X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Sender: To: lml@lancaironline.net Date: Thu, 29 Dec 2005 10:12:30 -0500 Message-ID: X-Original-Return-Path: Received: from sfa.gami.com ([68.89.254.162] verified) by logan.com (CommuniGate Pro SMTP 5.0.5) with ESMTP id 905005 for lml@lancaironline.net; Thu, 29 Dec 2005 08:50:31 -0500 Received-SPF: none receiver=logan.com; client-ip=68.89.254.162; envelope-from=gwbraly@gami.com Received: from localhost (localhost.localdomain [127.0.0.1]) by sfa.gami.com (Postfix) with ESMTP id 9858429C070 for ; Thu, 29 Dec 2005 07:49:14 -0600 (CST) Received: from sdf1.mail.taturbo.com (unknown [10.10.10.173]) by sfa.gami.com (Postfix) with ESMTP id 4C76F29C06E for ; Thu, 29 Dec 2005 07:49:13 -0600 (CST) X-MimeOLE: Produced By Microsoft Exchange V6.0.6603.0 content-class: urn:content-classes:message MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----_=_NextPart_001_01C60C7F.4D60059A" Subject: RE: [LML] Re: Where has all the power gone? X-Original-Date: Thu, 29 Dec 2005 07:53:52 -0600 X-Original-Message-ID: X-MS-Has-Attach: X-MS-TNEF-Correlator: Thread-Topic: [LML] Re: Where has all the power gone? Thread-Index: AcYMK6Q+c0MnLOZ3QjKURhcejv+p1QAUS2Uw From: "George Braly" X-Original-To: "Lancair Mailing List" X-Virus-Scanned: by amavisd-new-2.3.2 (20050629) (Debian) at gami.com This is a multi-part message in MIME format. ------_=_NextPart_001_01C60C7F.4D60059A Content-Type: text/plain; charset="us-ascii" Content-Transfer-Encoding: quoted-printable =20 Rob, =20 >>Consider a perfect ottocycle engine with perfect fuel. Mechanically the engine can survive whatever pressure the fuel can generate and the fuel burns infinitely fast. In this perfect world, with no cylinder leakage and no heat transfer, the cylinder pressure at any theta after ignition on the power stroke is always the same, no matter where ignition takes place. If the fuel is ignited at TDC, then the entire charge is combusted at TDC and Theta PP is also at TDC. Now, the piston sees the maximum possible pressure force at every inch of downward travel. Energy is force times distance so this has to extract the most shaft power from the fuel charge.<< =20 That is an interesting concept. =20 But I don't think it is even theoretically true. I think if you apply some boundary condition analysis you will come to the same conclusion. The configuration you suggests does not appear to take into account the effects of the connecting rod-crankshaft geometry. =20 It is not the simple area under the expansion curve that results in the power. It is the correct integration of the expanding pressure curve, with the contribution of each point on the curve a function of sin(theta-crankangle). =20 It is not a desirable design to arrange the combustion event so that the maximum pressure point in the combustion cycle gets multiplied by zero (0=3Dsin (zero degrees) in the integration of the = pressure-expansion curve to arrive at the torque applied to the crank shaft. =20 =20 Regards, George =20 ------_=_NextPart_001_01C60C7F.4D60059A Content-Type: text/html; charset="us-ascii" Content-Transfer-Encoding: quoted-printable

Rob,

>>Consider a perfect ottocycle engine with = perfect fuel. Mechanically the engine can survive whatever pressure the fuel can generate and the fuel burns infinitely fast. In this perfect world, with = no cylinder leakage and no heat transfer, the cylinder pressure at any theta after ignition on the power stroke is always the same, no = matter where ignition takes place. If the fuel is ignited at TDC, then the = entire charge is combusted at TDC and Theta PP is also at TDC. Now, the piston = sees the maximum possible pressure force at every inch of downward = travel. Energy is force times distance so this has to extract the most shaft = power from the fuel charge.<<

That is an interesting = concept.

But I don’t think it is even = theoretically true. I think if you apply some boundary condition analysis you will come to = the same conclusion.

The configuration you suggests does not appear to take into = account the effects of the connecting rod-crankshaft = geometry.

It is not the simple area under the expansion = curve that results in the power. It is the correct integration of the = expanding pressure curve, with the contribution of each point on the curve a = function of sin(theta-crankangle).

It is not a desirable design to arrange the = combustion event so that the maximum pressure point in the combustion cycle = gets multiplied by zero (0=3Dsin (zero degrees) in the integration of the = pressure-expansion curve to arrive at the torque applied to the crank shaft. =

Regards, George =

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