X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Sender: To: lml@lancaironline.net Date: Fri, 07 Jun 2013 10:43:06 -0400 Message-ID: X-Original-Return-Path: Received: from hrndva-omtalb.mail.rr.com ([71.74.56.122] verified) by logan.com (CommuniGate Pro SMTP 6.0.5) with ESMTP id 6312893 for lml@lancaironline.net; Fri, 07 Jun 2013 10:06:58 -0400 Received-SPF: none receiver=logan.com; client-ip=71.74.56.122; envelope-from=Wolfgang@MiCom.net X-Original-Return-Path: X-Authority-Analysis: v=2.0 cv=fZsvOjsF c=1 sm=0 a=MHZY6FYWMEQOp7S43i2QIw==:17 a=3Zlka_XeuxsA:10 a=ttCsPuSJ-FAA:10 a=rTjvlri0AAAA:8 a=dg00J6WOx4YA:10 a=Ia-xEzejAAAA:8 a=hOpmn2quAAAA:8 a=o1OHuDzbAAAA:8 a=6JcevbYce2e7R5yNIFIA:9 a=wPNLvfGTeEIA:10 a=EzXvWhQp4_cA:10 a=hUswqBWy9Q8A:10 a=ILCZio5HsAgA:10 a=CjxXgO3LAAAA:8 a=OuyeD9zSBC8w_XvRY4YA:9 a=_W_S_7VecoQA:10 a=rC2wZJ5BpNYA:10 a=9So9MtSjaCnrPA7U:21 a=MHZY6FYWMEQOp7S43i2QIw==:117 X-Cloudmark-Score: 0 X-Authenticated-User: X-Originating-IP: 74.218.201.50 Received: from [74.218.201.50] ([74.218.201.50:1318] helo=lobo) by hrndva-oedge02.mail.rr.com (envelope-from ) (ecelerity 2.2.3.46 r()) with ESMTP id 25/00-10569-0E8E1B15; Fri, 07 Jun 2013 14:06:24 +0000 X-Original-Message-ID: <4F6F9A2E7025490A9D7B85FD41893822@lobo> From: "Wolfgang" X-Original-To: Subject: Fw: loss of power on takeoff X-Original-Date: Fri, 7 Jun 2013 10:06:19 -0400 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_004C_01CE6366.A7EF60A0" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2900.5512 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.5512 This is a multi-part message in MIME format. ------=_NextPart_000_004C_01CE6366.A7EF60A0 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable In a steady-state descent, the wing will be supporting 1.000 G. Descent = rate or angle has nothing to do with it. Understand that a G is a measure of acceleration - 32.1740ft/sec/sec.=20 In a steady-state descent there is no acceleration or deceleration = except the force of gravity which stays at 1.000 G. . . . unless you are in a turn where you induce your own additional = acceleration due to centrifical force. If you want to get technical, the force of gravity (1.000 G) may change = ever so slightly as the distance between the aircraft and the center of = mass of the earth changes but that change is so slight as to be = immeasurable. Wolfgang ----- Original Message -----=20 From: Gary Casey=20 To: lml@lancaironline.net=20 Sent: Thursday, June 06, 2013 4:44 PM Subject: Re: loss of power on takeoff 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.=20 And, Charles' comment is a bit off. In level flight, the wing AOA provides sufficient lift (wing loading) = to=20 equal the effect of the force of gravity (1 G) on the aircraft weight = (W). =20 Thrust overcomes drag to result in forward speed.=20 In a descent at the same speed used in level flight, lift is less than W = =20 and either power (thrust) is reduced or drag is increased. Remember = that G =20 is just for relative reference. Again, in level flight at the same power, but in a coordinated banked=20 turn, the wing AOA has been increased to add enough bank angle lift = necessary=20 to maintain 1 G with respect to the vertical. I.E. The wing load must = be=20 increased to keep the plane at the same altitude - The lift has to = equal the=20 weight divided by the cosine of the bank angle. To visualize: One could redraw this with force vectors to see it better. Of course, =20 because of increased load, the induced drag is also increased. Finally, in a coordinated banked turn without power and even further = drag=20 from other bits and pieces, descent (glide) will occur unless the AOA = could =20 be increased provide sufficient lift to offset the vertical component = (the =20 pull of gravity). But, there is a limit AOA at which a stall would = occur -=20 thus descent. In a banked turning descent at a certain speed (best = glide =20 for the conditions), less lift is required, thus less load on the wing,=20 thus a lower stall speed than a higher load. This supports the = statements =20 made by both Dave Morss and myself. Dave's point is that large bank = angle =20 conducted at a optimal speed shortens the time (distance) and lessens = the=20 altitude loss plus in the descent the stall speed is not as great as = that in=20 the same bank holding altitude. =20 An optimal speed is somewhere above stall speed. Factors affecting = stall=20 speed are load and drag (wheels, flaps, prop, etc.) - hence the = requirement=20 that you point the nose down making use of kinetic energy rather than=20 gasoline to keep up the speed. Uh, the Aeronautics for Naval Aviators is silent on powerless descending = =20 turns (maybe a glider tech manual would be more informative). I have =20 included the simplified Excel spreadsheet to give you a feel for some of = these=20 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, =20 troneill@charter.net writes: Angle, angle, angle. Angle of stall is constant, no matter what. = Simpler,=20 not requiring mental gymnastics. Terrence. Sent from my iPad On Jun 3, 2013, at 7:03 AM, Charles Brown <_browncc1@verizon.net_=20 (mailto:browncc1@verizon.net) > wrote: In a straight ahead descent, the wing is producing 1g lift and the = stall=20 speed is the same as in level flight. You guys may be thinking of the=20 change in stall speed when *initiating* a descent (pushover, less than = 1g for a=20 moment), or when *terminating* a descent (pull-up, or flare, = momentarily=20 more than 1g). ------=_NextPart_000_004C_01CE6366.A7EF60A0 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
In a steady-state descent, the = wing will be=20 supporting 1.000 G. Descent rate or angle has nothing to do with=20 it.
Understand that a G is a measure of = acceleration -=20 32.1740ft/sec/sec.
In a steady-state descent there = is no=20 acceleration or deceleration except the force of gravity which stays at = 1.000=20 G.
. . . unless you are in a turn where = you induce=20 your own additional acceleration due to centrifical force.
 
