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#1280 by Bob012345 on 06 Oct, 2016 19:03
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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.
Thanks. I don't easily understand your argument so let me ponder your words a bit before I respond. -
#1281 by dustinthewind on 07 Oct, 2016 05:11
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We should remind readers that there is a member of NSF with experience in General Relativity that posted not just a theory but a formula that predicts experimental results:
http://emdrive.wiki/@notsosureofit_Hypothesis
and he did so years ago, before many of the people conducting experiments were posting at NSF. His formula takes into account the electromagnetic mode shapes (unlike other often quoted theories that do not explicitly take into account the mode shapes).
I get the impression again that this hypothesis works by modifying the mass of the photon by modifying its frequency or wavelength.
Sort of the concept that a motor can also be a generator. If gravity can change the properties of light then modifying these properties of light may also induce gravitational effects. -
#1282 by TheTraveller on 07 Oct, 2016 07:44
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...Put another way, how do we know that all this resonance stuff is needed and that best results aren't simply generated by heating a magnet feeding into an asymmetrical can?...
Another EmDrive builder and myself measured the thrust curve versus freq and found it was narrower than the rtn loss curve. When we shared this data with Roger, he explained why the thrust curve was much narrower than the rtn loss curve as both Qu and forward power drop as you excite off freq.
Force = ( 1 Qu Pwr Df) / c
We also confirmed each others data that the peak thrust occurred slightly off the peak rtn loss freq, which we believe was caused by the peak rtn loss freq having a slight internal phase distortion that exciting slightly off freq pushed in the other way. -
#1283 by chongma on 07 Oct, 2016 12:01
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[...
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!
the builders definitely need to be seeing this kind of resonance in IR before they even begin taking any measurements on a balance. also why do the magnetrons burn out so quickly? in an oven the magnetron lasts ages -
#1284 by TheTraveller on 07 Oct, 2016 12:53
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...also why do the magnetrons burn out so quickly? in an oven the magnetron lasts ages...
Several reasons I can think of:
1) Oven cavity VSWR, reflected power back into the maggie, is limited to that which the maggie is designed to handle.
2) Oven maggie has fan forced cooling to limit max maggie temperature.
3) Oven cavity has very low Q and as a result there is little maggie freq pulling. Freq pulling (maggie raw freq pulled to the high Q cavity bandwidth freq) increases VSWR and maggie heating.
Solution is single freq excitation, freq adjustment to lowest reflected power and output power attenuation to allow freq adjustment at low power and then pulse to the desired higher power. -
#1285 by rfmwguy on 07 Oct, 2016 13:31
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Why magnetrons burn out in EmDrive testing vs oven.[...
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!
the builders definitely need to be seeing this kind of resonance in IR before they even begin taking any measurements on a balance. also why do the magnetrons burn out so quickly? in an oven the magnetron lasts ages
1. The EmDrive cavity is much smaller leading to higher temperatures transferring to the magnetron base.
2. A tighter RF coupling meaning the EmDrive cavity is a narrower bandpass resonance and there are higher reflection coeffecients leading to higher heating of the magnetron as it drifts in frequency.
3. An oven has a fan pushing air constantly over the yoke of the magnetron, leading to far better cooling. A fan is not used in EmDrive tests because the of mechanical errors that would be introduced on precision displacement measurements.
4. An oven's cooking cavity is also air vented leading to cooler operation of the magnetron.
Its all about higher yoke temps on the magnetron during EmDrive testing into a much narrower and smaller resonance cavity. Basic RF design principles.
I'm not sure what your point is regarding IR. If you want to see (imperfect) TM013 mode forming you can watch my video:
The discussion of this with monomorphic and Mr Li was here:
https://www.reddit.com/r/QThruster/comments/4sv7ni/1701a_emdrive_thermal_test_2_with_new_magnetron/
Note there is slight misalignment of endplates leading to imperfect mode forming on the small diameter. This is a critical discovery, meaning 18.4mN displacement was achieved by a less than perfect alignment and more is available.
Despite speculations by non-EmDrive builders, there is no single mode that has proven to be the only one which could yield displacement.
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#1286 by TheTraveller on 07 Oct, 2016 13:54
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Despite speculations by non-EmDrive builders, there is no single mode that has proven to be the only one which could yield displacement.
Correct.
My understanding is the mode alters the surface resistance losses in the cavity and thus Qu varies as the mode varies. There are paper online that describe the effect.
Also the mode alters where the eddy currents flow. It is desired to avoid eddy currents that flow from the side walls onto the end plates, well across any physical joint, as that requires a electrically conductive surface across the joint at 5x skin depth or there may be arching at higher power.
