Author Topic: Woodward's effect  (Read 288056 times)

Offline wallofwolfstreet

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Re: Woodward's effect
« Reply #720 on: 03/31/2016 08:22 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?

Sorry birchoff, that section is structured terribly on my part.  The key thing to take away there is the triple asterisk which points to my footnote, which demonstrates that mass flow across the system boundary in the form of gravienterial waves would yield thrust no greater than a perfectly collimated photon rocket (under some assumptions of course), which follows the relation between energy flux and mass flux.  I should have made that footnote the main thrust of the post and relegated the dm/dt=0 case to a footnote, or better yet left it out altogether. 

Here is question that I think would really help me understand:

Place a control volume around an MET so that only the MET is contained (the rest of the mass of the universe is on the other side of the control volume).  Let the MET experience a transient mass fluctuation so that it's mass goes from 1 kg to 1.0001 kg.  The delta(m) gained by the MET is thus 0.0001 kg.  Does this delta(m)=0.0001 kg come from the rest of the universe outside the control volume, so that somehow 0.0001 kg of mass has passed over the control surface, or does this delta(m) come from somewhere else?  If somewhere else, where?

Offline M.E.T.

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Re: Woodward's effect
« Reply #721 on: 03/31/2016 08:55 PM »
My understanding is that the interaction is with all the mass in the rest of the universe. So whatever the Mach Effect Spacecraft gains, is lost to a minute extent by every other bit of matter in the universe.

Offline wallofwolfstreet

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Re: Woodward's effect
« Reply #722 on: 03/31/2016 09:44 PM »
(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.



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.


But this is exactly my point Mezzenile, Woodward's expressions here are missing terms, so the derivation of d2G(t)/dt2!=0 isn't accurate.  For systems that do not have constant mass, you can't just sum up the Forces.  You have to take into account dm/dt terms. 

Remember that all three of Newton's "Laws" are only different expressions of conservation of momentum.  Starting from conservation of momentum, you can derive all three laws.  You can apply apply these laws to certain systems that meet certain criteria, and the necessity of a constant system mass is one of these criteria (as I hopefully demonstrated with my rocket example), 

Woodward's "Newton's Law of Action and Reaction" equation has essentially dropped a term.  Maybe there is a theoretical justification for this which I am unaware of. 
« Last Edit: 03/31/2016 09:47 PM by wallofwolfstreet »

Offline GeneralRulofDumb

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Re: Woodward's effect
« Reply #723 on: 04/01/2016 12:41 PM »
Place a control volume around an MET so that only the MET is contained (the rest of the mass of the universe is on the other side of the control volume).  Let the MET experience a transient mass fluctuation so that it's mass goes from 1 kg to 1.0001 kg.  The delta(m) gained by the MET is thus 0.0001 kg.  Does this delta(m)=0.0001 kg come from the rest of the universe outside the control volume, so that somehow 0.0001 kg of mass has passed over the control surface, or does this delta(m) come from somewhere else?  If somewhere else, where?

Matter and energy are interchangeable, for lack of a better word.
Increasing a mass' or body's internal energy state will change its mass.
The mass fluctuation in a MET is a direct consequence of 'pumping' (in this case: electrical) energy into it.
There is no mass being transferred past boundaries during a MET's operation.
« Last Edit: 04/01/2016 12:45 PM by GeneralRulofDumb »

Offline Mezzenile

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Re: Woodward's effect
« Reply #724 on: 04/01/2016 03:25 PM »
The change of mass value introduced by Woodward is considered to be the same in all isotropic directions. An image could be the one of a regular evaporation or condensation of mass everywhere on the body.
Whatever can be the amount of mass and its speed during evaporation or condensation, the geometrical integration of these isotropic speeds all over the body gives a null result (this is easy to see on a spherical body). So the total amount of exchanged momentum is also null.

Woodward has exposed a similar reasonning in his answer to Oak-ridge paper taking the example (if I remember well) of a bottle of water drilled by two identical holes diametrically opposite and installed on a moving cart.
If water escapes through gravity from one hole the bottle, this  has an impact on the movement of the cart. If water escapes through the two holes there is no impact..

Said differently a rocket motor with an omni-directionnal thruster would not move one centimeter whatever is its reserve of propergol ! 

Offline wallofwolfstreet

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Re: Woodward's effect
« Reply #725 on: 04/01/2016 05:53 PM »
The change of mass value introduced by Woodward is considered to be the same in all isotropic directions. An image could be the one of a regular evaporation or condensation of mass everywhere on the body.
Whatever can be the amount of mass and its speed during evaporation or condensation, the geometrical integration of these isotropic speeds all over the body gives a null result (this is easy to see on a spherical body). So the total amount of exchanged momentum is also null.

Do you mean that, given a spherical coordinate system at rest with respect to the MET, but instantaneously oriented so that the MET is located at the origin:

dm/dt=f(ρ,ϕ,θ)=f(ρ)
f(ρ0)=Const for every ρo?

Because if that is what you mean, then that isn’t a resolution to the problem.  A moving body which has mass isotopically (as defined above) evaporate or condense upon it does NOT stay at the same speed.  It slows down when mass condenses and it speeds up when the mass evaporates.

Think about it like this:
I have a moving body of mass m travelling at velocity v.  It has momentum mv.  Now isotopically condense some mass on this body, δm.  The condition of isotropic condensation means no momentum has entered the system.  However, we now have that the new momentum of the system post condensation is (m+δm)v, which is clearly impossible.  The issue is, when the mass condenses on the body, it gains momentum equal to δmv (remember that the isotopic condition gives δm an initial velocity of zero).  What isn’t being accounted for is that the body loses speed dues to the application of the “ force” (dδm/dt)v.  Remember that force is just the rate of change in momentum, so the “force” on the system is equal to the mass flow into the system times the change in velocity the mass inflow experiences.  This emerges from conservation of momentum.     

