X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from fed1rmmtao104.cox.net ([68.230.241.42] verified) by logan.com (CommuniGate Pro SMTP 5.2.14) with ESMTP id 3647463 for flyrotary@lancaironline.net; Tue, 19 May 2009 19:25:33 -0400 Received-SPF: none receiver=logan.com; client-ip=68.230.241.42; envelope-from=alventures@cox.net Received: from fed1rmimpo02.cox.net ([70.169.32.72]) by fed1rmmtao104.cox.net (InterMail vM.7.08.02.01 201-2186-121-102-20070209) with ESMTP id <20090519232456.BIBB17135.fed1rmmtao104.cox.net@fed1rmimpo02.cox.net> for ; Tue, 19 May 2009 19:24:56 -0400 Received: from BigAl ([72.192.135.181]) by fed1rmimpo02.cox.net with bizsmtp id tbQv1b0083uzsQg04bQvsc; Tue, 19 May 2009 19:24:55 -0400 X-Authority-Analysis: v=1.0 c=1 a=U9rFrUpSVtIA:10 a=WFRSiL8JFSQA:10 a=dtUaNHdrd6DvLKhTDLEA:9 a=w_kwybvJV-aDLFZIoAgA:7 a=6hXgk_SgKfIjFtrnXbJee5H2TF4A:4 a=aiZU6_R4eC6DQC3lfvwA:9 a=Fm9sQNfiYRiubaq2XMEA:7 a=Pqx6tvZmw0D_4rEGs4Xr2nIDFKUA: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 16:24:42 -0800 Message-ID: MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_000D_01C9D89E.50F22A50" X-Priority: 3 (Normal) X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook, Build 10.0.6838 Importance: Normal Thread-Index: AcnYqHQOcE7j01peS8iF2YvOXHBtawAOH28g X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.5579 In-Reply-To: This is a multi-part message in MIME format. ------=_NextPart_000_000D_01C9D89E.50F22A50 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. Just realized there is a typo in this part - should 0.5 diameter. I'm = sure you noticed that.=20 Al 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_000D_01C9D89E.50F22A50 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.

Just realized = there is a typo in this part – should 0.5 diameter. I’m sure you = noticed that.

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.

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Or am I missing something?

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Al G

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