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#1240 by SeeShells on 05 Oct, 2016 17:17
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I have some brushes and holders I can supply free that were designed to provide high power (2.5Kw) to a high frequency air-bearing spindle motor that are high silver content which makes them very slippery on a copper or brass commutator if anyone needs.
That is indeed a real possibility. The engineering solution of looping the wire is a good solution to allow for a dynamic movement of a contracting/expanding wire, but it does not prevent the generation of forces due to thermal expansion.
Rfmwguy's most likely source of force is the wire thermal expansion, same as Yang and Monomorphic. I noticed that Rfmwguy once relied on the lead rigidity to horizontally balance the beam by moving the off-beam lead position. Perfect leads should be very soft, and when off-beam lead position is shifted the balance should not be affected. His setting is the opposite. When the leads expand because of electrical heating, the balance moves as a result.
The problem is caused because you always end up with an external power wire ending perpendicular to the torsion balance setup .
You can only "nullify" those momentum forces when they run dead center through the suspending wire, but...the slightest offset will provoke a sideways momentum and ruin the setup.
That's why some time ago, I suggested the use of an electrical slip ring. To minimize friction, might be best to use copper wheels instead of brushes ? That would probably solve the issue with the thermal expansion momentum of the power wire.
Very Best,
Shell
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#1241 by Bob012345 on 05 Oct, 2016 17:22
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...(unrelated)
As promised, I looked at your paper and found issues with parts II and IV. In part II, I believe you are confusing two different powers. The power (F a t) is the instantaneous Mechanical power, for any force. It is a fundamental property of mechanics, the way things work and it has absolutely no relationship to the electrical, chemical or any kind of power assumed for the force. It can also be written as (F v) for constant acceleration and it is frame dependent. A rocket, observed at two different velocities in relation to two frames will have two different instantaneous mechanical powers for the same exact chemical power released. If you're trying to say the chemical power has to match with velocity in each separate frame to maintain the same acceleration across all frames you would be contradicting known physics.
In section II I did not talk about rocket at all, but talked about "Simple Mechanical System" where the force F is an external force, as had been exaggerated in Fig 1. In such a system, mass M did not change. This is also emphasized by Professor Woodward.QuoteIn part IV I believe you are applying Woodward's formula incorrectly. Woodward's formula is basic physics that applies to all systems. For a constant acceleration, one can always, irrespective of Mach devices, EmDrives or any such devices, break an interval into a series of steps. Each step has an co-moving inertial instantaneous rest frame where the acceleration starts from rest. In your case, for 10 steps, the kinetic energy of the first step is 2E5 but the correct way to sum up the10 steps is not to simply multiply the first step kinetic energy by 10. Since velocity increases linearly with constant acceleration, the velocity of each step increases by the same amount. For the nth step, the velocity is n times the first step velocity and the kinetic energy is n squared times the first step. Thus, after 10 steps the kinetic energy is 100*2E5 or 2E7. This works for 10 steps or 1000 steps or any number.
In the limit, you do end up with the physical concept that the power applied in the ships frame is always enough to create the force it was designed to create and it is always starting from zero velocity. That doesn't imply it won't move, it just means an infinite series of new frames and energy is not violated in any frame. This is what Woodward's formula actually implies for his device in the continuous regime because any movement at all is always new frame and is always below the 'over unity' condition for that frame. I hope this helps.
I do not know how to convince you that your understanding is not correct.
Whether or not you mentioned a rocket in section II does not mean I can't use it to illustrate my point. A rocket abides by all the same laws as any simple mechanical system. An EmDrive looked at from two different frames would also do. But you seem to be saying that F*v is the required instantaneous input power for any accelerating system. Yes? Assuming that the input power is real and invariant, different observers would say different power is required at the source. Does anyone else see a really big problem with that?
I am going to update my pdf file to include an analysis of the rocket equation. Your misconception comes from the fact that the rocket has constant acceleration with constant burning rate of chemical fuel, but you fail to acknowledge that the fuel at later times has more kinetic energy than earlier times, and that kinetic energy was provided by earlier burnt and ejected fuel. For MET or EMDriveThere is no dM with high kinetic energy to eject, so the analogy with an constant acceleration with constant consumed power rocket does not help them. My update will clarify just that.Quotet
We need other people to comment between us on whether your paper is spot on or has errors or if Woodward is right and you are mistaken. In the meantime I propose asking each other very straightforward and simple diagnostic questions. For example;
1) Forgetting Woodward's paper, do you at least agree with my comments regarding the kinetic energies of the ten intervals? This is a yes or no question.