Quote
Woodward has exposed a similar reasonning in his answer to Oak-ridge paper taking the example (if I remember well) of a bottle of water drilled by two identical holes diametrically opposite and installed on a moving cart.
If water escapes through gravity from one hole the bottle, this  has an impact on the movement of the cart. If water escapes through the two holes there is no impact..

Quote
If water escapes through gravity from one hole the bottle, this  has an impact on the movement of the cart
This is actually incorrect.  If the water is escaping solely through gravity (i.e. force applied is vertical) then the moving cart continues to move at the same velocity, unaffected.  This is because each particle of water carries away momentum exactly proportional to the mass it has also carried away relative to the cart.  The change in velocity of the mass flow across the system boundary is zero. 

Quote
 
Said differently a rocket motor with an omni-directionnal thruster would not move one centimeter whatever is its reserve of propergol !

This does not relate to your supposition above, because an omni-directional thruster with non-zero velocity does NOT emit isotropic mass flow.  Isotropic mass flow is only true for an observer at rest with respect to the thruster.   Given a moving thruster, the mass flow in the direction of the thruster's velocity has velocity equal to (velocity of rocket + exhaust velocity) and the mass flow in the direction opposite the velocity of the thruster has velocity equal to (velocity of rocket – exhaust velocity).  A rocket moving with non-zero velocity which is expelling mass flow across the system boundary isotopically WOULD in fact experience a net impulse.  Once again, times the mass flow rate across the system boundary by the change in velocity the mass flow rate experiences as it cross the boundary.  For a mass flow rate to become isotropic after being expelled from a moving body, their must be a change in velocity. 

It is this impulse which Woodward, if you say he is claiming isotropic mass flow, has effectively dropped in the calculation you highlighted.  Doing so does not appear to be valid in the framework he has developed for transient mass fluctuations.

I think what you might be trying to say is that, somehow, the mass inflow and outflow with respect to the MET is always such that v=0 in the dm/dt term not matter the inertial frame (so dm/dt is isotropic when the MET is a rest, but anisotropic when the MET is not a rest).  The problem with this is, why should anyone believe that?  I have seen nothing in Woodward's derivations that validate such a belief.  Maybe I just haven't found it yet, as I'm still learning. 
« Last Edit: 04/01/2016 06:04 PM by wallofwolfstreet »

Offline 93143

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Re: Woodward's effect
« Reply #726 on: 04/01/2016 09:51 PM »
The mass fluctuation in a MET is a direct consequence of 'pumping' (in this case: electrical) energy into it.

Wrong.  The mass fluctuations predicted by Woodward's theory are not E/c˛ mass; they are an inherently transient gravinertial effect that has not been previously observed, presumably due to the difficulty of exciting it.  The effect is supposed to be related to the second derivative of the energy, not to the energy itself, and supposedly only shows up during accelerations (I'm not clear on this part yet).

Think about it like this:
I have a moving body of mass m travelling at velocity v.  It has momentum mv.  Now isotopically condense some mass on this body, δm.  The condition of isotropic condensation means no momentum has entered the system.  However, we now have that the new momentum of the system post condensation is (m+δm)v, which is clearly impossible.

Why have you assumed "isotropic condensation" in the reference frame of a random observer rather than that of the thruster?  Have you never heard of Galilean relativity?

The equations Woodward presents are independent of the thruster's bulk velocity in any external reference frame.

Quote
If water escapes through gravity from one hole the bottle, this  has an impact on the movement of the cart
This is actually incorrect.  If the water is escaping solely through gravity (i.e. force applied is vertical) then the moving cart continues to move at the same velocity, unaffected.  This is because each particle of water carries away momentum exactly proportional to the mass it has also carried away relative to the cart.  The change in velocity of the mass flow across the system boundary is zero.

See - you do understand this principle.

What you don't understand is fluid mechanics.  A liquid reservoir with a small hole in the side will expel its contents sideways, not down, because while the liquid next to the hole is supported from below by more liquid, it is not supported from the side - the weight of the liquid above it means it's at a higher pressure than the gas next to it.  It is, in effect, a pressure-fed rocket motor.

Caveat - I haven't read Woodward's rebuttal lately, and a quick search hasn't uncovered it; I'm just going on the description given.
« Last Edit: 04/01/2016 10:12 PM by 93143 »

Offline wallofwolfstreet

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Re: Woodward's effect
« Reply #727 on: 04/01/2016 10:52 PM »
Why have you assumed "isotropic condensation" in the reference frame of a random observer rather than that of the thruster?  Have you never heard of Galilean relativity?

The equations Woodward presents are independent of the thruster's bulk velocity in any external reference frame.

I have heard of Galilean relativity.  Please explain to me what you think Galilean relativity is and what you think it means here. 

Consider this:  Let's say the condensation is isotropic in the rest frame of the thruster.  The thruster accelerates to some new velocity.  Is the condensation still isotropic in the rest frame of the thruster at this new velocity?  If so, how?  How has the condensation changed so that it is isotropic in both frames, despite the fact that one frame is boosted with respect to the other?

That's where Galilean relativity is working against you.  The frame of the thruster doesn't accelerate along with the thruster.  If the thruster is stationary with respect to the frame at some instant, then the thruster will be moving with respect to it at some later instance given non-zero acceleration.  How is the condensation staying isotropic, without accelerating the bulk velocity of the condensation along with the thruster?  What magical mechanism enables such an acceleration? 

