Hi Dave,
Thanks for the reply, and a well reasoned argument (argument being used
in the positive mathametical sense).
Thanks Bob, yes it is fun.
I think there is an error in your analysis however. In one revolution
of the e-shaft, all three faces are involved in a single Otto cycle.
Correct... the 3 faces, together, have completed the FOUR Otto cycles, acting on 3 packets of air, but only 2 of those affect the outside world (the one that came in, and the one that went out). If the rotor were, say, hexagonal it would not change things as long as one packet of air came in, and one left. It doesn't matter how many faces the rotor has, just that one turn of the shaft causes one packet to enter and one leave.
The size of each packet is 650cc.
The 13B inputs 2 packets and exhausts 2 packets with each rotation of the shaft. The same as would a 2.6 L 4-cylinder engine. If the rotors were hexagonal but the engine still input and output 2 packets of air with each rotation of the shaft, then it would still be a
2.6 L engine.
One face is on the intake, the second is on the compression and
expansion, and the third is on the exhaust. In that one revolution
face 1 draws in 650 cc, face 2 compresses it's 650 cc (drawn in on the
previous rotation of the e-shaft) to about 65cc (I just made that
number up - compression ratio is about 9.7:1), ignites it and expands
it back out again. face 3 is exhausting the burned gasses (drawn in
two revolutions of the e-shaft ago). That is one complete Otto cycle,
and we have only had one intake event so 650 cc X two rotors = 1.3L. I
don't think you can double or tripple the volume because more than one
face is involved in the process.
You are right, I was just presenting another way of looking at the rotary and explaining the two volumes of importance - the in and the out.
Just like a 2.6L 4 stroke engine, you have "displaced" 1.3L in one
revolution of the e-shaft.
exactly!
I can see how your argument would conclude that the 13B is a 1.3L
engine, but I still don't see 2.6L.
see above. :-)