It is a solid no.Quote2) Do you agree or not that the MET or EmDrive might work up to the time where 'over-unity' allegedly occurs?
Another solid no.Quote3) If yes to question 2, suppose we have two MET devices. One at rest, one coasting at the critical speed where 'over-unity' happens according to our frame. We turn both on. What happens to each? Why?
They each does not produce any thrust.Quote4) You can make simple direct statements about my statements about what you think is wrong in my understanding.
It is in the first answer above (the "rocket analogy" part).QuoteI'm enjoying the discussion so please don't give up! Thanks.
Thanks.
Tellmeagain, thanks for the answers!
For 1) Please tell me how, in your own example problem, you would break up the problem into ten steps.
What is the kinetic energy of subsequent steps and how do you add them and what is the total, since you say my numbers are incorrect?
For 2) I understand you are saying that the MET or EmDrive cannot work at all then. If it worked at all then you say it would always give a violation, yes? Can you give a reason why it couldn't work at least as well as simple light?
For 4) A few folks have asserted that I do not understand how a rocket works because I did not exhaustively mention the exhaust.
No, I know all about that. You seem to think that since the MET or EmDrive supposedly do not have an exhaust, then there is nothing to give that 'apparent' large gain in kinetic energy because there is nothing to borrow kinetic energy from. Yes?
Finally, are you saying that the power delivered to the device from a rocket's chemical fuel or an EmDrive' electrical energy is always exactly equal to F*V? -
#1242 by X_RaY on 05 Oct, 2016 17:25
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There it is. I increase the mesh density for this sims.[...
Could you also run TM212?
The mode at the port was predefined as TE01.
...
The overwhelming majority of tests by the NASA Eagleworks group has been in TM212.
Paul March is the only investigator in the world (to my recollection) that has actually verified the mode of operation, by experimentally verifying the mode shape (TM212) with measurements.
Furthermore, the TE012 Brady et.al. measurements were not reproducibly robust as the TM212 measurements.
Thanks!
Port impedance: 50 Ohm
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#1243 by gargoyle99 on 05 Oct, 2016 17:33
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We both say each case of the rocket uses the same energy in the burn and in the faster frames the kinetic energy ends up higher so where is my statement factually wrong?
BTW, I have a degree in physics also.
You have a degree in physics? Awesome! There are lots of parts of physics and mechanics may not have been your specialty.
The factually wrong part is that you say the kinetic energy ends up higher. The ship's payload kinetic energy ends up much higher in a reference frame with a higher initial velocity, but the ship's propellant exhaust kinetic energy ends up so much lower that it exactly cancels out all of the extra payload kinetic energy so that the change in kinetic energy of the system does not depend on reference frame.
Let's do some math together. (And then stop talking about this, forever.)
A rocket starts with mass of payload and a mass of propellant. It will use chemical potential energy to burn the propellant and eject it out at the back to gain velocity.
Initial kinetic energy = 1/2 (mass payload + mass propellant) * velocity initial^2.
Depending on the ISP of the propellant and rocket engine and other efficiencies, the rocket payload will gain a certain delta v and the exhaust will have a certain delta v exhaust. These velocities are relative to the rocket and do not depend on reference frame. You can look up the exhaust velocities for your rocket engine and calculate the delta v using the rocket equation and your mass fractions.
velocity final payload = velocity initial + delta v
For a real rocket, it's a bit more complicated because the absolute velocity in your chosen reference frame of the propellant varies between what's kicked out at the start of the burn and what's kicked out at the end, so you have to integrate over the velocity profile to get the total energy. To make the math easier, let's consider an instantaneous burn. It doesn't change the conclusions.
velocity final exhaust = velocity initial - delta v exhaust
The delta v of the exhaust is in the opposite direction of the delta v of the payload. By conservation of momentum, we know the relationship between these:
mass payload * delta v = mass exhaust * delta v exhaust
(We'll need that in a moment.)