Quote
The equations Woodward presents are independent of the thruster's bulk velocity in any external reference frame.

Again, that only works and makes any physical sense, if by some magic the bulk, averaged velocity of the mass inflow and outflow to the thruster is always the same as the velocity of the thruster in every frame of reference (including the inertial frames in which the thruster is seen as accelerating, and hence gains non-zero velocity with respect to an inertial observer who was initially at rest with the thruster), which is nonphysical, exactly because it doesn't respect Galilean (or Lorentz) invariance. 

Quote
See - you do understand this principle.

What you don't understand is fluid mechanics.  A liquid reservoir with a small hole in the side will expel its contents sideways, not down, because while the liquid next to the hole is supported from below by more liquid, it is not supported from the side - the weight of the liquid above it means it's at a higher pressure than the gas next to it.  It is, in effect, a pressure-fed rocket motor.

Caveat - I haven't read Woodward's rebuttal lately, and a quick search hasn't uncovered it; I'm just going on the description given.

Who said anything about a hole in the side?  I know I didn't.  Why not make the hole on the bottom? 
And yes, I can apply Bernoulli's equation along a streamline too; I understand fluid mechanics plenty, thanks.  You're point has nothing to do with the principle of the problem.  I have no idea why you felt the need to point it out, as I even clarified:

Quote
This is actually incorrect.  If the water is escaping solely through gravity (i.e. force applied is vertical) then the moving cart continues to move at the same velocity, unaffected.
« Last Edit: 04/01/2016 10:58 PM by wallofwolfstreet »

Offline 93143

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Re: Woodward's effect
« Reply #728 on: 04/03/2016 01:12 AM »
You seem to have some unstated concept or mental picture of how this "condensation" works, that may be at variance with the actual theory.  Why would you think that an effect related to inertia (which behaves exactly the same regardless of the velocity of the object exhibiting it) should show a dependence on an outside reference frame?  Wouldn't that seem to constitute a violation of Lorentz covariance?

Remember, the mass fluctuation isn't a delayed effect; it's instantaneous, based on the acceleration state and second-order rate of change of internal energy.  The transactional nature of the interaction with distant matter means the response to the local perturbation is locally instantaneous.  There's nothing to 'leave behind'.

And since Woodward's equations show no dependence on a universal reference frame, it can be safely assumed that if the theory is correct, the math describing the interaction with distant matter must work out in such a way as to eliminate any dependence on the average velocity of said distant matter.  There's a precedent - Sciama's Machian model of inertia does exactly that.

...

Also, you may not have read the post you were responding to carefully enough re: water leaking from a hole, since the idea of two holes drilled "diametrically opposite" and both leaking water would seem to rule out either hole being on the bottom.
« Last Edit: 04/03/2016 01:24 AM by 93143 »

Offline Mezzenile

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Re: Woodward's effect
« Reply #729 on: 04/03/2016 06:16 AM »
Do you mean that, given a spherical coordinate system at rest with respect to the MET, but instantaneously oriented so that the MET is located at the origin:

dm/dt=f(ρ,ϕ,θ)=f(ρ)
f(ρ0)=Const for every ρo?

Because if that is what you mean, then that isn’t a resolution to the problem.  A moving body which has mass isotopically (as defined above) evaporate or condense upon it does NOT stay at the same speed.  It slows down when mass condenses and it speeds up when the mass evaporates.

Think about it like this:
I have a moving body of mass m travelling at velocity v.  It has momentum mv.  Now isotopically condense some mass on this body, δm.  The condition of isotropic condensation means no momentum has entered the system.  However, we now have that the new momentum of the system post condensation is (m+δm)v, which is clearly impossible.  The issue is, when the mass condenses on the body, it gains momentum equal to δmv (remember that the isotopic condition gives δm an initial velocity of zero).  What isn’t being accounted for is that the body loses speed dues to the application of the “ force” (dδm/dt)v.  Remember that force is just the rate of change in momentum, so the “force” on the system is equal to the mass flow into the system times the change in velocity the mass inflow experiences.  This emerges from conservation of momentum.     
...
This is actually incorrect.  If the water is escaping solely through gravity (i.e. force applied is vertical) then the moving cart continues to move at the same velocity, unaffected.  This is because each particle of water carries away momentum exactly proportional to the mass it has also carried away relative to the cart.  The change in velocity of the mass flow across the system boundary is zero. 
...
This does not relate to your supposition above, because an omni-directional thruster with non-zero velocity does NOT emit isotropic mass flow.  Isotropic mass flow is only true for an observer at rest with respect to the thruster.   Given a moving thruster, the mass flow in the direction of the thruster's velocity has velocity equal to (velocity of rocket + exhaust velocity) and the mass flow in the direction opposite the velocity of the thruster has velocity equal to (velocity of rocket – exhaust velocity).  A rocket moving with non-zero velocity which is expelling mass flow across the system boundary isotopically WOULD in fact experience a net impulse.  Once again, times the mass flow rate across the system boundary by the change in velocity the mass flow rate experiences as it cross the boundary.  For a mass flow rate to become isotropic after being expelled from a moving body, their must be a change in velocity. 

It is this impulse which Woodward, if you say he is claiming isotropic mass flow, has effectively dropped in the calculation you highlighted.  Doing so does not appear to be valid in the framework he has developed for transient mass fluctuations.

