Quote from: wallofwolfstreet on 03/30/2016 01:56 AMThis 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:F_{net}=(dm/dt)v + ma0=(dm/dt)v + ma (F_{net}=0 because the system is closed)0=0 + ma (since there is no flow of mass across the system boundary)***a=0And 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?

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:F_{net}=(dm/dt)v + ma0=(dm/dt)v + ma (F_{net}=0 because the system is closed)0=0 + ma (since there is no flow of mass across the system boundary)***a=0And 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.

(F_{net}=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.

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?

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..

If water escapes through gravity from one hole the bottle, this has an impact on the movement of the cart

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

The mass fluctuation in a MET is a direct consequence of 'pumping' (in this case: electrical) energy into it.

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.

QuoteIf water escapes through gravity from one hole the bottle, this has an impact on the movement of the cartThis 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.

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.

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

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.

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.

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(ρ_{0})=Const for every ρ_{o}?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.

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.

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.

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 :QuoteA 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

The correct expression for Force on a system with non-constant mass is:F_{net}=dm/dtv+maWhere 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?

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...

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.

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.

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?

Quote from: Mezzenile on 04/01/2016 03:25 PMIf water escapes through gravity from one hole the bottle, this has an impact on the movement of the cartThis 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 from: 93143 on 04/03/2016 01:12 AMAlso, 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.

Quote from: wallofwolfstreet on 04/01/2016 05:53 PMQuote from: Mezzenile on 04/01/2016 03:25 PMIf water escapes through gravity from one hole the bottle, this has an impact on the movement of the cartThis 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 from: 93143 on 04/03/2016 01:12 AMAlso, 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.

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

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.

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.

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.

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.

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.

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.

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

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 ! 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).

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.

Hallo wallofwolfstreet, can you comment on this paper?Thanks.

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.

…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.

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