Next we will show that the change of kinetic energy in the system has no dependency at all on initial velocity, that is the change in kinetic energy does not depend on which reference frame we choose to use to measure velocity.
Final kinetic energy = 1/2 (mass payload) * (velocity initial + delta v)^2
+ 1/2 (mass exhaust) * (velocity initial - delta v exhaust)^2
The difference in kinetic energy is the difference between the two systems:
Delta kinetic energy = Final kinetic energy - Initial kinetic energy
= 1/2 (mass payload) * (velocity initial + delta v)^2
+ 1/2 (mass exhaust) * (velocity initial - delta v exhaust)^2
- 1/2 (mass payload + mass exhaust) * velocity initial^2.
Multiply it out...
Delta kinetic energy =
1/2 (mass payload)*(velocity initial)^2 + (mass payload)*(velocity initial)*(delta v) + 1/2 (mass payload)*(delta v)^2
+ 1/2 (mass exhaust)*(velocity initial)^2 - (mass exhaust)*(velocity initial)*(delta v exhaust) + 1/2 (mass exhaust)*(delta v exhaust)^2
- 1/2 (mass payload)*(velocity initial)^2 - 1/2 (mass exhaust)*(velocity initial)^2
Use the conservation of momentum relation from above and simplify:
Delta kinetic energy =
1/2 (mass payload)*(delta v)^2
+ 1/2 (mass exhaust payload)*(delta v exhaust)^2
Pretty neat! That makes sense and you will note that the initial velocity does not appear in this equation at all. Therefore, the change in kinetic energy is independent of the initial velocity and therefore independent of reference frame. While the final kinetic energy of just the payload may increase much more in some reference frames than in others, energy is fully conserved because of the corresponding change of kinetic energy of the exhaust. The change in energy is always dependent only on the potential chemical energy used and not on the reference frame.
Energy is conserved in all reference frames and all is well with Newtonian physics and the Universe. Whew.
For now.
Some people (for example, Shawyer) have claimed that the EmDrive does not work that way and there is a constant acceleration regardless of reference frame. In a closed system, that would clearly violate conservation of both energy and momentum. From my physics background, that is a clear indication to me that the EmDrive is NOT a closed system. The next question for me would be, what is it interacting with that makes it not a closed system?
The rest of the mass of the Universe (I am respectful of this opinion, but personally quite doubtful.)
The quantum vacuum (I am also doubtful of this, but it brings up interesting ideas, which generally don't seem to resolve the CoE problem.)
The local gravitational field (Now this would be intriguing if it panned out.)
Something boring that no one cares about (Local EM fields, thermal air currents, etc.)
Something else?
We won't know until there are reproducible examples that can be independently verified and tested in different environments and reference frames. God speed, builders.
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#1244 by Bob012345 on 05 Oct, 2016 17:40
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We both say each case of the rocket uses the same energy in the burn and in the faster frames the kinetic energy ends up higher so where is my statement factually wrong? You probably thought that I think that the 'extra' kinetic energy is somehow 'over unity'. No, I never said that or meant that. Perhaps you are interpreting them from a different frame of reference than the one I wrote them in.
Thanks. BTW, I have a degree in physics also.
P.S.
If you ignore the propellant and only talk about the accelerating rocket, then yes, its kinetic energy is increasing faster in reference frames moving relative to it (the faster the relative movement, the greater the increase over a unit of time). This does not make sense, and that's one of the reasons I doubt a truly propellentless drive is possible. There has to be something that offsets that kinetic energy. Moreover, this cannot be potential energy since it has to change as you pick faster moving reference frames. It can be the kinetic energy of a propellant, the energy of a photon (which also varies in different frames due to doppler shift), or something else that I can't think of right now... otherwise you get paradoxes like this (different rates of kinetic energy change in different frames). If I understand it correctly, in any "real" scenarios these paradoxes do not exist, and are easily resolved if you take into account the entire system.
I didn't ignore it, it just didn't mention it. But folks should remember that if momentum is conserved in a non obvious way, it's still conserved and if it is then there also must be some aspect of the greater universe that acts just like the exhaust. In other words, the MET or EmDrive must be borrowing kinetic energy from that source like the rocket borrows kinetic energy from its own exhaust.
