X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Sender: To: lml@lancaironline.net Date: Sun, 06 Oct 2013 19:32:11 -0400 Message-ID: X-Original-Return-Path: Received: from fmailhost05.isp.att.net ([207.115.11.55] verified) by logan.com (CommuniGate Pro SMTP 6.0.7) with ESMTP id 6504272 for lml@lancaironline.net; Sun, 06 Oct 2013 09:12:50 -0400 Received-SPF: none receiver=logan.com; client-ip=207.115.11.55; envelope-from=bbradburry@bellsouth.net Received: from desktop (adsl-98-85-93-150.mco.bellsouth.net[98.85.93.150]) by isp.att.net (frfwmhc05) with SMTP id <20131006131215H05008ptgce>; Sun, 6 Oct 2013 13:12:15 +0000 X-Originating-IP: [98.85.93.150] From: "Bill Bradburry" X-Original-To: "'Lancair Mailing List'" References: In-Reply-To: Subject: RE: [LML] Re: LNC2 flaps at Reflex X-Original-Date: Sun, 6 Oct 2013 09:12:21 -0400 X-Original-Message-ID: MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_000F_01CEC274.2A60E0B0" X-Mailer: Microsoft Office Outlook 11 Thread-Index: Ac7CkeNPDaMUO8gRSIap3cInCVXeTAAAoI+A X-MimeOLE: Produced By Microsoft MimeOLE V6.0.6002.18463 This is a multi-part message in MIME format. ------=_NextPart_000_000F_01CEC274.2A60E0B0 Content-Type: text/plain; charset="us-ascii" Content-Transfer-Encoding: 7bit Thanks guys! Some one is going to give it a try. How do you determine the correction to sea level from whatever altitude the test is performed? I assume, for example, if you started your descent rate test at 8000 ft and ended it at 7000 ft, you would use 7500 for your calculations? Same in reverse for the climb rate? Bill _____ From: Lancair Mailing List [mailto:lml@lancaironline.net] On Behalf Of Gary Casey Sent: Sunday, October 06, 2013 8:45 AM To: lml@lancaironline.net Subject: [LML] Re: LNC2 flaps at Reflex Wolfgang gave a correct partial answer - as opposed to a partially-correct answer :-). 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 descent 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 percent, 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 correction 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 power of your engine. Somebody out there should give it a try and report the results. Gary Sure you can. Wt x V / 33000 = HP Wt = weight in Lbs V = climb rate in Ft/Min 33000 = Ft-Lbs/Min per HP 1800 x 1000 / 33000 = 54.5 HP Keep in mind that's the NET HP . . . after engine efficiency, propeller efficiency and drag. Wolfgang ----- Original Message ----- From: Bill Bradburry To: lml@lancaironline.net Sent: Friday, October 04, 2013 7:32 AM Subject: RE: [LML] Re: LNC2 flaps at Reflex Hmmmm, If you don't know the HP that your engine is developing how would you go about discovering it? Can you determine HP by climb rate at a known weight? B2 ------=_NextPart_000_000F_01CEC274.2A60E0B0 Content-Type: text/html; charset="us-ascii" Content-Transfer-Encoding: quoted-printable

Thanks = guys!

 

Some one is going to give it a = try.  How do you determine the correction to sea level from whatever altitude the = test is performed?  I assume, for example, if you started your descent rate = test at 8000 ft and ended it at 7000 ft, you would use 7500 for your = calculations?  Same in reverse for the climb rate?

 

Bill

From: = Lancair Mailing List = [mailto:lml@lancaironline.net] On Behalf Of Gary Casey
Sent: Sunday, October 06, = 2013 8:45 AM
To: = lml@lancaironline.net
Subject: [LML] Re: LNC2 = flaps at Reflex

 

Wolfgang gave a correct partial answer - as opposed to a partially-correct answer :-).  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 descent 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 percent, 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 correction 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 power of your engine.  Somebody out there should give it = a try and report the results.

Gary

 

Sure you can.

 

Wt x V / 33000 =3D HP

 

Wt =3D weight in Lbs

V =3D climb 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 engine efficiency, propeller = efficiency and drag.

 

Wolfgang<= /o:p>

 

----- Original Message -----

Sent: Friday, October 04, 2013 7:32 AM

Subject: RE: [LML] Re: LNC2 flaps at Reflex

 

Hmmmm,

 

If you = don’t know the HP that your engine is developing how would you go about discovering it?  Can you determine HP by climb rate at a known = weight?

 

B2<= font size=3D2 color=3D"#454545" face=3DArial>

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