X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from omr-m02.mx.aol.com ([64.12.143.76] verified) by logan.com (CommuniGate Pro SMTP 6.0.5) with ESMTP id 6312811 for lml@lancaironline.net; Fri, 07 Jun 2013 09:37:18 -0400 Received-SPF: pass receiver=logan.com; client-ip=64.12.143.76; envelope-from=Sky2high@aol.com Received: from mtaomg-da06.r1000.mx.aol.com (mtaomg-da06.r1000.mx.aol.com [172.29.51.142]) by omr-m02.mx.aol.com (Outbound Mail Relay) with ESMTP id 9EE0F70055C06 for ; Fri, 7 Jun 2013 09:36:41 -0400 (EDT) Received: from core-mtb004a.r1000.mail.aol.com (core-mtb004.r1000.mail.aol.com [172.29.234.205]) by mtaomg-da06.r1000.mx.aol.com (OMAG/Core Interface) with ESMTP id CA6AAE000087 for ; Fri, 7 Jun 2013 09:36:40 -0400 (EDT) From: Sky2high@aol.com Full-name: Sky2high Message-ID: <124854.22ade20b.3ee33be8@aol.com> Date: Fri, 7 Jun 2013 09:36:40 -0400 (EDT) Subject: Re: [LML] Re: loss of power on takeoff To: lml@lancaironline.net MIME-Version: 1.0 Content-Type: multipart/mixed; boundary="part1_124854.22ade20b.3ee33be8_boundary" X-Mailer: AOL 9.6 sub 168 X-Originating-IP: [67.175.156.123] x-aol-global-disposition: G X-AOL-VSS-INFO: 5400.1158/91259 X-AOL-VSS-CODE: clean DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=mx.aol.com; s=20121107; t=1370612201; bh=dYabwnLs3WAapqoBId0DKj/BAST7HOu0qJM/eDj+hjw=; h=From:To:Subject:Message-ID:Date:MIME-Version:Content-Type; b=OMHreF1wqy+Ij+qMwKc3hfQkYwO09DkG+BtnzB3KSvFdpkUcltF7YhqzxU1x9rYX5 0n+ikIG31AhOS6jaD1hkb8wdIeM8eeNvzQfwJKFbA1IoLH5cDpcK1snzIuoEH5lROw gcmF1SpoRITgS+t1PlH+nLwu48nAiJ3Vdv7IcPhQ= X-AOL-SCOLL-SCORE: 1:2:479558304:93952408 X-AOL-SCOLL-URL_COUNT: 18 x-aol-sid: 3039ac1d338e51b1e1e82c7a --part1_124854.22ade20b.3ee33be8_boundary Content-Type: multipart/alternative; boundary="part1_124854.22ade20b.3ee33be8_alt_boundary" --part1_124854.22ade20b.3ee33be8_alt_boundary Content-Type: text/plain; charset="US-ASCII" Content-Transfer-Encoding: 7bit Gary, The numbers don't come out the same. Keep in mind that Dave Morss did a great article in Sport Aviation on this. My original point is in agreement with yours - Does an ordinary pilot in a slick high-wing loaded Lancair have the skill to go to 70 degrees and keep the nose DOWN to get back? There is still some minimum critical altitude and aircraft configuration that the pilot must have fixed in his mind that requires he land within 30 degrees of his heading and keeping the nose DOWN to maintain a speed above stall. Mine was 700 AGL in a 320. Definitely read the Wikipedia extract below with respect to instantaneous turn performance in high wing loaded airplanes. This was my experience too. Snapping into a 60 degree banked constant altitude turn in my 320 didn't require much for the first 15 or so degrees of the turn - then the AOA had to be significantly increased to address the fact that drag and increased wing loading had finally caught up. Attached is the spread sheet that was one I originally used for determining the best way to accomplish a cross country race turn over a pylon, The formulae are from "Aeronautics for Naval Aviators", a very useful reference. It does calcs for bank angles over 60 degrees. I included time and distance numbers for various speeds and bank angles to determine the best configuration for an efficient turn. I added some calculations for considering a power out descending turn back to the airport. I have still been unable to find performance formulae for descending turns but then I haven't tried very hard. A great short article on this stuff is in Wikipedia : http://en.wikipedia.org/wiki/Wing_loading This paragraph is a useful explanation: "Effect on turning performance [_edit_ (http://en.wikipedia.org/w/index.php?title=Wing_loading&action=edit§ion=6) ] To turn, an aircraft must _roll_ (http://en.wikipedia.org/wiki/Flight_dynamics_(aircraft)) in the direction of the turn, increasing the aircraft's _bank angle_ (http://en.wikipedia.org/wiki/Banked_turn#Aviation) . Turning flight lowers the wing's lift component against gravity and hence causes a descent. To compensate, the lift force must be increased by increasing the angle of attack by use of up _elevator_ (http://en.wikipedia.org/wiki/Elevator_(aircraft)) deflection which increases drag. Turning can be described as 'climbing around a circle' (wing lift is diverted to turning the aircraft) so the increase in wing _angle of attack_ (http://en.wikipedia.org/wiki/Angle_of_attack) creates even more drag. The tighter the turn _radius_ (http://en.wikipedia.org/wiki/Radius) attempted, the more drag induced, this requires that power (thrust) be added to overcome the drag. The maximum rate of turn possible for a given aircraft design is limited by its wing size and available engine power: the maximum turn the aircraft can achieve and hold is its sustained turn performance. As the bank angle increases so does the _g-force_ (http://en.wikipedia.org/wiki/G-force) applied to the aircraft, this has the effect of increasing the wing loading and also the _stalling speed_ (http://en.wikipedia.org/wiki/Stall_(flight)) . This effect is also experienced during level _pitching_ (http://en.wikipedia.org/wiki/Pitch_(flight)) maneuvers._[8]_ (http://en.wikipedia.org/wiki/Wing_loading#cite_note-8) Aircraft with low wing loadings tend to have superior sustained turn performance because they can generate more lift for a given quantity of engine thrust. The immediate bank angle an aircraft can achieve before drag seriously bleeds off airspeed is known as its instantaneous turn performance. An aircraft with a small, highly loaded wing may have superior instantaneous turn performance, but poor sustained turn performance: it reacts quickly to control input, but its ability to sustain a tight turn is limited. A classic example is the _F-104 Starfighter_ (http://en.wikipedia.org/wiki/F-104_Starfighter) , which has a very small wing and high wing loading. At the opposite end of the spectrum was the gigantic _Convair B-36_ (http://en.wikipedia.org/wiki/Convair_B-36) . Its large wings resulted in a low wing loading, and there are disputed claims[_who?_ (http://en.wikipedia.org/wiki/Wikipedia:Avoid_weasel_words) ] that this made the bomber more agile than contemporary jet fighters (the slightly later _Hawker Hunter_ (http://en.wikipedia.org/wiki/Hawker_Hunter) had a similar wing loading of 250 kg/m2) at high altitude. Whatever the truth of that, the delta winged _Avro Vulcan_ (http://en.wikipedia.org/wiki/Avro_Vulcan) bomber, with a wing loading of 260 kg/m2 could certainly be rolled at low altitudes._[9]_ (http://en.wikipedia.org/wiki/Wing_loading#cite_note-9) Like any body in _circular motion_ (http://en.wikipedia.org/wiki/Circular_motion) , an aircraft that is fast and strong enough to maintain level flight at speed v in a circle of radius R accelerates towards the centre at . That acceleration is caused by the inward horizontal component of the lift, , where is the banking angle." We all have the level flight AOA, wing load and the force of gravity concepts but details on descending turns - not yet. I defer to Dave. Scott In a message dated 6/7/2013 7:12:04 A.M. Central Daylight Time, casey.gary@yahoo.com writes: Scott, I think the difference when talking about this might be in the assumption of reference axes. I'm using the flight path for the creation of the lift, drag, thrust and gravity directions, but you could use absolute vertical and horizontal axes as well - the numbers come out the same. So in a descent the axis is tilted from horizontal by the angle of the descent (a "zero degree descent" would then be level flight). So the drag is slightly in the up direction, overcoming a tiny bit of gravity. The statement someone made that lift (or more accurately "wing load factor", if you will) is less than 1 G during a descent is technically correct, but practically it is close enough to 1 G to make the discussion moot. For wide deviations from level flight, such as during aerobatics or even the no-power turn back to the airport, the assumption that descent angle doesn't affect the load factor is less accurate. My comment about doing the turn referencing instruments is one borne of inexperience in the conditions of high bank angles, no power and close to ground. I just don't know how one (me, at least) could perform the turn-back accurately referencing outside the window. For instance, what does the ground look like during 60-degree no-power bank? How does prevailing wind affect the view? I know the instruments will look the same as when I practice that at altitude. I'm afraid I lost your spreadsheet, but I have created one in the past that I'm sure says about the same thing. The accuracy of mine decreases at extreme bank angles (over 60), so I don't know whether or not there is a true optimum bank angle. Regardless, I'm sure it is higher than a typical pilot will be able to accurately control. So the target then becomes the highest bank angle that will allow the pilot to maintain a "safe" margin above