DandelionTheory

Lol. The rails bro. Add the rails and they press on each other direcly away from each other and the slug because of the angle of interaction(you know, because of sin theta in the lorentz force). The magnetic field only points it, the current moving does the work. You still don't get it's a 3 body solve with rotation to a 2 body problem. External fields means it's not one person. If 2 equal masses are attached by a long non-rigid wire on equally opposite sides of an explosive, and I was holding the middle of the wire as far from them as possible without pulling. When the explosive goes off both masses will fly away at equal magnitudes while pulling me towards the explosion. If more energy is in the momentum of the masses after propelling me forward they would pull the wire to it's tensile limit and either brake or return with spring force, maybe even collide if conditions are right. I guess it depends on if I can let go of the wire. So would a Delta t variable be viable? "The result is a magnetic field between the two rails (B = 2x(u0/2pi)x(I/r)) and an intercepting field by the armature. The rails repel one another (F = u0I1I2L/2piR) and they both repel the armature (F = ILB). Since rails are both fixed the net result is a propulsive force on the armature, which will be accelerated forward by electromagnetic means (A=ILB/M)." From http://www.powerlabs.org/railgun.htm If a positive particle came into contact with 2 magnetic fields intersecting, the particle would be forced to sin theta and the magnetic fields would spin in place.

http://web.mit.edu/mouser/www/railgun/physics.html https://en.m.wikipedia.org/wiki/Ampère's_force_law https://isaacphysics.org/concepts/cp_lorentz_force Good enough yet? Current experiences force at right angles to the perpendicular magnetic field it's in.

I agree, but you only need to push off a surface not hold it. Once it's been pushed off of the action is done. I agree, but what's a rail gun? If you don't know, the rails fly away from each other while propelling the armature in one direction. The recoil isn't the opposite of the force on the armature. If a wire with a 90° bend in it has current running through it, the current, not the wire, will experience force... From itself. So it depends on how you "push" on the box. If you are current and you run along 3 edges of the box, your magnetic field you generate will do work on you as you make a 90° turn. Your magnetic field pushes on you.

It means you don't have to uphold an image to be correct. I do not have to be a certain way to find truth or play with physics. The current is the hinge. You don't think it applies why? Because you don't want it to? Hinge has attached parts to a fulcrum, the magnetic fields are attached. In the updated version of the machine design I make the internal angles smaller thereby making it a concave parallelogram. -DandelionTheory

Remember the 2 persons and the wall? If the wall had a hinge.... The wall wouldn't move much unless the hinge was open alot. *Bong toke* Equal but opposite haha

I added specifics to show you you missed the magnitude in your calculation that strictly processed angles. Then the definition needs to look at this 3 body idea to canceling magnitudes. not as an argument against Newton's 3rd law, but a work around by stearing 2 of the bodies, after the initial reaction, to hit together thereby adding another action to the system. That reaction's magnitude diminishes the closer their initial velocity angle gets to parallel.

Not reactionless, you just don't understand reactions can cancel and assume magnitude and angle are the same thing. I do not claim it's reactionless, you already have though. It's funny you did the math you wanted and "forgot" to do the real math involved with the reaction I'm specific to. You don't believe you can "push off a wall" that way, cool, did you know it's rotational torque and not exactly a "push"? So when I interacts with B1 rotational torque is experienced by both, when I interacts with B2 rotational torque is experienced by both. Good? Good. When B1 and B2 physically make contact with each other how much force do you think is absorbed? That is another action depending on the interaction angle. So we have another action reaction, and opposite actions cancel when they collide head on. So they need to be as "head on" as possible.

No you just refuse to understand physics is explained with vectors and can be translated between them. You insist 3 force vectors cancel out if one is opposite direction of the other 2 added together, I gave you a way to reduce one magnitude and you do not agree. Cool. I'm not trying to make magic reality, I'm arguing you can cancel an opposite force magnitude if you have 3 bodies, while one gets F x 1 the other 2 get F x 0.5 You say the other 2 cannot cancel out. I say they do if they hit each other "head on". How much? Well it depends on how "head on" they interact with the first body that took half the energy in the first reaction. Ie it's angle. You are arguing with Newton here not me. I can do some calculation and find vector angles with magnitudes which can always be applied to physical conditions. The fact that you argued that from that point of view means physics is still not understood by you. Conservation of momentum requires the magnetic fields of the interacting particles to be intact, not bent by steel. So did you forget to calculate interacting vector angles at each point? Or just assume you could do the math in your head...? I'm assuming you're mind is stuck on the magnetic fields of each current opposing each other, rightly so! How much and what direction? Or did you forget that counts... Lol the force the current puts on each interacting magnetic field is rotational, did you negate that too? Can rotating objects hit each other? Yes, can their momentum be changed? Yes with a magnetic field. Which one? The current.... Haha. If someone shoves you, your arms swinging causes you to absorb some energy rotationally when you make them come together. on the way down to the ground, if you clapp your hands wide around your body with enough energy you will wiggle in place before hitting the ground.

