X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Sender: To: lml@lancaironline.net Date: Fri, 24 Feb 2012 16:36:47 -0500 Message-ID: X-Original-Return-Path: Received: from imr-da02.mx.aol.com ([205.188.105.144] verified) by logan.com (CommuniGate Pro SMTP 5.4.4) with ESMTP id 5414005 for lml@lancaironline.net; Fri, 24 Feb 2012 16:23:40 -0500 Received-SPF: pass receiver=logan.com; client-ip=205.188.105.144; envelope-from=Sky2high@aol.com Received: from mtaomg-da04.r1000.mx.aol.com (mtaomg-da04.r1000.mx.aol.com [172.29.51.140]) by imr-da02.mx.aol.com (8.14.1/8.14.1) with ESMTP id q1OLMp5H029310 for ; Fri, 24 Feb 2012 16:22:51 -0500 Received: from core-mtd002c.r1000.mail.aol.com (core-mtd002.r1000.mail.aol.com [172.29.235.197]) by mtaomg-da04.r1000.mx.aol.com (OMAG/Core Interface) with ESMTP id 135C8E000086 for ; Fri, 24 Feb 2012 16:22:51 -0500 (EST) From: Sky2high@aol.com X-Original-Message-ID: <3594.4dafba5.3c7959aa@aol.com> X-Original-Date: Fri, 24 Feb 2012 16:22:50 -0500 (EST) Subject: Re: [LML] Re: 360 rudder balance X-Original-To: lml@lancaironline.net MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="part1_3594.4dafba5.3c7959aa_boundary" X-Mailer: AOL 9.6 sub 168 X-Originating-IP: [67.175.156.123] x-aol-global-disposition: G X-AOL-SCOLL-SCORE: 0:2:486348320:93952408 X-AOL-SCOLL-URL_COUNT: 0 x-aol-sid: 3039ac1d338c4f47ffab522e --part1_3594.4dafba5.3c7959aa_boundary Content-Type: text/plain; charset="UTF-8" Content-Transfer-Encoding: quoted-printable Content-Language: en Gents, =20 Chris is right. In other words: =20 Interesting. Think about a balance beam scale used by that Justice lady. = =20 First, if the weights were above the pivot point, the scale/system would= =20 be unstable. Notice that the weights hang much below the pivot making the= =20 system stable and, should it be perfect, the beam would stay in whatever= =20 position if the weights were equal. But, such scales have a very slight b= uilt=20 in bias to only be in balance when the beam is level. =20 Let's think about an LNC2 aileron that is hinged on top (pivot) - note that= =20 both the aft and fwd weight center is below the pivot point and is thus = =20 stable when balanced to a "level" position. The fore and aft weighting is= =20 not symmetrical, thus changing from the level position may have an unknown= =20 result because the weight CGs and thus its arms may not change in the same= =20 ratio as they were in the level balance, And, then we attach the push rod= =20 that throws off the level balance also. =20 The elevator on a small tail LNC2 is also hinged at the top and the weights= =20 are below the pivot (not very far), thus it's stable. When displaced, =20 the horn weight distribution CG can alter the fore/aft arm relationship. = =20 Also, the addition of the push rod certainly changes the balance that was d= one =20 at level and without push rods attached. =20 =20 The MKII tail is interesting since the center hinge may have as much fore = =20 and aft weight above the pivot as below - thus, not very stable and perhaps= =20 would stay in any displaced position (depending on friction and balance)=20 until the push rod is attached.