The Tom Bearden

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Subject: RE: Meg
Date: Sat, 29 Sep 2001 13:42:28 -0500


It's a little more complicated.

  The basic formula means that the magnitude of the produced field is directly dependent on the rate of change of the input signal.  If too fast, one gets a voltage in the output coil that promptly destroys the insulation.  If one gets too slow a signal, one does not get enough to go overunity.  The duty cycle is also involved.  Those are the ones I can discuss.  There are two other things involved that are proprietary.

  Tom Bearden

Subject: Meg
Date: Fri, 28 Sep 2001 15:13:11 -0700
A previous writer suggested his understanding of the sharpness of the current cutoff below.
"In the MEG: If I understand its workings correctly, it is very important to achieve a sharpest possible cutoff in the input coil current, to achieve the highest dPhi/dt from the permanent magnet through the output coils."
How sharp a dropoff is not sharp enough? What about the dropoff on sinewave. Is that sharp enough or not?