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#1040 by Rodal on 25 Feb, 2015 12:32
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.....
The mathematical equations I used are in the aforementioned paper, if you object to any of the equations and assumptions we would have a better basis for discussion.
I have to think again about all that. ...
To state that the buckling forces produces a force (towards the right in the picture), in the opposite direction to the measured movement and force of the EM Drive (towards the left in the picture) is, I think, untenable.
Here are two tenable positions that can be taken:
1) The buckling force cannot produce any motion of the center of mass of an unconstrained EM Drive free-floating in space. If one assumes simply supported boundary conditions at the edges of the big diameter plate, just before instability takes place, the end plate is flat and hence the membrane forces are balanced by the rigid circular rim. These membrane forces are directed radially from the rim towards the center, they are self-balanced. There are no forces perpendicular to the plate. If one assumes that the buckling instability takes place instantaneously , superluminally, in the buckled state, there are no membrane forces if the plate is infinitesimally thin, such that it will take whatever bent position is necessary to accommodate the required membrane thermal expansion. Since there are no membrane forces in the final buckled state, there are no forces at the simply supported edges at all in the buckled state. Hence there are no forces to move the EM Drive before and after the buckled state, and that's all, because the buckling motion is assumed to take place instantaneously (hence there are no forces to be considered to take place in between the initial flat state and the final buckled state).
2) If one assumes that buckling cannot take place instantaneously, but that its maximum speed has to be limited by the velocity of stress waves in the material, that is, by the speed of sound in the material, then the minimum time interval for buckling to take place is governed by the speed of sound. In the paper I derived a speed and acceleration for the buckling motion under some assumptions that are explicitly stated (see equations). Those assumptions can indeed be criticized. The equations can also be criticized, and you are invited to do so. The derived speed is below the speed of sound in the material, so it satisfies this physical condition (that the buckling speed should be lower than the speed of sound in the material). After the onset of buckling, and during this (short time) buckling motion, the membrane force decreases, from a maximum at the onset of buckling to a minimum at the final buckling state. During this (short time) the plate is in a bent condition, with increased bending shape with time. At the edges of the plate, the bending of the plate produces a slope at the simply supported edges. The membrane force at the edges can be decomposed into two directions: A) a component that is in the original flat direction. This component is self-balanced. B) The other force vector component is directed in the same direction (towards the left) as the motion of the EM Drive. This is a very short-time impulsive force, calculated to be of the same magnitude as the initial transient rise force in the experiments. -
#1041 by Mulletron on 25 Feb, 2015 12:36
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We're being talked about over there.
http://forums.sufficientvelocity.com/threads/em-drive-system-further-confirmed.14456/page-9
The thread starts with some mathematical analysis of the Dr. White conjecture which is pretty good. At page 9 they tear into me. They think the magnetic field is literally external to the copper frustum. Later they rip into MiHsC. Glad to see people are fans. -
#1042 by MazonDel on 25 Feb, 2015 14:18
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Greetings all, new to the forums.
I am a Robotics Engineer and so not nearly as well versed on physics as you all and I was kind of hoping for a clarification on the current discussion about the thermal buckling and such.
My current understanding of the how the experiment is measured (based on the image that Rodal has posted regarding the Areas of Thermal Expansion) is that any thrust that is generated will cause the torque pendulum to twist, and this is what is being measured (via the laser distance interferometer). I note that the center of mass of the frustum is centered over the toque pendulum.
From what I have read it seems as though the buckling of the frustum wall can be a source of some initial force, but cannot account for the constant thrust detected over the course of the experiment. What I have been wondering is if this buckling could cause enough of a shift of the COM towards the left to throw off the balance? As I have seen some discussions about the COM after the posts that made me think of this, I am trying to confirm that this is at some level what is being discussed.
Thanks! Keep up the good work! -
#1043 by Flyby on 25 Feb, 2015 14:56
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I've probably got the thickness of the frustum walls wrong but it's difficult to gauge because where the frustum walls flare out at the bottom and attach to the base cap they appear to be thinner than where they flare out at the top and attach to the top cap. I wish the photo was from a lower point of view so we could see beneath the top rim of the thing.
At any rate I can't get the dimensions of the inner cavity in the photo to jive with Shawyer's numbers. So I'm at a loss. I will say that it's a pretty remarkable coincidence that his numbers work precisely with the measurements in the photo if we use the diameters of the base and top caps, rather than the estimated diameters of the inner cavity, whatever they may be.
I'll post the chart just for the sake of having it visible. Like I said, the two diameter scales must be incorrect. My best guess for "overall" diameter of the top, would be 189mm.
