X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Sender: To: lml@lancaironline.net Date: Fri, 17 Aug 2007 15:36:25 -0400 Message-ID: X-Original-Return-Path: Received: from imo-d21.mx.aol.com ([205.188.144.207] verified) by logan.com (CommuniGate Pro SMTP 5.1.11) with ESMTP id 2268777 for lml@lancaironline.net; Fri, 17 Aug 2007 10:54:04 -0400 Received-SPF: pass receiver=logan.com; client-ip=205.188.144.207; envelope-from=REHBINC@aol.com Received: from REHBINC@aol.com by imo-d21.mx.aol.com (mail_out_v38_r9.2.) id q.d0a.16ec6e66 (30739) for ; Fri, 17 Aug 2007 10:53:01 -0400 (EDT) From: REHBINC@aol.com X-Original-Message-ID: X-Original-Date: Fri, 17 Aug 2007 10:53:00 EDT Subject: Re: [LML] Re: Apologies to the farmer's daughter X-Original-To: lml@lancaironline.net MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="-----------------------------1187362380" X-Mailer: 9.0 for Windows sub 5128 X-Spam-Flag: NO -------------------------------1187362380 Content-Type: text/plain; charset="US-ASCII" Content-Transfer-Encoding: 7bit Mike, I never studied parachutes, but I think round parachutes work primarily from velocity pressure. The formula is pressure = 1/2 * density * v^2. Pressure must be constant at equilibrium, so velocity would vary more or less with the inverse of the square root of density. This ignores viscous and aerodynamic effects, which I would expect to be small. Thus, for standard atmosphere, the speed at sea level would be about the square root of 2.048/2.377, or 92.8% of the velocity at 5000'. Based on the assumed 1500 fpm decent rate at 5000', this would yield a sea level decent of 1392 fpm or 23.2 fps. This is equivalent to stepping of a 8.36 ft ladder. Rob ************************************** Get a sneak peek of the all-new AOL at http://discover.aol.com/memed/aolcom30tour -------------------------------1187362380 Content-Type: text/html; charset="US-ASCII" Content-Transfer-Encoding: quoted-printable

Mike,

I never studied parachutes, but I think round parachutes work primarily= from velocity pressure. The formula is pressure =3D 1/2 * density * v^= 2. Pressure must be constant at equilibrium, so velocity would vary more or=20= less with the inverse of the square root of density. This ignores viscous an= d aerodynamic effects, which I would expect to be small.

Thus, for standard atmosphere, the speed at sea level would b= e about the square root of 2.048/2.377, or 92.8% of the velocity at 5000'. B= ased on the assumed 1500 fpm decent rate at 5000', this would yield a sea le= vel decent of 1392 fpm or 23.2 fps. This is equivalent to stepping of a 8.36= ft ladder.

Rob

Get a sneak peek of the=20= all-new AOL.com.

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