## Mach Effect progress

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

Moderators: tonybarry, MSimon

TallDave
Posts: 3114
Joined: Wed Jul 25, 2007 7:12 pm
Contact:
In vacuum, there is. But I was trying to explain the situation where the rocket does not move, but needs to expend power. The explanation for that is simple: All of the energy wasted during those moments when the rocket remains on the pad, is converted into heat. It first needs to overcome drag before it can go anywhere.

In space, there is hardly any drag, so almost all of the energy thrown out the back is immediately converted into motion.
No, no, no. Gravity is holding it down, not air. Drag is relatively picayune by comparison.

Let's pretend for a moment the Earth is airless, or your rocket is on the surface of an airless chunk of Earth-sized mass. Surely you are not under the impression that just because there's no air, any arbitrarily small amount of thrust will be sufficient to lift an object of any arbitrary size?

Your thrust must first exceed the force of gravity holding the rocket down. Up to that point, all your rocket energy does not move the rocket one millimeter.
n*kBolt*Te = B**2/(2*mu0) and B^.25 loss scaling? Or not so much? Hopefully we'll know soon...

TallDave
Posts: 3114
Joined: Wed Jul 25, 2007 7:12 pm
Contact:
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.

That's why I say it will be very interesting to determine the actual possible efficiency. I suspect any theory that says efficiency can be arbitrarily high will eventually be found lacking, and we'll find that thrust must be well below the level at which an M-E flywheel can generate net power: useful as an alternative to propellant for interplanetary/interstellar travel, not world-changing.

The experiments thus far have apparently been somewhat contradictory.

http://en.wikipedia.org/wiki/Woodward_effect
n*kBolt*Te = B**2/(2*mu0) and B^.25 loss scaling? Or not so much? Hopefully we'll know soon...

GeeGee
Posts: 95
Joined: Fri Jun 10, 2011 7:00 pm
TallDave wrote:
The experiments thus far have apparently been somewhat contradictory.

http://en.wikipedia.org/wiki/Woodward_effect
Am I the only one spotting a ton of errors on this page?

"The local universe is the "system" in this device, and thus momentum is conserved. No explanation is given regarding the nonlocality of such an effect, and it may conflict with special relativity."

Woodward explains the nonlocality of the effect as a Wheeler-Feynman advanced/retarded wave interaction. This doesn't conflict with SR.

"The hypothesis is also related to the Nordtvedt effect proposed by Kenneth L. Nordtvedt from Montana State University, who observed that some theories of gravity suggest that massive bodies should fall at different rates depending upon their gravitational self-energy. This would violate the strong equivalence principle that the laws of gravitation are independent of velocity and location, a principle considered fundamental by many theoretical physicists.The Lunar Laser Ranging Experiment has shown that if the Nordtvedt effect exists at all, it is extremely weak."

The Nordtvedt effect they're quoting above is from a 1969 paper on a test of the equivalence principle

http://prola.aps.org/abstract/PR/v169/i5/p1014_1

The Nordtvedt paper Woodward often refers to is on the existence of gravitomagnetism, published in 1988.

93143
Posts: 1131
Joined: Fri Oct 19, 2007 7:51 pm
I take a single day off, and look what happens...

@Stoney3K:

The question we've been arguing about is whether or not the work done by an M-E thruster necessarily equals the energy input required to run it. You attempt to answer this question by calculating the work done by an M-E thruster in the laboratory reference frame, make the unstated and unsupported assumption that this equals the energy input required to run it, and then triumphantly declare that there is no difference between the value you've calculated and the value you've calculated.

Listen closely, because I'm tired of explaining this:

The reaction mass used by a Mach-effect thruster is not in the laboratory frame of reference. It therefore makes no sense at all to calculate the thruster's power requirements based on its velocity in the laboratory frame of reference.

What mental block is preventing you from understanding this?
An M/E thruster will operate on the same idea, but the 'road' is beyond the scope of our research. It will still have a limited efficiency ('grip') with which it can exert a force on that 'road', and thus generate thrust.
That is EXACTLY IT. Now, all you need to do is realize that the effective average relative velocity between the thruster and its 'road' is what determines the minimum power required to generate a certain amount of thrust (ie: the maximum possible "thrust efficiency"). It has nothing to do with how fast the thruster appears to be moving relative to the building it's in.

