X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Sender: To: lml@lancaironline.net Date: Thu, 15 Jul 2010 19:08:18 -0400 Message-ID: X-Original-Return-Path: Received: from proton.sasknet.sk.ca ([142.165.20.178] verified) by logan.com (CommuniGate Pro SMTP 5.3.8) with ESMTPS id 4394373 for lml@lancaironline.net; Thu, 15 Jul 2010 13:09:01 -0400 Received-SPF: none receiver=logan.com; client-ip=142.165.20.178; envelope-from=hjjohnson@sasktel.net Received: from pps.filterd (proton [127.0.0.1]) by proton.sasknet.sk.ca (8.14.3/8.14.3) with SMTP id o6FGweuJ001798 for ; Thu, 15 Jul 2010 11:08:24 -0600 Received: from bgmpomr1.sasknet.sk.ca (bgmpOMR1.sasknet.sk.ca [142.165.72.22]) by proton.sasknet.sk.ca with ESMTP id q25f1kmx2-1 for ; Thu, 15 Jul 2010 11:08:23 -0600 Received: from sasktel.net ([192.168.234.97]) by bgmpomr1.sasknet.sk.ca (SaskTel eMessaging Service) with ESMTP id <0L5L008HTZLZ3U10@bgmpomr1.sasknet.sk.ca> for lml@lancaironline.net; Thu, 15 Jul 2010 11:08:23 -0600 (CST) Received: from [192.168.234.24] (Forwarded-For: [24.89.93.210]) by cgmail1.sasknet.sk.ca (mshttpd); Thu, 15 Jul 2010 11:08:23 -0600 X-Original-Date: Thu, 15 Jul 2010 11:08:23 -0600 From: H & J Johnson Subject: Re: [LML] Re: Small tail, MK II tail, CG range X-Original-To: Lancair Mailing List X-Original-Message-id: <46d0c4fd30e6.4c3eec27@sasktel.net> MIME-version: 1.0 X-Mailer: Sun Java(tm) System Messenger Express 6.1 HotFix 0.20 (built Feb 27 2006) Content-type: text/html; charset=iso-8859-1 Content-language: en Content-transfer-encoding: quoted-printable Content-disposition: inline X-Accept-Language: en Priority: normal X-Proofpoint-Virus-Version: vendor=fsecure engine=2.50.10432:5.0.10011,1.0.148,0.0.0000 definitions=2010-07-15_04:2010-07-15,2010-07-15,1970-01-01 signatures=0 X-Proofpoint-Spam-Details: rule=outbound_notspam policy=default score=0 spamscore=0 ipscore=0 phishscore=0 bulkscore=0 adultscore=0 classifier=spam adjust=0 reason=mlx engine=5.0.0-1005130000 definitions=main-1007150074 =3CBODY=3E=3CP=3E=3C/P=3E=3CB=3E=3C/B=3E =3CP=3E=3C/P=3E =3CP=3E=3C/P=3E =3CP=3E=3CSTRONG=3EChris=2C do you have any references as to the Neutral= Point being the same as the MAC =5Bmean aerodynamic chord=5D=3F=3C/STRO= NG=3E=3C/P=3E =3CP=3E=3CSTRONG=3EAll the book=27s I=27ve owned tell me the MAC is calc= ulated on the wing platform/shape while the NP is derived from this and = the tail volume =5Bwhich=26nbsp=3Bis composed of=26nbsp=3Ban arm=2C MAC = and a area value=5D=2E =3C/STRONG=3E=3C/P=3E =3CP=3E=3CSTRONG=3EJust wanted to confirm what is correct and incorrect=2E= Also=2C do you have any feedback as to how the Aerodynamic Center