Return-Path: Received: from [24.25.9.100] (HELO ms-smtp-01-eri0.southeast.rr.com) by logan.com (CommuniGate Pro SMTP 4.3c1) with ESMTP id 723641 for flyrotary@lancaironline.net; Thu, 10 Feb 2005 11:10:16 -0500 Received-SPF: pass receiver=logan.com; client-ip=24.25.9.100; envelope-from=eanderson@carolina.rr.com Received: from edward2 (cpe-024-074-185-127.carolina.rr.com [24.74.185.127]) by ms-smtp-01-eri0.southeast.rr.com (8.12.10/8.12.7) with SMTP id j1AG9Tbo006902 for ; Thu, 10 Feb 2005 11:09:29 -0500 (EST) Message-ID: <004601c50f8a$e83f8e70$2402a8c0@edward2> From: "Ed Anderson" To: "Rotary motors in aircraft" Subject: Rotary AirFlow Equation? Help? Date: Thu, 10 Feb 2005 11:09:32 -0500 MIME-Version: 1.0 Content-Type: multipart/related; type="multipart/alternative"; boundary="----=_NextPart_000_0042_01C50F60.FF2E0410" 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_0042_01C50F60.FF2E0410 Content-Type: multipart/alternative; boundary="----=_NextPart_001_0043_01C50F60.FF2E0410" ------=_NextPart_001_0043_01C50F60.FF2E0410 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable In a previous posting, I alluded to my suspicions that a commonly = accepted formula for calculating airflow for the rotary engine might be = flawed. I was asked off-line to expound on that suspicion. Well, I can think of no better audience to present my suspicions to and = see what you think. Last time I tried this on another list I was = berated, beat upon and chastised for daring to raise the question - but, = never one to be daunted for long and knowing this list has some = experienced and clear thinkers, here goes again. 1. The racing crowd was thrown into a turmoil when the rotary appeared = on the scene - what displacement class should it be place in?? Well eventually the following was apparently (If not agreed to was at = least accepted) that the 1300 cc 13B rotary was equivalent to a 160 CID = 4 stroke engine. Then the standard air flow formula could be applied - = which is Air Flow (CFM) =3D Displacement*rpm/(2 *1728)*Ve (Volumetric efficiency = (Ve) is generally assume to be 100% or 1, so we can drop it for this = discussion as well as the effects of compression ratios) How did they arrive at 160 CID displacement, well they apparently = decided to reference the rotary to the standard 720 Deg 4 stroke = rotation cycle even though for all six faces of the two rotors to go = through their cycle requires 1080 deg of e shaft rotation - well that is = 360 deg more than the standard of 720Deg. So if you limit consideration = to only 720deg of rotation only 4 of the six rotor faces have occurred, = so since the displacement by a single face is 40 CID then 4* 40 =3D 160 = CID. So that appears to be how the decision of 160 CID equivalent = displacement was arrived at. No problem with that part - but, continuing = on: So taking the formula and assuming 6000 rpm, we have Air Flow =3D = 160*6000/(2*1728) =3D 277.77 CFM (assuming 100% Ve). 2. So what's the problem? Well, unfortunately, I like to fully = understand how and why something works rather than just plugging in = numbers to a formula (although, heaven knows I do enough of that).=20 So here is how I started sliding down the slipper slope {:>). We know that the rotors turn at a speed 3 times less than the E shaft. = So if the e shaft is turning at 6000 rpm then the rotors are turning at = 6000/3 =3D 2000 rpm. Its the rotor of course that actually suck in the = air - not the eccentric shaft. So we should be able to arrive at the = same answer from either reference point (rotor or e shaft). Well the = formula above uses the e shaft reference. So I looked at what the rotor = reference would produce (expecting the same answer). If the rotors are spinning at 2000 rpm, we know that 2 of the rotor = faces have gone through their cycle for each 360 deg revolution of the e = shaft or 4 will have for 720 deg. Looking at it from the rotor = reference if the e shaft has turned 720 deg then the rotor has turned = 720/3 =3D 240 degrees. 240/360 =3D 0.6666 * 6 faces =3D 4 faces. So = that confirms that 4 faces go through their cycle in 720 degs of e shaft = rotation. So if 4 faces of the two rotors complete their cycle in = 720Deg eshaft then we have Air Flow =3D 4(number of faces) *40 (displacement for each face) * rpm = (rotor rpm)/(1728) note: we drop the division by 2 which in the = original 4 stroke reciprocating engine equation takes into account that = only 1/2 of the cylinders are sucking air on each revolution. =20 3. This Logic gives us rotary Air Flow =3D 4*40*2000/(1728) =3D 185 = CFM! This is ,of course, NOT the 277 CFM arrived at by the approved = formula. As you can see the implications of this (if correct) are fairly = significant and that is why even though I can't spot the error, I must = have made one. I would greatly appreciate any assistance in helping me understand where = I have gone wrong in logic or math. Please don't just quote an = authority or formula - explain it and show me your calculations on how = you got there. Its things like this that keep me awake at night {:>). Help Meee..! =20 Thanks Ed Ed Anderson Rv-6A N494BW Rotary Powered Matthews, NC eanderson@carolina.rr.com ------=_NextPart_001_0043_01C50F60.FF2E0410 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
In a previous posting, I alluded to my = suspicions=20 that a commonly accepted formula for calculating airflow for the rotary=20 engine might be flawed. I was asked off-line to expound on = that=20 suspicion.
 
