Can you move in space?

[MPMin]

17 minutes ago, J.C.MacSwell said:

I will give you an example:

If say, I tell you I have a system inside a black box, and I tell you I can get more energy out from it than I put in on a continuous basis, with no loss of energy to the system over time, you should be able to tell me that is impossible based on the known laws of physics.

You don't need to know how it is designed, or what is going on inside the box.

 You could dismiss it out of hand, and no one here would dispute your claim that my design will not work.

 Without some major change in the laws of physics it simply will not work.

I get your example but it needs to relate to my design to have relevance, Ive said my system is an open system, it needs energy in and it loses energy out, my design is not in a box and I’m not saying you are saying that but if you are going to say my design is violation of newtons law tell me how? The argument that the wires are on the same rig doesn’t contravene newtons laws because you aren’t considering the detachment of the emp from wire A. Once the emp leaves wire A it is no longer part of the rig so when the detached emp from wire A interacts with wire B, wire B is effectively interacting with an outside field.

Edited July 22, 2019 by MPMin

[Mordred]

1 hour ago, MPMin said:

 

Are you saying that if one of the wires in parallel carries 10 amps and the other has no current that the force of attraction or repulsion (depending on direction of current) will be equal between the two wires?

Absolutely finally a glimmer of hope on seeing the problem.

What else can equal  but opposite  reaction of a force mean roflmao.

The link shows  this detail. I even quoted the specific line. 

Edited July 22, 2019 by Mordred

[MPMin]

6 minutes ago, Mordred said:

Absolutely finally a glimmer of hope on seeing the problem.

What else can equal  but opposite  reaction of a force mean roflmao.

 The link shows  this detail. I even quoted the specific line. 

 Edited 1 minute ago by Mordred

I think if one of the wires has no current there will be no force in any direction between the wires 

[Mordred]

Pretend the wire A is a body. (Yourself ) wire B is a wall. When you push against  an unmovable wall What happens to your body ?

The equal  but opposite force moves you from the wall.

Do you think wire A pushing against  wire B is any different ?

(Your wires are fixed mounted cannot move past the mount constraints )

Edited July 22, 2019 by Mordred

[MPMin]

5 minutes ago, Mordred said:

Pretend the wire A is a body. (Yourself ) wire B is a wall. When you push against  an unmovable wall What happens to your body ?

Do you think wire A pushing against  wire B is any different ?

I get what you are saying but your example is not the same kind of pushing, the emp will only cause moment if a current is presenting the wire.

 

[Mordred]

Really how is it different when wire B cannot separate  from wire A and this is where you are applying  your force ?

How can you possibly get thrust from that ?

(You are pretty clear in not involving any other influence  such as radiation pressure on walls or ionized gas) even though a closed box is a major factor in radiation pressure behavior.

Do how  do you gain thrust upon a technically immovable object ?

And most important how do you prevent wire B from obeying the laws of nature in preventing an equal but opposite force on wire  A.

Please look at that link I posted it so you know I'm not pulling fluff out my a@@@#. I also specifically quoted the relevant  section.

Edited July 22, 2019 by Mordred

[MPMin]

24 minutes ago, Mordred said:

Really how is it different when wire B cannot separate  from wire A and this is where you are applying  your force ?

I’ll try to explain in stages, let me know which part you disagree with:

Wire A and B are both physically connected to the same craft, you’re right there; but that fact alone doesn’t mean they are exerting a force on each other. 

Wire A and B will only produce a force on each other only when a current is present in both wires at the same time, then the forces will cancel each other out and the craft won’t move as have you described many times but that’s not the situation I’m describing in my design.

If there is a current in only one of the wires, no force will occurs between the wires. This point isn’t that important at this stage but it might help you understand the next point.

This is the crucial point: If you pulse a current through one of the wires, the emp detaches from the wire. Even though the emp originated from one of the wires on the craft, the detached emp is no longer part of the craft so when it interacts with the other wire, the other is effectively acting on a force that isn’t part of the craft anymore thus momentum of the whole craft can occur by the other wire acting on the detached emp. 