BTW, it's very funny that I made a statement I got from rocket experts "the upper stage can generate much more usable kinetic energy than the total chemical energy of the propellants it carries." a perfectly factual statement of which there was a rush to "correct" me on.
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#1245 by wicoe on 05 Oct, 2016 17:49
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But folks should remember that if momentum is conserved in a non obvious way, it's still conserved and if it is then there also must be some aspect of the greater universe that acts just like the exhaust. In other words, the MET or EmDrive must be borrowing kinetic energy from that source like the rocket borrows kinetic energy from its own exhaust.
OK, it's understood that it has to "borrow" it from some source, but the tricky part is that the amount of the "borrowed" energy must depend on the reference frame of the observer, i.e. it will have to "borrow" more from the point of view of an external moving observer (the faster the relative movement, the more "borrowing" will occur). In other words, this has to be a very tricky mechanism. If it was "pushing" against "the rest of the universe" in a primitive way, this wouldn't be possible (i.e. it would be harder and harder for it to accelerate as it gains speed relative to the CMB or distant galaxies). -
#1246 by Bob012345 on 05 Oct, 2016 18:29
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We both say each case of the rocket uses the same energy in the burn and in the faster frames the kinetic energy ends up higher so where is my statement factually wrong?
BTW, I have a degree in physics also.
You have a degree in physics? Awesome! There are lots of parts of physics and mechanics may not have been your specialty.
The factually wrong part is that you say the kinetic energy ends up higher. The ship's payload kinetic energy ends up much higher in a reference frame with a higher initial velocity, but the ship's propellant exhaust kinetic energy ends up so much lower that it exactly cancels out all of the extra payload kinetic energy so that the change in kinetic energy of the system does not depend on reference frame.
Let's do some math together. (And then stop talking about this, forever.)
A rocket starts with mass of payload and a mass of propellant. It will use chemical potential energy to burn the propellant and eject it out at the back to gain velocity.
Initial kinetic energy = 1/2 (mass payload + mass propellant) * velocity initial^2.
Depending on the ISP of the propellant and rocket engine and other efficiencies, the rocket payload will gain a certain delta v and the exhaust will have a certain delta v exhaust. These velocities are relative to the rocket and do not depend on reference frame. You can look up the exhaust velocities for your rocket engine and calculate the delta v using the rocket equation and your mass fractions.
velocity final payload = velocity initial + delta v
For a real rocket, it's a bit more complicated because the absolute velocity in your chosen reference frame of the propellant varies between what's kicked out at the start of the burn and what's kicked out at the end, so you have to integrate over the velocity profile to get the total energy. To make the math easier, let's consider an instantaneous burn. It doesn't change the conclusions.
velocity final exhaust = velocity initial - delta v exhaust
The delta v of the exhaust is in the opposite direction of the delta v of the payload. By conservation of momentum, we know the relationship between these:
mass payload * delta v = mass exhaust * delta v exhaust
(We'll need that in a moment.)
Next we will show that the change of kinetic energy in the system has no dependency at all on initial velocity, that is the change in kinetic energy does not depend on which reference frame we choose to use to measure velocity.
Final kinetic energy = 1/2 (mass payload) * (velocity initial + delta v)^2
+ 1/2 (mass exhaust) * (velocity initial - delta v exhaust)^2
The difference in kinetic energy is the difference between the two systems:
Delta kinetic energy = Final kinetic energy - Initial kinetic energy
= 1/2 (mass payload) * (velocity initial + delta v)^2
+ 1/2 (mass exhaust) * (velocity initial - delta v exhaust)^2
- 1/2 (mass payload + mass exhaust) * velocity initial^2.
Multiply it out...
Delta kinetic energy =
1/2 (mass payload)*(velocity initial)^2 + (mass payload)*(velocity initial)*(delta v) + 1/2 (mass payload)*(delta v)^2
+ 1/2 (mass exhaust)*(velocity initial)^2 - (mass exhaust)*(velocity initial)*(delta v exhaust) + 1/2 (mass exhaust)*(delta v exhaust)^2
- 1/2 (mass payload)*(velocity initial)^2 - 1/2 (mass exhaust)*(velocity initial)^2
Use the conservation of momentum relation from above and simplify:
Delta kinetic energy =
1/2 (mass payload)*(delta v)^2
+ 1/2 (mass exhaust payload)*(delta v exhaust)^2
Pretty neat! That makes sense and you will note that the initial velocity does not appear in this equation at all. Therefore, the change in kinetic energy is independent of the initial velocity and therefore independent of reference frame. While the final kinetic energy of just the payload may increase much more in some reference frames than in others, energy is fully conserved because of the corresponding change of kinetic energy of the exhaust. The change in energy is always dependent only on the potential chemical energy used and not on the reference frame.
