Author Topic: Woodward's effect  (Read 285260 times)

Offline Robotbeat

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Re: Woodward's effect
« Reply #220 on: 02/19/2013 03:46 AM »
I understand maxwell just fine. Just explain to me why I should read a ton of what will almost surely end up being a crackpot theory?
Chris  Whoever loves correction loves knowledge, but he who hates reproof is stupid.

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Offline 93143

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Re: Woodward's effect
« Reply #221 on: 02/19/2013 06:26 AM »
Ek = 1/2 m v^2

F = m a

Those two equations are all you need to show that Woodward's device would produce free energy.

We've been over this (and over it, and over it, for years).  M-E thrusters (if they work) do not violate conservation of momentum or energy.  They exchange momentum and energy with the rest of the observable universe.

The claim that an M-E thruster violates the entropy condition is at least plausible, though IMO it is (for a couple of reasons) on pretty thin ice as a reason to discount the whole concept.  But the claim that the concept necessarily violates conservation of energy based simply on Newtonian mechanics cannot be seriously maintained; anyone capable of drawing a free-body diagram shouldn't even need to do so to understand this.

Let's take a thruster at a velocity v1, and a large quantity of mostly distant matter (the Far-Off Active Mass, or FOAM, the gravitational potential of which is what gives the thruster and its payload their inertia) at an average velocity v2.  The thruster produces a thrust F, which results in the FOAM experiencing some distributed force pattern that integrates to -F.  The thruster accelerates at an acceleration a1 = F/m1, and everything else accelerates at a mean acceleration of a2 = -F/m2 (note that |m2| >> |m1|, and accordingly |a2| << |a1|).  Momentum is conserved.

The rate of gain of kinetic energy of the thruster and its payload is of course P1 = Fv1.  For the FOAM, we get P2 = -Fv2.  This means that the input power to the thruster, assuming no extra energy from non-obvious cosmological effects, needs to be at least Pin = P1+P2, or Pin = F(v1-v2).

In the controversial case where the operating principle of the thruster manages to maintain an effective reaction velocity v2 = v1 irrespective of the value of v1, by somehow weighting its interaction with the rest of the observable universe, or perhaps by calling in some funky cosmological weirdness to balance the energy books, the thrust efficiency ηF = |F|/Pin [N/W] is independent of v1 and in principle unlimited.  This case gives you the flywheel-type pseudo-free-energy machine, possibly violating or at least circumventing the entropy condition in the process.  In the opposite case, where v2 is essentially constant and no cosmological weirdness occurs, the maximum value of ηF is the inverse of the velocity difference between the thruster and the rest of the observable universe in the axis of thrust.  This case gives you the linear-brake-type pseudo-free-energy machine (which works on the same principle as a windmill), and the entropy condition is classically respected.  (I don't think this second case is tenable given the form of Woodward's equations, but I haven't pored over all his papers in detail and I'm not very far into his book yet, so...)  Neither of these cases results in true "free energy" in the sense of a global conservation violation, nor does any member of the family of cases that can be imagined along these lines.

BTW, the only way to get a frame-invariant kinetic energy - in other words a real energy - is to sum the values of mivivi in the centre-of-mass frame of reference Σmivi / Σmi = 0.  The value mvv for a single object of mass m and velocity v is not a real energy, though it can be used for bookkeeping if you know what you're doing.

Just explain to me why I should read a ton of what will almost surely end up being a crackpot theory?

Because you insist on trying to criticize it.  What you're doing is called "contempt prior to investigation".  As Jim would put it, know something before posting.
« Last Edit: 02/19/2013 08:41 AM by 93143 »

Offline Robotbeat

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Re: Woodward's effect
« Reply #222 on: 02/19/2013 06:38 AM »
Give me a single paper to critique and I will.
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 JohnFornaro

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Re: Woodward's effect
« Reply #223 on: 02/19/2013 01:30 PM »
Start with the "easy" paper:
Sometimes I just flat out don't get it.

Offline antiquark

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Re: Woodward's effect
« Reply #224 on: 02/19/2013 02:29 PM »
M-E thrusters (if they work) do not violate conservation of momentum or energy.  They exchange momentum and energy with the rest of the observable universe.

