X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Sender: To: lml@lancaironline.net Date: Wed, 11 Aug 2010 19:21:46 -0400 Message-ID: X-Original-Return-Path: Received: from web36902.mail.mud.yahoo.com ([209.191.85.70] verified) by logan.com (CommuniGate Pro SMTP 5.3.8) with SMTP id 4427099 for lml@lancaironline.net; Wed, 11 Aug 2010 15:49:54 -0400 Received-SPF: none receiver=logan.com; client-ip=209.191.85.70; envelope-from=chris_zavatson@yahoo.com Received: (qmail 21283 invoked by uid 60001); 11 Aug 2010 19:49:19 -0000 DomainKey-Signature:a=rsa-sha1; q=dns; c=nofws; s=s1024; d=yahoo.com; h=Message-ID:X-YMail-OSG:Received:X-Mailer:References:Date:From:Subject:To:In-Reply-To:MIME-Version:Content-Type; b=aCexOfFmQPu6FoA3w3Dt12nT3ktkgfZwAU2ITbv2YMiI7DmhS4Wtf/L+qoWIBQaRNX4dLbQnN7GKOGGd1KGUGtcgAL65xs1QuT8E7zIpNT+nq0LjrfXOaUFnYdWThYzV6tQb7tRC/pkuKpM5sNjoFVOVd6nXYtGelclP1gToYGM=; X-Original-Message-ID: <409833.20958.qm@web36902.mail.mud.yahoo.com> X-YMail-OSG: TKeY3FcVM1lCb6jaFPiCaBr25BP8i4z4gUrNcDTgCm0Mmh8 drqDB..3JbzePxBPb_Nm94DDSDIFD3VBW83ICJbRkTDQREhOfLaSmJYvLZlV CBfdQrRPvC9ILAqj5fXh3FO4YqzHJfGnrsblO1XEZr4jciTQPlgHHH2LNKkh A6OHH.ILrxjA8wgK6x7FR7REyzMxBt.k4cTATNy9Knguy4UhO9GT75P4G5j3 64IhZCy4XOOrz8jPhQwavFu48nGIMIr4XEkQG33OcGd2P1xFXqyLbftblpG9 IdOA_megBfsZrInT84.teiia7jnX39sEQjbZDF2_g9VU9RlhYO2sp64ryFRS pRv2Zhlx_KcwS58eXlsmH7A9.ihmZFLCt87jkVu0- Received: from [149.32.224.33] by web36902.mail.mud.yahoo.com via HTTP; Wed, 11 Aug 2010 12:49:19 PDT X-Mailer: YahooMailRC/459 YahooMailWebService/0.8.105.279950 References: X-Original-Date: Wed, 11 Aug 2010 12:49:19 -0700 (PDT) From: Chris Zavatson Subject: Re: [LML] Re: flap coupling, LNC2 flap implications X-Original-To: Lancair Mailing List In-Reply-To: MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="0-1709958904-1281556159=:20958" --0-1709958904-1281556159=:20958 Content-Type: text/plain; charset=iso-8859-1 Content-Transfer-Encoding: quoted-printable Scott,=0AWhen the airfoil was tested, M 0.1 was likely used to eliminate co= mpressibility =0Aeffects.=A0 Note that the Reynolds numbers are in the appr= opriate range for our =0Aapplication.=A0 There is also a chart showing some= compressibility effects.=A0=A0 When =0Atesting, temperature and/or pressur= e are changed to vary Reynolds numbers=A0while =0Amaintaining a constant Ma= ch number.=A0 The 0.1 Mach number is not meant to =0Aimply=A0usage at 66 kt= s=0Ahttp://en.wikipedia.org/wiki/Reynolds_number=0AChris=0A=0A=0AChris Zava= tson=0AN91CZ=0A360std=0Awww.N91CZ.com=0A=0A=0A=0A=0A_______________________= _________=0AFrom: "Sky2high@aol.com" =0ATo: lml@lancairon= line.net=0ASent: Tue, August 10, 2010 1:52:05 PM=0ASubject: [LML] Re: flap = coupling, LNC2 flap implications=0A=0AChris and Wolfgang,=0A=0AThe airfoil = report only deals with a section, not the whole wing with its =0Awashout or= the whole airplane,=A0etc.=A0 Furthermore, while figure 12 indeed =0Ademon= strates drag reduction for a section in reflex, the notation claims it=A0is= =0Aat .1 Mach or about 66 Knots.=A0 I can't get my head around that since = that is =0Atouchdown speed where the flaps=A0should be out of reflex at lea= st.=A0 =0A=0A=0AExperience with my airplane=A0tells me that full reflex (pe= rhaps up to -10 degrees =0Ain flight out at the twisted flap end) minimizes= drag in the speed range above =0Aabout=A0160 KIAS.=A0 Below those speeds= =A0a nose up=A0attitude (and AOA) begins to creep =0Ain to maintain lift.