Ghideon Posted July 22, 2019 Posted July 22, 2019 33 minutes ago, MPMin said: The thing that separates itself from the craft is the emp, once it leaves the wire it is no longer part of the craft.I think you might be thinking of the emp as a physical force rather than a magnetic force. where F = I x L x B then -F = -I x L x B EMP leaves craft carrying momentum, not carrying force. You are (still) using wrong formula.
Strange Posted July 22, 2019 Posted July 22, 2019 (edited) 1 hour ago, MPMin said: The thing that separates itself from the craft is the emp, once it leaves the wire it is no longer part of the craft.I think you might be thinking of the emp as a physical force rather than a magnetic force. OK. But if the EMP leaves the craft symmetrically, in all directions, then there is no change in the momentum of the craft (ie. no motion, no thrust). So the only way of applying some momentum to the craft is to make the EMP leave the craft asymmetrically (the more asymmetrically the better). So the only reason your idea may be able to generate some thrust is because it causes the EMP to leave the craft asymmetrically. The only reason it can cause the EMP to leave the craft asymmetrically is because wire B blocks part of the pulse. The wire will block that force equally whether it is carrying current in one direction, the other direction, or not at all. (It is slightly more complicated because if there is a current than that will contribute another EMP pulse which is also radiating away from the craft and carrying momentum. If things are symmetrical, then this will cancel out the effect of the first pulse.) [Edit: and it is even more complicated than that because there will be all sorts of other metal infrastructure that will also block the EMP pulse in random directions.] Edited July 22, 2019 by Strange
Mordred Posted July 22, 2019 Posted July 22, 2019 (edited) The EMP is still contained within the craft so counts as part of the craft. Just like the atmosphere within a shuttle. As Strange mentioned if you allow the EM field to leave the craft then it can act externally. Edited July 22, 2019 by Mordred
Strange Posted July 22, 2019 Posted July 22, 2019 2 minutes ago, Mordred said: The EMP is still contained within the craft so counts as part of the craft. Just like the atmosphere within a shuttle. If none of the EMP leaves the craft (which is entirely plausible in a real craft) then there is zero change in momentum, zero thrust, zero motion.
Mordred Posted July 22, 2019 Posted July 22, 2019 Correct even though we cross posted on edit hehe had to add a bit
Strange Posted July 22, 2019 Posted July 22, 2019 11 minutes ago, Strange said: If none of the EMP leaves the craft (which is entirely plausible in a real craft) then there is zero change in momentum, zero thrust, zero motion. And, in a real craft, most of the EMP will be blocked by things other than the single wire B. I have been considering only the effect of that (with a small diversion to talk about solar panels) because that is the only information provided.
Ghideon Posted July 22, 2019 Posted July 22, 2019 (edited) Back in the early pages of the thread there was an analysis of the momentum from EMPs. Here is a modified version again addressing the discussion of radiation leaving the craft vs radiation in the craft. There is still no analysis of how possible it is to create the device, only an analysis of momentum is done. There still seems to be an agreement that ”particles carrying momentum” is reasonable model for the pulse as long as no detailed calculations are needed for the interaction at cable B. Einstein looked at the special case of the energy being carried by a light flash. According to Feynman the argument would work as well for particles. Einsteins though experiment, more pictures and a mathematical rigorous walkthrough of momentum of (escaping) electromagnetic radiation can be found at http://www.feynmanlectures.caltech.edu/II_27.html. Here is the initial situation at time t=0. A and B are mounted in a rig, cables can't be moved independently. Cables are straight and seen running straight into the paper. Current is switched on in cable A.The rig is big so time of travel at speed of light across the rig is not possible to neglect. At some small time t1some amount of electromagnetic radiation have radiated away and the current is switched of. There will be magnetic and electric energy. We chose to model the radiation using particles. There are probably other usable models. Important thing is that cables A and B are not attracted by force (Ampere’s force law) but act more like antennas far apart. Here is a crucial simplification that other readers