I'm at work now, so I can't really get in depth, so I'll just ask one question here:  when the rest of the universe acts as a reaction mass, is that simultaneous, or propagating at the speed of light?

If simultaneous, then it violates the speed of light. If it propagates, then the Woodward drive would have problems operating in the depths of space (no mass nearby).

Offline mrmandias

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Re: Woodward's effect
« Reply #225 on: 02/19/2013 05:17 PM »
No one cares if you read it or not.  What is peculiar is repeatedly opining, over and over again, on something you don't know much about and have announced is not worth your time to learn.

I understand maxwell just fine. Just explain to me why I should read a ton of what will almost surely end up being a crackpot theory?

Offline KelvinZero

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Re: Woodward's effect
« Reply #226 on: 02/19/2013 07:32 PM »
I realized there is something basic I dont understand:

http://en.wikipedia.org/wiki/Woodward_effect :
Thus, if the mass of a given object can be varied while being oscillated either in a linear or orbital path, such that the mass is high while the mass is moving in one direction and low while moving back, then the net effect should be acceleration in one direction as the inertial drag of the universe upon the object varies as its mass varies. Woodward claims the mass variation has been accomplished by demonstrating that the initial mass of a capacitor will increase with the energy stored in its electrical charge (m=E/c2).

If you can vary a mass like this, wouldnt you expect it to work without any new physics? I mean, you push a small mass to the left, you push a large mass to the right, repeat. That is like pushing (large minus small) mass to the right, right?
The devil is in the detail of how you are increasing the energy/mass of the capacitor without undoing the push you just gave yourself. For example you can't just let the mass flow as electricity in the reverse direction since that would give you a reverse push. What am I (and perhaps the wiki article) missing?

One possible difference is that in standard physics it would be the mass as you accelerated it at each end that mattered. That could have just been bad wording though.

Offline Robotbeat

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Re: Woodward's effect
« Reply #227 on: 02/19/2013 07:45 PM »
Of course it's new physics. If this effect is replicated convincingly, Woodward will get a Nobel Prize.
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 93143

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Re: Woodward's effect
« Reply #228 on: 02/19/2013 09:50 PM »
I'm at work now, so I can't really get in depth, so I'll just ask one question here:  when the rest of the universe acts as a reaction mass, is that simultaneous, or propagating at the speed of light?

If simultaneous, then it violates the speed of light. If it propagates, then the Woodward drive would have problems operating in the depths of space (no mass nearby).

It supposedly uses the underlying principle of inertia.  We observe that inertial reaction forces are instantaneous and independent of the proximity of large gravitating masses.  Mach's principle states that inertia is due to the action (in Sciama and Woodward's theory, the gravitational potential) of the rest of the matter in the observable universe.  Therefore the drive's operation is likely to be unaffected by the lightspeed limit in the way you're suggesting, though the use of the word "observable" would seem to imply that the limit isn't right out the window.

It has been proposed that the gravinertial radiation that carries the momentum and energy to and from the distant matter incorporates a reversed-time component, similar to Wheeler-Feynman absorber theory and Cramer's transactional interpretation of quantum mechanics.  This allows for instantaneous inertial reaction forces while technically maintaining the lightspeed limit.

Hence the slogan on the door to Woodward's lab: "Tomorrow's momentum today!"...
« Last Edit: 02/20/2013 12:17 AM by 93143 »

Offline D_Dom

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Re: Woodward's effect
« Reply #229 on: 02/20/2013 04:34 AM »
... I have long held intuition that the speed of light is not necessarily a limiting factor to causality.  I don't mention this in polite society.

I couldn't agree more, fortunately we are among friends here!
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 #230 on: 02/20/2013 02:26 PM »
It [the Woodward "flux capacitor"] supposedly uses the underlying principle of inertia.  We observe that inertial reaction forces are instantaneous and independent of the proximity of large gravitating masses.  Mach's principle states that inertia is due to the action (in Sciama and Woodward's theory, the gravitational potential) of the rest of the matter in the observable universe.  Therefore the drive's operation is likely to be unaffected by the lightspeed limit ...

Hence the slogan on the door to Woodward's lab: "Tomorrow's momentum today!"...

If you kick a basketball sized rock on Earth, you'll break your toe.  Same if you kick the same rock on ISS or on the Moon.

