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#2700 by meberbs on 10 Nov, 2017 16:12
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I have to agree with you. This image is confusing to the extreme. I was taught, if I push on a wall, there is an equal and opposite force acting on my hand, and the wall does not move. The amount of "work" done is zero. Work is NET Force x Distance and in this case, the NET force is zero.
What are you saying here?
Alternatively, if I am exerting a force on an object that is free to accelerate, the NET force cannot be zero because work is being done to accelerate it. Work will still be the NET Force x Distance, so therefore the NET force is equal to the Applied force, and it is not zero.
When you push on something, whether it is a wall or something that moves, in either case there is an equal and opposite force on your hand. This has nothing to do with why the wall doesn't move, which is instead due to the force exerted on the wall by the ground balancing the force you are applying. (I know you can take it from here with the rest of the circle of forces between you and the ground)
While I agree that D'Alembert's principle can cause confusion, as Rodal has pointed out, it is a mathematically useful formulation of dynamics. It basically says that an acceleration will exist that balances out any otherwise unbalanced force. It seems a bit inside out compared to the more common Newtonian description, but it is just as valid. -
#2701 by Bob012345 on 10 Nov, 2017 17:45
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...If the total force acting on the body were zero, the body wouldn't accelerate. This concept is a subtle one to explain what it looks like if you are that body and your relative velocity to yourself is of course zero but in reality, it's confusing and unnecessary. Mass is a measure of inertia. If inertia were zero acceleration would be infinite for any force applied. Newton is enough.
D'Alembert's principle is routinely used in Aerospace Engineering for dynamic design of aerospace vehicles. It was used in the Apollo program and continues to be used to this date in most Finite Element Analysis programs, including NASTRAN, ANSYS etc. to calculate dynamics of multi-degree of freedom deformable bodies.
That's how NASTRAN analyzed dynamics problems.
I get the impression that you have not been well exposed to D'Alembert's principle and why it is used. (Your comparison with Zeno and other comments are not pertinent.) People that have been involved with dynamic analysis of actual (multi-degree of freedom, deformable) aerospace vehicles understand the usefulness of D'Alembert's principle, and they do not see it as a competitor to Newton's analysis, it just facilitates the analysis in a similar fashion as Lagrange's formulation helps to analyze general problems.
If one is not going to be involved in the actual analysis of dynamics of aerospace vehicles, and just discusses simple lumped mass, single degree of freedom problems, then Newton's formulation suffices. Yes, for Physics 1.00, one starts with Newton's formulation, but for further dynamics classes one progresses to the use of D'Alembert's principle, and variational principles for very good reasons.
See https://www.colorado.edu/engineering/CAS/KCPark.d/KCParkHome.d/lectures.d/usnccm2007.pdf whether that helps (I did not go to Colorado, this is just the first thing that came in google )
Same reason why Lagrange's formulation is useful.
To appreciate the usefulness of Lagrange's formulation one has to analyze multi-degree of freedom general problems. Otherwise Lagrange's formulation may look like overkill.
I agree that to understand that Shawyer's analysis is flawed, D'Alembert's principle is overkill. Just Newton's law is enough.
Thanks. I get the concept for dynamic analysis but we were discussing a static problem and the slide was not quite accurate. The Zeno comment was meant to be a joke. The total force on the body can be set as zero for analysis purposes but it's not really zero so that statement in the slide should have more careful. As I pointed out, I think that the issue of what a scale shows was not necessarily wrong and that was the issue at hand. -
#2702 by Rodal on 10 Nov, 2017 17:53
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...
Thanks for your reply.
Thanks. I get the concept for dynamic analysis but we were discussing a static problem and the slide was not quite accurate. The Zeno comment was meant to be a joke. The total force on the body can be set as zero for analysis purposes but it's not really zero so that statement in the slide should have more careful. As I pointed out, I think that the issue of what a scale shows was not necessarily wrong and that was the issue at hand.
Concerning <<we were discussing a static problem>>.
If there is acceleration of the EM Drive involved, the problem is not static but dynamic.
If there is a pendulum involved of any kind (torsional, swinging, etc.) the problem is not static either, involving acceleration as well, as the pendulum accelerates from the zero position, achieves maximum velocity and eventually the velocity goes to zero again, etc (the oscillations we see in the response).
