I need help with my math problem.

[Ghideon]

I think I made an error in the rotation, applying the wrong matrix at A and B,D. I'll post an update once i get some time. It does not affect my conclusion, but it affects the signs in the final matrices. 

 

[DandelionTheory]

Ive added sources to the magnetic fields with this new picture.

 

Edited January 22, 2020 by DandelionTheory

[Ghideon]

Here is the corrected last step using the matrices. 

At point A we apply 270 ccw degree rotation of the vector by using matrix 

[math]\begin{bmatrix}0&1\\-1&0\end{bmatrix}[/math]

A 90 degree ccw rotation, is applied to the vector at point B and D

[math]\begin{bmatrix}0&-1\\1&0\end{bmatrix} [/math]

The resulting vectors "F" at each point A,B and D is:

[math]A: F_{1}=\begin{bmatrix}0&1\\-1&0\end{bmatrix} \cdot \begin{bmatrix}2B_{4x}\\0\end{bmatrix} = \begin{bmatrix}0\\-2B_{4x}\end{bmatrix} [/math]

[math]B: F_{2}=\begin{bmatrix}0&-1\\1&0\end{bmatrix}  \cdot \begin{bmatrix}B_{4x}+B_{6x}\\-B_{4y}\end{bmatrix} =\begin{bmatrix}B_{4y}\\B_{4x}+B_{6x}\end{bmatrix} [/math]

D:[math] F_{3}=\begin{bmatrix}0&-1\\1&0\end{bmatrix}  \cdot \begin{bmatrix}B_{6x}+B_{4x}\\B_{4y}\end{bmatrix} =\begin{bmatrix}-B_{4y}\\B_{6x}+B_{4x}\end{bmatrix} [/math]

The result is vectors F pointing in directions predicted by calculations in OP

 

1 hour ago, DandelionTheory said:

Ive added sources to the magnetic fields with this new picture.

Ok.

Is the expected outcome that the rig will lift from the ground and accelerate upwards? As drawn that seems not possible.

 

Edited January 22, 2020 by Ghideon

[DandelionTheory]

24 minutes ago, Ghideon said:

Ok.

Is the expected outcome that the rig will lift from the ground and accelerate upwards? As drawn that seems not possible.

yes. what else would i need to calculate? i have other versions of lifters.

Edited January 22, 2020 by DandelionTheory

[Ghideon]

2 hours ago, DandelionTheory said:

yes. what else would i need to calculate?

At this point it is not so much of a math problem but physics. I tried to ask earlier regarding conservation of momentum and how magnetic fields are created to nudge you in that direction. Now that the sketch and expected outcome is more complete we can analyse it further. Again, slightly more detailed:

If the rig is stationary on ground its momentum is zero. When started there will be internal stresses in the rig due to the currents and magnetic fields; there will be internal forces. So momentum is conserved. The rig's momentum is zero. The rig will not move. 

 

[DandelionTheory]

i would like to see how you concluded your internal force calculation. in one of the illustrations i attempted to "hide" like magnetic fields within the steel core, the bottom current interacts with the inside of the iron core through "fringing fields" which do not have a permeability of iron because its an air core.

[Ghideon]

42 minutes ago, DandelionTheory said:

i would like to see how you concluded your internal force calculation. in one of the illustrations i attempted to "hide" like magnetic fields within the steel core, the bottom current interacts with the inside of the iron core through "fringing fields" which do not have a permeability of iron because its an air core.

I do not calculate. I look at the rig and intuitively it tells me "reaction less drive" or "issues with conservation of momentum". I asked about it earlier. That leaves two* options as far as I can see:
1: I have misunderstood the setup completely; the lifter either ejects mass or it interacts with external things that is not in the pictures. 
2: the lifter is not following basic laws of physics and cannot work.

The forces that we calculated have equal and opposite forces in the rig. If a magnetic field pushing at something then the magnet will be pushed back. 
 

 

*) Number 3, new physics beyond current mainstream, is not probable. 

Edited January 23, 2020 by Ghideon

[DandelionTheory]

right, I1 will push on B4 and B5 inwards, I2 and I3 will push on B4, B5 and B6 outward. i wonder if they.... cancel out, but i cant know that till its calculated. I need help with that part. weather its fringe fields or B4, B5 & B6 acting on each other

[Ghideon]

39 minutes ago, DandelionTheory said:

 i wonder if they.... cancel out, but i cant know that till its calculated.

It looks like we are reasoning along the same lines. Just approaching it different. My point is that due to the deep principles of mechanics it has to cancel. All other options are incorrect or requires new physics. So no calculation is needed. That said, it is interesting and maybe educating, I may have time to do some calculations later today. 