appli= es to all of this=3F=3C/STRONG=3E=3C/P=3E =3CP=3E=3CSTRONG=3E=3C/STRONG=3E=26nbsp=3B=3C/P=3E =3CP=3E=3CSTRONG=3EThanks=3C/STRONG=3E=3C/P=3E =3CP=3E=3CSTRONG=3EJ=2E Johnson=3C/STRONG=3E=3C/P=3E =3CP=3E=3CSTRONG=3E235/320 55=25 =5Band holding=5D=3C/STRONG=3E=3C/P=3E =3CP=3E=26nbsp=3B=3C/P=3E =3CP=3E=26gt=3B Wolfgang=2C et al =3CBR=3E=26gt=3B The aircraft MAC (als= o called neutral point) relative to CG is the =3CBR=3E=26gt=3B key to =3C= BR=3E=26gt=3B evaluating aircraft longitudinal stability=2E=26nbsp=3B Th= is is independent =3CBR=3E=26gt=3B of whether the =3CBR=3E=26gt=3B tail=26= nbsp=3Bis providing an up or down force (either=26nbsp=3Bcan be stable)=2E= =26nbsp=3B =3CBR=3E=26gt=3B Longitudinal =3CBR=3E=26gt=3B stability is d= efined by the reaction of the entire airframe to a =3CBR=3E=26gt=3B dist= urbance =3CBR=3E=26gt=3B from equilibrium=2E=26nbsp=3B The size=2C locat= ion and pitching moment =3CBR=3E=26gt=3B characteristics =3CBR=3E=26gt=3B= of=26nbsp=3Beach component factors in (wing=2C tail=2C fuselage =3CBR=3E= =26gt=3B etc=2E)=2E=26nbsp=3B=26nbsp=3BEvaluating the =3CBR=3E=26gt=3B b= ehavior of=26nbsp=3Bjust the wing is not sufficient to describe the =3CB= R=3E=26gt=3B response of the =3CBR=3E=26gt=3B aircraft as a whole and ce= rtainly not to quantify the response=2E=26nbsp=3B =3CBR=3E=26gt=3B Actua= lly=2C a =3CBR=3E=26gt=3B wing section alone=26nbsp=3Bwill be unstable a= s the pitching moment is =3CBR=3E=26gt=3B negative=2E=26nbsp=3B It is =3C= BR=3E =26gt=3B stable when inverted - flying wings have negative camber for th= is =3CBR=3E=26gt=3B reason=2E=26nbsp=3B=26nbsp=3B=26nbsp=3B =3CBR=3E=26g= t=3B A stable=26nbsp=3Baircraft must=26nbsp=3Bhave a=26nbsp=3Bpositive p= itching moment when in =3CBR=3E=26gt=3B equilibrium=2E=26nbsp=3B=26nbsp=3B= In =3CBR=3E=26gt=3B order to be stable=2C the pitching moment coefficien= t=26nbsp=3Bmust=26nbsp=3Bhave a =3CBR=3E=26gt=3B negative=26nbsp=3Bslope= =3CBR=3E=26gt=3B with increasing angle of attack=2E=26nbsp=3B=26nbsp=3B= This provides an increasing =3CBR=3E=26gt=3B opposing=26nbsp=3Bmoment to= =3CBR=3E=26gt=3B an increasing disturbance=2E=26nbsp=3B=26nbsp=3B =3CBR= =3E=26gt=3B A=26nbsp=3Blarger tail increases the response when a disturb= ance occurs=2E=26nbsp=3B =3CBR=3E=26gt=3B It is a =3CBR=3E=26gt=3B funct= ion of the larger=26nbsp=3Barea producing more=26nbsp=3Brestoring force = for=26nbsp=3Bany =3CBR=3E=26gt=3B given=26nbsp=3Bangular =3CBR=3E=26gt=3B= disturbance=2E=26nbsp=3B=26nbsp=3BThe size of the horizontal stabilizer= =26nbsp=3Bfeeds into=26nbsp=3Ba =3CBR=3E=26gt=3B quantity called =3CBR=3E= =26gt=3B the tail volume ratio -=26nbsp=3Ba unit-less measure of relatin= g tail area =3CBR=3E=26gt=3B to wing area =3CBR=3E =26gt=3B and wing mean=26nbsp=3Bwing chord to distance to the horizontal= =3CBR=3E=26gt=3B stabilizer=2E=26nbsp=3B More area =3CBR=3E=26gt=3B or=26= nbsp=3Ba longer tail increase the effectiveness in terms of stability=2E= =3CBR=3E=26gt=3B The neutral point=26nbsp=3Bis fixed by=26nbsp=3Bthe co= nfiguration of the =3CBR=3E=26gt=3B aircraft=2E=26nbsp=3B=26nbsp=3BOnly = =3CBR=3E=26gt=3B configuration changes will move the neutral point=2E=26= nbsp=3B Lowering the =3CBR=3E=26gt=3B flaps=2C for =3CBR=3E=26gt=3B exam= ple=2C changes the airfoil=2C relative incidence angles=2C pitching =3CB= R=3E=26gt=3B moment of the =3CBR=3E=26gt=3B wing and so on=2E=26nbsp=3B = In all configurations the neutral point must =3CBR=3E=26gt=3B remain wel= l behind =3CBR=3E=26gt=3B the CG=2E=26nbsp=3B 10=25 of the mean chord le= ngth is a good starting minimum=2E=26nbsp=3B =3CBR=3E=26gt=3B Once the =3C= BR=3E=26gt=3B neutral point is known=2C the incidence angles and CG can = be set=2E=26nbsp=3B =3CBR=3E=26gt=3B What will fall =3CBR=3E=26gt=3B out= is the trim airspeed=2E=26nbsp=3B That is=2C=26nbsp=3Bin equilibrium th= e aircraft =3CBR=3E=26gt=3B will seek out a =3CBR=3E =26gt=3B specific angle of attack and the corresponding airspeed=2E=26nb= sp=3B One can =3CBR=3E=26gt=3B play around =3CBR=3E=26gt=3B with=26nbsp=3B= different combinations of=26nbsp=3Bincidence angles and CG locations =3C= BR=3E=26gt=3B to achieve both =3CBR=3E=26gt=3B a stable aircraft and min= imum trim drag at any desired airspeed=2E=26nbsp=3B =3CBR=3E=26gt=3B hop= e that helps=2C =3CBR=3E=26gt=3B Chris =3CBR=3E=26gt=3B =3CBR=3E=26gt=3B= =3CBR=3E=26gt=3B =3CBR=3E=26gt=3B Chris Zavatson =3CBR=3E=26gt=3B N91CZ= =3CBR=3E=26gt=3B 360std =3CBR=3E=26gt=3B www=2EN91CZ=2Ecom =3CBR=3E=26g= t=3B =3CBR=3E=26gt=3B =26nbsp=3B =3CBR=3E=26gt=3B =3CBR=3E=26gt=3B =3CBR= =3E=26gt=3B =3CBR=3E=26gt=3B =5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F= =5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F =3CBR=3E=26gt=3B Fro= m=3A Wolfgang =26lt=3BWolfgang=40MiCom=2Enet=26gt=3B =3CBR=3E=26gt=3B To= =3A lml=40lancaironline=2Enet =3CBR=3E=26gt=3B Sent=3A Wed=2C July 14=2C= 2010 10=3A37=3A18 AM =3CBR=3E=26gt=3B Subject=3A =5BLML=5D Re=3A Small = tail=2C MK II tail=2C CG range =3CBR=3E=26gt=3B =3CBR=3E=26gt=3B =3CBR=3E= =26gt=3B I=27m not familiar with MAC as applied to the entire airframe=2C= can =3CBR=3E=26gt=3B you elaborate=3F =3CBR=3E=26gt=3B I think there ma= y be a problem with that idea since the tail is =3CBR=3E=26gt=3B typical= ly =3CBR=3E =26gt=3B providing a down