You say there is power expended in maintaining this rock's gravitational field.  Is the total energy thus constantly increasing/decreasing in this theoretical universe?  How is this reconciled with CoE?

I see I am asking you a lot of questions, but what you are saying is completely foreign to me.   

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!

This is the crux of the preferred frame argument. "IFF" there were someplace in the universe that was free of influence from all fields and matter, no EM, no gravity, no nothing, that would be nice. However, since it can be shown that the fields must span the entire universe, no such preferred frame exists. Therefore, everything in GR is relative to everything else. You can change your gravitational potential, you can hover with the engine running or turn off the engine and continue in free-fall (inertial frame) at a higher constant inertia content than you started with. But your rulers and clocks are not absolute, they are ALWAYS relative to what you define them to be in your local space-time neighborhood. In other words, matter is not immutable, it inflates or contracts depending on its relative inertia content.

Todd



Offline kdhilliard

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If the EMDrive is not acceleration (in motor mode) nor opposing acceleration (in generator mode), some small fraction of cavity energy does turn into thermal energy, assuming constant input, due to wall losses.

Are you invoking Shawyer's claim that an EMDrive will generate no measurable force if it is not free to accelerate, as described in his Measurement Paper?
Quote from: Pg 3.
In each successful case, the EmDrive force data has been superimposed on an increasing or decreasing background force, generated by the test equipment itself.  Indeed, in the UK when the background force changes were eliminated, in an effort to improve force measurement resolution, no EmDrive force was measured.  This was clearly a result of attempting to measure the forces on a fully static thruster, where T and R cancel each other.  UK flight thruster measurements employ this principle to calibrate the background noise on the force balance prior to carrying out force measurements.

~Kirk

Offline ThinkerX

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Been contemplating photons the past few days.

Weird little buggers.  Particle or Wave?  Entanglement?  And when they hit something, they can transfer momentum despite not having any mass - repeatedly.

David Bae's laser driven two space ship concept involves 'recycling' photon's thousands of times, with a corresponding increase in the efficiency of photon rockets.   This design produces around 5000 times the thrust of a photon rocket, but...

...about 14-15 times less than claimed for the EM Drive. 

Still, photon's are durable little critters, and can transfer momentum for something on the order of 50,000 - 100,000 'bounces,' though if I understand such correctly, the kinetic energy drops a hair with each subsequent 'bounce.'

So, I have been wondering: suppose all the kinetic energy locked into a photon could be released all at once?  One super hard bounce that unleashes 50,000 - 100,000 times the amount of kinetic energy that would be transferred during a 'normal bounce.'  The photon's full potential converted to kinetic energy...or perhaps inertial energy.  (Which for some reason makes me think of Doctor McCulloch's pet theory).     
« Last Edit: 06/07/2015 12:40 AM by ThinkerX »

Online WarpTech

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Let's change the question up a bit.  After Stage 1 has been powered for long enough that it's maximum delta-v has been achieved, turn it off.  Don't detach it this time.  Your spaceship, made of the two identical emdrive stages, is now floating through space at delta-v plus whatever initially velocity it had.  Let's say that the spaceship is in deep space, negligible gravity.  In this inertial frame, the spaceship has no way of "knowing" whether it is moving or standing still.  How does Stage 2 "know" it is currently moving at it's limiting velocity, without invoking a preferred frame? ...

Because, when the engine was running, every sub-atomic particle of matter was accelerated and in dong so, their momentum increased. Relative to where it started from, the wavelength of every matter-wave has been reduced in size and this represents the real stored energy of inertia. Therefore, the matter that was accelerated "knows" it was accelerated because it possesses more inertia than when it started. In this regard, when the engine is turned off, it is equivalent to orbiting at a constant gravitational potential (v^2), in free-fall at a constant velocity, as opposed to hovering at this potential when the engine was running.

Einstein's Equivalence Principle still rocks!  ::)

But...
Stage 2 has gone through the same acceleration and momentum change and should have experience the same subatomic effects as Stage 1. So why would it operate?  In fact after reaching the first delta-v and turning off Stage 1, the two stages should be indistguishable. So why could Stage 2 accelerate and Stage 1 be "dead"?

I didn't say that. They would be indistinguishable. If after reaching delta-v with Stage-1, it is now "hovering" at a constant potential. Turning off Stage-1, it now continues with constant inertia in free-fall. If you turn on Stage-2, identical to Stage-1, it's now hovering again at the same potential. If you turn on both stages, you've doubled the Power so it will accelerate to 2*delta-v.

The difference will be in that when the engine is running, there is an acceleration toward the floor, when it's off, you're floating. The relative speed does not change, and just like pushing on a wall, no work is being done. So although the engine is running, it's is not increasing the inertia any higher. To increase the inertia further, a higher input power is required.

Todd

Offline wallofwolfstreet

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Inside the EMDrive frustum all it knows of the world outside is when acceleration causes internal Doppler shift, dropping Q and stored energy, which converts into Kinetic. Reverse happens if something tries to move it small end to big end. Acts like a Kinetic energy sink.

Push big end toward small end and it is a Kinetic source, push it the other and it is a Kinetic sink.

That doesn't resolve the energy paradox though.  The EMdrive doesn't have to be accelerating at all to act as a free energy machine, so converting EM energy to kinetic energy when an acceleration occurs solves nothing.  This has been fleshed out a few times now, so I'm just going to link to my previous post on the matter: http://forum.nasaspaceflight.com/index.php?topic=37642.msg1385685#msg1385685.

Free energy doesn't require acceleration, so CoE during acceleration just isn't enough.

