I can see Leon's point even though my heart is with Ed.  But to be fair,
since the lowest common denominator is 2-Stroke, we have to reduce to that.
So either a 250cc 2-stroke is regarded as a 500cc engine, or a 2.4L 4-stroke
must be regarded as a 1.2L engine.  Bringing Wankels into the group would
make the rotary a 3900cc engine, the Ford a 3.6L and the motorbike 750cc.
Decide what the rule is going to be (swept volume over 360, 720 or 1080deg)
and go from there.  In the end analysis, it's up to some dickweed bureaucrat
on the racing commission to decide how big your engine is and therefore
whether or not you can play in this or that league.  Beyond that, it's an
exercise in mental masturbation.
My .02

peon@pacific.net.au wrote:

> Hi Ed,
>
> I sorta KNEW this would create contovesy.  As I alluded to last
> time,  this whole capacity question is a can of worms, and it tends
> to just keep going around and aound and back and forth }:>) .
> <snip>
>
> > Hi Leon
> >
> >     Good to have you jump in.  First, there is no disagreement over
> >     the
> > equivalency of the 13B twin rotor engine with two stroke or four
> > stroke engines.  I agree that if you (for example) take a 160 (2616
> > cc) CID displacement 4 stroke engine through one cycle (i.e. 720 deg
> > of crankshaft rotation) you will end up with the same air displacement
> > (say at 6000 rpm) as you will with the twin rotor.  Similar for the 80

< SNIP>