Author Topic: Woodward's effect  (Read 287185 times)

Offline wallofwolfstreet

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Re: Woodward's effect
« Reply #700 on: 03/20/2016 06:19 PM »
I engaged with a length argument on this very thready about that issue. And after reading that paper over multiple times the only thing clear in my mind is that I dont believe that essay is meant to answer the question we want it to answer. Yes I know that sounds absurd. I can even admit it sounds like denial. But its the only logical conclusion I can reach after reading that essay and arguing about it here. That last paragraph in that paper in particular feels pretty weird. how can you spend 3/4 of an essay arguing that you cannot arbitrarily choose a time to restart acceleration. But then at the end argue that is exactly what you would need to do.

My gut tells me that if he actually believes a MET behaves in the manner described in that last paragraph then the only thing that he can be describing is the oscillating nature of the fluctuating mass in a MET device. which is constantly being accelerated and deccelerated as its mass fluctuates.

Unfortunately I have no way of confirming if my gut feeling is accurate as I have no line of communication with Woodward or Fern.

I think I see what you're saying, but the essay is pretty cut and dry in how it is worded in my opinion.  The last paragraph is funky as all hell, but it even opens:

Quote
To wrap this up, we ask: is it possible to do a correct calculation of the sort that
critics did that does not lead to wrong predictions of the violation of energy conservation?
By paying attention to the physics of the situation, yes, such a calculation is possible.

So it seems pretty clear he is talking about the energy calculation routinely done which demonstrates a constant thrust to power ratio eventually leads to over unity.  He even frames the "calculation of the sort that critics did" in the very first line of the essay:

Quote
We routinely hear a criticism of METs based upon an argument that claims: if a
MET is operated at constant power input for a sufficiently long time, it will acquire
enough kinetic energy to exceed the total input energy of operation

Taking those two lines together, it seems like he's definitely saying METs do not lead to a local gain in energy, and unfortunately he believes this by hand waving a physically nonsensical explanation into existence. Worse yet, he chose to write the whole essay in a startling arrogant tone with the whole "stupid and wrong" line.

Offline aceshigh

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Re: Woodward's effect
« Reply #701 on: 03/21/2016 02:48 PM »
Unfortunately I have no way of confirming if my gut feeling is accurate as I have no line of communication with Woodward or Fern.

there supposedly is a mailing list.

Paul March of NASA Eagleworks Lab should have access to it. You could PM him about it? (since he posts once every 6 months here :))

Offline wallofwolfstreet

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Re: Woodward's effect
« Reply #702 on: 03/21/2016 03:51 PM »
Or you could just email him directly and ask.  His professional email is on his web page:

http://physics.fullerton.edu/~jimw/

Offline dustinthewind

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Re: Woodward's effect
« Reply #703 on: 03/23/2016 06:18 PM »
...

Light changing the mass of a dielectric:
URL: Mass change of dielectric media induced by propagation of electromagnetic waves by C.Z. Tan

The diagram of the device I suggested to push on trapped light inside the cavity I believe would not work.  I should have realized this earlier.  Kudos to those that pointed this out.

It also appears the direction of momentum should not be being violated.  If the photon was made traveling towards the direction the cavity would move via the dielectric pushing, the cavity would get a kick from the photons creation in the opposite direction of dielectric push p=(hf/c).  The photon would then reflect pushing the cavity in the direction of the dielectric pushing on the light +p=(-2hf/c).  After this the photon strikes the back of the cavity, moving it in the opposite direction of the dielectric push +p=(2hf/c).  The photon then encounters the dielectric which pushes on the light and lets say the dielectric completely red-shifts the light +p=(-hf/c).  After this the cavity should have zero momentum.  Back to the drawing board. 

This was an attempt to make propulsion on the concept of the mach effect by somehow violating the direction of momentum.
« Last Edit: 03/23/2016 06:19 PM by dustinthewind »

Offline Mezzenile

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Re: Woodward's effect
« Reply #704 on: 03/28/2016 11:45 AM »
But the principle of the Mach Effect Thruster is that it is not the electrical energy of the onboard power source that is converted into acceleration. Instead, the powersource merely enables the creation of the gravinertial flux, which then exchanges momentum with the rest of the mass of the universe.

