X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Sender: To: lml@lancaironline.net Date: Thu, 11 May 2006 18:41:38 -0400 Message-ID: X-Original-Return-Path: Received: from haaga.com ([71.4.87.133] verified) by logan.com (CommuniGate Pro SMTP 5.0.9) with ESMTP id 1101283 for lml@lancaironline.net; Thu, 11 May 2006 18:26:02 -0400 Received-SPF: pass receiver=logan.com; client-ip=71.4.87.133; envelope-from=clkeller@utahweb.com Received: from utahweb.com [67.106.48.70] by haaga.com with ESMTP (SMTPD-8.22) id A9501AAC; Thu, 11 May 2006 16:23:12 -0600 X-Original-Message-ID: <4463B9D2.5020503@utahweb.com> X-Original-Date: Thu, 11 May 2006 16:25:22 -0600 From: Charles Keller User-Agent: Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.4) Gecko/20030624 Netscape/7.1 (ax) X-Accept-Language: en-us, en MIME-Version: 1.0 X-Original-To: Lancair Mail List Subject: Poor Man's Dyno Content-Type: multipart/alternative; boundary="------------090701010808020001090009" X-Declude-Sender: clkeller@utahweb.com [67.106.48.70] X-Spam-Tests-Failed: Whitelisted [0] This is a multi-part message in MIME format. --------------090701010808020001090009 Content-Type: text/plain; charset=us-ascii; format=flowed Content-Transfer-Encoding: 7bit On 8 May Rob wrote: -- >>Good concept, but I think you have made it more complicated than it needs to be. Rather than calculating a moment arm from the prop center to the wheel and then correcting the scale force to this moment arm, you could simply use the moment arm between the scales and the direct scale readings. Thus: T = (dR + dL) * d/2 where "T" is in lb ft, "dR" and "dL" are in lb and "d" is in ft. HP is then simply T * rpm / 5252. The answer should be the same either way.<< (Rob was referring to a method given by Paul Lipps in a pdf document attached to his 5 May LML posting.) Rob, the answer would be the same only if the prop shaft were at the same vertical level as the scales, an impractical situation. The problem is that the force vector at the scale due to the engine torque is normal to a line between the prop shaft and the scale, whereas the scale is reading only the vertical component of that vector. And torque being a force times a distance, the distance of importance in this case is that between the scale and the prop shaft rather than its horizontal component. As the thrust line goes higher, or the gear tread gets smaller, the difference between the two can be substantial. Calculating the actual vector and distance was the purpose of Paul's calculations (complications?). Using the figures in Paul's sample calculation in his attached pdf he calculated 159.4 hp. Your equations give 102.8 hp for the same figures, for an error of 35.5 per cent. The theoretical error between the two methods is 36.1 for his prop height and gear tread. Actually, Paul's calculations aren't that complicated. Even the mickey mouse calculator I got from my insurance broker does square roots. Charles Keller --------------090701010808020001090009 Content-Type: text/html; charset=us-ascii Content-Transfer-Encoding: 7bit On 8 May Rob wrote: -- >>Good concept, but I think you have made it more complicated than it needs to be. Rather than calculating a moment arm from the prop center to the wheel and then correcting the scale force to this moment arm, you could simply use the moment arm between the scales and the direct scale readings.

Thus: T = (dR + dL) * d/2 where "T" is in lb ft, "dR" and "dL" are in lb and "d" is in ft.

HP is then simply T * rpm / 5252.

The answer should be the same either way.<<

    (Rob was referring to a method given by Paul Lipps in a pdf document attached to his 5 May LML posting.)

    Rob, the answer would be the same only if the prop shaft were at the same vertical level as the scales, an impractical situation. The problem is that the force vector at the scale due to the engine torque is normal to a line between the prop shaft and the scale, whereas the scale is reading only the vertical component of that vector. And torque being a force times a distance, the distance of importance in this case is that between the scale and the prop shaft rather than its horizontal component. As the thrust line goes higher, or the gear tread gets smaller, the difference between the two can be substantial. Calculating the actual vector and distance was the purpose of Paul's calculations (complications?). Using the figures in Paul's sample calculation in his attached pdf he calculated 159.4 hp. Your equations give 102.8 hp for the same figures, for an error of 35.5 per cent. The theoretical error between the two methods is 36.1 for his prop height and gear tread.

    Actually, Paul's calculations aren't that complicated. Even the mickey mouse calculator I got from my insurance broker does square roots.

                   Charles Keller
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