X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Sender: To: lml@lancaironline.net Date: Thu, 21 May 2009 06:24:08 -0400 Message-ID: X-Original-Return-Path: Received: from imr-d04.mx.aol.com ([205.188.157.42] verified) by logan.com (CommuniGate Pro SMTP 5.2.14) with ESMTP id 3648760 for lml@lancaironline.net; Wed, 20 May 2009 18:00:52 -0400 Received-SPF: pass receiver=logan.com; client-ip=205.188.157.42; envelope-from=Sky2high@aol.com Received: from imo-ma02.mx.aol.com (imo-ma02.mx.aol.com [64.12.78.137]) by imr-d04.mx.aol.com (v107.10) with ESMTP id RELAYIN1-24a147d4d3e6; Wed, 20 May 2009 17:59:41 -0400 Received: from Sky2high@aol.com by imo-ma02.mx.aol.com (mail_out_v40_r1.5.) id q.d41.4002e11c (30740) for ; Wed, 20 May 2009 17:59:38 -0400 (EDT) From: Sky2high@aol.com X-Original-Message-ID: X-Original-Date: Wed, 20 May 2009 17:59:38 EDT Subject: Re: [LML] Re: power cower X-Original-To: lml@lancaironline.net MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="-----------------------------1242856778" X-Mailer: AOL 9.1 sub 5006 X-Spam-Flag:NO X-AOL-IP: 64.12.78.137 -------------------------------1242856778 Content-Type: text/plain; charset="UTF-8" Content-Transfer-Encoding: quoted-printable Content-Language: en Gary, OK. But - in your second formulation, the max calculated power is only 80% when I know that, with everything fire walled at sea level and dry cold = air, it is 100% - unless I am flying at 200 KIAS, then the new 100% is probabl= y 105% times the old. :-) Thank heavens the SAE has a different standard fo= r calculating HP. Perhaps an Excel app on your I-phone would help you in the air. Grayhawk In a message dated 5/20/2009 1:59:20 P.M. Central Daylight Time, casey.gary@yahoo.com writes: Grayhawk, Too late {:-) I already looked at it. I compared it to the rough formula= I have always used - % power =3D RPM/RPMmax * MAP/MAPmax - and it comes= within 1% of it. I stopped there and didn't try to "reverse engineer" th= e formula, except I have hunch it has to do with assuming the max RPM is 25= 00 (the 2.5) and the max MAP is 35 inches (the 3.5). Regardless, the formula= s are not quite right. As rpm increases, engine friction increases and volumetric efficiency decreases so the increase in power won't go up dire= ctly with RPM. The opposite is true of MAP. It takes a certain MAP just to power= the engine (overcome friction) so power will increase more than the increase= in MAP. How much are these effects? Without real data it would be hard= to say, but in the RPM and MAP ranges that are useful in aircraft engines th= e factor for rpm might be something like 0.8. For MAP there is a subtractiv= e term of about 10 inches. The equation would then look like %power =3D (RPM/RPMmax*0.8) * (MAP-10)/MAPmax-10). Gets hard to do in your head (we= ll, mine anyway), so I just remember that a 10% reduction in RPM will drop the power less than 10%, but a 10% drop in MAP will drop the power by more th= an 10%. Then there is the effect of altitude.. Gary ____________________________________ From: Lancair Mailing List Se Ignore the formula. I got it from a very old email and it is wrong. Perhaps Walter will correct it. Grayhawk the brainless In a message dated 5/18/2009 7:39:13 P.M. Central Daylight Time, Sky2high@aol.com writes: Finally, thanks to Walter at GAMI, %HP =3D 100-((max RPM/100-RPM)*2.5+(Max MP-MP)*3.5) **************Dell Inspiron 15 Laptop: Now in 6 vibrant colors! Shop Dell= =E2=80=99s full line of laptops. (http://pr.atwola.com/promoclk/100126575x1222399266x1201456865/aol?redir= =3Dhttp:%2F%2Fad.doubleclick.net%2Fclk%3B215073777%3B3703434 3%3Bf) -------------------------------1242856778 Content-Type: text/html; charset="US-ASCII" Content-Transfer-Encoding: quoted-printable
Gary,
 
OK.
 
But - in your second formulation, the max calculated power= is only 80% when I know that, with everything fire walled at sea level and dr= y cold air, it is 100% - unless I am flying at 200 KIAS, then the new 100%= is probably 105% times the old. :-) Thank heavens the SAE has a different sta= ndard for calculating HP.
 
Perhaps an Excel app on your I-phone would help you in the air.
 
Grayhawk
 
In a message dated 5/20/2009 1:59:20 P.M. Central Daylight Time, casey.gary@yahoo.com writes:
Grayhawk,
Too late {:-) I already looked at it.  I compared it to the ro= ugh formula I have always used - % power =3D RPM/RPMmax * MAP/MAPmax  -= and it comes within 1% of it.  I stopped there and didn't try to "reverse= engineer" the formula, except I have hunch it has to do with assuming th= e max RPM is 2500 (the 2.5) and the max MAP is 35 inches (the 3.5).  Regardless, the formulas are not quite right.  As rpm increas= es, engine friction increases and volumetric efficiency decreases so the inc= rease in power won't go up directly with RPM.  The opposite is true of MA= P.  It takes a certain MAP just to power the engine (overcome friction= ) so power will increase more than the increase in MAP.  How much are th= ese effects?  Without real data it would be hard to say, but in the RPM= and MAP ranges that are useful in aircraft engines the factor for rpm might= be something like 0.8. For MAP there is a subtractive term of about 10 inch= es.  The equation would then look like  %power =3D (RPM/RPMmax*0.8= ) * (MAP-10)/MAPmax-10).  Gets hard to do in your head (well, mine anyw= ay), so I just remember that a 10% reduction in RPM will drop the power less= than 10%, but a 10% drop in MAP will drop the power by more than 10%.  T= hen there is the effect of altitude..
Gary

From: Lancair Mailing Li= st <lml@lancaironline.net>
Se

Ignore the formula.  I got it from a very old email and it is= wrong.  Perhaps Walter will correct it.
 
Grayhawk the brainless
 
In a message dated 5/18/2009 7:39:13 P.M. Central Daylight Time, Sky2high@aol.com writes:
 
Finally, thanks to Walter at GAMI,
 
%HP =3D 100-((max RPM/100-RPM)*2.5+(Max MP-MP)*3.5)
 

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