...

Hence, if the Wheeler-Feynman absorber aspect of Woodward's theory holds true, the mechanism for the instantaneous exchange of energy with the rest of the Universe can be explained, and hence, indeed, a Mach Effect Thruster also doubles as a generator of local energy, leeched from the distant universe.

...

Hence, from a local point of view, it does indeed represent a free energy device, but only because it is sourcing this energy from the closed system that is the entire universe. Thus, the surplus energy that is extracted here, thanks to the Mach Effect Thruster, is taken away from the entire rest of the Universe.

So the net energy gain is zero. The system in which this takes place is just colossal - it is the entire Universe.

I agree at 99% with the analysis which is presented herabove, but I am not sure that to present the entire universe as a closed system is really adequate.
In fact we should talk instead of the "closed entire universe" of "the causally connected universe" for  which the Mach Principle and its traduction by the Wheeler-Feynman absorber theoretical approach, predict an instantaneous interaction with any massive observer.
The "causally connected universe"  is in expansion (at least for its observable part) and so does  not present an overall time symmetry. This fact, according Emmy Noether theorem, has for consequence that most probably the causally connected universe is not governed by an overall law of energy conservation.
So it would be in my opinion preferrable to talk about a Woodward massive observer who exchanges energy with a causally connected universe whose overall energy increases thanks to its expansion dynamic.

Moreover the LIGO/VIRGO recent experimental direct discovery of gravitationnal waves re-inforces the radiative status of gravitation in the General Relativity Theory and so the Mach/Sciama/Papapetrou/Woodward idea that inertia forces have a gravitational origin.
« Last Edit: 03/28/2016 12:09 PM by Mezzenile »

Offline wallofwolfstreet

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Re: Woodward's effect
« Reply #705 on: 03/29/2016 07:10 PM »
Does anyone have a digital copy of Woodward's book?

I'm looking for a specific section of the book where he says he addresses an issue with his thrust derivation.  I have not seen him address this issue elsewhere.  I am in particular looking for pages 77 and 127.

I'd like to be able to see how he addressed this issue, but I'm not going to buy a book to read two or three pages of interest.

Thanks in advance. 

Offline birchoff

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Re: Woodward's effect
« Reply #706 on: 03/29/2016 09:01 PM »
Does anyone have a digital copy of Woodward's book?

I'm looking for a specific section of the book where he says he addresses an issue with his thrust derivation.  I have not seen him address this issue elsewhere.  I am in particular looking for pages 77 and 127.

I'd like to be able to see how he addressed this issue, but I'm not going to buy a book to read two or three pages of interest.

Thanks in advance.

I have the kindle version. Unfortunately those pages are the book page numbers not the ebook one so I couldnt provide you the information you want.

Offline wallofwolfstreet

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Re: Woodward's effect
« Reply #707 on: 03/30/2016 01:56 AM »
I have the kindle version. Unfortunately those pages are the book page numbers not the ebook one so I couldnt provide you the information you want.

Thanks for the consideration anyway.

The reason I am asking origintates in the followng quote from woodward's recent monograph

Quote
In the case of a rocket motor, the thing to observe is that there is one invariant
velocity involved: that of the exhaust plume with respect to the motor. All observers,
irrespective of their own motions, agree on both the magnitude and direction of this
velocity. And it is the velocity that yields momentum conservation. An argument based
on an incorrect application of Newton’s second law to METs was advanced as a criticism
of Mach Effects by an Oak Ridge scientist many years ago. It is dealt with on pages 77
and 127 of Making Starships and Stargates: the Science of Interstellar Propulsion and
Absurdly Benign Wormholes.
It will not be discussed further here.

As I understand it, the argument advanced by the Oak Ridge scientist was that F=ma is only correct at the system level for systems where the mass of the system is constant in time (which is clearly not true for METs).  This is because the more general expression for Force is correctly given by (I leave it to the reader to distinguish vectors from scalars):

Fnet=(d/dt)(p)
=d/dt(mv)
=(dm/dt)v + m(dv/dt)
=(dm/dt)v + ma

where:
F=external forces (i.e. forces applied across the system boundary)
p=momentum
m=mass
v=velocity
a=acceleration
d/dt= partial derivative with respect to time

Obviously we can return to the more familiar F=ma expression by just setting mass constant so that dm/dt=0.

