Return-Path: Received: from smtpauth09.mail.atl.earthlink.net ([209.86.89.69] verified) by logan.com (CommuniGate Pro SMTP 4.3c1) with ESMTP id 723784 for flyrotary@lancaironline.net; Thu, 10 Feb 2005 12:27:05 -0500 Received-SPF: pass receiver=logan.com; client-ip=209.86.89.69; envelope-from=jerryhey@earthlink.net Received: from [65.176.160.213] (helo=earthlink.net) by smtpauth09.mail.atl.earthlink.net with asmtp (Exim 4.34) id 1CzI55-0002Cc-2L for flyrotary@lancaironline.net; Thu, 10 Feb 2005 12:26:21 -0500 DomainKey-Signature: a=rsa-sha1; q=dns; c=simple; s=test1; d=earthlink.net; h=Date:Subject:Content-Type:Mime-Version:From:To:In-Reply-To:Message-Id:X-Mailer; b=HDK47BHYN6S5RtFKSvpe2+1ewnkdVViMIlyziVMy2fLQJ5OhnAWgcsNh3GJbZt79; Date: Thu, 10 Feb 2005 12:27:18 -0500 Subject: Re: [FlyRotary] Rotary AirFlow Equation? Help? Content-Type: multipart/alternative; boundary=Apple-Mail-22-900219503 Mime-Version: 1.0 (Apple Message framework v552) From: Jerry Hey To: "Rotary motors in aircraft" In-Reply-To: Message-Id: <037A1BA4-7B89-11D9-B3AA-0003931B0C7A@earthlink.net> X-Mailer: Apple Mail (2.552) X-ELNK-Trace: 8104856d7830ec6b1aa676d7e74259b7b3291a7d08dfec7956065b039edb633fde06f1e83ffa0cfd350badd9bab72f9c350badd9bab72f9c350badd9bab72f9c X-Originating-IP: 65.176.160.213 --Apple-Mail-22-900219503 Content-Transfer-Encoding: quoted-printable Content-Type: text/plain; charset=ISO-8859-1; format=flowed On Thursday, February 10, 2005, at 11:09 AM, Ed Anderson wrote: > In a previous posting, I alluded to my suspicions that a commonly=20 > accepted formula for calculating airflow for the rotary engine=A0might=20= > be flawed.=A0I was asked off-line to expound on that suspicion. > =A0 > =A0Well, I can think of no better audience to present my suspicions to=20= > and see what you think.=A0 Last time I tried this on another list I = was=20 > berated, beat upon and chastised for daring to raise the question -=20 > but, never one to be daunted for long and knowing this list has some=20= > experienced and clear thinkers, here goes again. > =A0 > 1.=A0=A0 The racing crowd was thrown into a turmoil when the rotary=20 > appeared on the scene - what displacement class should it be place > = in?? > =A0 > Well eventually the following was apparently (If not agreed to was at=20= > least accepted) that the 1300 cc 13B rotary was equivalent to a 160=20 > CID 4 stroke engine.=A0 Then the standard air flow formula could be=20 > applied - which is > =A0 > Air Flow (CFM) =3D Displacement*rpm/(2 *1728)*Ve=A0 (Volumetric = efficiency=20 > (Ve) is=A0generally assume to be 100% or 1, so we can drop it for this=20= > discussion as well as the effects of compression ratios) > =A0 > How did they arrive at 160 CID displacement, well they apparently=20 > decided to reference the rotary=A0 to the standard 720 Deg 4 stroke=20 > 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=20= > is 360 deg more than the standard of 720Deg.=A0 So if you limit=20 > consideration to only 720deg of rotation only 4 of the six rotor faces=20= > have occurred, so since the displacement by a single face is 40 CID=20 > then 4* 40 =3D 160 CID.=A0 So that appears to be how the decision of = 160=20 > CID equivalent displacement was arrived at. No problem with that part=20= > - but, continuing on: > =A0 > So taking the formula and assuming 6000 rpm, we have Air Flow =3D=20 > 160*6000/(2*1728) =3D 277.77 CFM (assuming 100% Ve). > =A0 > 2.=A0 So what's the problem?=A0 Well, unfortunately, I like to fully=20= > understand how and why something works rather than just plugging in=20 > numbers to a formula (although, heaven knows I do enough of that).=A0 > =A0 > =A0So here is how I started sliding down the slipper slope {:>). > =A0 > We know that the rotors turn at a speed 3 times less than the E=20 > shaft.=A0 So if the e shaft is turning at 6000 rpm then the rotors are=20= > turning at 6000/3 =3D 2000 rpm.=A0 Its the rotor of course that = actually=20 > suck in the air - not the eccentric shaft.=A0 So we should be able to=20= > arrive at the same answer from either reference point (rotor or e=20 > shaft).