Return-Path: Sender: (Marvin Kaye) To: lml Date: Mon, 29 Sep 2003 20:29:13 -0400 Message-ID: X-Original-Return-Path: Received: from imo-m08.mx.aol.com ([64.12.136.163] verified) by logan.com (CommuniGate Pro SMTP 4.1.4) with ESMTP id 2608323 for lml@lancaironline.net; Mon, 29 Sep 2003 17:36:23 -0400 Received: from Sky2high@aol.com by imo-m08.mx.aol.com (mail_out_v36_r1.1.) id q.18b.200db405 (3310) for ; Mon, 29 Sep 2003 17:35:04 -0400 (EDT) From: Sky2high@aol.com X-Original-Message-ID: <18b.200db405.2ca9ff87@aol.com> X-Original-Date: Mon, 29 Sep 2003 17:35:03 EDT Subject: =?UTF-8?Q?For=20a=20Few=20Knots=20More!=20Or,=20A=20Fistful=20of?= =?UTF-8?Q?=20Knots=E2=80=A6=20=20?= X-Original-To: lml@lancaironline.net MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="-----------------------------1064871303" X-Mailer: 9.0 for Windows sub 1100 -------------------------------1064871303 Content-Type: text/plain; charset="UTF-8" Content-Transfer-Encoding: quoted-printable Content-Language: en Gentlemen, =20 <8=3D=3D=3DV=3D^=3D=3D=3D=3D=3D=3D=3Dv=3D =20 Where: < is the spinner 8 is the prop V is the CG ^ is the wing center of lift (Cl) v is the inverted wing (h-stab) center of lift and =3D" is the fuselage and, most importantly, it is flying from right to left. =20 So, in level flight at a constant speed, the =1C"h-stab=E2=80=9C is exertin= g a=20 negative lift (v) to balance the airplane against the snapshot CG (V) and th= e current=20 wing center of lift (^). =20 =20 At first, let us keep the conversation strictly to the 320/360 Lancair (it= =E2=80=99s=20 the only one upon which I can perform experiments) with the small tail, shor= t=20 engine mount. =20 The horizontal stabilizer is not symmetric =E2=80=93 it is an inverted wing.= In my=20 airplane the angle of incidence is about negative 1* (negative to the=20 longeron, not the main wing angle of incidence which is a positive angle(abo= ut 2.5* in=20 the flap area and washes out to about 1* at the aileron) to the longeron). =20 The Aspect Ratio of the small tail is 3.45, similar to a jet fighter wing, b= ut=20 not as low as a flying saucer. Low AR wings induced drag is sensitive to the= =20 lift load. Induced drag increases with a decrease in speed. =20 The main wing NLF (Natural Laminar Flow) flapped airfoil has some interestin= g=20 drag reduction, lift and pitching moment characteristics =E2=80=93 especiall= y when=20 the reflexed flap is moved out of reflex. Thanks to Ian for pointing to the=20= NASA=20 technical paper describing the wing with tests at .1 mach (about 64 kts). =20 Suppose we have a 53# weight distributed on the seat next to me and the=20 header is full with about 10 gallons in each wing for a gross weight of 1645= =20 pounds. For my plane, this would result is a CG of about 25.2 or 17% MAC. =20= If I=20 moved the weight to the back of the baggage compartment at about FS 84, the=20= CG=20 moves back to 26.7 or 20% MAC. According to popular thought, this would red= uce=20 the negative lift load on the tail (and, consequently, the =E2=80=9Cweight= =E2=80=9D the main=20 wing was carrying), the induced tail drag and thus allow the plane to go=20 faster. Note that if the h-stab negative center of lift is located 7 feet f= urther=20 back, the weight movement is equivalent to about 20# less exerted on the ta= il=20 and 20# less the wing has to carry. The questions to be answered are: =20 With the expected trim change, does this reduce the angle of attack of the=20 main wing and increase the AOA of the h-stab even though both are carrying l= ess=20 aerodynamic weight (above example is 20#)?=20 Is this significant enough to be reflected in an increase in speed? If the speed increases, is the main wing AOA further decreased? =20 Relative to query 3 above, we all note that as speed increases, more nose=20 down trim is added. This means less negative h-stab lift, less aerodynamic=20 weight on all the wings still attached. If you think upside down for a mome= nt, the=20 extreme nose down trim at max cruise speed is actually beginning to put the=20 h-stab wing flap, the elevator, in reflex! This is, of course, reducing the= =20 negative lift on the h-stab. =20 =20 Why does the pitch change occur? Must it be due to the speed related=20 pitching moments of the main wing with the flaps in reflex i.e. nose up, tai= l down. =20 Or, is it a change in the center of lift (center of pressure) closer to the=20 CG? If it is a change in the Cl towards the CG, that would reduce the load=20= on=20 the tail. If we reduce the load on the tail enough, moving the CG rearward=20= may=20 have less of an effect on the tail download, or maybe more leading to=20 potential instability? Should we install Jim Frantz=E2=80=99s AOA on someb= ody=E2=80=99s tail=20 (upside down, of course)? Do I see any hands raised? Volunteers to calibra= te it? =20 In any event, the experiment is set up with 53 pounds of weight as above. A= =20 digital level is affixed parallel to the longeron and my airplane is equippe= d=20 with the digital AOA. =20 In level flight (under auto pilot control, smooth air) at 8500 feet=20 (Baro=3D30.19, temp=3D2*C), WOT 23.3=E2=80=9D MAP, 2500 RPM, 9.3 gph, 74% po= wer =20 175 IAS, 197 TAS AOA=3D2.4*, Longeron (L)=3D0* and=20 =20 then the weight is transferred to the FS 84 Location. The auto pilot request= s=20 trim assist and the Reichel trim wheel is rotated forward (nose down) about=20 1/6th of a turn. =20 176 IAS, 197 TAS AOA=3D 2.2* to 2.4*, L=3D0* to -.2*. =20 Leaving the weight in the back, the power was reduced to 21=E2=80=9D MAP and= 2300=20 RPM, 7.1 gph, 62% power =20 161 IAS, 180 TAS AOA=3D 2.7* to 2.9*, L=3D.4* =20 The weight was moved back to the front and the nose was trimmed up =20 161 IAS, 181 TAS AOA=3D2.7* to 2.9*, L =3D.4* to .6* =20 Conclusion: I won=E2=80=99t be abnormally shifting weight to try to increase= speed. =20 Let=E2=80=99s see, during the Air Venture Cup Race, Mark=E2=80=99s child was= heavier than=20 Larry=E2=80=99s child (moving CG to the rear) yet Larry was faster. I ran a= bout as=20 expected even though I had all my =E2=80=9Cstuff=E2=80=9D in the back. Mayb= e I am fast because my=20 engine thrust line was shifted upward which reduces tail down load? =20 Other interesting numbers: WOT @ 7500 feet 2700 RPM, 11.3 gph, 83% power =20 185 IAS, 204 TAS AOA=3D2.1* L=3D-.2* =20 Climbing out at 130 IAS (about 2000 MSL) AOA=3D4.1* L=3D7.2* AI=3D5* and, Still climbing at 8000 MSL, 140 IAS AOA=3D3.8* L=3D4.6* AI=3D2* Scott Krueger Sky2high@aol.com II-P N92EX IO320 Aurora, IL (KARR) -------------------------------1064871303 Content-Type: text/html; charset="UTF-8" Content-Transfer-Encoding: quoted-printable Content-Language: en