= =A0 By the time 120 KIAS is reached=A0my flaps are in takeoff =0Aposition (= 10 degrees down from full reflex).=A0 Why?=A0 Because the nose up =0Aattitu= de=A0would be perhaps 6 degrees higher if no flaps were used.=A0 I have nev= er =0Ameasured the power or speed difference needed to maintain level fligh= t=A0between =0Athe two flap positions (drag indicators for the same lift) b= ecause speeds below =0A120 are only flown on an approach, in the pattern or= at OSH.=A0 =0A=0A=0AI continue to claim that the LNC2 flaps are a fourth c= ontrol surface that has =0Aimportant=A0drag implications and significant pi= tch consequences - certainly more =0Apronounced than that of=A0any spam can= .=A0 =A0=A0=A0=0A=0AScott=0A=0APS - I certainly like the LNC2 continuous fl= ap position capability rather than =0Afixed detents.=A0 Continuous position= ing=A0fits its role as another control surface.=0A=0AIn a message dated 8/9= /2010 6:33:38 P.M. Central Daylight Time, =0Achris_zavatson@yahoo.com write= s:=0AWolfgang,=0A>The MkII tail uses a servo to drive the trim tab.=A0 If i= nstalled, the bob-weight =0A>is the only thing contributing to any static f= orce in rear half of the system.=A0 =0A>If an autopilot is fighting an out = of trim condition that would=A0add =0A>some=A0force-much like the old trim = system does.=A0 Outside of that all trim forces =0A>for the large tail are = self-contained in the elevator.=0A>Attached are some charts showing the ben= efits of reflex relative to the =0A320/360.=0A>=0A>Chris Zavatson=0A>N91CZ= =0A>360std=0A>www.N91CZ.com=0A>=0A>=0A>=0A>=0A_____________________________= ___=0AFrom: Wolfgang =0A>To: lml@lancaironline.net=0A>S= ent: Mon, August 9, 2010 12:04:51 PM=0A>Subject: [LML] Re: flap coupling=0A= >=0A>=0A>The push rod forces are definitely there. =0A>The trim system that= keeps those forces from showing up at the control stick.=0A>. . . . unless= you're using servo tabs . . . .=0A>=A0=0A>Yes, I want to map the drag buck= et for various flap conditions. =0A>NASA tech paper 1865 shows it's effect.= I want to expand that on the 300 =0Aseries.=0A>I believe it can add some e= fficiency points if utilized.=0A>=A0=0A>Wolfgang=0A>=0A____________________= ____________=0A=0A>From: Sky2high@aol.com =0A>Sender: =0A>Subject: Re: [LML] Re: flap coupling =0A>Date: Mon, 09 Aug 2010 07:= 33:19 -0400 =0A>To: lml@lancaironline.net =0A>Uh, the push rod forces sho= uld be zero when trimmed.=A0 If one cannot reach a =0A>trimmed configuratio= n, then force will be required to reach sustained level =0A>flight.=A0 One = can only wonder about the position of the trimming device (there =0A>are so= many different methods) when one then calculates=A0forces necessary for = =0A>level flight at different airspeeds/configurations.=0A>=0A>Wolfgang is = seeking the "drag bucket" for different flight regimes.=A0 The purpose =0A>= is unknown.=A0 Each configuration change affects either lift (induced drag)= =A0or =0A>parasitic drag or both.=A0 Faster =3D less induced drag, more par= asitic drag.=A0 =0A>Slower =3D more induced drag, less parasitic.=A0 Parasi= tes are everywhere.=0A>=0A>http://www.charlesriverrc.org/articles/asfwpp/le= lke_airfoilperf.htm=A0clarifies =0A>the "drag bucket" concept.=A0 Good (an = extra "o" converts God to good) Is only of =0A>concern at cruise configurat= ions.=A0 Why?=A0 Because anything else is confounded by =0A>other variables= - density altitude, wind, efficiency, etc.