stall. This discussion has been very useful to me, as are most of the inputs to the List. Gary Gary, Please go back and re-read what I wrote. Where did you find the formulation that the reduction in wing load is related to the cosine of the descent angle in a wings level descent at some arbitrary speed? After all, then the "cosine" of a zero degree descent should be 1. Remember that in a banked turn there are two components, wing loading (perpendicular to the aircraft) and vertical due to gravity. Do not be confused by a term like G as it is only used as a substitute for the weight of the airplane. Please plug some numbers in the spread sheet. Note that as turns become more shallow the turn rate slows, the aircraft travels a greater distance and there is a greater loss of altitude. If you are below a critical altitude, a 30 degree banked turn will not get you back. Instrument reference is best used for engine failures at night or that risky takeoff in 0-0 conditions. Good Luck, Scott In a message dated 6/6/2013 3:44:42 P.M. Central Daylight Time, _casey.gary@yahoo.com_ (mailto:casey.gary@yahoo.com) writes: Terrence, Scott and George are all correct - sort of. George said in a steady-state descent the wing is still supporting 1 G, but Scott says it is supporting less than 1 G because of the descent. For a typical descent of 500 ft/min at a speed of 180 statute miles per hour the wing is supporting 99.95 percent of the weight (the cosine of the descent angle, which is 1.8 degrees). So for all normal climbs and descents George is essentially correct. Of course, for a vertical climb or descent the wing supports nothing. For a higher descent angle, such as for turning back with no engine, is the descent angle significant enough to change the stall speed? I haven't run the numbers, but I suspect it is a very small factor compared to the increase in lift required for the bank. Here's the technique I think is theoretically correct, and one that I have practiced. As Terrence said, "Angle, angle, angle." When power failure is first perceived, simultaneously roll into a steep bank while keeping the AOA at the optimum value with back pressure on the stick. Initially, that will require forward stick movement - remember, just because the plane is banked doesn't mean the G force goes up. Now as the airspeed increases, increase back pressure to hold the same AOA. With an AOA-indicator-equipped plane you only control 2 things - the bank angle and AOA. When you are again pointed at the runway (at an angle, but don't be picky) immediately level the wings. What bank angle? I haven't run the numbers, but as Dave has said, it is a steep angle. The steeper the angle the more difficult the maneuver, so I have picked 45 degrees as my personal target. 60, 70, or even more might be the theoretical optimum, but that requires more skill than I think I would have in a crisis situation. The completion of the turn could happen quite close to the ground, but the extra speed required for the turn will be used to arrest the rapid descent and return to the "normal" glide speed (remember to hold the AOA after the wings are level). What to do if your plane is not AOA-indicator-equipped? The maneuver is still the same, but you have to control G loading as a function of airspeed. Of course, you likely don't have a G meter either, so you have to use your own derriere for that purpose. The mental gymnastics get to be a real challenge and I suspect that very few pilots would be able to accurately control AOA during the maneuver. The result is that the bank angle has to be reduced to maintain some degree of accuracy in AOA. I would guess that 30 degrees bank might be a good target for most non test pilots. If you get the AOA too high you will certainly arrive at crash scene much sooner - too low and you will lose more altitude than necessary. I'm sure that the maneuver can best be performed in reference solely to instruments, as the view of the ground, close-up, oddly angled and rapidly rotating would be a huge distraction. Practicing at altitude doesn't really prepare one for that. However, it does prepare you to concentrate on the instruments, and that might help. In principle, the turn is exactly similar (my favorite words) to the Chandelle performed for the Commercial ticket, except done without power. Just my 2 cents worth, Gary Casey --part1_124854.22ade20b.3ee33be8_alt_boundary Content-Type: text/html; charset="US-ASCII" Content-Transfer-Encoding: quoted-printable
Gary,
 
