Return-Path: Received: from [24.25.9.101] (HELO ms-smtp-02-eri0.southeast.rr.com) by logan.com (CommuniGate Pro SMTP 4.3c1) with ESMTP id 724175 for flyrotary@lancaironline.net; Thu, 10 Feb 2005 16:18:47 -0500 Received-SPF: pass receiver=logan.com; client-ip=24.25.9.101; envelope-from=eanderson@carolina.rr.com Received: from edward2 (cpe-024-074-185-127.carolina.rr.com [24.74.185.127]) by ms-smtp-02-eri0.southeast.rr.com (8.12.10/8.12.7) with SMTP id j1ALHxed021706 for ; Thu, 10 Feb 2005 16:17:59 -0500 (EST) Message-ID: <003201c50fb6$014375f0$2402a8c0@edward2> From: "Ed Anderson" To: "Rotary motors in aircraft" References: Subject: Solved!!!! => Rotary AirFlow Equation? Help? Date: Thu, 10 Feb 2005 16:18:02 -0500 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_002F_01C50F8C.1836A680" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2800.1106 X-MIMEOLE: Produced By Microsoft MimeOLE V6.00.2800.1106 X-Virus-Scanned: Symantec AntiVirus Scan Engine This is a multi-part message in MIME format. ------=_NextPart_000_002F_01C50F8C.1836A680 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable =20 > 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=20 > 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 = > 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=20 > that the rotors haven't completed their cycle. You're just ignoring=20 > 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. =20 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 {:>). =20 =20 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_002F_01C50F8C.1836A680 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
 
> Ed, I think you've = blinded yourself=20 with science 8*)
>
> Each face of the rotor will cycle = 40cid of air=20 on each revolution.  You
> can ignore the eshaft.  = Every time=20 the rotor turns around, each face
> will have processed 40cid. If = a rotor=20 is turning at 2000 RPM, then three
> faces will each cycle = 2000*40cid of=20 air.  You bring in referencing off
> the eshaft, and all of = a sudden=20 only 4 faces are processing air.  I
> think that is the=20 mistake.
>
> You reference 720 degrees of eshaft = rotation. =20 But all that means is
> that the rotors haven't completed their=20 cycle.  You're just ignoring
> every third rotor = face.
>=20
>
 
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 160=20 CID 4 stroke engine with e shaft at 6000 rpm - but that is for a = FULL=20 360 deg rotation of the rotor.  See part 2 next=20 {:>). 
 
2.  Example
 
160*6000/(2*1728)  =3D 277.77 = CFM which=20 agrees with the above but if the 160 CID  figure is = correct then=20 that means that 160/4 =3D 4 , only 4 rotor faces worth of = displacement have=20 been considered in the formula.  Which if that is correct then the = rotor=20 have only turned 240 degs  and not 360 deg.  But if the rotor = has only=20 turned 240 deg (240*3 =3D 720Deg Checks!) then that means that to make a = valid=20 comparison with 1 above, we need to reduce the amount of rotation in = statement 1=20 to 240 deg instead of 360 which means that we only have 277.77*4/6 =3D = 185 CFM=20 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 not=20 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 = believe)=20 its simply the amount of displacement occurring in the rotary at 720 deg = of e=20 shaft rotation  which is the standard of comparison with other=20 engines.
 
At 720 deg of e shaft rotation then = indeed 160 CID=20 of displace has occurred.  If I calculate using the 240 deg of = rotor=20 rotation I again get 240/360 * 6 =3D 4 rotor faces.  But if that is = indeed=20 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 = consistency=20 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 problem,=20 my problem was a reference transformation = problem!
 
 
          &nbs= p; =20 Using the rotor representation we have Air flow =3D 240 * 2000/1728 = =3D 277.77=20 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 720=20 deg standard.  So I can compare apples and = apples.
 
Here's what happens. 
 
1.   First the e shaft rpm is = the e shaft=20 rpm and is the standard reference point for rpm of engines.  So we = can=20 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 =3D=20 277.77 by reference system transformation.  To do = that:
 
1.   We multiply the = rotor rpm by 3=20 - the rpm does not care whether its 720 or 1080 referenced, one = revolution/min=20 is one revolution/min.
2.   The 720 deg standard = DOES affect the=20 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 factor=20 0.6666* 240 =3D 159.9999 =3D 160 and we have
 
  Airflow =3D 160CID*6000/1728 =3D = 277.77  So=20 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 me=20 to be a difference when looking at it from two different = perspectives.  If=20 the "reference" transformation is applied to the rotor equation then it=20 translates into the  720 reference state.
 
 Best Regards=20 and thanks for your input
 
Ed
 
 
 
 
 
 
 
 
 
 
------=_NextPart_000_002F_01C50F8C.1836A680--