X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Sender: To: lml@lancaironline.net Date: Fri, 07 Jun 2013 08:11:13 -0400 Message-ID: X-Original-Return-Path: Received: from nm16-vm2.bullet.mail.ne1.yahoo.com ([98.138.91.92] verified) by logan.com (CommuniGate Pro SMTP 6.0.5) with ESMTPS id 6312484 for lml@lancaironline.net; Fri, 07 Jun 2013 07:42:08 -0400 Received-SPF: none receiver=logan.com; client-ip=98.138.91.92; envelope-from=casey.gary@yahoo.com Received: from [98.138.226.180] by nm16.bullet.mail.ne1.yahoo.com with NNFMP; 07 Jun 2013 11:41:32 -0000 Received: from [98.138.89.192] by tm15.bullet.mail.ne1.yahoo.com with NNFMP; 07 Jun 2013 11:41:32 -0000 Received: from [127.0.0.1] by omp1050.mail.ne1.yahoo.com with NNFMP; 07 Jun 2013 11:41:32 -0000 X-Yahoo-Newman-Property: ymail-3 X-Yahoo-Newman-Id: 513742.14499.bm@omp1050.mail.ne1.yahoo.com Received: (qmail 11500 invoked by uid 60001); 7 Jun 2013 11:41:32 -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=z79iq1llab/hj+FzoTGnHGsslGLFwIcRRZEbxBARCvmt4wh08CfaNl74jzFXejPAekBEbUbotMuhzUkmM9oc9l5qTLZX22QrnPNxNHD0EdA9tZIRD8cvCatnRHpTdS774boPROR08lsCw6743jn4aA8dYWzlrLh3IdtWBFhLtRY=; X-YMail-OSG: 3_PX2CIVM1npVMsLnXX9cdc6CpwdzDCuMc1o4HOV_g97GIf YuXS5xqN_oWpQAE0fdJh0shLrrCvRoIu7Rm3u.RKx5vctDJaWScUwLaUFqX2 Sje3v7SbFRc1VBM9Sfx.fL3PWoJ7bKQW_hM_Kjgtr5MX9Ku0l08GUbAUEoM1 EEg3wBuxphC7_3qvrumRvK7ihMFB5XNOb85NUWhTxvoOpuUjYN65FRIYm07f C4hlGF4RqF6J8UAbKd_0Un5Wi8sng0Ecj7Eda9e0tnvOzWPQNAFq.kBBMv3U lmMXXgI3lBraz9lqc_FmhCp3viFLpwayvSNPB8KtJkH9Yjc4_.DGK1Ub_byr 2EXDssm5vEB74sMaGe33RyxukfHwmqbUKpekAKSiCDe4Y_XmnjYfsVvvp.7v N5sowMCopzQZC2fWIVx_sRjDk3nHAhQdkZHsCC8D5wvMmDiF71FI0wNduh20 hn.aOlfDTedb5sTQwiJ0Rp0vor3hKBU6N32imNU90.SqQc5FvF3rhX1SbOFM A.3Nfeu00sa815gT5UPwqpTvmXgfbzD.ooXFciCJnMJtPPn_OhsSxClJHROM EmVs- Received: from [97.92.63.83] by web120103.mail.ne1.yahoo.com via HTTP; Fri, 07 Jun 2013 04:41:32 PDT X-Rocket-MIMEInfo: 002.001,U2NvdHQsCkkgdGhpbmsgdGhlIGRpZmZlcmVuY2Ugd2hlbiB0YWxraW5nIGFib3V0IHRoaXMgbWlnaHQgYmUgaW4gdGhlIGFzc3VtcHRpb24gb2YgcmVmZXJlbmNlIGF4ZXMuIMKgSSdtIHVzaW5nIHRoZSBmbGlnaHQgcGF0aCBmb3IgdGhlIGNyZWF0aW9uIG9mIHRoZSBsaWZ0LCBkcmFnLCB0aHJ1c3QgYW5kIGdyYXZpdHkgZGlyZWN0aW9ucywgYnV0IHlvdSBjb3VsZCB1c2UgYWJzb2x1dGUgdmVydGljYWwgYW5kIGhvcml6b250YWwgYXhlcyBhcyB3ZWxsIC0gdGhlIG51bWJlcnMgY29tZSBvdXQgdGhlIHNhbWUBMAEBAQE- X-Mailer: YahooMailWebService/0.8.145.547 References: X-Original-Message-ID: <1370605292.91263.YahooMailNeo@web120103.mail.ne1.yahoo.com> X-Original-Date: Fri, 7 Jun 2013 04:41:32 -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="-955686164-457786725-1370605292=:91263" ---955686164-457786725-1370605292=:91263 Content-Type: text/plain; charset=iso-8859-1 Content-Transfer-Encoding: quoted-printable Scott,=0AI think the difference when talking about this might be in the ass= umption of reference axes. =A0I'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. =A0So= in a descent the axis is tilted from horizontal by the angle of the descen= t (a "zero degree descent" would then be level flight). =A0So the drag is s= lightly in the up direction, overcoming a tiny bit of gravity. =A0The state= ment 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 practic= ally it is close enough to 1 G to make the discussion moot. =A0For wide dev= iations from level flight, such as during aerobatics or even the no-power t= urn back to the airport, the assumption that descent angle doesn't affect t= he load factor is less accurate.