may have opinions about: the radiated pulse is symmetrical and we do not care about the exact composition; photons, electrons or other. We only care about the momentum carried by the radiated particles. Each individual particle carries momentum directed away from the center of the cable A. There are two components of the momentum, x and y. Since momentum in y direction is never part of an interaction and has total sum of zero we simplify and analyse x-component only. We will analyse the particles that will later interact with a field around cable B and the particles radiated in 180 degree opposite direction. This simplification allows for a new diagram showing two pulses traveling in opposite directions carrying momentum P and -P. Note the convention, I prefer the pulse later reaching cable B to have positive sign of momentum P. At time t=t2 the pulse reaches cable B. Again a simplification. The cable is surrounded by a symmetrical field interacting with incoming pulse. Then, the maximum amount of momentum gained by interaction with the pulse is P. We do not need to model the exact interaction of the fields. There is no magical way to gain more momentum than what the incoming pulse carries. The result is that the complete rig with both cables starts moving to the left. Note that I discuss only the x components of the circular pulse. The movement of the rig is not due to momentum P of left pulse. It it is momentum -P of right pulse that does the job. -P is just prevented from doing so until cable B stops the left pulse momentum P. If cable A had sent only the burst -P the movement would have begun at t=0. Now the rig is stationary until time t=t2. Had cable A sent only pulse P then the rig would have moved to the right for a short while and then stopped when cable B interact with left pulse. Left pulse is not adding momentum P, left pulse prevents the rig from moving until cable B stops the left pulse. Note: above is a zoomed in look at the best possible outcome only. The interaction at B happens in such a way that all of the incoming momentum P is used to generate an impulse that push at cable B. As Strange pointed out there are other possibilities. Let's draw some and clarify, I’ll draw the complete rig with cables A and B this time. First t=t2. An electromagnetic pulse, modelled as a set of particles traveling away from cable A is about to reach cable B. Pulse is a circular set of particles but I simplify and draw only particles along X axis carrying momentum P and -P (as I did before) Next: First case, the optimal one. B manages to prevent any particles from leaving the system. Exactly how the interaction works is not important. This is the only case i mentioned in the earlier post. Result, due to conservation of momentum is that the rig gains momentum P to the left. As Strange has pointed out several times, an extremely inefficient way to propel the rig compared to use energy to send only pulse -P. P and -P are, also noted by Strange, just a tiny part of the energy sent from cable A since A sends the pulse 360 degrees. next: Second case: This is an interaction with a pulse/field around cable B in such a way that total momentum of any particles leaving the rig is zero. Next: Third case supplied as reference; if cable B fails to interact at all and particles continue past/through cable B. Total sum of momentum of EMP pulse leaving system is zero. Rig will not move. Again: Analysis using Ampere force law of force from cable A on B and reactive force from B to A is not useful for calculating momentum. It is the small "momentum surplus" allowed to leave the rig asymmetrically, carrying away momentum, that may allow for an extremely small push in opposite direction due to conservation of momentum. Concentrate on calculating momentum of radiation since calculating forces between cables is not the same thing and and incorrect method. If analogies are preferred we talk about two distant antennas causally disconnected. Or the effect of a solar flare on a distant planet's magnetic field. Again; this is likely far from possible to build or will be so inefficient that it cannot be used, as was also shown. Edited July 22, 2019 by Ghideon grammar
iNow Posted July 22, 2019 Posted July 22, 2019 On 7/18/2019 at 4:55 PM, Strange said: Jesus H Christ. Still? <...> How many times does this need to be explained? On 7/18/2019 at 4:56 PM, iNow said: At least for another ten pages Woot! Merely 4 days later and we’re already half way there
J.C.MacSwell Posted July 22, 2019 Posted July 22, 2019 5 hours ago, MPMin said: If you are referring to the emp as being part of (or remaining attached to) the craft... I'm not. Best of luck.
