X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from [24.25.9.103] (HELO ms-smtp-04-eri0.southeast.rr.com) by logan.com (CommuniGate Pro SMTP 4.3c5) with ESMTP id 936742 for flyrotary@lancaironline.net; Sun, 08 May 2005 16:50:47 -0400 Received-SPF: pass receiver=logan.com; client-ip=24.25.9.103; 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-04-eri0.southeast.rr.com (8.12.10/8.12.7) with SMTP id j48KnwL5021390 for ; Sun, 8 May 2005 16:49:59 -0400 (EDT) Message-ID: <000701c5540f$8736b800$2402a8c0@edward2> From: "Ed Anderson" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Re: 4-port intake measurements Date: Sun, 8 May 2005 16:50:11 -0400 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0004_01C553ED.FFE4B3A0" 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_0004_01C553ED.FFE4B3A0 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable MessageActually, 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_0004_01C553ED.FFE4B3A0 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Message
Actually, Ian, I think you will find I = stated that=20 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 for=20 all six faces of the rotor to go past the intake.  However, it is = not the e=20 shaft which sucks air its the rotors, so I always use rotor revolution = which of=20 course is 1/3 of the e shaft rotation.  So at 6000 rpm e shaft the = rotor is=20 revolving at 6000/3 =3D 2000 rpm giving a rotation period of .030 sec or = 30=20 millisecond.  So when the ROTOR has rotated 360 deg (=3D 1080 deg e = shaft=20 rotation) all six faces have gone past the intake so there is 30/6 =3D 5 = milliseconds between intake events. 
 
 I have no experience with plenum = design, but=20 from what I have read your  plenum would need to have a volume = between=20 40%-60% of your displacement - depending of course onthe rpm you are = going to be=20 operating at. So your suggestion seems reasonable to me.
 
Ed
 
=3D ----- Original Message ----- =
From:=20 Ian = Dewhirst=20
Sent: Sunday, May 08, 2005 4:33 = PM
Subject: [FlyRotary] Re: 4-port = intake=20 measurements

I=20 think that 6 pulses for each 360 degree of e shaft rotation = would be=20 too good to be true :-).  You would get three pulses per rotor = revolution=20 which is equal to three e shaft revolutions, so for every 1080 degrees = of e=20 shaft revolution you complete 6 intake cycles.  I might be wrong = about=20 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=20 hoping that the same intake and exhaust manifold theory holds for = both.  When calculating the intake runner I think that=20 you should include the length of the inlet port, also as far as I = know=20 the intake runner sees the opening to the plenum, the throttle body = just=20 controls the pressure in the plenum, nothing more, WOT is a good = example of=20 this 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 or=20 throttle body, I don't know how this works properly in practice, I = would have=20 expected best performance to be achieved with a plenum that has a = volume=20 between 1 and two litres. -- = 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 are=20 some thoughts on the topic.
 
The two primary tubes would create = intake=20 pulses every 60 deg rotary (or 180 deg = e shaft)=20 revolution.  Same for the secondaries.  = The=20 primary and secondary of the same rotor would be near simultaneous = in=20 sucking air.  So those two pulses I would think tend to = reinforce each=20 other.  Now the pulse theory states that when a FAW pulse = hits an=20 "Open" (relative speaking) it reflects an opposite wave back down = the=20 tube.  So if a "negative" pulse hits the TB opening, then it = would=20 reflect a positive wave (helping to push the air toward the=20 intake).  However that wave could be opposing = the following set of=20 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=20 I presume) such that the positive reflected wave would reach=20 the intake port area in time to help shove more air in=20 - before running into the next pair of pulses in the=20 tube.  
 
No! No! no more manifolds (at = least not=20 this summer {:>))  Well, maybe a few = calculations.
 
Lets see - at 6000 rpm the rotation = period of=20 the rotor is 60/2000 =3D .03 sec =3D 30 millsec.  6 pulses = created(one for=20 each of the 3 faces *2)  during each revolution so we have 30/6 = =3D 5=20 millisec between each pulse sets.  If you assume sea level = speed of=20 sound approx 1100 ft/sec, then for a pulse to be generated by the = intake=20 opening, travel to the TB and back to the next set of intake port = openings=20 would be approx .005* 1100 =3D 5.5 ft. or 1/2 of that for length of = runner=20 (trip one way and trip back) =3D 2.5 ft or apprx 29".  But that = was at=20 6000 rpm and 1100 fps. 
 
For 7000 rpm 60/3500 =3D .017 = sec.  =20 .017/6 =3D .00286 sec or 2.86 millsec.   .00286 *1100 =3D = 3.14 ft or=20 1/2 of that would be 1.5 ft or 18" for a runner length, at least if = my math=20 is correct.  This is only an approximation but should give you = an idea=20 of lengths.
 
 
Ed 
 
 
----- Original Message -----
From:=20 Russell=20 Duffy
To: Rotary motors in = aircraft=20
Sent: Sunday, May 08, 2005 = 1:07=20 PM
Subject: [FlyRotary] Re: = 4-port=20 intake measurements

When I combined my=20 primary and secondary on one of my intake designs I used a 1.75" = dia tube,=20 so sounds like your dimensions closely agree.  I = currently have=20 a 1 1/2 and 1 1/4 for the secondary and primary = respectively.  So=20 that gives me a total of 2.99 sq inch runner area per rotor. So = approx .45=20 sq inches more per rotor than you currently have, so I would say = your=20 current design could be restrictive.
 
 
Thanks=20 Ed,  That's pretty much my conclusion too.  While it's = certainly=20 making good power, it could clearly be better, so it's = probably=20 worth the hassle of making another = intake.  
 
I'm not=20 sure if we ever cleared up this business about whether you = need a=20 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, it=20 comes to an equivalent tube of about 63mm ID, which is right in = line with=20 the recent discussion of TB = sizes.  
 
Here's=20 another wacky idea for your entertainment- I'm familiar with=20 the scavenging concept used in exhaust = systems.  For one=20 bank of a V-8, you start with 4 pipes, then combine them = to two=20 bigger pipes, and finally one even bigger pipe.  The = thought is=20 that active flow from one pipe is creating a suction on the = others to=20 help pull out exhaust of a cylinder that's almost done with it's = exhaust=20 cycle. 
 
Will=20 this work in reverse?  It almost seems like it would to=20 me.  What would happen if you start with two = primaries, and=20 two secondaries, then combine them to make two larger pipes (one = for each=20 rotor), then combine again for a single larger pipe with a TB on = the end=20 of it?     
 
Cheers,
Rusty=20 (more power Scotty)
<= /SPAN> ------=_NextPart_000_0004_01C553ED.FFE4B3A0--