=A0 The designer defined =0A>the= cruise range as the best conditions (altitude, power, etc) where the =0A>l= ongeron was level.=A0 Other things can affect drag, engine cooling, laminar= flow =0A>because of smooth surfaces, weight (lift-induced drag), wax (para= sitic drag), =0A>etc. etc. etc.=A0 =0A>=0A>=0A>Who cares at other speeds le= ss than cruise=A0- we know that max efficiency can be =0A>reached when para= sitic drag and induced drag cross at some minima.=A0 Uh, the old =0A>max ra= nge vs max endurance question. =A0 Frequently, best efficiency occurs at = =0A>best glide speed (like 107 KIAS in a half loaded=A0320).=A0 So what?=A0= Do I care if I =0A>can reach Austin, TX in=A08 hours using only 20 gallons= or 4.3 hours using 30 =0A>gallons or=A04.8 hours at best power requiring a= fuel stop to maintain minimums =0A>(43 gal tank).=A0 Of course.=A0 But I d= on't need anything more than ROP/LOP fuel =0A>burns and associated TAS - fo= rtunately for=A0my very slick bird, there is only a =0A>loss of 6 or 7 knot= s for a drop of 2 gph from ROP to LOP at some useful =0A>altitude.=A0=A0So,= I get >1 hour more=A0endurance=A0at LOP and I can see if that 28 NM =0A>di= fference (4 hours)=A0is worth the 1 hour refueling stop.=A0 Uh, Austin is a= flip =0A>of the coin at 820 NM (wind and weather depending).=0A>=0A>Scott = Krueger=0A>LNC2 320=A0=A0=A0=A0=A0=0A>=0A>In a message dated 8/8/2010 6:46:= 31 P.M. Central Daylight Time, =0A>chris_zavatson@yahoo.com writes:=0A>The = MKII tail is a little different.=A0=A0Push rod forces are zero for all trim= med =0A>conditions.=0A>>=0A>>Chris Zavatson=A0=A0=A0 =0A>>N91CZ=A0=A0=A0 = =0A>>360std=0A>>www.N91CZ.com=0A>>=0A>>=A0=0A>>=0A>>=0A>>=0A_______________= _________________=0AFrom: Wolfgang =0A>>To: lml@lancair= online.net=0A>>Sent: Fri, August 6, 2010 10:06:44 PM=0A>>Subject: [LML] Re:= flap coupling=0A>>=0A>>=0A>>I have taken elevator pushrod force measuremen= ts and was surprised.=0A>>Elevator pushrod forces to stick forces are about= 6.5 to 1=0A>>The trim system, when dialed in, provides these forces.=0A>>= =A0=0A>>At=A0190 imph and -7=BA flaps, there is a 60lb forward force.=0A>>A= t 80 imph and 10=BA flaps, there is about zero force.=0A>>At 80 imph and 20= =BA flaps, there is a slight (-1lb)=A0rearward force.=0A>>=A0=0A>>These num= bers are with the horizontal stabilizer built at -1.2=BA=0A>>- - - plans ra= nge is -0.5=BA to -1.0=BA=0A>>=A0=0A>>An=A0input from the flap bellcrank of= about 20-40 lb at -7=BA would be good,=0A>>=A0tapering down to zero lbs at= 10=BA flaps=0A>>=A0=0A>>A horizontal stabilizer built at -0.5=BA would, of= course, change these numbers.=0A>>=A0=0A>>Comments ?=0A>>=A0=0A>>Wolfgang= =0A>>=A0 =0A>=0A>=0A>--=0A>For archives and unsub http://mail.lancaironline= .net:81/lists/lml/List.html=0A>=0A=0A=0A --0-1709958904-1281556159=:20958 Content-Type: text/html; charset=iso-8859-1 Content-Transfer-Encoding: quoted-printable
Scott,
=0A
When the airfoil was tested, M 0.1 was likely= used to eliminate compressibility effects.  Note that the Reynolds nu= mbers are in the appropriate range for our application.  There is also= a chart showing some compressibility effects.   When testing, te= mperature and/or pressure are changed to vary Reynolds numbers while m= aintaining a constant Mach number.  The 0.1 Mach number is not meant t= o imply usage at 66 kts
=0A
http://en.wikipedia.org/wiki/Reynolds_number=0A
Chris
=0A
 
