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Gary,
The numbers don't come out the same.
Keep in mind that Dave Morss did a great article in Sport Aviation on
this. My original point is in agreement with yours - Does an ordinary
pilot in a slick high-wing loaded Lancair have the skill to go to 70 degrees and
keep the nose DOWN to get back? There is still some minimum critical
altitude and aircraft configuration that the pilot must have fixed in
his mind that requires he land within 30 degrees of his heading and keeping the
nose DOWN to maintain a speed above stall. Mine was 700 AGL in a
320. Definitely read the Wikipedia extract below with respect to
instantaneous turn performance in high wing loaded airplanes. This
was my experience too. Snapping into a 60 degree banked constant
altitude turn in my 320 didn't require much for the first 15 or
so degrees of the turn - then the AOA had to be significantly increased to
address the fact that drag and increased wing loading had finally caught
up.
Attached is the spread sheet that was one I originally used for
determining the best way to accomplish a cross country race turn over a
pylon, The formulae are from "Aeronautics for Naval Aviators", a very
useful reference. It does calcs for bank angles over 60 degrees.
I included time and distance numbers for various speeds and bank angles to
determine the best configuration for an efficient turn. I added some
calculations for considering a power out descending turn back to the
airport. I have still been unable to find performance formulae for
descending turns but then I haven't tried very hard.
This paragraph is a useful explanation:
"Effect on turning
performance [edit]
To turn, an aircraft must roll in the
direction of the turn, increasing the aircraft's bank angle. Turning
flight lowers the wing's lift component against gravity and hence causes a
descent. To compensate, the lift force must be increased by increasing the angle
of attack by use of up elevator deflection
which increases drag. Turning can be described as 'climbing around a circle'
(wing lift is diverted to turning the aircraft) so the increase in wing angle of attack creates
even more drag. The tighter the turn radius attempted, the more drag
induced, this requires that power (thrust) be added to overcome the drag. The
maximum rate of turn possible for a given aircraft design is limited by its wing
size and available engine power: the maximum turn the aircraft can achieve and
hold is its sustained turn performance. As the bank angle increases so
does the g-force applied to the aircraft,
this has the effect of increasing the wing loading and also the stalling speed. This
effect is also experienced during level pitching maneuvers.[8]
Aircraft with low wing loadings tend to have superior sustained turn
performance because they can generate more lift for a given quantity of engine
thrust. The immediate bank angle an aircraft can achieve before drag seriously
bleeds off airspeed is known as its instantaneous turn performance. An
aircraft with a small, highly loaded wing may have superior instantaneous turn
performance, but poor sustained turn performance: it reacts quickly to control
input, but its ability to sustain a tight turn is limited. A classic example is
the F-104 Starfighter,
which has a very small wing and high wing loading. At the opposite end of the
spectrum was the gigantic Convair B-36. Its large
wings resulted in a low wing loading, and there are disputed claims[who?]
that this made the bomber more agile than contemporary jet fighters (the
slightly later Hawker Hunter had a
similar wing loading of 250 kg/m2) at high altitude. Whatever
the truth of that, the delta winged Avro Vulcan bomber, with a
wing loading of 260 kg/m2 could certainly be rolled at low
altitudes.[9]
Like any body in circular motion, an
aircraft that is fast and strong enough to maintain level flight at speed
v in a circle of radius R accelerates towards the centre at  .
That acceleration is caused by the inward horizontal component of the lift,  ,
where 
is the banking angle."
We all have the level flight AOA, wing load and the force of gravity
concepts but details on descending turns - not yet. I defer to Dave.
Scott
In a message dated 6/7/2013 7:12:04 A.M. Central Daylight Time,
casey.gary@yahoo.com writes:
Scott,
I think the difference when talking about this might be in the assumption
of reference axes. I'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. So in a
descent the axis is tilted from horizontal by the angle of the descent (a
"zero degree descent" would then be level flight). So the drag is
slightly in the up direction, overcoming a tiny bit of gravity. The
statement 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
practically 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 doesn't
affect the load factor is less accurate.
My
comment about doing the turn referencing instruments is one borne 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 the
turn-back accurately referencing outside the window. For instance, what
does the ground look like during 60-degree no-power bank? How does
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 know 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 been very useful to me, as are most of the
inputs to the List.
Gary
Gary,
Please
go back and re-read what I wrote. Where did you find
the
formulation that the reduction in wing load is related to
the cosine of the descent
angle in a wings level descent at
some arbitrary speed? After all, then
the "cosine" of a
zero degree descent should be 1.
Remember that in a banked turn there
are two components, wing loading
(perpendicular to the
aircraft) and vertical due to 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. Note that as turns become
more shallow the
turn rate slows, the aircraft travels a greater distance
and
there is a greater loss of altitude. If you are below a
critical
altitude, a 30 degree banked turn will not get you
back.
Instrument 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,
casey.gary@yahoo.com writes:
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
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