[DandelionTheory]

3 hours ago, Ghideon said:

It looks like we are reasoning along the same lines. Just approaching it different. My point is that due to the deep principles of mechanics it has to cancel. All other options are incorrect or requires new physics. So no calculation is needed. That said, it is interesting and maybe educating, I may have time to do some calculations later today. 

I appreciate your help and patience. 😊 

I expected fields to interact, I do not have the ability to explain the balanced forces of I1, I2 & I3 on B4, B5 & B6. I only suspect they do due to some fancy right hand rule gestures. 

I was going with the fact that steel beams are used as passive shielding in MRI machines.

http://mriquestions.com/how-to-reduce-fringe.html

Edited January 23, 2020 by DandelionTheory

[Ghideon]

12 hours ago, DandelionTheory said:

I appreciate your help and patience. 😊 

Thanks. I also note that this thread has stayed away from non-mainstream claims, so we can pursue this topic in a more collaboratory fashion than what is possible in a speculations thread. 😀

I will use the latest picture you posted, with the triangular core, and try some ideas before calculating. In a speculations this would have been a more blunt and short question but this time some more details will be provided, it will be easier to spot if my understanding of the setup is wrong.
 

First we so a series of simplifications.

-Let’s assume the the rig is large enough so that interference between the points a, B and D are neglectable. Hence we can look at one corner at a time.
-It is not known yet how the magnetic field will look in detail at the corners. The magnetic flux will likely have some radius, there are no sharp turns so the vector calculations I did do not apply. But that does not matter. What matter is that the point A is located where the magnetic field B is horizontal. That is all we care about.
Now we reverse the reasoning. We do not look at the suggested rig in drawings above, we look at an idealised rig, where we have managed to create the magnetic field we want and it affects the cable at point A pushing it upwards. How we managed to build it is not important.

Here is corner A with force F pushing on the cable due to the magnetic field B.

 

Now we simplify further. It does not matter what creates the magnetic field or what it pushes. AFAIK there is no fundamental differences between an electromagnet and a permanent magnet for the properties we want to investigate. We are interested in a force and a magnetic field and the interaction, nothing more at this point. So let’s replace the magnetic field with a permanent magnet. And since the force F is directed in positive y direction we can use another magnet instead of a wire. (since a piece of iron would be pulled towards the magnet)

Now we have reduced the setup to its basics. The force F pointing up will have an equal force pointing down. If the magnets are attached to the same solid frame, as the cable and the iron core in your drawings, there is no way there can be an acceleration. The forces cancel internally in the rig.

Does this show how any kind of magnetic fields, acting on internal parts of the rig, will be unable to generate accelerating force?

It does not matter how the magnets are angled, if they have some unusual material or some unusual geometrical shapes. The forces cancel.

 

 

Edited January 23, 2020 by Ghideon
clarified a few words

[DandelionTheory]

9 hours ago, Ghideon said:

Thanks. I also note that this thread has stayed away from non-mainstream claims, so we can pursue this topic in a more collaboratory fashion than what is possible in a speculations thread. 😀

I will use the latest picture you posted, with the triangular core, and try some ideas before calculating. In a speculations this would have been a more blunt and short question but this time some more details will be provided, it will be easier to spot if my understanding of the setup is wrong.
 

First we so a series of simplifications.

-Let’s assume the the rig is large enough so that interference between the points a, B and D are neglectable. Hence we can look at one corner at a time.
-It is not known yet how the magnetic field will look in detail at the corners. The magnetic flux will likely have some radius, there are no sharp turns so the vector calculations I did do not apply. But that does not matter. What matter is that the point A is located where the magnetic field B is horizontal. That is all we care about.
Now we reverse the reasoning. We do not look at the suggested rig in drawings above, we look at an idealised rig, where we have managed to create the magnetic field we want and it affects the cable at point A pushing it upwards. How we managed to build it is not important.

Here is corner A with force F pushing on the cable due to the magnetic field B.

 

Now we simplify further. It does not matter what creates the magnetic field or what it pushes. AFAIK there is no fundamental differences between an electromagnet and a permanent magnet for the properties we want to investigate. We are interested in a force and a magnetic field and the interaction, nothing more at this point. So let’s replace the magnetic field with a permanent magnet. And since the force F is directed in positive y direction we can use another magnet instead of a wire. (since a piece of iron would be pulled towards the magnet)

Now we have reduced the setup to its basics. The force F pointing up will have an equal force pointing down. If the magnets are attached to the same solid frame, as the cable and the iron core in your drawings, there is no way there can be an acceleration. The forces cancel internally in the rig.

Does this show how any kind of magnetic fields, acting on internal parts of the rig, will be unable to generate accelerating force?

It does not matter how the magnets are angled, if they have some unusual material or some unusual geometrical shapes. The forces cancel.

 

 

What happens when you apply force on 2 sides of a triangle? Some of it is applied to it's center axis and cancels. Did you forget the source of B is larger than I? I talked about fringe fields and you attempted to proof one axis when I said you had some canceling on the sides. How many sides are there? Do you think 2 evens and an odd make an odd?