The numbers don't come out the same.
 
Keep in mind that Dave Morss did a great article in Sport Aviation on= =20 this.  My original point is in agreement with yours - Does an ordinary= =20 pilot in a slick high-wing loaded Lancair have the skill to go to 70 degree= s and=20 keep the nose DOWN to get back?  There is still some minimum critical= =20 altitude and aircraft configuration that the pilot must have fixe= d in=20 his mind that requires he land within 30 degrees of his heading and keeping= the=20 nose DOWN to maintain a speed above stall.  Mine was 700 AGL in a=20 320.  Definitely read the Wikipedia extract below with respect to= =20 instantaneous turn performance in high wing loaded airplanes.  Th= is=20 was my experience too.  Snapping into a 60 degree banked con= stant=20 altitude turn in my 320 didn't require much for the first 15 or=20 so degrees of the turn - then the AOA had to be significantly increase= d to=20 address the fact that drag and increased wing loading had finally caug= ht=20 up.
 
Attached is the spread sheet that was one I originally used = for=20 determining the best way to accomplish a cross country race turn over a=20 pylon,  The formulae are from "Aeronautics for Naval Aviators", a very= =20 useful reference.  It does calcs for bank angles over 60 degrees.= =20  I included time and distance numbers for various speeds and bank angl= es to=20 determine the best configuration for an efficient turn.  I added some= =20 calculations for considering a power out descending turn back to the=20 airport.  I have still been unable to find performance formulae f= or=20 descending turns but then I haven't tried very hard.
 
A great short article on this stuff is in Wikipedia : http://en.wikipedia.org/= wiki/Wing_loading
 
This paragraph is a useful explanation:
 
"Effect o= n turning=20 performance [edit]

To turn, an aircraft must roll i= n the=20 direction of the turn, increasing the aircraft's bank angle. = Turning=20 flight lowers the wing's lift component against gravity and hence causes a= =20 descent. To compensate, the lift force must be increased by increasing the = angle=20 of attack by use of up elevator defl= ection=20 which increases drag. Turning can be described as 'climbing around a circle= '=20 (wing lift is diverted to turning the aircraft) so the increase in wing angle of attack c= reates=20 even more drag. The tighter the turn radius attempted, the more= drag=20 induced, this requires that power (thrust) be added to overcome the drag. T= he=20 maximum rate of turn possible for a given aircraft design is limited by its= wing=20 size and available engine power: the maximum turn the aircraft can achieve = and=20 hold is its sustained turn performance. As the bank angle increases = so=20 does the g-force applied to the ai= rcraft,=20 this has the effect of increasing the wing loading and also the stalling speed. Th= is=20 effect is also experienced during level pitching maneuvers= .[8]