force which would move the =22airframe MAC=22=26= nbsp=3Bto the =3CBR=3E=26gt=3B front=2C not the =3CBR=3E=26gt=3B rear=2E= =3CBR=3E=26gt=3B =26nbsp=3B =3CBR=3E=26gt=3B Wolfgang =3CBR=3E=26gt=3B = =3CBR=3E=26gt=3B =5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F= =5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F =3CBR=3E=26gt=3B =3CBR=3E=26gt=3B= ----- Original Message ----- =3CBR=3E=26gt=3B =26gt=3BFrom=3A Chris Zav= atson =3CBR=3E=26gt=3B =26gt=3BTo=3A lml=40lancaironline=2Enet =3CBR=3E=26= gt=3B =26gt=3BSent=3A Tuesday=2C July 13=2C 2010 8=3A35 PM =3CBR=3E=26gt= =3B =26gt=3BSubject=3A Re=3A =5BLML=5D Small tail=2C MK II tail=2C CG ra= nge =3CBR=3E=26gt=3B =26gt=3B =3CBR=3E=26gt=3B =26gt=3B =3CBR=3E=26gt=3B= =26gt=3BWolfgang=2C et al=2C =3CBR=3E=26gt=3B =26gt=3B=26lt=3B=26lt=3BA= ny more to the rear and you get negative stability at cruise =3CBR=3E=26= gt=3B and a larger =3CBR=3E=26gt=3B =26gt=3Btail doesn=27t help much wit= h that anyway=2E=26gt=3B=26gt=3B =3CBR=3E=26gt=3B =26gt=3B =3CBR=3E=26gt= =3B =26gt=3BA=26nbsp=3Blarger tail moves the MAC rearward allowing the C= G to move =3CBR=3E=26gt=3B farther aft while =3CBR=3E=26gt=3B =26gt=3Bma= intaining the same level of stability=2E =3CBR=3E=26gt=3B =26gt=3BThere = has been=26nbsp=3Ba lot of discussion about Cm=2E=26nbsp=3B We need to b= e =3CBR=3E=26gt=3B careful to =3CBR=3E =26gt=3B =26gt=3Bdistinguish between the Cm for the wing=2C tail and tot= al =3CBR=3E=26gt=3B aircraft=2E=26nbsp=3B It is the =3CBR=3E=26gt=3B =26= gt=3Blater that is critical to stability and this is where the larger =3C= BR=3E=26gt=3B tail influences =3CBR=3E=26gt=3B =26gt=3Bthe situation=2E=26= nbsp=3B The large tail moves the MAC to the rear approx=2E =3CBR=3E=26gt= =3B 1=2E5 inches=2E=26nbsp=3B =3CBR=3E=26gt=3B =26gt=3BFor the same CG=2C= the more rearward MAC produces a greater =3CBR=3E=26gt=3B restoring for= ce if the =3CBR=3E=26gt=3B =26gt=3Bplane is disturbed from level flight=2E= =26nbsp=3B The practical benefit for =3CBR=3E=26gt=3B us is that it =3CB= R=3E=26gt=3B =26gt=3Ballows=26nbsp=3Ba lot more baggage to be thrown the= rear of the plane =3CBR=3E=26gt=3B before =3CBR=3E=26gt=3B =26gt=3Bsuff= ering=26nbsp=3Bstability problems=2E=26nbsp=3B You pointed out the other= benefit =3CBR=3E=26gt=3B of increased =3CBR=3E=26gt=3B =26gt=3Bcontrol = authority at slow speed with full flaps=2E =3CBR=3E=26gt=3B =26gt=3B =3C= BR=3E=26gt=3B =26gt=3BChris Zavatson =3CBR=3E=26gt=3B =26gt=3BN91CZ =3CB= R=3E=26gt=3B =26gt=3B360std =3CBR=3E=26gt=3B =26gt=3Bwww=2EN91CZ=2Ecom =3C= BR=3E=26gt=3B =26gt=3B =3CBR=3E=26gt=3B =26gt=3B =3CBR=3E=26gt=3B =26gt=3B= =3CBR=3E=26gt=3B =26gt=3B =3CBR=3E =26gt=3B =5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F= =5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F =3CBR=3E=26gt=3B From=3A Wolfgang =26l= t=3BWolfgang=40MiCom=2Enet=26gt=3B =3CBR=3E=26gt=3B =26gt=3BTo=3A lml=40= lancaironline=2Enet =3CBR=3E=26gt=3B =26gt=3BSent=3A Tue=2C July 13=2C 2= 010 2=3A51=3A23 AM =3CBR=3E=26gt=3B =26gt=3BSubject=3A =5BLML=5D Small t= ail=2C MK II tail=2C CG range =3CBR=3E=26gt=3B =26gt=3B =3CBR=3E=26gt=3B= =26gt=3B =3CBR=3E=26gt=3B =26gt=3BThe quest continues=2E =3CBR=3E=26gt=3B= =26gt=3B =3CBR=3E=26gt=3B =26gt=3BI=27m checking further into the data = on these questions and am =3CBR=3E=26gt=3B coming to question =3CBR=3E=26= gt=3B =26gt=3Bthe need for a larger tail=2E I=27m not sure a larger tail= by itself =3CBR=3E=26gt=3B will solve the =3CBR=3E=26gt=3B =26gt=3Bprob= lem=2E After doing some static and in flight measurements=2C it =3CBR=3E= =26gt=3B looks like the =3CBR=3E=26gt=3B =26gt=3Btail authority is not a= big problem=2C if a problem at all=2E =3CBR=3E=26gt=3B =26gt=3B =3CBR=3E= =26gt=3B =26gt=3BStatic measurements of N31161 have shown =22vanilla=22 = parameters=2E =3CBR=3E=26gt=3B 2=2E5=BA incidence =3CBR=3E=26gt=3B =26gt= =3Bbetween the wing root at full reflex and the tail and a 1=2E3=BA =3CB= R=3E=26gt=3B washout=2E Put the =3CBR=3E=26gt=3B =26gt=3Bflaps at 0=BA a= nd you get an additional AoA of 1=2E8=BA at the root for =3CBR=3E =26gt=3B a total =3CBR=3E=26gt=3B =26gt=3Bincidence of 4=2E3=BA =2E =2E = =2E =2E not radical at all=2E =3CBR=3E=26gt=3B =26gt=3B =3CBR=3E=26gt=3B= =26gt=3BWhat is interesting is the POH (Dec=2E 1994 pg=2E VI-3) gives t= he CG =3CBR=3E=26gt=3B range as 24=2E5=22 =3CBR=3E=26gt=3B =26gt=3Bto 30= =2E3=22 aft of the rear face of the fire wall and the MAC at 15=25 =3CBR= =3E=26gt=3B to 20=25 =3CBR=3E=26gt=3B =26gt=3B =3CBR=3E=26gt=3B =26gt=3B= =2E =2E =2E well =2E =2E =2E no =2E =2E =2E that range is more like a MA= C range of =3CBR=3E=26gt=3B 15=25 to 30=25 - - =3CBR=3E=26gt=3B =26gt=3B= - a good range made touchy only by the small size of the air frame=2E =3C= BR=3E=26gt=3B =26gt=3B =3CBR=3E=26gt=3B =26gt=3BAfter going over the pla= n view kit drawings=2C I come up with a CG =3CBR=3E=26gt=3B range of =3C= BR=3E=26gt=3B =26gt=3B23-1/4=22 to 29-1/4=22 for a MAC range of 15=25 to= 30=25 =3CBR=3E=26gt=3B =26gt=3BThat range is about 1-1/4=22 forward of = the book and fits better =3CBR=3E=26gt=3B with first hand =3CBR=3E=26gt=3B= =26gt=3Bflight experience=2E =3CBR=3E=26gt=3B =26gt=3B =3CBR=3E=26gt=3B= =26gt=3B =3CBR=3E=26gt=3B =26gt=3BAny more to the rear and you get nega= tive stability at cruise and =3CBR=3E=26gt=3B a larger tail =3CBR=3E=26g= t=3B =26gt=3Bdoesn=27t help much with that anyway=2E =3CBR=3E=26gt=3B =26= gt=3B =3CBR=3E =26gt=3B =26gt=3BNegative stability makes pitch control a real chore=2E = As Scott K=2E =3CBR=3E=26gt=3B has indicated=2C =3CBR=3E=26gt=3B =26gt=3B= going to 0=BA flaps helps under that loading condition=2E =3CBR=3E=26gt=3B= =26gt=3B =3CBR=3E=26gt=3B =26gt=3BToo far forward and landing becomes =22= interesting=22=2E A larger tail =3CBR=3E=26gt=3B can help there =3CBR=3E= =26gt=3B =26gt=3B=2E =2E =2E or don=27t use as much flaps=2E =3CBR=3E=26= gt=3B =26gt=3B =3CBR=3E=26gt=3B =26gt=3BI think understanding these cond= itions can help everyone=2E =3CBR=3E=26gt=3B =26gt=3B =3CBR=3E=26gt=3B =26= gt=3B=2E =2E =2E The quest continues =2E =2E =2E Comments welcome=2E =3C= BR=3E=26gt=3B =26gt=3B =3CBR=3E=26gt=3B =26gt=3BWolfgang =3CBR=3E=26gt=3B= =26gt=3B =3CBR=3E=26gt=3B =26gt=3B=26nbsp=3B =3CBR=3E=26gt=3B =26gt=3B = =3CBR=3E=26gt=3B =5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F= =5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F =3CBR=3E=26gt=3B =3CBR=3E=26gt=3B= =26gt=3BFrom=3A =22Wolfgang=22 =26lt=3BWolfgang=40MiCom=2Enet=26gt=3B =3C= BR=3E=26gt=3B =26gt=3BSender=3A =26lt=3Bmarv=40lancaironline=2Enet=26gt=3B= =3CBR=3E=26gt=3B =26gt=3BSubject=3A Small tail=2C MK II tail=2C CG rang= e =3CBR=3E=26gt=3B =26gt=3BDate=3A Sat=2C 10 Jul 2010 21=3A01=3A11 -0400= =3CBR=3E=26gt=3B =26gt=3BTo=3A lml=40lancaironline=2Enet=26nbsp=3B=26nb= sp=3B =3CBR=3E=26gt=3B =26gt=3BThe LNC2 uses the NLF(1)-0215F airfoil=2E= A lot can be found by =3CBR=3E =26gt=3B doing a Google =3CBR=3E=26gt=3B =26gt=3Bsearch on that number=2E= =3CBR=3E=26gt=3B =26gt=3BMore detail can be found by=26nbsp=3Bgoing to = Google for =22NASA Technical =3CBR=3E=26gt=3B Paper 1865=22=2E =3CBR=3E=26= gt=3B =26gt=3B =3CBR=3E=26gt=3B =26gt=3BI have not taken the time to rev= erse engineer the CG range of the =3CBR=3E=26gt=3B LNC2 but let =3CBR=3E= =26gt=3B =26gt=3Bme offer some observations=2E =3CBR=3E=26gt=3B =26gt=3B= =3CBR=3E=26gt=3B =26gt=3BThe airfoil used has long been touted as =22th= e greatest thing =3CBR=3E=26gt=3B since sliced bread=22 =3CBR=3E=26gt=3B= =26gt=3Bfor General Aviation and it definitely has some advantages=2E B= ut =3CBR=3E=26gt=3B it=27s not new=2E =3CBR=3E=26gt=3B =26gt=3BCompare t= his airfoil to the P-51 airfoil and you will see some =3CBR=3E=26gt=3B c= lose =3CBR=3E=26gt=3B =26gt=3Bsimilarities=2E The LNC2 being composite c= onstruction instead of =3CBR=3E=26gt=3B aluminum lets the =3CBR=3E=26gt=3B= =26gt=3Bairfoil show more of it=27s theoretical advantages=2E =3CBR=3E=26= gt=3B =26gt=3B =3CBR=3E=26gt=3B =26gt=3BIt=27s a laminar shape with a go= od drag bucket=2E That bucket can be =3CBR=3E=26gt=3B made to move to =3C= BR=3E=26gt=3B =26gt=3Bthe lower Cl (lift coefficient) ranges with reflex= allowing =3CBR=3E =26gt=3B noticeably lower =3CBR=3E=26gt=3B =26gt=3Bdrag at higher cruise= speeds=2E Along with reflex=2C the Cm (moment =3CBR=3E=26gt=3B coeffici= ent) =3CBR=3E=26gt=3B =26gt=3Bgoes positive=2C the center of lift of the= wing travels forward =3CBR=3E=26gt=3B giving a nose up =3CBR=3E=26gt=3B= =26gt=3Bforce requiring down trim=2E This is in addition to the usual n= ose =3CBR=3E=26gt=3B up force that =3CBR=3E=26gt=3B =26gt=3Bgoes with mo= st all airfoils=26nbsp=3Bat high speed before considering flaps=2E =3CBR= =3E=26gt=3B =26gt=3B =3CBR=3E=26gt=3B =26gt=3BWith down flap=2C the drag= bucket will move to higher Cl=27s making =3CBR=3E=26gt=3B slower flight= =3CBR=3E=26gt=3B =26gt=3Bmore efficient=2E And=2C of course=2C the Cm g= oes negative giving a =3CBR=3E=26gt=3B nose down force =3CBR=3E=26gt=3B = =26gt=3Brequiring up trim=2E =3CBR=3E=26gt=3B =26gt=3B =3CBR=3E=26gt=3B = =26gt=3B=2E =2E =2E and appropriate variations in-between =2E =2E =2E =3C= BR=3E=26gt=3B =26gt=3B =3CBR=3E=26gt=3B =26gt=3B =3CBR=3E=26gt=3B =26gt=3B= So=2C the rear CG limit is determined by high speed flight and =3CBR=3E=26= gt=3B available control =3CBR=3E=26gt=3B =26gt=3Bauthority=2C =3CBR=3E=26= gt=3B =26gt=3Band the forward CG is determined by low speed / landing fl= ight =3CBR=3E=26gt=3B and available =3CBR=3E =26gt=3B =26gt=3Bcontrol authority=2E =3CBR=3E=26gt=3B =26gt=3B =3CBR=3E= =26gt=3B =26gt=3BWhat is becoming clear here is that the center of lift = does quite =3CBR=3E=26gt=3B a bit of =3CBR=3E=26gt=3B =26gt=3Btraveling = fore and aft which is exaggerated by allowing negative =3CBR=3E=26gt=3B = or =22cruise=22 =3CBR=3E=26gt=3B =26gt=3Bflaps=2E Since you can=27t shif= t the CG during flight=2C you need a =3CBR=3E=26gt=3B large amount of =3C= BR=3E=26gt=3B =26gt=3Bpitch authority from the tail to cover that range = of lift travel=2E =3CBR=3E=26gt=3B =26gt=3B=26nbsp=3B =3CBR=3E=26gt=3B =26= gt=3BYou have two choices in the LNC2=2C live with the limitations or =3C= BR=3E=26gt=3B install a larger =3CBR=3E=26gt=3B =26gt=3Btail to give tha= t extra pitch authority=2E =3CBR=3E=26gt=3B =26gt=3B=2E =2E =2E A larger= tail area can also help with=26nbsp=3Babnormal =3CBR=3E=26gt=3B attitud= e=26nbsp=3Brecovery=2E=26gt=3B =3CBR=3E=26gt=3B =26gt=3BWolfgang =3CBR=3E= =26gt=3B =26gt=3B =3CBR=3E=26gt=3B =3CBR=3E=26gt=3B =3CBR=3E=26gt=3B=26n= bsp=3B=26nbsp=3B=26nbsp=3B=26nbsp=3B=26nbsp=3B =3C/P=3E=3C/BODY=3E