It makes no difference what frame of reference the power supply is in, either, because electrical power transmission doesn't care about piddly little velocities like these, so the power leaving the power supply in the lab frame will be basically the same as the power arriving at the thruster in the thruster's frame. And once the thruster has been supplied with electrical power, its mode of operation has nothing further to do with the lab frame or anything in it. It's not interacting with the floor or the atmosphere; it's interacting with the gravinertial field, which (if it is in fact the cause of inertia, and if M-E thrusters work we can assume that it is) has been observed to behave the same independent of velocity state.
In essence, an M/E thruster that causes some object to get into an equilibrium state will, theoretically, require zero power. For example, if you have an airship that uses M/E thrusters to counteract its weight and stay afloat at a constant altitude, it would, in theory, require no power to stay put because the change in potential energy (as a result of changing altitude) is zero.
Right, just like a helicopter can hover without expending any energy. Wait...
The reason conventional 'engines' (e.g. jets or props) require continuous power to keep an aircraft aloft in hover is because they're pushing air to work against gravity. Since air is notoriously inefficient at that and has a tendency to disperse, you will have to keep pushing to maintain a constant upward force.
A helicopter needs to expend energy to hover because the reaction mass the helicopter is pushing on is a fluid, and is thus unable to sustain a static stress. So far your handwaving is more or less on the mark. But the real reason, resulting from the above fact, is this: the reaction mass is moving relative to the helicopter in the axis of thrust, so finite power is required to exert a force on it. (You can "hover" by walking up a down escalator, but this does require power even in the ideal case, because the relative velocity between you and the escalator is nonzero, so you need to do work in the course of exerting a force. Conversely, you can 'windmill' a helicopter in a sufficiently strong updraft.)

So what is the M-E thruster pushing on and how fast is it going?
gravity is used to generate power because it moves water down a drop. It's not the gravity that does the work, but the drop
Wrong. It's gravity that's doing the work, because the object it's exerting a force on is moving in the axis of the force vector. It's that simple - in the laboratory frame of reference. The amount of work done by a force is frame-dependent. (Don't get too scared of this; it always ends up working out...)
That's where the 1N/W figure comes from. It's the maximum in a static thrust situation, e.g. when there is equilibrium, the thing requires 1W for each Newton it has to push. This doesn't mean it's necessarily the same for a dynamic (moving) situation as well, since that has not been tested.
Wrong.

Here's the misunderstanding.

The solution (which should be obvious by now) is that according to Lorentz invariance, a particular Mach-effect thruster fed a certain quantity of power should produce the same thrust no matter what inertial reference frame it's in. There is no 'preferred' reference frame. So the idea of "static" vs. "dynamic" thrust, which is derived from aeronautical engine testing, has no basis here.

In other words, if inertia is the same regardless of velocity (and it is), M-E thrust efficiency should be too, because the field it's interacting with behaves the same.

If this were not true, you would then have to explain why the preferred velocity state of the M-E thruster just happens to follow the testing lab around the Earth as it rotates and orbits the Sun, galactic centre, etc., not to mention why the inertial effect it's supposedly taking advantage of shows no such preference...

...

Oh, and both you and Carl miscalculated the rocket-on-flywheel problem. His is worse, but they're both pretty bad. (For one thing, his heat of combustion is off by a factor of 1000, and you didn't correct him, choosing instead to neglect a factor of 9 in a follow-on calculation...) Has either of you ever heard of specific impulse?

At 62.8 m/s tangential velocity, almost all of the energy expended by the rockets will end up in the ~4 km/s exhaust stream, not the motion of the generator.

And that should be a clue to where the 'extra' energy comes from in my M-E flywheel example...

EDIT: I just thought of a great example. Take a rocket engine, in a vacuum. At full throttle, it uses the same amount of energy per second to produce the same thrust, completely independent of how fast it's moving or how much mass it's pushing. You can calculate a frame-independent thrust efficiency value for it, even though the amount of work it does NEGLECTING THE KINETIC ENERGY OF THE PROPELLANT is frame-dependent.

The above is exactly true of a Mach-effect thruster.
Last edited by 93143 on Sun Jul 17, 2011 9:58 pm, edited 2 times in total.