As you already started on this before i could actually find time, i did no longer bother to continue on this, but applying perspective rules on a picture needs some corrections first...
The image suffers from 3 type of distortions :
-there is a distinct "barrel" distortion, very noticeable on the back-wall horizontal join, which shows as a curve instead of a straight line. the further away from the center of the image, the more severe the distortion. There exists software to correct this with 100% accuracy (PTLens), but the problem is that you need to know the camera type and lens type...
-There is a tilt-down distortion (up-down), causing all verticals to skew together. The further down of the horizon line, the bigger the distortion. Again, you can correct this with software, but due to the yawn (left/right movement) of the camera the vertical lines cant be corrected symmetrically. But I think i could come up with an approximation to solve that problem
-Then there is a yawn-distortion, this causes the horizontal axis to shift. Moving the camera target to the right (out of the center of the image) cause the horizontal axis to shift left-down and right-up. So your axis of the ellipses (circle in perspective) should not be horizontal, but slightly angled , due to the camera target not being in the dead center.
If you're interested, I'll finish up a corrected image this evening which should be a "close enough" approximation. Too bad we do not have any information about the camera and the used lens... -
#1044 by frobnicat on 25 Feb, 2015 15:15
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(Bold added for emphasis)....
That's incorrect.
Later on thread 2 (related to spaceflight) while explaining why the buckling analysed by Rodal ... would make a ... thrust in opposite direction to the one observed by the pendulum, ...
Maybe, but it appears Paul March in this post says that it is opposite, and I don't see that this disagreement was acknowledged ...
The fact that the thermal instability (buckling force) cannot be used to nullify the EM Drive thrust, was explicitly acknowledged by me and explained here http://forum.nasaspaceflight.com/index.php?topic=36313.msg1336196#msg1336196:QuoteDr. Rodal analyzed possible thermal instability (thermal buckling of the flat ends) as a cause for the measured thrust and reported this at NSF and at ResearchGate (https://www.researchgate.net/publication/268804028_NASA%27S_MICROWAVE_PROPELLANT-LESS_THRUSTER_ANOMALOUS_RESULTS_CONSIDERATION_OF_A_THERMO-MECHANICAL_EFFECT). A thermo-mechanical effect (thermal buckling) is shown that occurs in less than 1 second (for the copper thickness employed for the microwave cavity), with a temperature increase of a degree C or less and that results in forces of the same magnitude as reportedly measured by NASA. Moreover, this thermal instability produces forces in the same direction as measured, and it will occur in a vacuum (since the heating can be due either to induction heating from the axial magnetic field in a TE mode or resistive heating due to the axial electric field in a TM mode). However, this effect can only explain the initial impulsive force and cannot explain the longer 30 to 40 second measured force. Thus the thrust force measured for up to 40 second is not nullified by this explanation either.
(Bold and color added for emphasis)
Thermal expansion effect as posited by a team from Oak Ridge National Labs for another propellant-less set of experiments was also eliminated as a possible source by the NSF contributors because it would result in forces in the complete opposite direction as the forces measured by NASA.
That "the thermal instability (buckling force) cannot be used to nullify the EM Drive thrust" was acknowledged by everyone, I agree, but on the ground of the duration of the observed thrust alone. Disagreement (respectful) is not on that. Maybe it would be best if Paul March himself confirmed or restated, or some other native Newtonian and English reader could confirm the same reading ( I may have misinterpreted ) but appears to me (and makes sense) that the actual disagreement is explicit enough :...
Now Newton's third law still states that for every action there is an equal and opposite reaction. So when the copper frustum's large OD end-cap's prompt and inward oil canning action, followed by the slower frustum cone thermal expansions, they both push the copper frustum's Center of Mass (CoM) to the left as viewed from the front of the Eagleworks' vacuum chamber looking back at the test article and torque pendulum, while noting how the copper frustum is bolted on to the T.P.. These thermally induced actions to the left requires the torque pendulum's arm to move to the right to maintain the balance of the torque pendulum's arm in the lab's 1.0 gee gravity field, since we also use the Earth's g-field to help null the pendulum's movements.
...
Don't you read like me that Paul March says that the effect to be expected (by a short duration buckling) by a displacement of the centre of mass of the end-cap to the left is a displacement of the arm to the right, at least when things start to move ? Doesn't it make sense from an action/reaction momentum conservation standpoint ? So what I say is that there is disagreement not on the fact that buckling can't account for duration of observed pendulum's displacement (induced by forces or apparent transient forces when the frustum centre of mass moves relative to fixation point on pendulum's arm) but on the fact that the very direction of such effect would be expected to push the pendulum's arm to the right, at least initially : that would mean a dip (step down) on the chart, not a rise as observed (step up, increasing distance, arm moving to left).
