X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from [24.25.9.102] (HELO ms-smtp-03-eri0.southeast.rr.com) by logan.com (CommuniGate Pro SMTP 4.3c5) with ESMTP id 936887 for flyrotary@lancaironline.net; Sun, 08 May 2005 21:32:14 -0400 Received-SPF: pass receiver=logan.com; client-ip=24.25.9.102; envelope-from=eanderson@carolina.rr.com Received: from edward2 (cpe-024-074-189-178.carolina.res.rr.com [24.74.189.178]) by ms-smtp-03-eri0.southeast.rr.com (8.12.10/8.12.7) with SMTP id j491VQY5019718 for ; Sun, 8 May 2005 21:31:27 -0400 (EDT) Message-ID: <001601c55436$d1881210$2402a8c0@edward2> From: "Ed Anderson" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Re: 4-port intake measurements Date: Sun, 8 May 2005 21:31:26 -0400 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0013_01C55415.4A446590" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2900.2180 X-MIMEOLE: Produced By Microsoft MimeOLE V6.00.2900.2180 X-Virus-Scanned: Symantec AntiVirus Scan Engine This is a multi-part message in MIME format. ------=_NextPart_000_0013_01C55415.4A446590 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable MessageI understand, Ian. It took my old brain quite a while to get the = difference internalized. In the end everything is translatable from = rotor to e shaft and back - its just easier for me to mentally grasp the = fact that the rotors are doing the rotating and sucking and exhausting. = Then if necessary I can translate the results to e shaft degrees. =20 ----- Original Message -----=20 From: Ian Dewhirst=20 To: Rotary motors in aircraft=20 Sent: Sunday, May 08, 2005 7:51 PM Subject: [FlyRotary] Re: 4-port intake measurements Hi Ed, that makes sense; bare with me, for 22 years I have measured = everything in crank degrees, having the combustion faces rotate is a new = concept... -----Original Message----- From: Rotary motors in aircraft = [mailto:flyrotary@lancaironline.net]On Behalf Of Ed Anderson Sent: Sunday, May 08, 2005 4:50 PM To: Rotary motors in aircraft Subject: [FlyRotary] Re: 4-port intake measurements Actually, Ian, I think you will find I stated that you would get 6 = pulse per 360deg of ROTOR revolution - not e shaft revolution - you are = correct it would take 1080 of e shaft rotation for all six faces of the = rotor to go past the intake. However, it is not the e shaft which sucks = air its the rotors, so I always use rotor revolution which of course is = 1/3 of the e shaft rotation. So at 6000 rpm e shaft the rotor is = revolving at 6000/3 =3D 2000 rpm giving a rotation period of .030 sec or = 30 millisecond. So when the ROTOR has rotated 360 deg (=3D 1080 deg e = shaft rotation) all six faces have gone past the intake so there is 30/6 = =3D 5 milliseconds between intake events.=20 I have no experience with plenum design, but from what I have read = your plenum would need to have a volume between 40%-60% of your = displacement - depending of course onthe rpm you are going to be = operating at. So your suggestion seems reasonable to me. Ed =3D ----- Original Message -----=20 From: Ian Dewhirst=20 To: Rotary motors in aircraft=20 Sent: Sunday, May 08, 2005 4:33 PM Subject: [FlyRotary] Re: 4-port intake measurements I think that 6 pulses for each 360 degree of e shaft rotation = would be too good to be true :-). You would get three pulses per rotor = revolution which is equal to three e shaft revolutions, so for every = 1080 degrees of e shaft revolution you complete 6 intake cycles. I = might be wrong about that but thats how I see it in my head. I am just = learning about rotaries, however I do have a lot of experience with = piston engines, I am hoping that the same intake and exhaust manifold = theory holds for both. When calculating the intake runner I think that = you should include the length of the inlet port, also as far as I know = the intake runner sees the opening to the plenum, the throttle body just = controls the pressure in the plenum, nothing more, WOT is a good example = of this technically the TB is not there. I have seen a number of = aircraft rotary intake manifolds with next to no plenum and a single = carb or throttle body, I don't know how this works properly in practice, = I would have expected best performance to be achieved with a plenum that = has a volume between 1 and two litres. -- Ian=20 -----Original Message----- From: Rotary motors in aircraft = [mailto:flyrotary@lancaironline.net]On Behalf Of Ed Anderson Sent: Sunday, May 08, 2005 2:42 