The smaller the angle, the smaller the magnitude. Oh look, a smaller angle. Daaamn that magnitude is still not opposite. Again smaller. So 3 bodies... Do I have a little credit now? -DandelionTheory It looks like you don't know how to add vectors properly. May I suggest going back and paying attention to angles and magnitudes. Because pushing off the center is a thing you don't understand when balancing forces.

Cool. Thanks for the doubt.

Yes I'm attempting to produce lift from imbalanced internal forces. Those internal forces do not balance if you use 3 bodies and some fancy vector math. The momentum from the ions in the wire is transferred to the wire via the lorentz force. "When a wire carrying an electric current is placed in a magnetic field, each of the moving charges, which comprise the current, experiences the Lorentz force, and together they can create a macroscopic force on the wire" ie, the magnetic fields do not to work on the wire they do work on the ions in the wire. so the "ions momentum" is changed, not the lifts momentum initially. It's after the force is transferred to the wire does the lift have enough momentum to move. I notice the imbalance of total momentum is towards the side of the lift with the smallest angle due to the "reaction" of I1 on B4 and B5 being pointed towards the center and losing magnitude.

Also this.

Does this make a better argument? I tried to show you it's the angle between B1 and B2 that determines the magnitude of F2 and F3 added together. Doesn't the lorentz force state the ions in the current being acted on transfers momentum to the wire?? "When a wire carrying an electric current is placed in a magnetic field, each of the moving charges, which comprise the current, experiences the Lorentz force, and together they can create a macroscopic force on the wire"

like this. i miss labled B1 and B2 as B4 and B5, my bad.

its 0 if B4 and B5 were <1T, 90, 90> & <1T, 270, 90> respectively, and I1 was between them <10A, 0, 180> and Z origin is towards the observer, what is the net force on B4 and B5?

i edited the statement, i am sorry. i have alot on my mind. you can push off the center of something. if B4 and B5 were <1T, 90, 90> & <1T, 270, 90> respectively, and I1 was between them <10A, 0, 180> and Z origin is towards the observer, what is the net force on B4 and B5?

correct, it will be greater the greater the angle of the wall, so if the wall angle is very small you get less net motion in the downward direction. because I's magnetic field is round and each interacting B has its own angle..... wait do you know how to add vectors? because opposite angles does not mean opposite force. if you press towards the center some force is canceled out i don't know how else to say it. no i expect it to transfer momentum to the rig in the upwards direction. no i expecting B4 and B5 to push on I1 like normal, and im expecting I1 to push on B4 and B5 and that force to lose magnitude because each force angle is pointing opposite the other internally. right, you forgot forces cancel the more they become perpendicular with the center axis. im not, im pushing off the center.

i gave you an example of the 2 person problem because you wanted to know how to apply 2 forces and get a lower net force magnitude on the center. you now need to apply the force the "wall would apply to the parsons" if it did in fact "apply force", which it does in the physics issue we have been debating. if the total B field acting on I is perpendicular to its velocity, I experiences force in the plane perpendicular to both. period. so it matters what angles they are at when you calculate opposing forces. the magnetic fields that add up to total B field are not at the same angles of reaction as total B field.

"pushing off the center"

this assumes a Lorentz force calculation for F1. i do not know how to calculate F2 and F3 but i know the magnitude gets smaller the closer B4 and B5 are to touching opposite ends. also i modified the structure and did away with I6 to utilize this new information. i dare you to calculate fringe fields. -DandelionTheory

can 2 people push oppositely on the same wall and nothing happen? what if the wall was an angle and 2 people push on the outer sides? some of the force would push the wall away from the 2 people. the force on the wall/angle is proportional to the angle size.

the Lorentz force says otherwise. https://en.wikipedia.org/wiki/Lorentz_force the angles cancel, the magnitudes don't if the triangle is isosceles. okay, you agree the currents are inside and attached to the rig, you agree both B4 and B5 are interacting with I1. your logic in the example is for after the current has left the rig, not while it is in it. because you think it will be "attracted" you negated the Lorentz force calculation while its in the rig completely. You point out B4 and B5 experience force from I1, i agree, how much? because its not exactly opposite, due to the force on B4 from I1 is towards the center axis and the force on B5 from I1 is towards the canter axis, some magnitude is lost. (i just realized only if the triangle is isosceles with the smallest angel at the vertex, magnitude on B4 and B5 is lost the smaller the angle because force is towards the center)

B4 and B5 are 2 different sides with 2 different vector angles. You say the force on I1 due to B4 and B5 is the exact opposite the force on B4 and B5 due to I1. They are 3 bodies not 2. From I1 rotational torque is applied to B4 and B5 at axis A. Rotational torque towards and away from the median of B4 and B5. A median can be something to push against if it is done oppositely. I am familiar, I'm also talking bout a 3 body problem.

So fringe fields arn't a thing... Cool. I'm curious as to the magnitude of each of your F1 & F2 assumptions, it's interesting that you see the angle as opposite and not the magnitude. In the picture with the structure, I6 is the odd man out when external fringe fields are at play against I4 and I5. Your "calculation" assumes the force on A is exactly opposite in the neg y direction. You're assuming the vector calculation for total B assumes opposite reaction on 3 bodies.