Return-Path: Received: from [24.25.9.102] (HELO ms-smtp-03-eri0.southeast.rr.com) by logan.com (CommuniGate Pro SMTP 4.3c1) with ESMTP id 724499 for flyrotary@lancaironline.net; Thu, 10 Feb 2005 20:12:01 -0500 Received-SPF: pass receiver=logan.com; client-ip=24.25.9.102; envelope-from=eanderson@carolina.rr.com Received: from edward2 (cpe-024-074-185-127.carolina.rr.com [24.74.185.127]) by ms-smtp-03-eri0.southeast.rr.com (8.12.10/8.12.7) with SMTP id j1B1BFkc028298 for ; Thu, 10 Feb 2005 20:11:15 -0500 (EST) Message-ID: <001101c50fd6$98144390$2402a8c0@edward2> From: "Ed Anderson" To: "Rotary motors in aircraft" References: Subject: Solved!!! Done!!!! Date: Thu, 10 Feb 2005 20:11:17 -0500 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_000E_01C50FAC.ADE22E00" 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_000E_01C50FAC.ADE22E00 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable 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. =20 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) =3D 6 * 40 * 2000/1728 =3D 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 =3D 160 * 6000/(2*1728) =3D 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 =3D 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.=20 That transforms 240* 2000/1728 directly into 160*6000/(2*1728) both = =3D 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 =20 ----- Original Message -----=20 From: To: "Rotary motors in aircraft" Sent: Thursday, February 10, 2005 7:41 PM Subject: [FlyRotary] Re: Solved!!!! =3D> 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 >=20 >=20 > In a message dated 2/10/2005 4:18:02 PM Eastern Standard Time, "Ed = Anderson" writes: >=20 > > > >> 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 =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 {:>). > > > >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. > >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. > > > >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. > > > >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 > > > > [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 > > > > > > > > > > > > > > > > > > > > > > >=20 > >> Homepage: http://www.flyrotary.com/ > >> Archive: http://lancaironline.net/lists/flyrotary/List.html ------=_NextPart_000_000E_01C50FAC.ADE22E00 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Hi Bill,
 
     Actually it = looks there is=20 a relationship between the formula and what can be figured out from = rotation=20 speed of the rotor and their displacement.
 
 
     Here it is = in  a nut=20 shell. 
 
     Taken the = rotor reference=20 we have 6 faces of 40 CID displacement in one revolution of the = rotor.  So=20 at 6000 e shaft rpm (2000 rotor rpm) we have
 
    Air Flow (rotor Ref) = =3D 6 * 40 *=20 2000/1728 =3D  277.77 CFM
 
The problem I was having was how this = equation got=20 transformed (and that's the key term) in to the "traditional" one used = for the=20 rotary
 
   Air Flow =3D 160 * = 6000/(2*1728) =3D =20 277.77 CFM
 
The answer was a reference = transformation of the=20 first "real" equation from a rotor reference to an=20 eccentric shaft reference - of which 720deg is the = accepted=20 standard for all 4 stroke engines.
 
The rpm translation is simply  = Rotary rpm * 3=20 =3D eccentric shaft rpm
 
The displacement transition factor is = the ratio=20 between the standard 720 deg and the rotary's 1080deg cycle
 
So displacement transition factor is = 720/1080 or=20 2/8 or 0.6666 * total rotor displacement.
 
That transforms    240* = 2000/1728=20 directly into 160*6000/(2*1728)  both =3D 277.77.  The 2 is = because we=20 are now treating the rotary as a traditional 4 stroke engine of which = only 1/2=20 of the cylinders are filled on each revolution.
 
I think everyone including myself has = had enough of=20 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=20 makes some sense to me and I can continue to use the formula with a = higher=20 degree of confidence.
 
Best Regards
 
Ed A
 
 
 
----- Original Message -----
From: <WRJJRS@aol.com>
To: "Rotary motors in aircraft" = <flyrotary@lancaironline.net>
Sent: Thursday, February 10, 2005 7:41=20 PM
Subject: [FlyRotary] Re: Solved!!!! = =3D> Rotary=20 AirFlow Equation? Help?

