[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

# r-component of force in problems with axisymmetric geometry

- To: <femm@yahoogroups.com>
- Subject: r-component of force in problems with axisymmetric geometry
- From: "James Rabchuk" <ja-rabchuk@wiu.edu>
- Date: Tue, 2 Jul 2002 14:55:46 -0500
- Importance: Normal
- Reply-to: <JA-Rabchuk@wiu.edu>

    Hi,

    I've just begun using FEMM, and have found it to be very useful and
    accurate. What a great tool! Not only so, but the archive of discussions on
    the program have been very useful as well. Thanks so much.

    I have a question regarding the calculation of forces on objects lying along
    the axis of symmetry using the stress tensor integral. It's clear from the
    Roters' example provided on the website that to find the z-component of the
    force on such an object, I need to create a contour with beginning and
    ending points on the axis (r = 0) line which "encloses" the object, like so:

    -------|
    | ^
    -| | | z I hope this picture is preserved!
    -| | -> r
    -------|

    Doing so has given me very believable results for the force, if I use a
    small enough mesh, though the magnitude of the force is quite small (on the
    order of 10^-4 Newtons). However, just as a check, since I am looking for an
    equilibrium solution, I wanted to see what the program would say about any
    forces acting in the r direction. Of course, I expected that they should
    cancel out because of the symmetry of the problem (allowing for possible
    numerical error). Surprisingly, the program predicted a quite large,
    positive force in the r-direction (on the order of 10^-1 N), which varied
    smoothly with vertical position of the object.

    As I was doing this calculation using lua scripting, I just changed the
    expression found in the Roters' example to

    fr,p,fz=lineintegral(3)

    where fr and fz should pick out the real parts of the r and z components of
    the 1x force, as I understand it.

    Can anybody explain to me what is going on? Even if I were to explain away
    the positive force in the r-direction as being equivalent for every value of
    theta, and therefore cancelling out as expected, I expect and know that
    there should in fact be a restoring force pushing the object back toward the
    center line, so that the sign of the force should be negative in this case.
    Not only so, but the magnitude of the force is just too big.

    I did notice when I graphed B dot n along an upward vertical line, the
    program automatically gave the sign of the dot product to be positive, when
    the field should have a negative r-component, as verified by the plot of B
    dot t for a horizontal contour moving away from the axis. Does the normal
    always point 90 degrees ccw from the direction of the line segment?

    I should also mention that in this case, the stress tensor proved much MORE
    useful than did the coenergy method that is usually recommended by the
    program author. I guess that it is because in my problem, the object which
    is moving is so small compared to the entire solution domain. I could never
    resolve a clear trend in coenergy vs. position.

    Any insight into what I may be missing would be greatly appreciated. Thanks
    in advance.

    *************************************
    James Rabchuk
    Associate Professor of Physics

- Follow-Ups:
  - Re: [femm] r-component of force in problems with axisymmetric geometry
    - From: David Meeker <dmeeker@ieee.org>

- Prev by Date: Re: [femm] current density
- Next by Date: Re: [femm] r-component of force in problems with axisymmetric geometry
- Previous by thread: current density
- Next by thread: Re: [femm] r-component of force in problems with axisymmetric geometry
- Index(es):
  - Date
  - Thread