=0A=0AMy comment about doing the turn refer= encing instruments is one borne of inexperience in the conditions of high b= ank angles, no power and close to ground. =A0I just don't know how one (me,= at least) could perform the turn-back accurately referencing outside the w= indow. =A0For instance, what does the ground look like during 60-degree no-= power bank? =A0How does prevailing wind affect the view? =A0I know the inst= ruments will look the same as when I practice that at altitude. =A0=0A=0AI'= m afraid I lost your spreadsheet, but I have created one in the past that I= 'm sure says about the same thing. =A0The accuracy of mine decreases at ext= reme bank angles (over 60), so I don't know whether or not there is a true = optimum bank angle. =A0Regardless, I'm sure it is higher than a typical pil= ot will be able to accurately control. =A0So the target then becomes the hi= ghest bank angle that will allow the pilot to maintain a "safe" margin abov= e stall. =A0This discussion has been very useful to me, as are most of the = inputs to the List.=0AGary=0A=0AGary,=0A=0APlease go back and re-read what = I wrote.=A0 Where did you find the=A0=A0=0Aformulation that the reduction i= n wing load is related to the cosine of the=A0 descent=A0=0Aangle in a wing= s level descent at some arbitrary speed?=A0 After all,=A0 then=A0=0Athe "co= sine" of a zero degree descent should be 1.=0A=0ARemember that in a banked = turn there are two components, wing loading=A0=A0=0A(perpendicular to the a= ircraft) and vertical due to gravity.=A0 Do not be=A0=A0=0Aconfused by a te= rm like G as it is only used as a substitute for the weight=A0 of the=A0=0A= airplane.=A0=A0=0A=0APlease plug some numbers in the spread sheet.=A0 Note = that as turns=A0 become=A0=0Amore shallow the turn rate slows, the aircraft= travels a=A0 greater distance=A0=0Aand there is a greater loss of altitude= .=A0 If you are below=A0 a critical=A0=0Aaltitude, a 30 degree banked turn = will not get you back.=A0=A0=0A=0AInstrument reference is best used for eng= ine failures at night or that=A0=A0=0Arisky takeoff in 0-0 conditions.=0A= =0AGood Luck,=0A=0AScott=0A=0A=0A=0AIn a message dated 6/6/2013 3:44:42 P.M= . Central Daylight Time,=A0=A0=0Acasey.gary@yahoo.com=A0writes:=0A=0A=0ATer= rence, Scott and George are all correct - sort of.=A0 George said=A0 in a= =A0=0Asteady-state descent the wing is still supporting 1 G, but Scott says= it=A0 is=A0=0Asupporting less than 1 G because of the descent.=A0 For a ty= pical=A0 descent of=A0=0A500 ft/min at a speed of 180 statute miles per hou= r the wing is=A0 supporting=A0=0A99.95 percent of the weight (the cosine of= the descent angle, which=A0 is 1.8=A0=0Adegrees).=A0 So for all normal cli= mbs and descents George is=A0 essentially=A0=0Acorrect.=A0 Of course, for a= vertical climb or descent the wing=A0 supports nothing.=A0=A0=0AFor a high= er descent angle, such as for turning back=A0 with no engine, is=A0=0Athe d= escent angle significant enough to change the stall=A0 speed?=A0 I haven't= =A0=0Arun the numbers, but I suspect it is a very small=A0 factor compared = to the=A0=0Aincrease in lift required for the bank.=0A=0A=0AHere's=A0 the t= echnique I think is theoretically correct, and one that I have=A0=0Apractic= ed.=A0 As Terrence said, "Angle, angle, angle."=A0 When power failure is=A0= =0Afirst perceived, simultaneously roll into a steep bank while keeping the= =A0=0AAOA=A0 at the optimum value with back pressure on the stick.