Mach Effect progress

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

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GIThruster
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Postby GIThruster » Sun Jul 17, 2011 10:14 pm

tomclarke wrote:(3) Conservation of energy can still be kept if you assume M-E cools down the local universe in some unclear fashion. After all, if momentum transfer is nonlocal energy transfer can also be.

This is the point of Jim's most recent work with absorber theory, to make this more clear. For anyone interested:

http://en.wikipedia.org/wiki/Wheeler–Feynman_absorber_theory
"Courage is not just a virtue, but the form of every virtue at the testing point." C. S. Lewis

93143
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Postby 93143 » Sun Jul 17, 2011 10:20 pm

...actually, I should clarify. If there's momentum transfer happening between the thruster and the distant universe, then in most frames of reference (like the laboratory frame in the flywheel example), there is also work going on.

In the flywheel case, the work done by the thruster on the distant universe by the very act of thrusting would be negative (I'd say "in the lab frame", but since it describes a 360° path under constant thrust the result is the same no matter what inertial frame you pick). This balances the positive net work on the flywheel and allows you to generate net power locally without violating conservation of energy.

This only works if the effective averaged reference frame of the distant matter somehow follows the thruster around...
Last edited by 93143 on Sun Jul 17, 2011 11:16 pm, edited 4 times in total.

tomclarke
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Postby tomclarke » Sun Jul 17, 2011 10:21 pm

93143 wrote:I'm still not seeing it, but oh well...


I don't think it matters! I would just say there is no relationship between drive power and power added to driven object

GeeGee
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Postby GeeGee » Sun Jul 17, 2011 10:25 pm

TallDave wrote:
Seriously, it might not be a bad idea. Wiki is where a lot of people first stop to learn about something new. A pity if they should get the wrong impression...


The worst part of that article is the Wormhole section. It doesn't explain how it is related to the idea of transient mass fluctuations. It just kinda says "well if the device is powerful enough, then we can make wormholes." Anyone who reads that will likely just giggle.

GeeGee
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Postby GeeGee » Mon Jul 18, 2011 12:25 am

GIThruster wrote:This is the point of Jim's most recent work with absorber theory, to make this more clear. For anyone interested:

http://en.wikipedia.org/wiki/Wheeler–Feynman_absorber_theory


He's writing a paper on this?

Grumalg
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Postby Grumalg » Mon Jul 18, 2011 12:47 am

Stoney3K wrote:That's only in an ideal, theoretical situation. Yet we live on a practical planet, and if that rocket doesn't budge an inch on its first seconds when the engines are started, but does expend a lot of energy, it's only because it needs to move its reaction mass (water, for an LH2/LO2 rocket) against a pile of air underneath. All of the energy in those moments is converted into heat through friction between the water vapor out of the rocket and the air surrounding it.


This brings to mind a bit of trivia from the old days of launches when you'd see it start firing and then sit there a bit before it began moving upwards. The reason for this is that they were deliberately loaded with more fuel weight than the engine could lift. Doing so insured that when you lifted off you had the maximum fuel load you could lift. Those first few stationary moments are burning off the excess fuel load.

GIThruster
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Postby GIThruster » Mon Jul 18, 2011 12:49 am

GeeGee wrote:
GIThruster wrote:This is the point of Jim's most recent work with absorber theory, to make this more clear. For anyone interested:

http://en.wikipedia.org/wiki/Wheeler–Feynman_absorber_theory


He's writing a paper on this?


He's including it in the book he's writing this summer.
"Courage is not just a virtue, but the form of every virtue at the testing point." C. S. Lewis

GIThruster
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Postby GIThruster » Mon Jul 18, 2011 12:54 am

Grumalg wrote:
Stoney3K wrote:That's only in an ideal, theoretical situation. Yet we live on a practical planet, and if that rocket doesn't budge an inch on its first seconds when the engines are started, but does expend a lot of energy, it's only because it needs to move its reaction mass (water, for an LH2/LO2 rocket) against a pile of air underneath. All of the energy in those moments is converted into heat through friction between the water vapor out of the rocket and the air surrounding it.


This brings to mind a bit of trivia from the old days of launches when you'd see it start firing and then sit there a bit before it began moving upwards. The reason for this is that they were deliberately loaded with more fuel weight than the engine could lift. Doing so insured that when you lifted off you had the maximum fuel load you could lift. Those first few stationary moments are burning off the excess fuel load.


