> Help me out here Paul. This is a point that could assist newcomers.
    > Perhaps
    > a rough explanation proof of this 'equivalent displacement' would
   be a
    > statement akin to:
    >
    >   " and so if at similar comparative VE the Wankel rotary engine will
    > have
    > to run at about twice the rpm compared to an RPE to pump (displace)
    > the same
    > amount of air."
    >
   Doug,
   I wasn't going to say anything anymore.
   However, regarding your concern for the new people, above is also not
   correct.

   Both engine run at the same rpm. It is the number one aspect to have
   equal speed.
   Second we have to count two chamber not two rotations of the Rotary to
   be equal to the RPE.
   I know multiplying it out results in the same, but this would be
   incorrect as the rpm number would be double. It is the displacement of
   one rotation that needs to be doubled to make things equals with the
   RPE. That is the same as counting two chamber per rotor.

   There is difference between saying we we have to run the rotary at twice
   the speed and saying we have to calculate its displacement at two shaft
   rotations.

   Your sentence could read:
     " and so if at similar comparative VE the Wankel rotary engine
   will have
   to run at the same rpm compared to an RPE and we have to count two
   chambers per rotor to pump (displace) the same
   amount of air."

   Rolf



I haven't heard anybody talk about why RPE engines measure displacement
to start with..

I think "displacement" is a way to predict how much air (volume and
mass) are going to be moved at a given operating condition (output
RPM).  And this is useful because the amount of air moved in a given
amount of time is a way to predict engine output.

Thus, it makes sense to look at how much air is moved through a wankel
at a given RPM.  From this come up with a method of comparing chamber
volume to piston engine displacement.  Anything else is mental gyrations
for no good purpose.


Matt-



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