Author Topic: Woodward's effect  (Read 288018 times)

Offline HMXHMX

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Re: Woodward's effect
« Reply #580 on: 11/21/2015 03:35 AM »
HMXHMX : has you forwarded Gargoyle's criticism to Woodward, particularly post #570?

I did forward earlier comments. Professor Woodward feels he's addressed the matter sufficiently in both the book and the white paper, so I doubt he's going to add to those writings, preferring to concentrate his attention on experiments currently underway.  But he did add a comment to me and a few others yesterday in a private email that I doubt he'd mind me sharing:

"I spelled out both how and why a correct calculation is to be done in the last paragraph of the paper.  The ... mistake is to assume that you can set the integral of the FOM equation over some arbitrarily long time equal to the integral of the work equation for the same time.  You have to set the integrals so that the integration time is short enough that the kinetic energy generated is less that the input energy to the thruster.  Then you repeat this calculation for all subsequent intervals of the same length until you reach whatever time of operation you are interested in.  This guarantees that the kinetic energy sum never exceeds the input energy sum in the two sums.  This must be done because the only invariant velocity is Zero at the start of each integration interval and you know for a fact that simple mechanical systems do NOT violate energy conservation.  There is nothing magical about METs."

That's all he's going to say, I'm certain, so I'm gong to respect his position and leave matters there.

Offline Paul451

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Re: Woodward's effect
« Reply #581 on: 11/21/2015 12:07 PM »
This must be done because the only invariant velocity is Zero at the start of each integration interval

And that's a weirdly elementary mistake. He's changing the frame of reference for each interval, but then trying to sum linearly. That's not a real thing.

The ... mistake is to assume that you can set the integral of the FOM equation over some arbitrarily long time equal to the integral of the work equation for the same time.

The mistake (well, his other mistake) is that he believes his "FoM" applies to classical systems, and therefore classical systems are "over-unity" if assessed with the same method as MET and other reactionless thrusters; thus "proving" that the method is flawed. However, no classical system has a FoM that is constant over a long enough period for it to reach an over-unity state. That's the very trait that defines a reactionless thruster, the constant acceleration per unit power input.

That's the thing so many people can't seem to get their heads around... classical systems don't have a constant FoM. Not an object on a frictionless surface. Not a rocket on a rotor. Not a gravitational sling-shot. Not a solar sail (or a regular sail). Not an anti-matter engine. Not an ion-drive. Not a photon drive. None of them. Only MET, EMDrives, and similar reactionless thrusters.

[edit: " Not a photon drive." Oops, got carried away. Actually that one is the only classical system that does have a constant thrust/watt ("figure of merit"), and could therefore potentially reach an overunity state... And the velocity at which the perfect photon drive (ie, the actual emission of a photon from an atom) achieves break-even is... the speed of light.]
« Last Edit: 11/21/2015 11:57 PM by Paul451 »

Offline Povel

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Re: Woodward's effect
« Reply #582 on: 11/21/2015 12:12 PM »
HMXHMX : has you forwarded Gargoyle's criticism to Woodward, particularly post #570?

I did forward earlier comments. Professor Woodward feels he's addressed the matter sufficiently in both the book and the white paper, so I doubt he's going to add to those writings, preferring to concentrate his attention on experiments currently underway.  But he did add a comment to me and a few others yesterday in a private email that I doubt he'd mind me sharing:

"I spelled out both how and why a correct calculation is to be done in the last paragraph of the paper.  The ... mistake is to assume that you can set the integral of the FOM equation over some arbitrarily long time equal to the integral of the work equation for the same time.  You have to set the integrals so that the integration time is short enough that the kinetic energy generated is less that the input energy to the thruster.  Then you repeat this calculation for all subsequent intervals of the same length until you reach whatever time of operation you are interested in.  This guarantees that the kinetic energy sum never exceeds the input energy sum in the two sums.  This must be done because the only invariant velocity is Zero at the start of each integration interval and you know for a fact that simple mechanical systems do NOT violate energy conservation.  There is nothing magical about METs."

