Hi Bill,
Actually it looks there is
a relationship between the formula and what can be figured out from rotation
speed of the rotor and their displacement.
Here it is in a nut
shell.
Taken the rotor reference
we have 6 faces of 40 CID displacement in one revolution of the rotor. So
at 6000 e shaft rpm (2000 rotor rpm) we have
Air Flow (rotor Ref) = 6 * 40 *
2000/1728 = 277.77 CFM
The problem I was having was how this equation got
transformed (and that's the key term) in to the "traditional" one used for the
rotary
Air Flow = 160 * 6000/(2*1728) =
277.77 CFM
The answer was a reference transformation of the
first "real" equation from a rotor reference to an
eccentric shaft reference - of which 720deg is the accepted
standard for all 4 stroke engines.
The rpm translation is simply Rotary rpm * 3
= eccentric shaft rpm
The displacement transition factor is the ratio
between the standard 720 deg and the rotary's 1080deg cycle
So displacement transition factor is 720/1080 or
2/8 or 0.6666 * total rotor displacement.
That transforms 240* 2000/1728
directly into 160*6000/(2*1728) both = 277.77. The 2 is because we
are now treating the rotary as a traditional 4 stroke engine of which only 1/2
of the cylinders are filled on each revolution.
I think everyone including myself has had enough of
this by now, so I am going to go back to designing my PC board
But, I am glad I finally have the connection that
makes some sense to me and I can continue to use the formula with a higher
degree of confidence.
Best Regards
Ed A
----- Original Message -----
Sent: Thursday, February 10, 2005 7:41
PM
Subject: [FlyRotary] Re: Solved!!!! => Rotary
AirFlow Equation? Help?
> Ed, > The problem here is that the formula used for
displacement. The real solution is that the racing sanctioning bodies want the
rotary to be at a slight disadvantage to piston engines! It, (the formula), has
no relation to actual reality! > Bill Jepson > > > In
a message dated 2/10/2005 4:18:02 PM Eastern Standard Time, "Ed Anderson"
<eanderson@carolina.rr.com>
writes: > > > > >> 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 > > > > > > > > >
> > > > > > > > > > > >
> > > >> Homepage: http://www.flyrotary.com/ >
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