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Date: Fri, 07 Jun 2013 08:11:13 -0400
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References: <listdigest-6312397@logan.com>
From: Gary Casey <casey.gary@yahoo.com>
Reply-To: Gary Casey <casey.gary@yahoo.com>
Subject: Re: loss of power on takeoff
In-Reply-To: <listdigest-6312397@logan.com>
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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
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<html><body><div style=3D"color:#000; background-color:#fff; font-family:bo=
okman old style, new york, times, serif;font-size:12pt"><div>Scott,</div><d=
iv>I think the difference when talking about this might be in the assumptio=
n of reference axes. &nbsp;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. &nbsp;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). &nbsp;So the drag i=
s slightly in the up direction, overcoming a tiny bit of gravity. &nbsp;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. &nbsp;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.</div><div><br></div><div st=
yle=3D"color: rgb(0, 0, 0); font-size: 16px; font-family: 'bookman old styl=
e', 'new york', times, serif; background-color: transparent; font-style: no=
rmal; ">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. &nbsp;I just don't know how one (me, at least) could perform th=
e turn-back accurately referencing outside the window. &nbsp;For instance, =
what does the ground look like during 60-degree no-power bank? &nbsp;How do=
es prevailing wind affect the view? &nbsp;I know the instruments will look =
the same as when I practice that at altitude. &nbsp;</div><div style=3D"col=
or: rgb(0, 0, 0); font-size: 16px; font-family: 'bookman old style', 'new y=
ork', times, serif; background-color: transparent; font-style: normal;
 "><br></div><div style=3D"color: rgb(0, 0, 0); font-size: 16px; font-famil=
y: 'bookman old style', 'new york', times, serif; background-color: transpa=
rent; font-style: normal; ">I'm afraid I lost your spreadsheet, but I have =
created one in the past that I'm sure says about the same thing. &nbsp;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. &nbsp;Regardless, I'm =
sure it is higher than a typical pilot will be able to accurately control. =
&nbsp;So the target then becomes the highest bank angle that will allow the=
 pilot to maintain a "safe" margin above stall. &nbsp;This discussion has b=
een very useful to me, as are most of the inputs to the List.</div><div sty=
le=3D"color: rgb(0, 0, 0); font-size: 16px; font-family: 'bookman old style=
', 'new york', times, serif; background-color: transparent; font-style: nor=
mal; ">Gary</div><div style=3D"color: rgb(0, 0, 0); font-size: 16px;
 font-family: 'bookman old style', 'new york', times, serif; background-col=
or: transparent; font-style: normal; "><br></div><div style=3D"color: rgb(0=
, 0, 0); font-size: 16px; font-family: 'bookman old style', 'new york', tim=
es, serif; background-color: transparent; font-style: normal; "><span class=
=3D"Apple-style-span" style=3D"color: rgb(69, 69, 69); font-family: arial, =
helvetica, sans-serif; font-size: 13px; ">Gary,<br><br>Please go back and r=
e-read what I wrote.&nbsp; Where did you find the&nbsp;&nbsp;<br>formulatio=
n that the reduction in wing load is related to the cosine of the&nbsp; des=
cent&nbsp;<br>angle in a wings level descent at some arbitrary speed?&nbsp;=
 After all,&nbsp; then&nbsp;<br>the "cosine" of a zero degree descent shoul=
d be 1.<br><br>Remember that in a banked turn there are two components, win=
g loading&nbsp;&nbsp;<br>(perpendicular to the aircraft) and vertical due t=
o gravity.&nbsp; Do not be&nbsp;&nbsp;<br>confused by a term like G as it
 is only used as a substitute for the weight&nbsp; of the&nbsp;<br>airplane=
.&nbsp;&nbsp;<br><br>Please plug some numbers in the spread sheet.&nbsp; No=
te that as turns&nbsp; become&nbsp;<br>more shallow the turn rate slows, th=
e aircraft travels a&nbsp; greater distance&nbsp;<br>and there is a greater=
 loss of altitude.&nbsp; If you are below&nbsp; a critical&nbsp;<br>altitud=
e, a 30 degree banked turn will not get you back.&nbsp;&nbsp;<br><br>Instru=
ment reference is best used for engine failures at night or that&nbsp;&nbsp=
;<br>risky takeoff in 0-0 conditions.<br><br>Good Luck,<br><br>Scott<br><br=
><br><br>In a message dated 6/6/2013 3:44:42 P.M. Central Daylight Time,&nb=
sp;&nbsp;<br><a ymailto=3D"mailto:casey.gary@yahoo.com" href=3D"mailto:case=
