X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Sender: To: lml@lancaironline.net Date: Sun, 06 Oct 2013 08:44:57 -0400 Message-ID: X-Original-Return-Path: Received: from nm4-vm6.bullet.mail.gq1.yahoo.com ([98.136.218.165] verified) by logan.com (CommuniGate Pro SMTP 6.0.7) with ESMTPS id 6504153 for lml@lancaironline.net; Sun, 06 Oct 2013 06:30:06 -0400 Received-SPF: none receiver=logan.com; client-ip=98.136.218.165; envelope-from=casey.gary@yahoo.com Received: from [216.39.60.180] by nm4.bullet.mail.gq1.yahoo.com with NNFMP; 06 Oct 2013 10:29:30 -0000 Received: from [98.137.12.244] by tm16.bullet.mail.gq1.yahoo.com with NNFMP; 06 Oct 2013 10:29:30 -0000 Received: from [127.0.0.1] by omp1052.mail.gq1.yahoo.com with NNFMP; 06 Oct 2013 10:29:30 -0000 X-Yahoo-Newman-Property: ymail-3 X-Yahoo-Newman-Id: 498569.42493.bm@omp1052.mail.gq1.yahoo.com Received: (qmail 94381 invoked by uid 60001); 6 Oct 2013 10:29:29 -0000 DomainKey-Signature:a=rsa-sha1; q=dns; c=nofws; s=s1024; d=yahoo.com; h=X-YMail-OSG:Received:X-Rocket-MIMEInfo:X-Mailer:References:Message-ID:Date:From:Reply-To:Subject:To:In-Reply-To:MIME-Version:Content-Type; b=sOFQcjHemRsSfUGTWI/R1v+KYhcdULwHhT6cgHjIxvyFimFbcJmOziV7+Iu8skkL3G2KVtkL/9L4AXcBMzEMGdikRIMVDWjusRUp7Kbj1kPMRKnvNjEKD35puguyrT+5LZxianoWVMfAEK8g5c6mz52wzVIGvbR+BYus9Q0j8wo=; X-YMail-OSG: 5kxOvSMVM1ng4OGd3ejHJvmD1Yw0ammq9OqKRiED3UsID3k zg4O2zBKpUGQF1.I8ujHTFAtJMCb1OB5_MLhfbf5fE209by.IyMBHdUbRmMY Q8WzrP7FjqF78vkpIuILFSyTO4r8A05.XpUzpfj_uFnUulIFYRDiHkDX4PdG DY1Af0iR3hbwfISErcsd.VkJNHTmnhbdGkjNYlb1hsdFWt8OTks7MvQDJ.uf J_1bM.TJf3xRfUAifH6wp_q8ELOAzN1uNNeKuQKs2XkXMbktgblWl5FFEYD9 FJcY1nmXWZbDha3xvwUVIdyq6AEPr_GenOS9DlAQge4DR4pHsq8jdi3dwe3M w0oRZPqFZbzLQO_W3LSax5dxsDtgV2ZKQXeKiOuyRF60wBQZIyTwtxW9iFVD hJThjKtfBvuyIk1qu0Yw_JVZ0..caQoUZB_1TUJJagebotyPmUETMQFdeei7 BjwFSN3Surf5247B2.ZiVKAAyayBO18VRV5ziMmSoeWVTsMtZ2eLmQ2G_JQ2 M0USduvLR7S8f9ESy0iFmbHfbjYwhIpI2unwaAOf6W_gT_1_brLvZEOA3fHo 1B48fGciLpH9MWQ6Zo6t1U4Ovv_YpNRiV9wqMAmEPxWYeVtQvEfRfcRnnqUJ PxeWrGH7qgbc10AKlHRB0IxdpK2pDadBa80tD Received: from [97.92.63.83] by web120101.mail.ne1.yahoo.com via HTTP; Sun, 06 Oct 2013 03:29:29 PDT X-Rocket-MIMEInfo: 002.001,V29sZmdhbmcgZ2F2ZSBhIGNvcnJlY3QgcGFydGlhbCBhbnN3ZXIgLSBhcyBvcHBvc2VkIHRvIGEgcGFydGlhbGx5LWNvcnJlY3QgYW5zd2VyIDotKS4gwqBUaGUgY2xpbWIgcmF0ZSB5b3UgYXJlIGxvb2tpbmcgZm9yIGlzIHRoZSBESUZGRVJFTkNFIGJldHdlZW4gdGhlICJjbGltYiIgcmF0ZXMgYXQgZnVsbCBwb3dlciBhbmQgYXQgbm8gcG93ZXIuIMKgQXQgdGhlIHNhbWUgYWlyc3BlZWQsIGNsaW1iIGF0IGZ1bGwgcG93ZXIgYW5kIHRoZW4gZ2xpZGUgd2l0aCBubyBwb3dlciAoYXQgdGhlIHNhbWUgYWx0aXQBMAEBAQE- X-Mailer: YahooMailWebService/0.8.160.587 References: X-Original-Message-ID: <1381055369.93952.YahooMailNeo@web120101.mail.ne1.yahoo.com> X-Original-Date: Sun, 6 Oct 2013 03:29:29 -0700 (PDT) From: Gary Casey Reply-To: Gary Casey Subject: Re: LNC2 flaps at Reflex X-Original-To: Lancair Mailing List In-Reply-To: MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="1072509510-661600908-1381055369=:93952" --1072509510-661600908-1381055369=:93952 Content-Type: text/plain; charset=utf-8 Content-Transfer-Encoding: quoted-printable Wolfgang gave a correct partial answer - as opposed to a partially-correct = answer :-). =C2=A0The climb rate you are looking for is the DIFFERENCE betw= een the "climb" rates at full power and at no power. =C2=A0At the same airs= peed, climb at full power and then glide with no power (at the same altitud= e, of course). =C2=A0Say you weigh 1800 pounds, can climb at 1000 ft/min an= d have a descent rate of 1000 ft/min at no power (so I can do the math in m= y head). =C2=A0The difference is 2000 ft/min, giving an engine power (no, l= et's call it "thrust horsepower") of 109 hp. =C2=A0The propeller efficiency= might be 80 percent, so the crank power would be 136 hp (109 divided by 0.