So to make my position explicit in regard to observables : I predict that the consequence of an "inward oil canning" is a sharp plunge on the record charts (or to add a sharp plunge term to other possible effects that move the arm). And I also think Paul March sees that the same way. Will take time to further reply of your next post, unless other contributors feel it clutters the main thread and we could discuss that by PM.
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#1045 by lasoi on 25 Feb, 2015 15:24
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@Flyby
The second type distortion you refer to is a perspective distortion. All of that is true and in addition the camera wasn't held on a flat level when the photo was taken. I've rotated it a few degrees clockwise to compensate for this problem but I might still be off of the horizontal by a fraction of a degree. At any rate my experience is that these distortions are so small that they don't significantly affect the accuracy of the estimate, which is a just that, a rough estimate give or take a couple mms.
If you have the software and are able to correct for some of the distortion, I'll recalculate. I'm not sure that it matters much but I'd be happy to do so. -
#1046 by Rodal on 25 Feb, 2015 15:29
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....
So to make my position explicit in regard to observables : I predict that the consequence of an "inward oil canning" is a sharp plunge on the record charts (or to add a sharp plunge term to other possible effects that move the arm). And I also think Paul March sees that the same way. Will take time to further reply of your next post, unless other contributors feel it clutters the main thread and we could discuss that by PM.
I explcitly gave you analytical arguments that stating that the buckling force would be to the right, as you maintain, is untenable, in this post: http://forum.nasaspaceflight.com/index.php?topic=36313.msg1337106#msg1337106
Tenable positions are that for an EM Drive unrestrained in space either 1) there is no buckling force on the center of mass or 2) there is a very short-time minute buckling force to the left, in the same direction as measured.
You are not addressing my arguments in this post http://forum.nasaspaceflight.com/index.php?topic=36313.msg1337106#msg1337106 but instead re-stating your reading of Paul March (he can argue this for himself, if he wants) and re-stating your belief. It would be more constructive to analytically address the arguments that I posted here http://forum.nasaspaceflight.com/index.php?topic=36313.msg1337106#msg1337106 and/or the equations and assumptions in my paper, rather than repeatedly stating your prediction. -
#1047 by frobnicat on 25 Feb, 2015 15:34
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Quickly, before I address the rest of your post in more depth... I wouldn't insist on that point if I didn't feel that the opposite is untenable.....
The mathematical equations I used are in the aforementioned paper, if you object to any of the equations and assumptions we would have a better basis for discussion.
I have to think again about all that. ...
To state that the buckling forces produces a force (towards the right in the picture), in the opposite direction to the measured movement and force of the EM Drive (towards the left in the picture) is, I think, untenable.
...
I state that the buckling forces impart a (transient) momentum transfer from the frustum to balance's arm and that this transient momentum transfer is to the right, as a result the balance arm would like to recoil (move) to the right. If this buckling was the only action taking place, the observed step would be a fall, not a rise. Since what is observed is a rise, there must be other action taking place and imparting more momentum to the left than the buckling imparts to the right. Force is only "transferred momentum flux" but maybe my point is more clear if you see it as momentum transfer rather than force.
Sorry, partial answers encourages posting past each other, I will address you points as analytically as possible...
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#1048 by Notsosureofit on 25 Feb, 2015 16:20
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FYI
Anytime I hear the term "Two-Photon Loss"...........
http://arxiv.org/pdf/1412.4633v1.pdf
"In conclusion, we have realized a non-linearly driven-dissipative oscillator which spontaneously evolves towards the quantum manifold spanned by two coherent states."
And
"This was achieved by attaining the regime in which the photon pair exchange rate is of the same order as the single photon decay rate. The ratio between these two rates can be further improved within the present technology by using a higher Q oscillator and increasing its on-linear coupling to the bath."
For whatever it's worth. (requires non-linear elements. surface ? also ignored some terms ?) -
#1049 by Rodal on 25 Feb, 2015 17:11
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FYI
Anytime I hear the term "Two-Photon Loss"...........
http://arxiv.org/pdf/1412.4633v1.pdf
"In conclusion, we have realized a non-linearly driven-dissipative oscillator which spontaneously evolves towards the quantum manifold spanned by two coherent states."
And
"This was achieved by attaining the regime in which the photon pair exchange rate is of the same order as the single photon decay rate. The ratio between these two rates can be further improved within the present technology by using a higher Q oscillator and increasing its on-linear coupling to the bath."
For whatever it's worth. (requires non-linear elements. surface ? also ignored some terms ?)
Excellent point. I completely agree.
