X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from wproxy.gmail.com ([64.233.184.203] verified) by logan.com (CommuniGate Pro SMTP 5.0.6) with ESMTP id 934573 for flyrotary@lancaironline.net; Fri, 20 Jan 2006 12:52:22 -0500 Received-SPF: pass receiver=logan.com; client-ip=64.233.184.203; envelope-from=russell.duffy@gmail.com Received: by wproxy.gmail.com with SMTP id i22so542741wra for ; Fri, 20 Jan 2006 09:51:37 -0800 (PST) DomainKey-Signature: a=rsa-sha1; q=dns; c=nofws; s=beta; d=gmail.com; h=received:from:to:subject:date:message-id:mime-version:content-type:x-priority:x-msmail-priority:x-mailer:importance:x-mimeole; b=qgjJcRQ7jiPsmEThJHpVopNiNVxcB7RUekq3j4JJWsqvo9/SAQjm1cfEeMigOZtlHI0CXiZxB0YTDiKi/QTXZhCWh5USuCj7UjOHkpDE2m1y2vOQbqzAPvZPAKgr93xLaSb6Lp19dfg/MonvT0Hs0sFXdJ68MrTilwvoKM8xO/Y= Received: by 10.54.117.12 with SMTP id p12mr2714489wrc; Fri, 20 Jan 2006 09:51:36 -0800 (PST) Return-Path: Received: from rd ( [65.6.194.9]) by mx.gmail.com with ESMTP id 24sm3946510wrl.2006.01.20.09.51.36; Fri, 20 Jan 2006 09:51:36 -0800 (PST) From: "Russell Duffy" To: "Flyrotary List" Subject: flywheel weight? Date: Fri, 20 Jan 2006 11:51:36 -0600 Message-ID: <000c01c61dea$28867110$6101a8c0@rd> MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_000D_01C61DB7.DDEC0110" X-Priority: 3 (Normal) X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook, Build 10.0.6626 Importance: Normal X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.2527 This is a multi-part message in MIME format. ------=_NextPart_000_000D_01C61DB7.DDEC0110 Content-Type: text/plain; charset="us-ascii" Content-Transfer-Encoding: quoted-printable Greetings, =20 Yes, I'm trying to avoid spending too much time on this, but I'm tempted = to try adding more significant flywheel weight to the single rotor, to see = it it will behave better. I realized this isn't an ideal solution to an already heavy engine, but it might still be worth trying. =20 =20 My question is how much weight should I add to make sure this will get = rid of the torque reversal issues? =20 =20 I understand that weight added farther out from the center has more = effect than weight close to the center. Ideally, I would plan to make a ring, = with about an 11" OD (or as large as I can without interfering with the = starter ring), and about 7" ID. This would bolt to the redrive side of the = dampener plate, using the existing 7/16" bolts. I made an estimate of the = weight of steel, by measuring the volume of a 4130 plate, and weighing it. The = result was 4.5 oz per cu in. My best estimate is that I can get close to 10 = lbs, using a half inch thick piece of steel. Unfortunately, this will be = quite time consuming, and tough to fabricate, balance, etc. =20 =20 The other option is to try to add more weight than I did before at the 4 bolt locations. This is a much easier plan to implement, but I'm afraid I'll be lucky to get 5 lbs added using this method. =20 =20 Anyone want to take a swag at how well this would work? =20 =20 Thanks, Rusty (resistance is futile) ------=_NextPart_000_000D_01C61DB7.DDEC0110 Content-Type: text/html; charset="us-ascii" Content-Transfer-Encoding: quoted-printable Message
Greetings,
 
Yes, I'm trying = to avoid=20 spending too much time on this, but I'm tempted to try adding more = significant=20 flywheel weight to the single rotor, to see it it will behave = better.  I=20 realized this isn't an ideal solution to an already heavy engine, but it = might=20 still be worth trying. 
 
My question is = how much=20 weight should I add to make sure this will get rid of the torque = reversal=20 issues? 
 
I understand = that weight=20 added farther out from the center has more effect than weight close to = the=20 center.  Ideally, I would plan to make a ring, with about an 11" OD = (or as=20 large as I can without interfering with the starter ring), and about 7"=20 ID.  This would bolt to the redrive side of the dampener plate, = using the=20 existing 7/16" bolts.   I made an estimate of the weight of = steel, by=20 measuring the volume of a 4130 plate, and weighing it.  The result = was 4.5=20 oz per cu in.  My best estimate is that I can get close to 10 lbs, = using a=20 half inch thick piece of steel.  Unfortunately, this will be quite = time=20 consuming, and tough to fabricate, balance, etc.  =
 
The other = option is to try=20 to add more weight than I did before at the 4 bolt locations.  This = is a=20 much easier plan to implement, but I'm afraid I'll be lucky to get 5 lbs = added=20 using this method. 
 
Anyone want to = take a swag=20 at how well this would work? 
 
Thanks,
Rusty = (resistance is=20 futile)
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