I need help with my math problem.

[DandelionTheory]

I figure "magnetic lifter" was appropriate name for this problem.

The question is: "What is the total Lorentz force on all currents? Is my math correct for total force on all currents?

 

Edited January 19, 2020 by DandelionTheory
Better title

[Ghideon]

27 minutes ago, DandelionTheory said:

Is my math correct for total force on all currents?

That may depend. Are the calculations based on physically correct assumptions? 
There seems to be things missing from the image (maybe formatting) so it is not easy to tell what the math is describing.

[DandelionTheory]

13 minutes ago, Ghideon said:

That may depend. Are the calculations based on physically correct assumptions? 

Yes. 

Do I need to specify the magnitude of each arbitrary B to be proportional to the sides of ∆ABD?

[Strange]

1 hour ago, DandelionTheory said:

The question is: "What is the total Lorentz force on all currents? Is my math correct for total force on all currents?

I would suggest taking a few minutes to learn the basics of Latex so you can post the equations here. That will make them easier (possible) to read, make it possible for people to quote and correct them, etc.  (And is a valuable life skill for anyone interested in math or physics.)

[DandelionTheory]

1 hour ago, Strange said:

I would suggest taking a few minutes to learn the basics of Latex so you can post the equations here. That will make them easier (possible) to read, make it possible for people to quote and correct them, etc.  (And is a valuable life skill for anyone interested in math or physics.)

Thank you. I will take the time to learn Latex and come back with editable equations.

Until then, I have respecified the defined magnitude of B4 , B5 and B6 to be equal to the length of each respective parallel side of ∆ABD.

The total force seems to always be within 180° of the line between the 2 currents with the same Z angles.

Edited January 19, 2020 by DandelionTheory

[Ghideon]

6 hours ago, DandelionTheory said:

figure "magnetic lifter" was appropriate name for this problem.

The question is: "What is the total Lorentz force on all currents? Is my math correct for total force on all currents?

The word "lifter" seems to suggest that some force will accelerate something so that it lifts. It may help with an overview of what the magnetic lifter is before going into the details of the calculations. 

Edit: Pictures above are cropped. Instead of having to go to an external site each time I'll insert the picture here:

Edited January 19, 2020 by Ghideon
picture

[Strange]

A diagram would help too: where is the current carrying wire, what is the source of the magnetic field, what are the directions of the three forces ([math]\vec{F_1}[/math], [math]\vec{F_2}[/math], [math]\vec{F_3}[/math]), what is the angle [math]\alpha[/math], etc.

Sorry, just noticed there is a diagram!

[studiot]

This has become a mess.

1) Which diagram are we supposed to be discussing ?

2) The diagrams show an xy plane (no z axis) with three currents represented by dots. Are you using the dot and cross convention for currents normal to the plane or what?
If so please enclose the currents in a circle.

3) The pictures are necessarily shrunk to fit in the post but enlarge to be readable if downloaded.

 

Thanks

[DandelionTheory]

2 hours ago, studiot said:

This has become a mess.

1) Which diagram are we supposed to be discussing ?

2) The diagrams show an xy plane (no z axis) with three currents represented by dots. Are you using the dot and cross convention for currents normal to the plane or what?
If so please enclose the currents in a circle.

3) The pictures are necessarily shrunk to fit in the post but enlarge to be readable if downloaded.

 

Thanks

I'm sorry, this one. I added mass and gravity to show the definition of "lifter"

[Ghideon]

44 minutes ago, DandelionTheory said:

I added mass and gravity to show the definition of "lifter"

Thanks for the picture. By what principle does the lifter work? Interaction with earth magnetic field, ejection of particles, other? Or is it is a proposal for a reactionless drive?

[DandelionTheory]

7 hours ago, Ghideon said:

Thanks for the picture. By what principle does the lifter work? Interaction with earth magnetic field, ejection of particles, other? Or is it is a proposal for a reactionless drive?

Each current has a force calculation based on the total magnetic field at the intersection of xy plane. if they are the same structure, total force is applied to the whole structure in one direction if one current is 180° from the other 2 on the Z plane.