Concerning the dielectric: perhaps it is not just a question of the well knonw dielectric constant: relative permittivity. Perhaps there are nonlinear terms involving material constants characterizing the nonlinear material behavior of the dielectric that are not usually measured, (because such nonlinear terms are usually neglected) and hence we have not taken into account. Helical, chiral anisotropy of the dielectric, as mentioned by Mulletron, is always an issue, as the analysis has so far been isotropic. Anisotropic analysis would involve many more material constants. -
#1050 by Flyby on 25 Feb, 2015 17:31
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@Flyby
Well, it might be interesting to pretend we do not know the real dimensions and see if such a photo-touch up can be used to make more accurate measures.. or... that it proves to be totally useless...
The second type distortion you refer to is a perspective distortion. All of that is true and in addition the camera wasn't held on a flat level when the photo was taken. I've rotated it a few degrees clockwise to compensate for this problem but I might still be off of the horizontal by a fraction of a degree. At any rate my experience is that these distortions are so small that they don't significantly affect the accuracy of the estimate, which is a just that, a rough estimate give or take a couple mms.
If you have the software and are able to correct for some of the distortion, I'll recalculate. I'm not sure that it matters much but I'd be happy to do so.
I do feel a bit uncomfortable about the tilt-correction, but ignoring it would mean you have to resort to a 3 point perspective, which is highly uncomfortable (horizon points far far away) considering you have a near 1-point perspective...
but... let's give it a spin...if it hits a wall, then we'll know...
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#1051 by RanulfC on 25 Feb, 2015 17:51
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FYI
Anytime I hear the term "Two-Photon Loss"...........
Strangely enough every time I hear/see that my mind switches it to:
"Two Photon's enter... One Photon leaves" and Tina Turner singing in the background...
Weird
Randy -
#1052 by Rodal on 25 Feb, 2015 18:40
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@Flyby
Well, it might be interesting to pretend we do not know the real dimensions and see if such a photo-touch up can be used to make more accurate measures.. or... that it proves to be totally useless...
The second type distortion you refer to is a perspective distortion. All of that is true and in addition the camera wasn't held on a flat level when the photo was taken. I've rotated it a few degrees clockwise to compensate for this problem but I might still be off of the horizontal by a fraction of a degree. At any rate my experience is that these distortions are so small that they don't significantly affect the accuracy of the estimate, which is a just that, a rough estimate give or take a couple mms.
If you have the software and are able to correct for some of the distortion, I'll recalculate. I'm not sure that it matters much but I'd be happy to do so.
I do feel a bit uncomfortable about the tilt-correction, but ignoring it would mean you have to resort to a 3 point perspective, which is highly uncomfortable (horizon points far far away) considering you have a near 1-point perspective...
but... let's give it a spin...if it hits a wall, then we'll know...
What software did you use to do that?
I have in the past used Mathematica to transform images like that but one has to write Mathematica code to get it just right like you did. -
#1053 by frobnicat on 25 Feb, 2015 18:51
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.....
The mathematical equations I used are in the aforementioned paper, if you object to any of the equations and assumptions we would have a better basis for discussion.
I have to think again about all that. ...
Well first, thanks for taking time trying to answer my concerns, and it's only natural you link back to your previous work on the topic for me to work my argument. But frankly I would have a hard time following all this work (thermal buckling of end caps) on details. I'll try an illustrated qualitative attempt at explaining my view and we will see if it's worth going on to equations.Quote from: RodalTo state that the buckling forces produces a force (towards the right in the picture), in the opposite direction to the measured movement and force of the EM Drive (towards the left in the picture) is, I think, untenable.
Here are two tenable positions that can be taken:
1) The buckling force cannot produce any motion of the center of mass of an unconstrained EM Drive free-floating in space. If one assumes simply supported boundary conditions at the edges of the big diameter plate, just before instability takes place, the end plate is flat and hence the membrane forces are balanced by the rigid circular rim. These membrane forces are directed radially from the rim towards the center, they are self-balanced. There are no forces perpendicular to the plate. If one assumes that the buckling instability takes place instantaneously , superluminally, in the buckled state, there are no membrane forces if the plate is infinitesimally thin, such that it will take whatever bent position is necessary to accommodate the required membrane thermal expansion. Since there are no membrane forces in the final buckled state, there are no forces at the simply supported edges at all in the buckled state. Hence there are no forces to move the EM Drive before and after the buckled state, and that's all, because the buckling motion is assumed to take place instantaneously (hence there are no forces to be considered to take place in between the initial flat state and the final buckled state).
