X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from zproxy.gmail.com ([64.233.162.206] verified) by logan.com (CommuniGate Pro SMTP 5.0c5) with ESMTP id 771591 for flyrotary@lancaironline.net; Mon, 17 Oct 2005 21:23:28 -0400 Received-SPF: pass receiver=logan.com; client-ip=64.233.162.206; envelope-from=wdleonard@gmail.com Received: by zproxy.gmail.com with SMTP id z3so1504705nzf for ; Mon, 17 Oct 2005 18:22:45 -0700 (PDT) DomainKey-Signature: a=rsa-sha1; q=dns; c=nofws; s=beta; d=gmail.com; h=received:message-id:date:from:to:subject:in-reply-to:mime-version:content-type:references; b=Lf8gD2NQ52+oaQ8niKrvT/DLWv9RyLsWWlg2HVg+5zzGfV1tod7tCsVVLU2/J1yxgr4a8/M2w3UAOI/WAvXl4IzLiTErpzfFO/U5BGVflDLL++b6E5WkjMg2FZIVYfCvXN8BnSlvPnOjADlLxbKrad1Hye5bcLmCZLJEZRIoCy0= Received: by 10.36.227.19 with SMTP id z19mr3248642nzg; Mon, 17 Oct 2005 18:22:44 -0700 (PDT) Received: by 10.36.222.63 with HTTP; Mon, 17 Oct 2005 18:22:44 -0700 (PDT) Message-ID: <1c23473f0510171822g9b410d0m14d3fb9b15c1a1d1@mail.gmail.com> Date: Mon, 17 Oct 2005 18:22:44 -0700 From: David Leonard To: Rotary motors in aircraft Subject: Re: [FlyRotary] Re: flyrotary_Web_Archive Re: Banishment In-Reply-To: MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_Part_1708_29589518.1129598564736" References: ------=_Part_1708_29589518.1129598564736 Content-Type: text/plain; charset=ISO-8859-1 Content-Transfer-Encoding: quoted-printable Content-Disposition: inline Hi Dave, > > Thanks for the reply, and a well reasoned argument (argument being used > in the positive mathametical sense). Thanks Bob, yes it is fun. I think there is an error in your analysis however. In one revolution > of the e-shaft, all three faces are involved in a single Otto cycle. Correct... the 3 faces, together, have completed the FOUR Otto cycles, acting on 3 packets of air, but only 2 of those affect the outside world (the one that came in, and the one that went out). If the rotor were, say, hexagonal it would not change things as long as one packet of air came in, and one left. It doesn't matter how many faces the rotor has, just that one turn of the shaft causes one packet to enter and one leave. The size of each packet is 650cc. The 13B inputs 2 packets and exhausts 2 packets with each rotation of the shaft. The same as would a 2.6 L 4-cylinder engine. If the rotors were hexagonal but the engine still input and output 2 packets of air with each rotation of the shaft, then it would still be a 2.6 L engine. One face is on the intake, the second is on the compression and > expansion, and the third is on the exhaust. In that one revolution > face 1 draws in 650 cc, face 2 compresses it's 650 cc (drawn in on the > previous rotation of the e-shaft) to about 65cc (I just made that > number up - compression ratio is about 9.7:1), ignites it and expands > it back out again. face 3 is exhausting the burned gasses (drawn in > two revolutions of the e-shaft ago). That is one complete Otto cycle, > and we have only had one intake event so 650 cc X two rotors =3D 1.3L. I > don't think you can double or tripple the volume because more than one > face is involved in the process. You are right, I was just presenting another way of looking at the rotary and explaining the two volumes of importance - the in and the out. > Just like a 2.6L 4 stroke engine, you have "displaced" 1.3L in one > revolution of the e-shaft. exactly! I can see how your argument would conclude that the 13B is a 1.3L > engine, but I still don't see 2.6L. see above. :-) -- Dave Leonard Turbo Rotary RV-6 N4VY http://members.aol.com/_ht_a/rotaryroster/index.html http://members.aol.com/_ht_a/vp4skydoc/index.html ------=_Part_1708_29589518.1129598564736 Content-Type: text/html; charset=ISO-8859-1 Content-Transfer-Encoding: quoted-printable Content-Disposition: inline

Hi Dave,

Thanks for the r= eply, and a well reasoned argument (argument being used
in the positive = mathametical sense).
 
Thanks Bob, yes it is fun.

I think there is an error in you= r analysis however.  In one revolution
of the e-shaft, all thr= ee faces are involved in a single Otto cycle.
 
Correct...  the 3 faces, together, have completed the FOUR Otto c= ycles, acting on 3 packets of air, but only 2 of those affect the outside w= orld (the one that came in, and the one that went out).  If the rotor = were, say, hexagonal it would not change things as long as one packet of ai= r came in, and one left.  It doesn't matter how many faces the rotor h= as, just that one turn of the shaft causes one packet to enter and one leav= e.
 
The size of each packet is 650cc.
 
The 13B inputs 2 packets and exhausts 2 packets with each rotation of = the shaft.  The same as would a 2.6 L 4-cylinder engine.  If the = rotors were hexagonal but the engine still input and output 2 packets of ai= r with each rotation of the shaft, then it would still be a=20 2.6 L engine.

One face is on the intake, the s= econd is on the compression and
expansion, and the third is on the exhau= st.  In that one revolution
face 1 draws in 650 cc, face 2 compresses it's 650 cc (drawn in on the<= br>previous rotation of the e-shaft) to about 65cc (I just made that
num= ber up - compression ratio is about 9.7:1), ignites it and expands
it ba= ck out again. face 3 is exhausting the burned gasses (drawn in
two revolutions of the e-shaft ago). That is one complete Otto cycle,and we have only had one intake event so 650 cc X two rotors =3D 1.3L.&nb= sp; I
don't think you can double or tripple the volume because more= than one
face is involved in the process.
 
You are right, I was just presenting another way of looking at the rot= ary and explaining the two volumes of importance - the in and the out. = ;
 
Just like a 2.6L 4 stroke engine= , you have "displaced" 1.3L in one
revolution of the e-shaft.
 
exactly!

I can see how your argument woul= d conclude that the 13B is a 1.3L
engine, but I still don't see 2.6L.
 
see above.  :-)
 
 
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