To demonstrate that this is the correct expression, consider the case of a rocket expelling exhaust in free space (i.e. no gravity; the only thing in this universe is the rocket and it's exhaust).  Now draw a control volume around the rocket plus a portion of the exhaust stream that has been expelled from the rocket.  Let this control volume be fixed to the rocket so it moves with it (i.e. a lagrangian control volume). 

Now as is always the case, internal forces (f) are always equal and opposite to one another, and so they sum to zero.  Because the "rocket + some exhaust" system is in free space, it experiences no external forces either, so F=0.  Remember that the force applied to the rocket by the exhaust is equal and opposite the force on the exhaust applied by the rocket.  Since we have included this portion of the exhaust in our control volume, these forces are themselves internal, and sum to zero (see attached figure). 

Now if we applied F=ma we would get that 0=ma so a=0.  Apparently rockets don't work. 

But if we apply the correct force derivation, F=d/dt(p) then we get:

Fnet=(dm/dt)v + ma
0=(dm/dt)v + ma
a=((-dm/dt)v)/m

which is the correct equation of motion for a rocket.  Essentially, when we choose a control volume where a portion of the exhaust is in the control volume and that portion has been fully accelerated (such that any particle of exhaust has a=0), then the acceleration of a rocket is NOT in fact from a "force" at all, but from the expulsion of mass across the system boundary that envelops the control volume (at least in the lagrangian interpretation).

This preamble has to do with the MET because mass fluctuations only lead to an apparent force within a closed system IF you take the INCORRECT F=ma approach as opposed to the more general F=d/dt(p) approach.  If you have a closed system (i.e. nothing transfers across the system boundary) and you apply cyclic mass variations within the system, you would see that:

Fnet=(dm/dt)v + ma
0=(dm/dt)v + ma                       (Fnet=0 because the system is closed)
0=0 + ma                                   (since there is no flow of mass across the system boundary)***
a=0

And so (following this derivation) cyclic mass variation won't actually yield any propulsive benefits.

Basically I want to see how Woodward addresses this issue, and whether it is done so in such a way as to make METs viable.  I want to see why Woodward finds this argument "incorrect" as he says.  Welcome other posters to point out any issues with this derivation as well of course. 



***If there is energy flux across the system boundary in the form of gravinertial waves, then those would account for the dm/dt term and presumably gravinertial waves travel at the speed of light, so v=c.  However, if that were the case, then the MET would be unable to achieve thrust greater than a perfectly collimated photon rocket, which can be derived as follows:

E=mc2
m=E/c2
(d/dt)m=(d/dt)E/c2
dm/dt=P/c2

E=energy
P=power
c=speed of light

Sub this expression for dm/dt into the rocket equation (realizing that we have now replaced the exhaust flow with photons/gluons/W,Z bosons, gravitons, gravienterial waves, magic energy waves - literally anything that has energy but no rest mass) and you get:

0=(P/c2)c + ma
ma=P/c

So we are back to the thrust of a perfectly collimated photon rocket.
« Last Edit: 03/30/2016 02:13 AM by wallofwolfstreet »

Offline Mezzenile

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Re: Woodward's effect
« Reply #708 on: 03/30/2016 07:08 PM »
... 

Now as is always the case, internal forces (f) are always equal and opposite to one another, and so they sum to zero.  Because the "rocket + some exhaust" system is in free space, it experiences no external forces either, so F=0.  Remember that the force applied to the rocket by the exhaust is equal and opposite the force on the exhaust applied by the rocket.  Since we have included this portion of the exhaust in our control volume, these forces are themselves internal, and sum to zero (see attached figure). 

Now if we applied F=ma we would get that 0=ma so a=0.  Apparently rockets don't work.
...