PM To: Rotary motors in aircraft Subject: [FlyRotary] Re: 4-port intake measurements I have given some thought to the idea you mentioned. I am not = certain what the results would be. Here are some thoughts on the topic. The two primary tubes would create intake pulses every 60 deg = rotary (or 180 deg e shaft) revolution. Same for the secondaries. The = primary and secondary of the same rotor would be near simultaneous in = sucking air. So those two pulses I would think tend to reinforce each = other. Now the pulse theory states that when a FAW pulse hits an "Open" = (relative speaking) it reflects an opposite wave back down the tube. So = if a "negative" pulse hits the TB opening, then it would reflect a = positive wave (helping to push the air toward the intake). However that = wave could be opposing the following set of pulses which it could = interfere with. So would have to do some calculations to see how long = the manifold would need to be at some rpm (high I presume) such that the = positive reflected wave would reach the intake port area in time to help = shove more air in - before running into the next pair of pulses in the = tube. =20 No! No! no more manifolds (at least not this summer {:>)) Well, = maybe a few calculations. Lets see - at 6000 rpm the rotation period of the rotor is = 60/2000 =3D .03 sec =3D 30 millsec. 6 pulses created(one for each of = the 3 faces *2) during each revolution so we have 30/6 =3D 5 millisec = between each pulse sets. If you assume sea level speed of sound approx = 1100 ft/sec, then for a pulse to be generated by the intake opening, = travel to the TB and back to the next set of intake port openings would = be approx .005* 1100 =3D 5.5 ft. or 1/2 of that for length of runner = (trip one way and trip back) =3D 2.5 ft or apprx 29". But that was at = 6000 rpm and 1100 fps. =20 For 7000 rpm 60/3500 =3D .017 sec. .017/6 =3D .00286 sec or = 2.86 millsec. .00286 *1100 =3D 3.14 ft or 1/2 of that would be 1.5 ft = or 18" for a runner length, at least if my math is correct. This is = only an approximation but should give you an idea of lengths. Ed =20 ----- Original Message -----=20 From: Russell Duffy=20 To: Rotary motors in aircraft=20 Sent: Sunday, May 08, 2005 1:07 PM Subject: [FlyRotary] Re: 4-port intake measurements When I combined my primary and secondary on one of my intake = designs I used a 1.75" dia tube, so sounds like your dimensions closely = agree. I currently have a 1 1/2 and 1 1/4 for the secondary and primary = respectively. So that gives me a total of 2.99 sq inch runner area per = rotor. So approx .45 sq inches more per rotor than you currently have, = so I would say your current design could be restrictive. Thanks Ed, That's pretty much my conclusion too. While it's = certainly making good power, it could clearly be better, so it's = probably worth the hassle of making another intake. =20 I'm not sure if we ever cleared up this business about whether = you need a single TB big to allow for both rotors, or just one at a = time. That was at the heart of the Ellison debate. Interestingly, if = you add the sizes of the ports on both rotors, it comes to an equivalent = tube of about 63mm ID, which is right in line with the recent discussion = of TB sizes. =20 Here's another wacky idea for your entertainment- I'm familiar = with the scavenging concept used in exhaust systems. For one bank of a = V-8, you start with 4 pipes, then combine them to two bigger pipes, and = finally one even bigger pipe. The thought is that active flow from one = pipe is creating a suction on the others to help pull out exhaust of a = cylinder that's almost done with it's exhaust cycle. =20 Will this work in reverse? It almost seems like it would to = me. What would happen if you start with two primaries, and two = secondaries, then combine them to make two larger pipes (one for each = rotor), then combine again for a single larger pipe with a TB on the end = of it? =20 Cheers, Rusty (more power Scotty) ------=_NextPart_000_0013_01C55415.4A446590 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Message
I understand, Ian.  It took my old = brain quite=20 a while to get the difference internalized.  In the end everything = is=20 translatable from rotor to e shaft and back - its just easier for me to = mentally=20 grasp the fact that the rotors are doing the rotating and sucking and=20 exhausting.  Then if necessary I can translate the results to e = shaft=20 degrees. 
 
