I need help with my math problem.

[DandelionTheory]

11 minutes ago, Ghideon said:

Thanks.

You have three bodies: the wall and two persons. Note that you have drawn the forces from the walls perspective, not the total sum of forces for the three bodies. The persons are external in your picture, not part of the wall, not attached to the wall. What do you predict regarding acceleration of the bodies in the bottom picture? What relative motion of the bodies do you expect?

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.

Edited January 24, 2020 by DandelionTheory

[Ghideon]

13 minutes ago, DandelionTheory said:

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.

You have provided a drawing where two people pushing the wall and since there is a net force on the wall the wall will be pushed away, downwards. The distance between wall and individuals will increase. 

Now look at the sketches for your rig. You  have drawn the force in the same way as in the wall example, forces are from the cables perspective. The cable is embedded the in the iron structure. Do you expect the cable to brake loose and accelerate upwards? Or do you expect the iron structure to push back at the cable with an equal and opposite force? These questions are quite crucial to the understanding of the issues. You can't calculate forces on an individual internal part of a structure and apply that force as an accelerating force on the whole structure. 

Edited January 24, 2020 by Ghideon
spelling

[DandelionTheory]

22 minutes ago, Ghideon said:

You have provided a drawing where two people pushing the wall and since there is a net force on the wall the wall will be pushed away, downwards. The distance between wall and individuals will increase. 

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.

Quote

Do you expect the cable to brake loose en accelerate upwards?

no i expect it to transfer momentum to the rig in the upwards direction.

Quote

 Or do you expect the iron structure to push back at the cable with an equal and opposite force?

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.

Quote

These questions are quite crucial to the understanding of the issues.

right, you forgot forces cancel the more they become perpendicular with the center axis.

Quote

You can't calculate forces on an individual internal part of a structure and apply that force as an accelerating force on the whole structure.

im not, im pushing off the center.

Edited January 24, 2020 by DandelionTheory

[Ghideon]

5 minutes ago, DandelionTheory said:

wait do you know how to add vectors?

Yes.
Do you?

9 minutes ago, DandelionTheory said:

isn't that what a commercial airliner is?

No.
 

Idea: If you look at the wall and the individuals again, the last one on the page, what will it look like after a while? Do you know about the equation F=mA from Newtonian mechanics?

[DandelionTheory]

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?

Edited January 24, 2020 by DandelionTheory

[Ghideon]

3 minutes ago, DandelionTheory said:

i edited the statement, i am sorry.

Ok! No problem. Again:

11 minutes ago, Ghideon said:

Idea: If you look at the wall and the individuals again, the last one on the page, what will it look like after a while? Do you know about the equation F=mA from Newtonian mechanics?

Adding to the above, since your drawing has a net force on the wall it is not a static situation. What will it look like at some time later? What does Newton say?

[DandelionTheory]

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?

Edited January 24, 2020 by DandelionTheory

[DandelionTheory]

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

Edited January 24, 2020 by DandelionTheory

[DandelionTheory]

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"

Edited January 24, 2020 by DandelionTheory

[DandelionTheory]

Also this.

[Ghideon]

2 hours ago, DandelionTheory said:

Does this make a better argument?

I think you might want to slow down and address the more basic issues first. 
Note: you are not arguing against me, you are arguing against Newton. I'm just the messenger*. 

So I'll drop the details and go back to basics for a while until that is sorted out.

Situation at the top. Newton's third again: "For every action, there is an equal and opposite reaction." The wall is pushed by two equal forces that cancels each other. Put person P1 and P2? They push against the wall and the wall pushes back with the same force. You have not drawn those forces. 

I will skip middle image since it is in principle the same as the bottom."For every action, there is an equal and opposite reaction." So the wall pushes back on the persons with force F1 and F2. And since you have nothing to balance the force FTotal F=ma tells us that the wall will accelerate. OR are there other forces you are not drawing? Are there a force from the ground, holding back the wall? And the individuals, are they standing firmly on the ground or are they sliding backwards while pushing? What you are drawing is a free body diagram of the wall, not a diagram for a system of two persons, wall, and ground. Note: the vector sum (F1 + F2 = Ftotal) seems ok. So yes FTotal becomes larger. As will the force from the ground holding back the wall, unless you accept the persons to push away the wall.

You seem to constantly confuse components of a system and its internal forces with the sum of forces acting on the complete system. And each time we get close to the core you throw in a set of magnetic fields that seems to get the discussion of track.

To further reduce the possibility for misunderstandings about the intention with the lifter, here is again a question about its basic workings:

Let's put the lifter in a concealed box. The box has low mass to not interfere with the lifters capacity. "The lifter" means everything that is required for it to operate. Power sources, cables etc. There is no connection to the outside of the box; no electro magnetics, cables, noting. Note that I draw a simple 3d drawing. This is not a cross section of a larger structure. The lifter is started (via a timer since it is not readable from outside the box). If the lifter works as you intend it to do, will the box lift into the air, pushed by the lifter inside it?

This is as simple and abstract as it can get I think. It does not matter how the lifter is constructed only that you managed to build it somehow. 
From my earlier questions and your responses I think you intend the lifter to work in this manner but I want to be sure.

 

 

 

Standard disclaimer: There is always the possibility that I'm wrong. I'm trying to argue from a mainstream perspective but my knowledge is of course limited.

[DandelionTheory]

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.

 

[Ghideon]

47 minutes ago, DandelionTheory said:

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.

Thanks! Then guess the initial question regarding the vector math have been answered: "I need help with my math problem". The vectors I helped verify is correct. But due to a misunderstanding of the physics you end up with something physically impossible, the vectors does not describe a physical valid configuration. By the way, "fancy vector math", when applied to the correct set of forces, will show that internal forces cancel*. Unfortunately you have been unable to follow along in my attempts at explaining this.

