Return-Path: Sender: (Marvin Kaye) To: lml@lancaironline.net Date: Thu, 13 Nov 2003 19:40:29 -0500 Message-ID: X-Original-Return-Path: Received: from mxsf30.cluster1.charter.net ([209.225.28.230] verified) by logan.com (CommuniGate Pro SMTP 4.1.6) with ESMTP id 2730822 for lml@lancaironline.net; Thu, 13 Nov 2003 18:21:42 -0500 Received: from erics1200mhz (cpe-68-186-243-158.ma.charter.com [68.186.243.158]) by mxsf30.cluster1.charter.net (8.12.10/8.12.8) with SMTP id hADNJ56i039245 for ; Thu, 13 Nov 2003 18:19:05 -0500 (EST) (envelope-from emjones@charter.net) X-Original-Message-ID: <007f01c3aa3d$5953d8a0$0300a8c0@erics1200mhz> From: "Eric M. Jones" X-Original-To: "Lancair Mailing List" Subject: Fw: [LML] Re: Braking Energy X-Original-Date: Thu, 13 Nov 2003 18:24:56 -0500 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_007C_01C3AA13.703AAD20" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2800.1158 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2800.1165 This is a multi-part message in MIME format. ------=_NextPart_000_007C_01C3AA13.703AAD20 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable OOPS....Can't get it back If we assume W=3D10^3 lbs., V=3D10^2 knots, N=3D1 Then the braking = energy per wheel by that formula is E=3D0.44 X 10^3 X 10^2 X 10^2=3D0.44 X 10^7=3D4.4 X 10^6 lb-ft If we look at the kinetic energy per wheel, then the braking energy per = wheel is E=3D1/2 WV^2 W=3Dpounds "mass"=3D10^3 V=3D100 knots=3D170 feet/second E=3D1/2 X 10^3 X 170 X 170=3D14.5 X 10^6 (over 3 X more) What gives? Eric ----- Original Message -----=20 From: Tubamanflies@aol.com=20 To: Lancair Mailing List=20 Sent: Thursday, November 13, 2003 3:24 PM Subject: [LML] Re: Braking Energy Jeff, Braking energy per wheel can be calculate as follows Kinetic Energy =3D (.044 x W x V x V) / N W =3D Landing weight in lbs V =3D Landing speed in knots N =3D Number of wheels with brakes I had a lot of trouble with the original brakes on my LNC360MKII until = I replaced them with "Grove" brakes. They make all the difference in = the world. I don't know what the IVP has but if it anything like the = 360 replacement may be a good solution. Talk to Grove 1-619-562-1268 = and see what they have for the IVP. Good luck Ray ------=_NextPart_000_007C_01C3AA13.703AAD20 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
OOPS....Can't get it back

If we assume W=3D10^3 lbs., V=3D10^2 = knots, N=3D1 Then=20 the braking energy per wheel by that formula is
 
E=3D0.44 X 10^3 X 10^2 X 10^2=3D0.44 X = 10^7=3D4.4 X 10^6=20 lb-ft
 
If we look at the kinetic energy per = wheel, then=20 the braking energy per wheel is
 
E=3D1/2 WV^2
 
W=3Dpounds "mass"=3D10^3
V=3D100 knots=3D170 = feet/second
E=3D1/2 X 10^3 X 170 X 170=3D14.5 X = 10^6  =20 (over 3 X more)
 
What gives?
Eric
----- Original Message -----
From:=20 Tubamanflies@aol.com
Sent: Thursday, November 13, = 2003 3:24=20 PM
Subject: [LML] Re: Braking = Energy

Jeff,

Braking energy per wheel can be = calculate as=20 follows

Kinetic Energy =3D (.044 x  W x V x V) / = N

W =3D=20 Landing weight in lbs
V =3D Landing speed in knots
N =3D Number = of wheels=20 with brakes

I had a lot of trouble with the original brakes on = my=20 LNC360MKII until I replaced them with "Grove" brakes.  They make = all the=20 difference in the world.  I don't know what the IVP has but if it = anything like the 360 replacement may be a good solution.  Talk = to Grove=20 1-619-562-1268 and see what they have for the IVP.

Good = luck

Ray=20 ------=_NextPart_000_007C_01C3AA13.703AAD20--