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PMsquare thing.



Stefan,

How does the force picture change with the track elongated? I attached a screenshot and the .fem file. Do you still get some sideways force with an infinitely long track, and steel backing?

In the screenshot (Alt+PrintScrn would not capture all of it), the force values are

Force in x-direction = -1.061e+001 N/m
Force in y-direction = 1.195e+001 N/m

with the red line starting in the lower left hand corner and circuiting counter-clockwise. 

(Interesting, that you might - I have things similar to this PMsquare, thatwill give some unusual inductive effect gliding on a track of arbitrary length, where the magnetic field intensity never changes, but once you bend that track into a circle and cause the gliding motion to be continuous and rotary, once you remove the ends, so to speak, the effects disappear... it is as if the magnetic field is superfluid, and to maintain the effects you need a boundary that halts its inertialess circulation, and instead forces motion coincident with the boundary...)

By the way, the simulation goes a lot faster if you "fake" certain materials.

For example, for the flux return plate you used in your simulation, you specified the material Steel. This makes the computer solve the problem many times, to converge at some point where the nonlinearity is close to what youshould practically expect. 

Materials that FEMM interprets as linear, like air or the magnets themselves, only have to calculate through one "conjugate gradient solver" interation, instead of many. In your simulation, it's not of primary importance whatthe magnetic characteristics of the back flux return circuit are - you just want the flux returned, at any rate. I can recommend going into the materials library, and creating a "fake" material that has some permeability (like 3,000 or so) and is simply linear (specifying >1 mu while leaving the nonlinear box unchecked). You can use this fake material in any situation where you need to carry flux, but how that flux gets carried is not important.This saves processor time, and speeds up solving the problems when you're drafting ideas.

Also, under the standard boundary conditions in FEMM you can treat the boundaries of your drawing as infinitely permeable. For instance, I imagine thethings I draw in FEMM to exist in little air bubbles inside some kind of infinitely permeable steel. Instead of drawing flux return circuits (if I amdoodling in FEMM and just testing ideas), it appears OK to just stick the magnets to the edges of the "air bubble", or the outer boundary; the boundary itself then serves as the flux return circuit. (I hope people feel free to correct me if something is wrong with this.)

Graham

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<DIV><FONT face="Courier New">Stefan,</FONT></DIV>
<DIV><FONT face="Courier New"></FONT>&nbsp;</DIV>
<DIV><FONT face="Courier New">How does the force picture change with the track 
elongated? I attached a screenshot and the .fem file. Do you still get some 
sideways force with an infinitely long track, and steel backing?</FONT></DIV>
<DIV><FONT face="Courier New"></FONT>&nbsp;</DIV>
<DIV><FONT face="Courier New">In the screenshot (Alt+PrintScrn would not capture 
all of it), the force&nbsp;values&nbsp;are</FONT></DIV>
<DIV>&nbsp;</DIV>
<DIV><FONT face="Courier New">Force in x-direction = -1.061e+001 N/m<BR>Force in 
y-direction = 1.195e+001 N/m</FONT></DIV>
<DIV>&nbsp;</DIV>
<DIV><FONT face="Courier New">with the red line starting in the lower left hand 
corner and circuiting counter-clockwise.</FONT>&nbsp;</DIV>
<DIV><FONT face="Courier New"></FONT>&nbsp;</DIV>
<DIV><FONT face="Courier New">(Interesting, that you might - I have things 
similar to this PMsquare, that will give some unusual inductive effect gliding 
on a track of arbitrary length, <EM>where the magnetic field intensity</EM> 
<EM>never changes</EM>, but once you bend that track into a circle and cause the 
gliding motion to be continuous and rotary, once you remove the ends, so to 
speak, the effects disappear... it is as if&nbsp;the magnetic field is 
superfluid, and to maintain the effects you need a boundary that halts its 
inertialess circulation, and instead forces motion coincident with the 
boundary...)</FONT></DIV>
<DIV><FONT face="Courier New"></FONT>&nbsp;</DIV>
<DIV><FONT face="Courier New">By the way, the simulation goes a lot faster if 
you "fake" certain materials.</FONT></DIV>
<DIV><FONT face="Courier New"></FONT>&nbsp;</DIV>
<DIV><FONT face="Courier New">For example,&nbsp;for the flux return plateyou 
used in your simulation, you specified the material Steel. This makes the 
computer solve the problem many times, to converge at some point where the 
nonlinearity is close to what you should practically expect. </FONT></DIV>
<DIV><FONT face="Courier New"></FONT>&nbsp;</DIV>
<DIV><FONT face="Courier New">Materials that FEMM interprets as linear, like air 
or the magnets themselves,&nbsp;only have to calculate through one "conjugate 
gradient solver" interation, instead of many. In your simulation, it's not of 
primary importance what the magnetic characteristics of the back flux return 
circuit are - you just want the flux returned, at any rate. I can recommend 
going into the materials library, and creating a "fake" material that has some 
permeability (like 3,000 or so) and is&nbsp;simply linear (specifying &gt;1mu 
while leaving the nonlinear box unchecked). You can use this fake material in 
any situation where you need to carry flux, but how that flux gets carried is 
not important. This saves processor time, and speeds up solving the problems 
when you're drafting ideas.</FONT></DIV>
<DIV><FONT face="Courier New"></FONT>&nbsp;</DIV>
<DIV><FONT face="Courier New">Also, under the standard boundary conditions in 
FEMM you can treat the boundaries of your drawing as infinitely permeable. For 
instance, I imagine the things I draw in FEMM to exist in little air bubbles 
inside some kind of infinitely permeable steel. Instead of drawing flux return 
circuits (if I am doodling in FEMM and just testing ideas), it appears OK to 
just stick the magnets to the edges of the "air bubble", or the outer boundary; 
the boundary itself then&nbsp;serves as the flux return circuit. (I hope people 
feel free to correct me if something is wrong with this.)</FONT></DIV>
<DIV><FONT face="Courier New"></FONT>&nbsp;</DIV>
<DIV><FONT face="Courier New">Graham</FONT></DIV></BODY></HTML>

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