That's all he's going to say, I'm certain, so I'm gong to respect his position and leave matters there.

I've been following Woodward's work for a couple of years.
I'm not qualified in advanced physics, but, as previously noted, this is basic Newtonian mechanics, and I have to agree with gargoyle's argument here.
I'm sorry if this sounds harsh, it's not my intention, but that paper is counterproductive if its objective was to close once for all the diatribe, and its only contribute is to casting further doubts on Woodward himself.

I find Woodward's argument flawed at the level of first assumptions, but I'm willing to listen if he actually discusses the point raised, instead of just repeating his own.

Since his whole point stems from this, could you please ask Woodward about an example of an existing classical system with a constant figure of merit? I can't think of any, since they violate CoE.

Also, in the overunity paper Woodward doesn't seem to be consistent with what is still written on his own page, the one quoted by gargoyle. At the end of he wrote:

Quote
The net momentum flux is accompanied by a net energy flux, so although our impulse engine, considered locally, appears to violate energy conservation, that need not necessarily be the case. The extraction of useful work from matter that may be completely thermalized raises interesting questions. Boosting, rather than borrowing, from the future, however, seems to be the nature of the process involved.

I understand this was written many years ago, and that maybe his ideas about changed in the meanwhile. But if so, why is this still written there?
« Last Edit: 11/21/2015 12:20 PM by Povel »

Offline dustinthewind

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Re: Woodward's effect
« Reply #583 on: 11/26/2015 07:53 PM »
HMXHMX : has you forwarded Gargoyle's criticism to Woodward, particularly post #570?

I did forward earlier comments. ...


...

Since his whole point stems from this, could you please ask Woodward about an example of an existing classical system with a constant figure of merit? I can't think of any, since they violate CoE.

...

I am assuming your talking about systems that don't use reaction mass to speed up.  One system that comes to mind is a swing.  Also I remember this same subject came up in the EM drive section and I pointed out a device that may qualify as a classical system of this type.  The video is on YouTube here: .  I don't think it is over unity but it certainly doesn't appear to use reaction mass to speed up.  How it works is like a swing.  When the mass wants to swing out to the largest radius you just twist the weights against the force of the mass.  When the mass is coming in to the center you can reverse the twist and assist against the force pulling it out and the energy transfers to the system as a whole.  Basically F(x).dx=E .  This is angular velocity rather than linear but the argument is that there is this break even speed where the energy you put in becomes more than you put in.  My argument is if this is true then you might be able to argue the same for this device.  Explaining this systems conservation of energy may be instructive in similar systems, though in this case it is mechanical instead of electro-magnetic.
« Last Edit: 11/27/2015 10:52 AM by dustinthewind »

Offline ppnl

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Re: Woodward's effect
« Reply #584 on: 11/30/2015 08:53 PM »
HMXHMX : has you forwarded Gargoyle's criticism to Woodward, particularly post #570?

I did forward earlier comments. ...


...

Since his whole point stems from this, could you please ask Woodward about an example of an existing classical system with a constant figure of merit? I can't think of any, since they violate CoE.

...

I am assuming your talking about systems that don't use reaction mass to speed up.  One system that comes to mind is a swing.  Also I remember this same subject came up in the EM drive section and I pointed out a device that may qualify as a classical system of this type.  The video is on YouTube here: .  I don't think it is over unity but it certainly doesn't appear to use reaction mass to speed up.  How it works is like a swing.  When the mass wants to swing out to the largest radius you just twist the weights against the force of the mass.  When the mass is coming in to the center you can reverse the twist and assist against the force pulling it out and the energy transfers to the system as a whole.  Basically F(x).dx=E .  This is angular velocity rather than linear but the argument is that there is this break even speed where the energy you put in becomes more than you put in.  My argument is if this is true then you might be able to argue the same for this device.  Explaining this systems conservation of energy may be instructive in similar systems, though in this case it is mechanical instead of electro-magnetic.


I commented on this in the other thread. This device is just reacting against the earth and is really no different than a car except that it is angular velocity rather than linear velocity. It will require ever increasing power for constant acceleration just like a car. If it didn't it would be over unity and could power the world.

