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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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