Mach Effect progress

Point out news stories, on the net or in mainstream media, related to polywell fusion.

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hanelyp
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Re: Mach Effect progress

Postby hanelyp » Thu Sep 12, 2013 12:46 am

GIThruster wrote:If you take any thrust efficiency figure of merit in force/input power (N/W) and allow the thruster to accelerate at any rate over an arbitrarily long period of time, you can show that all thrusters will appear to violate conservation at some time in the future.

If you're finding an apparent conservation violation, you're doing it wrong. Likely either not doing the reference frame transforms right, or not handling energy-momentum of the exhaust right.

If your thruster can produce 1N/W regardless of how fast it's going, it becomes very easy to produce a device that looks like its violating conservation of energy.

A point to consider: So long as the reaction mass is a zillion times larger that the ship being accelerated, it can absorb a large amount of momentum without absorbing detectable energy or delta-V. This works in Newtonian mechanics or SR, the latter with the limit that the ship not pick up so much energy that it has inertia an appreciable fraction of the reaction mass.
The daylight is uncomfortably bright for eyes so long in the dark.

GIThruster
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Re: Mach Effect progress

Postby GIThruster » Thu Sep 12, 2013 3:51 am

hanelyp wrote:If you're finding an apparent conservation violation, you're doing it wrong. Likely either not doing the reference frame transforms right, or not handling energy-momentum of the exhaust right.


I agree. If you do a transform at all of the exhaust plume (power in) then you have a chance to do it right. I think this is what both chris and 93143 have done. It's when you don't do a transform that you get the wrong answer.
"Courage is not just a virtue, but the form of every virtue at the testing point." C. S. Lewis

pbelter
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Re: Mach Effect progress

Postby pbelter » Mon Sep 16, 2013 3:23 pm

There is a new article about Gliese 1214b most likely having water rich atmosphere

http://nextbigfuture.com/2013/09/gliese ... -rich.html

It lists the estimated mass and diameter of the planet being respectively 7 and 2.7 times Earth.

With those proportions we get 96% of Earth gravity and roughly 7 times the surface area. Not bad...

Now we need to find the surface composition (any land there?), rotational period and temperature to know how much the real estate is worth there.

AcesHigh
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Re: Mach Effect progress

Postby AcesHigh » Mon Sep 16, 2013 8:15 pm

I do not want to offend anyone here, but it SEEMS to me that Paul March has a better understanding of the kind of physics related to ME than anyone on this forum.

I would really like to see his input in the whole discussion between GI and the rest of the forumers.

GIThruster
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Re: Mach Effect progress

Postby GIThruster » Mon Sep 16, 2013 8:50 pm

I'm certainly not offended. I think it's very likely Paul knows much more than anyone else in this forum. I would just note to you though, that he works full time for Eagleworks now, which probably explains why he posts far less than he used to. There are likely many issues he cannot discuss.

For instance, if you look up warp drive at wiki, and pull up Sonny's "Warp 101" paper, near the end of the ppt following the paper, you'll see a photo of a 2.45 Ghz "QVPT" thruster. I had thought that the QVPT title was used exclusively for rebranding the MLT, but the photo appears to be that of a Shawyer resonator. Looks like since Sonny wants to explain thrust from both the MLT and the Shawyer resonator, he renamed BOTH of them under his QVPT title.

If this is so, and Sonny has already tested a 2.45 Ghz resonator, obviously there is a lot he is not releasing to the public. And that's his choice. Paul has fewer choices. Since he works for Sonny he needs to avoid whatever topics instructed, which I am presuming is why we have heard so much less from him the last year.

And I would note to you too, at the Starship Congress in Dallas last August, Sonny continued deliberately misrepresenting his work by claiming Paul's test results from 9-10 years ago were predicted by his QVF model, which in fact followed many years after. Since Paul has no choice (if he wants to keep his job) in supporting the chicanery, I know I would not want to be caught on a public forum and asked to answer the obvious questions. It's just sensible for Paul to stay out of the public eye so long as his boss is a bullshit artist.
"Courage is not just a virtue, but the form of every virtue at the testing point." C. S. Lewis

AcesHigh
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Re: Mach Effect progress

Postby AcesHigh » Mon Sep 16, 2013 11:36 pm

I suppose he can answer the physics related issues of the argument here at Talk Polywell while at the same time avoiding answering sensitive issues that he cannot discuss due to the work at Eagleworks?

