Strange Posted July 23, 2019 Share Posted July 23, 2019 Just now, MPMin said: This only refers to the momentum of the emp. You haven’t addressed the magnetic field component of the emp which is what causes the force on wire B when wire B carries a current. That is like saying that I am only considering the momentum of the rock leaving the craft instead of the force on the wall it was thrown at. Because it is (still) irrelevant. Link to comment Share on other sites More sharing options...
Ghideon Posted July 23, 2019 Share Posted July 23, 2019 6 hours ago, MPMin said: You are asking for a reference that uses the emp the way i am proposing which hasn’t been done before. Does that mean that new physical formulas are required? Or that you believe there are no experiments done that resembles your idea? 7 hours ago, MPMin said: But I have posted a reference for every part of my system and each part has been supported. No. You have an idea requiring ultra short EMPs between wires. The pulses described are so short that the time for current running along the length of the wire might need to be addressed*. Please provide some evidence that F = I x L x B is applicable for such a short pulse. *) I think @swansont pointed this out already Link to comment Share on other sites More sharing options...
MPMin Posted July 23, 2019 Author Share Posted July 23, 2019 1 minute ago, Strange said: Because it is (still) irrelevant. It’s not irrelevant, Because in the rock analogy as for the emp, the rocks also have a magnetic field, I’ve explained and given references to how a magnetic fields effect wires carrying a current and Ive given references to show that emp has a magnetic field component, how is this irrelevant? Link to comment Share on other sites More sharing options...
Strange Posted July 23, 2019 Share Posted July 23, 2019 Just now, MPMin said: It’s not irrelevant, Because in the rock analogy as for the emp, the rocks also have a magnetic field, I’ve explained and given references to how a magnetic fields effect wires carrying a current and Ive given references to show that emp has a magnetic field component, how is this irrelevant? Because the only thing that is relevant is the momentum of the rock (or EMP) leaving the craft. This is why it makes a difference if the whole thing is in a box. I know you don't understand this. But I have tried to explain it every way I can think of.... Link to comment Share on other sites More sharing options...
MPMin Posted July 23, 2019 Author Share Posted July 23, 2019 (edited) 11 minutes ago, Ghideon said: Does that mean that new physical formulas are required? Or that you believe there are no experiments done that resembles your idea? It was said somewhere that everything has been tried before, I can’t find any experiments or discussions about this. 11 minutes ago, Ghideon said: You have an idea requiring ultra short EMPs between wires. The pulses described are so short that the time for current running along the length of the wire might need to be addressed*. Please provide some evidence that F = I x L x B is applicable for such a short pulse. The emp carries a magnetic field as referenced already, the force on a wire carrying a current in a magnetic field is determined with F = I x L x B, as it would seem this hasn’t been tried before, there is enough evidence from related studies with continuous currents to try this 5 minutes ago, Strange said: But I have tried to explain it every way I can think of.... You haven’t actually explained why it won’t work in a box Edited July 23, 2019 by MPMin Link to comment Share on other sites More sharing options...
Strange Posted July 23, 2019 Share Posted July 23, 2019 3 minutes ago, MPMin said: You haven’t actually explained why it won’t work in a box I LITERALLY JUST EXPLAINED IT. For the bazillionth(*) time. (*) Some exaggeration may have been used for effect. Let me explain again: Because the only thing that is relevant is the momentum of the rock (or EMP) leaving the craft. (If it is in a box, then the rock (or pulse) cannot leave.) Link to comment Share on other sites More sharing options...
MPMin Posted July 23, 2019 Author Share Posted July 23, 2019 7 minutes ago, Strange said: Let me explain again: Because the only thing that is relevant is the momentum of the rock (or EMP) leaving the craft. (If it is in a box, then the rock (or pulse) cannot leave.) And what about the magnetic field effect on the current carrying wire? Is this some kind of closed loop logic because its been assumed to not work in a box? Let’s assume it does work in a box, does the magnetic field component produce a force on the current carrying wire now? Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted July 23, 2019 Share Posted July 23, 2019 18 minutes ago, Strange said: I LITERALLY JUST EXPLAINED IT. For the bazillionth(*) time. (*) Some exaggeration may have been used for effect. Let me explain again: Because the only thing that is relevant is the momentum of the rock (or EMP) leaving the craft. (If it is in a box, then the rock (or pulse) cannot leave.) ...but getting to be less and less of an exaggeration as we go...will be understatement very soon... Link to comment Share on other sites More sharing options...
Mordred Posted July 23, 2019 Share Posted July 23, 2019 First off your assuming an incredibly short pulse is going to stay with the same current and length as the current carrying wire that generates the field. You have disconnected the field source yet apply a formula that is specific to a continuous source. Working off that assumption. Once you disconnect the source you have no path to ground which enables the current flow from the source So how can the pulse possibly have the same current ? Link to comment Share on other sites More sharing options...
