Mailing List lml@lancaironline.net Message #65729
From: Gary Casey <casey.gary@yahoo.com>
Sender: <marv@lancaironline.net>
Subject: Re: loss of power on takeoff
Date: Thu, 06 Jun 2013 16:44:15 -0400
To: <lml@lancaironline.net>
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 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) >  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).

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