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On Mon, 17 Oct 2005 18:22:44 -0700
David Leonard <wdleonard@gmail.com> wrote:
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.
Ah! Now I understand what you are saying. But I think this is an
error. You are counting the same packet twice. All you can count is
how much air (fuel/air mixture) goes in. You can't count it agin when
it comes out.
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!
OK, now we are narrowing this down to semantics problems. Just as a
1.3L 2 stroke engine is not a 2.6L engine because it's "equivalent" to
a 2.6L 4 stroke engine, the rotary isn't a 2.6L engine because it's
"equivalent" to a 2.6L 4 stroke engine. BTW, I agree with your statement in another post:
"That is why I claim that the rotary is a 2.6 L equavelent." as long as the 2.6L is a 4 stroke. :)
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. :-)
--
Dave Leonard
Turbo Rotary RV-6 N4VY
http://members.aol.com/_ht_a/rotaryroster/index.html
http://members.aol.com/_ht_a/vp4skydoc/index.html
Bob W. (No more displacement talk for me. I think I've located a
source for D581 coil connectors.)
--
http://www.bob-white.com
N93BD - Rotary Powered BD-4 (real soon)
Prewired EC2 Cables - http://www.roblinphoto.com/shop/
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