=0A
 
=0A
Chris= Zavatson
=0A
N91CZ
=0A
360std
=0A=0A

=0A
=0A
=0ATo: lml@lancaironline.net
Sent: Tue, August 10, 2010 1:52:05= PM
Subject: [LML] Re: f= lap coupling, LNC2 flap implications

=0A
Chris and Wolfgang,
=0A=
 
=0A
The airfoil report only deals with a section, not = the whole wing with its washout or the whole airplane, etc.  Furt= hermore, while figure 12 indeed demonstrates drag reduction for a section i= n reflex, the notation claims it is at .1 Mach or about 66 Knots. = ; I can't get my head around that since that is touchdown speed where the f= laps should be out of reflex at least. 
=0A
 =0A
Experience with my airplane tells me that full reflex (perhap= s up to -10 degrees in flight out at the twisted flap end) minimizes drag i= n the speed range above about 160 KIAS.  Below those speeds = a nose up attitude (and AOA) begins to creep in to maintain lift. = ; By the time 120 KIAS is reached my flaps are in takeoff position (10= degrees down from full reflex).  Why?  Because the nose up attit= ude would be perhaps 6 degrees higher if no flaps were used.  I h= ave never measured the power or speed difference needed to maintain level f= light between the two flap positions (drag indicators for the same lif= t) because speeds below 120 are only flown on an approach, in the pattern o= r at OSH. 
=0A
 
=0A
I continue to claim that = the LNC2 flaps are a fourth control surface that has important drag im= plications and significant pitch consequences - certainly more pronounced t= han that of any spam can.     
=0A
 =
=0A
Scott
=0A
 
=0A
PS - I certainly like = the LNC2 continuous flap position capability rather than fixed detents.&nbs= p; Continuous positioning fits its role as another control surface.=0A
 
=0A
=0A
In a message dated 8/9/2010 6:33:38 = P.M. Central Daylight Time, chris_zavatson@yahoo.com writes:
=0A=0A
=0A
=0A
Wolfgang,
=0A
The MkII tail uses a serv= o to drive the trim tab.  If installed, the bob-weight is the only thi= ng contributing to any static force in rear half of the system.  If an= autopilot is fighting an out of trim condition that would add some&nb= sp;force-much like the old trim system does.  Outside of that all trim= forces for the large tail are self-contained in the elevator.
=0AAttached are some charts showing the benefits of reflex relative to the 32= 0/360.
=0A
 
=0A
Chris Zavatson
=0A
N91CZ=0A
360std
=0A
www.N91CZ.com
=0A

=0A
=0A
=0A<= SPAN style=3D"FONT-WEIGHT: bold">From: Wolfgang <Wolfgang@MiC= om.net>
To: lml@lanca= ironline.net
Sent: Mon, = August 9, 2010 12:04:51 PM
Subject:= [LML] Re: flap coupling

=0A=0A=0A<= DIV>The push rod forces are definitely there. <= /FONT>
=0A
The trim system that keeps = those forces from showing up at the control stick.
=0A
. . . . unless you're using servo tabs . . . .
=0A
 
=0A
Yes, I want to map the drag bucket for various flap= conditions.
=0A
NASA tech pap= er 1865 shows it's effect. I want to expand that on the 300 series.<= /DIV>=0A
I believe it can add some efficien= cy points if utilized.
=0A
 
