Hi Tom,

As I say, my comment isn't directed at the calculations at all. There
has been much discussion about what the displacement is, and how much of
the rotation to count.  Mazda claiming 1.3 L for example.

I stand by my statement about the displacement.  That seems to be
exactly the way Ed has used it.

Quote from Ed's post:

     "Taken the rotor reference we have 6 faces of 40 CID displacement
in one revolution of the rotor.  So at 6000 e shaft rpm (2000 rotor rpm)
we have

Air Flow (rotor Ref) = 6 * 40 * 2000/1728 =  277.77 CFM"

The 6 * 40 is the 240 cu. in. displacement, 2000 RPM, and 1728 to
convert to CFM.  What am I missing?

Bob White


On Thu, 10 Feb 2005 20:33:43 -0800 (PST)
Tom <tomtugan@yahoo.com> wrote:

> Sorry,
> Additonally,  Bob you say "the rotary (uses it's displacement ) in 3
> revolutions of the E-shaft or one revolution of the rotors."     While
> all of the displacement of all the rotors is disturbed during one
> revolution of all of the rotors, Ed's work attempts to determine how
> much combustion charge is consumed (factoring in displacement and
> other factors) per revolution of the rotors and then determine CFM
> based on consumed combustion charges and rpm. Apparently it isn't as
> simple as what you said about 'uses it's displacement'  Or so I think.
>  Tom
>  
>
>
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