Your rocket works perfectly !  ;) You have just forgotten that you have included in your rocket system all the plume delivered by its propulsion motor, so in fact the center of gravity of your whole rocket system does not move and so its acceleration a is zero even if the massive plume goes in one direction and the usefull rocket in the opposite direction !  ;)

One point that you seem to miss : The laws of galilean dynamics are valid only when expressed in a galilean referential (the natural referential attached to the rocket is not galilean as the rocket is accelerated and so is prone to fictious forces)

To discriminate True forces from Fictious forces is rather simple : True forces are invariant when observed from different galilean referentials. By opposition Fictious forces change with the galilean referential from which they are observed.

   
« Last Edit: 03/30/2016 07:41 PM by Mezzenile »

Offline wallofwolfstreet

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Re: Woodward's effect
« Reply #709 on: 03/30/2016 08:20 PM »
... 

Now as is always the case, internal forces (f) are always equal and opposite to one another, and so they sum to zero.  Because the "rocket + some exhaust" system is in free space, it experiences no external forces either, so F=0.  Remember that the force applied to the rocket by the exhaust is equal and opposite the force on the exhaust applied by the rocket.  Since we have included this portion of the exhaust in our control volume, these forces are themselves internal, and sum to zero (see attached figure). 

Now if we applied F=ma we would get that 0=ma so a=0.  Apparently rockets don't work.
...

Your rocket works perfectly !  ;) You have just forgotten that you have included in your rocket system all the plume delivered by its propulsion motor, so in fact the center of gravity of your whole rocket system does not move and so its acceleration a is zero even if the massive plume goes in one direction and the usefull rocket in the opposite direction !  ;)

One point that you seem to miss : The laws of galilean dynamics are valid only when expressed in a galilean referential (the natural referential attached to the rocket is not galilean as the rocket is accelerated and so is prone to fictious forces)

To discriminate True forces from Fictious forces is rather simple : True forces are invariant when observed from different galilean referentials. By opposition Fictious forces change with the galilean referential from which they are observed.

 

Thanks for the response Mezzenile, but I think this misses the mark of my argument.  I'll try to address each of your paragraphs as follows:

Quote
Your rocket works perfectly !  ;) You have just forgotten that you have included in your rocket system all the plume delivered by its propulsion motor, so in fact the center of gravity of your whole rocket system does not move and so its acceleration a is zero even if the massive plume goes in one direction and the useful rocket in the opposite direction !  ;)
     

In my control volume, I only included part of the rocket plume; particularly the part experiencing acceleration from the propulsion motor.  The system boundary is at the point where the exhaust has been fully accelerated and no longer experiences any forces.  This means there is mass flux across the system boundary in the form of constant velocity rocket exhaust.  You are correct that if the control volume contained all of the rocket exhaust, then the center of mass of the total system (rocket + all exhaust) would experience no acceleration.  But in my control volume mass escapes across the system boundary (anisotropically), and so the center of mass contained in the control volume must in fact accelerate.

Quote
One point that you seem to miss : The laws of galilean dynamics are valid only when expressed in a galilean referential (the natural referential attached to the rocket is not galilean as the rocket is accelerated and so is prone to fictious forces)

This is true, but doesn't relate to my example.  A control volume and a referential (frame of reference) are not the same thing.  An accelerating frame of reference is an issue and prone to fictitious forces, but the frame of reference is not accelerating my example.  .  In my example the referential is taken to be instantaneously at rest with respect to the rocket prior to acceleration (i.e. the referential does not accelerate with the rocket; it is an inertial frame of reference).  Furthermore, the analysis only needs to apply instantaneously.  You could fix the control volume in space (you seem to not like that I fixed the control volume to the rocket, which on second thought I shouldn't have done), and as long as some of the exhaust plume is passing across the control volume boundary, you would have the exact same problem.  A fixed control volume can still have the center of mass contained within accelerate. 

Quote
To discriminate True forces from Fictitious forces is rather simple : True forces are invariant when observed from different galilean referentials. By opposition Fictitious forces change with the galilean referential from which they are observed.

Likewise true, but not relevant to the example at hand because none of the forces examined are fictitious because the forces are examined from the perspective of an inertial frame.  They are dependent on how we choose the control volume (i.e. how we define a 3D volume to encompass our system), not on the reference frame.