If the EMDrive is not acceleration (in motor mode) nor opposing acceleration (in generator mode), some small fraction of cavity energy does turn into thermal energy, assuming constant input, due to wall losses.

How is this a Free Energy Source?

Did you read the post I linked?  It doesn't require a perfectly efficient cavity, where no power is lost to resistive heating effects.  Such power losses can be, and are free to be, quite large.  All that matters for the analysis to hold is that the Thrust to Power ratio is greater than 1/c, which has been true for all EMdrives to date.

As long as the thrust to power ratio is greater than 1/c, there will be a break even velocity where, when held at constant velocity (so no "motor mode" effects), the emdrive creates more power than it takes in.  It doesn't matter if the power it takes in is converted to heat or unicorn dust. 

Please, read the link.  You will see that thermal energy losses in the cavity don't play into this issue at all. 

(note: I reserve the right for the preceding analysis, props to @frobnicat, to potentially be made irrelevant by theories like @warptech's.)

Offline ThinkerX

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I didn't say that. They would be indistinguishable. If after reaching delta-v with Stage-1, it is now "hovering" at a constant potential. Turning off Stage-1, it now continues with constant inertia in free-fall. If you turn on Stage-2, identical to Stage-1, it's now hovering again at the same potential. If you turn on both stages, you've doubled the Power so it will accelerate to 2*delta-v.

The difference will be in that when the engine is running, there is an acceleration toward the floor, when it's off, you're floating. The relative speed does not change, and just like pushing on a wall, no work is being done. So although the engine is running, it's is not increasing the inertia any higher. To increase the inertia further, a higher input power is required.

Hmmm...

Imagine a road climbing a hill with an ever steepening grade.  At the bottom of the hill is a giant, heavy SUV...out of gas.  Next to the SUV are two muscular men - Bill and Carl.

Bill starts pushing the car up the hill.  He's tough, so at first no problem - one step at a time. 

Then the grade gets steeper.  Bill is tough, but not that tough - push though he might, the most he can do is keep the car in place...until Carl decides to help out.  Together they can push the car.

Probably a pretty badly flawed analogy - but maybe close enough?


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Because, when the engine was running, every sub-atomic particle of matter was accelerated and in dong so, their momentum increased. Relative to where it started from, the wavelength of every matter-wave has been reduced in size and this represents the real stored energy of inertia. Therefore, the matter that was accelerated "knows" it was accelerated because it possesses more inertia than when it started. In this regard, when the engine is turned off, it is equivalent to orbiting at a constant gravitational potential (v^2), in free-fall at a constant velocity, as opposed to hovering at this potential when the engine was running.

Einstein's Equivalence Principle still rocks!  ::)

Todd

So let me get this right. The real reference frame isn't the gravitational field but spacetime? Since that is how an object "knows" it's speed from the stored compression of the matter it's made of? So if we backwards think this and look at the Emdrive removing that spacetime compression while it's running...

The real reference frame is the rest frame the mass started from. As work is done to accelerate it, matter acquires inertia which is physically stored as a reduction of the wavelength of matter waves, leading to relativistic length contraction and time dilation as physical effects as v -> c. For v << c, the effect is still there, but we perceive it as inertial mass, or total energy content of the body. The inertial mass of an object moving at velocity v is greater than it was in the frame in which it started.

Again, this is not a Lorentz transformation anymore, this is GR not SR. This is that part of SR that they leave out when trying to explain the Twin Paradox. Then they say "it's because one of the twins was accelerated". You must integrate along the world-line of the matter being accelerated to get the right answer. It's not simply comparing 2 identical inertial frames, it is comparing 2 inertial frames at different gravitational potentials.

Todd

Offline Prunesquallor

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Let's change the question up a bit.  After Stage 1 has been powered for long enough that it's maximum delta-v has been achieved, turn it off.  Don't detach it this time.  Your spaceship, made of the two identical emdrive stages, is now floating through space at delta-v plus whatever initially velocity it had.  Let's say that the spaceship is in deep space, negligible gravity.  In this inertial frame, the spaceship has no way of "knowing" whether it is moving or standing still.  How does Stage 2 "know" it is currently moving at it's limiting velocity, without invoking a preferred frame? ...

Because, when the engine was running, every sub-atomic particle of matter was accelerated and in dong so, their momentum increased. Relative to where it started from, the wavelength of every matter-wave has been reduced in size and this represents the real stored energy of inertia. Therefore, the matter that was accelerated "knows" it was accelerated because it possesses more inertia than when it started. In this regard, when the engine is turned off, it is equivalent to orbiting at a constant gravitational potential (v^2), in free-fall at a constant velocity, as opposed to hovering at this potential when the engine was running.

Einstein's Equivalence Principle still rocks!  ::)

But...
Stage 2 has gone through the same acceleration and momentum change and should have experience the same subatomic effects as Stage 1. So why would it operate?  In fact after reaching the first delta-v and turning off Stage 1, the two stages should be indistguishable. So why could Stage 2 accelerate and Stage 1 be "dead"?

I didn't say that. They would be indistinguishable. If after reaching delta-v with Stage-1, it is now "hovering" at a constant potential. Turning off Stage-1, it now continues with constant inertia in free-fall. If you turn on Stage-2, identical to Stage-1, it's now hovering again at the same potential. If you turn on both stages, you've doubled the Power so it will accelerate to 2*delta-v.

The difference will be in that when the engine is running, there is an acceleration toward the floor, when it's off, you're floating. The relative speed does not change, and just like pushing on a wall, no work is being done. So although the engine is running, it's is not increasing the inertia any higher. To increase the inertia further, a higher input power is required.

Todd

Hmmm.. So t becomes an artificial gravity machine upon reaching delta-v?
Retired, yet... not

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