TE01x modes form eddy currents that are circular around the centre of the end plate and never flow onto the side walls. Thus for end plates that have a mechanical joint and can be removed, TE01x modes are highly desirable.
Example of eddy currents crossing and not crossing into the side walls from the end plate are attached
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#1287 by Rodal on 07 Oct, 2016 14:37
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...Example of eddy currents crossing and not crossing into the side walls from the end plate are attached
Are you now running FEKO Boundary Element simulations of the mode shapes, or is somebody else (Monomorphic ?) to be credited for those nice computer runs you posted?
Thanks -
#1288 by TheTraveller on 07 Oct, 2016 14:53
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To those who seek to learn how to build an EmDrive
1st warning. Microwave Radiation can be health threatening and possibly LETHAL.
2nd warning. Magnetron power supplies ARE LETHAL.
3rd warning: If you are not experienced in dealing with high power microwave systems and their power supplies,
DO NOT ATTEMPT TO BUILD AN EMDRIVE.
Info sources from NSF are the following builders. Both their posts and attachments are available.
StarDrive (Paul March):
http://forum.nasaspaceflight.com/index.php?action=profile;area=showposts;u=2074
RfMwGuy:
http://forum.nasaspaceflight.com/index.php?action=profile;u=47785
Monomorphic:
http://forum.nasaspaceflight.com/index.php?action=profile;u=49802
Seeshells:
http://forum.nasaspaceflight.com/index.php?action=profile;u=48229
TheTraveller:
https://forum.nasaspaceflight.com/index.php?action=profile;u=47641
EmDrive Wiki:
http://emdrive.wiki/Main_Page
RfMwGuy YouTube channel:
https://www.youtube.com/channel/UCm54FS3u2aDeutnMsV0cITg/videos
Monomorphic YouTube channel:
https://www.youtube.com/user/monomorph1/videos
TheTraveller EmDrive Research forum and archive:
https://groups.google.com/forum/#!forum/emdriveresearch
https://drive.google.com/drive/u/0/folders/0B7kgKijo-p0ifk9EakZfbW9aZGMwNWZMQ01xVnBON0tkM2w0Q1NLbmtjRFFwMXBuNVlVN0U
Roger Shawyer's web site:
www.emdrive.com with 2 excellent detailed engineering reports.
EagleWorks 1st EmDrive paper with many force generation results:
https://drive.google.com/open?id=0B7kgKijo-p0iS3hvZzV5Rzl6Rlk
If I have missed any, please let me know so I can update this list.
There are more than enough information in the public space to assist the build of a working EmDrive. I suggest designing for mode TE013, using a variable single freq generator at 2.45GHz, a solid state Rf amp, with monitoring of forward and reflected power and the ability to vary output power over at least a 31dBm range.
Please note that while both Rfmwguy and Monomorphic started with a magnetron, they are both moving to single freq Rf amp builds. This should give prospective EmDrive builders a good pointer as to how to excite their frustums.
Also note Dave's 18.4mN of force was achieved with a very highly polished and solid Copper frustum, so as to obtain max Q and max force generation. This result should also guide prospective EmDrive builders about how to build their frustums.
====================
4th warning. Microwave Radiation can be health threatening and possibly LETHAL.
5th warning. Magnetron power supplies ARE LETHAL.
6th warning: If you are not experienced in dealing with high power microwave systems and their power supplies,
DO NOT ATTEMPT TO BUILD AN EMDRIVE. -
#1289 by TheTraveller on 07 Oct, 2016 15:04
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...Example of eddy currents crossing and not crossing into the side walls from the end plate are attached
Are you now running FEKO Boundary Element simulations of the mode shapes, or is somebody else (Monomorphic ?) to be credited for those nice computer runs you posted?
Thanks
Have a LOT of images in my archive. Don't remember where each came from. They may be from Monomorphic or from others who have done similar work.
Should add that Monomorphic did the FEKO analysis of my TE013 spherical end plate frustum design and the antenna / coupler placement. Attached is an early run.
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#1290 by WarpTech on 07 Oct, 2016 15:55
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Despite speculations by non-EmDrive builders, there is no single mode that has proven to be the only one which could yield displacement.
Correct.
My understanding is the mode alters the surface resistance losses in the cavity and thus Qu varies as the mode varies. There are paper online that describe the effect.
Also the mode alters where the eddy currents flow. It is desired to avoid eddy currents that flow from the side walls onto the end plates, well across any physical joint, as that requires a electrically conductive surface across the joint at 5x skin depth or there may be arching at higher power.