 
----- Original Message -----
From:=20 Ian = Dewhirst=20
Sent: Sunday, May 08, 2005 7:51 = PM
Subject: [FlyRotary] Re: 4-port = intake=20 measurements

Hi=20 Ed, that makes sense; bare with me, for 22 years I have measured = everything in=20 crank degrees, having the combustion faces rotate is a new=20 concept...
-----Original Message-----
From: Rotary motors in = aircraft=20 [mailto:flyrotary@lancaironline.net]On Behalf Of Ed=20 Anderson
Sent: Sunday, May 08, 2005 4:50 PM
To: = Rotary=20 motors in aircraft
Subject: [FlyRotary] Re: 4-port intake=20 measurements

Actually, Ian, I think you will = find I stated=20 that you would get 6 pulse per 360deg  of ROTOR revolution - not = e=20 shaft revolution - you are correct it would take 1080 of e shaft = rotation=20 for all six faces of the rotor to go past the intake.  However, = it is=20 not the e shaft which sucks air its the rotors, so I always use = rotor=20 revolution which of course is 1/3 of the e shaft rotation.  So = at 6000=20 rpm e shaft the rotor is revolving at 6000/3 =3D 2000 rpm giving a = rotation=20 period of .030 sec or 30 millisecond.  So when the ROTOR has = rotated=20 360 deg (=3D 1080 deg e shaft rotation) all six faces have gone past = the=20 intake so there is 30/6 =3D 5 milliseconds between intake=20 events. 
 
 I have no experience with = plenum design,=20 but from what I have read your  plenum would need to have a = volume=20 between 40%-60% of your displacement - depending of course onthe rpm = you are=20 going to be operating at. So your suggestion seems reasonable to=20 me.
 
Ed
 
=3D ----- Original Message = -----
From:=20 Ian = Dewhirst=20
To: Rotary motors in = aircraft=20
Sent: Sunday, May 08, 2005 = 4:33=20 PM
Subject: [FlyRotary] Re: = 4-port=20 intake measurements

I think that 6 pulses for each 360 degree of e=20 shaft rotation would be too good to be true :-).  You = would get=20 three pulses per rotor revolution which is equal to three e shaft=20 revolutions, so for every 1080 degrees of e shaft revolution you = complete=20 6 intake cycles.  I might be wrong about that but thats how I = see it=20 in my head.  I am just learning about rotaries, however I do = have a=20 lot of experience with piston engines, I am hoping that = the same=20 intake and exhaust manifold theory holds for both. =20 When calculating the intake runner I think that = you should=20 include the length of the inlet port, also as far as I know the = intake=20 runner sees the opening to the plenum, the throttle body just = controls the=20 pressure in the plenum, nothing more, WOT is a good example of = this=20 technically the TB is not there.  I have seen a number = of=20 aircraft rotary intake manifolds with next to no plenum and a = single carb=20 or throttle body, I don't know how this works properly in = practice, I=20 would have expected best performance to be achieved with a plenum = that has=20 a volume between 1 and two litres. --=20 Ian 
 
 -----Original=20 Message-----
From: Rotary motors in aircraft=20 [mailto:flyrotary@lancaironline.net]On Behalf Of Ed=20 Anderson
Sent: Sunday, May 08, 2005 2:42 = PM
To: Rotary=20 motors in aircraft
Subject: [FlyRotary] Re: 4-port = intake=20 measurements

I have given some thought to = the idea you=20 mentioned.  I am not certain what the results would = be.  Here=20 are some thoughts on the topic.
 
