X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Sender: To: lml@lancaironline.net Date: Thu, 16 Aug 2007 21:19:17 -0400 Message-ID: X-Original-Return-Path: Received: from vms040pub.verizon.net ([206.46.252.40] verified) by logan.com (CommuniGate Pro SMTP 5.1.11) with ESMTP id 2265424 for lml@lancaironline.net; Wed, 15 Aug 2007 22:53:25 -0400 Received-SPF: pass receiver=logan.com; client-ip=206.46.252.40; envelope-from=tom.gourley@verizon.net Received: from jacky0da39824a ([71.97.246.237]) by vms040.mailsrvcs.net (Sun Java System Messaging Server 6.2-6.01 (built Apr 3 2006)) with ESMTPA id <0JMU007DVINIIN2Q@vms040.mailsrvcs.net> for lml@lancaironline.net; Wed, 15 Aug 2007 21:52:33 -0500 (CDT) X-Original-Date: Wed, 15 Aug 2007 19:52:25 -0700 From: "Tom Gourley" Subject: Re: [LML] Re: Apologies to the farmer's daughter X-Original-To: "Lancair Mailing List" Reply-to: "Tom Gourley" X-Original-Message-id: <001d01c7dfb0$7a3aa3d0$650610ac@jacky0da39824a> MIME-version: 1.0 X-MIMEOLE: Produced By Microsoft MimeOLE V6.00.2900.3138 X-Mailer: Microsoft Outlook Express 6.00.2900.3138 Content-type: multipart/alternative; boundary="----=_NextPart_000_001A_01C7DF75.CD326C70" X-Priority: 3 X-MSMail-priority: Normal References: This is a multi-part message in MIME format. ------=_NextPart_000_001A_01C7DF75.CD326C70 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable The equation for distance is 0.5 * a * t^2 where a is the acceleration = due to gravity, very close to 32ft/sec^2. From 0 to 25ft/sec requires = 0.78 sec. So the distance covered is 0.5 * 32 * 0.78^2 =3D 9.73 ft. = And, as Rick has already pointed out, 25ft/sec is 17 mph. That's going = to be hard on the landing gear. Tom Gourley ------=_NextPart_000_001A_01C7DF75.CD326C70 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable

The equation for distance is 0.5 * a *=20 t^2 where a is the acceleration due to gravity, very close to=20 32ft/sec^2. From 0 to 25ft/sec requires 0.78 sec. So the = distance=20 covered is 0.5 * 32 * 0.78^2 =3D 9.73 ft. And, as Rick has already = pointed=20 out, 25ft/sec is 17 mph. That's going to be hard on the landing=20 gear.

Tom Gourley

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