=A0 Initi= ally, that=A0=A0=0Awill require forward stick movement - remember, just bec= ause the plane is=A0=A0=0Abanked doesn't mean the G force goes up.=A0 Now a= s the airspeed increases,=A0=A0=0Aincrease back pressure to hold the same A= OA.=A0 With an=A0 AOA-indicator-equipped=A0=0Aplane you only control 2 thin= gs - the bank angle and=A0 AOA.=A0 When you are=A0=0Aagain pointed at the r= unway (at an angle, but don't be=A0 picky) immediately level=A0=0Athe wings= .=A0 What bank angle?=A0 I haven't run=A0 the numbers, but as Dave has=A0= =0Asaid, it is a steep angle.=A0 The steeper the=A0 angle the more difficul= t the=A0=0Amaneuver, so I have picked 45 degrees as my=A0 personal target.= =A0 60, 70, or even=A0=0Amore might be the theoretical optimum,=A0 but that= requires more skill than I=A0=0Athink I would have in a crisis situation.= =A0 The completion of the turn could=A0=0Ahappen quite close to the ground,= but=A0 the extra speed required for the turn=A0=0Awill be used to arrest t= he rapid descent=A0 and return to the "normal" glide=A0=0Aspeed=A0 (remembe= r to hold the AOA after=A0 the wings are level).=0A=0A=0AWhat=A0 to do if y= our plane is not AOA-indicator-equipped?=A0 The maneuver is=A0=0Astill=A0 t= he same, but you have to control G loading as a function of airspeed.=A0=0A= Of course, you likely don't have a G meter either, so you have to use=A0=A0= =0Ayour own derriere for that purpose.=A0 The mental gymnastics get to be a= =A0 real=A0=0Achallenge and I suspect that very few pilots would be able to= accurately=A0=A0=0Acontrol AOA during the maneuver.=A0 The result is that = the bank angle has=A0 to be=A0=0Areduced to maintain some degree of accurac= y in AOA.=A0 I would guess=A0 that 30=A0=0Adegrees bank might be a good tar= get for most non test pilots.=A0 If=A0 you get=A0=0Athe AOA too high you wi= ll certainly arrive at crash scene much sooner=A0 - too=A0=0Alow and you wi= ll lose more altitude than necessary.=0A=0A=0AI'm=A0 sure that the maneuver= can best be performed in reference solely to=A0=A0=0Ainstruments, as the v= iew of the ground, close-up, oddly angled and rapidly=A0=A0=0Arotating woul= d be a huge distraction.=A0 Practicing at altitude doesn't=A0 really=A0=0Ap= repare one for that.=A0 However, it does prepare you to concentrate=A0 on t= he=A0=0Ainstruments, and that might help.=A0 In principle, the turn is=A0 e= xactly=A0=0Asimilar (my favorite words) to the Chandelle performed for the= =A0 Commercial ticket,=A0=0Aexcept done without power.=0A=0A=0AJust=A0 my 2= cents worth,=0AGary=A0 Casey=0A ---955686164-457786725-1370605292=:91263 Content-Type: text/html; charset=iso-8859-1 Content-Transfer-Encoding: quoted-printable
Scott,
I think the difference when talking about this might be in the assumptio= n of reference axes.  I'm using the flight path for the creation of th= e lift, drag, thrust and gravity directions, but you could use absolute ver= tical 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 descen= t (a "zero degree descent" would then be level flight).  So the drag i= s slightly in the up direction, overcoming a tiny bit of gravity.  The= statement someone made that lift (or more accurately "wing load factor", i= f you will) is less than 1 G during a descent is technically correct, but p= ractically 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 d= oesn't affect the load factor is less accurate.