TE01x modes form eddy currents that are circular around the centre of the end plate and never flow onto the side walls. Thus for end plates that have a mechanical joint and can be removed, TE01x modes are highly desirable.
Example of eddy currents crossing and not crossing into the side walls from the end plate are attached
What is the relative current density (A/m^2) in these simulations?
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#1291 by RERT on 07 Oct, 2016 16:05
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A while ago someone posted that E=cP was mandated by special relativity for any massless particle moving at c.
Since that would mean ditching special relativity in order for the EMdrive to have a massless exhaust plume with P>(E/c), I thought I'd check.
The actual requirement of the Lorentz transformation is that E=±cP, not that E=cP.
Probably a nit, but nice to know there is a tiny crack there. Here's the algebra. For a Lorentz transformation of energy-momentum to a primed frame moving at velocity v:
p’=γ(p-vE/c²) and E’=γ(E-pv)
If we require E=λp for some constant λ, we must also have E’=λp’, or we could tell the frames apart. So:
λ=(E’/p’) = (λp-pv)/(p-vλp/c²)=(λ-v)/(1-λv/c²) ⇒λ-λ²v/c²=λ-v ⇒λ²v/c²=v ⇒λ=±c
If this has any bearing at all, it would be that an 'exhaust' might consist of pairs of particles with the same momentum and opposite energy, and so the ratio of momentum to energy could be arbitrarily high.
(edit to correct 2nd Para) -
#1292 by Bob012345 on 07 Oct, 2016 16:54
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A while ago someone posted that E=cP was mandated by special relativity for any massless particle moving at c.
Since that would mean ditching special relativity in order for the EMdrive to have a massless exhaust plume with P>(E/c), I thought I'd check.
The actual requirement of the Lorentz transformation is that E=±cP, not that E=cP.
Probably a nit, but nice to know there is a tiny crack there. Here's the algebra. For a Lorentz transformation of energy-momentum to a primed frame moving at velocity v:
p’=γ(p-vE/c²) and E’=γ(E-pv)
If we require E=λp for some constant λ, we must also have E’=λp’, or we could tell the frames apart. So:
λ=(E’/p’) = (λp-pv)/(p-vλp/c²)=(λ-v)/(1-λv/c²) ⇒λ-λ²v/c²=λ-v ⇒λ²v/c²=v ⇒λ=±c
If this has any bearing at all, it would be that an 'exhaust' might consist of pairs of particles with the same momentum and opposite energy, and so the ratio of momentum to energy could be arbitrarily high.
(edit to correct 2nd Para)
The Finnish group suggests that paired photons where the fields cancel but not the momentum are involved.
I'm not sure what you mean by 'opposite' energy. -
#1293 by X_RaY on 07 Oct, 2016 17:48
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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.
...
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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#1294 by MrFrankenverse on 07 Oct, 2016 17:55
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Here's a simple thought experiment:
I have a recoilless gun suspended rigidly in a closed free moving cavity. The gun is fired and the bullet hits the cavity wall at one end.
Hypothesis:
Since there is no recoil felt by the cavity all the energy in the bullet's motion is transformed into movement of the box.
A recoilless gun is a way of allowing the gun itself to remain relatively fixed in space while some gas is ejected out the back to counter the kick of the bullet. If you account for the gas all the momentum will cancel. So, no net motion beyond adjusting the center of mass a bit. Nice try though.
In this simple thought experiment the gun is actually recoilless, so it's meant as more of a tautology aimed at bringing other closed cavity experiments into focus. For example in the emdrive the shape of the cavity combined with the standing waves of varying group velocities and increased Q may dislocate the recoil of the originating waves from the system thereby creating the necessary phantom forces; unlike effects in fluid mechanics. in gun manufacture engineers seek to reduce the effect of recoil by using various designs to redirect gas flow and by engaging moving counterbalances, increasing mass, and so on. -
#1295 by X_RaY on 07 Oct, 2016 18:34
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For future calculations I will need some defined data:
- frustum: shape/ dimensions, material
- possible dielectric: shape/ dimensions, dielectric constants
- frequency or frequency band (start-stop)
- source: type, position and direction, for wg-port mode shape and impedance
- input power
- defined requests about measurements, for example: magnitude current at plate X for a defined frequency Y
- maybe some more informations, I will ask for if needed
I doesn't make sense to ask for results without define some input data.
Thanks
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#1296 by Bob012345 on 07 Oct, 2016 19:06
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"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.
That's what they say. They don't say one can't have a constant force acting as long as you want. That's basic mechanics. They worry about how much energy it takes to generate that force. I've never seen that worry expressed in any text on classical mechanics.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.