Carl White
Posts: 321
Joined: Mon Aug 24, 2009 10:44 pm
93143 wrote:Oh, and both you and Carl miscalculated the rocket problem. His is worse, but they're both pretty bad. Has either of you ever heard of specific impulse?
What does it matter? So long as there is net thrust, according to how you calculate it, you can pick parameters to make it over unity, right? Try it with an ion thruster or so, then.
Last edited by Carl White on Sun Jul 17, 2011 9:38 pm, edited 2 times in total.

GIThruster
Posts: 4686
Joined: Tue May 25, 2010 8:17 pm
GeeGee wrote: Am I the only one spotting a ton of errors on this page?
Yes, the page is full of errors. Jim has always asserted that Einstein's Equivalence Principle must be true, that GR must the true, SR must be true and has recently (the last 1 1/2 years) been saying that Wheeler-Feynman Absorber Theory is true as well. That web page predates Woodward's stance on absorber theory, but the rest are just factual errors someone made about Jim's claims and work.
"Courage is not just a virtue, but the form of every virtue at the testing point." C. S. Lewis

93143
Posts: 1131
Joined: Fri Oct 19, 2007 7:51 pm
No, you can't. I explained this. You need to accelerate up the propellant to the speed of the engines.

If the engines are moving at less than the exhaust velocity, then some energy is wasted in the exhaust stream. If they are moving at more than the exhaust velocity, it takes more energy to accelerate the propellant up to the engines than you get back from the thrust. (Think of jet engines instead of rockets if you want; the principle is the same - the momentum transfer to the propellant is effectively negative in this case, so the engines are generating net drag, not thrust.) The rotor edge velocity has an optimum of sorts somewhere between zero and the exhaust velocity, but this device can never break even, never mind go over unity.

The advantage of the M-E thruster is that you don't have to accelerate the propellant - it's already there, because the M-E thruster carries its own reference frame with it, and inertia works the same in all reference frames. How exactly this translates to a real-world effect on the distant universe is the complicated part, but we don't need to worry about it; it's out of scope.
Last edited by 93143 on Sun Jul 17, 2011 9:45 pm, edited 4 times in total.

TallDave
Posts: 3114
Joined: Wed Jul 25, 2007 7:12 pm
Contact:
GIThruster wrote:
GeeGee wrote: Am I the only one spotting a ton of errors on this page?
Yes, the page is full of errors.
If only there were some knowledgable people around who could update the page (nudge, nudge)...

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...
n*kBolt*Te = B**2/(2*mu0) and B^.25 loss scaling? Or not so much? Hopefully we'll know soon...

Carl White
Posts: 321
Joined: Mon Aug 24, 2009 10:44 pm
93143 wrote:No, you can't. I explained this. You need to accelerate up the propellant to the speed of the engines.

If the engines are moving at less than the exhaust velocity, then some energy is wasted in the exhaust stream. If they are moving at more than the exhaust velocity, it takes more energy to accelerate the propellant up to the engines than you get back from the thrust. (Think of jet engines instead of rockets if you want; the principle is the same - the momentum transfer to the propellant is effectively negative in this case, so the engines are generating net drag, not thrust.) The rotor edge velocity has an optimum of sorts somewhere between zero and the exhaust velocity, but this device can never break even, never mind go over unity.

The advantage of the M-E thruster is that you don't have to accelerate the propellant - it's already there, because the M-E thruster carries its own reference frame with it, and inertia works the same in all reference frames. How exactly this translates to a real-world effect on the distant universe is the complicated part, but we don't need to worry about it; it's out of scope.
I understand this, and in the example I tried to build, unsuccessfully I admit, it included a calculation for accelerating the propellant.

Since our flywheel can perform at 100%, why not assume an ion thruster that can spit out ions at relativistic speeds at nearly 100% efficiency. Maybe the thrust itself sucks (depending on the power in), but the flywheel is going to be having to move awfully fast before the effort of accelerating more ions to reach the thruster becomes a major component of the power drain.

tomclarke
Posts: 1683
Joined: Sun Oct 05, 2008 4:52 pm
Location: London
Contact:
93143 wrote:I take a single day off, and look what happens...

@Stoney3K:

The question we've been arguing about is whether or not the work done by an M-E thruster necessarily equals the energy input required to run it. You attempt to answer this question by calculating the work done by an M-E thruster in the laboratory reference frame, make the unstated and unsupported assumption that this equals the energy input required to run it, and then triumphantly declare that there is no difference between the value you've calculated and the value you've calculated.