Gentlemen,

 

<8=3D=3D=3DV=3D^=3D= =3D=3D=3D=3D=3D=3Dv=3D

 

Where:

< is the spinner

8 is the prop

V is the CG

^ is the wing center of l= ift (Cl)

v is the inverted wing (h= -stab) center of lift and

=3D is the fuselage and,=20= most importantly, it is flying from right to left.

 

So, in level flight at a=20= constant speed, the =E2=80=9Ch-stab=E2=80=9C is exerting a negative lift (v)= to balance the airplane against the snapshot CG (V) and the current wing ce= nter of lift (^). 

 

At first, let us keep the= conversation strictly to the 320/360 Lancair (it=E2=80=99s the only one upo= n which I can perform experiments) with the small tail, short engine mount.<= /P>

 

The horizontal stabilizer= is not symmetric =E2=80=93 it is an inverted wing.  In my airplane the angle of incidence is about  negative 1* (negative to the longer= on, not the main wing angle of incidence which is a positive angle(about 2.5= * in the flap area and washes out to about 1* at the aileron) to the longero= n).  The Aspect Ratio of the s= mall tail is 3.45, similar to a jet fighter wing, but not as low as a flying= saucer. Low AR wings induced drag is sensitive to the lift load.  Induced drag increases with a decrease=20= in speed.

 

The main wing NLF (Natura= l Laminar Flow) flapped airfoil has some interesting drag reduction, lift an= d pitching moment characteristics =E2=80=93 especially when the reflexed fla= p is moved out of reflex. Thanks to Ian for pointing to the NASA technical p= aper describing the wing with tests at .1 mach (about 64 kts).