=0A
Wolfgang
=0A=
=0A
=0A
=0A
=0A=0A=0A=0A=0A=0A
=0A=0A=0A=0A
=0A= =0A=0A=0A=0A=0A= =0A=0A=0A=0A=0A= =0A=0A=0A=0A=0A=0A
From:Sky2high@aol.com
Sender:<ma= rv@lancaironline.net>
Subjec= t:Re: [LML] Re: flap coupling
Date:M= on, 09 Aug 2010 07:33:19 -0400
= To:lml@lancaironline.net
=0A
Uh, the push rod forces should be zero = when trimmed.  If one cannot reach a trimmed configuration, then force= will be required to reach sustained level flight.  One can only wonde= r about the position of the trimming device (there are so many different me= thods) when one then calculates forces necessary for level flight at d= ifferent airspeeds/configurations.
=0A
 
=0A
Wolfgan= g is seeking the "drag bucket" for different flight regimes.  The purp= ose is unknown.  Each configuration change affects either lift (induce= d drag) or parasitic drag or both.  Faster =3D less induced drag,= more parasitic drag.  Slower =3D more induced drag, less parasitic.&n= bsp; Parasites are everywhere.
=0A
 
=0A
http://www.charlesriverrc.or= g/articles/asfwpp/lelke_airfoilperf.htm clarifies the "drag= bucket" concept.  Good (an extra "o" converts God to good) Is only of= concern at cruise configurations.  Why?  Because anything else i= s confounded by other variables - density altitude, wind, efficiency, etc.&= nbsp; The designer defined the cruise range as the best conditions (altitud= e, power, etc) where the longeron was level.  Other things can affect = drag, engine cooling, laminar flow because of smooth surfaces, weight (lift= -induced drag), wax (parasitic drag), etc. etc. etc. 
=0A
&n= bsp;
=0A
Who cares at other speeds less than cruise - we know= that max efficiency can be reached when parasitic drag and induced drag cr= oss at some minima.  Uh, the old max range vs max endurance question. =   Frequently, best efficiency occurs at best glide speed (like 107 KIA= S in a half loaded 320).  So what?  Do I care if I can reach= Austin, TX in 8 hours using only 20 gallons or 4.3 hours using 30 gal= lons or 4.8 hours at best power requiring a fuel stop to maintain mini= mums (43 gal tank).  Of course.  But I don't need anything more t= han ROP/LOP fuel burns and associated TAS - fortunately for my very sl= ick bird, there is only a loss of 6 or 7 knots for a drop of 2 gph from ROP= to LOP at some useful altitude.  So, I get >1 hour more = endurance at LOP and I can see if that 28 NM difference (4 hours) = ;is worth the 1 hour refueling stop.  Uh, Austin is a flip of the coin= at 820 NM (wind and weather depending).
=0A
 
=0A
Scot= t Krueger
=0A
LNC2 320     
=0A
=  
=0A
=0A
In a message dated 8/8/2010 6:46:31 P.M. Centra= l Daylight Time, chris_zavatson@yahoo.com writes:
=0A
=0A
=0A
The MKII tail is a l= ittle different.  Push rod forces are zero for all trimmed condit= ions.
=0A
 
=0A
Chris Zavatson    =0A
N91CZ   
=0A
360std
=0A=0A
 
=0A

=0A
=0A
=0AFrom: Wolfgang <Wolfgang@MiCom.net>
To: lml@lancaironline.net
Sent: Fr= i, August 6, 2010 10:06:44 PM
Subject: [LML] Re: fla= p coupling

=0A
I have taken e= levator pushrod force measurements and was surprised.
=0A
<= FONT face=3DArial size=3D2>Elevator pushrod forces to stick forces are abou= t 6.5 to 1
=0A
The trim system,= when dialed in, provides these forces.
=0A
 
=0A
At = ;190 imph and -7=BA flaps, there is a 60lb forward force.
=0AAt 80 imph and 10=BA flaps, there is about z= ero force.
=0A
At 80 imph and 2= 0=BA flaps, there is a slight (-1lb) rearward force.
=0A 
=0A
These numbers are with the horizontal stabilizer built at -1.2= =BA
=0A
- - - plans range is -0= .5=BA to -1.0=BA
=0A
&nb= sp;
=0A
An input from the flap be= llcrank of about 20-40 lb at -7=BA would be good,
=0A
 tapering down to zero lbs at 10=BA flaps
=0A
 
=0A
A horizontal stabilizer built at -0.5=BA would, of c= ourse, change these numbers.
=0A
 
=0A
Comments ?<= /DIV>=0A
 
=0A
Wolfgang
=0A
 
=



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