=20 =20 The rudder - hmmmmm.... The recommendation was to add X weight to the horn= =20 and half round lead bars to the leading edge. Then, If I remember =20 correctly, balance it to at least 80% (whatever that meant). When mounted= ,=20 exactly what is the angled pivot for the rudder and at what pitch orientat= ion of=20 the fuselage? After all, when the rudder is mounted, most of the fwd=20 weight is above the pivot points. What happens to the balance when the r= udder=20 cables are attached? What about those that installed the highly curved=20 tubing as in the plans and introduced a tremendous amount of friction in t= he=20 rudder system? Note that those of us that straightened the rudder cable= =20 tubing have very little friction in the system. =20 Finally, the control rods and cables actually provide a stabilizing effect = =20 whereas an unbalanced horizontal control surface can induce flutter because= =20 the bias may exacerbate the action/reaction of induced vibration. =20 =20 Balance the control surfaces in whatever way and to whatever degree the =20 construction manual calls for and do not exceed Vne. =20 Grayhawk =20 =20 In a message dated 2/22/2012 11:44:49 A.M. Central Standard Time, =20 chris_zavatson@yahoo.com writes: =20 Consider the vertical position of the CG of the surface relative to the=20 point from which it is being supported during balancing. Assume a surface= =20 that is perfectly symmetric in the z direction. If you remove a hinge pin= =20 (1/4" bolt on the elevator, for example) and hang the the surface by wire r= un=20 through the hinge hole, the CG is below the support point and a balanced= =20 surface will tend to return to level. If the same surface was balanced by= =20 setting it down on hinge pins (360 rudder, for example), the CG would now= =20 be above the support point and the surface will be difficult to keep level= . =20 If truly balanced on the hinge center line, it will stay in any position. = =20 In reality the vertical CG position is not always known. If the surface= =20 will not stay in position, look at which direction it wants to fall when= =20 released from a level position. If the surface does attain a stable posit= ion=20 it is easy to see which end is heavier.=20 =20 Chris Zavatson N91CZ 360std _www.N91CZ.net_ (http://www.n91cz.net/)=20 =20 =20 =20 =20 From: Jim Nordin To: lml@lancaironline.net=20 Sent: Tuesday, February 21, 2012 8:30 AM Subject: [LML] 360 rudder balance =20 =20 =20 Hummmm =E2=80=A6 correct me if I=E2=80=99m wrong. Balance of control surfaces is attained when the component (aileron or=20 elevator for example) hung at the center of rotation and given a perturbat= ion=20 (trailing edge pushed down) minimally returns to the level condition or=20 better a small nose down condition. Any other condition (leading edge high= =20 above level, trailing edge below level) warrants adding weight to make the= =20 nose (leading edge) to settle below level. Nose heavy is the point here. Any other condition may result in flutter. Jim =20 =20 From: Lancair Mailing List [mailto:lml@lancaironline.net] On Behalf Of=20 Charles Brown Sent: Tuesday, February 21, 2012 8:02 AM To: lml@lancaironline.net Subject: [LML] Re: 360 rudder balance =20 =20 They should stay in any position where you let them go. They should not= =20 return to level. If they are out of balance, they will have some preferred= =20 orientation (as you say, will move up or down). =20 =20 Charley Brown =20 MS Aero/Astro Engineering =20 =20 =20 On Feb 20, 2012, at 6:51 AM, Bill Bradburry wrote: I have a question about balancing the flight surfaces. If they are in=20 balance, should they return to level from where ever you move them to, or= =20 should they just stay where you put them. =20 I assume that if they are out of balance they will either move up or down= =20 from level depending on whether you need to add or remove weight. =20 Bill B =20 =20 --part1_3594.4dafba5.3c7959aa_boundary Content-Type: text/html; charset="UTF-8" Content-Transfer-Encoding: quoted-printable Content-Language: en
Gents,
 