[Ghideon]

1 hour ago, DandelionTheory said:

Each current has a force calculation based on the total magnetic field at the intersection of xy plane. if they are the same structure, total force is applied to the whole structure in one direction if one current is 180° from the other 2 on the Z plane.

Ok! Just so I follow: due to the force from the magnetic fields the rig will accelerate. How is momentum conserved?

[DandelionTheory]

2 hours ago, Ghideon said:

Ok! Just so I follow: due to the force from the magnetic fields the rig will accelerate. How is momentum conserved?

Conservation of momentum is normally calculated as loose particles and their magnetic fields acting on the other particle. 

If a magnetic field independent of I is at the correct angle, acceleration in the B angle plus or minus 90° is "felt" by the current. I have not brought up how the uniform B fields are made initially, that's not the point of the problem.

I have given values to magnitudes and some angles to show what I mean at the bottom right of the picture.

 

PS, I'm currently unemployed. Lol. *Bong toke*

-DandelionTheory

Edited January 20, 2020 by DandelionTheory
Force is in Newtons, duh.

[Strange]

4 hours ago, DandelionTheory said:

Each current has a force calculation based on the total magnetic field at the intersection of xy plane.

What is the source of the magnetic field?

[DandelionTheory]

43 minutes ago, Strange said:

What is the source of the magnetic field?

I cannot answer that at the moment. I'm currently being tutored to express magnetic fields induced in iron due to a current loop.

[Ghideon]

  

4 hours ago, DandelionTheory said:

Conservation of momentum is normally calculated as loose particles and their magnetic fields acting on the other particle. 

If a magnetic field independent of I is at the correct angle, acceleration in the B angle plus or minus 90° is "felt" by the current. I have not brought up how the uniform B fields are made initially, that's not the point of the problem.

I have given values to magnitudes and some angles to show what I mean at the bottom right of the picture.

Are the magnetic fields generated in the lift or by another device, not onboard the lift? 

I assume that Ampère's force law can be neglected, correct? The currents do not generate a strong enough magnetic field to have an effect on the outcome.

[DandelionTheory]

2 hours ago, Ghideon said:

  

Are the magnetic fields generated in the lift or by another device, not onboard the lift? 

I assume that Ampère's force law can be neglected, correct? The currents do not generate a strong enough magnetic field to have an effect on the outcome.

I drew another example I need to proof.

Its just an idea, but I need some help figuring out how to calculate B3 if Q is steel.

 

Edited January 20, 2020 by DandelionTheory

[Ghideon]

42 minutes ago, DandelionTheory said:

I drew another example I need to proof.

Its just an idea, but I need some help figuring out how to calculate B3 if Q is steel.

Why a new lifter, did you abandon the first one? 

 

[DandelionTheory]

35 minutes ago, Ghideon said:

Why a new lifter, did you abandon the first one? 

 

I do not know how to calculate magnetic field due to current yet. The new lifter is to illustrate how the currents can be "attached" to the original lifter and be inline with the initial magnetic fields.

[Ghideon]

1 hour ago, DandelionTheory said:

The new lifter is to illustrate how the currents can be "attached" to the original lifter and be inline with the initial magnetic fields.

Please provide straight answers. Have you abandoned the first version of the lifter and wish to discuss only the new one? I am still looking at first version:

5 hours ago, Ghideon said:

Are the magnetic fields generated in the lift or by another device, not onboard the lift? 

You state that there is an acceleration. Acceleration relative to what?

 

 

 

[DandelionTheory]

7 hours ago, Ghideon said:

Please provide straight answers. Have you abandoned the first version of the lifter and wish to discuss only the new one? I am still looking at first version:

You state that there is an acceleration. Acceleration relative to what?

 

I set up a g vector at the top. Breh read all the variables please. Gravity is constant and in the y<270°,0°> direction.

The original is still viable, it is still here and I am working on it still with you. The magnetic fields experience torque, I don't know how much and what exact direction yet till my tutor helps me or one of you are kind enough to help me solve for B3 in the second lifter.