Offline dustinthewind

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Re: Woodward's effect
« Reply #585 on: 12/10/2015 12:10 AM »
... If it didn't it would be over unity and could power the world.

I don't think anyone here is claiming this device is overunity.
« Last Edit: 12/10/2015 12:11 AM by dustinthewind »

Offline ppnl

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Re: Woodward's effect
« Reply #586 on: 12/12/2015 07:30 PM »
... If it didn't it would be over unity and could power the world.

I don't think anyone here is claiming this device is overunity.

People are making claims that inevitably lead to an over unity result. That is not only true of some people here but is true of both Shawyer and Woodward. That is somewhat excusable for people here because we live on an effectively limitless and massive plain where Galilean  relativity isn't immediately apparent. But there is no excuse for Shawyer and Woodward. I simply cannot wrap my head around how they could fail so totally.

Offline M.E.T.

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Re: Woodward's effect
« Reply #587 on: 12/12/2015 07:58 PM »
My understanding has always been that the Mach Effect Thruster is not generating  energy. Instead, it is tapping into the energy of the rest of the universe, similar to a sailing ship tapping into the wind to generate forward motion.

The controversial part here is the mechanism by which this transfer takes place instantaneously with the mass of the distant universe. But Woodward has covered that with the Wheeler-Feynman Absorber theory, as far as I understand. I have not yet seen that theoretically debunked.

And according to my understanding, it is this extraction of energy from the mass of the rest of the Universe that makes the Mach Effect Thruster increase the entropy in the rest of the universe each time that it is used.

Going even further, the argument that I have seen put forward by some who understand the theory better than I do, is that the extensive use of Mach Effect Thrusters elsewhere in space and time could be one potential explanation for the accelerating expansion of the Universe. And ultimately, could be accelerating the heat death of the Universe.

So far from generating more energy than is put into it, the Mach Effect Thruster is simply extracting energy from elsewhere and tapping into it locally. With the entire Universe as the system, there is no net energy being created. It is simply being channeled from one place to the next.

That is how I read the theory, anyway.

Offline Paul451

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Re: Woodward's effect
« Reply #588 on: 12/12/2015 08:53 PM »
My understanding has always been that the Mach Effect Thruster is not generating  energy. Instead, it is tapping into the energy of the rest of the universe

That doesn't change the free-energy issue. The energy-input is constant with thrust, therefore increases linearly with the change in velocity. The energy output (the change in kinetic energy) varies with the square of change in velocity. Therefore there must be a velocity at which energy-out increases faster than energy-in. At that point, you have a free energy machine. This is regardless of where that input energy comes from.

(And over-unity occurs when velocity (in m/s) is greater than P/F. Where power is in watts (joules/second) and thrust is in newtons.)

Before anyone brings Woodward's claimed solution up again: The rate of energy increase exceeds the rate of energy applied. The "over-unity" is continuous, not in sum.

similar to a sailing ship tapping into the wind to generate forward motion.

No, with a sailing ship, or any conventional device, the thrust varies with velocity. That's what stops them from becoming free-energy machines.

it is this extraction of energy from the mass of the rest of the Universe that makes the Mach Effect Thruster increase the entropy in the rest of the universe each time that it is used.

Errr, that's actually the definition of decreasing entropy. Taking a broad low grade, relatively uniform energy and concentrating it to do work. Maxwell's demon and all that.

Offline M.E.T.

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Re: Woodward's effect
« Reply #589 on: 12/12/2015 09:07 PM »
Apologies on the last point.

Regarding the former though, I think wikipedia does a much better job than me of explaining the conservation issues:

Although the momentum and energy exchange with distant matter guarantees global conservation of energy and momentum, this field exchange is supplied at no material cost, unlike the case with conventional fuels. For this reason, when the field exchange is ignored, a propellantless thruster behaves locally like a free energy device. This is immediately apparent from basic Newtonian analysis: if constant power produces constant thrust, then input energy is linear with time and output (kinetic) energy is quadratic with time. Thus there exists a break-even time (or distance or velocity) of operation, above which more energy is output than is input. The longer it is allowed to accelerate, the more pronounced will this effect become, as simple Newtonian physics predicts.