That means we have reached the end of this thread. From here I see two options, up to you.
1: continue to argue that Newton (and Einstein and others) were wrong, move to speculations and led the moderators close the thread**. You have no evidence that your physics works and seem to object to any attempts to show you how it works or where there are misunderstandings. 

2: Continue asking about how physics works, how to analyse systems, add forces, implications of newtons laws etc (maybe a separate thread?) Since we seem to be able to communicate rather well using pictures such a discussion could be productive. But that is only if you are interested in learning, it's up to you.

 

55 minutes ago, DandelionTheory said:

"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"

True. And used all the time in electric motors for instance. But have you any reference that this rule is allowed to break the rules of momentum conservation? You can't pick one rule from physics, apply the math in isolation, and expect a valid result.

 

1 hour ago, DandelionTheory said:

I notice the imbalance of total momentum

That is due to imaginations or misunderstandings. The laws of physics seems to disagree with your statement.

 

By the way, maybe there is a third option:

On 1/21/2020 at 2:39 AM, DandelionTheory said:

my tutor helps me

Maybe they could help you with understanding the basics about mechanics? 

 

 

*unless the stricture breaks apart of course, accelerating cables, iron cores and what not, all around.
** It would be cool to be proven wrong, that evidence supporting this lifter can be provided. But I doubt that.

[DandelionTheory]

Cool. Thanks for the doubt.

[Ghideon]

25 minutes ago, DandelionTheory said:

Cool. Thanks for the doubt.

Have you selected one of the options?

Edited January 25, 2020 by Ghideon
Double content

[DandelionTheory]

13 minutes ago, Ghideon said:

Have you selected one of the options?

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.

Edited January 25, 2020 by DandelionTheory
Mistyped magnitude.

[Ghideon]

 

10 minutes ago, DandelionTheory said:

The smaller the angle, the smaller the magnitude.

I think you missed my comment:

1 hour ago, Ghideon said:

 The vectors I helped verify is correct. But due to a misunderstanding of the physics you end up with something physically impossible, the vectors does not describe a physical valid configuration. By the way, "fancy vector math", when applied to the correct set of forces, will show that internal forces cancel*. Unfortunately you have been unable to follow along in my attempts at explaining this.

 

[DandelionTheory]

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.

Edited January 25, 2020 by DandelionTheory

[Ghideon]

1 hour ago, DandelionTheory said:

you just refuse to understand physics

 

1 hour ago, DandelionTheory said:

physics is still not understood by you.

 

1 hour ago, DandelionTheory said:

I'm assuming you're mind is stuck

That's always a possibility! But when you verified that the lifter will work in the concealed box, accelerating with it upwards, that was the final confirmation that the physics you provide is flawed. 

1 hour ago, DandelionTheory said:

Or just assume you could do the math in your head...?

To he honest, yes of course! It just took a while to write down the latex notation a few pages back. The vector algebra is really simple in this case. Too bad my explanations fail to show that your math may be correct but does not seem to describe forces of any physics. 

My first post:

On 1/19/2020 at 10:14 AM, Ghideon said:

Are the calculations based on physically correct assumptions? 

I had this initial feeling that you were pursuing a reactionless drive of some sort but the lack of details made it hard to make that statement in the first post. So it was better to address the math question since it was interesting and contained no baseless claims. 

[DandelionTheory]

3 minutes ago, Ghideon said:

I had this initial feeling that you were pursuing a reactionless drive of some sort but the lack of details made it hard to make that statement in the first post. So it was better to address the math question since it was interesting and contained no baseless claims. 

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.

[Ghideon]

13 minutes ago, DandelionTheory said:

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.

The lifter, working in the box, is by definition reactionless. 

14 minutes ago, DandelionTheory said:

It's funny you did the math you wanted and "forgot" to do the real math involved with the reaction I'm specific to.

It's funny I did the math you wanted and you choose to ignore to deal with all relevant forces involved in the system. 

[DandelionTheory]

4 minutes ago, Ghideon said:

It's funny I did the math you wanted and you choose to ignore to deal with all relevant forces involved in the system.

I added specifics to show you you missed the magnitude in your calculation that strictly processed angles.

7 minutes ago, Ghideon said:

The lifter, working in the box, is by definition reactionless

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.

[Ghideon]

4 minutes ago, DandelionTheory said:

I added specifics to show you you missed the magnitude in your calculation that strictly processed angles.

I do not question the addition of vectors. I question the fact that there are vector missing. Vectors that are important to be able to predict the outcome of the systems behaviour if it was built and observed. The reasoning is to be found above as a comment on your angled wall example.

Another attempt at explaining, reasoning top down from abstraction:

We look at the big picture. You have your lifter in the box and it is able to propel itself up, OK? Then replace the lifter with a person. The person pushes his hands against the top of the box, from the inside. From the outside there is no difference, so the internal forces on the box will make it lift and accelerate. Does this sound plausible? 

Then apply that reasoning (iteratively if you wish) layer by layer through the internals of the system, applying physical laws and formulas  as you go, mechanics, electrics, magnetics, down to quantum level if necessary. Always there will be a force and a counter force similar to the person in the box, but applied to smaller  and smaller parts of the structure. There will never be a level where the total forces don't add.

If there is a net force on some level, that means that there are forces (vectors) that were neglected. 

 

[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

Edited January 25, 2020 by DandelionTheory

[Ghideon]

25 minutes ago, DandelionTheory said:

hinge

If the wall has a hinge* so that it can be folded then that can be taken into account by adding relevant force vector to the picture. But since you fail to comment on my analysis it's hard to know.

 

 

*)