=A0 Well the formula above uses the e shaft reference.=A0 So I=20= > looked at what the rotor reference would produce (expecting the same=20= > answer). > =A0 > If the rotors are spinning at 2000 rpm, we know that 2 of the rotor=20 > faces have gone through their cycle for each 360 deg revolution of the=20= > e shaft or 4 will have for 720 deg.=A0 Looking at it from the rotor=20 > reference if the e shaft has turned 720 deg then the rotor has turned=20= > 720/3 =3D 240 degrees.=A0 240/360 =3D 0.6666 * 6 faces =3D 4 faces.=A0 = So that=20 > confirms that 4 faces go through their cycle in 720 degs of e shaft=20 > rotation.=A0 So if 4 faces of the two rotors complete their cycle in=20= > 720Deg eshaft=A0then we have > =A0 > Air Flow =3D 4(number of faces) *40 (displacement for each face) * rpm=20= > (rotor rpm)/(1728)=A0=A0=A0 note: we drop the division by 2 which in = the=20 > original 4 stroke reciprocating engine equation takes into account=20 > that only 1/2 of the cylinders are sucking air on each revolution.=A0 > =A0 > 3.=A0 This Logic =A0gives us rotary=A0=A0Air Flow =3D 4*40*2000/(1728) = =3D 185=20 > CFM!=A0 This is ,of course,=A0NOT the 277 CFM arrived at by the = approved=20 > formula. > =A0 > As you can see the implications of this (if correct) are fairly=20 > significant=A0and that is why even though I can't spot the error, I = must=20 > have made one. > =A0 > I would greatly appreciate any assistance in helping me understand=20 > where I have gone wrong in logic or math.=A0 Please don't just quote = an=20 > authority or formula - explain it and show me your calculations on how=20= > you got there.=A0 Its things like this that keep me awake at night = {:>). > =A0 > Help Meee..!=A0=A0 > =A0 > Thanks > =A0 > Ed Ed, I passed your analysis on to Rolf Pfeiffer. Here is his =20 response. Jerry I quote Ed: 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. The answer is simple. On one rotation of the a rotor, all three faces came to "work". Having two rotors the total faces or chambers displaced is 6. So the formula should read: Air Flow =3D 6*40*2000/(1728) =3D 277 CFM! The operation of the rotary was established as a four cycle engine, which it is. In order to have the same displacement as a four cycle RPE and to be comparable therewith, the displacement at "two" shaft rotations was considered to be "equal". So, at equal output shaft rotations the engines can be compared in displacement and therewith in performance. I personally think that is fair, although it is not quite so as the rotary can easily turn at higher speeds, and therefore make more power for the "same" displacement of only two chambers counted. One must also consider that at equal speeds the duration of a power stroke in the Rotary is 1.5 times slower and therewith the rotary has an advantage in a longer duration and better combustion. The disadvantage the Rotary has over the RPE is the large surface area and poor combustion chamber shape that lead to a poorer combustion process. Both factors are offsetting somewhat, so that in overall, the "counting" of only two chambers is the most realistic in comparing the Rotary to the RPE. Besides, one can use the RPE formula for throughput, namely "displacement" * shaft rpm * the same factors. Rolf > =A0 --Apple-Mail-22-900219503 Content-Transfer-Encoding: quoted-printable Content-Type: text/enriched; charset=ISO-8859-1 On Thursday, February 10, 2005, at 11:09 AM, Ed Anderson wrote: ArialIn a previous posting, I alluded to my suspicions that a commonly accepted formula for calculating airflow for the rotary engine=A0might be flawed.=A0I was asked off-line to expound on that suspicion. =A0 Arial=A0Well, I can think of no better audience to present my suspicions to and see what you think.=A0 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. =A0 Arial1.=A0=A0 The racing crowd was thrown into a turmoil when the rotary appeared on the scene - what displacement class should it be place in?? =A0 ArialWell 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.