There's also the fact that rockets like Shuttle are clamped for the first couple seconds burn, to ensure the engines are running at full throttle at liftoff. A problem with ignition in the first couple seconds will cause a rocket that's just left the ground to fall back, strike and explode. This has happened plenty of times in the past. Most recent I think was a SeaLaunch Zenit in 2007.
"Courage is not just a virtue, but the form of every virtue at the testing point." C. S. Lewis

Aero
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Postby Aero » Mon Jul 18, 2011 3:33 am

Vanguard, the first attempt by the United States to orbit a satellite, comes to mind. It underwent "rapid unplanned disassembly shortly after launch." It made an altitude of 1.2 meters. Here is a clip showing the launch.
http://www.youtube.com/watch?v=zVeFkakURXM
We had better rockets at the time but they were military, and politics dictated that this experimental rocket, satellite and launch should be a civilian effort. Hence the above clip was broadcast live for the world to see. The Soviet Union had already launched Sputnik, so the Vanguard launch failure was a clear message that the US was behind in the space race.

For those who like detailed information, see:
http://en.wikipedia.org/wiki/Vanguard_%28rocket%29
Aero

GIThruster
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Postby GIThruster » Mon Jul 18, 2011 8:41 am

Yeah. Here's from 4 years ago:

http://www.youtube.com/watch?v=kHXfWRTf8Bs

Rockets are not safe. . .
"Courage is not just a virtue, but the form of every virtue at the testing point." C. S. Lewis

Stoney3K
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Postby Stoney3K » Mon Jul 18, 2011 12:51 pm

TallDave wrote:
The whole pressure differential analog is only true for an ideal M/E thruster, with zero losses.


Okay... that kinda sorta makes sense, but you should realize an ideal M/E thruster (i.e. one that can generate thrust at arbitrarily high energy efficiency) is far more valuable as an energy generator, as 93143 went into.


It would not be. Even an ideal thruster would comply with the law of conservation of energy: It may be able to supply a tremendous amount of static thrust, but once you start things moving and the thruster is under load, it will draw just as much energy as the work being effected on the target object.

Energy in means energy out, no strange physics here.

It's the same fallacy made by some 'free energy' enthousiasts out there who confuse 'voltage' with 'power'. Just because an electric motor generates a certain amount of back EMF, does not mean you can extract power from that back EMF. For that, you need an element that can supply current as well.

Physically, force (thrust) is actually the analog of voltage, and if it has no path for current (momentum) to carry, or has no means of supplying that momentum as well, power output is zero. Practical example:

Suppose you have an ideal M/E thruster that gives you 500N of thrust at 12V of voltage across it. It's happily pushing against a wall, not doing any work, so the current (and therefore power) through it is zero.

Now we remove that wall and let the M/E thruster push a 1kg cart forward. It will do so with the same 500N, accelerating the cart with 500m/s each second. In the first second, the power draw will be (on average) 250W, in the next, it will increase, up to the point where air resistance takes the cart to terminal velocity. At which point, the amount of power drawn will be equal to the amount of power spent in drag at the front.

The actual thrust on a moving object now only depends on the power source upstream. As long as it can source enough current through the thruster at the same voltage, thrust will sustain. Hence the term ideal truster, with an energy conversion efficiency of 100%.

Remember the 'Newtons per Watt' figure is only a static measure of the amount of losses when the thruster is not doing any work. If it is doing any work, all of the energy required for that will be added to that. However, dynamic situations have not been tested.

You can calculate the same values for a wheel or other mover, just bolt some cart down to a track and apply power to the wheel while its target cannot move. A wheel with a lot of slip will have a lousy figure of motive force versus power, while an ideal wheel will never slip and the motive force for 1W will reach infinite.

Release the brakes, and the amount of power required will just be dependent on the efficiency of your wheel. A non-slipping wheel will require just as much power as is needed to keep the vehicle going, where a lousy wheel will require a crap-load of power more to keep the vehicle going and slip at the same time.
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TallDave
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Postby TallDave » Mon Jul 18, 2011 3:01 pm

It would not be. Even an ideal thruster would comply with the law of conservation of energy: It may be able to supply a tremendous amount of static thrust, but once you start things moving and the thruster is under load, it will draw just as much energy as the work being effected on the target object.


