X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Sender: To: lml@lancaironline.net Date: Tue, 28 Jun 2005 01:09:46 -0400 Message-ID: X-Original-Return-Path: <210flyer@earthlink.net> Received: from smtpauth05.mail.atl.earthlink.net ([209.86.89.65] verified) by logan.com (CommuniGate Pro SMTP 4.3.5) with ESMTP id 1024387 for lml@lancaironline.net; Tue, 28 Jun 2005 00:56:33 -0400 Received-SPF: pass receiver=logan.com; client-ip=209.86.89.65; envelope-from=210flyer@earthlink.net DomainKey-Signature: a=rsa-sha1; q=dns; c=nofws; s=dk20050327; d=earthlink.net; b=ZNgVy8cwpj0fJe//1PyXzQH3JHsG1tEt3n+d/nUSHNWODfQ/Yx9PCuRozE34cZL4; h=Received:From:To:Subject:Date:MIME-Version:Content-Type:X-Mailer:Thread-Index:X-MimeOLE:Message-ID:X-ELNK-Trace:X-Originating-IP; Received: from [69.3.123.106] (helo=dune) by smtpauth05.mail.atl.earthlink.net with asmtp (Exim 4.34) id 1Dn88R-00079q-CR for lml@lancaironline.net; Tue, 28 Jun 2005 00:55:47 -0400 From: "Mike Hutchins" <210flyer@earthlink.net> X-Original-To: "Lancair Mailing List" Subject: Seatbelt photos X-Original-Date: Mon, 27 Jun 2005 22:55:27 -0600 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_00AC_01C57B6B.4FBD72A0" X-Mailer: Microsoft Office Outlook, Build 11.0.6353 Thread-Index: AcV7nZm6OS+Jpp/MR1GKo0D4SFXmbg== X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.2180 X-Original-Message-ID: X-ELNK-Trace: 96606e18df264d301aa676d7e74259b7b3291a7d08dfec7990d435ac4caf68b7dae92f615f277d59350badd9bab72f9c350badd9bab72f9c350badd9bab72f9c X-Originating-IP: 69.3.123.106 This is a multi-part message in MIME format. ------=_NextPart_000_00AC_01C57B6B.4FBD72A0 Content-Type: text/plain; charset="US-ASCII" Content-Transfer-Encoding: 7bit Hi Bill, I believe the tension in the belt itself is not effected by the mere existence of the pivot, but rather, by the angle between the pivot point and the CG of the torso and relative position of the shoulder, torso weight and the peak deceleration (in your case, 6Gs). The pivot anchor, on the other hand, would be subjected to nearly twice the tension in the belt since it is effectively acted upon by two belts pulling on the pivot simultaneously at a very acute angle. If I remember my mechanics and trig, the problem is similar to that of a suspension bridge. The tension in the cables is much greater than merely the weight supported by the main cables. Similarly, the compression load of the support towers, as in the Golden Gate Bridge, is more than twice that of the weight supported (2 x sin(angle) x tension), but not so nearly as much as twice the tension, due to the rather large angle of the cables as they cross over the towers. Lastly, when I diagram the force vectors, it appears that the lap belt contributes a much greater proportion of the total belt tension than the shoulder belt. Then again, what do I know? I'm not a real engineer, just a wannabe. Best Regards, Mike ------=_NextPart_000_00AC_01C57B6B.4FBD72A0 Content-Type: text/html; charset="US-ASCII" Content-Transfer-Encoding: quoted-printable

Hi Bill,

 

I believe the tension in the belt itself is not = effected by the mere existence of the pivot, but rather, by the angle between the = pivot point and the CG of the torso and relative position of the shoulder, = torso weight and the peak deceleration (in your case, 6Gs). The pivot anchor, = on the other hand, would be subjected to nearly twice the tension in the belt = since it is effectively acted upon by two belts pulling on the pivot = simultaneously at a very acute angle.

 

 If I remember my mechanics and trig, the = problem is similar to that of a suspension bridge. The tension in the cables is = much greater than merely the weight supported by the main cables. Similarly, = the compression load of the support towers, as in the Golden Gate Bridge, is = more than twice that of the weight supported (2 x sin(angle) x tension), but = not so nearly as much as twice the tension, due to the rather large angle of = the cables as they cross over the towers.

 

Lastly, when I diagram the force vectors, it appears = that the lap belt contributes a much greater proportion of the total belt = tension than the shoulder belt. Then again, what do I know? I’m not a real engineer, just a wannabe.

 

Best Regards,

Mike

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