Considering those conservation issues, a Mach effect thruster relies on Mach's principle, hence it is not an electrical to kinetic transducer, i.e. it does not convert electric energy to kinetic energy. Rather, a MET is a gravinertial transistor that controls the flow of gravinertial flux, in and out of the active mass of the thruster. The primary power into the thruster is contained in the flux of the gravitational field, not the electricity that powers the device. Failing to account for this flux, is much the same as failing to account for the wind on a sail.[62] Mach effects are relativistic by nature, and considering a spaceship accelerating with a Mach effect thruster, the propellant is not accelerating with the ship, so the situation should be treated as an accelerating and therefore non-inertial reference frame, where F does not equal ma. Keith H. Wanser, professor of physics at California State University, Fullerton, published a paper in 2013 concerning the conservation issues of Mach effect thrusters


End quote

So in short, the electrical energy going into the device is not what produces its acceleration.
« Last Edit: 12/12/2015 09:19 PM by M.E.T. »

Offline Paul451

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Re: Woodward's effect
« Reply #590 on: 12/12/2015 10:00 PM »
the electrical energy going into the device is not what produces its acceleration.

It doesn't change the over-unity result. Unless the amount of energy stolen from the universe-at-large (by reversing entropy) is somehow varying with device's velocity (and, to quote an old Benny Hill routine, "how does it know"), the device will still have a velocity at which energy created exceeds energy used or transferred.

Quote
"Failing to account for this flux, is much the same as failing to account for the wind on a sail."

<sigh> Frakkin' sailin' boats, you guys al'ys with the frakkin' sailin' boats.

Offline ppnl

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Re: Woodward's effect
« Reply #591 on: 12/12/2015 11:43 PM »
My understanding has always been that the Mach Effect Thruster is not generating  energy. Instead, it is tapping into the energy of the rest of the universe, similar to a sailing ship tapping into the wind to generate forward motion.

In principle there could be a mechanism to do that. But that does not get you constant acceleration with constant power. For example if you are in a car you can accelerate by pushing against the earth. But the faster you go the more power is needed to keep a constant acceleration. If there is a mechanism that allows you to push against a distant mass then how much you can accelerate with a given power depends on how fast you are moving compared to that distant mass and what direction you try to accelerate. For example our motion with respect to the cosmic background is about 10% of the speed of light. So accelerating in one direction would require huge amounts of power for undetectable amounts of acceleration. Accelerating in the other direction would allow you to extract enough energy to melt down the earth.


Quote
The controversial part here is the mechanism by which this transfer takes place instantaneously with the mass of the distant universe. But Woodward has covered that with the Wheeler-Feynman Absorber theory, as far as I understand. I have not yet seen that theoretically debunked.

It is a real shame that neither Feynman nor Wheeler are alive to offer their opinion. In any case nothing here allows violations of conservation of momentum and so the above analysis stands. Also if real energy and real momentum are being transferred then it's hard to see how you could avoid real information being sent. FTL communication. Wheeler-Feynman Absorber theory does not allow that either. 

Also Shawyer does not believe any such mechanism is needed since he believes ordinary physics explains it. His claim that there is no violation of conservation of momentum is about as wrong as you can get. Woodward makes a very similar mistake and seemingly refuses to address it. It's like neither can wrap their head around Galilean relativity.

Quote
And according to my understanding, it is this extraction of energy from the mass of the rest of the Universe that makes the Mach Effect Thruster increase the entropy in the rest of the universe each time that it is used.

A car can extract energy from the motion of the road under it or it can react against the road and accelerate. What it cannot do is get constant acceleration with constant power.


Quote
Going even further, the argument that I have seen put forward by some who understand the theory better than I do, is that the extensive use of Mach Effect Thrusters elsewhere in space and time could be one potential explanation for the accelerating expansion of the Universe. And ultimately, could be accelerating the heat death of the Universe.