Aircraft with low wing loadings tend to have superior sustained turn=20 performance because they can generate more lift for a given quantity of eng= ine=20 thrust. The immediate bank angle an aircraft can achieve before drag seriou= sly=20 bleeds off airspeed is known as its instantaneous turn performance. = An=20 aircraft with a small, highly loaded wing may have superior instantaneous t= urn=20 performance, but poor sustained turn performance: it reacts quickly to cont= rol=20 input, but its ability to sustain a tight turn is limited. A classic exampl= e is=20 the F-104 Starfighter,=20 which has a very small wing and high wing loading. At the opposite end of t= he=20 spectrum was the gigantic Convair B-36. Its la= rge=20 wings resulted in a low wing loading, and there are disputed claims[who?]=20 that this made the bomber more agile than contemporary jet fighters (the=20 slightly later Hawker Hunter had a= =20 similar wing loading of 250 kg/m2) at high altitude. Whatev= er=20 the truth of that, the delta winged Avro Vulcan bomber, w= ith a=20 wing loading of 260 kg/m2 could certainly be rolled at low= =20 altitudes.[9]

Like any body in circular motion, = an=20 aircraft that is fast and strong enough to maintain level flight at speed= =20 v in a circle of radius R accelerates towards the centre at <= IMG=20 class=3Dtex alt=3D"\scriptstyle\frac{v^2} {R}"=20 src=3D"http://upload.wikimedia.org/math/f/e/6/fe6f2e7175254cb8c29f31971c513= 27b.png">.=20 That acceleration is caused by the inward horizontal component of the lift,= ,=20 where 3D\theta=20=20 is the banking angle."

We all have the level flight AOA, wing load and the force of gravity= =20 concepts but details on descending turns - not yet.  I defer to Dave.<= /DIV>
 
Scott
 
 
In a message dated 6/7/2013 7:12:04 A.M. Central Daylight Time,=20 casey.gary@yahoo.com writes:
=
Scott,
I think the difference when talking about this might be in the assum= ption=20 of reference axes.  I'm using the flight path for the creation of th= e=20 lift, drag, thrust and gravity directions, but you could use absolute ver= tical=20 and horizontal axes as well - the numbers come out the same.  So in = a=20 descent the axis is tilted from horizontal by the angle of the descent (a= =20 "zero degree descent" would then be level flight).  So the drag is= =20 slightly in the up direction, overcoming a tiny bit of gravity.  The= =20 statement someone made that lift (or more accurately "wing load factor", = if=20 you will) is less than 1 G during a descent is technically correct, but= =20 practically it is close enough to 1 G to make the discussion moot.  = For=20 wide deviations from level flight, such as during aerobatics or even the= =20 no-power turn back to the airport, the assumption that descent angle does= n't=20 affect the load factor is less accurate.

My=20 comment about doing the turn referencing instruments is one borne of=20 inexperience in the conditions of high bank angles, no power and close to= =20 ground.  I just don't know how one (me, at least) could perform the= =20 turn-back accurately referencing outside the window.  For instance, = what=20 does the ground look like during 60-degree no-power bank?  How does= =20 prevailing wind affect the view?  I know the instruments will look t= he=20 same as when I practice that at altitude.  

I'm=20 afraid I lost your spreadsheet, but I have created one in the past that I= 'm=20 sure says about the same thing.  The accuracy of mine decreases at= =20 extreme bank angles (over 60), so I don't know whether or not there is a = true=20 optimum bank angle.  Regardless, I'm sure it is higher than a typica= l=20 pilot will be able to accurately control.  So the target then become= s the=20 highest bank angle that will allow the pilot to maintain a "safe" margin = above=20 stall.  This discussion has been very useful to me, as are most of t= he=20 inputs to the List. Gary
Gary,

Please=20 go back and re-read what I wrote.  Where did you find=20 the  
formulation that the reduction in wing load is related= to=20 the cosine of the  descent 
angle in a wings level descent a= t=20 some arbitrary speed?  After all,  then 
the "cosine" o= f a=20 zero degree descent should be 1.

Remember that in a banked turn th= ere=20 are two components, wing loading  
(perpendicular to the=20 aircraft) and vertical due to gravity.  Do not be  
con= fused=20 by a term like G as it is only used as a substitute for the weight  = of=20 the 
airplane.  

Please plug some numbers in the= =20 spread sheet.  Note that as turns  become 
more shallow= the=20 turn rate slows, the aircraft travels a  greater distance 
a= nd=20 there is a greater loss of altitude.  If you are below  a=20 critical 
altitude, a 30 degree banked turn will not get you=20 back.  