This would be a model very far from reality. In all cause, if a part mass is sent to movement (relative to a free floating assembly) instantaneously rather than progressively, as a limit case, then one would observe the same total recoil for the rest of assembly (at equal end relative speed) only that the force would be concentrated in a Dirac delta of same integrated momentum value. Not understanding where you are heading with that one. We are (well, I am) concerned by the first initial answer of the pendulum from the initial part of buckling (when there is a growing acceleration of mass of plate relative to frustum). The limit case of instantaneous move would make hard to see that we have first a recoil (opposite direction from that the part is leaving) then a recoil in the same direction as the part's relative speed when it breaks (if it is not forever leaving the assembly, at some point it will have to break). The integrated sum of the observed force (of the moving object on rest of assembly) for a bound object between an initial rest position and a final rest position would be 0, is it what you are saying in green ?Quote2) If one assumes that buckling cannot take place instantaneously, but that its maximum speed has to be limited by the velocity of stress waves in the material, that is, by the speed of sound in the material, then the minimum time interval for buckling to take place is governed by the speed of sound. In the paper I derived a speed and acceleration for the buckling motion under some assumptions that are explicitly stated (see equations). Those assumptions can indeed be criticized. The equations can also be criticized, and you are invited to do so. The derived speed is below the speed of sound in the material, so it satisfies this physical condition (that the buckling speed should be lower than the speed of sound in the material). After the onset of buckling, and during this (short time) buckling motion, the membrane force decreases, from a maximum at the onset of buckling to a minimum at the final buckling state. During this (short time) the plate is in a bent condition, with increased bending shape with time. At the edges of the plate, the bending of the plate produces a slope at the simply supported edges. The membrane force at the edges can be decomposed into two directions: A) a component that is in the original flat direction. This component is self-balanced. B) The other force vector component is directed in the same direction (towards the left) as the motion of the EM Drive. This is a very short-time impulsive force, calculated to be of the same magnitude as the initial transient rise force in the experiments.
If necessary I will try to dive in you equations but please try to understand what I'm saying, the specifics don't make difference for the argument.
On attached pictures I will use a simplified model of buckling end Plate as a pure mass P connected to massless springs S all around. Those springs hook on the Rim R (lighter blue) that is part of a solid bloc comprised of the frustum + arm (where I assume all in blue has no deformation). This could be freely floating in deep space but I figure the connection to Chamber C through a spring B (Balance's flexure pivots that tend to keep the system at a given rest position figured by 0). This model's springs S are pure tension/compression with no flexure stiffness, this is obviously oversimplified for the final rest position but right now I'm not interested in the final rest position (when buckling is ended). This system does exhibit buckling.
That's what I have in mind when I say that inward buckling to the left will appear as a kick to the right for the system :
1. System at rest before power on, Springs at their rest length, no forces (gravity neglected).
2. Heating of springs increase their rest length, they are now too short, they work in compression, but the symmetry makes opposite forces equilibrate, with no horizontal components. Note the forces of mass P on rim R are equal but opposite to the forces of rim R on mass P (point of application shifted for better clarity). No movement of arm yet, d2=d1.
3. The symmetry breaks leftward, the buckling starts. Mass P move to the left, in order for this move to occur, at least initially (regardless of quantitative equations or details of model) the mass P has to accelerate to the left (red arrow to the left with high magnitude). This is consistent with the now inclined springs still working in compression : the sum of the Frp forces (rim on plate CoM P) cancel radially but are indeed axially oriented to the left. Conversely the sum of the Fpr forces (plate's CoM P on rim) cancel radially but are axially oriented to the right. Action/reaction, a mass starts to leave leftward because the system push it leftward, the leaving mass "pushes back" on the system rightward. The whole balance system sees a force exerted on it that is oriented to the right. The balance departs from it's initial rest position to the right. The restoring torque of the balance flexure pivots will start to oppose this movement ( Fcr chamber on arm ) but significant displacement of balance has to occur first, Fcr < Frp in magnitude (Fcr is proportional to the deviation of balance relative to rest position).
d3<d2
4. As the buckling continues at its finite speed (not only because of inertia of mass P but also because the heating, ie increase of springs' rest length, is progressive and accompanies the movement) the acceleration of P to the left will slow down as the effective length of springs start to match their (ever increasing) rest length. Horizontal sum of Fpr that pushes to the right decreases, and the pendulum's arm travelled enough to the right that the restoring force Fcr now can be higher magnitude than Fpr, the arm is still travelling to the right but is now slowing down (acceleration to the left, as per the vector sum Fpr+Fcr)
5. ... I'm exhausted, to be continued (if worth)
From 3 to 4 : the consequence of an "inward oil canning" alone is a kick to the right initially. Do I have to write down equations ? For me this is more a matter of proper forces orientation conventions than anything else. I hope this is clear enough as to why I said that buckling would kick opposite to the observed thrust. So, what's wrong with this way of seeing orientations ?