I think the bottom line is that you can fix the control volume in space, where at a given instant the rocket plus accelerating exhaust is contained within said control volume while constant velocity exhaust flowing across the control surface, and the problem is exactly the same, with no concerns as to non-inertial frames of reference or accelerating control volumes.                 
« Last Edit: 03/30/2016 09:09 PM by wallofwolfstreet »

Offline 93143

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Re: Woodward's effect
« Reply #710 on: 03/30/2016 10:02 PM »
I gotta say, your derivation of a=0 is downright hilarious.  You start by assuming the net force is zero, completely ignore the term you just got through saying was important (by setting it to zero too, in brazen disregard for the claimed operating principle of the device), and then declare that since there's no force (something you assumed a priori), there's no acceleration.

That's not what the Oak Ridge guys did, by the way.

An M-E thruster is not supposed to be a closed system.  You can't declare Fnet to be zero arbitrarily.  This means you now have to address what v is, and since the transactional interaction is in all directions and largely due to the action of the distant matter rather than the power input to the device, you can't assign v=c.

I'd assign v=0, actually, under the assumption for the sake of argument that the theory is correct, but I have yet to arrive at a satisfactory theoretical demonstration of this.  I think I've managed to show numerically that a naive application of the Doppler effect results in v being linear with the peculiar velocity of the thruster for small values thereof, but even if I haven't screwed up somewhere that's a pretty weak result...  and I'm rather busy these days so I can't devote a lot of effort to this right now...
« Last Edit: 03/30/2016 10:13 PM by 93143 »

Offline wallofwolfstreet

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Re: Woodward's effect
« Reply #711 on: 03/30/2016 11:48 PM »
I gotta say, your derivation of a=0 is downright hilarious.  You start by assuming the net force is zero, completely ignore the term you just got through saying was important (by setting it to zero too, in brazen disregard for the claimed operating principle of the device), and then declare that since there's no force (something you assumed a priori), there's no acceleration.

Thanks for the feedback.

I don't believe that is how the operating principle of the device works.  My (potentially wrong) read of the device's operating principle is:

You have a reaction mass, actuator and fluctuating mass (see Fig. 3.1 in Woodward's book if possible).  When the fluctuating mass is high, you use the actuator to push the reaction mass and fluctuating mass away from one another.  When the fluctuating mass is low, you use the actuator to pull them back together.  This should give net momentum to the bulk device.  The equal and opposite momentum is to be found as momentum flux in the gravinertial field.  This would imply to me that you could place a control volume around the device and the radiating gravinertial field, and within this control volume Fnet=0 because the Force on the MET is equal and opposite the force applied to the local gravinertial field (and hence it can be contained in a finite control volume).  I also only set dm/dt=0 contingent on my footnote (marked with ***), which demonstrates that non-zero mass flux just brings us back to the case of a regular rocket

I think one way to help clear this up, is to ask:

Is the MET like a photon rocket, emitting gravinertial waves in place of photons, or is it like a charge in space, experiencing a force caused by some other, far off charge?  Clearly I subscribe to the former from reading Woodward's stuff.

Quote
       
That's not what the Oak Ridge guys did, by the way.

Okay, I'll have to do some more looking then.

Quote
An M-E thruster is not supposed to be a closed system.  You can't declare Fnet to be zero arbitrarily.  This means you now have to address what v is, and since the transactional interaction is in all directions and largely due to the action of the distant matter rather than the power input to the device, you can't assign v=c.

Like I said, I set Fnet=0 because I thought an MET plus it's radiating gravinertial wave would, while the wave is contained in the control volume, form a closed system.  Maybe this is wrong. 

I don't understand how v=c logically follows from the above argument though.   

Quote
I'd assign v=0, actually, under the assumption for the sake of argument that the theory is correct, but I have yet to arrive at a satisfactory theoretical demonstration of this.  I think I've managed to show numerically that a naive application of the Doppler effect results in v being linear with the peculiar velocity of the thruster for small values thereof, but even if I haven't screwed up somewhere that's a pretty weak result...  and I'm rather busy these days so I can't devote a lot of effort to this right now...