Under that understanding of the M-E effect, it can't hover without using energy. That requires us to expend energy equivalent to the force of gravity, just like a hovering rocket would. Thrust is never free (and if it ever is, you have a machine that is much more useful for making free energy than for hovering) regardless of work performed.

The basis of your confusion seems to be the difference between energy as measured by the work done on an object and the energy actually expended to do the work. They are not the same thing, nor do they need to be equal. Forces can always be applied and do as little as zero work; that does NOT mean no energy is required to apply the force.

http://en.wikipedia.org/wiki/Work_%28physics%29
Last edited by TallDave on Mon Jul 18, 2011 3:39 pm, edited 1 time in total.
n*kBolt*Te = B**2/(2*mu0) and B^.25 loss scaling? Or not so much? Hopefully we'll know soon...

Stoney3K
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Postby Stoney3K » Mon Jul 18, 2011 3:36 pm

TallDave wrote:They are not the same thing, nor do they need to be equal. Forces can always be applied and do as little as zero work; that does NOT mean no energy is required to apply the force.

http://en.wikipedia.org/wiki/Work_%28physics%29


They are, however, equal in the situation of an ideal mover/thruster (M/E or otherwise). It's the very definition of an ideal mover, 100% of the work in or out results in the same amount of energy on the other side.

Obviously, such a thing does not exist, but if it would, for example, to maintain a hover, one does not, in theory, have to apply any work to the object you want to hover (as neither its kinetic, nor potential energy changes) and therefore, no power has to be spent.

That is, however, in theory. A real mover would always need to spend more energy making it work than can actually be transferred into useful work (or the other way round), therefore, even if it needs to do zero work but does need to maintain a constant force (e.g. in hover), it will need to spend power on it.

It's a practical limitation that is causing this, not the physical theory of a prime mover.
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TallDave
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Postby TallDave » Mon Jul 18, 2011 3:50 pm

They are, however, equal in the situation of an ideal mover/thruster (M/E or otherwise).


Nope, that is still only true in the absence of other forces.

This is trivially obvious: have your "ideal mover" apply vertical thrust less than the force of gravity to something held down by gravity -- it won't move. You are expending energy but not doing work.

You might argue "Okay, but how do you know it's using energy?" Easy -- the net force on the object has changed, reducing the force the object applies to whatever's underneath it. That can't happen for free, or you're back to free energy again.
n*kBolt*Te = B**2/(2*mu0) and B^.25 loss scaling? Or not so much? Hopefully we'll know soon...

Stoney3K
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Postby Stoney3K » Mon Jul 18, 2011 4:22 pm

TallDave wrote:
They are, however, equal in the situation of an ideal mover/thruster (M/E or otherwise).


This is trivially obvious: have your "ideal mover" apply vertical thrust less than the force of gravity to something held down by gravity -- it won't move. You are expending energy but not doing work.


OK, then tell me, into what is that energy being converted? It can't disappear into oblivion, since the law of conservation of energy always holds.

And since this is an ideal case we're talking about with no energy wasted in the device....

If you apply a current to a blocked DC motor it will develop a static torque, and a voltage drop will occur only to serve losses inside the motor's windings. An ideal DC motor will draw have no voltage drop (and hence zero resistance) when it's blocked, but will develop a torque proportional to its winding's voltage. It's a direct result of the ideal DC motor equation.

I think I'm starting to see your missed thought pattern here:

You can't just throw energy at something.

Just as it is impossible to stuff 100W of power down a 60W light bulb, you can't just put energy into your M/E thruster and expect it to do something. You will need an element that can potentially induce energy flow, but will only do so as your energy converter requests it.

In the case of the light bulb, you have a fixed 110V grid voltage and, because of the light bulb's properties, it will draw 0,55 amps from your grid. You can't make it draw 1 amp or 2 amps, it just makes power flow as it is designed.

For the M/E thruster, I suspect it will have some kind of V/I behavior dependent on its load, which is, the amount of work it can possibly do. If it's blocked up against something, the theoretical load is zero. It will not be able to do any work, so it won't draw any power from your power source, no matter how many Volts you want to throw at it.

If it is not blocked, however, it will draw power (and convert energy into work) up to the point where either the upstream power source can't deliver anymore power to it (in layman's terms, you pop the breaker), or it reaches an equilibrium where it reaches the limit of work it can do in a second. (layman's term: You gunned the throttle, and your car maxes out at 88MPH because of drag) In that case, it will draw that amount of power in a sustained manner indefinitely.
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