I think what you might be trying to say is that, somehow, the mass inflow and outflow with respect to the MET is always such that v=0 in the dm/dt term not matter the inertial frame (so dm/dt is isotropic when the MET is a rest, but anisotropic when the MET is not a rest).  The problem with this is, why should anyone believe that?  I have seen nothing in Woodward's derivations that validate such a belief.  Maybe I just haven't found it yet, as I'm still learning.
First the image of evaporation or condensation of matter on a body, to try to understand in a classical way the mass change introduced by Woodward, is only an image and I think that the analogy should not be pushed too far. The right approach of this mass variation phenomena is to be found in the mathematical study of the properties of the relativistic “Gravito-Magnetic” field as performed by Woodward. Moreover it seems that the special role of acceleration in the apparition of this change of mass can only be seeen througt the Gravito-Magnetic field modelisation of reality. We should never forget that a correct thinking can only be performed using the right concepts as intellectual tools.

Now let us go back to the simplified model of evaporation/condensation of matter on a body. Let us analyse your statement :

Quote
A moving body which has mass isotopically (as defined above) evaporate or condense upon it does NOT stay at the same speed. It slows down when mass condenses and it speeds up when the mass evaporates.

Let us suppose that the body is at rest in a Galilean frame. I imagine that you have no difficulty to consider that, as observed from this rest frame, the body will not suddenly accelerate in an arbitrary direction if some isotropic condensation or evaporation phenomena occurs. Now you seems to say that an other observer in an other galilean frame with a relative speed V will see a momentum change and so that evaporation/condensation is the source of a real force applied to the body. Moreover this force being real it should be present, as invariant, in any other galilean frame with the same magnitude (not same direction naturally).

In fact this is an apparent paradox which comes from the consideration of wrong forces by the false relation : F(t) = dP(t)/dt = M(t) dV(t)/dt + V(t) dM(t)/dt which should be replaced by F(t) = dP(t)/dt = M(t) dV(t)/dt - U(t) dM(t)/dt  where U(t) is the velocity of the ejected/accreted mass as seen in the body's galilean rest frame (*). For an isotropic condensation or evaporation the average value of U(t) over all the directions is zero and dM(t)/dt is constant, so the average value of the U(t) dM(t)/dt term for the whole body is zero. Woodward is right, Oak-Ridge was wrong  ;)

We should not push too far this simplified model of condensation/evaporation in a newtonian mechanic context ! It has the advantage to give some insight to the situation but only insight...

(*) The fact that the classical relation F(t) = dP(t)/dt = M(t) dV(t)/dt + V(t) dM(t)/dt cannot be applied comes from the fact that it supposes that the mechanical system is closed to be valid. This is not the case for a body on which matter either condense or evaporate.  The correct relation takes into account the fact that the system is initialy composed of two masses M and dM each having different velocities before becoming one body of mass M + dM.
« Last Edit: 04/04/2016 01:27 PM by Mezzenile »

Offline wallofwolfstreet

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Re: Woodward's effect
« Reply #730 on: 04/12/2016 01:32 AM »
You seem to have some unstated concept or mental picture of how this "condensation" works, that may be at variance with the actual theory.

Sorry for the late response.

Well the condensation concept is something I’m working with from Mezzenile’s post, it’s not something I came up with.  It certainly could be at variance with the theory.  I’m just trying to work within the framework of “condensation” to demonstrate that there are terms missing in the force derivations I have seen so far from Woodward, and the missing term isn't peculiar to the "condensation" framework.

Quote
Why would you think that an effect related to inertia (which behaves exactly the same regardless of the velocity of the object exhibiting it) should show a dependence on an outside reference frame?  Wouldn't that seem to constitute a violation of Lorentz covariance?

Because inertia and momentum go hand in hand in all the frames of reference where the MET is not at rest.  Inertia is invariant, but momentum is not.  You can’t just add mass to the MET system and not have an impact on momentum.  THAT would constitute a violation of Lorentz covariance.

Quote
Remember, the mass fluctuation isn't a delayed effect; it's instantaneous, based on the acceleration state and second-order rate of change of internal energy.  The transactional nature of the interaction with distant matter means the response to the local perturbation is locally instantaneous.  There's nothing to 'leave behind'.

I'm not sure what you mean by this, you'll have to help me out. 

Quote
And since Woodward's equations show no dependence on a universal reference frame, it can be safely assumed that if the theory is correct, the math describing the interaction with distant matter must work out in such a way as to eliminate any dependence on the average velocity of said distant matter.  There's a precedent - Sciama's Machian model of inertia does exactly that.

I think in the preceding discussion, and I’m at fault here, it’s not really clear exactly what my argument is because I switch between inertial reference frames at rest with respect to the MET and boosted frames, but I also switch between different velocity states of the MET in a single frame, and it comes off muddled.

Here is the entire thrust of my argument:

The correct expression for Force on a system with non-constant mass is:
Fnet=dm/dtv+ma

Where as Mezzenile correctly points out, the v term is really the difference in velocity between the center of mass of the system and ejected mass crossing the system boundary.

So let’s say this v term is zero in an inertial reference frame in which the MET is a rest (this is equivalent to saying mass condensation is isotropic in the condensation model)  Now, let’s accelerate the MET under its own power in this frame of reference so it is no longer at rest.

How is it possible that the v term is still zero despite the fact that the MET has accelerated?  How is it that the mass always shows up with just the right momentum to always cancel the vdm/dt?  Ignoring the v term is the mathematical equivalent of saying that the mass fluctuations, and the momentum that must attend them BY Lorentz invariance (not in spite of it), is somehow just right for the vdm/dt term to be dropped.  Why is it that as the MET is accelerated, the momentum contained in the transient mass fluctuations is somehow accelerating along with it?

