X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from fed1rmmtao101.cox.net ([68.230.241.45] verified) by logan.com (CommuniGate Pro SMTP 5.2.14) with ESMTP id 3647005 for flyrotary@lancaironline.net; Tue, 19 May 2009 13:37:12 -0400 Received-SPF: none receiver=logan.com; client-ip=68.230.241.45; envelope-from=alventures@cox.net Received: from fed1rmimpo02.cox.net ([70.169.32.72]) by fed1rmmtao101.cox.net (InterMail vM.7.08.02.01 201-2186-121-102-20070209) with ESMTP id <20090519173633.OYXY17670.fed1rmmtao101.cox.net@fed1rmimpo02.cox.net> for ; Tue, 19 May 2009 13:36:33 -0400 Received: from BigAl ([72.192.135.181]) by fed1rmimpo02.cox.net with bizsmtp id tVca1b0053uzsQg04VcaZM; Tue, 19 May 2009 13:36:34 -0400 X-Authority-Analysis: v=1.0 c=1 a=U9rFrUpSVtIA:10 a=WFRSiL8JFSQA:10 a=dtUaNHdrd6DvLKhTDLEA:9 a=PTrOjJOnHppMMg1mMvcA:7 a=TgEZ9BaczxRBrk5uo5s3iVCgj7YA:4 a=-4K2CqF1siauUVpaU2oA:9 a=PTGA9YhdZoiukXSvcMIA:7 a=u3ddH_ekLTSSiRmUXLouvr5JMqkA:4 X-CM-Score: 0.00 From: "Al Gietzen" To: "'Rotary motors in aircraft'" Subject: RE: [FlyRotary] Re: Phonic Band gap filter Date: Tue, 19 May 2009 10:36:21 -0800 Message-ID: <6C882814877C40B5818492602A656794@BigAl> MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0002_01C9D86D.A6D6CE20" X-Priority: 3 (Normal) X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook, Build 10.0.6838 Thread-Index: AcnYj0NumLsUpEcfT2aoxFMOAlhcFgAIASUg In-Reply-To: X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.5579 Importance: Normal This is a multi-part message in MIME format. ------=_NextPart_000_0002_01C9D86D.A6D6CE20 Content-Type: text/plain; charset="us-ascii" Content-Transfer-Encoding: quoted-printable > > =20 Area of filler rods divided by total area. =20 I probably computed it wrong. ---- So on that basis I'd estimate your filling ratio at about .13. For a filling ratio of 0.5 with .05 diameter tubes on a square pitch you'd = need a center to center distance on the tubes of 0.63. =20 One square, drawn from the centers of 4 tubes includes the area of one = tube, which is about 0.2. So the total area of the square needs to be 0.4, so = the length of a side is the square root of 0.4, =3D 0.6325. The tubes would = be little more than 1/8" apart. =20 Or am I missing something? =20 Al G =20 =20 ------=_NextPart_000_0002_01C9D86D.A6D6CE20 Content-Type: text/html; charset="us-ascii" Content-Transfer-Encoding: quoted-printable

Area of filler rods divided by total area.

I probably computed it wrong.

----

So on that = basis I’d estimate your filling ratio at about .13. For a filling ratio of = 0.5 with .05 diameter tubes on a square pitch you’d need a center to center distance on the tubes of 0.63.

<= /font>

One square, = drawn from the centers of 4 tubes includes the area of one tube, which is about = 0.2. So the total area of the square needs to be 0.4, so the length of a side is = the square root of 0.4, =3D 0.6325. The tubes would be little more = than 1/8” apart.

<= /font>

Or am I = missing something?

<= /font>

Al = G

<= /font>

------=_NextPart_000_0002_01C9D86D.A6D6CE20--