Return-Path: Received: from mail07.syd.optusnet.com.au ([211.29.132.188] verified) by logan.com (CommuniGate Pro SMTP 4.3c1) with ESMTP-TLS id 724277 for flyrotary@lancaironline.net; Thu, 10 Feb 2005 17:15:55 -0500 Received-SPF: none receiver=logan.com; client-ip=211.29.132.188; envelope-from=lendich@optusnet.com.au Received: from george (d211-31-119-41.dsl.nsw.optusnet.com.au [211.31.119.41]) by mail07.syd.optusnet.com.au (8.12.11/8.12.11) with SMTP id j1AMF6OW030395 for ; Fri, 11 Feb 2005 09:15:08 +1100 Message-ID: <004a01c50fbe$4834f440$29771fd3@george> From: "George Lendich" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Rotary AirFlow Equation? Help? Date: Fri, 11 Feb 2005 08:17:17 +1000 MIME-Version: 1.0 Content-Type: multipart/related; type="multipart/alternative"; boundary="----=_NextPart_000_0046_01C51012.19760D70" 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 This is a multi-part message in MIME format. ------=_NextPart_000_0046_01C51012.19760D70 Content-Type: multipart/alternative; boundary="----=_NextPart_001_0047_01C51012.19760D70" ------=_NextPart_001_0047_01C51012.19760D70 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Ed, I love this stuff! I think when you made this point on another discussion group it was too = early in rotary engine knowledge to understand it - thanks for taking = the time and trouble! George (down under) 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_0047_01C51012.19760D70 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Ed,
I love this stuff!
I think when you made this point on = another=20 discussion group it was too early in rotary engine knowledge to = understand it -=20 thanks for taking the time and trouble!
George (down under)
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=20 on 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=20 some experienced and clear thinkers, here goes again.
 
1.   The racing = crowd was=20 thrown into a turmoil when the rotary appeared on the scene - what=20 displacement class 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=20 equivalent to a 160 CID 4 stroke engine.  Then the standard air = flow=20 formula could be applied - which is
 
Air Flow (CFM) =3D = Displacement*rpm/(2=20 *1728)*Ve  (Volumetric efficiency (Ve) is generally assume = to be=20 100% 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=20 4 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=20 deg more than the standard of 720Deg.  So if you limit = consideration to=20 only 720deg of rotation only 4 of the six rotor faces have occurred, = so since=20 the displacement by a single face is 40 CID then 4* 40 =3D 160 = CID.  So=20 that appears to be how the decision of 160 CID equivalent displacement = was=20 arrived 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%=20 Ve).
 
2.  So what's the = problem? =20 Well, unfortunately, I like to fully understand how and why something = works=20 rather than just plugging in numbers to a formula (although, heaven = knows I do=20 enough 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=20 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=20 turned 720/3 =3D 240 degrees.  240/360 =3D 0.6666 * 6 faces =3D 4 = faces. =20 So that 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=20 for each face) * rpm (rotor rpm)/(1728)    note: we = drop the=20 division by 2 which in the original 4 stroke reciprocating engine = equation=20 takes into account that only 1/2 of the cylinders are sucking air on = each=20 revolution. 
 
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 = calculations on how you got there.  Its things like this that = keep me=20 awake 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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