Author Topic: Woodward's effect  (Read 284774 times)

Offline Robotbeat

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Re: Woodward's effect
« Reply #160 on: 02/12/2013 09:46 PM »
The current model is that you can't just change the inertia in the way Woodward supposes, there's no long-distance field (other than the usual inverse squared ones). In the current model, there is no way to "push" against the whole of the universe and develop propellantless propulsion.

The current model says that momentum and energy are both conserved locally.
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Offline Elmar Moelzer

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Re: Woodward's effect
« Reply #161 on: 02/12/2013 09:56 PM »
I am not saying that Woodward is right, but you counter argumentation is wrong.
then you now have a way to determine your absolute direction and velocity, which goes against current physics.
How?

Offline cuddihy

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Re: Woodward's effect
« Reply #162 on: 02/12/2013 10:07 PM »
The current model is that you can't just change the inertia in the way Woodward supposes, there's no long-distance field (other than the usual inverse squared ones). In the current model, there is no way to "push" against the whole of the universe and develop propellantless propulsion.

The current model says that momentum and energy are both conserved locally.

Woodward claims (on the basis of Sciama) that the long-distance field that transmits inertial forces is precisely the long distance inverse square one...i.e. gravity...that you admit is part of the current model. Might want to read the actual Woodward papers, they're not terribly long.

Your second statement is semantic--in that statement "local" is taken to mean any particles that are interacting. QED if your test particle is interacting with the entire universe (as it must for Mach's Principle to hold), the entire universe is local. So momentum and energy are conserved.

Anyway, you're focusing your fire on the wrong point. The only really incredible part of the theory is that it requires time-traveling (as it were) gravity waves to make inertia work the way it does instantly. That's clearly the actual incredible part without resorting to blaming the effect rather than the theory itself.

Offline D_Dom

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Re: Woodward's effect
« Reply #163 on: 02/12/2013 10:22 PM »
Some ten months ago in email exchange with Star-Drive he described his effort to

 "pin down the reality of the effect(s) we are seeing and if real AND scalable, then determine what math model best fits the data that will be obtained".

 I consider that effort to be very interesting and all other "incredible"  discussion is lost in the noise. I hope to understand enough of the math eventually and build a "flightworthy" experiment.
 Rejecting the concept out of hand because of my lack of understanding will never result in controlled flight. I choose to concentrate my effort on understanding the signal, rejecting the noise is part and parcel of good design of experiments.
« Last Edit: 02/12/2013 10:49 PM by cygnusX1 »
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Offline Robotbeat

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Re: Woodward's effect
« Reply #164 on: 02/12/2013 10:38 PM »
The current model is that you can't just change the inertia in the way Woodward supposes, there's no long-distance field (other than the usual inverse squared ones). In the current model, there is no way to "push" against the whole of the universe and develop propellantless propulsion.

The current model says that momentum and energy are both conserved locally.

Woodward claims (on the basis of Sciama) that the long-distance field that transmits inertial forces is precisely the long distance inverse square one...i.e. gravity......
Well if it's inverse squared, then the bulk mass of the Universe should affect me barely at all (and would only exert maybe a dozen or so micronewtons... which it does by the way, but it's not the mechanism for inertia), and the affect due to local objects (like, say, the EARTH) would overwhelm it.

Nope, the coupling field that is posited is much more than inverse squared.
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Offline D_Dom

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Re: Woodward's effect
« Reply #165 on: 02/12/2013 10:56 PM »
So the bulk mass of the universe does affect us in the micro-newtons range.
I accept the earth overwhelms due to distance.
I wish I understand the mechanism for inertia, can you explain the coupling field? Not the posited "Woodward effect" but the micronewtons of the bulk mass of the universe.
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Offline Robotbeat

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Re: Woodward's effect
« Reply #166 on: 02/12/2013 10:58 PM »
So the bulk mass of the universe does affect us in the micro-newtons range.
I accept the earth overwhelms due to distance.
I wish I understand the mechanism for inertia, can you explain the coupling field? Not the posited "Woodward effect" but the micronewtons of the bulk mass of the universe.
To get a rough order-of magnitude estimate, just use the law of gravitation:

Force= (Gravitational Constant)*massofuniverse*massofme/(roughlyradiusofuniverse)^2
Chris  Whoever loves correction loves knowledge, but he who hates reproof is stupid.

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Offline antiquark

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Re: Woodward's effect
« Reply #167 on: 02/13/2013 12:32 AM »
I am not saying that Woodward is right, but you counter argumentation is wrong.
then you now have a way to determine your absolute direction and velocity, which goes against current physics.
How?