Edited January 21, 2020 by DandelionTheory

[DandelionTheory]

My tutor told me I need to understand more calculus before derivatives can be understood. They also explained my Z angles were off for my B vectors, and I needed to specify my vector notation.

So arbitrary vector V = V<magnitude, angle from X origin, angel from Z origin>

I still don't know how to calculate induced magnetic field in steel like the ?? State in the second picture below. If I can answer that, maybe I can answer how the initial B fields of the first picture are generated

1st

2nd

-DandelionTheory

[Ghideon]

7 hours ago, DandelionTheory said:

The original is still viable,

Is it OK to assume that ABD is an equilateral triangle? 

[DandelionTheory]

36 minutes ago, Ghideon said:

Is it OK to assume that ABD is an equilateral triangle? 

Equilateral or isosceles with the odd current at the odd angle vertex.

The E vectors have no X angle.

I made a correction.

Edited January 21, 2020 by DandelionTheory

[Ghideon]

13 hours ago, DandelionTheory said:

Equilateral or isosceles with the odd current at the odd angle vertex.

Ok!

Here are some quick calculations regarding the original question. NOTE: I’ll handle this as a mathematics problem only. I do intentionally not mention any electromagnetics since there is not enough information provided to know if this setup have any physical meaning or is physically possible.

I’ll write out conclusions made from the notes above to give the whole picture, maybe unnecessary detailed. It’s a little tricky to read all the indexes in the original image but the intention is to keep the numbering and letters from other pictures posted so far. 

Let’s start with an equilateral triangle ABD. ABD is placed in a cartesian coordinate system xy. Initially there is no need for a z-axis. The corners are associated with vectors and the vectors have x and y components identified by subscript.

Since ABD is equilateral and the vectors having identical length that lead to some symmetries and one zero component.

[math]B_{4x}=B_{5x}\\B_{4y}=-B_{5y}\\B_{6y}=0[/math]

At each point A,B,D, add* the pair of vectors associated with that point. Use symmetries to simplify:

[math]A: B_{1}=\begin{bmatrix}B_{4x}\\B_{4y}\end{bmatrix} +\begin{bmatrix}B_{5x}\\B_{5y}\end{bmatrix}=\begin{bmatrix}B_{4x}+B_{5x}\\B_{4y}+B_{5y}\end{bmatrix} =\begin{bmatrix}2B_{4x}\\0\end{bmatrix} 
[/math]

[math]B: B_{2}=\begin{bmatrix}B_{5x}\\B_{5y}\end{bmatrix}+\begin{bmatrix}B_{6x}\\B_{6y}\end{bmatrix}=\begin{bmatrix}B_{5x}+B_{6x}\\B_{5y}+B_{6y}\end{bmatrix} =\begin{bmatrix}B_{4x}+B_{6x}\\-B_{4y}\end{bmatrix} [/math]

D:[math] B_{3}=\begin{bmatrix}B_{6x}\\B_{6y}\end{bmatrix}+\begin{bmatrix}B_{4x}\\B_{4y}\end{bmatrix}=\begin{bmatrix}B_{6x}+B_{4x}\\B_{6y}+B_{4y}\end{bmatrix} =\begin{bmatrix}B_{6x}+B_{4x}\\B_{4y}\end{bmatrix} [/math]

These directions seems to match the hand-drawn picture by OP. 

To check how “force” vectors “F” would look like, we apply rotation** by 90 degrees ccw by using matrix

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

at point B and D. 

At point A we apply 270 degrees of rotation** to match original picture.

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

[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]

[math]D : 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]

Result is vector F1 in y direction at A and vectors F2 and F3 pointing inwards into the triangle at points B and D. 
This seems to match the directions of "F" vectors in OP's picture. Due to symmetry the x-components at B and D cancels. So the total sum of F1,F2 and F3 is in y direction. 

Note again: this is a set of vector calculations. It could possibly represent some setup of magnetic fields, currents and forces. But so far the thread lacks the responses to address that possibility.

 

*) If this would have been magnetic fields, they add as vectors.
**) If there would have been currents, right hand rule tells vector directions. 

Edited January 21, 2020 by Ghideon
latex mess. more latex mess. grammar. removed edit tag