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#1054 by lasoi on 25 Feb, 2015 19:02
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In straightening up the vertical lines, the program created more distortion than it corrected. The bottom left side of the engine is pulled down to the left, noticeably.
I would do more calculations but unfortunately my day job is preventing me at the moment. I'd be interested to see any estimates you might get from the altered photo however. -
#1055 by Rodal on 25 Feb, 2015 19:19
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...
Get rid of the pendulum and everything else for the time being, as it further muddles the picture.
That's what I have in mind when I say that inward buckling to the left will appear as a kick to the right for the system :
....
From 3 to 4 : the consequence of an "inward oil canning" alone is a kick to the right initially. Do I have to write down equations ? For me this is more a matter of proper forces orientation conventions than anything else. I hope this is clear enough as to why I said that buckling would kick opposite to the observed thrust. So, what's wrong with this way of seeing orientations ?
Just consider a free-body diagram for the EM Drive. http://en.wikipedia.org/wiki/Free_body_diagram
Just consider the flat plate and a rigid ring around it and ask yourself in which direction you have to push the plate's center for the plate's center to move towards the left. Of course that one has to have a force moving to the left.
All those words and images above are associated with a discussion of a reaction from the pendulum.
You could use all those words to also talk about a reaction from the pendulum concerning EM Drive's thrust, and get into a similar mambo jambo as Shawyer's discussion of thrust in EM Drives.
First you have to address in which direction is the movement of the buckled shape (it is towards the left), and when it moves towards the left , in which direction it is pulling the supports (towards the left). The buckled material cannot move towards the left without pulling on the supports towards the left: it is all connected.
What you are discussing instead is what is the effect of the buckling force pulling the supports to the left, on the pendulum. That is an entirely different question. You can ask yourself exactly the same question regarding what is the effect of the EM Drive moving towards the left, with a force pulling towards the left, on the pendulum.
After you have accepted that fact, you can talk about the reaction, which involves elastic deformation, and it is also present for the EM Drive thrust.
And, it is unclear why you think that this is so relevant, because the flat plate can theoretically buckle towards the left or towards the right (if there would be no plastic insulation on the outside and if the copper would be very thin) depending on initial imperfections. An even if you think that when the buckled plate moves towards the left it gives a force towards the right (which is a misscommunication based on the fact that you are focusing on the pendulum's reaction rather than the buckling force itself, but for argument sake's) then even per your admission then if the plate buckles towards the right according to you would produce a force towards the left.
It seems to me that if you wanted to argue about this for intellectual reasons, a much safer ground would be to say that the buckling force is self-balanced and that it produces no forces on the center of mass, either to the left or to the right (for the aforementioned reasons in my above post).
EDIT: And having said that, the thermal expansion explanation by Oak Ridge does not make any sense in this case for the HD PE (to think that an unrestrained, homogeneous, isotropic, free to expand material, will produce a force when expanding. Thermal expansion changes the VOLUME of a material. The mass stays the same. If free to expand, then the density of the HD PE will change (larger volume, same mass = lower density). Thermal expansion produces forces only when there is a thermal gradient through the material or the material is anisotropic, or the most general case: when the material is constrained so that it cannot expand !!!!! )
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#1056 by frobnicat on 25 Feb, 2015 23:37
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...
Get rid of the pendulum and everything else for the time being, as it further muddles the picture. Just consider the flat plate and a rigid ring around it and ask yourself in which direction you have to push the plate's center for the plate's center to move towards the left. Of course that one has to have a force moving to the left.
That's what I have in mind when I say that inward buckling to the left will appear as a kick to the right for the system :
....
From 3 to 4 : the consequence of an "inward oil canning" alone is a kick to the right initially. Do I have to write down equations ? For me this is more a matter of proper forces orientation conventions than anything else. I hope this is clear enough as to why I said that buckling would kick opposite to the observed thrust. So, what's wrong with this way of seeing orientations ?
To make the plate start to move to the left, I have to push the plate to the left. Certainly.
Now, on what else am I pushing when I push on the plate ? Either there is nothing else, in this case I am only "pushing on my inertia" i.e. recoil to the right. Or I am also pushing with my feet on a wall of the spacecraft and in this case me+spacecraft recoils to the right (a bit less, but still).QuoteAll those words and images above are associated with a discussion of a reaction from the pendulum.
You could use all those words to also talk about a reaction from the pendulum concerning EM Drive's thrust, and get into a similar mambo jambo as Shawyer's discussion of thrust in EM Drives.