GIThruster
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Re: Mach Effect progress

Postby GIThruster » Tue Sep 17, 2013 12:56 am

The physics related issues here always turn up as conservation complaints and Paul doesn't have any training in GR. I think you need a physicist for the kinds of answers you're looking for. And even so, if you had such a person, don't you think the engineers here would continue right on making objections anyway?

No offense intended, but you can tell an engineer 15X that they can't do the math that way, and many of them will go right on arguing that they can. This is what Andrew Palfreyman has been doing for all the years I've known him.

Hence why I am at times frustrated over the inability of the engineering community to take the word of the physics community. It is always engineers who make the conservation complaints. Never the physicists. I dunno what to think that each engineer who approaches the issue afresh tells himself he's seen something no physicist has ever noticed--the most obvious kinds of objections one could make--as if this stuff had never been peer reviewed.

Hence why real objections don't belong online where people bear no personal responsibility for their complaints, but in peer reviewed literature where their careers literally hang in the balance.
"Courage is not just a virtue, but the form of every virtue at the testing point." C. S. Lewis

paulmarch
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Re: Mach Effect progress

Postby paulmarch » Sun Sep 22, 2013 3:19 pm

AcesHigh wrote:I suppose he can answer the physics related issues of the argument here at Talk Polywell while at the same time avoiding answering sensitive issues that he cannot discuss due to the work at Eagleworks?


AcesHigh:

I wish I could say more on this forum, but as GI-Thruster (Ron) points out, I'm now covered by a number of Non-Disclosure Agreements (NDA) that make divulging technical details on what the Eagleworks Lab is currently working on problematic at best. I can say though that we've not had any earth shattering experimental results to date, but have pursued several interesting test campaigns over the last year that have produced data that could support either Dr. Woodward's Mach Effect (M-E) or Dr. White's Quantum Vacuum Fluctuation (QVF) thruster conjectures. I think though that our upcoming home-grown 2.45 GHz resonant cavity test series we are just now starting has the potential of showing which one of these thruster conjectures is the more inclusive one. However as in all things experimental, once you have your data, someone always comes along and points out what you didn't do to clinch the deal either way.

BTW, GI-Thruster (Ron) is right that I'm not an expert in GRT and in calculating relativistic energy and momentum transformations, and I gladly leave that chore to the experts in the field such as Jim Woodward. However after being witness to the long running debate between Woodward, Sarfatti and Zelinsky on this topic over the last few years on Woodward’s M-E e-mail forum, I’m not surprised that almost all of us engineers get confused at times on what the proper way to calculate energy and momentum transfers between inertial and accelerated frames of references actually is. I also think 93143’s comment that “getting energy conservation in an analysis of a thruster of any kind requires one to account for energy transferred to or from what the thruster is pushing on” is spot on. The real trick of course in this process is how does one go about the energy & momentum accounting process in a consistent way, and what is the nature of the two bodies involved in this reaction cycle.

In classical rockets this problem is well defined in that we know in detail the mass of the rocket & engine, the mass of the onboard propellant and the mass and energy contained in the onboard chemical and/or nuclear power supplies that will provide the energy to perform the required work needed to accelerate the propellant away from the rocket at velocities up to and including light speed c in the case of a photon rocket.

On the other hand, in a Gravity/Inertial (G/I) field mediated thruster such as the M-E units, some of these thruster system details are rather more ambiguous since the thruster’s propellant reaction mass is posited to be the very large in comparison to the thruster & local power supply “rest of the mass/energy of the causally connected universe”, AKA the mass/energy contained in the Hubble Volume. We also have the very obtuse issues surrounding the M-E’s action/reaction mechanism that uses instantaneous (no-time) Wheeler/Feynman radiation reaction forces. Try consistently accounting for momentum and energy transfers between the thruster and the Hubble Volume when the time elements in these equations go to zero. I’m still waiting for someone to tell me how to perform that calculation from first principles without infinites cropping up. Not renormalizations I hope...

Best,
Paul March
Friendswood, TX

AcesHigh
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Re: Mach Effect progress

Postby AcesHigh » Mon Sep 23, 2013 12:59 am

Thanks Paul. It´s always good to see your input here even if you can´t enter into detail due to your work at Eagleworks.

GIThruster
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Re: Mach Effect progress

Postby GIThruster » Mon Sep 23, 2013 1:51 am

I don't really understand the issue here, except to say that I do understand the need to transfer between inertial and non-inertial systems.