Energy is conserved in all reference frames and all is well with Newtonian physics and the Universe. Whew.
For now.
Some people (for example, Shawyer) have claimed that the EmDrive does not work that way and there is a constant acceleration regardless of reference frame. In a closed system, that would clearly violate conservation of both energy and momentum. From my physics background, that is a clear indication to me that the EmDrive is NOT a closed system. The next question for me would be, what is it interacting with that makes it not a closed system?
The rest of the mass of the Universe (I am respectful of this opinion, but personally quite doubtful.)
The quantum vacuum (I am also doubtful of this, but it brings up interesting ideas, which generally don't seem to resolve the CoE problem.)
The local gravitational field (Now this would be intriguing if it panned out.)
Something boring that no one cares about (Local EM fields, thermal air currents, etc.)
Something else?
We won't know until there are reproducible examples that can be independently verified and tested in different environments and reference frames. God speed, builders.
Thanks but everything you've said I already know and have said in different words many times myself. I do wonder sometimes if folks read what I write and/or if they are predisposed to find objections? I assume you meant by your comment that I may not be well versed in this area that you are. How well versed are you?
One thing, you state "The factually wrong part is that you say the kinetic energy ends up higher." And then you prove that the kinetic energy of the rocket ends up higher. The point you take such great pains to correct me on I have no disagreement with, that the actual total kinetic energy gain of the system as a whole is the same as computed from any observer frame. I know that full well but it's irrelevant to the problem I am debating. We are not riding the system as a whole to our destination! The effect is a form of the Oberth effect. It's very real and very useful.
I agree with your comment "Energy is conserved in all reference frames and all is well with Newtonian physics and the Universe". The Work-Energy Theorem shows how any device accelerating under a constant force would meet this requirement in any frame.
Also, I find this a very interesting discussion with many facets. I don't think it should be just shut down. I am glad you are open to the possibility the EmDrive or variants may really work. Thanks.
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#1247 by Fan Boi on 05 Oct, 2016 18:38
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As for the wires heating up and causing thrust by stiffening: Why can't you put the conductors in the exact center axis of the balance beam? You could use button contacts, or perhaps mercury.
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#1248 by Bob012345 on 05 Oct, 2016 18:39
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But folks should remember that if momentum is conserved in a non obvious way, it's still conserved and if it is then there also must be some aspect of the greater universe that acts just like the exhaust. In other words, the MET or EmDrive must be borrowing kinetic energy from that source like the rocket borrows kinetic energy from its own exhaust.
OK, it's understood that it has to "borrow" it from some source, but the tricky part is that the amount of the "borrowed" energy must depend on the reference frame of the observer, i.e. it will have to "borrow" more from the point of view of an external moving observer (the faster the relative movement, the more "borrowing" will occur). In other words, this has to be a very tricky mechanism. If it was "pushing" against "the rest of the universe" in a primitive way, this wouldn't be possible (i.e. it would be harder and harder for it to accelerate as it gains speed relative to the CMB or distant galaxies).
I disagree. The rocket 'borrows' from its own exhaust as seen by observers and all observers see a different amount depending on the frame they observe in. It shouldn't be fundamentally different for a MET or EmDrive. The fundamental unit is the force the device is able to generate in its own reference frame. The rest of the universe will adjust. -
#1249 by Tellmeagain on 05 Oct, 2016 18:42
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Tellmeagain, thanks for the answers!
You helped me to improve my pdf file. Thank you!QuoteFor 1) Please tell me how, in your own example problem, you would break up the problem into ten steps.
What is the kinetic energy of subsequent steps and how do you add them and what is the total, since you say my numbers are incorrect?
For 2) I understand you are saying that the MET or EmDrive cannot work at all then. If it worked at all then you say it would always give a violation, yes? Can you give a reason why it couldn't work at least as well as simple light?