Yes. He says you can define the proper interval up to the point where the energies are equal.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.
I think you are missing something here. Woodward is merely stating that in each interval, the total input energy must equal the kinetic energy as seen from a frame where the acceleration started from rest. He's basically saying one can always find such an interval for a given energy input and force. He's saying that's all that matters and his critics are simply wrong about CoE. Of course when you add up the effect of the frames the overall kinetic energy builds exponentially while the energy input grows only linearly. But Woodward correctly points out that this is not unique to MET's but is a property built into Classical mechanics for any constant force. It doesn't typically come up since problems usually involve an outside constant force applied in the observer frame and not a force generated in and by the accelerating ship.
The critics have to show how the ship frame knows what it's velocity is and what mechanism is invoked by nature to reduce the force and thus acceleration to comply with the critics view of CoE. If that's been done please point me to it. I don't think it has. At least you are consistent in you belief that because of CoE, the device would not work at all in any frame under any circumstance. But then your problem is to justify why won't work at least as well as radiation pressure.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".
There might be some confusion because of the example you gave of the cannon. In that example, the whole problem was already assumed to be within the bounds of Woodward's condition. In other words, the total input energy does equal the total kinetic energy. Breaking up that problem into steps is not necessary but I did it only to illustrate that it can be done. Yet you seemed to object how I did that which had nothing to do with MET's or EmDrives but just simply mechanics. Each frame is faster than the previous and when you take that into account it all ads up. Woodward says when the input energy is less than the kinetic energy, consider an interval where they are equal. It's just a way of looking at the problem.
In general, equating energy input in the ships frame with kinetic energy in the observers frame is confusing frames in my view. -
#1297 by Bob012345 on 07 Oct, 2016 19:29
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Here's a simple thought experiment:
I have a recoilless gun suspended rigidly in a closed free moving cavity. The gun is fired and the bullet hits the cavity wall at one end.
Hypothesis:
Since there is no recoil felt by the cavity all the energy in the bullet's motion is transformed into movement of the box.
A recoilless gun is a way of allowing the gun itself to remain relatively fixed in space while some gas is ejected out the back to counter the kick of the bullet. If you account for the gas all the momentum will cancel. So, no net motion beyond adjusting the center of mass a bit. Nice try though.
In this simple thought experiment the gun is actually recoilless, so it's meant as more of a tautology aimed at bringing other closed cavity experiments into focus. For example in the emdrive the shape of the cavity combined with the standing waves of varying group velocities and increased Q may dislocate the recoil of the originating waves from the system thereby creating the necessary phantom forces; unlike effects in fluid mechanics. in gun manufacture engineers seek to reduce the effect of recoil by using various designs to redirect gas flow and by engaging moving counterbalances, increasing mass, and so on.
If the gun was actually, literally recoilless (impossible of course) then it would work since you assume you are generating new momentum from the bullet which transmits it to the cavity. In fact if you let the bullet out one end it would not work. But even the adherents of EmDrives or MET devices believe momentum is conserved but debate as to how. -
#1298 by meberbs on 08 Oct, 2016 06:31
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Bob012345, I think you are are missing some important details about what is being said about CoE and why Woodward's paper is very wrong. First from a previous post, you asked:
...
You said you are a physics major, but you clearly have forgotten some of the basic physics relevant to this discussion. P = F*v is literally one form of the definition of power in classical mechanics, so yes, that is true as long as you include all relevant energy.
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?
(Note: I have an engineering degree, which came with a physics heavy curriculum)
There is no problem with constant force, there is a problem with trying to claim that the power required to generate that force is constant. Every text on classical mechanics shows as I stated above, the power to apply constant force is P=F*v. After this it is trivially obvious that the power required to apply a constant force must increase with time."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.
That's what they say. They don't say one can't have a constant force acting as long as you want. That's basic mechanics. They worry about how much energy it takes to generate that force. I've never seen that worry expressed in any text on classical mechanics.
(emphasis mine)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.
Yes. He says you can define the proper interval up to the point where the energies are equal.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.
I think you are missing something here. Woodward is merely stating that in each interval, the total input energy must equal the kinetic energy as seen from a frame where the acceleration started from rest. He's basically saying one can always find such an interval for a given energy input and force. He's saying that's all that matters and his critics are simply wrong about CoE. Of course when you add up the effect of the frames the overall kinetic energy builds exponentially while the energy input grows only linearly. But Woodward correctly points out that this is not unique to MET's but is a property built into Classical mechanics for any constant force. It doesn't typically come up since problems usually involve an outside constant force applied in the observer frame and not a force generated in and by the accelerating ship.