y.gary@yahoo.com" style=3D"text-decoration: underline; color: rgb(40, 98, 1=
97); outline-width: 0px; outline-style: initial; outline-color: initial; ">=
casey.gary@yahoo.com</a>&nbsp;writes:<br><br><br>Terrence, Scott and
 George are all correct - sort of.&nbsp; George said&nbsp; in a&nbsp;<br>st=
eady-state descent the wing is still supporting 1 G, but Scott says it&nbsp=
; is&nbsp;<br>supporting less than 1 G because of the descent.&nbsp; For a =
typical&nbsp; descent of&nbsp;<br>500 ft/min at a speed of 180 statute mile=
s per hour the wing is&nbsp; supporting&nbsp;<br>99.95 percent of the weigh=
t (the cosine of the descent angle, which&nbsp; is 1.8&nbsp;<br>degrees).&n=
bsp; So for all normal climbs and descents George is&nbsp; essentially&nbsp=
;<br>correct.&nbsp; Of course, for a vertical climb or descent the wing&nbs=
p; supports nothing.&nbsp;&nbsp;<br>For a higher descent angle, such as for=
 turning back&nbsp; with no engine, is&nbsp;<br>the descent angle significa=
nt enough to change the stall&nbsp; speed?&nbsp; I haven't&nbsp;<br>run the=
 numbers, but I suspect it is a very small&nbsp; factor compared to the&nbs=
p;<br>increase in lift required for the
 bank.<br><br><br>Here's&nbsp; the technique I think is theoretically corre=
ct, and one that I have&nbsp;<br>practiced.&nbsp; As Terrence said, "Angle,=
 angle, angle."&nbsp; When power failure is&nbsp;<br>first perceived, simul=
taneously roll into a steep bank while keeping the&nbsp;<br>AOA&nbsp; at th=
e optimum value with back pressure on the stick.&nbsp; Initially, that&nbsp=
;&nbsp;<br>will require forward stick movement - remember, just because the=
 plane is&nbsp;&nbsp;<br>banked doesn't mean the G force goes up.&nbsp; Now=
 as the airspeed increases,&nbsp;&nbsp;<br>increase back pressure to hold t=
he same AOA.&nbsp; With an&nbsp; AOA-indicator-equipped&nbsp;<br>plane you =
only control 2 things - the bank angle and&nbsp; AOA.&nbsp; When you are&nb=
sp;<br>again pointed at the runway (at an angle, but don't be&nbsp; picky) =
immediately level&nbsp;<br>the wings.&nbsp; What bank angle?&nbsp; I haven'=
t run&nbsp; the numbers, but as Dave has&nbsp;<br>said, it is a
 steep angle.&nbsp; The steeper the&nbsp; angle the more difficult the&nbsp=
;<br>maneuver, so I have picked 45 degrees as my&nbsp; personal target.&nbs=
p; 60, 70, or even&nbsp;<br>more might be the theoretical optimum,&nbsp; bu=
t that requires more skill than I&nbsp;<br>think I would have in a crisis s=
ituation.&nbsp; The completion of the turn could&nbsp;<br>happen quite clos=
e to the ground, but&nbsp; the extra speed required for the turn&nbsp;<br>w=
ill be used to arrest the rapid descent&nbsp; and return to the "normal" gl=
ide&nbsp;<br>speed&nbsp; (remember to hold the AOA after&nbsp; the wings ar=
e level).<br><br><br>What&nbsp; to do if your plane is not AOA-indicator-eq=
uipped?&nbsp; The maneuver is&nbsp;<br>still&nbsp; the same, but you have t=
o control G loading as a function of airspeed.&nbsp;<br>Of course, you like=
ly don't have a G meter either, so you have to use&nbsp;&nbsp;<br>your own =
derriere for that purpose.&nbsp; The mental gymnastics get to be
 a&nbsp; real&nbsp;<br>challenge and I suspect that very few pilots would b=
e able to accurately&nbsp;&nbsp;<br>control AOA during the maneuver.&nbsp; =
The result is that the bank angle has&nbsp; to be&nbsp;<br>reduced to maint=
ain some degree of accuracy in AOA.&nbsp; I would guess&nbsp; that 30&nbsp;=
<br>degrees bank might be a good target for most non test pilots.&nbsp; If&=
nbsp; you get&nbsp;<br>the AOA too high you will certainly arrive at crash =
scene much sooner&nbsp; - too&nbsp;<br>low and you will lose more altitude =
than necessary.<br><br><br>I'm&nbsp; sure that the maneuver can best be per=
formed in reference solely to&nbsp;&nbsp;<br>instruments, as the view of th=
e ground, close-up, oddly angled and rapidly&nbsp;&nbsp;<br>rotating would =
be a huge distraction.&nbsp; Practicing at altitude doesn't&nbsp; really&nb=
sp;<br>prepare one for that.&nbsp; However, it does prepare you to concentr=
ate&nbsp; on the&nbsp;<br>instruments, and that might help.&nbsp; In
 principle, the turn is&nbsp; exactly&nbsp;<br>similar (my favorite words) =
to the Chandelle performed for the&nbsp; Commercial ticket,&nbsp;<br>except=
 done without power.<br><br><br>Just&nbsp; my 2 cents worth,<br>Gary&nbsp; =
Casey<br></span></div>  </div></body></html>
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