MPMin

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 

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? 

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 

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 

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 

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? 

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. 

51 minutes ago, Strange said:

If you were correct, this would violate the conservation of momentum. You claim that the craft will move to the right. This means an increase in momentum to the right. To meet the conservation law (because there is no external force) there must be something, separate from the craft, that carries momentum to the left. There isn't anything, so it can't happen. 

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 

39 minutes ago, J.C.MacSwell said:

I said it could be considered as part of the craft/system. Do you see the advantage of being able to do that?

If you are referring to the emp as being part of (or remaining attached to) the craft, I don’t understand how this would be an advantage in my design? 

16 minutes ago, Strange said:

This is 100% wrong.

I disagree, for the same reason the magnetic field from A and wire B repel each other as shown in my animated example (move to the left) reversing the current in wire B would cause wire B to move into the emp from A (move to the right) 

38 minutes ago, Strange said:

You will get the same thrust with no current, or the current flowing in the opposite direction. The thrust is not caused by the force on the wire.

 

Think about the two wires in parallel and imagine them both carrying a continuous current for a moment; what’s the force pushing or pulling them apart in the continuous current situation? Further more, the force can be reversed by reversing one of the currents.

If you were to stop the current in one of the wires, does the force between the two wires stop the very instant one of the currents stops, or does the force between the wires continue until the magnetic field from the wire that was switched off finishes traveling towards the wire with the continuous current? I’m sure it’s the later. 

The emp is essentially a segment of the magnetic field traveling towards the wire with the current, therefore: 

F = I x L x B (where B is the magnetic field in Tesla’s and not the wire B)

Which is the equation for the force on a wire carrying a current in a magnetic field, no current, no force

Further more, reversing the current in wire B would cause wire B to be attracted to the magnetic field from wire A thus moving the craft to the right. 

50 minutes ago, Strange said:

That will still happen in a sealed metal box. And yet ... no thrust.

That will happen with any propulsion system and doesn’t conclude anything.

The disk analogy isn’t exactly the same. For example, you can have no current in B and no thrust, that’s like opening the window in that direction and letting the disk fly straight through, but closing the window will only generate as much thrust as the disk can transfer to the craft, where as the current in wire B can be increased beyond the current that produced the emp in wire A thus generating more thrust at B: I still maintain that the force occurs at the wires when the emp interacts with the wire carrying a current as the amount of current in the receiving wire is variable which can vary the amount of thrust generated where as a disk hitting the wall at B has only potential of the momentum in the disk hitting it. 

Thank you Strange for understanding my design, I appreciate you acknowledging that it could work.

I’m definitely not trying to start a new argument unnecessarily but I don’t agree that it’s the radiation going out the back that’s producing the momentum in the craft when using emps. The reason I think it’s the force at wire B is increasing the amperage in wire B will increase the force exerted on the emp from wire A without pushing more emp energy out the back from wire A; If this is incorrect (and I’m pretty sure it isn’t incorrect) then it’s still just a ‘six of one half dozen of the other’ argument as to where the force comes from as at the end of the day it’s action and reaction. Having said that though, you are right that the system loses energy in all directions except wire to wire and your analogy of throwing physical objects is correct. However, as the system can use renewable energy from the sun to generate emps, its inefficiency isn’t a deal breaker as it will produce more thrust per m2 than solar radiation pressure on the solar panels will and as technology improves the wires will be able to be mounted closer together which will increase the efficiency of the system.

i don’t know why the font size is different in the text above I didn’t indent that to happen 

I agree with Newtown law, I just don’t agree that my design is violating Newton’s law 

A continuous (simultaneous) force will cancel each other out thus generating no propulsion - claiming otherwise would be in violation of Newton’s third law. Pulsing the current creates a detached force that the system can then act on that detached force as though it’s not attached to the system because it’s not actually attached to the system. As the system is acting on a force that’s not attached to the system, Newton’s third law is not being violated 

The technology is the next hurdle but thanks for listening and understanding 

8 minutes ago, Mordred said:

 As I mentioned you can experiment the Lorentz force at home to see how two wires respond. 

I have no problem with this and agree this would work with a continuous current in both wires. But I do hope you now understand what I’ve been trying to say with regard to pulsing currents. The technology can be worked on if the principal is worth testing. 

I think we are moving away from the principal on which the design works on and moving towards the technology to make it work which is not my concern at this stage. I’m just want to know if the principle of my design is at least test worthy before moving to the next phase of development. 

If the wires are only 0.1m apart, and the emp is traveling out of the wire at the speed of light, and you need to pulse the current quickly enough so that the emp only emerges by 0.01m before stopping the current (because if you let the emp reach the other wire before stopping the current it will create an attachment or simultaneous force), how quickly does the switch need to go on and off to allow this to happen... too quick for anything I have access to that’s for sure. 

