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# Re: [femm] PMsquare dead end...

- To: femm@yahoogroups.com
- Subject: Re: [femm] PMsquare dead end...
- From: Dave Squires <djsquires@plix.com>
- Date: Wed, 13 Jun 2001 15:57:15 -0700
- References: <9g8h1r+fuft@eGroups.com>

    Stefan,
    The trick is to use more than one magnetic
    circuit acting at once. Think parallel magnetic
    circuits and then also use the fact that force
    is inversely proportional to gap spacing (distance).
    Increase the gap on the exit half in the direction
    of motion by using a notch. This way you can
    create a force asymmetry. You need the force
    patterns spaced apart and overlapped such that
    the positive is enhanced and the negative partially
    cancelled. Then you can get somewhere.
    Otherwise you will see the usual integration of zero
    net force.

    Dave Squires

    harti@harti.com wrote:

    > Hi All,
    > thanks a lot for all your help.
    > I see now, that the PMSquare as suggested
    > is a "dead horse".
    > It seems only to generate horizontal forces
    > due to the end conditions of the stator magnets....
    > It is also too complicated to be build for a real
    > linear motor.
    > Thus I am working now on a kind of Howard Johnson
    > like linear magnet motor, where the runner magnet
    > flies over a track of steel plates.
    > So the difference to the Howard Johnson design is,
    > that the stator track does not use any magnets but just steel
    > plates. The runner magnet above it uses special core pieces
    > so the magnet never locks in any position, but always
    > wants to move on to the side.
    > It just depends on the steel plate spacings versus air gaps
    > versus runner magnet dimension.
    >
    > I am trying now to find the right dimensions, so that there is
    > always a positive force into the x-direction.
    >
    > Regards, Stefan.
    >
    > --- In femm@y..., David Meeker <dmeeker@f...> wrote:
    > >
    > >
    > > Dave Squires wrote:
    > >
    > > > David,
    > > > I don't think I quite agree with your explanation.
    > > > If the two magnets are locked together and held
    > > > in the position shown you then have a
    > > >
    > > > Net force = +B*Q +(-B*Q) = 0
    > > >
    > > > if embedded in a uniform field. Isn't that correct?
    > > >
    > > > Dave Squires
    > >
    > > That's exactly right--there is no net force on a permanent magnet
    > in a uniform
    > > field.
    > >
    > > Dave.
    >
    >
    >
    > Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/

- References:
  - PMsquare dead end...
    - From: harti@harti.com

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