X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from omr-d10.mx.aol.com ([205.188.108.134] verified) by logan.com (CommuniGate Pro SMTP 6.0.5) with ESMTP id 6311972 for lml@lancaironline.net; Thu, 06 Jun 2013 18:45:44 -0400 Received-SPF: pass receiver=logan.com; client-ip=205.188.108.134; envelope-from=Sky2high@aol.com Received: from mtaomg-ma03.r1000.mx.aol.com (mtaomg-ma03.r1000.mx.aol.com [172.29.41.10]) by omr-d10.mx.aol.com (Outbound Mail Relay) with ESMTP id D0F3E700496C7 for ; Thu, 6 Jun 2013 18:45:07 -0400 (EDT) Received: from core-mta001c.r1000.mail.aol.com (core-mta001.r1000.mail.aol.com [172.29.234.129]) by mtaomg-ma03.r1000.mx.aol.com (OMAG/Core Interface) with ESMTP id 9945AE000085 for ; Thu, 6 Jun 2013 18:45:07 -0400 (EDT) From: Sky2high@aol.com Full-name: Sky2high Message-ID: <11ab2d.2b1dcc4c.3ee26af3@aol.com> Date: Thu, 6 Jun 2013 18:45:07 -0400 (EDT) Subject: Re: [LML] Re: loss of power on takeoff To: lml@lancaironline.net MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="part1_11ab2d.2b1dcc4c.3ee26af3_boundary" X-Mailer: AOL 9.6 sub 168 X-Originating-IP: [67.175.156.123] x-aol-global-disposition: G DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=mx.aol.com; s=20121107; t=1370558707; bh=tUoIgAhIMEJocJa8OL8DAVcpqZAEz4Ldajqw6ItZGJA=; h=From:To:Subject:Message-ID:Date:MIME-Version:Content-Type; b=Za1+25VaKnZK6kXcvFGtkzUgqi28O49zfNsHYMCaWW/G/eIhKQm7WLVI+LBHOBjB2 wroiZp5CNG8/YU6IR089lyt/2dt5lCChi+9zEX6+rnKRg360CO2+jsIAg0k/6gD4yL xqpYkwSFdkw+my7iKRw31qR0WD/CTEU0QxxtC0eA= X-AOL-SCOLL-SCORE: 0:2:461474592:93952408 X-AOL-SCOLL-URL_COUNT: 0 x-aol-sid: 3039ac1d290a51b110f368d6 --part1_11ab2d.2b1dcc4c.3ee26af3_boundary Content-Type: text/plain; charset="US-ASCII" Content-Transfer-Encoding: 7bit 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 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 Yes Terrence, AOA. No mental exercise necessary if one has an AOA sensor. And, Charles' comment is a bit off. In level flight, the wing AOA provides sufficient lift (wing loading) to equal the effect of the force of gravity (1 G) on the aircraft weight (W). Thrust overcomes drag to result in forward speed. In a descent at the same speed used in level flight, lift is less than W and either power (thrust) is reduced or drag is increased. Remember that G is just for relative reference. Again, in level flight at the same power, but in a coordinated banked turn, the wing AOA has been increased to add enough bank angle lift necessary to maintain 1 G with respect to the vertical. I.E. The wing load must be increased to keep the plane at the same altitude - The lift has to equal the weight divided by the cosine of the bank angle. To visualize: One could redraw this with force vectors to see it better. Of course, because of increased load, the induced drag is also increased. Finally, in a coordinated banked turn without power and even further drag from other bits and pieces, descent (glide) will occur unless the AOA could be increased provide sufficient lift to offset the vertical component (the pull of gravity). But, there is a limit AOA at which a stall would occur - thus descent. In a banked turning descent at a certain speed (best glide for the conditions), less lift is required, thus less load on the wing, thus a lower stall speed than a higher load. This supports the statements made by both Dave Morss and myself. Dave's point is that large bank angle conducted at a optimal speed shortens the time (distance) and lessens the altitude loss plus in the descent the stall speed is not as great as that in the same bank holding altitude. An optimal speed is somewhere above stall speed. Factors affecting stall speed are load and drag (wheels, flaps, prop, etc.) - hence the requirement that you point the nose down making use of kinetic energy rather than gasoline to keep up the speed. Uh, the Aeronautics for Naval Aviators is silent on powerless descending turns (maybe a glider tech manual would be more informative). I have included the simplified Excel spreadsheet to give you a feel for some of these parameters before testing at high altitudes. Blue Skies, Scott Krueger In a message dated 6/3/2013 8:29:23 A.M. Central Daylight Time, _troneill@charter.net_ (mailto:troneill@charter.net) writes: Angle, angle, angle. Angle of stall is constant, no matter what. Simpler, not requiring mental gymnastics. Terrence. Sent from my iPad On Jun 3, 2013, at 7:03 AM, Charles Brown <__browncc1@verizon.net__ (mailto:_browncc1@verizon.net_) (mailto:_browncc1@verizon.net_ (mailto:browncc1@verizon.net) ) > wrote: In a straight ahead descent, the wing is producing 1g lift and the stall speed is the same as in level flight. You guys may be thinking of the change in stall speed when *initiating* a descent (pushover, less than 1g for a moment), or when *terminating* a descent (pull-up, or flare, momentarily more than 1g). --part1_11ab2d.2b1dcc4c.3ee26af3_boundary Content-Type: text/html; charset="US-ASCII" Content-Transfer-Encoding: quoted-printable
Gary,
 