My comment about doing the turn referencing instruments is one born= e 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 th= e turn-back accurately referencing outside the window.  For instance, = what does the ground look like during 60-degree no-power bank?  How do= es 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 kno= w 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 b= een very useful to me, as are most of the inputs to the List.
Gary

Gary,

Please go back and r= e-read what I wrote.  Where did you find the  
formulatio= n that the reduction in wing load is related to the cosine of the  des= cent 
angle in a wings level descent at some arbitrary speed? = After all,  then 
the "cosine" of a zero degree descent shoul= d be 1.

Remember that in a banked turn there are two components, win= g loading  
(perpendicular to the aircraft) and vertical due t= o 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.  No= te that as turns  become 
more shallow the turn rate slows, th= e aircraft travels a  greater distance 
and there is a greater= loss of altitude.  If you are below  a critical 
altitud= e, a 30 degree banked turn will not get you back.  

Instru= ment 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,&nb= sp; 
= casey.gary@yahoo.com writes:


Terrence, Scott and George are all correct - sort of.  George said  in a 
st= eady-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 mile= s per hour the wing is  supporting 
99.95 percent of the weigh= t (the cosine of the descent angle, which  is 1.8 
degrees).&n= bsp; So for all normal climbs and descents George is  essentially = ;
correct.  Of course, for a vertical climb or descent the wing&nbs= p; supports nothing.  
For a higher descent angle, such as for= turning back  with no engine, is 
the descent angle significa= nt enough to change the stall  speed?  I haven't 
run the= numbers, but I suspect it is a very small  factor compared to the&nbs= p;
increase in lift required for the bank.


Here's  the technique I think is theoretically corre= ct, 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 th= e 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 t= he same AOA.  With an  AOA-indicator-equipped 
plane you = only control 2 things - the bank angle and  AOA.  When you are&nb= sp;
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.&nbs= p; 60, 70, or even 
more might be the theoretical optimum,  bu= t that requires more skill than I 
think I would have in a crisis s= ituation.  The completion of the turn could 
happen quite clos= e to the ground, but  the extra speed required for the turn 
w= ill be used to arrest the rapid descent  and return to the "normal" gl= ide 
speed  (remember to hold the AOA after  the wings ar= e level).


What  to do if your plane is not AOA-indicator-eq= uipped?  The maneuver is 
still  the same, but you have t= o control G loading as a function of airspeed. 
Of course, you like= ly 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 b= e able to accurately  
control AOA during the maneuver.  = The result is that the bank angle has  to be 
reduced to maint= ain some degree of accuracy in AOA.  I would guess  that 30 =
degrees bank might be a good target for most non test pilots.  If&= nbsp; 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 per= formed in reference solely to  
instruments, as the view of th= e ground, close-up, oddly angled and rapidly  
rotating would = be a huge distraction.  Practicing at altitude doesn't  really&nb= sp;
prepare one for that.  However, it does prepare you to concentr= ate  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
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