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> 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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