X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Sender: To: lml@lancaironline.net Date: Mon, 09 Aug 2010 07:33:19 -0400 Message-ID: X-Original-Return-Path: Received: from imr-db02.mx.aol.com ([205.188.91.96] verified) by logan.com (CommuniGate Pro SMTP 5.3.8) with ESMTP id 4423188 for lml@lancaironline.net; Sun, 08 Aug 2010 20:54:04 -0400 Received-SPF: pass receiver=logan.com; client-ip=205.188.91.96; envelope-from=Sky2high@aol.com Received: from imo-da02.mx.aol.com (imo-da02.mx.aol.com [205.188.169.200]) by imr-db02.mx.aol.com (8.14.1/8.14.1) with ESMTP id o790rFLK023273 for ; Sun, 8 Aug 2010 20:53:15 -0400 Received: from Sky2high@aol.com by imo-da02.mx.aol.com (mail_out_v42.9.) id q.cf1.781d116d (43989) for ; Sun, 8 Aug 2010 20:53:11 -0400 (EDT) Received: from magic-d27.mail.aol.com (magic-d27.mail.aol.com [172.19.146.161]) by cia-dd05.mx.aol.com (v129.4) with ESMTP id MAILCIADD056-abd54c5f51762d8; Sun, 08 Aug 2010 20:53:10 -0400 From: Sky2high@aol.com X-Original-Message-ID: X-Original-Date: Sun, 8 Aug 2010 20:53:10 EDT Subject: Re: [LML] Re: flap coupling X-Original-To: lml@lancaironline.net MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="part1_c50c9.882ffe1.3990ab76_boundary" X-Mailer: AOL 9.5 sub 155 X-AOL-ORIG-IP: 24.15.17.119 X-AOL-IP: 172.19.146.161 X-Spam-Flag:NO X-AOL-SENDER: Sky2high@aol.com --part1_c50c9.882ffe1.3990ab76_boundary Content-Type: text/plain; charset="ISO-8859-1" Content-Transfer-Encoding: quoted-printable Uh, the push rod forces should be zero when trimmed. If one cannot reach= =20 a trimmed configuration, then force will be required to reach sustained = =20 level flight. One can only wonder about the position of the trimming dev= ice=20 (there are so many different methods) when one then calculates forces=20 necessary for level flight at different airspeeds/configurations. =20 Wolfgang is seeking the "drag bucket" for different flight regimes. The= =20 purpose is unknown. Each configuration change affects either lift (induc= ed=20 drag) or parasitic drag or both. Faster =3D less induced drag, more=20 parasitic drag. Slower =3D more induced drag, less parasitic. Parasites= are=20 everywhere. =20 _http://www.charlesriverrc.org/articles/asfwpp/lelke_airfoilperf.htm_=20 (http://www.charlesriverrc.org/articles/asfwpp/lelke_airfoilperf.htm) cla= rifies =20 the "drag bucket" concept. Good (an extra "o" converts God to good) Is = =20 only of concern at cruise configurations. Why? Because anything else is= =20 confounded by other variables - density altitude, wind, efficiency, etc.= The=20 designer defined the cruise range as the best conditions (altitude, power= ,=20 etc) where the longeron was level. Other things can affect drag, engine= =20 cooling, laminar flow because of smooth surfaces, weight (lift-induced dr= ag),=20 wax (parasitic drag), etc. etc. etc. =20 =20 Who cares at other speeds less than cruise - we know that max efficiency= =20 can be reached when parasitic drag and induced drag cross at some minima.= =20 Uh, the old max range vs max endurance question. Frequently, best=20 efficiency occurs at best glide speed (like 107 KIAS in a half loaded 320= ). So=20 what? Do I care if I can reach Austin, TX in 8 hours using only 20 gallo= ns or=20 4.3 hours using 30 gallons or 4.8 hours at best power requiring a fuel=20 stop to maintain minimums (43 gal tank). Of course. But I don't need=20 anything more than ROP/LOP fuel burns and associated TAS - fortunately fo= r my very=20 slick bird, there is only a loss of 6 or 7 knots for a drop of 2 gph from= =20 ROP to LOP at some useful altitude. So, I get >1 hour more endurance at= =20 LOP and I can see if that 28 NM difference (4 hours) is worth the 1 hour= =20 refueling stop. Uh, Austin is a flip of the coin at 820 NM (wind and wea= ther =20 depending). =20 Scott Krueger LNC2 320 =20 =20 =20 In a message dated 8/8/2010 6:46:31 P.M. Central Daylight Time, =20 chris_zavatson@yahoo.com writes: =20 The MKII tail is a little different. Push rod forces are zero for all=20 trimmed conditions. =20 Chris Zavatson =20 N91CZ =20 360std _www.N91CZ.com_ (http://www.n91cz.com/)=20 =20 ____________________________________ From: Wolfgang To: lml@lancaironline.net Sent: Fri, August 6, 2010 10:06:44 PM Subject: [LML] Re: flap coupling I have taken elevator pushrod force measurements and was surprised. Elevator pushrod forces to stick forces are about 6.5 to 1 The trim system, when dialed in, provides these forces. =20 At 190 imph and -7=BA flaps, there is a 60lb forward force. At 80 imph and 10=BA flaps, there is about zero force. At 80 imph and 20=BA flaps, there is a slight (-1lb) rearward force. =20 These numbers are with the horizontal stabilizer built at -1.2=BA - - - plans range is -0.5=BA to -1.0=BA =20 An input from the flap bellcrank of about 20-40 lb at -7=BA would be good= , tapering down to zero lbs at 10=BA flaps =20 A horizontal stabilizer built at -0.5=BA would, of course, change these= =20 numbers. =20 Comments ? =20 Wolfgang =20 --part1_c50c9.882ffe1.3990ab76_boundary Content-Type: text/html; charset="ISO-8859-1" Content-Transfer-Encoding: quoted-printable <= FONT id=3Drole_document color=3D#000000 size=3D2 face=3DArial>
Uh, the push rod forces should be zero when trimmed.  If one can= not=20 reach a trimmed configuration, then force will be required to reach sustai= ned=20 level flight.  One can only wonder about the position of the trimming= =20 device (there are so many different methods) when one then=20 calculates forces necessary for level flight at different=20 airspeeds/configurations.
 