The two primary tubes would = create intake=20 pulses every 60 deg rotary (or 180 = deg e=20 shaft) revolution.  Same for the = secondaries. =20 The primary and secondary of the same rotor would be near = simultaneous=20 in sucking air.  So those two pulses I would think tend to=20 reinforce each other.  Now the pulse theory states that = when=20 a FAW pulse hits an "Open" (relative speaking) it reflects = an=20 opposite wave back down the tube.  So if a "negative" pulse = hits=20 the TB opening, then it would reflect a positive wave = (helping to=20 push the air toward the intake).  However that wave could = be=20 opposing the following set of pulses which it could = interfere=20 with.  So would have to do some calculations to see how = long the=20 manifold would need to be at some rpm (high I presume) such that = the=20 positive reflected wave would reach the intake port = area in=20 time to help shove more air in - before running into = the=20 next pair of pulses in the tube.  
 
No! No! no more manifolds = (at least=20 not this summer {:>))  Well, maybe a few=20 calculations.
 
Lets see - at 6000 rpm the = rotation period=20 of the rotor is 60/2000 =3D .03 sec =3D 30 millsec.  6 = pulses=20 created(one for each of the 3 faces *2)  during each = revolution so=20 we have 30/6 =3D 5 millisec between each pulse sets.  If = you assume=20 sea level speed of sound approx 1100 ft/sec, then for a pulse to = be=20 generated by the intake opening, travel to the TB and back to = the next=20 set of intake port openings would be approx .005* 1100 =3D 5.5 = ft. or 1/2=20 of that for length of runner (trip one way and trip back) =3D = 2.5 ft or=20 apprx 29".  But that was at 6000 rpm and 1100 fps. =20
 
For 7000 rpm 60/3500 =3D .017=20 sec.   .017/6 =3D .00286 sec or 2.86 = millsec.   .00286=20 *1100 =3D 3.14 ft or 1/2 of that would be 1.5 ft or 18" for a = runner=20 length, at least if my math is correct.  This is only an=20 approximation but should give you an idea of = lengths.
 
 
Ed 
 
 
----- Original Message -----
From:=20 Russell Duffy
To: Rotary motors in=20 aircraft
Sent: Sunday, May 08, = 2005 1:07=20 PM
Subject: [FlyRotary] = Re: 4-port=20 intake measurements

When I=20 combined my primary and secondary on one of my intake designs = I used a=20 1.75" dia tube, so sounds like your dimensions closely=20 agree.  I currently have a 1 1/2 and 1 1/4 for the = secondary and=20 primary respectively.  So that gives me a total of 2.99 = sq inch=20 runner area per rotor. So approx .45 sq inches more per rotor = than you=20 currently have, so I would say your current design could be=20 restrictive.
 
 
Thanks Ed,  That's pretty much = my=20 conclusion too.  While it's certainly making good = power,=20 it could clearly be better, so it's probably worth the = hassle of=20 making another intake.  
 
I'm=20 not sure if we ever cleared up this business about = whether you=20 need a single TB big to allow for both rotors, or just one at = a=20 time.  That was at the heart of the Ellison = debate. =20 Interestingly, if you add the sizes of the ports on both = rotors,=20 it comes to an equivalent tube of about 63mm ID, which is = right in=20 line with the recent discussion of TB=20 sizes.  
 
Here's another wacky idea for your=20 entertainment- I'm familiar with the scavenging concept = used in=20 exhaust systems.  For one bank of a V-8, you=20 start with 4 pipes, then combine them to two bigger = pipes, and=20 finally one even bigger pipe.  The thought is=20 that active flow from one pipe is creating a suction on = the=20 others to help pull out exhaust of a cylinder that's almost = done with=20 it's exhaust cycle. 
 
Will this work in reverse?  It = almost=20 seems like it would to me.  What would happen if you = start=20 with two primaries, and two secondaries, then combine = them to=20 make two larger pipes (one for each rotor), then combine again = for a=20 single larger pipe with a TB on the end of=20 it?     
 
Cheers,
Rusty (more power = Scotty)
<= /BLOCKQUOTE>
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