 

Suppose we have a 53# wei= ght distributed on the seat next to me and the header is full with about 10=20= gallons in each wing for a gross weight of 1645 pounds.  For my plane, this would result is a CG of about=20= 25.2 or 17% MAC.  If I moved t= he weight to the back of the baggage compartment at about FS 84, the CG move= s back to 26.7 or 20% MAC.  Ac= cording to popular thought, this would reduce the negative lift load on the=20= tail (and, consequently, the =E2=80=9Cweight=E2=80=9D the main wing was carr= ying), the induced tail drag and thus allow the plane to go faster.  Note that if the h-stab negative cent= er of lift is located 7 feet further back, the weight movement is equivalent= to about 20#  less exerted on= the tail and 20# less the wing has to carry.  The questions to be answered are:

 

  1. With the expected trim change, does this reduce the=20= angle of attack of the main wing and increase the AOA of the h-stab even tho= ugh both are carrying less aerodynamic weight (above example is 20#)?
  2. Is this significant enough to be reflected in an inc= rease in speed?
  3. If the speed increases, is the main wing AOA further= decreased?

 

Relative to query 3 above= , we all note that as speed increases, more nose down trim is added.  This means less negative h-stab lift= , less aerodynamic weight on all the wings still attached.  If you think upside down for a moment, the ext= reme nose down trim at max cruise speed is actually beginning to put the h-s= tab wing flap, the elevator, in reflex!&nb= sp; This is, of course, reducing the negative lift on the h-stab. 

 

Why does the pitch change= occur?  Must it be due to the= speed related pitching moments of the main wing with the flaps in reflex i.= e. nose up, tail down.  Or, is= it a change in the center of lift (center of pressure) closer to the CG?  If it is a change in the Cl tow= ards the CG, that would reduce the load on the tail.  If we reduce the load on the tail enough, moving the= CG rearward may have less of an effect on the tail download, or maybe more=20= leading to potential instability? &nb= sp; Should we install Jim Frantz=E2=80=99s AOA on somebody=E2=80=99s=20= tail (upside down, of course)?  Do I see any hands raised?  = Volunteers to calibrate it?

 

In any event, the experim= ent is set up with 53 pounds of weight as above.  A digital level is affixed parallel to the longeron and=20= my airplane is equipped with the digital AOA.

 

In level flight (under au= to pilot control, smooth air) at 8500 feet (Baro=3D30.19, temp=3D2*C), WOT 2= 3.3=E2=80=9D MAP, 2500 RPM, 9.3 gph, 74% power

 

175 IAS, 197 TAS AOA=3D2.= 4*, Longeron (L)=3D0* and

 

then the weight is transf= erred to the FS 84 Location. The auto pilot requests trim assist and the Rei= chel trim wheel is rotated forward (nose down) about 1/6th of a t= urn.

 

176 IAS, 197 TAS AOA=3D 2= .2* to 2.4*, L=3D0* to -.2*.

 

Leaving the weight in the= back, the power was reduced to 21=E2=80=9D MAP and 2300 RPM, 7.1 gph, 62% p= ower

 

161 IAS, 180 TAS AOA=3D 2= .7* to 2.9*, L=3D.4*

 

The weight was moved back= to the front and the nose was trimmed up

 

161 IAS, 181 TAS AOA=3D2.= 7* to 2.9*, L =3D.4* to .6*

 

Conclusion: I won=E2=80= =99t be abnormally shifting weight to try to increase speed.  Let=E2=80=99s see, during the Air Venture Cu= p Race, Mark=E2=80=99s child was heavier than Larry=E2=80=99s child (moving=20= CG to the rear) yet Larry was faster. = ; I ran about as expected even though I had all my =E2=80=9Cstuff=E2= =80=9D in the back.  Maybe I a= m fast because my engine thrust line was shifted upward which reduces tail d= own load?

 

Other interesting numbers= :

WOT @ 7500 feet 2700 RPM,= 11.3 gph, 83% power

 

185 IAS, 204 TAS AOA=3D2.= 1* L=3D-.2*

 

Climbing out at 130 IAS (= about 2000 MSL) AOA=3D4.1* L=3D7.2* AI=3D5* and,

Still climbing at 8000 MS= L, 140 IAS AOA=3D3.8* L=3D4.6* AI=3D2*

 
Scott Krueger
Sky2high@aol.com
II-P N92EX IO320 Aurora, IL (KARR)
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