Wolfgang is seeking the "drag bucket" for different flight regimes.&n= bsp;=20 The purpose is unknown.  Each configuration change affects either lif= t=20 (induced drag) or parasitic drag or both.  Faster =3D less induc= ed drag,=20 more parasitic drag.  Slower =3D more induced drag, less parasitic.&n= bsp;=20 Parasites are everywhere.
 
http://www.charlesriverrc.org/articles/asfwpp/lelke_airfoilpe= rf.htm clarifies=20 the "drag bucket" concept.  Good (an extra "o" converts God to good)= Is=20 only of concern at cruise configurations.  Why?  Because anythin= g else=20 is confounded by other variables - density altitude, wind, efficiency,=20 etc.  The designer defined the cruise range as the best conditions=20 (altitude, power, etc) where the longeron was level.  Other things ca= n=20 affect drag, engine cooling, laminar flow because of smooth surfaces, weig= ht=20 (lift-induced drag), wax (parasitic drag), etc. etc. etc. 
 
Who cares at other speeds less than cruise - we know that max=20 efficiency can be reached when parasitic drag and induced drag cross at so= me=20 minima.  Uh, the old max range vs max endurance question.  =20 Frequently, best efficiency occurs at best glide speed (like 107 KIAS in= a half=20 loaded 320).  So what?  Do I care if I can reach Austin, TX= =20 in 8 hours using only 20 gallons or 4.3 hours using 30 gallons or&nbs= p;4.8=20 hours at best power requiring a fuel stop to maintain minimums (43 gal=20 tank).  Of course.  But I don't need anything more than ROP/LOP= fuel=20 burns and associated TAS - fortunately for my very slick bird, there= is=20 only a loss of 6 or 7 knots for a drop of 2 gph from ROP to LOP at some us= eful=20 altitude.  So, I get >1 hour more endurance at LOP= and I=20 can see if that 28 NM difference (4 hours) is worth the 1 hour refuel= ing=20 stop.  Uh, Austin is a flip of the coin at 820 NM (wind and weather= =20 depending).
 
Scott Krueger
LNC2 320     
 
In a message dated 8/8/2010 6:46:31 P.M. Central Daylight Time,=20 chris_zavatson@yahoo.com writes:
www.N91CZ.com

 
From: Wolfgang=20 <Wolfgang@MiCom.net>
T= o: lml@lancaironline.net
Sent: Fri, August 6, 2010 10:06:44=20 PM
Subject: [LML] Re:= flap=20 coupling

I have taken elevator pushrod force mea= surements=20 and was surprised.
Elevator pushrod forces to stick forces= are about=20 6.5 to 1
The trim system, when dialed in, provid= es these=20 forces.
 
At 190 imph and -7=BA flaps, there= is a 60lb=20 forward force.
At 80 imph and 10=BA flaps, there is ab= out zero=20 force.
At 80 imph and 20=BA flaps, there is a= slight=20 (-1lb) rearward force.
 
These numbers are with the horizontal= stabilizer=20 built at -1.2=BA
- - - plans range is -0.5=BA to -1.0=BA=
 
An input from the flap bellcrank= of about=20 20-40 lb at -7=BA would be good,
 tapering down to zero lbs at 10= =BA=20 flaps
 
A horizontal stabilizer built at -0.5= =BA would, of=20 course, change these numbers.
 
Comments ?
 
Wolfgang
 
=
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