Chris is right.  In other words:
 
Interesting.  Think about a balance beam scale used by that Justi= ce=20 lady.  First, if the weights were above the pivot point, the=20 scale/system would be unstable.  Notice that the weights hang much bel= ow=20 the pivot making the system stable and, should it be perfect, the beam woul= d=20 stay in whatever position if the weights were equal.  But, such scales= have=20 a very slight built in bias to only be in balance when the beam is=20 level.
 
Let's think about an LNC2 aileron that is hinged on top (pivot) - note= that=20 both the aft and fwd weight center is below the pivot point and is thu= s=20 stable when balanced to a "level" position.  The fore and aft=20 weighting is not symmetrical, thus changing from the level positi= on=20 may have an unknown result because the weight CGs and thus its arms ma= y not=20 change in the same ratio as they were in the level balance,  And,= then=20 we attach the push rod that throws off the level balance also.
 
The elevator on a small tail LNC2 is also hinged at the top and the we= ights=20 are below the pivot (not very far), thus it's stable.  When displ= aced,=20 the horn weight distribution CG can alter the fore/aft arm relationship.&nb= sp;=20 Also, the addition of the push rod certainly changes the balance that was d= one=20 at level and without push rods attached. 
 
The MKII tail is interesting since the center hinge may have as much f= ore=20 and aft weight above the pivot as below - thus, not very stable and perhaps= =20 would stay in any displaced position (depending on friction and balance) un= til=20 the push rod is attached. 
 
The rudder - hmmmmm.... The recommendation was to add X weight to= the=20 horn and half round lead bars to the leading edge.  Then, If I remembe= r=20 correctly, balance it to at least 80% (whatever that meant).  Whe= n=20 mounted, exactly what is the angled pivot for the rudder and at what p= itch=20 orientation of the fuselage?  After all, when the rudder is mount= ed,=20 most of the fwd weight is above the pivot points.   What hap= pens=20 to the balance when the rudder cables are attached?  What about those = that=20 installed the highly curved tubing as in the plans and introduced a tremend= ous=20 amount of friction in the rudder system?  Note that those of us that= =20 straightened the rudder cable tubing have very little friction in the=20 system.
 
Finally, the control rods and cables actually provide a stabilizing ef= fect=20 whereas an unbalanced horizontal control surface can induce flutter be= cause=20 the bias may exacerbate the action/reaction of induced=20 vibration.  
 
Balance the control surfaces in whatever way and to whatever degree th= e=20 construction manual calls for and do not exceed Vne.
 
Grayhawk
 
In a message dated 2/22/2012 11:44:49 A.M. Central Standard Time,=20 chris_zavatson@yahoo.com writes:
=
Consider the verti= cal=20 position of the CG of the surface relative to the point from which it is = being=20 supported during balancing.  Assume a surface that is= =20 perfectly symmetric in the z direction.  If you remove a hinge = pin=20 (1/4" bolt on the elevator, for example) and hang the the surface by wire= run=20 through the hinge hole, the CG is below the support point and a= =20 balanced surface will tend to return to level.  If the same surface = was=20 balanced by setting it down on hinge pins (360 rudder, for example),= the=20 CG would now be above the support point and the surface will be= =20 difficult to keep level.  If truly balanced on the hinge center line, it will stay in any position. 
In reality the ver= tical CG=20 position is not always known.  If the surface will not stay in posit= ion,=20 look at which direction it wants to fall when released from a level position. = If the=20 surface does attain a stable position it is easy<= /VAR>=20 to see which end is heavier. 
 
Chris Zavatson
N91CZ
360std
 
From: Jim Nordin=20 <panelmaker@earthlink.net>
To: lml@lancaironline.net
<= SPAN=20 style=3D"FONT-WEIGHT: bold">Sent: Tuesday, February 21, 2012 8= :30=20 AM
Subject: [LML] 360 = rudder=20 balance

Hummmm =E2=80=A6 = correct me if=20 I=E2=80=99m wrong.
Balance of contro= l=20 surfaces is attained when the component (aileron or elevator for example)= hung=20 at the center of rotation and given a perturbation (trailing edge pushed = down)=20 minimally returns to the level condition or better a small nose down=20 condition. Any other condition (leading edge high above level, trailing e= dge=20 below level) warrants adding weight to make the nose (leading edge) to se= ttle=20 below level.
Nose heavy is the= point=20 here. Any other condition may result in flutter.
Jim=
=
From: Lancair=20 Mailing List [mailto:lml@lancaironline.net] On Behalf Of Charles Brown
Sent: Tuesday, Februa= ry 21,=20 2012 8:02 AM
To:=20 lml@lancaironline.net
Subject: [LML] Re: 36= 0=20 rudder balance
 
They should stay= in any=20 position where you let them go.  They should not return to level.=20  If they are out of balance, they will have some preferred orientati= on=20 (as you say, will move up or down).
 
Charley=20 Brown
MS Aero/Astro=20 Engineering
 
On Feb 20, 2012,= at 6:51=20 AM, Bill Bradburry wrote:

I have a quest= ion=20 about balancing the flight surfaces.  If they are in balance, should= they=20 return to level from where ever you move them to, or should they just sta= y=20 where you put them.
=  
I=20 assume that if they are out of balance they will either move up or down f= rom=20 level depending on whether you need to add or remove=20 weight.
=  
Bill=20 B
=  
 

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