Please go back and re-read what I wrote.  Where did you find= the=20 formulation that the reduction in wing load is related to the cosine of the= =20 descent angle in a wings level descent at some arbitrary speed?  After= all,=20 then the "cosine" of a zero degree descent should be 1.
 
Remember that in a banked turn there are two components, wing loading= =20 (perpendicular to the aircraft) and vertical due to gravity.  Do not b= e=20 confused by a term like G as it is only used as a substitute for the w= eight=20 of the airplane. 
 
Please plug some numbers in the spread sheet.  Note that as turns= =20 become more shallow the turn rate slows, the aircraft travels a= =20 greater distance and there is a greater loss of altitude.  If you are = below=20 a critical altitude, a 30 degree banked turn will not get you back.  <= /DIV>
 
Instrument reference is best used for engine failures at night or= that=20 risky takeoff in 0-0 conditions.
 
Good Luck,
 
Scott
 
 
In a message dated 6/6/2013 3:44:42 P.M. Central Daylight Time,=20 casey.gary@yahoo.com writes:
=
Terrence, Scott and George are all correct - sort of.  George s= aid=20 in a steady-state descent the wing is still supporting 1 G, but Scott say= s it=20 is supporting less than 1 G because of the descent.  For a typical= =20 descent of 500 ft/min at a speed of 180 statute miles per hour the wing i= s=20 supporting 99.95 percent of the weight (the cosine of the descent angle, = which=20 is 1.8 degrees).  So for all normal climbs and descents George is=20 essentially correct.  Of course, for a vertical climb or descent the= wing=20 supports nothing.  For a higher descent angle, such as for turning b= ack=20 with no engine, is the descent angle significant enough to change the sta= ll=20 speed?  I haven't run the numbers, but I suspect it is a very small= =20 factor compared to the increase in lift required for the bank.

Here's=20 the technique I think is theoretically correct, and one that I have pract= iced.=20  As Terrence said, "Angle, angle, angle."  When power failure i= s=20 first perceived, simultaneously roll into a steep bank while keeping the = AOA=20 at the optimum value with back pressure on the stick.  Initially, th= at=20 will require forward stick movement - remember, just because the plane is= =20 banked doesn't mean the G force goes up.  Now as the airspeed increa= ses,=20 increase back pressure to hold the same AOA.  With an=20 AOA-indicator-equipped plane you only control 2 things - the bank angle a= nd=20 AOA.  When you are again pointed at the runway (at an angle, but don= 't be=20 picky) immediately level the wings.  What bank angle?  I haven'= t run=20 the numbers, but as Dave has said, it is a steep angle.  The steeper= the=20 angle the more difficult the maneuver, so I have picked 45 degrees as my= =20 personal target.  60, 70, or even more might be the theoretical opti= mum,=20 but that requires more skill than I think I would have in a crisis situat= ion.=20  The completion of the turn could happen quite close to the ground, = but=20 the extra speed required for the turn will be used to arrest the rapid de= scent=20 and return to the "normal" glide speed  (remember to hold the AOA af= ter=20 the wings are level).