If vibration is necessary to excite the EM Drive (as apparently proposed by TT and Shawyer) the problem is not static either, as accelerometers are routinely used to measure amplitude of vibrations (since oscillations in speed imply acceleration). Hence the problem is dynamic, rather than static.
Actually one of the questions I and others have asked is "what is the amplitude of the vibrations that are required". Let's assume for discussion sake the hypothesis of TT and Shawyer: let's suspend our questioning and just accept the premise that vibration is needed. Then, any experimenter needs to quantify what is the amplitude of this vibration required in the experiment. Amplitude of vibration is usually measured with accelerometers. But I do not recall anybody answering what is the amplitude of the vibrations that are required for such excitation... -
#2703 by Bob012345 on 10 Nov, 2017 18:31
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...
Thanks for your reply.
Thanks. I get the concept for dynamic analysis but we were discussing a static problem and the slide was not quite accurate. The Zeno comment was meant to be a joke. The total force on the body can be set as zero for analysis purposes but it's not really zero so that statement in the slide should have more careful. As I pointed out, I think that the issue of what a scale shows was not necessarily wrong and that was the issue at hand.
Concerning <<we were discussing a static problem>>.
If there is acceleration of the EM Drive involved, the problem is not static but dynamic.
If there is a pendulum involved of any kind (torsional, swinging, etc.) the problem is not static either, involving acceleration as well, as the pendulum accelerates from the zero position, achieves maximum velocity and eventually the velocity goes to zero again, etc (the oscillations we see in the response).
If vibration is necessary to excite the EM Drive (as apparently proposed by TT and Shawyer) the problem is not static either, as accelerometers are routinely used to measure amplitude of vibrations (since oscillations in speed imply acceleration). Hence the problem is dynamic, rather than static.
Actually one of the questions I and others have asked is "what is the amplitude of the vibrations that are required". Let's assume for discussion sake the hypothesis of TT and Shawyer: let's suspend our questioning and just accept the premise that vibration is needed. Then, any experimenter needs to quantify what is the amplitude of this vibration required in the experiment. Amplitude of vibration is usually measured with accelerometers. But I do not recall anybody answering what is the amplitude of the vibrations that are required for such excitation...
Please let me clarify one point. My initial response was only related to the discussion around this one statement. Perhaps I should have said that I was discussing the static case. Thanks.QuoteNote that the reaction is either the acceleration a, or a force equal to Ma, but not both.
Clearly, in a static situation, where T and R both exist as forces, they will cancel out. Thus any attempt to measure them by simply placing the thruster vertically on a set of scales will fail. If however the thrust is sufficient such that a = -g, then the thruster could be made to hover above the scales.
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#2704 by qraal on 12 Nov, 2017 20:01
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Google translate has trouble with the Chinese character for 'newton' translating it as 'cow' or 'cattle'. Though there's something charming about the mental image of a 'micro-cow' as a unit of thrust.
I can't find the source. I will modify my post accordingly. Some anecdotes were found to be from oyzw. The best official news is still this one, as have been discussed multiple times here, http://digitalpaper.stdaily.com/http_www.kjrb.com/kjrb/html/2016-12/11/content_357004.htm. The best translation is by Baidu (usually better than google translation), as: http://fanyi.baidu.com/transpage?query=http%3A%2F%2Fdigitalpaper.stdaily.com%2Fhttp_www.kjrb.com%2Fkjrb%2Fhtml%2F2016-12%2F11%2Fcontent_357004.htm&source=url&ie=utf8&from=auto&to=zh&render=1 and the relevant quote is: "Chen Yue introduced, they have completed the development of test equipment for flight test, is being carried out in orbit validation."
You may ask him about the news. Thanks.
One such anecdotes: https://www.zhihu.com/question/53602370 in comments: The 73 agreed with the answer
In 2016, a small space experiment that was not very interesting could have an impact on the course of human history.
Several kinds of non working microwave thruster principle verification machine, by China Aerospace 5 Institute on the practice of 17 satellite on orbit test, the test results may be open next week.
This is the first time that the device has been tested in the near earth orbit after NASA announced that the Emdrive system measured the thrust in the laboratory vacuum.
Posted on 2016-12-15 -
#2705 by Rodal on 12 Nov, 2017 21:37
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Google translate has trouble with the Chinese character for 'newton' translating it as 'cow' or 'cattle'. Though there's something charming about the mental image of a 'micro-cow' as a unit of thrust.