Castles in the air dude. And if the Mach Effect Thrusters can visibly accelerate the expansion of the universe then surly you could use it to send a message faster than light - and so back in time.

Quote
So far from generating more energy than is put into it, the Mach Effect Thruster is simply extracting energy from elsewhere and tapping into it locally. With the entire Universe as the system, there is no net energy being created. It is simply being channeled from one place to the next.

That is how I read the theory, anyway.

So a mechanism for taping and draining the power of the entire universe. That is truly a marvelous castle you have built there. I... don't think I trust the foundation. All I see is air.

Re: Woodward's effect
« Reply #592 on: 12/22/2015 07:46 PM »
Of Woodward’s effect and Fluid space drive
I wish (very egotistical of me) to offer an alternative method.

Let us say that what James Woodward has is a box that can become heavier or lighter at command.

Now visualize a large cylinder floating in space, if we place the box at the inner “forward” end the cylinder and give it a strong push (with a spring to simplify).

The cylinder will accelerate “forward” while the box travels to the rear of the cylinder, the moment the box collides with the inner rear wall of the cylinder, the cylinder comes to a stop because the force now excreted in the backwards hull will be equal to the force that was exerted forward.

This is because the FORCE = MASS of the box X VELOCITY of the box.

BUT, if we really can make the box lighter (less mass) at command, then the resulting force in the backward direction will be less and the cylinder will gain velocity every cycle.

And there is another way.

If instead of reducing the boxes mass we reduce its velocity, by the same equation (F=MxV), the resulting force in the backwards direction will also be diminished.

The big difference is that while modifying a box’s mass is complex, modifying its velocity is a simple matter, all we need is an air brake or parachute if you will (did I mention that the cylinder is pressurized).

Another difference is that while demonstrating the Woodward effect is difficult, requires complex lab equipment to show a very (difficult to measure) small result compared to the random error Inherent in any experiment.
The results obtained by reducing the velocity can be demonstrated using the same set up available in physics classrooms for demonstrating of the Law of Conservation of Momentum, and the magnitudes of the acceleration obtained is sufficient to be observed without need for motion sensors o other means
Description of the experiment can be found here: http://wjetech.cl/

I am aware that the results seem to contradict that very important Law of Conservation of Linear Momentum, but that is precisely what makes the experiment interesting.


Offline Paul451

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Re: Woodward's effect
« Reply #593 on: 12/23/2015 04:13 AM »
This is because the FORCE = MASS of the box X VELOCITY of the box.

It really doesn't.

Momentum = Mass * Velocity. Force = Mass * Acceleration.

The big difference is that while modifying a box’s mass is complex, modifying its velocity is a simple matter, all we need is an air brake or parachute if you will (did I mention that the cylinder is pressurized).

The momentum of the object is transferred to the air that slows it, giving the air a slight motion in the same direction. So the net momentum transferred to the cylinder is the same on each pass.

There's always a second effect like this. That's why MET/EMDrive/etc are so unique (and why skepticism should be so high), they are claimed to be unlike any system ever studied. From mundane Earthly objects (like your box-in-cylinder) to extreme astronomical objects, like blackholes and neutron stars and decaying binary stars.

Offline Elmar Moelzer

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Re: Woodward's effect
« Reply #594 on: 12/23/2015 08:41 AM »
People are making claims that inevitably lead to an over unity result. That is not only true of some people here but is true of both Shawyer and Woodward.
I always was under the impression that Woodward required increasing energy input for a constant acceleration, but I might have misunderstood something.

Offline birchoff

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Re: Woodward's effect
« Reply #595 on: 12/23/2015 06:42 PM »
People are making claims that inevitably lead to an over unity result. That is not only true of some people here but is true of both Shawyer and Woodward.
I always was under the impression that Woodward required increasing energy input for a constant acceleration, but I might have misunderstood something.

I think there are alot of un answered questions about the ramifications of Woodwards METs that have yet to be answered. For now the only fact is a demonstrably working experiment with a solid theoretical justification for why the experiment works they way they believe it to work.

