Return-Path: Received: from mailout1.pacific.net.au ([61.8.0.84] verified) by logan.com (CommuniGate Pro SMTP 4.3c1) with ESMTP id 724705 for flyrotary@lancaironline.net; Fri, 11 Feb 2005 00:37:55 -0500 Received-SPF: none receiver=logan.com; client-ip=61.8.0.84; envelope-from=peon@pacific.net.au Received: from mailproxy2.pacific.net.au (mailproxy2.pacific.net.au [61.8.0.87]) by mailout1.pacific.net.au (8.12.3/8.12.3/Debian-7.1) with ESMTP id j1B5b7A6022095 for ; Fri, 11 Feb 2005 16:37:07 +1100 Received: from ar1 (ppp211C.dyn.pacific.net.au [61.8.33.28]) by mailproxy2.pacific.net.au (8.12.3/8.12.3/Debian-7.1) with SMTP id j1B5b2wE015253 for ; Fri, 11 Feb 2005 16:37:03 +1100 Message-ID: <005101c50ffb$439d7e10$1c21083d@ar1> From: "Leon" To: "Rotary motors in aircraft" References: Subject: Last Word on Displacement was Re: [FlyRotary] Solved!!!! => Rotary AirFlow Equation? Help? Date: Fri, 11 Feb 2005 16:33:48 +1100 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_004E_01C51057.76677CE0" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2800.1409 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2800.1409 This is a multi-part message in MIME format. ------=_NextPart_000_004E_01C51057.76677CE0 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable 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. =20 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. =20 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.=20 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! (}:>)), =20 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 -----=20 From: Ed Anderson=20 To: Rotary motors in aircraft=20 Sent: Friday, February 11, 2005 8:18 AM Subject: [FlyRotary] Solved!!!! =3D> Rotary AirFlow Equation? Help? > Ed, I think you've blinded yourself with science 8*) >=20 > Each face of the rotor will cycle 40cid of air on each revolution. = You=20 > 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=20 > faces will each cycle 2000*40cid of air. You bring in referencing = off=20 > the eshaft, and all of a sudden only 4 faces are processing air. I=20 > think that is the mistake. >=20 > 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. >=20 > Ernest, I think you are making my very point.=20 1. Each rotor has 3*40 =3D 120 cid of volume displace per 360 deg = rotation and with two rotor that is 2*120 =3D 240 CID each revolution. so we have 2000*240 cid/1728 =3D 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 {:>). =20 2. Example 160*6000/(2*1728) =3D 277.77 CFM which agrees with the above but if = the 160 CID figure is correct then that means that 160/4 =3D 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 =3D 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 =3D 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. =20 That gives 40*6*2000/1728 =3D 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 =3D 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. =20 It was not a math problem or logic problem, my problem was a reference = transformation problem! Using the rotor representation we have Air flow =3D 240 * = 2000/1728 =3D 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.=20 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 =3D 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 =3D 160 CID = referenced to the standard. So now the formula should migrate from rotor reference of = 240*2000/1728 =3D 277.77 to e shaft reference of 160*6000/1728 =3D = 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 =3D 160 CID. Taking the rotor reference formula above we can directly transform it = to the e shaft reference standard. Doing so gives us =20 [720/1080*240]*[3*2000]/1728 =3D [0.666*240]*[6000]/1729, taking = the first factor 0.6666* 240 =3D 159.9999 =3D 160 and we have Airflow =3D 160CID*6000/1728 =3D 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 ------=_NextPart_000_004E_01C51057.76677CE0 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Hi Guys,
 
I have that feeling of "de ja = vue"  I've been=20 here before.  Here's how to understand it all without a million = pages of=20 marthematics. 
 
The face of a rotary piston presents = itself to the=20 inlet port each revolution.  (forget about the fact that the = rotor has=20 3 faces - it only confuses the issue!!).  Each piston of a 4 stroke = piston=20 engine go through an "Otto Cycle" every TWO revolutions.  =
 
With a rotary,  you get ONE = suck,  per=20 ROTOR,  per REV.  With a 4 stroke,  you get ONE suck per=20 PISTON,  per  (get this) TWO revs!!!
 