Since we use horses as a unit of power, a cow as unit of thrust is not that outlandish
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#2706 by Augmentor on 12 Nov, 2017 22:27
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Rodal,
How much cowpower is equal to 1 hp which is equal to 746 watts?
You brought it up...so what is the conversion factor.
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#2707 by Rodal on 12 Nov, 2017 22:44
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Rodal,
One unit of cow-power appears to be equivalent to 100 watts.
How much cowpower is equal to 1 hp which is equal to 746 watts?
You brought it up...so what is the conversion factor.
David
This can be scientifically derived from the paper “Design of farm waste-centered demand-driven supply side infrastructure for data centers,” delivered at the 2010 ASME International Conference on Energy Sustainability.
Link to the paper by Hewlett Packard Laboratories: http://www.hpl.hp.com/news/2010/apr-jun/HP_ASME_PAPER.pdf
The paper described how a dairy farm with 10,000 head of cattle could supply a megawatt (MW) of electricity.
10^6 watts/10^4=100 watts per cow
Since a horsepower is 746 watts, this means that:
It takes 7.46 units of cowpower to be equivalent to 1 unit horsepower or a conversion factor of 7.46...roughly 7 and a half cows per every horse.
Notice that cowpower refers to electric power derivable from biogas (methane), while horsepower refers to mechanical power. I imagine that we could come up with a unit of rocket propulsion thrust based on average ejection of biogas (methane) from cows as a propellant.
Raptor is a family of cryogenic, methane-fueled rocket engines under development by SpaceX, designed for colonization of Mars.
However, there seems to be no advantage to such a unit of cow thrust, so I propose we should continue using Newtons as a unit of force in this thread. -
#2708 by LowerAtmosphere on 13 Nov, 2017 01:40
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Don't let it be said that there was no progress in 2017
Edit: Here's an interesting verification that opacity of medium does not change the mean path length for incident photons https://phys.org/news/2017-11-path-length-opaque-media.html -
#2709 by chongma on 13 Nov, 2017 08:10
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Don't let it be said that there was no progress in 2017
Now that 2017 is nearly over...I predict 2018 will be the year of the EmDrive
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#2710 by Peter Lauwer on 13 Nov, 2017 10:21
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Don't let it be said that there was no progress in 2017
Now that 2017 is nearly over...I predict 2018 will be the year of the EmDrive
Yes, I agree. If it doesn't become clear whether the EmDrive works or not (or, at least generate an anomalous force, which doesn't have to mean that it is practically usable as spacedrive) in the coming year, that would be rather strange. Several government labs are involved now, and even private researchers like Monomorphic and me should be able to have results within a year. -
#2711 by Bob Woods on 13 Nov, 2017 15:59
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It is a well established fact that cows follow a ballistic trajectory, and are not suitable for sustained thrust.
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#2712 by JonathanD on 13 Nov, 2017 17:15
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I love that moooovie!
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#2713 by spupeng7 on 13 Nov, 2017 23:33
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PotomacNeuron,
Furthermore, if my recollection is correct, I seem to recall Potomac Neuron writing that the test in space was not an unqualified success.
There were news that two new electrical thrusters ("电推")would be tested on Shijian-17 experimental satellite. People on several Chinese forums decoded that one is this device and the other is an ion thruster. After a few weeks, when other devices tested successfully had news about their successes (including the ion thruster), this device did not have any news. That was a sign of failure from experience about the way things were reported in China. Furthermore, there were anecdotes that it failed on some Chinese forums. Oyzw, a member of this forum, said in a Chinese forum that his messenger in the same institute with Chen Yue (the leader of the project) told him Chen blamed electrical or mechanical ("机电") problems for the failure. Oyzw is still active. His last post is the first one on page 127.
do you have details of the source of this news? I found it difficult to get reliable translations. It seems to me that this information is crucial to funding applications in the US and in Europe.