Put Woodward drives on two windowless trains, one moving, one stationary. The one on the moving train will not accelerate as fast as the one on the stationary train, indicating the velocity of the train.

Offline KelvinZero

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Re: Woodward's effect
« Reply #168 on: 02/13/2013 12:54 AM »
You dont need to build a working vehicle to prove it. You just need to make an accurate prediction that no other theory predicts and then reproduce it. If you claim 1.234 jiggaboos and that is what you see, then essentially you dont need to worry that it is some random effect you have not eliminated.

Then if other people can reproduce your 1.234 jiggaboos result, for a while the physics community will go wild trying to find a way to tear it down, then if they can't it will become accepted.

Offline QuantumG

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Re: Woodward's effect
« Reply #169 on: 02/13/2013 12:57 AM »
I am not saying that Woodward is right, but you counter argumentation is wrong.
then you now have a way to determine your absolute direction and velocity, which goes against current physics.
How?


Put Woodward drives on two windowless trains, one moving, one stationary. The one on the moving train will not accelerate as fast as the one on the stationary train, indicating the velocity of the train.

That's just using a Woodward drive as a "rest of the universe" detector. You can just substitute any sort of detector. The point of the word "windowless" in the thought experiment is to exclude all forms of detection of the outside world from consideration.
« Last Edit: 02/13/2013 12:58 AM by QuantumG »
Jeff Bezos has billions to spend on rockets and can go at whatever pace he likes! Wow! What pace is he going at? Well... have you heard of Zeno's paradox?

Offline JohnFornaro

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Re: Woodward's effect
« Reply #170 on: 02/13/2013 01:40 AM »
So you don't know about Sciama's gravelectric equation either?

If the equation predicts free energy, then sorry, I don't plan on reading it.

It does not.  That much is clear.

Sciama is defining inertia according to an interpretation of Mach's principle.  His gravelectric equation has been peer reviewed and not disproved in 60 years.

I have no idea if it's right or wrong, but my sense is that not many people understand it.  This thread confirms that observation so far.
Sometimes I just flat out don't get it.

Offline Cinder

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Re: Woodward's effect
« Reply #171 on: 02/13/2013 07:21 PM »
Does not Feynman (too) leave the question of inertia fairly open in his Lectures on Physics?  Saying something like "that question is curiously unresolved" (despite the rest of the model apparently not suffering from it).
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Offline Robotbeat

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Re: Woodward's effect
« Reply #172 on: 02/13/2013 07:36 PM »
So you don't know about Sciama's gravelectric equation either?

If the equation predicts free energy, then sorry, I don't plan on reading it.

It does not.  That much is clear.

Sciama is defining inertia according to an interpretation of Mach's principle.  His gravelectric equation has been peer reviewed and not disproved in 60 years.

I have no idea if it's right or wrong, but my sense is that not many people understand it.  This thread confirms that observation so far.
We are talking about woodward's purported effect, not sciama.
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Offline JohnFornaro

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Re: Woodward's effect
« Reply #173 on: 02/14/2013 12:40 PM »
We are talking about woodward's purported effect, not sciama.

I get it.  You don't understand his math either.
Sometimes I just flat out don't get it.

Offline Robotbeat

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Re: Woodward's effect
« Reply #174 on: 02/14/2013 01:35 PM »
We are talking about woodward's purported effect, not sciama.

I get it.  You don't understand his math either.
Give me a good reason to try.
Chris  Whoever loves correction loves knowledge, but he who hates reproof is stupid.

To the maximum extent practicable, the Federal Government shall plan missions to accommodate the space transportation services capabilities of United States commercial providers. US law http://goo.gl/YZYNt0

Offline D_Dom

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Re: Woodward's effect
« Reply #175 on: 02/14/2013 02:54 PM »
From one of many papers
http://arxiv.org/ftp/arxiv/papers/1301/1301.6178.pdf
in the previous thread
 http://forum.nasaspaceflight.com/index.php?topic=13020.1785
I find this interesting:
4. Conclusions
We have shown in Section 2 how, using a Mach Effect Thruster (MET) it is possible to produce a linear thrust with no propellant. We have utilized the Mach Principle which says in brief, that the inertial mass of a body is determined by its gravitational interaction with the rest of the matter and energy flow in the universe. We sought to prove that we had managed to eliminate all vibration effects from our data and attempted a null experiment. We attached equal size reaction masses to each end of the active PZT stack, this would cause the induced mass fluctuation to push and pull in both directions at once, and the device should not produce a net thrust. In section 3 we have shown that by using equal masses at both ends of our device we can indeed eliminate the net thrust.