What exactly are you applying the Doppler effect to here?  The gravinertial wave emanating from the MET? 
« Last Edit: 03/30/2016 11:58 PM by wallofwolfstreet »

Offline 93143

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Re: Woodward's effect
« Reply #712 on: 03/31/2016 01:58 AM »
This would imply to me that you could place a control volume around the device and the radiating gravinertial field, and within this control volume Fnet=0 because the Force on the MET is equal and opposite the force applied to the local gravinertial field (and hence it can be contained in a finite control volume).

Exactly the same argument applies as with the rocket - what happens when that gravinertial radiation leaves the control volume?  And you can't claim the thruster itself isn't generating force until then, because you can always draw the box smaller so that the wavefront in question has already left.

Quote
Is the MET like a photon rocket, emitting gravinertial waves in place of photons, or is it like a charge in space, experiencing a force caused by some other, far off charge?

More like the latter, I think.  Remember; this device is supposed to be tapping into the mechanism underlying inertia.  If you view it as a simple non-transactional radiator, you are ignoring the Machian interaction and thus missing the whole reason it's supposed to work.

Quote
I don't understand how v=c logically follows from the above argument though.

That's from your attempt to equate it with a photon rocket.  Again, this seems to proceed from viewing the device as a pure gravity wave generator in a potentially empty universe, rather than something that exploits the mechanism of inertia to generate a large transactional (therefore locally instant) response from all matter within its cosmic horizon.

Quote
What exactly are you applying the Doppler effect to here?  The gravinertial wave emanating from the MET?

Yes.  The cosmic redshift is isotropic, so it nulls out.  But the Doppler effect due to the peculiar velocity of the source does not.  So an M-E thruster should preferentially interact with matter in its direction of motion relative to the local comoving velocity, which in an expanding universe with a cosmic horizon might mean (I haven't proved this) that the mechanism is self-compensating with respect to the relative-velocity objection on the grounds of energy conservation.  (I need to do some more thinking about how exactly this stuff works in the context of cosmic expansion; I may be barking up the wrong tree entirely.  I should also read the latest papers and see if any of this is addressed...)
« Last Edit: 03/31/2016 04:10 AM by 93143 »

Offline birchoff

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Re: Woodward's effect
« Reply #713 on: 03/31/2016 04:13 AM »
This conversation feels a bit sureal.

That's from your attempt to equate it with a photon rocket.  Again, this seems to proceed from viewing the device as a pure gravity wave generator in a potentially empty universe, rather than something that exploits the mechanism of inertia to generate a large transactional (therefore locally instant) response from all matter within its cosmic horizon.
Given the theoretical justifications offered so far for MET's working they cannot work in an empty universe. Really for a MET there are two processes at work.

1. Being able to fluctuate the mass of some matter
2. pushing on that fluctuating mass when it is Massive and pulling on it when it is less massive.

Yes.  The cosmic redshift is isotropic, so it nulls out.  But the Doppler effect due to the peculiar velocity of the source does not.  So an M-E thruster should preferentially interact with matter in its direction of motion relative to the local comoving velocity, which in an expanding universe with a cosmic horizon might mean (I haven't proved this) that the mechanism is self-compensating with respect to the relative-velocity objection on the grounds of energy conservation.  (I need to do some more thinking about how exactly this stuff works in the context of cosmic expansion; I may be barking up the wrong tree entirely.  I should also read the latest papers and see if any of this is addressed...)

Not sure I get why this would be a problem?


Offline 93143

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Re: Woodward's effect
« Reply #714 on: 03/31/2016 05:39 AM »
Given the theoretical justifications offered so far for MET's working they cannot work in an empty universe.

Isn't that what I just said?

Quote
Not sure I get why this would be a problem?