I can't find the source. I will modify my post accordingly. Some anecdotes were found to be from oyzw. The best official news is still this one, as have been discussed multiple times here, http://digitalpaper.stdaily.com/http_www.kjrb.com/kjrb/html/2016-12/11/content_357004.htm. The best translation is by Baidu (usually better than google translation), as: http://fanyi.baidu.com/transpage?query=http%3A%2F%2Fdigitalpaper.stdaily.com%2Fhttp_www.kjrb.com%2Fkjrb%2Fhtml%2F2016-12%2F11%2Fcontent_357004.htm&source=url&ie=utf8&from=auto&to=zh&render=1 and the relevant quote is: "Chen Yue introduced, they have completed the development of test equipment for flight test, is being carried out in orbit validation."
You may ask him about the news. Thanks.
Thank You Potomac,
Obligada.
May these in-orbit tests be the first of many
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#2714 by RERT on 14 Nov, 2017 06:47
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Hmm... Guess the search for the holy grail really is about where we're at...
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#2715 by Bob012345 on 14 Nov, 2017 18:37
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In falling from h1 to h2, the atom lost energy. In this distant observer's frame, the atom's ground state energy is lower at h2 than it was at h1.
...
If in your view, as I understand it, the ZPF essentially supports all processes, is the field emanating from a charge or even a magnet a propagating field that must be continuously supplied by the ZPF? In other words, in your model is there really a continual flux from the ZPF keeping up appearances even in static situations? Thanks.
It's not just my view. It's part of QED;
In section 3.3 of The Quantum Vacuum, Milonni [2] writes,
“The fact that an accelerating charge loses energy by radiating implies, according to classical ideas, that an electron should spiral into the nucleus and that atoms should not be stable. The balancing of the effects of radiation reaction and the vacuum field..., however, suggest that the stability of atoms might be attributable to the influence on the atom of the vacuum field.... We now know that the vacuum field is in fact formally necessary for the stability of atoms in quantum theory. As we saw..., radiation reaction will cause canonical commutators [x, px] to decay to zero unless the fluctuating vacuum field is included, in which case commutators are consistently preserved.”
In my own words: Atoms are in equilibrium with the vacuum fields. Where, the ZPF is the driving field that inflates them, and RR and/or the fields of all mater in the universe, is the damping field that contracts them. Any imbalance in these two fields will cause matter to seek a new equilibrium by moving in the direction which reduces its self-energy. This is the direction of increased damping, which results in gravitational length contraction and time dilation.
In answer to a question Dr. Rodal asked of one presenter: In the manner I described above, the two fields might have infinite energy and we are taking the difference between these two infinities, as is done elsewhere in QED. The imbalance is what "gravitates". If matter is in equilibrium, it is an inertial reference frame. However, the "strength" of the frame, as was discussed by Marc Mills, I equate to the spectral energy density of the fields in equilibrium. Hence, we can have gravitational effects such as time dilation and length contraction, in an inertial frame (of any strength), and we can have gravitational effects at the order of magnitude we are familiar with, without the fields themselves causing a 120 order of magnitude catastrophe for the ZPF. It's all in how I/we interpret what we have for data.
Thanks. It seems a static magnetic field say from a bar magnet is ultimately due to ZPF sustaining the electron motion which causes the currents that produce the field and that field continually propagates and is refreshed.
If we considered that the energy contained in the field of typical bar magnet can be on the order of a Joule and most of that is contained within the volume of a sphere around the magnet of one light nanosecond or 30cm, which has to be refreshed each nanosecond or the field disappears, then the power emanating from the magnet is on the order of a GW. That seems untenable.
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#2716 by WarpTech on 14 Nov, 2017 22:01
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Thanks. It seems a static magnetic field say from a bar magnet is ultimately due to ZPF sustaining the electron motion which causes the currents that produce the field and that field continually propagates and is refreshed.
If we considered that the energy contained in the field of typical bar magnet can be on the order of a Joule and most of that is contained within the volume of a sphere around the magnet of one light nanosecond or 30cm, which has to be refreshed each nanosecond or the field disappears, then the power emanating from the magnet is on the order of a GW. That seems untenable.
Ferromagnetism is due to the spin of the electron AND the fact that they can align with the same N-S axis, across large domains within the iron. The amount of reactive power the vacuum contributes to electron spin doesn't change simply because it's part of a magnet. The electron is still in equilibrium with the vacuum, so there is no way to extract work from it. In the end, the magnetic field of a bar magnet doesn't oscillate at observable frequencies, so it can't do any work either. -
#2717 by spupeng7 on 15 Nov, 2017 00:12
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Thanks. It seems a static magnetic field say from a bar magnet is ultimately due to ZPF sustaining the electron motion which causes the currents that produce the field and that field continually propagates and is refreshed.