Quote
Also, you may not have read the post you were responding to carefully enough re: water leaking from a hole, since the idea of two holes drilled "diametrically opposite" and both leaking water would seem to rule out either hole being on the bottom.

Fair enough

Offline wallofwolfstreet

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Re: Woodward's effect
« Reply #731 on: 04/12/2016 01:47 AM »
Sorry for the lat reply.

First the image of evaporation or condensation of matter on a body, to try to understand in a classical way the mass change introduced by Woodward, is only an image and I think that the analogy should not be pushed too far. The right approach of this mass variation phenomena is to be found in the mathematical study of the properties of the relativistic “Gravito-Magnetic” field as performed by Woodward. Moreover it seems that the special role of acceleration in the apparition of this change of mass can only be seeen througt the Gravito-Magnetic field modelisation of reality. We should never forget that a correct thinking can only be performed using the right concepts as intellectual tools.

Is this regular Gravitomagnetism (GM)?  The GM I am (limitedly) familiar with is just a vanilla linearization of classical GR in the limit of weak field.  There is nothing within GM (that I know of) that has anything to do with a inertial fluctuations, any more than classical EM has to do with charge fluctuations.

Quote
 
Now let us go back to the simplified model of evaporation/condensation of matter on a body. Let us analyse your statement :


Quote
A moving body which has mass isotopically (as defined above) evaporate or condense upon it does NOT stay at the same speed. It slows down when mass condenses and it speeds up when the mass evaporates.

Let us suppose that the body is at rest in a Galilean frame. I imagine that you have no difficulty to consider that, as observed from this rest frame, the body will not suddenly accelerate in an arbitrary direction if some isotropic condensation or evaporation phenomena occurs. Now you seems to say that an other observer in an other galilean frame with a relative speed V will see a momentum change and so that evaporation/condensation is the source of a real force applied to the body.

No, I didn’t say that (and if it seems like I did, then sorry for the miscommunication).  What you’re missing is that the condensation is not isotropic in both frames.  Isotropic condensation in an frame of reference (FOR) where an observer is at rest is NOT isotropic as viewed by a moving observer.  Isotropic/anisotropic is NOT an invariant.  You can’t boost the frame and still talk about isotropic condensation in both frames.   

We have an inertial frame.  A body that is at rest with respect to frame has isotropic condensation.  As you say, it doesn't move.  If there is a body in this same frame which is not a rest, but it experiences isotropic condensation with respect to the frame of reference, it will slow down.  To continue along at the same speed, the condensation would have to be anisotropic as viewed by a stationary observer in the frame of reference.     

Quote
Moreover this force being real it should be present, as invariant, in any other galilean frame with the same magnitude (not same direction naturally).

In fact this is an apparent paradox which comes from the consideration of wrong forces by the false relation : F(t) = dP(t)/dt = M(t) dV(t)/dt + V(t) dM(t)/dt which should be replaced by F(t) = dP(t)/dt = M(t) dV(t)/dt - U(t) dM(t)/dt  where U(t) is the velocity of the ejected/accreted mass as seen in the body's galilean rest frame (*). For an isotropic condensation or evaporation the average value of U(t) over all the directions is zero and dM(t)/dt is constant, so the average value of the U(t) dM(t)/dt term for the whole body is zero. Woodward is right, Oak-Ridge was wrong  ;)

Okay, now we’re getting somewhere.  You are exactly right, your expression is even clearer.  Here is the problem though, and to save time I’m just going to post my response from above:

Here is the entire thrust of my argument:
Quote
The correct expression for Force on a system with non-constant mass is:
Fnet=dm/dtv+ma

Where as Mezzenile correctly points out, the v term is really the difference in velocity between the center of mass of the system and ejected mass crossing the system boundary.

So let’s say this v term is zero in an inertial reference frame in which the MET is a rest (this is equivalent to saying mass condensation is isotropic in the condensation model)  Now, let’s accelerate the MET under its own power in this frame of reference so it is no longer at rest.

How is it possible that the v term is still zero despite the fact that the MET has accelerated?  How is it that the mass always shows up with just the right momentum to always cancel the vdm/dt?  Ignoring the v term is the mathematical equivalent of saying that the mass fluctuations, and the momentum that must attend them BY Lorentz invariance (not in spite of it), is somehow just right for the vdm/dt term to be dropped.  Why is it that as the MET is accelerated, the momentum contained in the transient mass fluctuations is somehow accelerating along with it?

Coming back to this isotropic condescension idea; you are right, for isotropic condensation U(t) is zero when integrated over the whole control surface.  But when we accelerate the MET, the condensation does not stay isotropic.  For your explanation to work, something must be accelerating this mass along with the MET such that the condensation is always isotropic despite the acceleration of the MET.  My question is: what property allows the mass to accelerate along with the MET so that it is always just travelling fast enough to be ignored (ie. so that it can always condense isotropically)?  This something that I have not seen Woodward address in the things I have read so far.

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We should not push too far this simplified model of condensation/evaporation in a newtonian mechanic context ! It has the advantage to give some insight to the situation but only insight...

As I mentioned above, the model doesn't matter.  Somehow, when mass fluctuate itself into existence on the MET, is it always just magically going at a velocity equal to the MET, so that we can ignore dm/dtv terms, despite the fact that MET's don't have a constant velocity.  Can you point me to anything in Woodwards work that specifically addresses this point? 

Offline Paul451

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Re: Woodward's effect
« Reply #732 on: 04/12/2016 09:03 AM »
My understanding is that the interaction is with all the mass in the rest of the universe. So whatever the Mach Effect Spacecraft gains, is lost to a minute extent by every other bit of matter in the universe.