No, I am very clear and very standard as for the conventions of usual solid mechanics. When an object A and an object B exchange a force, the force A on B is noted Fab. The observable acceleration of object B (assuming no other forces) will be in the same direction as Fab, this leaves no mambo jambo to speak of. Is it a test ?
Now Newton's third law still states that for every action there is an equal and opposite reaction.
We have a clear unambiguous way to tell the orientations conventions from mambo jambo (observable acceleration) and we have Fab = -Fba (vectors).
If the pendulums restoring spring creates confusion, the very same arguments as for the direction of the start of the apparent displacement of rim of a free floating frustum are valid, see attached picture with pendulum's spring removed.
Note how a displacement of green mass (representing the plate) to the left implies at least initially an acceleration to the left. Being only linked to the rim, this acceleration of the plate must be due to a force F_rim_on_plate oriented to the left, and by Newton's third law there is a force F_plate_on_rim that is oriented to the right, and by definition (of the above mentioned usual convention) the rim and rest of frustum (assumed solid) accelerates to the right. Hope some passer-bys will enjoy.QuoteFirst you have to address in which direction is the movement of the buckled shape (it is towards the left), and when it moves towards the left , in which direction it is pulling the supports (towards the left). The buckled material cannot move towards the left without pulling on the supports towards the left: it is all connected.
Yes the movement is toward the left, and when it move toward the left it is not pulling the support, because if it would be pulling the support when departing (to the left) of its initial rest position then it would not be departing from its rest position in the first place ! That would be the case if the springs where in tension, and a coplanar plate were a stable situation (like for a drum). Only because the rise in temperature increases the rest length of the springs the springs will be compressed, and the situation becomes unstable as the spring want to push.
Now all right, this is only a toy model, the real situation is much more complex because plate's weight is distributed on the whole surface, because plate is very stiff in compression and more supple in flexure so that for instance when a static equilibrium is reached (when there is no more buckling movement) there are still radial forces (but no longer axial ones, that we agree) while in my toy model the complete absence of flexure stiffness leaves no radial forces when the static equilibrium is reached. + copper/epoxy left/right asymmetry of the sheet... damping...
But even if considering a quasi-static evolution in the real situation, where the rise of temperature is slow enough for the plate to always be near static equilibrium, near equilibrium is not exactly equilibrium. That's the whole point : however small is the deviation from perfect static equilibrium, this is this deviation that makes inertial things move in the first place. And whenever a part of mass m of a free floating system moves to the left a distance x, the rest of the system of mass M recoils to the right a distance X=x*m/M (where x and X are measured relative to initial inertial rest frame of the free floating system before the part started to move).QuoteWhat you are discussing instead is what is the effect of the buckling force pulling the supports to the left, on the pendulum. That is an entirely different question. You can ask yourself exactly the same question regarding what is the effect of the EM Drive moving towards the left, with a force pulling towards the left, on the pendulum.
? What I was discussing is the effect of the buckling recoil (momentum...) pushing the supports to the right, transiently. Main point. Then the plate would slow down, giving an equal amount of momentum in the opposite direction (to the left) and neutralise the speed acquired by the initial kick. That's for a free floating device.QuoteAfter you have accepted that fact, you can talk about the reaction, which involves elastic deformation, and it is also present for the EM Drive thrust.
And, it is unclear why you think that this is so relevant, because the flat plate can theoretically buckle towards the left or towards the right (if there would be no plastic insulation on the outside and if the copper would be very thin) depending on initial imperfections. An even if you think that when the buckled plate moves towards the left it gives a force towards the right (which is a misscommunication based on the fact that you are focusing on the pendulum's reaction rather than the buckling force itself, but for argument sake's) then even per your admission then if the plate buckles towards the right according to you would produce a force towards the left.
I do think that for the buckled plate to move (have a speed) toward the left it has to accelerate to the left, and that when the buckled plate accelerates towards the left it gives to the frustum (and attached mass) an apparent force towards the right (the frustum accelerates to the right, regardless of later dynamics due to it being attached to position restoring torque, ie a pendulum, so please forget pendulum if you want, my argument still holds).
This is so relevant at least because we are discussing here a potentially groundbreaking device that has an enigmatic relation to momentum conservation : the least we can do is to put such mundane engineering matter straight, on a forum read by people who breath (conventional) action/reaction... and contributions are useful if they stand correct (or can be read correctly if it is case of miscommunication).
If the buckling effect adds an initial downward step to the other effects, this is good to know, no ? An other interesting effect would look differently (ie. more square) when this negative term would be subtracted...QuoteIt seems to me that if you wanted to argue about this for intellectual reasons, a much safer ground would be to say that the buckling force is self-balanced and that it produces no forces on the center of mass, either to the left or to the right (for the aforementioned reasons in my above post).