If we considered that the energy contained in the field of typical bar magnet can be on the order of a Joule and most of that is contained within the volume of a sphere around the magnet of one light nanosecond or 30cm, which has to be refreshed each nanosecond or the field disappears, then the power emanating from the magnet is on the order of a GW. That seems untenable.
Ferromagnetism is due to the spin of the electron AND the fact that they can align with the same N-S axis, across large domains within the iron. The amount of reactive power the vacuum contributes to electron spin doesn't change simply because it's part of a magnet. The electron is still in equilibrium with the vacuum, so there is no way to extract work from it. In the end, the magnetic field of a bar magnet doesn't oscillate at observable frequencies, so it can't do any work either.
Cannot bring myself to agree with either of you. Occam's razor suggests that a charge imbalance longitudinal to the North / South axis of a magnet is a simpler explanation. Near either end of that imbalance electrical interactions explain the force and near the plane between the ends the force of those interactions sum to the longitudinal. Why complicate your appreciation further? -
#2718 by WarpTech on 15 Nov, 2017 01:31
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Thanks. It seems a static magnetic field say from a bar magnet is ultimately due to ZPF sustaining the electron motion which causes the currents that produce the field and that field continually propagates and is refreshed.
If we considered that the energy contained in the field of typical bar magnet can be on the order of a Joule and most of that is contained within the volume of a sphere around the magnet of one light nanosecond or 30cm, which has to be refreshed each nanosecond or the field disappears, then the power emanating from the magnet is on the order of a GW. That seems untenable.
Ferromagnetism is due to the spin of the electron AND the fact that they can align with the same N-S axis, across large domains within the iron. The amount of reactive power the vacuum contributes to electron spin doesn't change simply because it's part of a magnet. The electron is still in equilibrium with the vacuum, so there is no way to extract work from it. In the end, the magnetic field of a bar magnet doesn't oscillate at observable frequencies, so it can't do any work either.
Cannot bring myself to agree with either of you. Occam's razor suggests that a charge imbalance longitudinal to the North / South axis of a magnet is a simpler explanation. Near either end of that imbalance electrical interactions explain the force and near the plane between the ends the force of those interactions sum to the longitudinal. Why complicate your appreciation further?
There is no NET charge dipole across a bar magnet.
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#2719 by Bob012345 on 15 Nov, 2017 17:28
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Thanks. It seems a static magnetic field say from a bar magnet is ultimately due to ZPF sustaining the electron motion which causes the currents that produce the field and that field continually propagates and is refreshed.
If we considered that the energy contained in the field of typical bar magnet can be on the order of a Joule and most of that is contained within the volume of a sphere around the magnet of one light nanosecond or 30cm, which has to be refreshed each nanosecond or the field disappears, then the power emanating from the magnet is on the order of a GW. That seems untenable.
Ferromagnetism is due to the spin of the electron AND the fact that they can align with the same N-S axis, across large domains within the iron. The amount of reactive power the vacuum contributes to electron spin doesn't change simply because it's part of a magnet. The electron is still in equilibrium with the vacuum, so there is no way to extract work from it. In the end, the magnetic field of a bar magnet doesn't oscillate at observable frequencies, so it can't do any work either.
Magnets certainly can do work but what people really mean is if a magnet can be the energy source doing that work of which the answer classically is no. We can invest potential energy into a classical system involving magnets raising the potential energy which is reduced by the magnets then doing work. The myth than magnets call no work comes from the fact that a charged particle traveling in a constant magnetic field only has it's direction changed but not its momentum.
If I understand you, basically you are saying I'm correct regarding all that flux pouring out of a magnet or an electron charge or any EM field source but simply put, that energy can't be measured or used for any gain. If the ZPF continuously has to support each electron from decaying, it is delivering huge energies to support each electron in the universe to both keep it from decaying and to refresh the continuously departing EM fields emanating from each particle, exactly equal to the energy that would be released if the electrons were allowed to decay or not be a source of fields. So the ZPF does all the work to maintain the universe in that view which you assert is now the standard QED view. If one cannot measure that flux, how is that view falsifiable? Thanks.