However, relative to any arbitrary non-accelerating observer, the MET can achieve a velocity where the rate of increase in kinetic energy exceeds that of the rate of energy input into the device.

Since the mass of the universe doesn't have a particularly strong anisotrophy, there's no outside net force to transfer energy from. Ie, the device can't steal energy from another system. So the increase in energy (output vs input) isn't accounted for by your (or Woodward's) explanation.

Whether it transfers momentum to the rest of the universe (by whatever magical means) or not, it isn't transferring momentum from the rest of the universe, therefore it doesn't solve the dilemma that the energy produced would exceed the energy input.

Offline Mezzenile

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Re: Woodward's effect
« Reply #733 on: 04/12/2016 09:56 AM »
Sorry for the lat reply.
...

For your explanation to work, something must be accelerating this mass along with the MET such that the condensation is always isotropic despite the acceleration of the MET.
You right, if we absolutely want to stay within the strict paradigm of the Newtoniak mechanics we can consider small daemons at rest in the instantaneous galilean frame tangent to the accelerating body and which smoothly deliver or remove increment of mass to the moving body ! :D
Now of course, I think there is no way to incorporate further these small daemeons within the frame of Newtonian mechanics. We have to follow rather now an other path shown by Woodward (have a look to the relevant chapter of his book) : Mach Principle and its traduction by General Relativity with the existence of advanced and delayed gravitational waves emitted by any accelerated body (energy, momentum) or its linearized/simplified version using Gravito-Magnetic field (as considered by Sciama, A. Papapetrou, Hawkings, Woodward if I am right).

The interest of the Newtonian approach is that it gives access with a good approximation to the equation of the movement of the body (see my previous posts with the beginning of relevant modelisation). With it we can quantify the influence on this movement of :

  - the respective values of active and inert mass,
  - the length of the link betwen the active and inert mass,
  - the frequency of operation.

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when mass fluctuate itself into existence on the MET, is it always just magically going at a velocity equal to the MET, so that we can ignore dm/dtv terms, despite the fact that MET's don't have a constant velocity. Can you point me to anything in Woodwards work that specifically addresses this point?

Cf Part I of Making Starships and Stargates written by Woodward and edited by Springer.

The big merit of Woodward is that he presents at the same time both a theoretical frame and an experimental frame to his thesis.

« Last Edit: 10/04/2016 04:21 AM by Mezzenile »

Offline flux_capacitor

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Re: Woodward's effect
« Reply #734 on: 04/25/2016 12:58 PM »
If water escapes through gravity from one hole the bottle, this  has an impact on the movement of the cart
This is actually incorrect.  If the water is escaping solely through gravity (i.e. force applied is vertical) then the moving cart continues to move at the same velocity, unaffected.  This is because each particle of water carries away momentum exactly proportional to the mass it has also carried away relative to the cart.  The change in velocity of the mass flow across the system boundary is zero. 
Also, you may not have read the post you were responding to carefully enough re: water leaking from a hole, since the idea of two holes drilled "diametrically opposite" and both leaking water would seem to rule out either hole being on the bottom.

Folks, apparently you already understood what has caused that transient misunderstanding ;) but I will point out the fundamentals.
Woodward's experiment as an answer to the Oak Ridge paper involved the following apparatus, i.e. water escaping from two opposite holes on the side of a bottle, on a moving PASCO Dynamics demonstration cart:





Please find attached below Woodward's short (3 pages) answer. I suggest reading it before commenting on its content.

Offline birchoff

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Re: Woodward's effect
« Reply #735 on: 04/25/2016 08:37 PM »
If water escapes through gravity from one hole the bottle, this  has an impact on the movement of the cart
This is actually incorrect.  If the water is escaping solely through gravity (i.e. force applied is vertical) then the moving cart continues to move at the same velocity, unaffected.  This is because each particle of water carries away momentum exactly proportional to the mass it has also carried away relative to the cart.  The change in velocity of the mass flow across the system boundary is zero. 
Also, you may not have read the post you were responding to carefully enough re: water leaking from a hole, since the idea of two holes drilled "diametrically opposite" and both leaking water would seem to rule out either hole being on the bottom.

Folks, apparently you already understood what has caused that transient misunderstanding ;) but I will point out the fundamentals.
Woodward's experiment as an answer to the Oak Ridge paper involved the following apparatus, i.e. water escaping from two opposite holes on the side of a bottle, on a moving PASCO Dynamics demonstration cart:





Please find attached below Woodward's short (3 pages) answer. I suggest reading it before commenting on its content.

thank you very much for finding that. it is the response to a question that has been asked many times on this thread.

Offline wallofwolfstreet

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Re: Woodward's effect
« Reply #736 on: 05/09/2016 03:59 PM »

...



Please find attached below Woodward's short (3 pages) answer. I suggest reading it before commenting on its content.

Except that this paper by Woodward is incorrect, and doesn’t offer any insight into distinguishing between the correct equation of motion offered by the Oak Ridge Scientists (ORS) and Woodward’s wrong one.  All it does is offer some insight as to where Woodward went wrong in his analysis and why he fails to properly account for the necessary terms to accurately express the physics.

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But the ORS are wrong. Inclusion of the vdm/dt term in the equation of motion of M in the rest frame of B when calculating the force on B communicated through A is simply incorrect. This is not to say that including a vdm/dt term in the equation of motion for M is inappropriate in all circumstances. For example, if the mass fluctuation in M is achieved by the expulsion and recovery of some mass, that acts as a propellant, in the direction of the motion produced by A, then a force arising from vdm/dt would be communicated from M to B through A. But v in this case would be the invariant velocity of the propellant plume with respect to M, not the velocity of M relative to B.