The same goes for Power Pulses.  A = rotary gets=20 ONE power pulse per REV,  per ROTOR.  With a 4 stroke piston=20 engine,  you get ONE power pulse,  per PISTON,  per TWO=20 revs.
 
1.   The "SWEPT VOLUME" of a = 6.5B/13B/20B=20 rotary is 654 cc per rotor.
2.   For EACH revolution of = the=20 E-Shaft,  EACH rotor sucks 654 cc.
3.   So,  for a = 6.5B,  it sucks=20 654 cc,  for a 13B,  it sucks 1308 cc PER REV,  and for a = 20B it=20 sucks 1962 cc.
 
I'll leave you to do the conversion = into Cu.In.=20
 
Now,  a 4 stroke engine completes = it's Otto=20 Cycle in TWO revolutions.  So to equate a rotary to a 4 stroke = "Otto Cycle"=20 piston engine,  we must compare how much a rotary sucks = in TWO=20 revolutions.
 
According to my limited mathematical = ability, (I=20 could never "run the numbers" with the alacrity of the blokes = on ACRE!=20 (}:>)), 
 
A 6.5B is equivalent to a 1308 cc = 4 stroke=20 piston (Otto Cycle) engine.  (two times 654 cc per = rev)
A 13B is equivalent to a 2616 cc 4 = stroke piston=20 (Otto Cycle) engine.   (two times 1308 cc per = rev)
A 20B is equivalent to a 3924 cc 4 = stroke piston=20 (Otto Cycle) engine.   (two times 1962 cc per = rev)
 
It is THAT simple.  Compare Granny = Smith=20 Apples with Granny Smith Apples (not mangoes or bananas!).
 
Now there are a couple of = caveats.
 
1.   The rotary,  when = ported, =20 has the ability to easily run 125% to 130% volumetric efficiency.  = So it=20 has a slightly higher ability to draw air than a piston engine. =20 Runners,  throttle bodies,  and injectors must be sized=20 accordingly.
 
2.   This being so,  it = will put out=20 slightly (up to 20%) more power than the equivalent swept volume of a 4 = cycle=20 piston engine of the same SWEPT capacity.
 
So there you have it. It's THAT=20 simple!
 
Enjoy!
 
Leon
----- Original Message -----
From:=20 Ed=20 Anderson
Sent: Friday, February 11, 2005 = 8:18=20 AM
Subject: [FlyRotary] Solved!!!! = =3D>=20 Rotary AirFlow Equation? Help?


> Ed, I think you've blinded = yourself with=20 science 8*)
>
> Each face of the rotor will cycle 40cid = of air on=20 each revolution.  You
> can ignore the eshaft.  Every = time=20 the rotor turns around, each face
> will have processed 40cid. = If a=20 rotor is turning at 2000 RPM, then three
> faces will each = cycle=20 2000*40cid of air.  You bring in referencing off
> the = eshaft, and=20 all of a sudden only 4 faces are processing air.  I
> = think that=20 is the mistake.
>
> You reference 720 degrees of eshaft=20 rotation.  But all that means is
> that the rotors haven't = completed their cycle.  You're just ignoring
> every third = rotor=20 face.
>
>
 
Ernest, I think you are making my = very=20 point. 
 
1.  Each rotor has 3*40 =3D 120 = cid of volume=20 displace per 360 deg rotation and with two rotor that is 2*120 =3D 240 = CID each=20 revolution.
 so we have 2000*240 cid/1728 = =3D 277.77 CFM=20 exactly what quantity the formula gives  when treating the rotary = as a=20 160 CID 4 stroke engine with e shaft at 6000 rpm - but that is for a=20 FULL 360 deg rotation of the rotor.  See part=20 2 next {:>). 
 
2.  Example
 
160*6000/(2*1728)  =3D 277.77 = CFM which=20 agrees with the above but if the 160 CID  figure is = correct=20 then that means that 160/4 =3D 4 , only 4 rotor faces worth of = displacement=20 have been considered in the formula.  Which if that is correct = then the=20 rotor have only turned 240 degs  and not 360 deg.  But if = the rotor=20 has only turned 240 deg (240*3 =3D 720Deg Checks!) then that means = that to make=20 a valid comparison with 1 above, we need to reduce the amount of = rotation in=20 statement 1 to 240 deg instead of 360 which means that we only have = 277.77*4/6=20 =3D 185 CFM which leaves me back where I started.
 