If this is not the science you are looking for then move along.
« Last Edit: 02/14/2013 02:56 PM by cygnusX1 »
Space is not merely a matter of life or death, it is considerably more important than that!

Offline JohnFornaro

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Re: Woodward's effect
« Reply #176 on: 02/14/2013 02:58 PM »
We are talking about woodward's purported effect, not sciama.

I get it.  You don't understand his math either.
Give me a good reason to try.

I see that you can run as well as hide. 

You're not going to get away with brushing away Woodward without understanding Sciama.  You have no idea if Woodward has misinterpreted Sciama.

Instead, you focus your effort on Woodward's unverified energy claim, and make no effort to understand the math behind his experimental apparatus, which precedes the "claim" by a number of years.

The reason these two threads continue is because nobody understands the math.

On the other thread, Steven Fuesrt started educating the thread readers in June of 2011, but he stopped.  Another poster Blazotron, almost started explaining the math in May and June of 2009.  He debunked Shawyer's EM drive, but did not get around to examining Woodward's work.

It would be nice if these two would help out.


Happy to be able to share.  Glad everyone enjoyed them.

Quote

Who said anything about breaking conservation?  M-E doesn't.  If the EM-Drive works (which I am not claiming), whatever makes it work can be assumed to also not break conservation unless very good evidence shows up that it does.

M-E does.  Its math depends on a vector theory of gravity.  The reason everyone else uses the more complex tensor theory known as GR is because vector theories break energy-momentum conservation.


gross generalization. Tensor theory =/= GR.

Newtonian theories are all kinds of inconsistent. That doesn't invalidate every calculation or derivation done in Newtonian calculus either. Tensor theories are shown to not work with certain parts of quantum physics. That doesn't invalidate tensor GR math either.

What specific part of Woodward's derivation are you alleging cannot be calculated in vector form and why?


The Woodward derivation requires the existence of a mass-energy dipole.  This is possible with a vector theory (which is what they use).  With a tensor theory, like GR, it is impossible.  The lowest multipole order is a quadrapole.  This matters because the emitted power from a quadrapole is much much less than a dipole by many orders of magnitude due to the additional G/c^2 factor.

So why don't other physicists use vector gravity theories?  The reason is that they don't conserve energy-momentum.  In effect Woodward is assuming that momentum is not conserved, constructing a device, and then noticing that that device doesn't conserve momentum.  It is the physics version of "begging the question".

This particular problem is exercise 7.2 in MTW Gravitation.

The above explanation is still too advanced for me, and I would ask for an easier approach to tensor theory.
Sometimes I just flat out don't get it.

Offline antiquark

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Re: Woodward's effect
« Reply #177 on: 02/14/2013 03:04 PM »
You have no idea if Woodward has misinterpreted Sciama.

Sciama: no free energy.
Woodward: free energy.

Woodward has misinterpreted Sciama.

Offline JohnFornaro

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Re: Woodward's effect
« Reply #178 on: 02/14/2013 03:18 PM »
You have no idea if Woodward has misinterpreted Sciama.

Sciama: no free energy.
Woodward: free energy.

Woodward has misinterpreted Sciama.

Well yeah, that is the assertion.  But this is Woodward's fundamental equation:

E = -V phi - (phi/c^^2) (dv/dt)

Which doesn't get into "free" energy, that I can tell.
« Last Edit: 02/14/2013 03:19 PM by JohnFornaro »
Sometimes I just flat out don't get it.

Offline GeeGee

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Re: Woodward's effect
« Reply #179 on: 02/14/2013 05:37 PM »



This is possible with a vector theory (which is what they use).  With a tensor theory, like GR, it is impossible. 

Woodward's conjecture does not hinge on Sciama's vectory theory of gravity. He has stated before that Sciama's model is only an approximation to GR, and the phi=c^2 result can be obtained in GR using Nordtvedt's PPN formalism.

Here's a quote of his I found explaining this distinction

"I am not claiming (nor have I claimed) that Sciama's 1953 theory is exactly correct.  What I do claim is that the formalism is the vector approximation to GR -- especially the dA/dt term in the gravelectric field equation.  The same term, in the interpretation of this effect, shows up in the PPN version as Nordtvedt shows later as "linear accelerative frame dragging". And when the rigidly accelerating body producing the frame dragging is the observable universe, rigid frame dragging results (and, up to a constant factor of order unity, phi = c^2).  The point is that whether you treat this as frame dragging or inertial force, the distant matter in the universe affects the inertial behavior of local objects by producing the reaction force when local objects are forced out of geodesic motion."
« Last Edit: 02/14/2013 05:55 PM by GeeGee »