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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.
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 -----
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.
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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