[Mordred]

15 minutes ago, MPMin said:

I’ll try to explain in stages, let me know which which part you disagree with

Wire A and B will only produce a force on each other only when a current is present in both wires at the same time.

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

See link for reference  if that isn't sufficient  I can post thousands of other examples from textbooks etc.

[latex]f_{12}=F_{21}[/latex] third law in vector form.

The fact wire B has no net current means diddly squat Newtons third law still applies. 

" For   EVERY force there is an equal  but opposite force"

EVERY MEANS EVERY FORCE. ....

Wire B does not need to generate a field to oppose a force

Edited July 22, 2019 by Mordred

[MPMin]

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 

Edited July 22, 2019 by MPMin

[J.C.MacSwell]

1 hour ago, MPMin said:

I’ll try to explain in stages, let me know which part you disagree with:

Wire A and B are both physically connected to the same craft, you’re right there; but that fact alone doesn’t mean they are exerting a force on each other. 

Wire A and B will only produce a force on each other only when a current is present in both wires at the same time, then the forces will cancel each other out and the craft won’t move as have you described many times but that’s not the situation I’m describing in my design.

If there is a current in only one of the wires, no force will occurs between the wires. This point isn’t that important at this stage but it might help you understand the next point.

This is the crucial point: If you pulse a current through one of the wires, the emp detaches from the wire. Even though the emp originated from one of the wires on the craft, the detached emp is no longer part of the craft so when it interacts with the other wire, the other is effectively acting on a force that isn’t part of the craft anymore thus momentum of the whole craft can occur by the other wire acting on the detached emp. 

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...

 

 

Compare this to picking a reference frame. Regardless of which one you pick you should get the same result...some are simply more convenient and allow things to be more clear.

[Mordred]

1 hour ago, MPMin said:

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 

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

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.

Good luck on that challenge

It does not matter how you generate a force the third law applies.

After how many pages of repeating the same thing I've grown tired of this.

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

The EM field will push back on wire A as it pushes against wire  B.

Go ahead take two Newtonian scales and attach them to two electromagnet. As you pulse on one electromagnet they both will repel or attract depending on polarity. You can readily do that experiment at home.

 

 

Edited July 22, 2019 by Mordred

[MPMin]

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? 

[Mordred]

Take two magnets for Christ sake. Try and push them together in repulse alignment. Can you not feel magnet one resisting you as you push it against magnet two ? I'm positive you have two fridge magnets you can use.

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

Both magnets certainly radiate their field in all directions but you will still encounter resistance as you push against magnet two.

Better yet take two wires hang them on a mount with enough slack to move. Connect a battery with a switch in circuit. Both wires will move toward each other or away from each other.

Another household experiment you can perform see for yourself.

Edited July 22, 2019 by Mordred

[MPMin]

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   

[Mordred]

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.

No need to take my word for it. Test it for yourself.

Edited July 22, 2019 by Mordred

[MPMin]

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:

 

Edited July 22, 2019 by MPMin

[Mordred]

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

[MPMin]

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.

[Mordred]

Yes 

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

I did see the link then look at the last formula on that link notice that the force acts upon the radius of seperation between the wires and that in the third formula they combined the force of both wires.

Edited July 22, 2019 by Mordred

[MPMin]

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.   

Edited July 22, 2019 by MPMin

[Mordred]

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

It is an easy at home experiment. I recommend a battery for DC voltage and safety reasons.

Edited July 22, 2019 by Mordred

[MPMin]

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 

[Mordred]

So you claim I for one know different. Do the experiment take a video and prove me wrong if you can.

If you want a pulse generator use an RC circuit instead of a switch. Though quickly flicking the switch does the same. Though momentary switches are a dime a dozen.

Edited July 22, 2019 by Mordred

[MPMin]

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. 

[Mordred]

Sure you can use an RC circuit. How do you think we generate DC pulses ?

 If I recall your gap was 0.1 with 10 amps. How did you plan on delivering a 10 amp current with a pulse ? That's a pretty impressive inductive kick 

Edited July 22, 2019 by Mordred