For those interested, I pose the following:

http://en.wikipedia.org/wiki/Linear-quadratic_regulator

And just to say true--the real transfers for these occasions occur in GR. If you haven't been trained in GR (as I have not) then you ought to humbly accept that you don't get it.

Understanding and accepting that you "don't get it" is a requirement of GR.

If you think you can work in GR with no GR tools, you're a self-aggrandizing asshole.
"Courage is not just a virtue, but the form of every virtue at the testing point." C. S. Lewis

93143
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Re: Mach Effect progress

Postby 93143 » Mon Sep 23, 2013 10:08 am

paulmarch wrote:I also think 93143’s comment that “getting energy conservation in an analysis of a thruster of any kind requires one to account for energy transferred to or from what the thruster is pushing on” is spot on.

Thanks for the support...

GIThruster wrote:http://en.wikipedia.org/wiki/Linear-quadratic_regulator

What has dynamic system control got to do with the issue at hand?

GIThruster wrote:If you don't believe me, take the case of an Ion thruster and do the calcs.

All right, you asked for it. Watch this:

Let us posit a thruster with an input power of P, an efficiency of η, and an exhaust velocity of v_exh, on a spacecraft with total initial mass M0 and initial velocity V0.

The thruster is turned on at t = 0 and begins expending propellant at a rate mdot. Without loss of generality, we specify that the thrust is parallel to the initial velocity with respect to the observer, so as to avoid the unnecessary complication of vector math (recall that in accordance with Pythagoras' Theorem, momenta in different axes do not sum linearly, but kinetic energies do).

The static jet power is:

  η*P = 0.5*mdot*v_exh^2

Therefore

  mdot = 2*η*P/v_exh^2

After Tsiolkovsky,

  deltaV = v_exh*ln(M0/(M0-mdot*t))

Kinetic energy E of the vehicle at time t is

  E = 0.5*(M0-mdot*t)*(V0+deltaV)^2

and the change in vehicle kinetic energy since t = 0 is

  deltaE = 0.5*(M0-mdot*t)*(V0+deltaV)^2 - 0.5*M0*V0^2
       = 0.5*M0*V0^2 + M0*V0*deltaV + 0.5*M0*deltaV^2 - 0.5*mdot*t*V0^2 - mdot*t*V0*deltaV - 0.5*mdot*t*deltaV^2 - 0.5*M0*V0^2
       = (M0-mdot*t)*V0*deltaV + 0.5*(M0-mdot*t)*deltaV^2 - 0.5*mdot*t*V0^2

This is a quadratic in V0. In order to characterize it, we now take its derivatives. For a given deltaV, the first and second derivatives of deltaE with respect to initial velocity V0 are

  ddeltaE/dV0 = (M0-mdot*t)*deltaV - mdot*t*V0
  d^2deltaE/dV0^2 = -mdot*t

Note that the second derivative is always negative. This means the vehicle kinetic energy delta as a function of starting velocity, for a given deltaV, is concave downward; that is, it exhibits a maximum. Where? Let the first derivative equal zero:

  (M0-mdot*t)*deltaV - mdot*t*V0 = 0

Therefore the maximum occurs at

  V0 = (M0-mdot*t)*deltaV/(mdot*t)

The limit of this expression as t -> 0 is

  V0 = M0*deltaV/(mdot*t)

which is singular. However, as the numerator and denominator are both zero to the same order, it can be regularized. The limit of deltaV as t -> 0 is

  dV/dt*t [@ t=0] = v_exh*mdot*t/(M0-mdot*t) [@ t=0] = v_exh*mdot*t/M0

So the maximum rate of increase of deltaE occurs at

  V0 = M0*v_exh*mdot*t/(M0*mdot*t)
  V0 = v_exh

What is the rate of change of E with respect to time at this point?

  E = 0.5*(M0-mdot*t)*(V0^2 + 2*V0*deltaV + deltaV^2)
  dE/dt = M0*V0*dVdt + M0*deltaV*dVdt - mdot*t*V0*dVdt - mdot*t*deltaV*dVdt - 0.5*mdot*(V0^2 + 2*V0*deltaV + deltaV^2)

  dE/dt[@t=0] = M0*V0*v_exh*mdot/(M0-0) + 0 - 0 - 0 - 0.5*mdot*(V0^2 + 0 + 0)
            = mdot*V0*v_exh - 0.5*mdot*V0^2

so if V0 = v_exh, as established above, then

  dE/dt[@t=0,V=V0=v_exh] = 0.5*mdot*v_exh^2 = η*P

which makes sense, as at V = v_exh the exhaust is observed to be stationary, so all the propulsive energy must be going into the vehicle.