 Well, I can think of no better = audience to=20 present my suspicions to and see what you think.  Last time I tried = this on=20 another list I was berated, beat upon and chastised for daring to raise = the=20 question - but, never one to be daunted for long and knowing this list = has some=20 experienced and clear thinkers, here goes again.
 
1.   The racing crowd = was thrown=20 into a turmoil when the rotary appeared on the scene - what displacement = class=20 should it be place in??
 
Well eventually the following was = apparently (If=20 not agreed to was at least accepted) that the 1300 cc 13B rotary was = equivalent=20 to a 160 CID 4 stroke engine.  Then the standard air flow formula = could be=20 applied - which is
 
Air Flow (CFM) =3D Displacement*rpm/(2=20 *1728)*Ve  (Volumetric efficiency (Ve) is generally assume to = be 100%=20 or 1, so we can drop it for this discussion as well as the effects of=20 compression ratios)
 
How did they arrive at 160 CID = displacement, well=20 they apparently decided to reference the rotary  to the standard = 720 Deg 4=20 stroke rotation cycle even though for all six faces of the two rotors to = go=20 through their cycle requires 1080 deg of e shaft rotation - well that is = 360 deg=20 more than the standard of 720Deg.  So if you limit consideration to = only=20 720deg of rotation only 4 of the six rotor faces have occurred, so since = the=20 displacement by a single face is 40 CID then 4* 40 =3D 160 CID.  So = that=20 appears to be how the decision of 160 CID equivalent displacement was = arrived=20 at. No problem with that part - but, continuing on:
 
So taking the formula and assuming 6000 = rpm, we=20 have Air Flow =3D 160*6000/(2*1728) =3D 277.77 CFM (assuming 100% = Ve).
 
2.  So what's the = problem?  Well,=20 unfortunately, I like to fully understand how and why something works = rather=20 than just plugging in numbers to a formula (although, heaven knows I do = enough=20 of that). 
 
 So here is how I started sliding = down the=20 slipper slope {:>).
 
We know that the rotors turn at a speed = 3 times=20 less than the E shaft.  So if the e shaft is turning at 6000 rpm = then the=20 rotors are turning at 6000/3 =3D 2000 rpm.  Its the rotor of course = that=20 actually suck in the air - not the eccentric shaft.  So we should = be able=20 to arrive at the same answer from either reference point (rotor or e=20 shaft).  Well the formula above uses the e shaft reference.  = So I=20 looked at what the rotor reference would produce (expecting the same=20 answer).
 
If the rotors are spinning at 2000 rpm, = we know=20 that 2 of the rotor faces have gone through their cycle for each 360 deg = revolution of the e shaft or 4 will have for 720 deg.  Looking at = it from=20 the rotor reference if the e shaft has turned 720 deg then the rotor has = turned=20 720/3 =3D 240 degrees.  240/360 =3D 0.6666 * 6 faces =3D 4 = faces.  So that=20 confirms that 4 faces go through their cycle in 720 degs of e shaft=20 rotation.  So if 4 faces of the two rotors complete their cycle in = 720Deg=20 eshaft then we have
 
Air Flow =3D 4(number of faces) *40 = (displacement for=20 each face) * rpm (rotor rpm)/(1728)    note: we drop the = division=20 by 2 which in the original 4 stroke reciprocating engine equation takes = into=20 account that only 1/2 of the cylinders are sucking air on each = revolution. =20
 
3.  This Logic  gives = us=20 rotary  Air Flow =3D 4*40*2000/(1728) =3D 185 CFM!  This = is ,of=20 course, NOT the 277 CFM arrived at by the approved=20 formula.
 
As you can see the implications of this = (if=20 correct) are fairly significant and that is why even though I can't = spot=20 the error, I must have made one.
 
I would greatly appreciate any = assistance in=20 helping me understand where I have gone wrong in logic or math.  = Please=20 don't just quote an authority or formula - explain it and show me your=20 calculations on how you got there.  Its things like this that keep = me awake=20 at night {:>).
 
Help = Meee..!   3D""=20
 
Thanks
 
Ed
 
Ed Anderson
Rv-6A N494BW Rotary=20 Powered
Matthews, NC
eanderson@carolina.rr.com
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