What=20 to do if your plane is not AOA-indicator-equipped?  The maneuver is = still=20 the same, but you have to control G loading as a function of airspeed.=20  Of course, you likely don't have a G meter either, so you have to u= se=20 your own derriere for that purpose.  The mental gymnastics get to be= a=20 real challenge and I suspect that very few pilots would be able to accura= tely=20 control AOA during the maneuver.  The result is that the bank angle = has=20 to be reduced to maintain some degree of accuracy in AOA.  I would g= uess=20 that 30 degrees bank might be a good target for most non test pilots. &nb= sp;If=20 you get the AOA too high you will certainly arrive at crash scene much so= oner=20 - too low and you will lose more altitude than necessary.
I'm=20 sure that the maneuver can best be performed in reference solely to=20 instruments, as the view of the ground, close-up, oddly angled and rapidl= y=20 rotating would be a huge distraction.  Practicing at altitude doesn'= t=20 really prepare one for that.  However, it does prepare you to concen= trate=20 on the instruments, and that might help.  In principle, the turn is= =20 exactly similar (my favorite words) to the Chandelle performed for the=20 Commercial ticket, except done without power.
Just=20 my 2 cents worth, Gary=20 Casey

Yes=20 Terrence, AOA.  No mental exercise necessary if one has an AOA = =20 sensor. 
And, Charles' comment is a bit off.

In level flig= ht,=20 the wing AOA provides sufficient lift (wing  loading) to 
eq= ual=20 the effect of the force of gravity (1 G) on the  aircraft weight=20 (W).  
Thrust overcomes drag to result in forward =20 speed. 

In a descent at the same speed used in level flight, = lift=20 is less than W  
and either power (thrust) is reduced or dra= g is=20 increased.  Remember that G  
is just for relative=20 reference.

Again, in level flight at the same power, but in a=20 coordinated  banked 
turn, the wing AOA has been increased t= o add=20 enough  bank angle lift necessary 
to maintain 1 G with resp= ect=20 to the vertical.  I.E. The wing load must be 
increased to k= eep=20 the plane at the same altitude  - The lift has to equal=20 the 
weight divided by the cosine of the bank  angle.  = To=20 visualize:



One could redraw this with force vectors to see= it=20 better.  Of course,  
because of increased load, the in= duced=20 drag is  also increased.

Finally, in a coordinated banked tur= n=20 without power and even further  drag 
from other bits and pi= eces,=20 descent (glide) will occur unless the AOA could  
be increas= ed=20 provide sufficient lift to offset the vertical component=20 (the  
pull of gravity).  But, there is a limit AOA at = which=20 a stall would occur - 
thus descent.  In a banked turning de= scent=20 at a certain speed (best glide  
for the conditions), less l= ift=20 is required, thus less load on the wing, 
thus a  lower stal= l=20 speed than a higher load.  This supports the=20 statements  
made by both Dave Morss and myself.  Dave'= s=20 point is that large bank angle  
conducted at a optimal spee= d=20 shortens the time (distance) and lessens  the 
altitude loss= plus=20 in the descent the stall speed is not as great as  that in 
= the=20 same bank holding altitude.  

An optimal speed is somewh= ere=20 above stall speed.  Factors affecting  stall 
speed are= load=20 and drag (wheels, flaps, prop, etc.) - hence the =20 requirement 
that you point the nose down making use of kinetic e= nergy=20 rather  than 
gasoline to keep up the speed.

Uh, the= =20 Aeronautics for Naval Aviators is silent on powerless=20 descending  
turns (maybe a glider tech manual would be more= =20 informative).  I have  
included the simplified Excel= =20 spreadsheet to give you a feel for some of  these 
parameter= s=20 before testing at high altitudes.

Blue Skies,

Scott=20 Krueger


In a message dated 6/3/2013 8:29:23 A.M. Central Dayli= ght=20 Time,  
troneill@charter.net wri= tes:

Angle,=20 angle, angle.  Angle of stall is constant, no matter what. =20 Simpler, 
not requiring mental gymnastics.
Terrence.

Se= nt=20 from my iPad

On Jun 3, 2013, at 7:03 AM, Charles Brown <_browncc1@verizon.net_ = ;
(mailto:browncc1@verizon.net) >&nb= sp;=20 wrote:




In a straight ahead descent, the wing is produc= ing=20 1g lift and the  stall 
speed is the same as in level=20 flight.  You guys may be  thinking of the 
change in st= all=20 speed when *initiating* a descent (pushover,  less than 1g for=20 a 
moment), or when *terminating* a descent (pull-up, or  fl= are,=20 momentarily 
more than=20 1g).
 --part1_11ab2d.2b1dcc4c.3ee26af3_boundary--