The switching would have to occur quickly enough to account for the emp moving at the speed of light - this is something I can’t do in the garage. 

3 minutes ago, Mordred said:

If it's continuous or pulsed makes no difference. Why do you think I told you to add a switch to the circuit ?

It does make a difference; a continuous (simultaneous) force will cancel each other out. The pulse creates a detached force, that system can then act on that detached force as though it’s not attached to the system because it’s not attached. As the system is acting on a force that’s not attached to the system, Newton’s law is not being violated 

5 minutes ago, Mordred said:

A applies force to B B applies an equal but opposite force on A both through the EM field. That is the third law.

Ok you didn’t show me, you just used another example of a continuous force. You still haven’t addressed that my design doesn’t use a continuous force. Perhaps it’s more pertinent to say a simultaneous forces.   

19 minutes ago, MPMin said:

You are effectively saying that once the emp has left the wire it can not do work independently of the source wire. That’s like saying the source for a pulse of light can’t move from its position after the pulse of light leaves the source until the pulse of light arrives at its destination - you are creating an attachment where there isn’t one 

3 minutes ago, Mordred said:

You have to complete all the Newton laws when you do the calculations. Skipping the third law isn't complete

Ok please show me.

15 minutes ago, Mordred said:

Perform the experiment it will prove you wrong. I have done this experiment back in high school. You can apply the current however you choose. Both wires will attract to each other or repel from each other. Even if no current flows through the second wire.

Well that’s a surprise, I guess on this point I’d like to see what others have to say about this particular issue. When ‘sticking to the science’ as they say,  the mathematics do not allow for there to be a force between the two wires if one of the wires has no current. Try the mathematics on this one I’m curious as to what you will find? 

There is also the issue of the emp detachment that hasn’t been addressed yet. Let me say again:

21 minutes ago, Mordred said:

Take two magnets for Christ sake. 

You will continue to frustrate yourself while you continually use references that don’t apply to my design. You are still using continuous magnetic fields as a reference when I’ve said multiple times it’s not a continuous magnetic field, the incontinuity of the magnetic field is what creates the detached emp! Whilst you ignore this point we’ll get nowhere. 

You have not addressed your lack of understanding of your own reference. 

27 minutes ago, Mordred said:

Your claim that the EM field no longer affects wire A is false just as it is in this experiment.

 

I have explained this claim: 

2 hours ago, MPMin said:

You are effectively saying that once the emp has left the wire it can not do work independently of the source wire. That’s like saying the source for a pulse of light can’t move from its position after the pulse of light leaves the source until the pulse of light arrives at its destination - you are creating an attachment where there isn’t one 

If you refuse to directly address the points I’m making then we are not talking about the same things 

55 minutes ago, Mordred said:

Your laser pulse pushes against the emitter and the target. Just like recoil  from any gun.

 

You have ignored that this recoil happens in both directions along the x axis and cancels it’s self out when emitting the emp   

7 minutes ago, Mordred said:

I don't have to create anything. I even provided a reference. Research it yourself try and prove me wrong.

 

You don’t have to create anything but you are anyway. You have misunderstood your own reference, to illustrate this point; you think that only one of the wires needs a current to produce a force between the wires. As you don’t understand your own reference, your understanding of my design is as flawed as your understanding of your own reference. 

15 minutes ago, Mordred said:

I also gave you the option on how to do so. Find one craft of any type  where the force is not applied external to the craft to generate thrust.

 

As I said, the detachment of the emp from wire A essentially makes the emp an external force because it is no longer attached to the craft when it interacts with the other wire.

You will have to explain how the detached emp is still attached to the craft when it interacts with the other wire to make your point valid. 

54 minutes ago, J.C.MacSwell said:

The advantage you have is that you could still consider it part of the system/craft. In so doing you can readily tell that you should not get the result that you seem to expect. As the physics is the same either way...the result cannot be any different...

The point is; I don’t consider the emp attached to the craft when it interacts with the other wire. 

Considering the emp has left the wire behind as it emanates outward, can you please explain how it’s still attached to the craft even though it’s no longer attached to the wire it came from? 

47 minutes ago, Mordred said:

This part  in accordance with Newtons third law.  Wire B must oppose wire A with the same force A makes  on B.

Only when the forces occur simultaneously

You are effectively saying that once the emp has left the wire it can not do work independently of the source wire. That’s like saying the source for a pulse of light can’t move from its position after the pulse of light leaves the source until the pulse of light arrives at its destination - you are creating an attachment where there isn’t one