X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Sender: To: lml@lancaironline.net Date: Thu, 06 Jun 2013 16:44:15 -0400 Message-ID: X-Original-Return-Path: Received: from nm46-vm2.bullet.mail.ne1.yahoo.com ([98.138.121.82] verified) by logan.com (CommuniGate Pro SMTP 6.0.5) with ESMTPS id 6310388 for lml@lancaironline.net; Thu, 06 Jun 2013 07:02:25 -0400 Received-SPF: none receiver=logan.com; client-ip=98.138.121.82; envelope-from=casey.gary@yahoo.com Received: from [98.138.90.54] by nm46.bullet.mail.ne1.yahoo.com with NNFMP; 06 Jun 2013 11:01:49 -0000 Received: from [98.138.89.175] by tm7.bullet.mail.ne1.yahoo.com with NNFMP; 06 Jun 2013 11:01:49 -0000 Received: from [127.0.0.1] by omp1031.mail.ne1.yahoo.com with NNFMP; 06 Jun 2013 11:01:49 -0000 X-Yahoo-Newman-Property: ymail-3 X-Yahoo-Newman-Id: 416939.81555.bm@omp1031.mail.ne1.yahoo.com Received: (qmail 8050 invoked by uid 60001); 6 Jun 2013 11:01:49 -0000 DomainKey-Signature:a=rsa-sha1; q=dns; c=nofws; s=s1024; d=yahoo.com; h=X-YMail-OSG:Received:X-Rocket-MIMEInfo:X-Mailer:References:Message-ID:Date:From:Reply-To:Subject:To:In-Reply-To:MIME-Version:Content-Type; b=p+/OCff3mLZcsB/RcyMSppb46UY4ce7RSr3XZF1moLIYWNu/IzkVVvkOIc2xsBkE2XB/mLXzRMhm5ektqOkB48757fFzw7U+6H2hNX9J71GzNsbIQDT2tO3Jf4Ba6X37T1ybHadogaT/yWMmqhRuZPrMBB4mO/SO7p41RHBwd6c=; X-YMail-OSG: OHGZzXIVM1nqw2sJ5prTEcbQL9XK6E8XCnMuBj7CaCCGUw_ 6dlDsIJhBapeKXsTR0TLM0bj50fIgZfPHXd1X4uoV_76RR5rol9M_bLP0RAc oBkPjSHy81NoDqCQB.EMw0BYpz7UdGkn3Iq52GkW1l.3JyucpFSK04O9H2lW erxE9Y53uX2dr6.nn0AjNpbUBITGKBFo0SZxx5CMhyhB80NRJKDqjHLfVaxn DNxKehLUZYRfB9UBZi.lS4R9aCZaJ0XPz4EHb8.s4FzohwdsHuHLMAqLYtet HY9XNE4a1MYIBKDLZSwj5O8Nm2T9q3ulkwxUD68B8kzrR3CPCnUnz6R6OaBx bdXJlHxhhdeW3jUi1RdIyeByyjD7ZbIGYAKuiU7vb91nRc0zBpL.HZnQLgcE vd6FxaRuOVwV9x7j20_fZh3nSgc23P_xPblFArP4xzmLwsPbqowP5kFXpNr6 2d9plp0N4NGVF11kDz3VCneXdbG1F6PEtlZdubmA7wti1t_RjSveuqUSt_GQ v3tQM1uByxItEUhK7gySdZX0JZlbvQbh4Yf_aLy2P65Q9WFiYi.lRss1gR1S 3NB4- Received: from [97.92.63.83] by web120103.mail.ne1.yahoo.com via HTTP; Thu, 06 Jun 2013 04:01:49 PDT X-Rocket-MIMEInfo: 002.001,VGVycmVuY2UsIFNjb3R0IGFuZCBHZW9yZ2UgYXJlIGFsbCBjb3JyZWN0IC0gc29ydCBvZi4gwqBHZW9yZ2Ugc2FpZCBpbiBhIHN0ZWFkeS1zdGF0ZSBkZXNjZW50IHRoZSB3aW5nIGlzIHN0aWxsIHN1cHBvcnRpbmcgMSBHLCBidXQgU2NvdHQgc2F5cyBpdCBpcyBzdXBwb3J0aW5nIGxlc3MgdGhhbiAxIEcgYmVjYXVzZSBvZiB0aGUgZGVzY2VudC4gwqBGb3IgYSB0eXBpY2FsIGRlc2NlbnQgb2YgNTAwIGZ0L21pbiBhdCBhIHNwZWVkIG9mIDE4MCBzdGF0dXRlIG1pbGVzIHBlciBob3VyIHRoZSB3aW5nIGlzIHMBMAEBAQE- X-Mailer: YahooMailWebService/0.8.144.546 References: X-Original-Message-ID: <1370516509.362.YahooMailNeo@web120103.mail.ne1.yahoo.com> X-Original-Date: Thu, 6 Jun 2013 04:01:49 -0700 (PDT) From: Gary Casey Reply-To: Gary Casey Subject: Re: loss of power on takeoff X-Original-To: Lancair Mailing List In-Reply-To: MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="-955686164-1736898827-1370516509=:362" ---955686164-1736898827-1370516509=:362 Content-Type: text/plain; charset=iso-8859-1 Content-Transfer-Encoding: quoted-printable Terrence, Scott and George are all correct - sort of. =A0George said in a s= teady-state descent the wing is still supporting 1 G, but Scott says it is = supporting less than 1 G because of the descent. =A0For a typical descent o= f 500 ft/min at a speed of 180 statute miles per hour the wing is supportin= g 99.95 percent of the weight (the cosine of the descent angle, which is 1.