Return-Path: Sender: (Marvin Kaye) To: lml Date: Wed, 25 Sep 2002 12:17:21 -0400 Message-ID: X-Original-Return-Path: Received: from www.dynacomm.ws ([198.22.63.66] verified) by logan.com (CommuniGate Pro SMTP 4.0b8) with ESMTP id 1790693 for lml@lancaironline.net; Wed, 25 Sep 2002 11:32:26 -0400 Received: from [10.0.1.201] (adsl-66-72-178-142.dsl.sfldmi.ameritech.net [66.72.178.142]) (authenticated) by www.dynacomm.ws (8.10.2/8.10.2) with ESMTP id g8PFWEH05717; Wed, 25 Sep 2002 11:32:14 -0400 Mime-Version: 1.0 X-Sender: lorn@pop.dynacomm.ws X-Original-Message-Id: In-Reply-To: References: X-Original-Date: Wed, 25 Sep 2002 11:33:19 -0400 X-Original-To: lml@lancaironline.net From: "Lorn H. Olsen" Subject: [LML] Re: RAM-air induction X-Original-Cc: Villi Seemann Content-Type: text/plain; charset="us-ascii" ; format="flowed" >Villi Seemann : >. >. >If we have a RAM air opening of 3 inch diameter, it has an area of appx 7 >sq.in ( 1/2 * 3^2 * PI ) . This means, if we did not have any inlet >losses, back pressure etc, the RAM air would enter our plenum chamber with >. >. >Regards >Villi H. Seemann >Senior Engineer >Infrastructure Network >Phone (+45) 3333 2101 >FAX (+45) 3333 1130 >CellPhn (+45)2220 7690 Villa, Thank you for the very clear and understandable post. However, I believe that you have a problem in that the area listed above sould be: PI * r^2. With a 3 inch diameter opening, the wrong calculation leads to an answer 2 times as large as it should be. eg. 1/2 * 3^2 * PI = 14.14 and PI * 1.5^2 = 7.07 -- /---------------------------------------------------------------\ | Lorn H. Olsen, MAA, President | mailto:lorn@michusa.com | | DynaComm, Corp. | Voice: 1-248-478-4301 | | 21042 Laurelwood Street | FAX: 1-248-478-4302 | | Farmington, MI 48336-5052 | | |---------------------------------------------------------------| | for info: mailto:info@dynacomm.ws or: http://www.dynacomm.ws/ | \---------------------------------------------------------------/