If you want to get technical, the force = of gravity=20 (1.000 G) may change ever so slightly as the distance between the = aircraft and=20 the center of mass of the earth changes but that change is so slight as = to be=20 immeasurable.
 
Wolfgang
 
 
----- Original Message -----=20
From: Gary = Casey=20
Sent: Thursday, June 06, 2013 4:44 PM
Subject: Re: loss of power on takeoff

Terrence, Scott and George are all correct - sort of.  George = said in=20 a steady-state descent the wing is still supporting 1 G, but Scott says = it is=20 supporting less than 1 G because of the descent.  For a typical = descent of=20 500 ft/min at a speed 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 1.8 = degrees).=20  So for all normal climbs and descents George is essentially = correct.=20  Of course, for a vertical climb or descent the wing supports = nothing.=20  For a higher descent angle, such as for turning back with no = engine, is=20 the descent angle significant enough to change the stall speed?  I = haven't=20 run the numbers, but I suspect it is a very small 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 = practiced.=20  As Terrence said, "Angle, angle, angle."  When power failure = is first=20 perceived, simultaneously roll into a steep bank while keeping the AOA = at the=20 optimum value with back pressure on the stick.  Initially, that = will=20 require forward stick movement - remember, just because the plane is = banked=20 doesn't mean the G force goes up.  Now as the airspeed increases, = increase=20 back pressure to hold the same AOA.  With an AOA-indicator-equipped = plane=20 you only control 2 things - the bank angle and AOA.  When you are = again=20 pointed at the runway (at an angle, but don't be picky) immediately = level the=20 wings.  What bank angle?  I haven't run the numbers, but as = Dave has=20 said, it is a steep angle.  The steeper the angle the more = difficult the=20 maneuver, so I have picked 45 degrees as my personal target.  60, = 70, or=20 even more might be the theoretical optimum, but that requires more skill = than I=20 think I would have in a crisis situation.  The completion of the = turn could=20 happen quite close to the ground, but the extra speed required for the = turn will=20 be used to arrest the rapid descent and return to the "normal" glide = speed=20  (remember to hold the AOA after 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. =  Of=20 course, you likely don't have a G meter either, so you have to use your = own=20 derriere for that purpose.  The mental gymnastics get to be a real=20 challenge and I suspect that very few pilots would be able to accurately = control=20 AOA during the maneuver.  The result is that the bank angle has to = be=20 reduced to maintain some degree of accuracy in AOA.  I would guess = that 30=20 degrees bank might be a good target for most non test pilots.  If = you get=20 the AOA too high you will certainly arrive at crash scene much sooner - = too low=20 and you will lose more altitude than necessary.
I'm=20 sure that the maneuver can best be performed in reference solely to = instruments,=20 as the view of the ground, close-up, oddly angled and rapidly rotating = would be=20 a huge distraction.  Practicing at altitude doesn't really prepare = one for=20 that.  However, it does prepare you to concentrate on the = instruments, and=20 that might help.  In principle, the turn is exactly similar (my = favorite=20 words) to the Chandelle performed for the Commercial ticket, except done = without=20 power.
Just=20 my 2 cents worth, Gary=20 Casey

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

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

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

Again, in level flight at the same power, but = in a=20 coordinated  banked 
turn, the wing AOA has been increased = to add=20 enough  bank angle lift necessary 
to maintain 1 G with = respect to=20 the vertical.  I.E. The wing load must be 
increased to = keep the=20 plane at the same altitude  - The lift has to equal = the 
weight=20 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 = induced=20 drag is  also increased.

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

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

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

Blue Skies,

Scott Krueger


In a = message dated=20 6/3/2013 8:29:23 A.M. Central Daylight 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.

Sent from=20 my iPad

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




In a straight ahead descent, the wing is = producing 1g=20 lift and the  stall 
speed is the same as in level = flight. =20 You guys may be  thinking of the 
change in stall speed = when=20 *initiating* a descent (pushover,  less than 1g for = a 
moment), or=20 when *terminating* a descent (pull-up, or  flare, = momentarily 
more=20 than 1g).
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