MPMin Posted July 22, 2019 Author Posted July 22, 2019 (edited) 7 hours ago, Ghideon said: EMP leaves craft carrying momentum, not carrying force. You are (still) using wrong formula. The emp leaves the wire carrying both momentum and magnetic force. You guys are analysing the momentum and not the magnetic force. It’s the magnetic force that pushes the wires and an emp is a segment of that magnetic force which becomes detached from the wire when the current is pulsed. F = I x L x B is the right formula to calculate the force on the wire from the magnetic force of the emp. I have disregarded the momentum of the emp as it’s effect is virtually insignificant compared to the magnetic force. The emp’s proximity to the craft is irrelevant because it is no longer attached to the craft, further more, as I mentioned earlier in the thread, an emp can pass through non conductive substances but will interact with a wire carrying a current. In the design the wires will need to be mounted on a nonconductive material so that the magnetic forces only interact with the wires when they carry a current. Edited July 22, 2019 by MPMin
J.C.MacSwell Posted July 22, 2019 Posted July 22, 2019 7 minutes ago, MPMin said: The emp leaves the wire carrying both momentum and magnetic force. You guys are analysing the momentum and not the magnetic force. It’s the magnetic force that pushes the wires and an emp is a segment of that magnetic force which becomes detached from the wire when the current is pulsed. F = I x L x B is the right formula to calculate the force on the wire from the magnetic force of the emp. I have disregarded the momentum of the emp as it’s effect is virtually insignificant compared to the magnetic force. The emp’s proximity to the craft is irrelevant because it is no longer attached to the craft, further more, as I mentioned earlier in the thread, an emp can pass through non conductive substances but will interact with a wire carrying a current. In the design the wires will need to be mounted on a nonconductive material so that the magnetic forces only interact with the wires when they carry a current. With an equal but opposite force on what...exactly...during this "carrying"?
MPMin Posted July 22, 2019 Author Posted July 22, 2019 1 minute ago, J.C.MacSwell said: With an equal but opposite force on what...exactly...during this "carrying"? You’ll need to add context to your question? -2
Mordred Posted July 23, 2019 Posted July 23, 2019 (edited) 2 hours ago, MPMin said: You’ll need to add context to your question? Seriously after how many pages on Newtons law ? Anytime any force pushes on something that object pushes back. If you prefer let's assume it pushes back the EM field if you want to keep that disconnected with everything else. Now you specified all the EM radiates out of the craft. However this wouldn't be true as wire B is a conductor...So now you have an imbalance in a different direction. A portion of that field will be absorbed just like an antenna. As your craft will most likely have conductive material Ie other electronics you had best provide shielding to those. Get the picture not all of the EM radiation will radiate evenly in a craft. Some will reflect off walls, some will penetrate but certainly only a small percentage. The penetration factor of the EM field will depend upon its frequency. However as we're dealing with a Pulse we can apply Beer-Lamnbert Law. https://en.m.wikipedia.org/wiki/Penetration_depth Like I said until the radiation leaves the ship it is still part of the ship as it will continue to act upon the craft in some manner either it penetrates or it reflects/refracts. LMAO your going to want shielding regardless of your pulse. You are flying in space outside the magnetosphere. "Cosmic rays have sufficient energy to alter the states of circuit components in electronic integrated circuits, causing transient errors to occur (such as corrupted data in electronic memory devices or incorrect performance of CPUs) often referred to as "soft errors." This has been a problem in electronicsat extremely high-altitude, such as in satellites, but with transistors becoming smaller and smaller, this is becoming an increasing concern in ground-level electronics as well" https://en.m.wikipedia.org/wiki/Cosmic_ray You might want to think again about designing your craft to radiate your EM pulse in every direction outside your craft... 2 hours ago, MPMin said: The emp leaves the wire carrying both momentum and magnetic force. You guys are analysing the momentum and not the magnetic force. Do you actually believe there is a difference? Have you looked at the Maxwell radiation pressure formula ? [latex] S=\frac{1}{\mu_0}E*B [/latex] where E and B are the electric and magnetic fields respectively. [latex]F_{rp}=dp/dr[/latex] where p is the momentum. If the radiation is absorbed [latex]F_{rp}=\frac{\langle S \rangle A}{c} cos \vartheta[/latex] If it is reflected [latex]F_{rp}=\frac{2\langle S \rangle A}{c} cos^2 \vartheta[/latex] Edited July 23, 2019 by Mordred