Its when we try to take it farther than that with very little data to back up our assertions where we get into trouble.

Offline ppnl

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Re: Woodward's effect
« Reply #596 on: 12/23/2015 09:31 PM »
People are making claims that inevitably lead to an over unity result. That is not only true of some people here but is true of both Shawyer and Woodward.
I always was under the impression that Woodward required increasing energy input for a constant acceleration, but I might have misunderstood something.

Woodward seems to allow a constant power for constant acceleration but only for a time period small enough to prevent over unity. Then you turn off the drive and turn it on again for another period of acceleration. Of course this is beyond silly. You still violate conservation of energy. Changing the frame of reference that you are calculating energy from is a really stupid slight of hand.

Reduce the period of acceleration to an arbitrarily small time and you can have the effect of accelerating constantly indefinitely. 

I simply cannot wrap my head around what these people are thinking. It's like freaky twilight zone thinking. They fail the Turing test.

Offline birchoff

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Re: Woodward's effect
« Reply #597 on: 12/24/2015 04:03 AM »
I gotta ask.

Does anyone have a reference for where Woodward claims AND shows that his METs offer constant acceleration for constant power? I know thats always been interpreted. But I dont recall any of the papers arguing that interpretation.

Offline ppnl

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Re: Woodward's effect
« Reply #598 on: 12/24/2015 06:16 AM »
I gotta ask.

Does anyone have a reference for where Woodward claims AND shows that his METs offer constant acceleration for constant power? I know thats always been interpreted. But I dont recall any of the papers arguing that interpretation.

Dude this has been done to death. A Woodward link is here:

http://ssi.org/epi/Over-Unity_Argument_&_Mach_Effect_Thrusters.pdf

gargoyle99 did a breakdown on it just a few messages up from this one. But one more time...

Woodward starts with this:

Quote
We know that, starting from t= 0, if we let the integration interval t get very large, the work equation integral will first equal and then exceed the energy calculated by the figure of merit equation.

Here he acknowledges that constant power for constant acceleration will eventually violate conservation of energy. His solution?

Quote
So we require that t be sufficiently small that this obvious violation of energy conservation does not happen.

Simple. You don't run the engine long enough to violate conservation of energy. So how do you get velocities larger than you can get in time t?

Quote
That is, we note what should be obvious physics for this situation: the energies added to the two sums in every differential time interval are always in the same ratio as they are in the very first interval because the only invariant velocity that exists in this case is the one of instantaneous rest at the outset of each interval.
If this prescription - the only one that makes physical sense in the circumstances – is followed, no energy
conservation violation follows from the calculation.  And elementary mechanics is not threatened by an obviously
wrong calculation.

See under relativity all inertial frames are equally valid. If I am in a rocket traveling through space at 10 miles per second compared to you I am entitled to say that my speed is zero and it is you who is moving. Speed is relative. So Woodward is claiming that he can turn his drive on for an interval t that is short enough it does not violate conservation of energy. Then turning it off he notes that it is valid to take his speed as zero since speed is relative. But then doing the same calculation again he can turn his drive on again for another time interval t and double his speed without violating conservation of energy. And continue time interval after time interval until you have whatever speed you want. You have effectively achieved constant acceleration with constant power.

But is just mind blowingly stupid. At the end you have still built up a very high speed so that the kinetic energy is much higher than the energy you put in. You cannot do energy calculations in different frames of reference and then add them as if they were in the same frame of reference. As I said before he is not only wrong - he fails my Turing test.

 

Offline birchoff

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Re: Woodward's effect
« Reply #599 on: 12/25/2015 02:08 AM »
I gotta ask.

Does anyone have a reference for where Woodward claims AND shows that his METs offer constant acceleration for constant power? I know thats always been interpreted. But I dont recall any of the papers arguing that interpretation.

Dude this has been done to death. A Woodward link is here:

http://ssi.org/epi/Over-Unity_Argument_&_Mach_Effect_Thrusters.pdf

gargoyle99 did a breakdown on it just a few messages up from this one. But one more time...