= 8 degrees). =A0So for all normal climbs and descents George is essentially = correct. =A0Of course, for a vertical climb or descent the wing supports no= thing. =A0For a higher descent angle, such as for turning back with no engi= ne, is the descent angle significant enough to change the stall speed? =A0I= haven't run the numbers, but I suspect it is a very small factor compared = to the increase in lift required for the bank.=0A=0AHere's the technique I = think is theoretically correct, and one that I have practiced. =A0As Terren= ce said, "Angle, angle, angle." =A0When power failure is first perceived, s= imultaneously roll into a steep bank while keeping the AOA at the optimum v= alue with back pressure on the stick. =A0Initially, that will require forwa= rd stick movement - remember, just because the plane is banked doesn't mean= the G force goes up. =A0Now as the airspeed increases, increase back press= ure to hold the same AOA. =A0With an AOA-indicator-equipped plane you only = control 2 things - the bank angle and AOA. =A0When you are again pointed at= the runway (at an angle, but don't be picky) immediately level the wings. = =A0What bank angle? =A0I haven't run the numbers, but as Dave has said, it = is a steep angle. =A0The steeper the angle the more difficult the maneuver,= so I have picked 45 degrees as my personal target. =A060, 70, or even more= might be the theoretical optimum, but that requires more skill than I think I would have in a crisis situation. =A0The complet= ion 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 =A0(remember to hold the AOA after the wings ar= e level).=0A=0AWhat to do if your plane is not AOA-indicator-equipped? =A0T= he maneuver is still the same, but you have to control G loading as a funct= ion of airspeed. =A0Of course, you likely don't have a G meter either, so y= ou have to use your own derriere for that purpose. =A0The mental gymnastics= get to be a real challenge and I suspect that very few pilots would be abl= e to accurately control AOA during the maneuver. =A0The result is that the = bank angle has to be reduced to maintain some degree of accuracy in AOA. = =A0I would guess that 30 degrees bank might be a good target for most non t= est pilots. =A0If you get the AOA too high you will certainly arrive at cra= sh scene much sooner - too low and you will lose more altitude than necessa= ry.=0A=0AI'm sure that the maneuver can best be performed in reference sole= ly to instruments, as the view of the ground, close-up, oddly angled and ra= pidly rotating would be a huge distraction. =A0Practicing at altitude doesn= 't really prepare one for that. =A0However, it does prepare you to concentr= ate on the instruments, and that might help. =A0In principle, the turn is e= xactly similar (my favorite words) to the Chandelle performed for the Comme= rcial ticket, except done without power.=0A=0AJust my 2 cents worth,=0AGary= Casey=0A=0A=0AYes Terrence, AOA.=A0 No mental exercise necessary if one ha= s an AOA=A0 sensor.=A0=0AAnd, Charles' comment is a bit off.=0A=0AIn level = flight, the wing AOA provides sufficient lift (wing=A0 loading) to=A0=0Aequ= al the effect of the force of gravity (1 G) on the=A0 aircraft weight (W).= =A0=A0=0AThrust overcomes drag to result in forward=A0 speed.=A0=0A=0AIn a = descent at the same speed used in level flight, lift is less than W=A0=A0= =0Aand either power (thrust) is reduced or drag is increased.=A0 Remember t= hat G=A0=A0=0Ais just for relative reference.