MPMin Posted July 23, 2019 Author Posted July 23, 2019 2 hours ago, Mordred said: Seriously after how many pages on Newtons law ? Anytime any force pushes on something that object pushes back. If you prefer let's assume it pushes back the EM field if you want to keep that disconnected with everything else. Now you specified all the EM radiates out of the craft. However this wouldn't be true as wire B is a conductor...So now you have an imbalance in a different direction. A portion of that field will be absorbed just like an antenna. As your craft will most likely have conductive material Ie other electronics you had best provide shielding to those. Get the picture not all of the EM radiation will radiate evenly in a craft. Some will reflect off walls, some will penetrate but certainly only a small percentage. The penetration factor of the EM field will depend upon its frequency. However as we're dealing with a Pulse we can apply Beer-Lamnbert Law. https://en.m.wikipedia.org/wiki/Penetration_depth Like I said until the radiation leaves the ship it is still part of the ship as it will continue to act upon the craft in some manner either it penetrates or it reflects/refracts. LMAO your going to want shielding regardless of your pulse. You are flying in space outside the magnetosphere. "Cosmic rays have sufficient energy to alter the states of circuit components in electronic integrated circuits, causing transient errors to occur (such as corrupted data in electronic memory devices or incorrect performance of CPUs) often referred to as "soft errors." This has been a problem in electronicsat extremely high-altitude, such as in satellites, but with transistors becoming smaller and smaller, this is becoming an increasing concern in ground-level electronics as well" https://en.m.wikipedia.org/wiki/Cosmic_ray You might want to think again about designing your craft to radiate your EM pulse in every direction outside your craft... Do you actually believe there is a difference? Have you looked at the Maxwell radiation pressure formula ? S=E∗H Thought about it, just mount the propulsion system out the back of the craft. I never specified it had to be inside. And yes I do believe there is a difference: quote: Classically, electromagnetic radiation consists of electromagnetic waves, which are synchronized oscillations of electric and magnetic fields that propagate at the speed of light, which, in a vacuum, is commonly denoted c. In homogeneous, isotropic media, the oscillations of the two fields are perpendicular to each other and perpendicular to the direction of energy and wave propagation, forming a transverse wave. reference to quote: https://en.m.wikipedia.org/wiki/Electromagnetic_radiation let’s examine this again: 3 hours ago, MPMin said: The emp leaves the wire carrying both momentum and magnetic force. You guys are analysing the momentum and not the magnetic force. It’s the magnetic force that pushes the wires and an emp is a segment of that magnetic force which becomes detached from the wire when the current is pulsed. F = I x L x B is the right formula to calculate the force on the wire from the magnetic force of the emp. I have disregarded the momentum of the emp as it’s effect is virtually insignificant compared to the magnetic force. The emp’s proximity to the craft is irrelevant because it is no longer attached to the craft, further more, as I mentioned earlier in the thread, an emp can pass through non conductive substances but will interact with a wire carrying a current. In the design the wires will need to be mounted on a nonconductive material so that the magnetic forces only interact with the wires when they carry a current. Yes there is a difference. As quoted above, the emp carries both the momentum of radiation pressure and a magnetic field. When the emp interacts with a non conductive material the radiation pressure of the emp is the prevailing force as the magnetic force of the emp does not interact with the non conductive material. However when the emp interacts with a conductive material it will produce a current or at least move electrons within the conductive material but when the electrons are already moving ie like a wire carrying a current then the emp will produce a force on the wire as with F = I x L x B