> Ed,
>  The problem here is that the formula = used for=20 displacement. The real solution is that the racing sanctioning bodies = want the=20 rotary to be at a slight disadvantage to piston engines! It, (the = formula), has=20 no relation to actual reality!
> Bill Jepson
>
> =
> In=20 a message dated 2/10/2005 4:18:02 PM Eastern Standard Time, "Ed = Anderson"=20 <eanderson@carolina.rr.com>=20 writes:
>
> >
> >> Ed, I think you've = blinded=20 yourself with science 8*)
> >>
> >> Each face of = the=20 rotor will cycle 40cid of air on each revolution. You
> >> = can=20 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=20 off
> >> the eshaft, and all of a sudden only 4 faces are = processing=20 air. I
> >> think that is the mistake.
> = >>
>=20 >> You reference 720 degrees of eshaft rotation. But all that = means=20 is
> >> that the rotors haven't completed their cycle. = You're just=20 ignoring
> >> every third rotor face.
> = >>
>=20 >>
> >
> >Ernest, I think you are making my very = 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 = exactly what=20 quantity the formula gives when treating the rotary as a 160 CID 4 = stroke engine=20 with e shaft at 6000 rpm - but that is for a FULL 360 deg rotation of = the rotor.=20 See part 2 next {:>).
> >
> >2. Example
>=20 >
> >160*6000/(2*1728) =3D 277.77 CFM which agrees with the = above but=20 if the 160 CID figure is correct then that means that 160/4 =3D 4 , only = 4 rotor=20 faces worth of displacement have been considered in the formula. Which = if that=20 is correct then the rotor have only turned 240 degs and not 360 deg. But = if the=20 rotor has only turned 240 deg (240*3 =3D 720Deg Checks!) then that means = that to=20 make a valid comparison with 1 above, we need to reduce the amount of = rotation=20 in statement 1 to 240 deg instead of 360 which means that we only have=20 277.77*4/6 =3D 185 CFM which leaves me back where I started.
> = >
>=20 >There is no question that if the two rotors have completed their 360 = deg=20 rotation - then all six faces have completed the cycle.
> >That = gives=20 40*6*2000/1728 =3D 277.77 CFM, so lets not debate that point - we both = agree!=20 .
> >
> >My problem has been the use of the 160 CID = equivalent=20 in the formula. I understand how it was arrived at ( I believe) its = simply the=20 amount of displacement occurring in the rotary at 720 deg of e shaft = rotation=20 which is the standard of comparison with other engines.
> = >
>=20 >At 720 deg of e shaft rotation then indeed 160 CID of displace has = occurred.=20 If I calculate using the 240 deg of rotor rotation I again get 240/360 * = 6 =3D 4=20 rotor faces. But if that is indeed the actually amount of displacement = of the=20 rotor then according to our calculations in 1 above we need to use 240 = deg not=20 360 and that gives 185 cfm not 277.77.
> >
> >I mean I = don't=20 care if we use 240 CID (the total displacement across 1080) or 160 CID = (the=20 total displacement across 720 deg - the standard for comparison) but = there=20 should be some logical consistency between the two.
> >
> = >WAIT! Stop the presses I Finally Understand! The key is the 240 CID = total=20 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=20 reference transformation problem!
> >
> >
> > = Using=20 the rotor representation we have Air flow =3D 240 * 2000/1728 =3D 277.77 = this is=20 rotary based on 360 deg of rotation. Now if we take that formula (which = I think=20 we agree on), I then want to convert it into a formula that conforms to = the 720=20 deg standard. So I can compare apples and apples.
> >
>=20 >Here's what happens.
> >
> >1. First the e shaft = rpm is=20 the e 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 = eshaft.=20 So 3* 2000 =3D 6000. That "transformation" gives us the rpm figure in = the=20 "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 = of=20 displacement IS affected by the choice of reference. This gives 720/1080 = * 240 =3D=20 160 CID referenced to the standard.
> >
> >So now the = formula=20 should migrate from rotor reference of 240*2000/1728 =3D 277.77 to e = shaft=20 reference of 160*6000/1728 =3D 277.77 by reference system = transformation. To do=20 that:
> >
> >1. We multiply the rotor rpm by 3 - the = rpm does=20 not care whether its 720 or 1080 referenced, one revolution/min is one=20 revolution/min.
> >2. The 720 deg standard DOES affect the = amount of=20 displacement so 720/1080 *240 =3D 160 CID.
> >
> = >Taking the=20 rotor reference formula above we can directly transform it to the e = shaft=20 reference standard. Doing so gives 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=20 160CID*6000/1728 =3D 277.77 So the rotor reference is transformed to the = e shaft=20 reference of 720 deg.
> >
> >Thanks all for helping me = out of=20 my problem area. I just knew there had to be a connection someway. If = this is=20 not correct - please refrain from informing me {:>).
> = >
>=20 >
> >Be heartened, Ernest, I have been faithfully using the = 160=20 formula - just wanted to understand what appeared to me to be a = difference when=20 looking at it from two different perspectives. If the "reference" = transformation=20 is applied to the rotor equation then it translates into the 720 = reference=20 state.
> >
> > Best Regards and thanks for your = input
>=20 >
> >Ed
> >
> >
> >
> = >
>=20 >
> >
> >
> >
> >
> = >
>=20 >
>
> >>  Homepage:  http://www.flyrotary.com/
>=20 >>  Archive:   http://lancaironline.net/lists/flyrotary/List.html ------=_NextPart_000_000E_01C50FAC.ADE22E00--