By self-balanced you mean quasi-static, always near equilibrium ?
Anyway, if you prefer (and yes, this is probably more elegant) :
Self repeat : whenever a part of mass m of a free floating system moves to the left a distance x, the rest of the system of mass M recoils to the right a distance X=x*m/M (where x and X are measured relative to initial inertial rest frame of the free floating system before the part started to move).
Because : the centre of mass of the whole system m+M can't depart from its initial inertial trajectory.QuoteEDIT: And having said that, the thermal expansion explanation by Oak Ridge does not make any sense in this case for the HD PE (to think that an unrestrained, homogeneous, isotropic, free to expand material, will produce a force when expanding. Thermal expansion changes the VOLUME of a material. The mass stays the same. If free to expand, then the density of the HD PE will change (larger volume, same mass = lower density). Thermal expansion produces forces only when there is a thermal gradient through the material or the material is anisotropic, or the most general case: when the material is constrained so that it cannot expand !!!!! )
When thermal expansion displaces a part of a system relative to the rest of a system, the rest of the system will recoil. Whatever displaces a part of a system relative to the rest of the system, this displacement implies a force F_system_part. The rest of the system will recoil. Recoil is the acceleration due to the opposite force (F_part_system). An unrestrained, homogeneous, isotropic, free to expand but in one direction material, will produce a force when expanding against the wall against which it rests.
Why is it so hard to reach consensus now ?
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#1057 by Rodal on 26 Feb, 2015 01:14
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I like your drawings with the buckled condition idealized as a beam with three hinges: one hinge at the center and a hinge at each end.
Thank you very much for taking the time to make those drawings.
I should have looked at that initially. I agree with your drawing. To the extent that my prior wording disagreed with your drawing, my prior words were incorrect, when and if they referred to the thermal force Fpr.
Let's then address what happens in the opposite case that the plate instead buckles to the right:
The flat plate can theoretically buckle towards the left or towards the right (if the copper is thin enough) depending on initial imperfections. If when the buckled plate moves towards the left it gives a force towards the right then if the plate buckles towards the right it would produce a force towards the left, in the same direction as the EM Drive's motion, do you agree?.
Then, if this buckling analysis is correct, it gives transient force that is towards the left when the plate buckles towards the right and it gives a transient force to the right when the plate buckles towards the left.
Now I have to give further thought to which direction the plate buckles when heated. The real plate has a neutral surface that is not at the middle of the cross-section. It is really a bi-material thermostat: with the epoxy expanding much more than the very thin copper coating. If this unsymmetric laminate would be exposed to a uniform temperature it would expand towards the outside, producing a force towards the left, like the EM Drive force. The question is what happens under the superposed thermal gradient through the thickness. The IR Camera shows very pronounced heating on the outside surface....
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#1058 by Rodal on 26 Feb, 2015 01:33
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With respect to the HD PE dielectric, now I see your thinking. I think you think of it as a dynamic problem, for example, if instead of the HD PE we would have an explosive inside the cavity glued to the left wall, the effect of the explosive would be to force the wall towards the left. When thermal expansion displaces a part of a system relative to the rest of a system, the rest of the system will recoil. Whatever displaces a part of a system relative to the rest of the system, this displacement implies a force F_system_part. The rest of the system will recoil. Recoil is the acceleration due to the opposite force (F_part_system). An unrestrained, homogeneous, isotropic, free to expand but in one direction material, will produce a force when expanding against the wall against which it rests.
Why is it so hard to reach consensus now ?
However, thermal expansion does not work like that. There are no dynamic forces caused by a material experiencing thermal expansion such that it is free to expand. Thermal expansion is a very slow process that does not involve second order derivatives with respect to time. If a uniaxial isotropic material is glued to a wall, it will just expand without producing any force whatsoever on the wall it is glued to.
If free, it will just expand, strain = alpha *deltaT.
No stress if it is free to expand.
Can't use thermal expansion as a form of propellant-less propulsion.
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#1059 by ThinkerX on 26 Feb, 2015 03:41
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Going from the last eight or ten posts, this device really needs to be tested in a vacuum with a different measuring device, just to put paid to these 'tilt' / thermal issues.
When thermal expansion displaces a part of a system relative to the rest of a system, the rest of the system will recoil. Whatever displaces a part of a system relative to the rest of the system, this displacement implies a force F_system_part. The rest of the system will recoil. Recoil is the acceleration due to the opposite force (F_part_system). An unrestrained, homogeneous, isotropic, free to expand but in one direction material, will produce a force when expanding against the wall against which it rests.