Mordred Posted July 23, 2019 Posted July 23, 2019 (edited) Then you should have specified that long ago. Did you not see the reflection formula ? Of course the EM reacts to a non conductive material. The EM field exerts the radiation pressure look at the formulas. You don't treat them as seperate entities when calculating the force exerted. The Poynting vector formula shows that. It would be foolish to count the magnetic force then count the radiation pressure. When all you need is to account for the pressure term. Especially since pressure is force per unit area. Radiation passes through conductive material far easier than non conductors. If you don't believe me how do you think capacitors work it is two plates separated by an insulator. Even if the radiation is first absorbed then emitted to the exterior it still exerts a force. See the absorption formula above. How much passes through without interaction, gets reflected or gets absorbed will depend on the materials but you can lay good money you will most likely involve all three regardless of material. Lol when you get right down to it you can get better thrust pulsing against a good reflective surface. The surface area would be far greater than a wire . (That is how the EM drive was designed that was tested by both NASA and US Navy.) AFIAK to null results. Edited July 23, 2019 by Mordred
MPMin Posted July 23, 2019 Author Posted July 23, 2019 24 minutes ago, Mordred said: The EM field exerts the radiation pressure look at the formulas. You don't treat them as seperate entities when calculating the force exerted. The Poynting vector formula shows that. It would be foolish to count the magnetic force then count the radiation pressure. When all you need is to account for the pressure term. You do treat them as seperate entities when calculating the force on a wire carrying a current as explained above. The magnetic field component of the emp cannot be ignored as it is part of the emp as referenced above, thus when the current I is 0 in the receiving wire the force generated from the magnetic field component is 0, however when the current I has a value so does the force as expressed by F = I x L x B
Mordred Posted July 23, 2019 Posted July 23, 2019 (edited) Sure but you claiming that the EM field has no further influence which is obviously wrong. It will continue to act upon other items than just wire B. Part of it will get reflected back to wire A. You cannot choose to ignore other factors. Even if it's external to the craft it will react to the ionized gases present there. Thus exerting counter forces to propulsion. Edited July 23, 2019 by Mordred
MPMin Posted July 23, 2019 Author Posted July 23, 2019 (edited) 21 minutes ago, Mordred said: Sure but you claiming that the EM field has no further influence which is obviously wrong. It will continue to act upon other items than just wire B. Part of it will get reflected back to wire A. You cannot choose to ignore other factors. Even if it's external to the craft it will react to the ionized gases present there. Thus exerting counter forces to propulsion. In terms of the emp interacting with the rest of the craft I agree which is why mounting it on the back of the craft is most likely the best place for it. As for the interaction of the non magnetic field component against the wires, as the currents can be reversed to produce the force in the wire in either left or right direction, I guess it would be more efficient to produce the magnetic force in the same direction as the radiation pressure on the wire. This might reduce the number of cycles the system can perform per second and would need a separate analysis to see if the reduced cycles Is worse than producing a magnetic force against the radiation pressure in exchange for the extra cycles Edited July 23, 2019 by MPMin
Mordred Posted July 23, 2019 Posted July 23, 2019 (edited) Agreed the best place is the back of the craft with as near a perfect reflector on three sides . Reflection produces twice the force as absorption. This will also allow reaction to any ionized gases with a more uniform directional force. Edit upon reflection pardon the pun a parabolic reflective dish would be a better option. PS now you will take advantage of other vectors and maximizing the effects from you pulses. Edited July 23, 2019 by Mordred
Ghideon Posted July 23, 2019 Posted July 23, 2019 5 hours ago, MPMin said: The emp leaves the wire carrying both momentum and magnetic force. You guys are analysing the momentum and not the magnetic force. It’s the magnetic force that pushes the wires and an emp is a segment of that magnetic force which becomes detached from the wire when the current is pulsed. Show a reference where force (measured in newtons) is detatched. 1
MPMin Posted July 23, 2019 Author Posted July 23, 2019 12 hours ago, Ghideon said: Back in the early pages of the thread there was an analysis of the momentum from EMPs. Here is a modified version again addressing the discussion of radiation leaving the craft vs radiation in the craft. There is still no analysis of how possible it is to create the device, only an analysis of momentum is done. There still seems to be an agreement that ”particles carrying momentum” is reasonable model for the pulse as long as no detailed calculations are needed for the interaction at cable B. Einstein looked at the special case of the energy being carried by a light flash. According to Feynman the argument would work as well for particles. Are we not talking about the same thing?