=0A=0AAgain, in level flight a= t the same power, but in a coordinated=A0 banked=A0=0Aturn, the wing AOA ha= s been increased to add enough=A0 bank angle lift necessary=A0=0Ato maintai= n 1 G with respect to the vertical.=A0 I.E. The wing load must be=A0=0Aincr= eased to keep the plane at the same altitude=A0 - The lift has to equal the= =A0=0Aweight divided by the cosine of the bank=A0 angle.=A0 To visualize:= =0A=0A=0A=0AOne could redraw this with force vectors to see it better.=A0 O= f course,=A0=A0=0Abecause of increased load, the induced drag is=A0 also in= creased.=0A=0AFinally, in a coordinated banked turn without power and even = further=A0 drag=A0=0Afrom other bits and pieces, descent (glide) will occur= unless the AOA could=A0=A0=0Abe increased provide sufficient lift to offse= t the vertical component (the=A0=A0=0Apull of gravity).=A0 But, there is a = limit AOA at which a stall would occur -=A0=0Athus descent.=A0 In a banked = turning descent at a certain speed (best glide=A0=A0=0Afor the conditions),= less lift is required, thus less load on the wing,=A0=0Athus a=A0 lower st= all speed than a higher load.=A0 This supports the statements=A0=A0=0Amade = by both Dave Morss and myself.=A0 Dave's point is that large bank angle=A0= =A0=0Aconducted at a optimal speed shortens the time (distance) and lessens= =A0 the=A0=0Aaltitude loss plus in the descent the stall speed is not as gr= eat as=A0 that in=A0=0Athe same bank holding altitude.=A0=A0=0A=0AAn optima= l speed is somewhere above stall speed.=A0 Factors affecting=A0 stall=A0=0A= speed are load and drag (wheels, flaps, prop, etc.) - hence the=A0 requirem= ent=A0=0Athat you point the nose down making use of kinetic energy rather= =A0 than=A0=0Agasoline to keep up the speed.=0A=0AUh, the Aeronautics for N= aval Aviators is silent on powerless descending=A0=A0=0Aturns (maybe a glid= er tech manual would be more informative).=A0 I have=A0=A0=0Aincluded the s= implified Excel spreadsheet to give you a feel for some of=A0 these=A0=0Apa= rameters before testing at high altitudes.=0A=0ABlue Skies,=0A=0AScott Krue= ger=0A=0A=0AIn a message dated 6/3/2013 8:29:23 A.M. Central Daylight Time,= =A0=A0=0Atroneill@charter.net=A0writes:=0A=0AAngle, angle, angle.=A0 Angle = of stall is constant, no matter what.=A0 Simpler,=A0=0Anot requiring mental= gymnastics.=0ATerrence.=0A=0ASent from my iPad=0A=0AOn Jun 3, 2013, at 7:0= 3 AM, Charles Brown <_browncc1@verizon.net_=A0=0A(mailto:browncc1@verizon.n= et) >=A0 wrote:=0A=0A=0A=0A=0AIn a straight ahead descent, the wing is prod= ucing 1g lift and the=A0 stall=A0=0Aspeed is the same as in level flight.= =A0 You guys may be=A0 thinking of the=A0=0Achange in stall speed when *ini= tiating* a descent (pushover,=A0 less than 1g for a=A0=0Amoment), or when *= terminating* a descent (pull-up, or=A0 flare, momentarily=A0=0Amore than 1g= ).=0A ---955686164-1736898827-1370516509=:362 Content-Type: text/html; charset=iso-8859-1 Content-Transfer-Encoding: quoted-printable
Terrence, Scot= t 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 degree= s).  So for all normal climbs and descents George is essentially corre= ct.  Of course, for a vertical climb or descent the wing supports noth= ing.  For a higher descent angle, such as for turning back with no eng= ine, is the descent angle significant enough to change the stall speed? &nb= sp;I haven't run the numbers, but I suspect it is a very small factor compa= red to the increase in lift required for the bank.