So Woodward does understand the principle of needing to account for momentum flux across system boundaries.  Despite that, he still manages to go off the rails:

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Pedagogically, the fact that the vdm/dt term in Newton’s second law must be treated with some care is often made in simple problem situations. For example, one such problem asks: Does a flatcar rolling down a smooth level track at constant velocity accelerate if a pile of sand it carries is allowed to fall through a hole in the floor of the car onto the track? The obvious answer, of course, is no. The present case, where an external force acts on an object with a changing mass, in the context of this example, leads to the question: If an external force is applied to the flatcar as the sand falls through the hole, aside from the fact that m in ma must be treated as a function of time, does the falling sand affect the acceleration of the car produced by external force? Woodward says no. The ORS, to be consistent with their vdm/dt based cancellation claim, must say yes.

The bold section is where Woodward goes wrong.  The ORS claim is not at all consistent with a “yes” answer.  Woodward has misinterpreted or maybe even misunderstood the claim he is refuting, and so his supposed experiment to distinguish between the two claims is completely bunk; in the specific experiment he is examining, the ORS equation of motion and Woodward’s equation of motion yield identical results because vdm/dt = 0.  This is because the net velocity of the water flow relative to the cart is zero (due to the holes being diametrically opposed).  Even in the case where there is only one hole, but the hole is still drilled orthogonal to the direction of motion of the cart, the acceleration of the cart is unaffected because the force from the vdm/dt term is orthogonal to the track of the cart.  Woodward is examining a case where both equations of motion are in fact equivalent, it is only his misuse of the ORS equation that seems to give results in favor of his incorrect equation.

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Simple, compelling arguments for the absence of vdm/dt effects in the circumstances under consideration can easily be made. For instance, should one use the (inertial) instantaneous rest frame of M as that in which the force on B is computed, the vdm/dt term vanishes. Plainly, if the vdm/dt “force” vanishes in this inertial frame, it cannot act on B in any other inertial frame, for real forces that produce accelerations are invariant under the Galilean transformation group. That is, they cannot be made to vanish by simple choice of inertial frame of reference. Several years ago Woodward advanced such arguments to the ORS (and others). Those arguments, obviously, at least at ORNL, fell on deaf ears.

It falls on deaf ears because it makes no sense.  Why in the world should the vdm/dt term vanish in the rest frame of M?  The whole point of MET’s is that they are claimed to accelerate, so the instantaneous rest frame at time t1 should be different than the instantaneous rest frame at time t2.  The v term associated with the transient mass flux into the MET system cannot be zero in both frames.

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The proposition to be tested is: If a Pasco dynamics demonstration cart, fitted with a one liter water bottle that discharges its contents through two opposed nozzles (in a little more than two seconds), is accelerated by an external force, does the equation of motion (in the lab frame) of the cart include the vdm/dt term that arises from its velocity and changing mass? That is, is the correct equation of motion of the cart F = m(t)dv/dt + vdm/dt ? Or is it F = m(t)dv/dt? The ORS, to be consistent with their cancellation claim, must say the former. Woodward says the latter.

Yes, the ORS must use the first and correct equation of motion.  It is Woodward’s incorrect application of this equation that leads to the trivially flawed figure 4.  The first equation of motion reduces to the second, so the red line is in fact coincident with the blue line. 

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The results of the tests of this system for twenty runs of the cart are shown in Figure 4, where the ORS’ prediction for the velocity of the cart as a function of time is shown in red, and the prediction for vdm/dt not contributing to the equation of motion of the cart is shown in blue. Reality is represented by the black data error bar.  The obvious need not be belabored. The ORS’ claim is wrong.
   

Yeah, no.  Woodward should have actually shown his work and demonstrated how he arrived at that red bar.  It’s certainly not the result anyone correctly applying basic physics would arrive at. 

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Indeed, the only counter-argument mentioned here unknown to the ORS by the spring of 2000 is the results of the Physics 100 level experiment reported above. That experiment was only carried out recently, and its results give the lie to the ORS claims. Those interested in “revolutionary” propulsion physics would be ill advised to give any credence whatsoever to the ORS claims related thereto. As shown here, they rest on bad physics and faulty analysis.

Bad physics and faulty analysis definitely occurred, but it wasn’t advanced by ORS.  Woodward doesn’t seem to have a grip on what it is he is trying to refute.

One way to demonstrate why Woodward’s equation of motion is wrong is to consider a flatcar rolling down a smooth level track at constant velocity, where a hopper from above is pouring sand onto the cart.  This is the exact inverse of Woodward’s thought experiment, so Woodward’s equation of motion, F=m(t)dv/dt, should be applicable:

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Does a flatcar rolling down a smooth level track at constant velocity accelerate if a pile of sand it carries is allowed to fall through a hole in the floor of the car onto the track? The obvious answer, of course, is no
   

In Woodward’s above example, if both Woodward’s equation of motion and the ORS equation of motion are correctly applied, they predict zero acceleration.
My question is this:  Does the flatcar in my example accelerate, or continue to travel at constant velocity when the sand is being poured onto the flatcar?

So people who read Woodward’s rebuttal and actually thought it made sense, please use Woodward’s equation of motion in this instance to make a prediction: does the flatcar slow down or not?     

