X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from ispmxmta05-srv.alltel.net ([166.102.165.166] verified) by logan.com (CommuniGate Pro SMTP 5.0c5) with ESMTP id 771718 for flyrotary@lancaironline.net; Mon, 17 Oct 2005 23:29:25 -0400 Received-SPF: pass receiver=logan.com; client-ip=166.102.165.166; envelope-from=montyr2157@alltel.net Received: from Thorstwin ([4.89.246.37]) by ispmxmta05-srv.alltel.net with SMTP id <20051018032837.TYED18995.ispmxmta05-srv.alltel.net@Thorstwin> for ; Mon, 17 Oct 2005 22:28:37 -0500 Message-ID: <000b01c5d394$07365930$d0f95904@Thorstwin> From: "Monty Roberts" To: Subject: flyrotary Displacement Date: Mon, 17 Oct 2005 22:27:55 -0500 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0008_01C5D36A.04FF6E10" 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 This is a multi-part message in MIME format. ------=_NextPart_000_0008_01C5D36A.04FF6E10 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Okay this one is for David, Ed, Bob and Ken because they are actually = having an adult discussion with me on this. Unlike the Pope. For = everybody else, If this hurts your head (it is beginning to hurt mine) = just skip down to the last few paragraphs. I am looking at this from a thermo standpoint as a mechanical engineer. = If I were going to analyze this and make the numbers work theoretically. = You have to break this down to it's smallest component. To do this you follow one INDIVIDUAL induction, compression, expansion = cycle from beginning to end. It does not matter what kind of hardware we = are looking at. When I look at a single cylinder 4 cycle engine. It takes two complete = crank revolutions to complete this cycle on some displacement of air. I draw a PV diagram for this and get the work required to induct, = compress, expand and exhaust. Some of these terms are negative, some are = positive. The balance is the work output for this ONE cycle and = individual cylinder. For a multi cylinder engine, these work outputs are all just added = together at different phase angles based on the crank throws. Now for the wankel: It takes one complete revolution of the ROTOR to complete a cycle for a = single rotor face (same charge) and perform the previous analysis. The = eshaft will rotate 3 times for each rotor rev.. The additional rotor faces are just like additional cylinders added on = to a single cylinder engine. The mechanism for extracting power is a bit = more arcane but it is the same thing. Stick on another rotor 180 degrees = out of phase and it is the same thing. 3 more cylinders. So you are correct the 3.9L 6 cyl equivalent engine will rotate at 2/3 = the wankel speed not 1/3 as I misstated. And that matches with practice, = a well tuned 240 cubic inch 4 cycle multi cylinder engine spinning at = 6000 rpm (8500*2/3) is capable of making around 240 hp. Likewise the = Wankel makes the same kind of power at 8500 shaft rpm. with the same = number of cycles performed on the same amount of air charges as a 6 cyl = engine for a given amount of time. Likewise, keeping the rpm the same: a 2.6 liter 4 cylinder at 8500 rpm = can make 240 hp. Or how about a two cylinder 1.3 liter 2 cycle, or a .65 = one cycle. Your argument of one packet in one packet out only works from a black = box standpoint, you cannot analyze what happens in the black box using = this approach. You could pick any combination of rpm, cylinders, and = cylinder volume that gives the same number of events per unit time on = the same mass flow and say it was an equivalent engine. So you are also = correct that you could have any number of rotor faces and get the same = result (there actually are such engines in theory using the same math as = the wankel with different ratios of gearing and rotor faces) The statement about not understanding pretend RPM is correct. It is not = revolutions per unit time that matter, or mass flow per unit time, but = volumetric and pressure changes on each packet per unit time. To have thermodynamic equivalence, the engine must be doing the same = thing to the working fluid per packet per time. If you do this you can = actually integrate the rate of change of volume per unit time and see = how the Otto process is performed in the rotary vs the crank and piston = engine. Varying rates of compression/expansion at different times lead = to different efficiencies and cylinder/chamber pressures-ie stroke to = rod ratio.=20 I have changed my mind- it is now a .325L half cycle engine. I'm all spent on this subject. Call it whatever you want as long as it = gets the desired result.=20 What is really important to all of us is to figure out what the heck to = do with the damnable packets after they are (vigorously!!!) expelled = using your equivalent cycle of choice. OR-how to make the stupid thing = sound more like a Merlin and less like the model airplane from hell with = something less than a 30 lb boat anchor or the size of a 30 gal drum. That is what I am working on now. I have little hope for the outcome. I = will however bare all at Tracy's in full view of the public, you can = witness my triumph, failure, or puzzled looks. Monty ------=_NextPart_000_0008_01C5D36A.04FF6E10 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Okay this one is for David, Ed, = Bob and=20 Ken because they are actually having an = adult discussion=20 with me on this. Unlike the Pope. For everybody else, If this hurts your = head=20 (it is beginning to hurt mine) just skip down to the last few=20 paragraphs.
 