Where's your math that shows F2? Did you do the vector calculation or are you assuming with "new physics"?

Damn now I gotta be the only one that cares. Science died when you all stopped dreaming.

Edited January 24, 2020 by DandelionTheory

[Ghideon]

1 hour ago, DandelionTheory said:

What happens when you apply force on 2 sides of a triangle?

Vector addition will give you the answer. Do you want an example, see my earlier posts where the B-vectors are added, resulting in new vectors with a different direction and magnitude.

1 hour ago, DandelionTheory said:

Did you forget the source of B is larger than I?

Why does that matter? If I change one of my magnets in the simplified analysis to a smaller magnet it does not change the outcome.Forces cancel in the rig.

1 hour ago, DandelionTheory said:

I talked about fringe fields and you attempted to proof one axis when I said you had some canceling on the sides. How many sides are there? Do you think 2 evens and an odd make an odd?

The number of forces foes not matter. If they are all internal, acting on the structure, not pushing against anything external, they cancel.

1 hour ago, DandelionTheory said:

Where's your math that shows F2? Did you do the vector calculation or are you assuming with "new physics"?

No calculation needed. But I could have been clearer: When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body. It is called Newton's third law*. Are you familiar with its implications and why it allows me to make generalized comments about the lifter? 

  

1 hour ago, DandelionTheory said:

Damn now I gotta be the only one that cares.

I've spent a considerable amount of time in this thread, sorry if that it does not count as caring.

  

1 hour ago, DandelionTheory said:

Science died when you all stopped dreaming.

When did that happen? I dream all the time, just not so much about perpetuum mobiles and reactionless drives. I think the same goes for engineers and scientists at LIGO and CERN. Gravitational wave astronomy and the experimental evidence for Higgs came from dreaming about extending what is possible, not dreaming about what is known to be impossible. 

 

https://en.wikipedia.org/wiki/Newton%27s_laws_of_motion#Newton's_third_law or https://en.wikipedia.org/wiki/Reaction_(physics) 

Edited January 24, 2020 by Ghideon
missing some end notes

[DandelionTheory]

Quote

 The number of forces does not matter. If they are all internal, acting on the structure, not pushing against anything external, they cancel.

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.

[Ghideon]

3 minutes ago, DandelionTheory said:

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.

Sorry, I do not follow you reasoning. Can you cite the exact assumptions or statements I made that you disagree with? 

Again, are you familiar with Newton's third law? 

 

 

[DandelionTheory]

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.

Edited January 24, 2020 by DandelionTheory

[Ghideon]

9 minutes ago, DandelionTheory said:

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

Please use the qoute function. Please quote from the simplified analysis I provided. Please point out in detail where my simplifications does not apply to your rig. 

 

 

 

[DandelionTheory]

Quote

It does not matter how the magnets are angled

the Lorentz force says otherwise.

https://en.wikipedia.org/wiki/Lorentz_force

Quote

The forces cancel.

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) 

Edited January 24, 2020 by DandelionTheory

[Ghideon]

31 minutes ago, DandelionTheory said:

the Lorentz force says otherwise.

Sorry, you completely missed the point. Maybe you should check out Newtons laws first? When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body. Why do you think this law does not apply? When the cable at one of the corners is pushed with some force , it cant go anywhere because it is embedded in the iron triangle, and the iron triangle holds it back with the same force. 

Again, from what you have provided in the drawings and math and descriptions, this is a basic physics issue. You can't by any means, accelerate something by using internal forces. Nature just does not work that way. Either you push against something extarnal, like when you jump off the ground, or you eject mass, like a rocket engine. Don't get entangled in detailed geometry or magnetics. By what principle, compatible with basic laws of Newton, does your lifter operate? 

 

Edited January 24, 2020 by Ghideon

[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.

Edited January 24, 2020 by DandelionTheory

[Ghideon]

10 minutes ago, DandelionTheory said:

 

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.

 

if you provide a drawing we can calculate the outcome. This may be a good approach since it is more general and does not go into details about magnetic fields.

[DandelionTheory]

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 

Edited January 24, 2020 by DandelionTheory

[Ghideon]

The above contains many issues that I do not have time to adress right now. you have not yet managed to get to the core of the problem. 

maybe if you provide a picture of the two people and the wall you mentioned we can do some basic mechanics calculations?
I might get some time later today to post a few other ideas.

 

[DandelionTheory]

15 minutes ago, Ghideon said:

maybe if you provide a picture of the two people and the wall you mentioned we can do some basic mechanics calculations?
I might get some time later today to post a few other ideas.

"pushing off the center"

[Ghideon]

10 minutes ago, DandelionTheory said:

"pushing off the center"

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?