Not a problem.  A solution, possibly.  Paul451 isn't the first to note that the energy conservation argument seems to depend on the average/effective relative velocity of the thruster with respect to the mass it's pushing on.  (I might be the first, but I hope not...)  If it could be shown that the effective mean velocity of the far-off active mass somehow tracks that of the thruster, energy conservation would no longer be a problem (aside from known issues like cosmological redshift).
« Last Edit: 03/31/2016 08:20 AM by 93143 »

Offline Mezzenile

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Re: Woodward's effect
« Reply #715 on: 03/31/2016 07:00 AM »
This preamble has to do with the MET because mass fluctuations only lead to an apparent force within a closed system IF you take the INCORRECT F=ma approach as opposed to the more general F=d/dt(p) approach.  If you have a closed system (i.e. nothing transfers across the system boundary) and you apply cyclic mass variations within the system, you would see that:

Fnet=(dm/dt)v + ma
0=(dm/dt)v + ma                       (Fnet=0 because the system is closed)
0=0 + ma                                   (since there is no flow of mass across the system boundary)***
a=0

And so (following this derivation) cyclic mass variation won't actually yield any propulsive benefits.

Basically I want to see how Woodward addresses this issue, and whether it is done so in such a way as to make METs viable.  I want to see why Woodward finds this argument "incorrect" as he says.  Welcome other posters to point out any issues with this derivation as well of course. 

(Fnet=0 because the system is closed) : You do not demonstrate this point.
(since there is no flow of mass across the system boundary) : You do not demonstrate this point.

And so (following this derivation) cyclic mass variation won't actually yield any propulsive benefits. : false affirmation based on non demonstrated hypothesis.

I propose you to make the modelisation of the Woodward device using the following scheme which relies only on the law of Force Equality betweeen Action and Reaction.


Offline birchoff

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Re: Woodward's effect
« Reply #716 on: 03/31/2016 02:37 PM »
Isn't that what I just said?

I am agreeing with you. just being explicit about it.

Offline birchoff

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Re: Woodward's effect
« Reply #717 on: 03/31/2016 02:44 PM »
This preamble has to do with the MET because mass fluctuations only lead to an apparent force within a closed system IF you take the INCORRECT F=ma approach as opposed to the more general F=d/dt(p) approach.  If you have a closed system (i.e. nothing transfers across the system boundary) and you apply cyclic mass variations within the system, you would see that:

Fnet=(dm/dt)v + ma
0=(dm/dt)v + ma                       (Fnet=0 because the system is closed)
0=0 + ma                                   (since there is no flow of mass across the system boundary)***
a=0

And so (following this derivation) cyclic mass variation won't actually yield any propulsive benefits.

Basically I want to see how Woodward addresses this issue, and whether it is done so in such a way as to make METs viable.  I want to see why Woodward finds this argument "incorrect" as he says.  Welcome other posters to point out any issues with this derivation as well of course. 

Why do you believe there is no mass flow across the system boundary? Assuming your system boundary only includes the MET device and not the rest of the universe?

Offline Mezzenile

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Re: Woodward's effect
« Reply #718 on: 03/31/2016 05:40 PM »
I propose you to make the modelisation of the Woodward device using the following scheme which relies only on the law of Force Equality betweeen Action and Reaction
....
Fundamental relations governing the dynamic of the Woodward device.

Offline wallofwolfstreet

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Re: Woodward's effect
« Reply #719 on: 03/31/2016 08:11 PM »
Exactly the same argument applies as with the rocket - what happens when that gravinertial radiation leaves the control volume?  And you can't claim the thruster itself isn't generating force until then, because you can always draw the box smaller so that the wavefront in question has already left.
I know the same arguement applies to rockets, that was the point.  When the gravierrtial radiayion leaves the control volume then the MET becomes a rocket that uses gravientertital radiation as it's propellant, which per my footnote resticts it's performace to that of a photon rocket since presumable gravinertial radiation obeys E2=(pc)2+(moc2)2 which fundamentally limits performance.  The point of the control volume (in the incorrect gravinterial rocket interpretation) is that it can be drawn to enclose a system, in which case the mass flow acorss the boundary accounts for accelration of the CoM. 

Admittadely, I should have reoragnized the logic flow in my original post because the justifictaion is weak, with the main thrust of the arguement being lost.

Quote
More like the latter, I think.  Remember; this device is supposed to be tapping into the mechanism underlying inertia.  If you view it as a simple non-transactional radiator, you are ignoring the Machian interaction and thus missing the whole reason it's supposed to work.

I suppose I fundamentally don't understand how METs are supposed to work then.  I guess I need to do some more reading.