For 4) A few folks have asserted that I do not understand how a rocket works because I did not exhaustively mention the exhaust. No, I know all about that. You seem to think that since the MET or EmDrive supposedly do not have an exhaust, then there is nothing to give that 'apparent' large gain in kinetic energy because there is nothing to borrow kinetic energy from. Yes?
Finally, are you saying that the power delivered to the device from a rocket's chemical fuel or an EmDrive' electrical energy is always exactly equal to F*V?
Bob, you have a physics degree but I do not. So it seems I should be the one who ask and you be the one who answer. This discussion rapidly gets boring because what I want to say will not be much different from that of gargoyle99 or wicoe. Your understanding of the rocket equation is right. It is just that you use that to justify MET and EmDrive is questionable because one has exhaust and the others do not. -
#1250 by wicoe on 05 Oct, 2016 18:48
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I disagree. The rocket 'borrows' from its own exhaust as seen by observers and all observers see a different amount depending on the frame they observe in. It shouldn't be fundamentally different for a MET or EmDrive. The fundamental unit is the force the device is able to generate in its own reference frame. The rest of the universe will adjust.
The observers in different frames see a different amount because it 'borrows' from the *kinetic* energy of the exhaust. This works because kinetic energy has the required property (i.e. it depends on the observer frame in the "right way"). The energy of a photon also has this property. What other forms of energy can have this property? I'm really curious... -
#1251 by rfmwguy on 05 Oct, 2016 18:53
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Yes, Mr Li, it was getting to the point that I almost thought all was due to thermal expansion of wires. I even took a thermal video of the power lead nearby the cavity. Its on my youtube channelThx for the info...
Then...I do tend to side with TT then (on Lorentz forces), because the Lorentz forces of 100µN are indeed considerably lower then the measured forces: 18.4mN equals 18400µN, which is 3 magnitudes bigger.
In that respect you could indeed say that they're negligible compared to the measured forces. So Lorentz forces do not account for the measured forces...
what else can be checked off ?
Buoyancy? linear expansion of power wires? thermal convection?
Problem with all these tests and test results is that they're fragmented over several posts and threads...
Not easy to get a concise overview and understand (or remember ) of what has been done so far...
added:
It might also be a good idea to keep units to the same level, to avoid confusion. Why not keep it all to µN ?
Makes it so much more obvious...
Rfmwguy's most likely source of force is the wire thermal expansion, same as Yang and Monomorphic. I noticed that Rfmwguy once relied on the lead rigidity to horizontally balance the beam by moving the off-beam lead position. Perfect leads should be very soft, and when off-beam lead position is shifted the balance should not be affected. His setting is the opposite. When the leads expand because of electrical heating, the balance moves as a result.
What you might not know is I stripped off the power leads from the torsion beam and then began to experiment with perpendicular then parallel loop locations in relation to the torsion beam. In the end, the magnetron began to fail and was unable to complete more than 5 or so observations. I saw little variation in beam displacement regardless of loop/harness location, but I was unable to continue more testing.
In the end, I could not 100% say I removed wire thermal expansion as an error source, but the 5 or so displacement tests seemed to indicate I was not getting displacement forces. Not scientific enough (too few samples) for a public test report to be issued.
p.s. While this chart could be considered the definitive test that thermal wire expansion is not the cause, there still wasn't enough data runs. This was probably my best chart, with seconds as the X axis and the red zone where power was "on". displacement is happening too rapidly for a temperature coefficient. I do note the peak variances where there is slight displacement differences. I attribute that probably to magnetron output stability. The average of this chart translated into 18.4mN
https://lh6.googleusercontent.com/-fNfLZjVCikc/AAAAAAAAAAI/AAAAAAAAASk/xom5HSdE44Y/photo.jpg -
#1252 by Bob012345 on 05 Oct, 2016 19:00
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I disagree. The rocket 'borrows' from its own exhaust as seen by observers and all observers see a different amount depending on the frame they observe in. It shouldn't be fundamentally different for a MET or EmDrive. The fundamental unit is the force the device is able to generate in its own reference frame. The rest of the universe will adjust.