There is no question that if the two = rotors have=20 completed their 360 deg rotation - then all six faces = have=20 completed the cycle. 
That gives 40*6*2000/1728 =3D 277.77 = CFM, so lets=20 not debate that point - we both agree! .
 
My problem has been the use of the = 160 CID=20 equivalent in the formula.  I understand how it was arrived at ( = I=20 believe) its simply the amount of displacement occurring in the rotary = at 720=20 deg of e shaft rotation  which is the standard of comparison with = other=20 engines.
 
At 720 deg of e shaft rotation then = indeed 160=20 CID of displace has occurred.  If I calculate using the 240 = deg of=20 rotor rotation I again get 240/360 * 6 =3D 4 rotor faces.  But if = that is=20 indeed the actually amount of displacement of the rotor then according = to our=20 calculations in 1 above we need to use 240 deg not 360 and that gives = 185 cfm=20 not 277.77.
 
I mean I don't care if we use 240 CID = (the total=20 displacement  across 1080) or 160 CID (the total displacement = across 720=20 deg - the standard for comparison) but there should be some logical=20 consistency between the two.
 
WAIT! Stop the presses I =  Finally=20 Understand!  The key is the 240 CID total displacement for 360 = rotor=20 degrees or for 1080 deg of e shaft rotation.  =
 
It was not a math problem or = logic=20 problem, my problem was a reference transformation=20 problem!
 
 
          &nbs= p; =20 Using the rotor representation we have Air flow =3D 240 * = 2000/1728 =3D=20 277.77 this is rotary based on 360 deg of rotation.  Now if=20 we  take that formula (which I think we agree on),=20  I then want to convert it into a formula that conforms = to the=20 720 deg standard.  So I can compare apples and=20 apples.
 
Here's what = happens. 
 
1.   First the e shaft rpm = is the e=20 shaft rpm and is the standard reference point for rpm of = engines.  So we=20 can simply multiply our 2000 rpm rotor rpm by 3 to reference = it to the=20 eshaft. So 3* 2000 =3D 6000.  That "transformation" = gives us=20 the rpm figure in the "Standard" equation.
 
2.  However, to =  reference the 240=20 CID displacement to the 720 deg standard we must use the ratio of the=20 transformation from 1080 to the 720 reference system.  Because = the amount=20 of displacement IS affected by the choice of = reference. =20  This gives 720/1080 * 240 =3D 160 CID referenced to the=20 standard.
 
So now the formula should migrate = from rotor=20 reference of 240*2000/1728 =3D 277.77 to e shaft reference of = 160*6000/1728=20 =3D 277.77 by reference system transformation.  To do=20 that:
 
1.   We multiply the = rotor rpm by=20 3 - the rpm does not care whether its 720 or 1080 referenced, one=20 revolution/min is one revolution/min.
2.   The 720 deg standard = DOES affect=20 the amount of displacement so 720/1080 *240 =3D 160 CID.
 
Taking the rotor reference formula = above we can=20 directly transform it to the e shaft reference standard.  Doing = so gives=20 us
   
    =20 [720/1080*240]*[3*2000]/1728 =3D [0.666*240]*[6000]/1729, taking the = first=20 factor 0.6666* 240 =3D 159.9999 =3D 160 and we have
 
  Airflow =3D 160CID*6000/1728 = =3D 277.77 =20 So the rotor reference is transformed to the e shaft reference of 720=20 deg.
 
Thanks all for helping me out of my = problem=20 area.  I just knew there had to be a connection someway.  If = this is=20 not correct - please refrain from informing me {:>).
 
 
Be  heartened, Ernest,  I = have been=20 faithfully using the 160 formula - just wanted to understand what = appeared to=20 me to be a difference when looking at it from two different=20 perspectives.  If the "reference" transformation is applied to = the rotor=20 equation then it translates into the  720 reference = state.
 
 Best Regards=20 and thanks for your input
 
Ed
 
 
 
 
 
 
 
 
 
 
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