X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from omr-d02.mx.aol.com ([205.188.109.194] verified) by logan.com (CommuniGate Pro SMTP 6.0.6) with ESMTPS id 6442272 for flyrotary@lancaironline.net; Mon, 26 Aug 2013 14:55:00 -0400 Received-SPF: pass receiver=logan.com; client-ip=205.188.109.194; envelope-from=Lehanover@aol.com Received: from mtaomg-mb03.r1000.mx.aol.com (mtaomg-mb03.r1000.mx.aol.com [172.29.41.74]) by omr-d02.mx.aol.com (Outbound Mail Relay) with ESMTP id D7B7270000084 for ; Mon, 26 Aug 2013 14:54:25 -0400 (EDT) Received: from core-mob005b.r1000.mail.aol.com (core-mob005.r1000.mail.aol.com [172.29.194.209]) by mtaomg-mb03.r1000.mx.aol.com (OMAG/Core Interface) with ESMTP id 9AF4EE000081 for ; Mon, 26 Aug 2013 14:54:25 -0400 (EDT) From: Lehanover@aol.com Full-name: Lehanover Message-ID: Date: Mon, 26 Aug 2013 14:54:25 -0400 (EDT) Subject: Re: [FlyRotary] Re: Injector Location To: flyrotary@lancaironline.net MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="part1_edb6e.47140976.3f4cfe61_boundary" X-Mailer: AOL 9.7 sub 1028 X-Originating-IP: [173.88.5.72] x-aol-global-disposition: G DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=mx.aol.com; s=20121107; t=1377543265; bh=8EiIy/K5qlRYaAXwHhQnoH0dQ7ltRSna4DXqZqqeUng=; h=From:To:Subject:Message-ID:Date:MIME-Version:Content-Type; b=oI242veehIoKuPJseEQRXcrg5YNTyCJIGnUAKDe23FIPojASwvxviBAK1s5PDX1yW YFKFC5DmllpaVcvWGxBm0dBer/yta3FaFOdCeQvNT+Z8+cv3wU+HKvip1ujk69xSQa aKLCVa+/B2wlkCqXoyyNn2oDeeKsMLXSVE92X+e4= x-aol-sid: 3039ac1d294a521ba4614e8e --part1_edb6e.47140976.3f4cfe61_boundary Content-Type: text/plain; charset="UTF-8" Content-Transfer-Encoding: quoted-printable Content-Language: en If done over some distance, it would. The shapes blended as if by a=20 sculptor.=20 =20 Lynn E. Hanover =20 =20 In a message dated 8/26/2013 2:38:56 P.M. Eastern Daylight Time, =20 tmann@n200lz.com writes: Could I assume that if my port opening (which is 2=E2=80=9D x .78=E2=80=9D = on the =20 primaries) consisted of a inside perimeter of 4.89=E2=80=9D and an area of = 1.43 square =20 inches and I used a tube with an internal diameter of 1.55 inches (for a to= tal =20 area of 1.9 square inches) and then changed the shape of the tube to mate= =20 with the port dimensions, that the reduction of area from 1.9 s.i. to 1.42= =20 s.i would in effect achieve the tapered effect you referred to below? =20 Tom Mann =20 --part1_edb6e.47140976.3f4cfe61_boundary Content-Type: text/html; charset="UTF-8" Content-Transfer-Encoding: quoted-printable Content-Language: en
If done over some distance, it= =20 would. The shapes blended as if by a sculptor.= =20
 
Lynn E. Hanover
 
In a message dated 8/26/2013 2:38:56 P.M. Eastern Daylight Time,=20 tmann@n200lz.com writes:
=
Could I assume that if my port opening (which is 2=E2=80=9D x .78=E2= =80=9D on the=20 primaries) consisted of a inside perimeter of 4.89=E2=80=9D and an area o= f 1.43 square=20 inches and I used a tube with an internal diameter of 1.55 inches (for a = total=20 area of 1.9 square inches) and then changed the shape of the tube to mate= with=20 the port dimensions, that the reduction of area from 1.9 s.i. to 1.42 s.i= =20 would in effect achieve the tapered effect you referred to below?
 
Tom Mann
 
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