 
I am looking at this from a thermo = standpoint as a=20 mechanical engineer. If I were going to analyze this and make the = numbers work=20 theoretically. You have to break this down to it's smallest=20 component.
 
To do this you follow=20 one INDIVIDUAL induction, compression, expansion cycle from = beginning=20 to end. It does not matter what kind of hardware we are looking = at.
 
When I look at a single cylinder 4 = cycle engine. It=20 takes two complete crank revolutions to complete this cycle on some = displacement=20 of air.
 
I draw a PV diagram for this and get = the work=20 required to induct, compress, expand and exhaust. Some of these terms = are=20 negative, some are positive. The balance is the work output for = this ONE=20 cycle and individual cylinder.
 
For a multi cylinder engine, these work = outputs are=20 all just added together at different phase angles based on the crank=20 throws.
 
 
Now for = the  wankel:
 
It takes one complete revolution of the = ROTOR to=20 complete a cycle for a single rotor face (same charge) and perform = the=20 previous analysis. The eshaft will rotate 3 times for each rotor=20 rev..
 
The additional rotor faces are = just like=20 additional cylinders added on to a single cylinder engine. The mechanism = for=20 extracting power is a bit more arcane but it is the same = thing. Stick on=20 another rotor 180 degrees out of phase and it is the same thing. = 3 more=20 cylinders.
 
So you are correct the 3.9L 6 cyl = equivalent engine=20 will rotate at 2/3 the wankel speed not 1/3 as I misstated. And that = matches=20 with practice, a well tuned 240 cubic inch 4 cycle multi = cylinder engine=20 spinning at 6000 rpm (8500*2/3) is capable of making around 240 hp. = Likewise the Wankel makes the same kind of power at 8500 shaft rpm. = with=20 the same number of cycles performed on the same amount of air charges as = a 6 cyl=20 engine for a given amount of time.
 
Likewise, keeping the rpm the = same: a 2.6=20 liter 4 cylinder at 8500 rpm can make 240 hp. Or how about a two = cylinder 1.3=20 liter 2 cycle, or a .65 one cycle.
 
Your argument of one packet in one = packet out only=20 works from a black box standpoint, you cannot analyze what happens in = the black=20 box using this approach. You could pick any combination of = rpm,=20 cylinders, and cylinder volume that gives the same number of=20 events per unit time on the same mass flow and say it was = an=20 equivalent engine. So you are also correct that you could have any = number of=20 rotor faces and get the same result (there actually are such engines in = theory=20 using the same math as the wankel with different ratios of gearing and = rotor=20 faces)
 
The statement about not understanding = pretend RPM=20 is correct. It is not revolutions per unit time that = matter, or mass=20 flow per unit time, but volumetric and pressure changes on each packet = per unit=20 time.
 
To have thermodynamic equivalence, = the=20 engine must be doing the same thing to the working fluid per packet = per=20 time. If you do this you can actually integrate the rate of change of = volume per=20 unit time and see how the Otto process is performed in the rotary vs the = crank=20 and piston engine. Varying rates of compression/expansion at different = times=20 lead to different efficiencies and cylinder/chamber pressures-ie stroke = to rod=20 ratio.
 
I have changed my mind- it is now = a=20 .325L half cycle engine.
 
I'm all spent on this subject. Call it = whatever you=20 want as long as it gets the desired result.
 
What is = really important to=20 all of us is to figure out what the heck to do with the damnable packets = after=20 they are (vigorously!!!) expelled using your equivalent cycle of choice. = OR-how=20 to make the stupid thing sound more like a Merlin and less like the = model=20 airplane from hell with something less than a 30 lb boat anchor or the = size of a=20 30 gal drum.
 
That is what I am working on now. I = have little=20 hope for the outcome. I will however bare all at Tracy's in full view of = the=20 public, you can witness my triumph, failure, or puzzled=20 looks.
 
 
Monty
 
 
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