=A0 Then the standard air flow formula could be applied - which = is =A0 ArialAir Flow (CFM) =3D Displacement*rpm/(2 *1728)*Ve=A0 (Volumetric efficiency (Ve) is=A0generally assume to be 100% or 1, so we can drop it for this discussion as well as the effects of compression = ratios) =A0 ArialHow did they arrive at 160 CID displacement, well they apparently decided to reference the rotary=A0 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.=A0 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.=A0 So = that appears to be how the decision of 160 CID equivalent displacement was arrived at. No problem with that part - but, continuing = on: =A0 ArialSo taking the formula and assuming 6000 rpm, we have Air Flow =3D 160*6000/(2*1728) =3D 277.77 CFM (assuming 100% Ve). =A0 Arial2.=A0 So what's the problem?=A0 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).=A0 =A0 Arial=A0So here is how I started sliding down the slipper slope {:>). =A0 ArialWe know that the rotors turn at a speed 3 times less than the E shaft.=A0 So if the e shaft is turning at 6000 rpm then the rotors are turning at 6000/3 =3D 2000 rpm.=A0= Its the rotor of course that actually suck in the air - not the eccentric shaft.=A0 So we should be able to arrive at the same answer from either reference point (rotor or e shaft).=A0 Well the formula above uses the e shaft reference.=A0 So I looked at what the rotor reference would produce (expecting the same = answer). =A0 ArialIf 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.=A0 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.=A0 = 240/360 =3D 0.6666 * 6 faces =3D 4 faces.=A0 So that confirms that 4 faces go through their cycle in 720 degs of e shaft rotation.=A0 So if 4 faces of the two rotors complete their cycle in 720Deg eshaft=A0then we = have =A0 ArialAir Flow =3D 4(number of faces) *40 (displacement for each face) * rpm (rotor rpm)/(1728)=A0=A0=A0 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.=A0 =A0 Arial3.=A0 This Logic =A0gives us rotary=A0=A0Air Flow =3D 4*40*2000/(1728) =3D 185 CFM!=A0 This is ,of course,=A0NOT the 277 CFM arrived at by the approved = formula. =A0 ArialAs you can see the implications of this (if correct) are fairly significant=A0and that is why even though I can't spot the error, I must have made = one. =A0 ArialI would greatly appreciate any assistance in helping me understand where I have gone wrong in logic or math.=A0 Please don't just quote an authority or formula - explain it and show me your calculations on how you got there.=A0 Its things like this that keep me awake at night = {:>). =A0 = ArialHelp Meee..!=A0=A0 < =A0 ArialThanks =A0 ArialEd Ed, I passed your analysis on to Rolf Pfeiffer. Here is his=20 response. Jerry I quote Ed: 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. The answer is simple. On one rotation of the a rotor, all three faces came to "work". Having two rotors the total faces or chambers displaced is 6. So the formula should read: Air Flow =3D 6*40*2000/(1728) =3D 277 CFM! The operation of the rotary was established as a four cycle engine, which it is. In order to have the same displacement as a four cycle RPE and to be comparable therewith, the displacement at "two" shaft rotations was considered to be "equal". So, at equal output shaft rotations the engines can be compared in displacement and therewith in performance. I personally think that is fair, although it is not quite so as the rotary can easily turn at higher speeds, and therefore make more power for the "same" displacement of only two chambers counted. One must also consider that at equal speeds the duration of a power stroke in the Rotary is 1.5 times slower and therewith the rotary has an advantage in a longer duration and better combustion. The disadvantage the Rotary has over the RPE is the large surface area and poor combustion chamber shape that lead to a poorer combustion process. Both factors are offsetting somewhat, so that in overall, the "counting" of only two chambers is the most realistic in comparing the Rotary to the RPE. Besides, one can use the RPE formula for throughput, namely "displacement" * shaft rpm * the same factors. Rolf =A0 = --Apple-Mail-22-900219503--