Remember, this is a maximum. In other words, the energy gained by an accelerating spacecraft cannot exceed the energy provided to its thrusters in any reference frame, assuming the thrusters use internally-carried propellant. This is because the loss of mass due to expenditure of propellant becomes a large kinetic energy sink as the initial velocity increases.

But, you will note, the energy gained by the spacecraft can easily be less than that expended. Where does the excess go? Into the exhaust, of course.

The rate of outflow of exhaust kinetic energy e at time t is given by

  de/dt = 0.5*mdot*(V0 + deltaV - v_exh)^2
       = 0.5*mdot*(V0^2 + 2*V0*deltaV - 2*V0*v_exh + deltaV^2 - 2*deltaV*v_exh + v_exh^2)

Earlier, we had

  dE/dt = M0*V0*dVdt + M0*deltaV*dVdt - mdot*t*V0*dVdt - mdot*t*deltaV*dVdt - 0.5*mdot*(V0^2 + 2*V0*deltaV + deltaV^2)
  dV/dT = v_exh*mdot/(M0-mdot*t)

which results in

  dE/dt = (M0-mdot*t)*V0*v_exh*mdot/(M0-mdot*t) + (M0-mdot*t)*deltaV*v_exh*mdot/(M0-mdot*t) - 0.5*mdot*(V0^2 + 2*V0*deltaV + deltaV^2)
       = mdot*(V0*v_exh + deltaV*v_exh) - 0.5*mdot*(V0^2 + 2*V0*deltaV + deltaV^2)

Sum the two rates of increase of kinetic energy (the vehicle and the exhaust) and we get:

  dE/dt + de/dt = mdot*(V0*v_exh + deltaV*v_exh) - 0.5*mdot*(V0^2 + 2*V0*deltaV + deltaV^2) + 0.5*mdot*(V0^2 + 2*V0*deltaV - 2*V0*v_exh + deltaV^2 - 2*deltaV*v_exh + v_exh^2)
  dE/dt + de/dt = 0.5*mdot*v_exh^2 = η*P

This result holds for all values of V0 and t. In other words, the kinetic energies gained by an accelerating spacecraft and its exhaust always sum to the value of the total energy productively expended in the operation of the drive. Any inefficiency generally results in the wasted energy appearing as heat in both the exhaust and the structure of the vehicle. It can be shown that thermal energy is invariant under Galilean transformations, which results in conservation being maintained for the full energy budget; this is left as an exercise for the reader.

GIThruster
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Re: Mach Effect progress

Postby GIThruster » Tue Sep 24, 2013 12:03 am

Remember, this is a maximum. In other words, the energy gained by an accelerating spacecraft cannot exceed the energy provided to its thrusters in any reference frame, assuming the thrusters use internally-carried propellant. This is because the loss of mass due to expenditure of propellant becomes a large kinetic energy sink as the initial velocity increases.

If that were true, then this case would not apply in the case of a thruster on a swing arm, and it does; so obviously the propellant is not the salient explanation.


This result holds for all values of V0 and t. In other words, the kinetic energies gained by an accelerating spacecraft and its exhaust always sum to the value of the total energy productively expended in the operation of the drive.

Quite so, but this is irrelevant. The thruster and the exhaust are always in the same reference frame. The question is if you can mix frames and still maintain conservation and you're not doing this. You've side-stepped the trouble people always encounter which is specifically, treating non-inertial frames as inertial. Case not made.

I do appreciate the solution, but you are not addressing the point made--that you cannot treat non-inertial frames as inertial.

Just to say again, you are not mixing frames. You have cleverly avoided the trouble commonly found in conservation complaints, because the thruster and the exhaust are in all instances found in the same frame. If you had used the thrust/power FOM for a stationary ion thruster, allowed it to accelerate, and then used the energy for the exhaust in the accelerating frame instead of the stationary one, you would have the precise conservation violation we commonly find here online. That operation does not yield conservation because the thruster is in a non-inertial frame and energy is not invariant.
"Courage is not just a virtue, but the form of every virtue at the testing point." C. S. Lewis

GIThruster
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Re: Mach Effect progress

Postby GIThruster » Tue Sep 24, 2013 12:12 am

Remember, this is a maximum. In other words, the energy gained by an accelerating spacecraft cannot exceed the energy provided to its thrusters in any reference frame, assuming the thrusters use internally-carried propellant. This is because the loss of mass due to expenditure of propellant becomes a large kinetic energy sink as the initial velocity increases.

