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 771318 for flyrotary@lancaironline.net; Mon, 17 Oct 2005 17:57:55 -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 z3so1481014nzf for ; Mon, 17 Oct 2005 14:57:11 -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=hGA+4gIGFGDL349HWy/kcKJkHxBeN43VicnkoBsi1G4PFKl5h4Dm0DueGYUfAMmA+Mlm4swT6XGn8aJPCpPhttCJVvCAyTUe7KCzWCKHPGen4Ocuno6UbHb/JtxlUbZ9sFB5DDKdBumjvby9qS49vzQ5JPb4Gwd5X207SpChz7Y= Received: by 10.36.222.80 with SMTP id u80mr290812nzg; Mon, 17 Oct 2005 14:57:11 -0700 (PDT) Received: by 10.36.222.63 with HTTP; Mon, 17 Oct 2005 14:57:11 -0700 (PDT) Message-ID: <1c23473f0510171457x72531599la1a6b6faefc7f1c3@mail.gmail.com> Date: Mon, 17 Oct 2005 14:57:11 -0700 From: David Leonard To: Rotary motors in aircraft Subject: Re: [FlyRotary] Re: Displacement - Again? Timing of the Work In-Reply-To: MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_Part_867_31739991.1129586231148" References: ------=_Part_867_31739991.1129586231148 Content-Type: text/plain; charset=ISO-8859-1 Content-Transfer-Encoding: quoted-printable Content-Disposition: inline Jesse, you are getting a little confused... wait, so am I.... On 10/17/05, jesse farr wrote: > > Wait just a minute, fellows. Don't we go through the same cycle that a > four > cycle does in two revolutions with only one third of a rotor turn, one > ninth > of an e shaft turn ? No, ONE shaft turn. You got reversed which is going faster... What does this do to all your calculations ? We have > three power stokes for every rotor revolution No, just one power stroke for every revolution of the e-shaft. And the rotors are not so much rotating as they are orbiting (at shaft speed) with = a 1/3 speed rotation in the OPPOSITE direction. It is what is is really.... > -- 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_867_31739991.1129586231148 Content-Type: text/html; charset=ISO-8859-1 Content-Transfer-Encoding: quoted-printable Content-Disposition: inline Jesse, you are getting a little confused...  wait, so am I....

On 10/17/05, jesse farr <jesse@jessfarr.co= m> wrote:
Wait just a minute, fellows. Don= 't we go through the same cycle that a four
cycle does in two revolution= s with only one third of a rotor turn, one ninth
of an e shaft turn ?
 
No, ONE shaft turn.  You got reversed which is going faster...
What does this do to all your c= alculations ? We have
three power stokes for every rotor revolution
 
No, just one power stroke for every revolution of the e-shaft.  A= nd the rotors are not so much rotating as they are orbiting (at shaft speed= ) with a 1/3 speed rotation in the OPPOSITE direction.
 
It is what is is really....
 

--
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=20 ------=_Part_867_31739991.1129586231148--