Listen closely, because I'm tired of explaining this:

The reaction mass used by a Mach-effect thruster is not in the laboratory frame of reference. It therefore makes no sense at all to calculate the thruster's power requirements based on its velocity in the laboratory frame of reference.

What mental block is preventing you from understanding this?
An M/E thruster will operate on the same idea, but the 'road' is beyond the scope of our research. It will still have a limited efficiency ('grip') with which it can exert a force on that 'road', and thus generate thrust.
That is EXACTLY IT. Now, all you need to do is realize that the effective average relative velocity between the thruster and its 'road' is what determines the minimum power required to generate a certain amount of thrust (ie: the maximum possible "thrust efficiency"). It has nothing to do with how fast the thruster appears to be moving relative to the building it's in.

It makes no difference what frame of reference the power supply is in, either, because electrical power transmission doesn't care about piddly little velocities like these, so the power leaving the power supply in the lab frame will be basically the same as the power arriving at the thruster in the thruster's frame. And once the thruster has been supplied with electrical power, its mode of operation has nothing further to do with the lab frame or anything in it. It's not interacting with the floor or the atmosphere; it's interacting with the gravinertial field, which (if it is in fact the cause of inertia, and if M-E thrusters work we can assume that it is) has been observed to behave the same independent of velocity state.
In essence, an M/E thruster that causes some object to get into an equilibrium state will, theoretically, require zero power. For example, if you have an airship that uses M/E thrusters to counteract its weight and stay afloat at a constant altitude, it would, in theory, require no power to stay put because the change in potential energy (as a result of changing altitude) is zero.
Right, just like a helicopter can hover without expending any energy. Wait...
The reason conventional 'engines' (e.g. jets or props) require continuous power to keep an aircraft aloft in hover is because they're pushing air to work against gravity. Since air is notoriously inefficient at that and has a tendency to disperse, you will have to keep pushing to maintain a constant upward force.
A helicopter needs to expend energy to hover because the reaction mass the helicopter is pushing on is a fluid, and is thus unable to sustain a static stress. So far your handwaving is more or less on the mark. But the real reason, resulting from the above fact, is this: the reaction mass is moving relative to the helicopter in the axis of thrust, so finite power is required to exert a force on it. (You can "hover" by walking up a down escalator, but this does require power even in the ideal case, because the relative velocity between you and the escalator is nonzero, so you need to do work in the course of exerting a force. Conversely, you can 'windmill' a helicopter in a sufficiently strong updraft.)

So what is the M-E thruster pushing on and how fast is it going?
gravity is used to generate power because it moves water down a drop. It's not the gravity that does the work, but the drop
Wrong. It's gravity that's doing the work, because the object it's exerting a force on is moving in the axis of the force vector. It's that simple - in the laboratory frame of reference. The amount of work done by a force is frame-dependent. (Don't get too scared of this; it always ends up working out...)
That's where the 1N/W figure comes from. It's the maximum in a static thrust situation, e.g. when there is equilibrium, the thing requires 1W for each Newton it has to push. This doesn't mean it's necessarily the same for a dynamic (moving) situation as well, since that has not been tested.
Wrong.

Here's the misunderstanding.

The solution (which should be obvious by now) is that according to Lorentz invariance, a particular Mach-effect thruster fed a certain quantity of power should produce the same thrust no matter what inertial reference frame it's in. There is no 'preferred' reference frame. So the idea of "static" vs. "dynamic" thrust, which is derived from aeronautical engine testing, has no basis here.

In other words, if inertia is the same regardless of velocity (and it is), M-E thrust efficiency should be too, because the field it's interacting with behaves the same.

If this were not true, you would then have to explain why the preferred velocity state of the M-E thruster just happens to follow the testing lab around the Earth as it rotates and orbits the Sun, galactic centre, etc., not to mention why the inertial effect it's supposedly taking advantage of shows no such preference...

...

Oh, and both you and Carl miscalculated the rocket-on-flywheel problem. His is worse, but they're both pretty bad. (For one thing, his heat of combustion is off by a factor of 1000, and you didn't correct him, choosing instead to neglect a factor of 9 in a follow-on calculation...) Has either of you ever heard of specific impulse?

At 62.8 m/s tangential velocity, almost all of the energy expended by the rockets will end up in the ~4 km/s exhaust stream, not the motion of the generator.

And that should be a clue to where the 'extra' energy comes from in my M-E flywheel example...