If that were true, then this case would not apply in the case of a thruster on a swing arm, and it does; so obviously the propellant is not the salient explanation. The propellent or lack thereof does not create a special case for conservation. BTW, if you really feel up to it, you can do a very similar calculation using the active mass of an M-E thruster, and calculating the action of the mass using it as propellant. After all, M-E thrusters are not technically "propellantless" but rather they use a "recycled propellant" -- the active mass of the thruster. So the inclusion of or exclusion of exhaust is not the salient issue. Also note, that if the exclusion of propellant were an explanation for a real conservation violation, then both linear and circular electric motors would go overunity and they do not.


This result holds for all values of V0 and t. In other words, the kinetic energies gained by an accelerating spacecraft and its exhaust always sum to the value of the total energy productively expended in the operation of the drive.

Quite so, but this is irrelevant. The thruster and the exhaust are always in the same reference frame. The question is if you can mix frames and still maintain conservation and you're not doing this. You've side-stepped the trouble people always encounter which is specifically, treating non-inertial frames as inertial. Case not made.

I do appreciate the solution, but you are not addressing the point made--that you cannot treat non-inertial frames as inertial.

Just to say again, you are not mixing frames. You have cleverly avoided the trouble commonly found in conservation complaints, because the thruster and the exhaust are in all instances found in the same frame. If you had used the thrust/power FOM for a stationary ion thruster, allowed it to accelerate, and then used the energy for the exhaust in the accelerating frame instead of the stationary one, you would have the precise conservation violation we commonly find here online, used to object to M-E thrusters. That operation does not yield conservation because the thruster is in a non-inertial frame and energy is not invariant.
"Courage is not just a virtue, but the form of every virtue at the testing point." C. S. Lewis

hanelyp
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Joined: Fri Oct 26, 2007 8:50 pm

Re: Mach Effect progress

Postby hanelyp » Tue Sep 24, 2013 2:20 am

GIThruster wrote:
Remember, this is a maximum. In other words, the energy gained by an accelerating spacecraft cannot exceed the energy provided to its thrusters in any reference frame, assuming the thrusters use internally-carried propellant. This is because the loss of mass due to expenditure of propellant becomes a large kinetic energy sink as the initial velocity increases.

If that were true, then this case would not apply in the case of a thruster on a swing arm, and it does; so obviously the propellant is not the salient explanation.


This result holds for all values of V0 and t. In other words, the kinetic energies gained by an accelerating spacecraft and its exhaust always sum to the value of the total energy productively expended in the operation of the drive.

Quite so, but this is irrelevant. The thruster and the exhaust are always in the same reference frame. The question is if you can mix frames and still maintain conservation and you're not doing this. ...

Of course you can't mix reference frames without applying the proper frame transforms. Such sloppy frame mixing would be a case of not doing it right.

And it is relevant in so far as it gives an upper bound on thruster performance if energy conservation is to be observed.

I do appreciate the solution, but you are not addressing the point made--that you cannot treat non-inertial frames as inertial.

has 93143 ever implied otherwise?
The daylight is uncomfortably bright for eyes so long in the dark.

GIThruster
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Joined: Tue May 25, 2010 8:17 pm

Re: Mach Effect progress

Postby GIThruster » Tue Sep 24, 2013 2:29 am

Personally I think we've been talking past one another. I do appreciate the effort to do the calculation he did, but it doesn't show anything. It doesn't come to the issue. The claim that energy conservation is the result of using propellant is hand-waving. There is nothing in his analysis to show a propellantless system will violate conservation. Just an empty claim. So what was the point in the exercise?

As I said--put the thruster on a swing arm and his illustration still works, so carrying propellant is not the issue. Also, if the lack of propellant would yield a conservation violation, we would see this in the case of electric motors and we do not.

Just saying, I see no evidence that M-E thrusters will ever violate conservation. Certainly, this is not coming from the maths supplied here.
"Courage is not just a virtue, but the form of every virtue at the testing point." C. S. Lewis


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