Woodward starts with this:

Quote
We know that, starting from t= 0, if we let the integration interval t get very large, the work equation integral will first equal and then exceed the energy calculated by the figure of merit equation.

Here he acknowledges that constant power for constant acceleration will eventually violate conservation of energy. His solution?

Quote
So we require that t be sufficiently small that this obvious violation of energy conservation does not happen.

Simple. You don't run the engine long enough to violate conservation of energy. So how do you get velocities larger than you can get in time t?

Quote
That is, we note what should be obvious physics for this situation: the energies added to the two sums in every differential time interval are always in the same ratio as they are in the very first interval because the only invariant velocity that exists in this case is the one of instantaneous rest at the outset of each interval.
If this prescription - the only one that makes physical sense in the circumstances – is followed, no energy
conservation violation follows from the calculation.  And elementary mechanics is not threatened by an obviously
wrong calculation.

See under relativity all inertial frames are equally valid. If I am in a rocket traveling through space at 10 miles per second compared to you I am entitled to say that my speed is zero and it is you who is moving. Speed is relative. So Woodward is claiming that he can turn his drive on for an interval t that is short enough it does not violate conservation of energy. Then turning it off he notes that it is valid to take his speed as zero since speed is relative. But then doing the same calculation again he can turn his drive on again for another time interval t and double his speed without violating conservation of energy. And continue time interval after time interval until you have whatever speed you want. You have effectively achieved constant acceleration with constant power.

But is just mind blowingly stupid. At the end you have still built up a very high speed so that the kinetic energy is much higher than the energy you put in. You cannot do energy calculations in different frames of reference and then add them as if they were in the same frame of reference. As I said before he is not only wrong - he fails my Turing test.

 

So I have read this recently published essay already, and since I believe my reading comprehension skills have not failed me. No where in this essay is Woodward actually claiming that a MET will produce constant acceleration for constant force. This is the very first paragraph 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
. Assuming this
argument to be correct, critcs assert that METs violate energy conservation as the ratio of
the acquired kinetic energy to total input energy exceeds “unity.”

The argument being disputed is not Woodward's but the argument made by critics. Now you could argue that I am splitting hairs, Because if MET's cannot provide constant acceleration for constant input power. Wouldn't the simple answer to the critique be to say they can't. I'll come back to that in a minute.

Since I have already stated, in a previous post, that prior to the last paragraph on page 5 of the essay you linked to. Woodward hasnt done anything stupid with the math. equation 1-9 is all accepted physics. while equation 10-15 is the result of a critic's argument that woodward is trying to show as being wrong. Which from everything you have said it seems like Woodward and you would be in agreement.

Now the only bone of contention really left with the essay begins towards the end of the essay where Woodward rhetorically asks the reader.

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?
...

That is where he is saying that the way to avoid the over unity issue with the argument made by critics is to bound the equation by choosing a suitable value for t.

After re reading the essay(n.b. it looks like it has been updated). I was tempted to agree with you. but something kept gnawing at me. I have a hard time believing that anyone could argue that you simply limit the time the MET runs for to some calculated t and that is enough to avoid over unity. As you have said if that decision is left to a human what if that human maliciously decides to let it run for t+1. If that decision is supposed to be left to the engine then how does the engine know what velocity it is moving at much less how does it know when to stop accelerating.

Now piggy backing off of Woodward's the idea that over unity if possible would exist for all mechanical systems. I am forced to ask the question how would any mechanical system absent friction or gravitational interaction know that it isnt supposed to go any faster. For rockets if you plot the velocity over time from lift off you get a graph that looks similar to



