X-Virus-Scanned: clean according to Sophos on Logan.com X-SpamCatcher-Score: 87 [XXXX] (26%) BODY: contains "ambien" obfuscated (14%) SPAMTRICK: obfuscated phone number (14%) SPAMTRICK: obfuscated phone number (14%) SPAMTRICK: obfuscated phone number (14%) RECEIVED: IP not found on home country list (-11%) BODY: obfuscated phone number adjustment (-11%) BODY: obfuscated phone number adjustment Return-Path: Sender: To: lml@lancaironline.net Date: Mon, 30 Apr 2007 23:17:13 -0400 Message-ID: X-Original-Return-Path: Received: from mail09.syd.optusnet.com.au ([211.29.132.190] verified) by logan.com (CommuniGate Pro SMTP 5.1.8) with ESMTPS id 2018661; Mon, 30 Apr 2007 23:13:32 -0400 Received-SPF: none receiver=logan.com; client-ip=211.29.132.190; envelope-from=fredmoreno@optusnet.com.au Received: from fred (optussatellitebintb.22bjipb002.optus.net.au [59.154.24.148]) (authenticated sender fredmoreno) by mail09.syd.optusnet.com.au (8.13.1/8.13.1) with ESMTP id l413CD4h009559; Tue, 1 May 2007 13:12:23 +1000 From: "Fred Moreno" X-Original-To: "Lancair Mail" X-Original-Cc: Subject: Cooling Drag - Error! X-Original-Date: Tue, 1 May 2007 11:13:17 +0800 X-Original-Message-ID: <002101c78b9e$b2ec1b20$c211140a@fred> MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0022_01C78BE1.C10F5B20" X-Priority: 3 (Normal) X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook, Build 10.0.6822 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.3028 Thread-Index: AceLnqmqpx/rVbbbTlmnm3MDNvBEHw== Importance: Normal This is a multi-part message in MIME format. ------=_NextPart_000_0022_01C78BE1.C10F5B20 Content-Type: text/plain; charset="us-ascii" Content-Transfer-Encoding: quoted-printable I warned you guys that I was getting old and forgetful.=20 =20 I wrote previously that a rough rule of thumb for small changes is that = a 2% drag decrease will yield a 1% speed increase. Wrongo! That is a simplification of the fact that drag goes as velocity squared which is = not the full story.=20 =20 What I wrote is true only when THRUST is constant, as with a jet engine. For our piston engines, POWER is constant (more or less, fixed = conditions). =20 Power =3D thrust times velocity. (This is a definition, or if you = prefer, one of the fundamental laws. Sorry.) =20 Recall Thrust =3D Drag =20 For constant Power,=20 =20 Thrust goes like 1/Velocity. =20 =20 When you go faster, you make less thrust but also make more drag. (When = you go faster, your variable pitch prop bites more air to advance farther on each rev, but in so doing it makes less thrust consistent with the law = that power =3D thrust times velocity.) =20 This means that for piston engine airplanes, power goes like velocity = cubed, not velocity squared which is the error implied by my rule of thumb.=20 =20 Simplified ("linearized" for you math buffs) this means for small = changes the correct statement is: =20 It takes a 3% drag reduction to get a 1% speed increase.=20 =20 Alas. =20 You get a tiny bit more power arising from the slight increase in ram pressure which translates into a tiny increase in manifold pressure, but = not much. I calculated that if you are at 9000 feet, ambient pressure about = 21 in. HG, and you get full ram pressure into the intake manifold (you are getting this, aren't you?) to give about 23 inches of manifold pressure, then a 3% speed increase gives you about 0.1% more power.=20 =20 If you slog through all the math and take out the linear effects and = assume I made no more errors (big assumption) the bottom line for a 10% drag decrease (our maximum goal for aspirated airplanes, 9000 feet, 230 knots TAS) you get a 3.3% increase in speed (which means the 3% rule is pretty good). =20 =20 Or about 7.7 knots. =20 =20 If you start with a particularly leaky cowl, lots of air gushing out the spinner-cowl gap, and set of leaky baffles, you might get a bit more. = But that seems to be it folks. =20 Fanatics will still persist for that last knot. Those extra knots get = more and more expensive, but you already knew that. :-) =20 Fred Moreno ------=_NextPart_000_0022_01C78BE1.C10F5B20 Content-Type: text/html; charset="us-ascii" Content-Transfer-Encoding: quoted-printable

I warned you guys that I was getting old and forgetful.

 

I wrote previously that a rough rule of thumb = for small changes is that a 2% drag decrease will yield a 1% speed = increase.  Wrongo!  That is a simplification of the fact that drag goes as velocity squared = which is not the full story.

 

What I wrote is true only when THRUST is = constant, as with a jet engine.  For our piston engines, POWER is constant (more = or less, fixed conditions). 

 

Power =3D thrust times velocity.  =  (This is a definition, or if you prefer, one of the fundamental laws.  = Sorry.)

 

Recall Thrust =3D Drag

 

For constant Power,

 

Thrust goes like  1/Velocity.  =

 

When you go faster, you make less thrust but = also make more drag.  (When you go faster, your variable pitch prop bites = more air to advance farther on each rev, but in so doing it makes less thrust = consistent with the law that power =3D thrust times velocity.)

 

This means that for piston engine airplanes, = power goes like velocity cubed, not velocity squared which is the error = implied by my rule of thumb.

 

Simplified (“linearized” for you = math buffs) this means for small changes the correct statement = is:

 

It takes a 3% = drag reduction to get a 1% speed increase.

 

Alas.

 

You get a tiny bit more power arising from the = slight increase in ram pressure which translates into a tiny increase in = manifold pressure, but not much.  I calculated that if you are at 9000 feet, ambient pressure about 21 in. HG, and you get full ram pressure into the = intake manifold (you are getting this, aren’t you?) to give about 23 = inches of manifold pressure, then a 3% speed increase gives you about 0.1% more = power.

 

If you slog through all the math and take out = the linear effects and assume I made no more errors (big assumption) the = bottom line for a 10% drag decrease (our maximum goal for aspirated airplanes, = 9000 feet, 230 knots TAS) you get a 3.3% increase in speed (which means the = 3% rule is pretty good). 

 

Or about 7.7 knots. 

 

If you start with a particularly leaky cowl, = lots of air gushing out the spinner-cowl gap, and set of leaky baffles, you = might get a bit more.  But that seems to be it folks.

 

Fanatics will still persist for that last = knot.  Those extra knots get more and more expensive, but you already knew = that. J

 

Fred Moreno

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