Here's the technique I think is t= heoretically correct, and one that I have practiced.  As Terrence said= , "Angle, angle, angle."  When power failure is first perceived, simul= taneously roll into a steep bank while keeping the AOA at the optimum value= with back pressure on the stick.  Initially, that will require forwar= d stick movement - remember, just because the plane is banked doesn't mean = the G force goes up.  Now as the airspeed increases, increase back pre= ssure 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 po= inted 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 d= ifficult the maneuver, so I have picked 45 degrees as my personal target. &= nbsp;60, 70, or even more might be the theoretical optimum, but that requir= es more skill than I think I would have in a crisis situation.  The co= mpletion 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 re= turn to the "normal" glide speed  (remember to hold the AOA after the = wings are level).

What to do i= f 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. &nb= sp;Of course, you likely don't have a G meter either, so you have to use yo= ur own derriere for that purpose.  The mental gymnastics get to be a r= eal challenge and I suspect that very few pilots would be able to accuratel= y 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 scen= e much sooner - too low and you will lose more altitude than necessary.

I'm sure that the man= euver 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 t= hat.  However, it does prepare you to concentrate on the instruments, = and that might help.  In principle, the turn is exactly similar (my fa= vorite words) to the Chandelle performed for the Commercial ticket, except = done without power.

Just my 2 = cents worth,
Gary Casey

In level flight, the wing AOA p= rovides sufficient lift (wing  loading) to 
equal the effect o= f the force of gravity (1 G) on the  aircraft weight (W).  <= br>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.

Agai= n, in level flight at the same power, but in a coordinated  banked&nbs= p;
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 sam= e altitude  - The lift has to equal the 
weight divided by the= cosine of the bank  angle.  To visualize:



One cou= ld redraw this with force vectors to see it better.  Of course, &= nbsp;
because of increased load, the induced drag is  also increase= d.

Finally, in a coordinated banked turn without power and even furt= her  drag 
from other bits and pieces, descent (glide) will oc= cur unless the AOA could  
be increased provide sufficient lif= t to offset the vertical component (the  
pull of gravity).  But, there is a limit AOA at which a stall would occur -&n= bsp;
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 high= er load.  This supports the statements  
made by both Dav= e Morss and myself.  Dave's point is that large bank angle  =
conducted at a optimal speed shortens the time (distance) and lessens&n= bsp; the 
altitude loss plus in the descent the stall speed is not = as great as  that in 
the same bank holding altitude. &nb= sp;

An optimal speed is somewhere above stall speed.  Factors a= ffecting  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 w= ould be more informative).  I have  
included the simplif= ied Excel spreadsheet to give you a feel for some of  these 
p= arameters before testing at high altitudes.

Blue Skies,

Scott= Krueger


In a message dated 6/3/2013 8:29:23 A.M. Central Daylig= ht Time,  
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@v= erizon.net_ 
(mailto:browncc1@verizon.net) >  wrote:


=

In a straight ahead descent, the wing is producing 1g lift and the&= nbsp; stall 
speed is the same as in level flight.  You guys m= ay 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).
---955686164-1736898827-1370516509=:362--