The observers in different frames see a different amount because it 'borrows' from the *kinetic* energy of the exhaust. This works because kinetic energy has the required property (i.e. it depends on the observer frame in the "right way"). The energy of a photon also has this property. What other forms of energy can have this property? I'm really curious...
Me too but I'm not going to pretend to know the answer.. If there were an exhaust in the form of photons with an enhanced momentum, it might come from those?
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#1253 by Bob012345 on 05 Oct, 2016 19:18
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Tellmeagain, thanks for the answers!
You helped me to improve my pdf file. Thank you!QuoteFor 1) Please tell me how, in your own example problem, you would break up the problem into ten steps.
What is the kinetic energy of subsequent steps and how do you add them and what is the total, since you say my numbers are incorrect?
For 2) I understand you are saying that the MET or EmDrive cannot work at all then. If it worked at all then you say it would always give a violation, yes? Can you give a reason why it couldn't work at least as well as simple light?
For 4) A few folks have asserted that I do not understand how a rocket works because I did not exhaustively mention the exhaust. No, I know all about that. You seem to think that since the MET or EmDrive supposedly do not have an exhaust, then there is nothing to give that 'apparent' large gain in kinetic energy because there is nothing to borrow kinetic energy from. Yes?
Finally, are you saying that the power delivered to the device from a rocket's chemical fuel or an EmDrive' electrical energy is always exactly equal to F*V?
Bob, you have a physics degree but I do not. So it seems I should be the one who ask and you be the one who answer. This discussion rapidly gets boring because what I want to say will not be much different from that of gargoyle99 or wicoe. Your understanding of the rocket equation is right. It is just that you use that to justify MET and EmDrive is questionable because one has exhaust and the others do not.
Thanks. I'm also not going to claim I'm some true rocket expert. Also, I don't think I'm saying the Oberth effect of the rocket justifies the MET or EmDrive per se. But it helps me envision it. Still, I don't quite understand why my treatment of your example was wrong to you. I don't get why you wrote that Woodward would just add the first increment of kinetic energy ten times. I never got that from his paper at all.
Also, I thought that the discussion of the possibility of a constant force acting under a constant input power went beyond MET or EmDrive to basic classical mechanics so that was interesting to debate.
Thanks again. -
#1254 by as58 on 05 Oct, 2016 19:28
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I disagree. The rocket 'borrows' from its own exhaust as seen by observers and all observers see a different amount depending on the frame they observe in. It shouldn't be fundamentally different for a MET or EmDrive. The fundamental unit is the force the device is able to generate in its own reference frame. The rest of the universe will adjust.
The observers in different frames see a different amount because it 'borrows' from the *kinetic* energy of the exhaust. This works because kinetic energy has the required property (i.e. it depends on the observer frame in the "right way"). The energy of a photon also has this property. What other forms of energy can have this property? I'm really curious...
Me too but I'm not going to pretend to know the answer.. If there were an exhaust in the form of photons with an enhanced momentum, it might come from those?
Relativistic energy-momentum relation doesn't seem to leave room for photons with enhanced momentum. Unless enhanced momentum means just higher energy. But that doesn't change anything, photon rocket is a photon rocket, no matter what wavelength it operates at. -
#1255 by WarpTech on 05 Oct, 2016 19:34
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There it is. I increase the mesh density for this sims.[...
Could you also run TM212?
The mode at the port was predefined as TE01.
...
The overwhelming majority of tests by the NASA Eagleworks group has been in TM212.
Paul March is the only investigator in the world (to my recollection) that has actually verified the mode of operation, by experimentally verifying the mode shape (TM212) with measurements.
Furthermore, the TE012 Brady et.al. measurements were not reproducibly robust as the TM212 measurements.
Thanks!
Port impedance: 50 Ohm
The impedance without the insert had a high degree of asymmetry. Capacitive at about 1 Ohm and Resistive at about 500 Ohms. Here with the insert, there is high resistance in both directions and less asymmetry.
Thanks X-RaY!
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#1256 by Tellmeagain on 05 Oct, 2016 19:35
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Thanks. I'm also not going to claim I'm some true rocket expert. Also, I don't think I'm saying the Oberth effect of the rocket justifies the MET or EmDrive per se. But it helps me envision it. Still, I don't quite understand why my treatment of your example was wrong to you. I don't get why you wrote that Woodward would just add the first increment of kinetic energy ten times. I never got that from his paper at all.