Ghideon Posted July 23, 2019 Share Posted July 23, 2019 24 minutes ago, MPMin said: The emp carries a magnetic field as referenced already, the force on a wire carrying a current in a magnetic field is determined with F = I x L x B, as it would seem this hasn’t been tried before, there is enough evidence from related studies with continuous currents to try this Why do you think the magnetic field is constant along the cable B if the pulse is so short that it is causally disconnected from the wire A? Please provide evidence that F = I x L x B is relevant for such short pulses and that the force can occur at cable B without having a counter force on cable A as you seem to suggest. Link to comment Share on other sites More sharing options...
Strange Posted July 23, 2019 Share Posted July 23, 2019 17 minutes ago, MPMin said: Let’s assume it does work in a box, Lets assume momentum is not conserved? No, lets not. 19 minutes ago, MPMin said: does the magnetic field component produce a force on the current carrying wire now? Ignoring the real practical problems that others have pointed out, yes. It produces a force in the box. It produces a force without the box. That is why it is not relevant. Link to comment Share on other sites More sharing options...
Mordred Posted July 23, 2019 Share Posted July 23, 2019 Lol I shudder to think of all the interference patterns of an EM wave inside a box. May as well predict Brownian motion. Link to comment Share on other sites More sharing options...
MPMin Posted July 23, 2019 Author Share Posted July 23, 2019 13 minutes ago, Mordred said: First off your assuming an incredibly short pulse is going to stay with the same current and length as the current carrying wire that generates the field. You have disconnected the field source yet apply a formula that is specific to a continuous source. As this hasn’t been tried before, I’m applying the most relevant formula, its fair to a assume that segment of the continuous magnetic field will retain the same properties. 16 minutes ago, Mordred said: Working off that assumption. Once you disconnect the source you have no path to ground which enables the current flow from the source So how can the pulse possibly have the same current ? I’m not sure what you mean here? Link to comment Share on other sites More sharing options...
Mordred Posted July 23, 2019 Share Posted July 23, 2019 (edited) Sending a pulse into space where the signal varies from that wire would be very similar to an antenna. There is a series of antenna formulas that show the power is not the same as the supplied power. There is power losses due to directivity etc. The closest type of antenna that matches your scenario would be an omnidirectional antenna. Go ahead Google the radiated power of an omnidirectional antenna you will see it has huge power losses. Secondly the waveform will have different directional components than a current carrying wire. The EM field of a current carrying wire is not the same as a radiated EM waveform. The loss of continuous power supply and ground path is but one factor as to why the two are different. The other being differences in the transverse and longitudinal components. Edited July 23, 2019 by Mordred Link to comment Share on other sites More sharing options...
MPMin Posted July 23, 2019 Author Share Posted July 23, 2019 6 minutes ago, Strange said: Lets assume momentum is not conserved? Ah .. here’s the trap, so if momentum isn’t conserved because your rock analogy doesn’t work inside the box, the magnetic field component does still work inside the box. As the force on a current carrying wire in a magnetic field is a known thing, it would now seem that the momentum of the emp is irrelevant factor and the only part that’s now relevant is the magnetic field acting on the current carrying wire, thanks Strange. 1 minute ago, Mordred said: Sending a pulse into space where the signal varies from that wire would be very similar to an antenna. There is a series of antenna formulas that show the power is not the same as the supplied power. There is power losses due to directivity etc. The closest type of antenna that matches your scenario would be an omnidirectional antenna. Go ahead Google the radiated power of an omnidirectional antenna you will see it has huge power losses. Secondly the waveform will have different directional components than a current carrying wire. The EM field of a current carrying wire is not the same as a radiated EM waveform. The loss of continuous power supply and ground path is but one factor as to why the two are different. Obviously the greater the distance the greater the distortion, Im talking about very short distance here -1 Link to comment Share on other sites More sharing options...
Strange Posted July 23, 2019 Share Posted July 23, 2019 2 minutes ago, MPMin said: Ah .. here’s the trap, so if momentum isn’t conserved It is conserved. The rest of your post is complete gibberish Link to comment Share on other sites More sharing options...
swansont Posted July 23, 2019 Share Posted July 23, 2019 1 hour ago, MPMin said: I haven’t ignored this i just haven’t addressed this yet. The wires do not necessarily have to be in a loop or coil formation. The wires could be aligned in a zig zag type formation where the wire feeds in from the top and comes out from the bottom. “Loop” is a topological term here, not a geometric one ! Moderator Note After discussion with other mods, we have decided that we’ve reached the point where this is no longer fruitful. We probably hit that point long ago, but there were occasional hints that progress might be in the works. Alas, that’s not the case, as we keep hitting the same wall, and the OP refuses to learn any physics. Looking stuff up on wikipedia is not an acceptable substitute. Don’t bring this topic up again. Link to comment Share on other sites More sharing options...
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