Re: [femm] Axi-symmetric arcs for FEMMVIEW

[Dave Squires <djsquires@xxxxxxxx>]

Dave,
Thanks for the explanation. I must have been thinking of
stress tensor force calculations instead of the B.n integral.

Also, Jiri's suggestion of pre-drawing the contour as an
arc in FEMM before solving the problem works too.
FEMMVIEW recognizes the arc when the contour icon
is selected. You just have to watch out where you start
it and end it to insure a clockwise contour or you get a
reversed sign for the torque. I was not aware of this
little trick.

Dave Squires

David Meeker wrote:

> Dave Squires wrote:
>
> > > > I want it for a different reason though,
> > > > because I want to find the total flux crossing an arbitrary
> > circular arc
> > > > where it is not possible to include an arc in the geometry.
> > >
> > > Regardless of the contour, the B.n integral is evaluated by taking the
> > > difference of the value of A at the endpoints of the contour. If I
> > > understand what you are trying to do correctly, there's no need for you
> > > to actually draw your arc--just evaluate A at the endponts and take the
> > > difference to get the total flux crossing your arc.
> >
> > But don't you need to know the path of the arc? If you have only end
> > points the path could be anything, straight line, arc, polygon path.
> > Who knows? If there is no defined contour what is the point of
> > integrating
> > over it? I must be missing something....a hidden assumption.
>
> The result for this particular integral is path-independent. There are
> two ways of looking at it: the mathematical way and the intuitive way.
>
> First, the mathematical way. Say that you have some contour connecting
> two points. Let the parameter s denote the distance down the contour
> from the first point. If you want the portion of the flux density, that
> passes perpendicular to the contour at any point on the contour, it
> turns out that it is dA/ds -- the derivative of vector potential with
> respect to position on the line. Getting the total flux means
> integrating dA/ds from s=0 to the value of s at the second point. Since
> the integral of dA/ds with respect to s is A, the total result is the
> difference of A at the endpoints, meaning that the path that you take
> between the two end points doesn't matter.
>
> Now, the more intuitive way. Think in terms of Gauss's law, which
> states that the total flux passing through any closed volume has to add
> up to zero (well, at least in the absence of magnetic monopoles....).
> First, imagine a straight line that links the two points of interest.
> Say that the total flux passing normal to that straight line is f. Now,
> imagine some other contour, say Keith's arc, that also connects the two
> points, which, with the straight line, forms a closed contour. Now, all
> the flux going into and out of the closed contour, by Gauss's law, has
> to add up to zero. That means that the flux passing through the arc
> must be of the same magnitude, f, as that passing through the straight
> line, so that all of the flux going into and out of the contour can add
> up to zero. Since we could have applied the same logic to any contour
> that links the ends of the straight line and creates a closed contour,
> the implication is that any contour at all that links the two points of
> interest must have the same flux, f, passing through it.
>
> >
> > I still think it should be fairly trivial to use the same arc code
> > block in
> > FEMM and apply it to circular contours in FEMMVIEW. After all
> > the code is already written, just copy it over and use it.
>
> Yes, it ought not be that hard to add in making an arc contour in the
> postprocessor.
>
> >
> > DRS
> >
>
>
>
>
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