I would like to interject a point here that I tried to make on the other
list with little effect.  All the bickering about what the "real"
displacement for the rotry is is based on trying to make it "equilivent"
to a 4-stroke, or a 2-stroke, or a ???.  If I want to know the
displacement of any piston engine I would calculate a volume based on
the bore and stroke, multiply by the number of the pistons, and I would
have the displacement.  Mazda decideded the engine was a 1.3 liter for
tax reasons.  Various racing orginizations have numbers they calculate,
etc.  Displacement doesen't have anything to do with RPM, it's just how
much does the engine displace!

In operation, the 2-stroke "uses" it's displacement in 1 revolution, the
4-stroke in 2 revolutions, and the rotary in 3 revolutions of the
E-shaft or one revolution of the rotors.  In all these cases, the engine
has gone thru one complete cycle and is in exactly the same state it
started in.

I'm not saying these discussions are not meaningful, because the intent
is to know how to compare the rotary with a different engine geometry.
But when the insurance company wanted to know what size my engine was, I
calculated the "actual" displacement at 238 cu. in.  If I had a 2.6 L
2-stroke engine, I wouldn't use an "equilivent" displacement so that it
would match a 4-stroke so why would I do so for the rotary?

That's my understanding, and Ed it looks to me like you've done a great
job sorting thru the math.

Bob White


On Thu, 10 Feb 2005 19:41:42 -0500
WRJJRS@aol.com wrote:

> Ed,
>  The problem here is that the formula used for displacement. The real
>  solution is that the racing sanctioning bodies want the rotary to be
>  at a slight disadvantage to piston engines! It, (the formula), has no
>  relation to actual reality!
> Bill Jepson
>
>




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