X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from mx2.magma.ca ([206.191.0.250] verified) by logan.com (CommuniGate Pro SMTP 4.3c5) with ESMTPS id 936818 for flyrotary@lancaironline.net; Sun, 08 May 2005 19:52:12 -0400 Received-SPF: none receiver=logan.com; client-ip=206.191.0.250; envelope-from=ianddsl@magma.ca Received: from mail4.magma.ca (mail4.magma.ca [206.191.0.222]) by mx2.magma.ca (8.13.0/8.13.0) with ESMTP id j48NpOd0017687 for ; Sun, 8 May 2005 19:51:25 -0400 Received: from binky (ottawa-hs-64-26-156-111.s-ip.magma.ca [64.26.156.111]) by mail4.magma.ca (8.13.0/8.13.0) with SMTP id j48NpMZr007051 for ; Sun, 8 May 2005 19:51:24 -0400 Reply-To: From: "Ian Dewhirst" To: "Rotary motors in aircraft" Subject: RE: [FlyRotary] Re: 4-port intake measurements Date: Sun, 8 May 2005 19:51:19 -0400 Message-ID: MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_02F5_01C55407.4DFC45D0" 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_02F5_01C55407.4DFC45D0 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: 7bit MessageHi 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 = 2000 rpm giving a rotation period of .030 sec or 30 millisecond. So when the ROTOR has rotated 360 deg (= 1080 deg e shaft rotation) all six faces have gone past the intake so there is 30/6 = 5 milliseconds between intake events. 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 = ----- Original Message ----- From: Ian Dewhirst To: Rotary motors in aircraft 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 -----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_02F5_01C55407.4DFC45D0 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Message
Hi Ed,=20 that makes sense; bare with me, for 22 years I have measured everything = in crank=20 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 for=20 all six faces of the rotor to go past the intake.  However, it is = not the=20 e shaft which sucks air its the rotors, so I always use rotor = revolution which=20 of course is 1/3 of the e shaft rotation.  So at 6000 rpm e shaft = the=20 rotor is revolving at 6000/3 =3D 2000 rpm giving a rotation period of = .030 sec=20 or 30 millisecond.  So when the ROTOR has rotated 360 deg (=3D = 1080 deg e=20 shaft rotation) all six faces have gone past the intake so there is = 30/6 =3D 5=20 milliseconds between intake 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 intake=20 measurements

I=20 think that 6 pulses for each 360 degree of e = shaft rotation would=20 be too good to be true :-).  You would get three pulses per = rotor=20 revolution which is equal to three e shaft revolutions, so for every = 1080=20 degrees of e shaft revolution you complete 6 intake cycles.  I = might be=20 wrong about that but thats how I see it in my head.  I am just = learning=20 about rotaries, however I do have a lot of experience with piston = engines,=20 I am hoping that the same intake and exhaust manifold = theory holds=20 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=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=20 have expected best performance to be achieved with a plenum that has = a=20 volume between 1 and two litres. --=20 Ian 
 
 -----Original = 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 in=20 sucking air.  So those two pulses I would think tend to = reinforce=20 each other.  Now the pulse theory states that when a FAW = pulse=20 hits an "Open" (relative speaking) it reflects an opposite wave = back down=20 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=20 intake).  However that wave could be opposing = the following set=20 of 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=20 (high 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=20 for each of the 3 faces *2)  during each revolution so we = have 30/6 =3D=20 5 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=20 math is correct.  This is only an approximation but should = give you=20 an idea 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=20 my primary and secondary on one of my intake designs I used a = 1.75" dia=20 tube, so sounds like your dimensions closely agree.  I = currently have a 1 1/2 and 1 1/4 for the secondary and primary=20 respectively.  So that gives me a total of 2.99 sq inch = runner area=20 per rotor. So approx .45 sq inches more per rotor than you = currently=20 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 need=20 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, it=20 comes to an equivalent tube of about 63mm ID, which is right in = line=20 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 = that active=20 flow from one pipe is creating a suction on the others to help = pull out=20 exhaust of a cylinder that's almost done with it's exhaust = cycle. =20
 
Will=20 this work in reverse?  It almost seems like it would to=20 me.  What would happen if you start with two = primaries,=20 and two secondaries, then combine them to make two larger pipes = (one for=20 each rotor), then combine again for a single larger pipe with a = TB on=20 the end of it?     
 
Cheers,
Rusty=20 (more power Scotty)
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