Thanks again.
"Critics" say if Professor Woodward provides some input energy to the spaceship with MET, and the spaceship ends up with much higher output energy, this violates CoE.
He says for each of the 10 intervals, in each of the 10 different frames, he provides 1/10 of the input energy, and end up with much lower output energy for each interval in each different frame, each does not violate CoE.
I say if he sums up the ten 1/10 input energy he ends up with the same input energy; but if he sum up the 10 much lower output energy for each interval and each frame, the summation, no more than the total input energy, does not match the total output energy, so his treatment is wrong.
You seems to believe if he sums up the ten 1/10 input energy he ends up with the same input energy, and the final output energy is much higher than my summation, indeed, is equal to the critics' calculation. It seems you defeat Professor Woodward's treatment and agree with the "critics".
This is my understanding of the discussion between us.
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#1257 by X_RaY on 05 Oct, 2016 19:54
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From the viewpoint of the em-field, there is more space to propagate when the dielectric is present, more similar to the big end without a dielectric (effective diameter = small diameter*sqrt(eps_r)).
There it is. I increase the mesh density for this sims.[...
Could you also run TM212?
The mode at the port was predefined as TE01.
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The overwhelming majority of tests by the NASA Eagleworks group has been in TM212.
Paul March is the only investigator in the world (to my recollection) that has actually verified the mode of operation, by experimentally verifying the mode shape (TM212) with measurements.
Furthermore, the TE012 Brady et.al. measurements were not reproducibly robust as the TM212 measurements.
Thanks!
Port impedance: 50 Ohm
The impedance without the insert had a high degree of asymmetry. Capacitive at about 1 Ohm and Resistive at about 500 Ohms. Here with the insert, there is high resistance in both directions and less asymmetry.
Thanks X-RaY!
I am waiting for feko to complete a 2 port sim to get S21 and S12 data too.
This will take a couple of hours
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#1258 by Flyby on 05 Oct, 2016 20:04
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ahh... I didn't know you had indeed scanned the wiring for temperature.
Yes, Mr Li, it was getting to the point that I almost thought all was due to thermal expansion of wires. I even took a thermal video of the power lead nearby the cavity. Its on my youtube channel
What you might not know is I stripped off the power leads from the torsion beam and then began to experiment with perpendicular then parallel loop locations in relation to the torsion beam. In the end, the magnetron began to fail and was unable to complete more than 5 or so observations. I saw little variation in beam displacement regardless of loop/harness location, but I was unable to continue more testing.
In the end, I could not 100% say I removed wire thermal expansion as an error source, but the 5 or so displacement tests seemed to indicate I was not getting displacement forces. Not scientific enough (too few samples) for a public test report to be issued.
p.s. While this chart could be considered the definitive test that thermal wire expansion is not the cause, there still wasn't enough data runs. This was probably my best chart, with seconds as the X axis and the red zone where power was "on". displacement is happening too rapidly for a temperature coefficient. I do note the peak variances where there is slight displacement differences. I attribute that probably to magnetron output stability. The average of this chart translated into 18.4mN
Rfmwguy, is that the same current/voltage we see there that comes down from the wall plug?
As there is no increase in temperature, one can conclude there is barely to no additional linear expansion on that part. So if that is indeed the same wire that comes to the torsion setup, we can check off that possible cause also.... (It's starting to look better and better ! )
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#1259 by Tellmeagain on 05 Oct, 2016 20:16
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ahh... I didn't know you had indeed scanned the wiring for temperature.
Rfmwguy, is that the same current/voltage we see there that comes down from the wall plug?
As there is no increase in temperature, one can conclude there is barely to no additional linear expansion on that part. So if that is indeed the same wire that comes to the torsion setup, we can check off that possible cause also.... (It's starting to look better and better ! )
Small temperature increase can cause big difference. So before you check that off, better quantify it. For example, measure the force when runing the same amount of current (a few amperes AC in two leads and 500mA DC in another) through the leads without magnetron.
No, I know all about that. You seem to think that since the MET or EmDrive supposedly do not have an exhaust, then there is nothing to give that 'apparent' large gain in kinetic energy because there is nothing to borrow kinetic energy from. Yes?