What does the Woodward equation of motion predict, what does the ORS equation of motion predict, and what actually happens?
« Last Edit: 05/11/2016 12:35 AM by wallofwolfstreet »

Offline wallofwolfstreet

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Re: Woodward's effect
« Reply #737 on: 05/09/2016 04:13 PM »
You right, if we absolutely want to stay within the strict paradigm of the Newtoniak mechanics we can consider small daemons at rest in the instantaneous galilean frame tangent to the accelerating body and which smoothly deliver or remove increment of mass to the moving body ! :D
Now of course, I think there is no way to incorporate further these small daemeons within the frame of Newtonian mechanics. We have to follow rather now an other path shown by Woodward (have a look to the relevant chapter of his book) : Mach Principle and its traduction by General Relativity with the existence of advanced and delayed gravitational waves emitted by any accelerated body (energy, momentum) or its linearized/simplified version using Gravito-Magnetic field (as considered by Sciama, Petroz, Hawkings, Woodward if I am right).

This doesn't really have anything to do with Newtonian Mechanics or limitations therein.  Nor does Woodward's "other path" refute the issues raised.  I have read the relevant work by Woodward, and he doesn't offer any solution.  In fact, the recent paper posted by flux capacitor indicates to me that he doesn't really understand the issue in the first place.  Woodward is dropping terms without meaningful theoretical justification.  There is nothing within Gravitomagnetism that supports dropping terms from an equation of motion.  There is nothing in "Mach's Principle and its traduction by General Relativity with the existence of advanced and delayed gravitational waves emitted by any accelerated body" that supports dropping terms from an equation of motion (that I have seen in any of Woodward's work).       

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Cf Part I of Making Starships and Stargates written by Woodward and edited by Springer.

The big merit of Woodward is that he presents at the same time both a theoretical frame and an experimental frame to his thesis.

I have read the work and don't see any coherent theoretical justification for Woodward's continuing use of the incorrect equation of motion (which was pointed out as incorrect to him almost 15 years ago).  If there is a specific page number or quote you could give me from the book that you think justifies generally ignoring vdm/dt terms, that would be very much appreciated.     

Offline mubahni

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Re: Woodward's effect
« Reply #738 on: 05/09/2016 05:42 PM »
Hallo wallofwolfstreet,

can you comment on this paper?

Thanks.
 

Offline wallofwolfstreet

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Re: Woodward's effect
« Reply #739 on: 05/09/2016 09:51 PM »
Hallo wallofwolfstreet,

can you comment on this paper?

Thanks.

Sure thing.  Unfortunately the pdf has content copying not allowed, so I can’t just copy and paste.  To save myself time I’ll just offer the first little bit of a section I want to quote and you can ctrl+f it.  Apologies in advance for any mistakes in copying. 

In general, the paper offers  a result that is completely consistent with the Oak Ridge Scientists (ORS) interpretation of the correct form of the equation of motion for a variable mass system (John Cramer of the University of Washington also advances this equation of motion, see the second paragraph of the intro of this 2004 paper).

Right in the abstract, Wanser writes:

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There is no rocket type thrust in the usual sense of ejecting propellant, since it is supposed that there is no relative velocity along the direction of motion associated with the mass changes.

Emphasis mine.

This assumption of zero relative velocity is critical, because it allows him to present F=ma+dmdtvrel as just F=ma (i.e. he has reduced the general equation of motion to the Woodward's equation of motion under the constraint that vrel=0).  To be honest, I kind of feel that this paper is presenting a bit of a trivial result.  It is simply showing that when the mass flow into and out of a variable mass system is contrived so that there is no net relative velocity, then the center of mass of the open system can be made to accelerate.  This is basically what Mezzenile has said in his preceding posts.  The issue is, in order for the mass that flows in and out of the system to have vrel=0 despite the acceleration of the MET, there must be some net force accelerating this variable mass, which Wanser correctly identifies in his Discussion:

(start reading at ‘the center of mass of the cart’)

Quote

This momentum must be made up by the rest of the universe allowing the isotropically ejected mass (in the rest frame of the accelerating body) to return isotropically to the body in its rest frame, that has in the mean time accelerated forward, to keep pace with the MET device, thus requiring the “spherical shell  of ejected mass to move forward.

Wanser also reiterates that the results are contingent on assuming vrel=0.  It’s this assumption that doesn’t have any theoretical motivation as being applicable to METs as far as I can see, and in fact Woodward routinely advances explanations as to how you can set vrel=0 which just aren’t consistent with physics.

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The unidirectional acceleration found here is a consequence of our primary assumption, the possibility of changing the mass of an isolated system with zero relative velocity associated with the net convective momentum flux which produces the changing mass


Essentially, what Wanser has done is what I consider to be a bit of slight of hand (at least in regard to METs).  He has moved the problem from within a control volume containing the MET to outside the control volume.  The only way the variable mass can consistently have vrel=0 despite the claimed acceleration of the MET is that somehow the variable mass which exists outside the control volume is being accelerated by some unknown force alongside the MET.  The question is simply reframed from “what causes the MET to accelerate?” to “what causes the variable mass outside the system boundary of the MET to accelerate with the MET, so the net velocity of this mass is always just right to ignore vrel dependent terms?”

So like I said, I think this result is a bit trivial, because it just flips the problem definition around, demonstrating that it is possible to self-accelerate a variable mass system if you are somehow able to accelerate the variable mass when it is outside your control volume boundary.  I don’t think that's a very surprising result. 

There are a few things in the paper that I disagree with (Newton’s third law more fundamental that Conservation of Momentum? – I don’t think so), but I don’t think they are particularity relevant to the question at hand.
« Last Edit: 05/10/2016 12:11 AM by wallofwolfstreet »