But you make one if I understand you, if not apologies. M-E says that inertial mass can be changed by altering local mass density. From this effect you can easily generate unidirectional thrust (actually acceleration, not thrust).

(1) conservation of momentum requires nonlocal transfer from object to local universe.

(2) The energy required to make M-E work has no relationship to the energy added to the local object by M-E. The M-E drive energy is not intrinsic, and relates to losses in the M-E device, it can in principle be small.

(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.

(4) Locally however M-E looks like free energy. The acceleration is constant for given drive system in the instantaneously stationary frame of the accelerated object. Acceleration of a moving object => power. So mount M-E object on flywheel etc, make acceleration tangential, power output is obvious. Further power (delivered to flywhhel axil for, from "local universe" can be increased arbitrarily by making flywheel angular velocity faster.
Last edited by tomclarke on Sun Jul 17, 2011 10:06 pm, edited 1 time in total.

93143
Posts: 1131
Joined: Fri Oct 19, 2007 7:51 pm
Carl White wrote:Since our flywheel can perform at 100%, why not assume an ion thruster that can spit out ions at relativistic speeds at nearly 100% efficiency. Maybe the thrust itself sucks (depending on the power in), but the flywheel is going to be having to move awfully fast before the effort of accelerating more ions to reach the thruster becomes a major component of the power drain.
True, but it will have to move just as fast before the fraction of the input power that's actually applied to the flywheel as work gets anywhere near the amount wasted in that relativistic exhaust stream. My explanation of why this scheme can't break even doesn't depend on the specific numbers involved.

This is why ion thrusters have such awful thrust-to-power ratios compared with chemical engines - it's a direct consequence of having high Isp.
Last edited by 93143 on Sun Jul 17, 2011 10:07 pm, edited 2 times in total.

GIThruster
Posts: 4686
Joined: Tue May 25, 2010 8:17 pm
TallDave wrote:
GIThruster wrote:
GeeGee wrote: Am I the only one spotting a ton of errors on this page?
Yes, the page is full of errors.
If only there were some knowledgable people around who could update the page (nudge, nudge)...

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...
I've never contributed to wiki but you have a valid point. The notion feels wrong somehow though. . .editing someone's work without their permission. Were it up to me, certainly I'd start by retitling the page. Jim has always objected very strongly with his name being connected to his work. The page ought to be called "Mach-Effect Physics" or some such, but how is that a legitimate edit of someone else's writing?
"Courage is not just a virtue, but the form of every virtue at the testing point." C. S. Lewis

93143
Posts: 1131
Joined: Fri Oct 19, 2007 7:51 pm
tomclarke wrote:But you make one if I understand you. M-E says that inertial mass can be changed by altering local mass density. From this effect you can easily generate unidirctional thrust (actually acceleration, not thrust).

(1) conservation of momentum requires nonlocal transfer from object to local universe.

(2) The energy required to make M-E work has no relationship to the energy added to the local object by M-E. The M-E drive energy is not intrinsic, and relates to losses in the M-E device, it can in principle be small.

(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.

(4) Locally however M-E looks like free energy. The acceleration is constant for given drive system in the instantaneously stationary frame of the accelerated object.
...so how is any of this in contradiction to anything I said?

tomclarke
Posts: 1683
Joined: Sun Oct 05, 2008 4:52 pm
Location: London
Contact:
93143 wrote:
tomclarke wrote:But you make one if I understand you. M-E says that inertial mass can be changed by altering local mass density. From this effect you can easily generate unidirctional thrust (actually acceleration, not thrust).

(1) conservation of momentum requires nonlocal transfer from object to local universe.

(2) The energy required to make M-E work has no relationship to the energy added to the local object by M-E. The M-E drive energy is not intrinsic, and relates to losses in the M-E device, it can in principle be small.

(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.

(4) Locally however M-E looks like free energy. The acceleration is constant for given drive system in the instantaneously stationary frame of the accelerated object.
...so how is any of this in contradiction to anything I said?
The reaction mass used by a Mach-effect thruster is not in the laboratory frame of reference. It therefore makes no sense at all to calculate the thruster's power requirements based on its velocity in the laboratory frame of reference.
I thought the implication here unwarranted, though the conclusion and premise are both correct.

93143
Posts: 1131
Joined: Fri Oct 19, 2007 7:51 pm
I'm still not seeing it, but oh well...