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Date: Mon, 17 Oct 2005 21:00:32 -0600
From: Bob White <bob@bob-white.com>
To: "Rotary motors in aircraft" <flyrotary@lancaironline.net>
Subject: Re: [FlyRotary] Re: flyrotary_Web_Archive Re: Banishment
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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
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