X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Sender: To: lml@lancaironline.net Date: Wed, 04 Jul 2007 14:43:40 -0400 Message-ID: X-Original-Return-Path: Received: from wind.imbris.com ([216.18.130.7] verified) by logan.com (CommuniGate Pro SMTP 5.1.10) with ESMTPS id 2158122 for lml@lancaironline.net; Wed, 04 Jul 2007 13:53:45 -0400 Received-SPF: none receiver=logan.com; client-ip=216.18.130.7; envelope-from=brent@regandesigns.com Received: from [192.168.1.100] (cbl-238-80.conceptcable.com [207.170.238.80] (may be forged)) (authenticated bits=0) by wind.imbris.com (8.12.11/8.12.11.S) with ESMTP id l64Hqomb037242 for ; Wed, 4 Jul 2007 10:53:06 -0700 (PDT) (envelope-from brent@regandesigns.com) X-Original-Message-ID: <468BDE6F.4070506@regandesigns.com> X-Original-Date: Wed, 04 Jul 2007 10:52:47 -0700 From: Brent Regan User-Agent: Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.7.2) Gecko/20040804 Netscape/7.2 (ax) X-Accept-Language: en-us, en MIME-Version: 1.0 X-Original-To: Lancair Mailing List Subject: Re: ADAHRS TSO Content-Type: multipart/alternative; boundary="------------050002060801030608050607" This is a multi-part message in MIME format. --------------050002060801030608050607 Content-Type: text/plain; charset=us-ascii; format=flowed Content-Transfer-Encoding: 7bit The answer is three. As often happens in real situations, this problem has plenty of extraneous, irrelevant and useless data. All that is needed is the balance beam scale and the pistons. Because the balance beam compares two items it is often viewed as binary device when in fact it is a tristate device where the three states are less than, greater than and equal. With this in mind, the 27 pistons are divided into three groups of nine. Two groups are placed on the scale where an imbalance will indicate the group containing the heavy piston and a balance will indicate the heavy piston is part of the group not on the scale. The heavy group of 9 is divided into three groups of three and the weighing process is repeated, identifying the group of three containing the heavy piston. For the third and final measurement, two of the three pistons from the heavy group are compared on the balance. The balance will indicate the heavy piston or, if it indicates balance, the heavy piston will be the one not on the balance. Wishing all a joyous Independence Day! Regards Brent Regan --------------050002060801030608050607 Content-Type: text/html; charset=us-ascii Content-Transfer-Encoding: 7bit The answer is three.

As often happens in real situations, this problem has plenty of extraneous, irrelevant and useless data. All that is needed is the balance beam scale and the pistons. Because the balance beam compares two items it is often viewed as binary device when in fact it is a tristate device where the three states are less than, greater than and equal. With this in mind, the 27 pistons are divided into three groups of nine. Two groups are placed on the scale where an imbalance will indicate the group containing the heavy piston and a balance will indicate the heavy piston is part of the group not on the scale. The heavy group of 9 is divided into three groups of three and the weighing process is repeated, identifying the group of three containing the heavy piston. For the third and final measurement, two of the three pistons from the heavy group are compared on the balance. The balance will indicate the heavy piston or, if it indicates balance, the heavy piston will be the one not on the balance.

Wishing all a joyous Independence Day!

Regards
Brent Regan

--------------050002060801030608050607--