Ghideon Posted July 23, 2019 Posted July 23, 2019 19 minutes ago, MPMin said: Are we not talking about the same thing? No. I talk about the momentum of particles and radiation. You have disregarded the only thing that could possibly create a tiny inefficient thrust; the momentum of particles and radiation. You seem to replace that with an invalid application of F = I x L x B, based on some concept of a "detected force" without reference. You try to find a way to get more thrust from magnetics than from the momentum of the asymmetrically escaping radiation, resulting in unsupported speculations about concepts that seems not to exist. 7 hours ago, MPMin said: The emp leaves the wire carrying both momentum and magnetic force. You guys are analysing the momentum and not the magnetic force. It’s the magnetic force that pushes the wires and an emp is a segment of that magnetic force which becomes detached from the wire when the current is pulsed. F = I x L x B is the right formula to calculate the force on the wire from the magnetic force of the emp. I have disregarded the momentum of the emp as it’s effect is virtually insignificant compared to the magnetic force. Show a reference where force (measured in newtons) is detached.
MPMin Posted July 23, 2019 Author Posted July 23, 2019 13 minutes ago, Ghideon said: Show a reference where force (measured in newtons) is detached. https://en.m.wikipedia.org/wiki/Electromagnetic_pulse quoted from reference: An electromagnetic pulse (EMP), also sometimes called a transient electromagnetic disturbance, is a short burst of electromagnetic energy. Such a pulse's origin may be a natural occurrence or man-made and can occur as a radiated, electric, or magnetic field or a conducted electric current, depending on the source. Also quoted from reference: EMP energy may be transferred in any of four forms: Electric field Magnetic field Electromagnetic radiation Electrical conduction Due to Maxwell's equations, a pulse of any one form of electromagnetic energy will always be accompanied by the other forms, however in a typical pulse one form will dominate. as you can see the emp contains magnetic field as represented by B in the equation F = I x L x B the resulting force is measured in Newton’s
Strange Posted July 23, 2019 Posted July 23, 2019 8 hours ago, MPMin said: The emp leaves the wire carrying both momentum and magnetic force. You guys are analysing the momentum and not the magnetic force. It’s the magnetic force that pushes the wires and an emp is a segment of that magnetic force which becomes detached from the wire when the current is pulsed. I (at least) am considering both. My example with the two spheres thrown in opposite directions, explains this exactly. There is a force on the wire that intercepts the left sphere (or pulse), no net force on the machine that throws the two spheres (equivalent to there being no reaction no force on wire A), the only thing that moves the craft is the sphere (pulse) leaving the craft to the right. The force has no effect on the craft (as you admit, because nothing happens if the entire system is encased in a metal mix). The only effect is the momentum that leaves the craft (as you admit, because nothing happens if the entire system is encased in a metal mix).
Ghideon Posted July 23, 2019 Posted July 23, 2019 16 minutes ago, MPMin said: as you can see the emp contains magnetic field as represented by B in the equation F = I x L x B the resulting force is measured in Newton’s The part above is not part of the reference you posted. It is something you added. Provide references for your claim of a ”detached force”, measured in newtons.
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