X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Sender: To: lml@lancaironline.net Date: Tue, 28 Apr 2009 10:00:48 -0400 Message-ID: X-Original-Return-Path: Received: from web33904.mail.mud.yahoo.com ([209.191.69.182] verified) by logan.com (CommuniGate Pro SMTP 5.2.13) with SMTP id 3599020 for lml@lancaironline.net; Tue, 28 Apr 2009 02:26:08 -0400 Received-SPF: none receiver=logan.com; client-ip=209.191.69.182; envelope-from=wfhannahan@yahoo.com Received: (qmail 18646 invoked by uid 60001); 28 Apr 2009 06:25:31 -0000 DomainKey-Signature:a=rsa-sha1; q=dns; c=nofws; s=s1024; d=yahoo.com; h=Message-ID:X-YMail-OSG:Received:X-Mailer:Date:From:Subject:To:MIME-Version:Content-Type; b=quhlfDFB6AMqmaOqZT0HDa+lAXZw3dlAvrHzo2OiZit02tGsqxAdJhB692VJIncoAz1TyBrFEnmnOowv9TJmKYfCQgaDbht2M6nIjxyldtRqCZW9QwXtdY9STX9qQivbiJdTF5ML2/TrsB02ya+j0U7hbSnNqIJ7DS/5ySg2F5I=; X-Original-Message-ID: <800029.17361.qm@web33904.mail.mud.yahoo.com> X-YMail-OSG: N6r0JX4VM1nc5kNHvZ0ZyaYtEXTKsZMkf3G293_y3AaJmIH_o2y5XLPwMGD54MkybhfdVfOIHomP6cfBbIjaNgPoMa_UdoLu.Rsco_9jntCTRJsIvrMcpudnImuyqFQuQHbuPmSM_TI_a9S.qZ6AFddRv2AkU1jWPjiU9rCeT4Iq7eBnNdNCzuw8u5PlIGlWoM6P2MFY03SAlaxEWv.JYZ3PIsZg482_3UqglLXUFIydu6fnEYwVJHoOXMD..7Al6gbTRy20.9UWAlOEr.IRXUx.LnfebM78MoWMXxESwnMa8t1C2npsFyvFVI8NMopZW2r4Ir6iGwFe4suakf68KVcEJsw- Received: from [71.208.5.43] by web33904.mail.mud.yahoo.com via HTTP; Mon, 27 Apr 2009 23:25:31 PDT X-Mailer: YahooMailClassic/5.2.19 YahooMailWebService/0.7.289.1 X-Original-Date: Mon, 27 Apr 2009 23:25:31 -0700 (PDT) From: Bill Hannahan Subject: Re: [LML] Re: Balancing MkII Elevators X-Original-To: Lancair Mailing List MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="0-496233247-1240899931=:17361" --0-496233247-1240899931=:17361 Content-Type: text/plain; charset=utf-8 Content-Transfer-Encoding: quoted-printable =0A=0A=0A=0AI requested from Lancair agreement to make this change in=0A199= 1. It was rejected without explanation. =0A=0A=C2=A0=0A=0AMaybe just CYA, o= r an issue with torsional stiffness of the=0Arudder, or a change in vertica= l fin resonant frequency, I don=E2=80=99t know. You are the engineer. Good = luck.=0A=0A=C2=A0=0A=0A Regards, Bill Hannahan =20 wfhannahan@yahoo.com --- On Thu, 4/23/09, Bob Smiley wrote: From: Bob Smiley Subject: [LML] Re: Balancing MkII Elevators To: lml@lancaironline.net Date: Thursday, April 23, 2009, 6:40 PM =0A=0A =0A =0A=0AA suggestion,=0A=C2=A0=0AI did this on my 360 on the verti= dal =0Astabilizer.=C2=A0 The couterweights were placed on the leading edge = of the =0Arudder and in the counterweight overhang on top of the rudder.=C2= =A0 The moment =0Aarm on the leadig edge was about 2 inches verses about 6 = inches thus you need =0Aonly about 1/3rd the weight on the top of the rudde= r.=C2=A0 My solution was to =0Aadd 1.5 inches to the height of the rudder a= nd vertical stab.=C2=A0 More room =0Afor more lead. overall weight was redu= ced.=0A=C2=A0=0AAn addditional benefit I received was I increased =0Athe de= monstrated cross wind component for landing.=C2=A0 I landed in a 24 knot = =0Acrosswind at Jackson Hole Wyoming one time. =0A=C2=A0=0ABob Smiley=0A=0A= -=0A=0A=0A --0-496233247-1240899931=:17361 Content-Type: text/html; charset=utf-8 Content-Transfer-Encoding: quoted-printable

=0A=0A= =0A=0A

I requested from Lancair agreement to make thi= s change in=0A1991. It was rejected without explanation.

=0A=0A

 

=0A=0A

Maybe just= CYA, or an issue with torsional stiffness of the=0Arudder, or a change in = vertical fin resonant frequency, I don=E2=80=99t know. You are the engineer= . Good luck.

=0A=0A

 

=0A=0A
Regards,
Bill Hannahan


--- On Thu, 4/23/09, Bob Smiley <rsmiley2@cent= urytel.net> wrote:

From: Bob Smi= ley <rsmiley2@centurytel.net>
Subject: [LML] Re: Balancing MkII El= evators
To: lml@lancaironline.net
Date: Thursday, April 23, 2009, 6:4= 0 PM

=0A=0A =0A =0A=0A
= A suggestion,
=0A
 
=0A
I did this on my 360 on the vertidal =0Astabilizer.  The couter= weights were placed on the leading edge of the =0Arudder and in the counter= weight overhang on top of the rudder.  The moment =0Aarm on the leadig= edge was about 2 inches verses about 6 inches thus you need =0Aonly about = 1/3rd the weight on the top of the rudder.  My solution was to =0Aadd = 1.5 inches to the height of the rudder and vertical stab.  More room = =0Afor more lead. overall weight was reduced.
=0A
 
=0A
An addditional benefit I received was I increased =0Athe demonstrate= d cross wind component for landing.  I landed in a 24 knot =0Acrosswin= d at Jackson Hole Wyoming one time.
=0A
 
=0A
Bo= b Smiley
=0A
=0A
-

=0A=0A = --0-496233247-1240899931=:17361--