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Hi Dave,
Thanks for the reply, and a well reasoned argument (argument being used
in the positive mathametical sense).
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.
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.
Just like a 2.6L 4 stroke engine, you have "displaced" 1.3L in one
revolution of the e-shaft.
I can see how your argument would conclude that the 13B is a 1.3L
engine, but I still don't see 2.6L.
The second objection I have to doing the calculation this way is that
the pieces in the engine aren't in the same position as they were when
we started. The rotor has only rotated 1/3 of the way around. One
could argue that one face is indistinguishable from another, so 1/3
rotation is the same as a full rotation (or 2/3 rotation for that
matter). But I _know_ that the rotor is in a different position, so I
do the measurement for a full rotation of the rotor.
I just read the analysis from Fred Swain. I believe he is saying the
same thing I am in a more refined manner. And I can see that I've been
semantically incorrect in my use of the terms cycle and stroke. I've
rewritten some of my comments above although I don't know if I have it
all correct. :)
There is an excellent graphic of the rotary at
http://science.howstuffworks.com/rotary-engine7.htm
It lets you run thru each step of the Otto cycle one at a time. Pretty
neat.
If nothing else, we're seeing a little action here on the list. It's
been so quiet the last few days.
Bob W.
On Mon, 17 Oct 2005 13:18:07 -0700
David Leonard <wdleonard@gmail.com> wrote:
> Bob,
> You are right about my error. I should have said "2 revolutions of the
> e-shaft as is done with reciprocating 4-strokes". Since it is a 4-cycle
> engine, shouldn't it be measured the same?
> Your point about how they measure 2-strokes vs. 4-strokes is well taken.
> Since the 2-stroke completes a cycle in only 1 revolution of the shaft they
> only use one revolution of the shaft to compute volume. So as you say, why
> not apply that technique to the rotary?
> My question is to you then becomes: why do you assume that the rotor faces
> are the "combustion element" that needs to complete its cycle? Shouldn't we
> be looking at the combustion chamber as the important combustion element?
> After all, it is the chamber that is determining the amount of displacement,
> not the rotor. Similarly, in a piston engine it is the cylinder size that
> determines the displacement not the pistons. As you say, in 2-stroke and
> 4-stroke engines we wait for all cylinders (not pistons) to complete one
> cycle. In the rotary shouldn't we be waiting for each rotor assembly (or
> combustion chamber) to complete one cycle? In that case, to complete a cycle
> each combustion chamber requires one revolution of the e-shaft, and the 13B
> is a 1.3 L displacement engine... or is it?
> This is where we think outside the box. Inside a rotor housing, yes there
> are 3 rotor faces, but there are only TWO cumulative 650 cc volumes being
> moved around (spread between the 3 rotor faces). In one revolution of the
> e-shaft those two volumes, working together, complete exactly one full Otto
> cycle!!! Counted together (because those volumes do exist at the same time),
> those 2 650 cc volumes are 1.3 L per rotor or 2.6 L for the engine.
> Hugh? Two volumes? Where did I come up with that you say. How would you
> measure displacement of a combustion chamber If you could just look at one
> combustion chamber with its compression element? What would you measure? You
> would move the compression element to the position where it creates the
> largest space in the chamber and measure the volume. Then you would subtract
> out any area that is never displaced by the compression element, and you are
> left with the displacement. Do this exercise with the rotary using a single
> rotor and rotor housing, and you are left with 1.3 L displacement per rotor
> and a completed 4-cycle in 1 revolution of the e-shaft.
> But I still think it is best to look at it Ed's way. :-)
> --
> 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
> On 10/16/05, Bob White <bob@bob-white.com> wrote:
> >
> > Hi Dave,
> >
> > OK, one revolution of the e-shaft is 1/3 revolution of the rotors. So
> > each rotor has had one intake event. Each face has a calculated
> > displacement of about 650 cc. Two X 650 cc = 1.3L. If you can explain
> > why it's 2.6L, maybe I can send Paul an apology. Or are you just
> > trying to get my goat? :)
> >
> > I'm not trying to create a big discussion on the displacement of the
> > rotary, I just want to understand where that 2.6L per revolution number
> > is comming from. I haven't been able to see it. I think Paul gets it
> > from comparing to a piston engine, and I agree that the 13B compares
> > closest to a 2.6L 4 cycle 4 cylinder engine.
> >
> > Bob W.
> >
> >
> > On Sun, 16 Oct 2005 16:27:59 -0700
> > David Leonard <wdleonard@gmail.com> wrote:
> >
> > > Monty, Glad to have you and you know you will always be welcome here.
> > > However, you are wrong and 'he' is right about the displacement of the
> > 13B.
> > > It is 2.6L or 159.6 cubic inches to be more exact.
> > > That is the volume of intake on one revolution of the e-shaft.
> > > But I think you knew that, you were just trying to get his goat. ;-)
> > >
> > > --
> > > 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
> > >
> > > On 10/16/05, Monty Roberts <montyr2157@alltel.net> wrote:
> > > >
> > > > The doctrine of immaculate ingestion. Whereby molecules of air and
> > fuel
> > > > magically migrate into a very small, very perfect engine,unsullied by
> > the
> > > > mere laws of physics, thereby creating the salvation of the world
> > through
> > > > massive power levels.
> > > > In the protestant tradition of placing the individual at the front of
> > the
> > > > line rather than at the bottom of the church hierarchy, I will
> > henceforth
> > > > place all replies at the TOP of each post.
> > > > Monty
> > > > Which doctrine was that Monty?
> > > >
> > > > Bob W.
> > > >
> > > >
> > >
> >
> >
> > --
> > http://www.bob-white.com
> > N93BD - Rotary Powered BD-4 (real soon)
> > Prewired EC2 Cables - http://www.roblinphoto.com/shop/
> >
> > --
> > Homepage: http://www.flyrotary.com/
> > Archive and UnSub: http://mail.lancaironline.net/lists/flyrotary/
> >
> >
>
--
http://www.bob-white.com
N93BD - Rotary Powered BD-4 (real soon)
Prewired EC2 Cables - http://www.roblinphoto.com/shop/
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