now there are some interesting differences. a rocket engine lifting off from earth is fighting at t=0, air friction + gravity + inertia. Since air friction reduces the higher up you go the acceleration portion of the graph actually would get steeper as it climbed higher. Now lets assume we are dealing with a Rocket with a near inexhaustible supply of fuel and oxidizer. Once it is past the earth's atmosphere the acceleration portion of the graph would stop getting steeper. Since air friction is reduced to zero and the only thing the rocket is fighting is gravity + inertia; which have been present since t=0. Now the rocket has barely made a dent in its supply of fuel and oxidizer so the first question I would have is why would the velocity graph begin to flatten? how does the rocket engine know that it isn't supposed to go any faster. The answer to me is the fuel+oxidizer reaction which powers the rocket is inherently limited. That reaction determines the maximum amount of force the rocket engine can generate. That said the interesting thing is, the rocket did undergo near constant acceleration for constant power input. It just couldn't sustain that constant acceleration beyond its limiting factor; the energy imparted by the fuel+oxidizer reaction. If we switched to a fusion rocket or even an antimatter rocket you eventually get to a terminal velocity, though in those cases due to the large amount of energy that those reactions release inertia begins to play a more important role in the limiting terminal velocity.

Now I could be wrong about this. But I don't believe we are arguing about whether or not a MET can under go constant acceleration for constant input power; because any Newtonian system can experience this for a relatively short period of time. I think what we really want to know is, for a MET what is the limiting factor. What keeps a MET from accelerating for all eternity or at the very least past the point where you have an over-unity problem. That is the question we want an answer to, and that is the question which isn't answered at all in the Essay because when we talk about constant acceleration for constant input power; since the MET seems to only be powered by electricity it looks like there would be no reason for it to stop accelerating.

Now Since Woodward's work is based on SR and GR I feel it safe to say that at the very least the speed of light would be the absolute limiting factor for a MET. But given the efficiency at which a MET seems to convert electricity into kinetic energy, That doesn't solve the real problem alluded to in the over-unity debate; which is being able to get more energy out of a MET than the electrical energy provided by its user. From my perspective I think the solution is that there is no problem. If you define the maximum energy input to a MET as the electrical energy provided by the user + something else. However, if you define the maximum energy input to a MET as being the electrical energy provided by the user, then there is an over-unity problem and a MET is magic pixie dust. Put another way the only reason it looks like over-unity is because we are only counting (observe) the electrical energy being put into a MET by the user.

Bringing this back to the essay. From my perspective after re-reading the essay and thinking about the proposed thought experiment. I think Woodward believes that MET's undergo constant acceleration for constant input. However, he doesn't believe that the constant acceleration occurs for as long as the MET is turned on. which is why I believe the argument he makes towards the end of the paper should really be read "there is a value for t beyond which you will no longer be accelerating". Because there is a limiting factor which will prevent it. Which is why I think the essay is more of a proof that there must be some limiting factor. Instead of an answer to the really interesting question, WHAT is that limiting factor.

<Random Speculation>
Personally I think the limiting factor is inertia, or more specifically the force inertia applies to a physical object. Now while there is no accepted theory on what the hell inertia is. We do know that it acts on a physical object to resist any change in its state of motion. To do this it must at all times be applying a force to the physical object or at the least apply that force when the object attempts to change is state of motion. Now a MET works by applying a voltage to a device to cause its Inertial Mass to oscillate from heavier to lighter. While at the same time applying a force to that same object in just the right way, so that it is pushed on in one direction when it is lighter and pushed on in the opposite direction when it is heavier; this results in a net directional force. Now if inertia is always applying a force to an object that is proportional to its inertial mass. Then the magnitude of that force when you push on the object while its inertial mass is reduced will be Flighter. When you push on the object while its inertial mass is increased it will be Fheavier. Which means by definition you should end up with

Fheavier - Flighter = Fnet

Where Fnet is greater than zero.

That means the total energy being fed into a MET device is the electrical energy used to trigger the controlled mass fluctuations and apply the external force which is pushing on the mass while it is lighter in one direction and pushing on the mass while it is heavier in the opposite direction. Plus Fnet which is the generated by inertia.
</Random Speculation>

My speculation only works if you can argue that inertia is a radiating field permeating the universe. Which is why I suspect Woodward/Fearn/Watsner have spent time updating Hoyle and Narlikar theory. Since it is a Machian version of Einsteins relativity. Which provides a mechanism for mass interactions in the local environment to communicate with the mass that is far away. This mechanism works so long as the expansion of the Universe is accelerating, which all observations seem to support.