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Re: About Ken's MEG switching model

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  • carbonprobe <carbonprobe@yahoo.com>
    Camelazur, I m glad to see someone is analyzing my stuff pretty intensively. At first glance I noticed that your revised diagrams show a meg state at t 2
    Message 1 of 2 , Jan 21, 2003
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      Camelazur, I'm glad to see someone is analyzing my stuff pretty
      intensively. At first glance I noticed that your revised diagrams
      show a meg state at t'2 (bottom left hand side) with all the magnet's
      flux (1.2T) switched to the right side. This is inconsistant with the
      diagrams above at t'2 which show that both inputs are zero and both
      outputs are at a static magnetic flux of 0.6T. And that would also go
      for the meg state at t'3 (bottom right hand side).

      Your right about how the L3 and L4 wave diagrams should show a static
      0.6T dotted line in the middle, I did this in my Meg state diagrams

      I'll explain why I made this conclusion:

      "When one input coil is turned on it forces all the flux to the
      opposite side. For a small change in Flux applied to the input, A
      larger change in Flux is available at the output."

      If L1 (left input coil) changes from 0 to 0.6 T, then it will cancel
      out the static B-field on the left hand side of the core (which is
      0.6T) to zero. Well where does that 0.6T go, it can't just disappear.
      The only place that it can go is to the right side of the core to
      fill it up to 1.2T (but only a change of 0.6T on the right side). So
      what just happened is the left side Changed from 0 to 0.6 T and the
      right side changed from 0.6T to 1.2T. So for an input change of 0.6T
      on the core, the left output coil sees a change of 0.6T and the right
      out put coil sees a change of 0.6 T. Total input dB is 0.6 T, total
      output dB is 1.2T.

      It may seem that only half of the magnet's flux is being switched,
      and that is true, only 0.6T of the magnet's flux is being switched.
      But when you subtract 0.6T from highly permeable sheet steel the
      magnetic atomic poles of the steel are being changed, picture them as
      little magnets. And they generate their own B field as the B field is
      being subtracted from it. So as the input is switched off and the
      magnet returns to it's quiescent state and you attach a load to the
      ouput coils the energy that is going into the load would really be
      coming from the core itself and not the magnet. The magnet acts as a
      catalyst to the magnetic poles in the core.


      --- In MEG_builders@yahoogroups.com, blanc2002@f... wrote:
      > Hi Ken and all !
      > After the reading of your post 1140, I started to work on a new
      version of my
      > MEG control board, wich would perform the switch timing you
      > After several hours on this design, I though again about your GIF
      named "MEG
      > flux switching diagram"... there was something wrong, but I
      couldn't find what
      > it was...
      > So, I turned on my computer, to look again, with attention. I must
      admit that
      > this diagram has been a revelation to me, because it was the first
      time I found
      > an explanation easy to understand and apparently valid.
      > The main idea is that an input coil acts just like a magnet, when
      > and that B field created cancel the magnet's B field , in the same
      C core.
      > I agree with this, but I don't with your L3 and L4 wave form, and
      with your
      > conclusions.
      > First, when L1 is on and L2 off, the left C core will see the B
      field going
      > down to zero as the B field in the input coil goes to 0,6 T...
      don't forget
      > that the magnet gives a static B field, and that L1 oppose it (in
      this case,
      > substract it). So L3 is affected by the dynamic B field change.
      > You can copy this statement for L1 off and L2 on, and L4 instead of
      > Secondly, as L1 "push" the magnet's field on the opposite C core,
      we should
      > have the same B field ( and not more ) induced in L4.
      > You can find my version of these timings in the attachment.
      > The third thing is your conclusions. If you consider the B field
      between the
      > time laps t'1, t'2 and t'3 ( instead of t1, t2 and t3), you don't
      find the same
      > result :
      > - for the lap t'1 / t'2, we have in L1, L3 and L4 a D B field of
      0,6 T, and in
      > L2 a D B field of zero.
      > - for the lap t'2 / t'3, we have in L2, L3 and L4 a D B field of
      0,6 T, and in
      > L1 a D B field of zero.
      > If we want to know the overall balance between inputs and outputs,
      we must take
      > in account the lap t'1 / t'3, and we find :
      > input overall D B = 0,6 + 0,6 = 1,2 T
      > output overall D B = 1,2 + 1,2 = 2,4 T
      > So, the cop is 2. Is there any flaw in my analysis ?
      > Best regards,
      > Camelazur.
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