Inertia is the same regardless of where you happen to be standing.  Inertia is instantaneous.  Where does it come from?  Why is it the same value regardless of gravitational field?  Is it the same value when the mass oscillates in a certain fashion?
Sometimes I just flat out don't get it.

Offline djolds1

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Re: Woodward's effect
« Reply #231 on: 02/20/2013 03:22 PM »
I don't have an official opinion on this "flux capacitor", but I have a long held intuition that the speed of light is not necessarily a limiting factor to causality.  I don't mention this in polite society.
Pharis Williams' thermodynamic interpretation of relativity might appeal to you. He analogizes the speed of light to absolute zero temperature as limiting quantities.

Either we live in one universe which had one cause, or else we live in a number of different universes which sprang from one cause, but which are now not attached by causality, because of the speed of light.  Somehow, no matter how far "out" you go, you can never get to the expanding boundary of the universe that we are in, therefore those other universes which sprang from this one cannot be proven to exist, since no signal can be received from them.

Somehow, after the Big Bang, during the "expansionary" period, these parts of the universe sped away faster than the speed of light?  They were causally connected with us, but now they're not?  Even tho, at the edge of our universe, they would seem to be causally connected per the relative speeds of galaxies at the "edge".  You'd think that the residents of UDFj-39546284 would be able to see those galaxies, no?

This makes no intuitive sense to my tiny brain.

Sheesh.  I picked a fine time to stop sniffing glue...
  ;D

« Last Edit: 02/20/2013 03:40 PM by djolds1 »

Offline JohnFornaro

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Re: Woodward's effect
« Reply #232 on: 02/20/2013 04:38 PM »
Pharis Williams' thermodynamic interpretation...

No mother names her child Pharis:

I'm not going to review a paper written by someone named Pharis.  Something about "contempt prior to investigation".
Sometimes I just flat out don't get it.

Offline djolds1

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Re: Woodward's effect
« Reply #233 on: 02/21/2013 04:12 AM »
Pharis Williams' thermodynamic interpretation...
No mother names her child Pharis:
He was born ~1941.
« Last Edit: 02/21/2013 04:13 AM by djolds1 »

Offline JohnFornaro

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Re: Woodward's effect
« Reply #234 on: 02/21/2013 12:46 PM »
Pharis Williams' thermodynamic interpretation...
No mother names her child Pharis:
He was born ~1941.

Huh?
Sometimes I just flat out don't get it.

Offline djolds1

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Re: Woodward's effect
« Reply #235 on: 02/21/2013 04:21 PM »
« Last Edit: 02/21/2013 04:34 PM by djolds1 »

Offline JohnFornaro

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Re: Woodward's effect
« Reply #236 on: 02/21/2013 04:41 PM »
No mother names her son Pharis? Different standards c.1940, I would expect.

I see.  Note that I do not use smilies.  It is more effective that way, as demonstrated.
Sometimes I just flat out don't get it.

Offline djolds1

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Re: Woodward's effect
« Reply #237 on: 02/21/2013 09:38 PM »
No mother names her son Pharis? Different standards c.1940, I would expect.
I see.  Note that I do not use smilies.  It is more effective that way, as demonstrated.
It's difficult to convey tone via text w/o emoticons.

Offline JohnFornaro

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Re: Woodward's effect
« Reply #238 on: 02/21/2013 10:33 PM »
That's what happened to the Egyptians.  It was all emoticons, and then it was all over...
Sometimes I just flat out don't get it.

Online cuddihy

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Re: Woodward's effect
« Reply #239 on: 03/08/2013 12:29 AM »
M-E thrusters (if they work) do not violate conservation of momentum or energy.  They exchange momentum and energy with the rest of the observable universe.

I'm at work now, so I can't really get in depth, so I'll just ask one question here:  when the rest of the universe acts as a reaction mass, is that simultaneous, or propagating at the speed of light?

If simultaneous, then it violates the speed of light. If it propagates, then the Woodward drive would have problems operating in the depths of space (no mass nearby).

speed of light, but propogates both forward in time and back. No, your second point is incorrect, local mass has a negligible effect on the inertial reaction force because no matter how giant and close your local mass it is isignificant next to the mass and distribution of the rest of the universe. This is actually explained in part 2 of the book.