Strange Posted May 16, 2018 Share Posted May 16, 2018 Just now, xyrth said: The sides have pressure like the sides have pressure in a glass of water under gravity. But it isn't water. It is a collection of (non-physical) spheres all pulled in one direction by (magical) springs. How can they exert pressure? What are they made of? How rigid or compressible are they? You would need to know these things to calculate the pressure they cause. Link to comment Share on other sites More sharing options...
xyrth Posted May 16, 2018 Author Share Posted May 16, 2018 1 minute ago, Strange said: But it isn't water. It is a collection of (non-physical) spheres all pulled in one direction by (magical) springs. How can they exert pressure? What are they made of? How rigid or compressible are they? You would need to know these things to calculate the pressure they cause. Even, it is a theoretical device, the sum of energy must be at 0. I can choose the perfect spheres in theory, to have the pressure. Molecules of a fluid does it so theoretical spheres can do the same. Link to comment Share on other sites More sharing options...
Strange Posted May 16, 2018 Share Posted May 16, 2018 18 minutes ago, xyrth said: I can choose the perfect spheres in theory, to have the pressure. What are "perfect" spheres? How do they interact? How do they behave when compressed? Quote Molecules of a fluid does it so theoretical spheres can do the same. But molecules of a fluid are not attached to springs. I guess this might be another source of errors: you treat it as if it were water, but also try and have the water molecules attached to springs. It doesn't make much sense. Link to comment Share on other sites More sharing options...
xyrth Posted May 16, 2018 Author Share Posted May 16, 2018 1 minute ago, Strange said: But molecules of a fluid are not attached to springs. Each is attracted by gravity. My springs are theoretical. I calculated several devices and each time I found the sum of the energy at 0. 3 minutes ago, Strange said: It doesn't make much sense. It is because you focused yourself on the technology not the theory. Link to comment Share on other sites More sharing options...
Strange Posted May 16, 2018 Share Posted May 16, 2018 1 minute ago, xyrth said: It is because you focused yourself on the technology not the theory. But you have so many non-physical things in the system that I don't know why you would expect to be able to apply standard physics. Link to comment Share on other sites More sharing options...
swansont Posted May 16, 2018 Share Posted May 16, 2018 3 hours ago, xyrth said: Why not the sides ? The sides have pressure like the sides have pressure in a glass of water under gravity. All the spheres are attracted in direction of the green line, so they pressure all they can, like molecules of water can do under gravity. But you don't have gravity. Balls on a spring don't act like water under gravity. Link to comment Share on other sites More sharing options...
xyrth Posted May 16, 2018 Author Share Posted May 16, 2018 3 minutes ago, swansont said: But you don't have gravity. Balls on a spring don't act like water under gravity. Ok,I understood. Link to comment Share on other sites More sharing options...
xyrth Posted June 14, 2018 Author Share Posted June 14, 2018 (edited) I translate a resume, maybe it could be clear like that. det2.pdf Edited June 14, 2018 by xyrth Link to comment Share on other sites More sharing options...
Strange Posted June 14, 2018 Share Posted June 14, 2018 8 hours ago, xyrth said: I translate a resume, maybe it could be clear like that. What is the point? You have created an example that is so complicated that (1) you are not able to find your mistake and (2) you are unable to explain it to anyone else so they can find your mistake (if you were able to explain it, you would probably find the error). I suspect the error is because you are making all sorts of physically impossible assumptions and so your example is meaningless. But I dont know and I don’t care (and I doubt anyone else does). Energy IS conserved so your broken example is pointless. So, again, what is the point of this thread? Shall we just ask the mods to close it as a a waste of everyone’s time? Link to comment Share on other sites More sharing options...
xyrth Posted June 15, 2018 Author Share Posted June 15, 2018 13 hours ago, Strange said: What is the point? You have created an example that is so complicated that (1) you are not able to find your mistake and (2) you are unable to explain it to anyone else so they can find your mistake (if you were able to explain it, you would probably find the error). I suspect the error is because you are making all sorts of physically impossible assumptions and so your example is meaningless. But I dont know and I don’t care (and I doubt anyone else does). Energy IS conserved so your broken example is pointless. So, again, what is the point of this thread? Shall we just ask the mods to close it as a a waste of everyone’s time? You asked me some questions, tell me details about springs, etc. I replied with that document. Have you read the 30 pages ? I explained all the device in detail. It is not because I took simplifications the device is false, in the contrary, it could be easier to find the mistake. The point of this thread is to break the law of conservation of the energy, with a counterexample, nothing else. Link to comment Share on other sites More sharing options...
Mordred Posted June 15, 2018 Share Posted June 15, 2018 (edited) conservation of energy applies to a closed and isolated system but you have mentioned an external device in 4. which isn't a closed system. Also your involving both conservation of linear and angular momentum from what I can translate in your document. Edited June 15, 2018 by Mordred Link to comment Share on other sites More sharing options...
xyrth Posted June 16, 2018 Author Share Posted June 16, 2018 (edited) Note: the device is not isolated but the external device that controls the device counts all the energy in/out so the sum of the energies in/out must be at 0. Edited June 16, 2018 by xyrth Link to comment Share on other sites More sharing options...
Strange Posted June 16, 2018 Share Posted June 16, 2018 21 hours ago, xyrth said: I explained all the device in detail. It is not because I took simplifications the device is false, in the contrary, it could be easier to find the mistake. The point of this thread is to break the law of conservation of the energy, with a counterexample, nothing else. As you obviously can't break the law of conservation of energy, you have equally obviously made an error in your description or your calculations. As you are either too lazy or too stupid to find your own mistakes, why should anyone here help you? Link to comment Share on other sites More sharing options...
xyrth Posted June 16, 2018 Author Share Posted June 16, 2018 36 minutes ago, Strange said: As you are either too lazy or too stupid to find your own mistakes, why should anyone here help you? Waouhh, you can close all forums on earth. I hope the life will help you to understand what you are. Link to comment Share on other sites More sharing options...
Mordred Posted June 16, 2018 Share Posted June 16, 2018 (edited) 7 hours ago, xyrth said: Note: the device is not isolated but the external device that controls the device counts all the energy in/out so the sum of the energies in/out must be at 0. You still can't combine two conservation laws at the same time then expect correct results. You have torsion with angular components mixed with linear components with the springs. Not to mention adding hydrostatic components via pressure.... there is two laws involved in your scenario not just one. A closed system is isolated in either strictly linear or strictly angular momentum. take a careful note on what a conserved quantity entails. https://en.wikipedia.org/wiki/Conserved_quantity For example the conservation of mass energy is specifically translations along the time axis. However in your paper you are involving the x,y and z axis. Conservation of linear momentum is specifically translation only. conservation of linear momentum is rotation. You have mixed several laws in your scenario and not isolated each specifically in your analysis. https://en.wikipedia.org/wiki/Conservation_law Edited June 16, 2018 by Mordred Link to comment Share on other sites More sharing options...
xyrth Posted June 16, 2018 Author Share Posted June 16, 2018 @Mordred: thanks for your help, yeah, I will study again my device with your advices, I think there is a problem too with the geometry when the device is deformed. Link to comment Share on other sites More sharing options...
Mordred Posted June 16, 2018 Share Posted June 16, 2018 the deformation will definitely cause problems as your adding mechanical work to the system Link to comment Share on other sites More sharing options...
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