Instrument reference is best used for engine fai= lures=20 at night or that  
risky takeoff in 0-0 conditions.

G= ood=20 Luck,

Scott



In a message dated 6/6/2013 3:44:42 P.M= .=20 Central Daylight Time,  
casey.gary@yahoo.com wri= tes:


Terrence,=20 Scott and George are all correct - sort of.  George said  in=20 a 
steady-state descent the wing is still supporting 1 G, but Sco= tt=20 says it  is 
supporting less than 1 G because of the=20 descent.  For a typical  descent of 
500 ft/min at a sp= eed=20 of 180 statute miles per hour the wing is  supporting 
99.95= =20 percent of the weight (the cosine of the descent angle, which  is=20 1.8 
degrees).  So for all normal climbs and descents George= =20 is  essentially 
correct.  Of course, for a vertical cl= imb=20 or descent the wing  supports nothing.  
For a higher= =20 descent angle, such as for turning back  with no engine, is the=20 descent angle significant enough to change the stall  speed?  I= =20 haven't 
run the numbers, but I suspect it is a very small = =20 factor compared to the 
increase in lift required for the=20 bank.


Here's  the technique I think is theoretically corr= ect,=20 and one that I have 
practiced.  As Terrence said, "Angle, a= ngle,=20 angle."  When power failure is 
first perceived, simultaneou= sly=20 roll into a steep bank while keeping the 
AOA  at the optimu= m=20 value with back pressure on the stick.  Initially,=20 that  
will require forward stick movement - remember, just= =20 because the plane is  
banked doesn't mean the G force goes= =20 up.  Now as the airspeed increases,  
increase back pre= ssure=20 to hold the same AOA.  With an =20 AOA-indicator-equipped 
plane you only control 2 things - the ban= k=20 angle and  AOA.  When you are 
again pointed at the run= way=20 (at an angle, but don't be  picky) immediately level 
the=20 wings.  What bank angle?  I haven't run  the numbers, but = as=20 Dave has 
said, it is a steep angle.  The steeper the  = angle=20 the more difficult the 
maneuver, so I have picked 45 degrees as= =20 my  personal target.  60, 70, or even 
more might be th= e=20 theoretical optimum,  but that requires more skill than I 
t= hink=20 I would have in a crisis situation.  The completion of the turn=20 could 
happen quite close to the ground, but  the extra spee= d=20 required for the turn 
will be used to arrest the rapid descent&n= bsp;=20 and return to the "normal" glide 
speed  (remember to hold t= he=20 AOA after  the wings are level).


What  to do if your= =20 plane is not AOA-indicator-equipped?  The maneuver=20 is 
still  the same, but you have to control G loading as a= =20 function of airspeed. 
Of course, you likely don't have a G meter= =20 either, so you have to use  
your own derriere for that=20 purpose.  The mental gymnastics get to be a  real 
chal= lenge=20 and I suspect that very few pilots would be able to=20 accurately  
control AOA during the maneuver.  The resu= lt is=20 that the bank angle has  to be 
reduced to maintain some deg= ree=20 of accuracy in AOA.  I would guess  that 30 
degrees ba= nk=20 might be a good target for most non test pilots.  If  you=20 get 
the AOA too high you will certainly arrive at crash scene mu= ch=20 sooner  - too 
low and you will lose more altitude than=20 necessary.


I'm  sure that the maneuver can best be perfor= med=20 in reference solely to  
instruments, as the view of the gro= und,=20 close-up, oddly angled and rapidly  
rotating would be a hug= e=20 distraction.  Practicing at altitude doesn't =20 really 
prepare one for that.  However, it does prepare you = to=20 concentrate  on the 
instruments, and that might help. = In=20 principle, the turn is  exactly 
similar (my favorite words)= to=20 the Chandelle performed for the  Commercial ticket, 
except = done=20 without power.


Just  my 2 cents worth,
Gary =20 Casey
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