No, he is the one that is wrong about CoE, and very blatantly.
First, he is adding energy across reference frames, which doesn't make sense as you have pointed out before, but you seem to think it is fine when Woodward does it.
Second, finding a single interval where things ad up doesn't change the fact that physics still has to work in between.
Third, "overall kinetic energy builds exponentially while the energy input grows only linearly. But Woodward correctly points out that this is not unique to MET's but is a property built into Classical mechanics for any constant force." is wrong on 2 counts:
-the overally kinetic energy grows quadratically, not exponentially, I assume this is a typo.
-energy input doesn't grow linearly in classical mechanics. There is no such thing as a constant force/power ratio by definition in classical mechanics. The introduction of a device that does directly contradicts classical mechanics.The critics have to show how the ship frame knows what it's velocity is and what mechanism is invoked by nature to reduce the force and thus acceleration to comply with the critics view of CoE. If that's been done please point me to it. I don't think it has. At least you are consistent in you belief that because of CoE, the device would not work at all in any frame under any circumstance. But then your problem is to justify why won't work at least as well as radiation pressure.
You keep asking for the critics to explain the mechanism by which the device could work when the whole point is that a device cannot work as described, or it is a source of free energy. It is up to the person claiming the device works to propose a way that it could obey conservation of energy.
How many times do I have to explain that it is the energy stored in the observer frame in the battery, so it is not confusing frames?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".
There might be some confusion because of the example you gave of the cannon. In that example, the whole problem was already assumed to be within the bounds of Woodward's condition. In other words, the total input energy does equal the total kinetic energy. Breaking up that problem into steps is not necessary but I did it only to illustrate that it can be done. Yet you seemed to object how I did that which had nothing to do with MET's or EmDrives but just simply mechanics. Each frame is faster than the previous and when you take that into account it all ads up. Woodward says when the input energy is less than the kinetic energy, consider an interval where they are equal. It's just a way of looking at the problem.
In general, equating energy input in the ships frame with kinetic energy in the observers frame is confusing frames in my view.
What Woodward does changing references frames every time that energy goes overunity in the last part of his paper is badly confusing frames. Also, just to be sure, energy conservation is broken completely, not just after a certain amount of time. Energy disappearing into nothing is also a problem, and that is what happens as soon as you turn the device on, unless it can be explained where that energy goes (and you can't just say losses, you have to be specific)
And one final time, since this is the point you seem to be missing in this last post: Constant force is not a problem at all in classical mechanics. Constant power providing a constant force is where things break. Woodward's entire paper is trying to show that a device can generate constant force using constant power, but this is simply and completely inconsistent with classical mechanics. (Special relativity tweaks the definition of energy, which in turn tweaks the definition of power, so a constant force/power ratio becomes allowed, but it is strictly limited to a ratio of 1/c, or lower with inefficiencies.) -
#1299 by RERT on 08 Oct, 2016 12:03
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There's been a lot of discussion about conservation of energy and constant acceleration. I was surprised when I looked it up that the fully relativistic treatment is quite tractable. So here it is, for the record.
Suppose a particle experiences a constant acceleration α in its rest frame. Then its position x, velocity v and acceleration a in the frame in which it is initially at rest are:
x=(c²/α)(γ-1) , v=αt/γ , a=α/γ³ , and as a result γ=√(1+α²t²/c²)
Now suppose that the constant proper acceleration is driven by a constant force in the rest frame, and that the force is proportional to the power drain on some battery in the particle:
Nτ=m0α Nτ=λWτ where the subscript Tau indicates quantities in the rest frame.
Now it remains to me peculiar that the mechanical power in the rest frame of the particle is always zero, and so it would seem that energy is always conserved in that frame. I guess that is the heart of the recent discussion here. But we can calculate what happens in the initial rest frame of the particle. Using subscript 't' for that frame we must have
Nt=λWt or the frames are distinguishable. So now we can calculate the mechanical power in the initial frame:
Pt=Ntv=λWtv
But for CoE we require Pt ≤ Wt, i.e. λWtv ≤ Wt, which requires λv≤ 1, or λ ≤ 1/v, and so to be correct in all frames it must be that λ ≤ 1/c
(I said I'd probably made at least one mistake, and I've now found and corrected it above. The problem was that there is an inconsistency between
Nt=Nτ/γ² , Wt=Wτ/γ (since dt/dτ=γ), and Nτ=λWτ
which means that one or both of the two transformations is incorrect - why isn't obvious to me.)
