X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from mx1.magmacom.com ([206.191.0.217] verified) by logan.com (CommuniGate Pro SMTP 4.3c5) with ESMTPS id 936733 for flyrotary@lancaironline.net; Sun, 08 May 2005 16:34:29 -0400 Received-SPF: none receiver=logan.com; client-ip=206.191.0.217; envelope-from=ianddsl@magma.ca Received: from mail3.magma.ca (mail3.magma.ca [206.191.0.221]) by mx1.magmacom.com (8.13.0/8.13.0) with ESMTP id j48KXiRe013117 for ; Sun, 8 May 2005 16:33:45 -0400 Received: from binky (ottawa-hs-64-26-156-111.s-ip.magma.ca [64.26.156.111]) by mail3.magma.ca (8.13.0/8.13.0) with SMTP id j48KXflH029393 for ; Sun, 8 May 2005 16:33:43 -0400 Reply-To: From: "Ian Dewhirst" To: "Rotary motors in aircraft" Subject: RE: [FlyRotary] Re: 4-port intake measurements Date: Sun, 8 May 2005 16:33:38 -0400 Message-ID: MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_02ED_01C553EB.B0228740" X-Priority: 3 (Normal) X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook IMO, Build 9.0.6604 (9.0.2911.0) X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.2180 In-Reply-To: Importance: Normal This is a multi-part message in MIME format. ------=_NextPart_000_02ED_01C553EB.B0228740 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: 7bit MessageI 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 -----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. 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 = .03 sec = 30 millsec. 6 pulses created(one for each of the 3 faces *2) during each revolution so we have 30/6 = 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 = 5.5 ft. or 1/2 of that for length of runner (trip one way and trip back) = 2.5 ft or apprx 29". But that was at 6000 rpm and 1100 fps. For 7000 rpm 60/3500 = .017 sec. .017/6 = .00286 sec or 2.86 millsec. .00286 *1100 = 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 ----- Original Message ----- From: Russell Duffy To: Rotary motors in aircraft 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. 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. 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. 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? Cheers, Rusty (more power Scotty) ------=_NextPart_000_02ED_01C553EB.B0228740 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Message
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 that=20 but thats how I see it in my head.  I am just learning about = rotaries,=20 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=20 both.  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 = runner=20 sees the opening to the plenum, the throttle body just controls the = pressure in=20 the plenum, nothing more, WOT is a good example of this technically the = TB is=20 not there.  I have seen a number of aircraft rotary intake = manifolds=20 with next to no plenum and a single carb or throttle body, I don't know = how this=20 works properly in practice, I would have expected best performance to be = achieved with a plenum that has a volume between 1 and two=20 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 pulses=20 every 60 deg rotary (or 180 deg e shaft) revolution.  Same for = the=20 secondaries.  The primary and secondary of the same rotor would = be near=20 simultaneous in sucking air.  So those two pulses I would think = tend to=20 reinforce each other.  Now the pulse theory states that when = a FAW=20 pulse hits an "Open" (relative speaking) it reflects an opposite wave = back=20 down the tube.  So if a "negative" pulse hits the TB opening, = then it=20 would reflect a positive wave (helping to push the air toward the = intake).  However that wave could be opposing the following = set of=20 pulses which it could interfere with.  So would have to do some=20 calculations to see how long the manifold would need to be at some rpm = (high I=20 presume) such that the positive reflected wave would reach=20 the intake port area in time to help shove more air in = - before=20 running into the next pair of pulses in the tube.  =20
 
No! No! no more manifolds (at = least not this=20 summer {:>))  Well, maybe a few calculations.
 
Lets see - at 6000 rpm the rotation = period of the=20 rotor is 60/2000 =3D .03 sec =3D 30 millsec.  6 pulses = created(one for each=20 of the 3 faces *2)  during each revolution so we have 30/6 =3D 5 = millisec=20 between each pulse sets.  If you assume sea level speed of sound = approx=20 1100 ft/sec, then for a pulse to be generated by the intake opening, = travel to=20 the TB and back to the next set of intake port openings would be = approx .005*=20 1100 =3D 5.5 ft. or 1/2 of that for length of runner (trip one way and = trip=20 back) =3D 2.5 ft or apprx 29".  But that was at 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 1/2=20 of that would be 1.5 ft or 18" for a runner length, at least if my = math is=20 correct.  This is only an approximation but should give you an = idea of=20 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 intake=20 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 a=20 1 1/2 and 1 1/4 for the secondary and primary respectively.  So = that=20 gives me a total of 2.99 sq inch runner area per rotor. So approx = .45 sq=20 inches more per rotor than you currently have, so I would say your = current=20 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 bank=20 of a V-8, you start with 4 pipes, then combine them to two = bigger=20 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 this=20 work in reverse?  It almost seems like it would to = me.  What=20 would happen if you start with two primaries, and two = secondaries, then=20 combine them to make two larger pipes (one for each rotor), then = combine=20 again for a single larger pipe with a TB on the end of=20 it?     
 
Cheers,
Rusty=20 (more power Scotty)
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