= 8). =C2=A0Let's say you did the test at about 8,000 feet - as an approximat= ion, the correction to sea level power is 133%, so the corrected sea level = hp is then 180 hp. =C2=A0Yes, there are too many corrections and approximat= ions to make it very accurate, but it can give you a rough idea of the actu= al power of your engine. =C2=A0Somebody out there should give it a try and report the resul= ts.=0AGary=0A=0ASure you can.=0A=C2=A0=0AWt x V / 33000 =3D HP=0A=C2=A0=0AW= t =3D weight in Lbs=0AV =3D climb rate in Ft/Min=0A33000 =3D Ft-Lbs/Min per= HP=0A=C2=A0=0A1800=C2=A0x 1000 / 33000 =3D 54.5 HP=0A=C2=A0=0AKeep in mind= that's the NET HP . . . after engine efficiency, propeller efficiency and = drag.=0A=C2=A0=0AWolfgang=0A=C2=A0=0A----- Original Message -----=0A>From:= =C2=A0Bill Bradburry=0A>To:=C2=A0lml@lancaironline.net=0A>Sent:=C2=A0Friday= , October 04, 2013 7:32 AM=0A>Subject:=C2=A0RE: [LML] Re: LNC2 flaps at Ref= lex=0A>=0A>=0A>Hmmmm,=0A>=C2=A0=0A>If you don=E2=80=99t know the HP that yo= ur engine is developing how would you go about discovering it?=C2=A0 Can yo= u determine HP by climb rate at a known weight?=0A>=C2=A0=0A>B2 --1072509510-661600908-1381055369=:93952 Content-Type: text/html; charset=utf-8 Content-Transfer-Encoding: quoted-printable
Wolfgang gave = a correct partial answer - as opposed to a partially-correct answer :-). &n= bsp;The climb rate you are looking for is the DIFFERENCE between the "climb= " rates at full power and at no power.  At the same airspeed, climb at= full power and then glide with no power (at the same altitude, of course).=  Say you weigh 1800 pounds, can climb at 1000 ft/min and have a desce= nt rate of 1000 ft/min at no power (so I can do the math in my head).  = ;The difference is 2000 ft/min, giving an engine power (no, let's call it "= thrust horsepower") of 109 hp.  The propeller efficiency might be 80 p= ercent, so the crank power would be 136 hp (109 divided by 0.8).  Let'= s say you did the test at about 8,000 feet - as an approximation, the corre= ction to sea level power is 133%, so the corrected sea level hp is then 180 hp.  Yes, there are too many corrections and approximations = to make it very accurate, but it can give you a rough idea of the actual po= wer of your engine.  Somebody out there should give it a try and repor= t the results.
Gary

Sure you can.
 
Wt x V / 33000 =3D HP=
 
Wt =3D weight in Lbs
V =3D clim= b rate in Ft/Min
33000 =3D Ft-Lbs/Min per HP
=  
= 1800 x 1000 / 33000 =3D 54.5 HP=
 
Keep in mind that's the NET HP . . . after engin= e efficiency, propeller efficiency and drag.
 
Wolfgang
 
-= ---- Original Message -----
Sent: Friday, October 04, 2013 7:32 A= M
Subject:&= nbsp;RE: [LML] Re: LNC2 flaps at Reflex

Hmmmm,
 
If you don=E2=80=99t know the HP that your engine i= s developing how would you go about discovering it?  Can you determine= HP by climb rate at a known weight?
 
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