From owner-freebsd-questions@FreeBSD.ORG Sat Jun 28 16:56:55 2008 Return-Path: Delivered-To: freebsd-questions@freebsd.org Received: from mx1.freebsd.org (mx1.freebsd.org [IPv6:2001:4f8:fff6::34]) by hub.freebsd.org (Postfix) with ESMTP id 1C4CF1065678 for ; Sat, 28 Jun 2008 16:56:55 +0000 (UTC) (envelope-from prad@towardsfreedom.com) Received: from pd2mo2so.prod.shaw.ca (idcmail-mo1so.shaw.ca [24.71.223.10]) by mx1.freebsd.org (Postfix) with ESMTP id E76658FC13 for ; Sat, 28 Jun 2008 16:56:54 +0000 (UTC) (envelope-from prad@towardsfreedom.com) Received: from pd4mr5so.prod.shaw.ca (pd4mr5so-qfe3.prod.shaw.ca [10.0.141.50]) by l-daemon (Sun ONE Messaging Server 6.0 HotFix 1.01 (built Mar 15 2004)) with ESMTP id <0K3600GV1N1UXXB0@l-daemon> for freebsd-questions@freebsd.org; Sat, 28 Jun 2008 10:56:18 -0600 (MDT) Received: from pn2ml10so.prod.shaw.ca ([10.0.121.80]) by pd4mr5so.prod.shaw.ca (Sun Java System Messaging Server 6.2-7.05 (built Sep 5 2006)) with ESMTP id <0K3600HBIN1UKK30@pd4mr5so.prod.shaw.ca> for freebsd-questions@freebsd.org; Sat, 28 Jun 2008 10:56:18 -0600 (MDT) Received: from gom.home ([70.67.160.176]) by l-daemon (Sun ONE Messaging Server 6.0 HotFix 1.01 (built Mar 15 2004)) with ESMTP id <0K3600K55N1TMK10@l-daemon> for freebsd-questions@freebsd.org; Sat, 28 Jun 2008 10:56:17 -0600 (MDT) Received: from gom.home (localhost [127.0.0.1]) by gom.home (Postfix) with ESMTP id 62F4DB842 for ; Sat, 28 Jun 2008 09:56:13 -0700 (PDT) Date: Sat, 28 Jun 2008 09:56:13 -0700 From: prad In-reply-to: <200806281302.20814.pieter@degoeje.nl> To: freebsd-questions@freebsd.org Message-id: <20080628095613.38e92182@gom.home> MIME-version: 1.0 X-Mailer: Claws Mail 3.3.1 (GTK+ 2.12.1; i386-portbld-freebsd7.0) Content-type: text/plain; charset=US-ASCII Content-transfer-encoding: 7bit References: <20080628005702.2137bb8c@gom.home> <200806281302.20814.pieter@degoeje.nl> Subject: Re: first pre-emptive raid X-BeenThere: freebsd-questions@freebsd.org X-Mailman-Version: 2.1.5 Precedence: list List-Id: User questions List-Unsubscribe: , List-Archive: List-Post: List-Help: List-Subscribe: , X-List-Received-Date: Sat, 28 Jun 2008 16:56:55 -0000 On Sat, 28 Jun 2008 13:02:20 +0200 Pieter de Goeje wrote: > Parity is calculated using the following formula: > pieter, that is absolutely beautiful!! it was really bothering me how you can recover data that really wasn't 'there'. my son and i just worked out the mechanism with some nibbles: 0110 d0 0011 d1 0010 d2 ==== 0111 p so 0111 p 0111 p 0111 p 0011 d1 0110 d0 0110 d0 0010 d2 0010 d2 0011 d1 ==== ==== ==== 0110 d0 0011 d1 0010 d2 and just extend the concept from nibbles to blocks. why in diagram 20-3 of the handbook do they show 2 parity blocks though for disk3 and disk4? why would you ever have more than 1 for any single disk? -- In friendship, prad ... with you on your journey Towards Freedom http://www.towardsfreedom.com (website) Information, Inspiration, Imagination - truly a site for soaring I's