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Hi Guys,
I have that feeling of "de ja vue" I've been
here before. Here's how to understand it all without a million pages of
marthematics.
The face of a rotary piston presents itself to the
inlet port each revolution. (forget about the fact that the rotor has
3 faces - it only confuses the issue!!). Each piston of a 4 stroke piston
engine go through an "Otto Cycle" every TWO revolutions.
With a rotary, you get ONE suck, per
ROTOR, per REV. With a 4 stroke, you get ONE suck per
PISTON, per (get this) TWO revs!!!
The same goes for Power Pulses. A rotary gets
ONE power pulse per REV, per ROTOR. With a 4 stroke piston
engine, you get ONE power pulse, per PISTON, per TWO
revs.
1. The "SWEPT VOLUME" of a 6.5B/13B/20B
rotary is 654 cc per rotor.
2. For EACH revolution of the
E-Shaft, EACH rotor sucks 654 cc.
3. So, for a 6.5B, it sucks
654 cc, for a 13B, it sucks 1308 cc PER REV, and for a 20B it
sucks 1962 cc.
I'll leave you to do the conversion into Cu.In.
Now, a 4 stroke engine completes it's Otto
Cycle in TWO revolutions. So to equate a rotary to a 4 stroke "Otto Cycle"
piston engine, we must compare how much a rotary sucks in TWO
revolutions.
According to my limited mathematical ability, (I
could never "run the numbers" with the alacrity of the blokes on ACRE!
(}:>)),
A 6.5B is equivalent to a 1308 cc 4 stroke
piston (Otto Cycle) engine. (two times 654 cc per rev)
A 13B is equivalent to a 2616 cc 4 stroke piston
(Otto Cycle) engine. (two times 1308 cc per rev)
A 20B is equivalent to a 3924 cc 4 stroke piston
(Otto Cycle) engine. (two times 1962 cc per rev)
It is THAT simple. Compare Granny Smith
Apples with Granny Smith Apples (not mangoes or bananas!).
Now there are a couple of caveats.
1. The rotary, when ported,
has the ability to easily run 125% to 130% volumetric efficiency. So it
has a slightly higher ability to draw air than a piston engine.
Runners, throttle bodies, and injectors must be sized
accordingly.
2. This being so, it will put out
slightly (up to 20%) more power than the equivalent swept volume of a 4 cycle
piston engine of the same SWEPT capacity.
So there you have it. It's THAT
simple!
Enjoy!
Leon
----- Original Message -----
Sent: Friday, February 11, 2005 8:18
AM
Subject: [FlyRotary] Solved!!!! =>
Rotary AirFlow Equation? Help?
> Ed, I think you've blinded yourself with
science 8*)
>
> Each face of the rotor will cycle 40cid of air on
each revolution. You
> can ignore the eshaft. Every time
the rotor turns around, each face
> will have processed 40cid. If a
rotor is turning at 2000 RPM, then three
> faces will each cycle
2000*40cid of air. You bring in referencing off
> the eshaft, and
all of a sudden only 4 faces are processing air. I
> think that
is the mistake.
>
> You reference 720 degrees of eshaft
rotation. But all that means is
> that the rotors haven't
completed their cycle. You're just ignoring
> every third rotor
face.
>
>
Ernest, I think you are making my very
point.
1. Each rotor has 3*40 = 120 cid of volume
displace per 360 deg rotation and with two rotor that is 2*120 = 240 CID each
revolution.
so we have 2000*240 cid/1728 = 277.77 CFM
exactly what quantity the formula gives when treating the rotary as a
160 CID 4 stroke engine with e shaft at 6000 rpm - but that is for a
FULL 360 deg rotation of the rotor. See part
2 next {:>).
2. Example
160*6000/(2*1728) = 277.77 CFM which
agrees with the above but if the 160 CID figure is correct
then that means that 160/4 = 4 , only 4 rotor faces worth of displacement
have been considered in the formula. Which if that is correct then the
rotor have only turned 240 degs and not 360 deg. But if the rotor
has only turned 240 deg (240*3 = 720Deg Checks!) then that means that to make
a valid comparison with 1 above, we need to reduce the amount of rotation in
statement 1 to 240 deg instead of 360 which means that we only have 277.77*4/6
= 185 CFM which leaves me back where I started.
There is no question that if the two rotors have
completed their 360 deg rotation - then all six faces have
completed the cycle.
That gives 40*6*2000/1728 = 277.77 CFM, so lets
not debate that point - we both agree! .
My problem has been the use of the 160 CID
equivalent in the formula. I understand how it was arrived at ( I
believe) its simply the amount of displacement occurring in the rotary at 720
deg of e shaft rotation which is the standard of comparison with other
engines.
At 720 deg of e shaft rotation then indeed 160
CID of displace has occurred. If I calculate using the 240 deg of
rotor rotation I again get 240/360 * 6 = 4 rotor faces. But if that is
indeed the actually amount of displacement of the rotor then according to our
calculations in 1 above we need to use 240 deg not 360 and that gives 185 cfm
not 277.77.
I mean I don't care if we use 240 CID (the total
displacement across 1080) or 160 CID (the total displacement across 720
deg - the standard for comparison) but there should be some logical
consistency between the two.
WAIT! Stop the presses I Finally
Understand! The key is the 240 CID total displacement for 360 rotor
degrees or for 1080 deg of e shaft rotation.
It was not a math problem or logic
problem, my problem was a reference transformation
problem!
Using the rotor representation we have Air flow = 240 * 2000/1728 =
277.77 this is rotary based on 360 deg of rotation. Now if
we take that formula (which I think we agree on),
I then want to convert it into a formula that conforms to the
720 deg standard. So I can compare apples and
apples.
Here's what happens.
1. First the e shaft rpm is the e
shaft rpm and is the standard reference point for rpm of engines. So we
can simply multiply our 2000 rpm rotor rpm by 3 to reference it to the
eshaft. So 3* 2000 = 6000. That "transformation" gives us
the rpm figure in the "Standard" equation.
2. However, to reference the 240
CID displacement to the 720 deg standard we must use the ratio of the
transformation from 1080 to the 720 reference system. Because the amount
of displacement IS affected by the choice of reference.
This gives 720/1080 * 240 = 160 CID referenced to the
standard.
So now the formula should migrate from rotor
reference of 240*2000/1728 = 277.77 to e shaft reference of 160*6000/1728
= 277.77 by reference system transformation. To do
that:
1. We multiply the rotor rpm by
3 - the rpm does not care whether its 720 or 1080 referenced, one
revolution/min is one revolution/min.
2. The 720 deg standard DOES affect
the amount of displacement so 720/1080 *240 = 160 CID.
Taking the rotor reference formula above we can
directly transform it to the e shaft reference standard. Doing so gives
us
[720/1080*240]*[3*2000]/1728 = [0.666*240]*[6000]/1729, taking the first
factor 0.6666* 240 = 159.9999 = 160 and we have
Airflow = 160CID*6000/1728 = 277.77
So the rotor reference is transformed to the e shaft reference of 720
deg.
Thanks all for helping me out of my problem
area. I just knew there had to be a connection someway. If this is
not correct - please refrain from informing me {:>).
Be heartened, Ernest, I have been
faithfully using the 160 formula - just wanted to understand what appeared to
me to be a difference when looking at it from two different
perspectives. If the "reference" transformation is applied to the rotor
equation then it translates into the 720 reference state.
Best Regards
and thanks for your input
Ed
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