2013 SEPTEMBER Vol. 5 / No. 9 /
GA
61
the aircraft is subject to a number of
forces, namely lift, resultant lift and
gravity (drag and thrust don't play
any major roles in this discussion).
Balanced by the load factor acting
perpendicular to the main plane
(centrifugal force), if all the forces are
in equilibrium, the aircraft will be in a
stable circular flight path. In a perfect
word, an aircraft performing a loop
would be able to maintain a constant
load factor and speed, the aircraft
flight path would then be a perfect
circle, in practice this drops below
the horizon, the aircraft accelerates
and by increasing the "g" load the
aircraft pulls out of the resulting dive
and most recoveries are displaced
horizontally forwards of the entry
point. This is due to the fact that
gravity works against the rate of turn
when pitching up into the vertical and
against the trust, while the opposite
occurs when the aircraft is recovering
in the vertical down plane. At any
stage in this looping manoeuvre,
increasing the angle of attack past
the critical angle will result in the
aircraft stalling, irrespective of speed
or attitude in relation to the ground.
One way of reducing the possibility
of a stall at the top of the loop is to
lower a notch of flap, If the wing is
at the critical angle of attack at the
top of the loop and flaps are lowered,
this will change the cord line of
the wing, effectively reducing the
angle of attack and increasing the lift
generated by the wing. If the aircraft
is inverted at the top of a loop and
flaps are lowered, the nose pitches
rapidly towards the canopy and the
aircraft accelerates. Using flaps when
inverted has some drawbacks so
don't try this without the guidance
of an aerobatic instructor.
AGL at about 150 KIAS, ATC
spoke to me and while talking to
them I selected the lever, the nose of
the Impala pitched violently towards
the canopy, great when you are the
right way up but not good when
inverted at 200 feet AGL. I pushed
the stick into the dash, slammed open
full throttle, and immediately rolled
out, the aircraft was juddering on the
verge of stalling and to save myself
I selected approach flap. The Impala
seemed to run into a brick wall and
the speed was now decaying though
100 kts and I was sinking towards
the toll plaza at the threshold of 29,
time to reduce the drag so I selected
gear up ... the lever was in the up
position. At 200 AGL inverted I
had selected approach flap instead
of gear down, the flap immediately
increased the lift vector, straight
into the ground, the flaps extending
now increased the effective angle of
attack to close to the inverted stall
angle and the aircraft descended. I
cleaned up the aircraft and continued
the show... I was lucky and my
training saved me from a fiery death.
Now lets increase the angle of
bank and "g:" to maintain level flight.
As the bank angle increases so does
the lift required to sustain level flight.
At a certain point the load factor
coupled with the indicated airspeed
will reach the critical angle of attack
for those conditions and the wing
will stall, the lift vector will decrease,
so will the centripetal force and the
aircraft will break out of the turn,
normally dropping a wing at the same
time. The higher the load factor at
the time of the break away, the more
violent the departure is, recovery
is the same, unload the aircraft,
stop the wing drop with rudder.
Lets now look at the same
exercise but in the vertical. Is
there any difference, no, the same
principles apply, assume the aircraft
has completed the first 180 degrees
of the loop, is now inverted in
relation to the ground. Gravity and
lift and weight are all now working
in the same direction, so why don't
we fall out of the sky... centrifugal
force is acting in the opposite
direction counterbalancing the lift/
weight couple, any increase in the
relative angle of attack will now
cause an airflow breakaway over
the top surface of the wing ( that's
the surface now pointing down at
the ground) and the lift/weight/
gravity couple will be reduced. The
centrifugal force remains the same
so the aircraft departs to the outside
of the circular flight path. A very
simple principle to demonstrate,
take a piece of string , tie a small
object to it and swing it around in a
circular motion, the centripetal force
is through the string to the centre
of the circle and centrifugal force,
the mass of the object is working at
a tangent to the circular flight path,
let go the string, no more centripetal
force ( lift in the case of a stalled
wing) and the object departs from the
circular flight path at a tangent to the
outside of the circle. This principle
is also very well demonstrated by
making the bicycle wheel spokes
with fishing line, if they are all in
balance the wheel does not collapse.
How do aircraft
stall inverted?
Exactly the same as in an erect
stall, there are just however a few
extra nasty surprises. The first
being most engines cut out when
subjected to between 0 and -1 "g".
With the engine at idle or even
shut down, depending on a number
of design factors, the inverted stalling
characteristics can change quite
drastically when departing from
normal flight. Firstly, unless it is an
unlimited class aerobatic competition
aircraft, the wing is not symmetrical
and the elevators do not have equal
up and down travel angles. Most
utility aircraft and aircraft certified for
aerobatics in that class have elevators
that travel typically 24 degrees up
and 12 degrees down, or around there
somewhere. The wings are also not
symmetrical and mounted on the
airframe at an incidence angle of
about 4 degrees. So before we can
stall inverted we need to know how
to fly straight and level inverted.
Simply roll the aircraft inverted with
the nose above the horizon ... how
much above, enough to generate the
same lift required when erect. Lets
see what we need. At 140 kts an
RV7 is flying straight and level with
the cowling just below the horizon,
now lets roll her inverted, firstly we
must counter the angle of incidence, 4
degrees, so that means the nose must
now be 8 degrees higher that when
erect. But the wing is not symmetrical
so we will need to increase the angle
of attack by a few more degrees to
generate the lift required for S &
L flight, say another 4 degrees. In
order to get a higher angle of attack
we need to deflect the elevator down
( stick forward) and now a whole
lot of other things come into play,
drag increases drastically due the
lift induced drag, and all the bumph
under the wing, or more correctly
now on top of the wing, reduces the
efficiency of the wing. The shape
of the leading edge changes inverted
and instead of a nice rounded upper
surface, the lower surface has a sharp
curve which results in an earlier
airflow breakaway. Combining all
these factors together, results in a
much higher inverted stall speed and
nose attitude above the horizon.
The impala at 250 kts inverted
requires an increase of 10% power
to maintain 250 kts and a nose
attitude of 14-18 degrees higher
than when flying "the right way
up". In level inverted flight below
120 kts, the Impala cannot be easily
stalled as the 12 degrees "up"
elevator is insufficient to get the
wing to the critical angle of attack
inverted. Next time you are flying,
try and stall your aircraft straight
and level, using only half the stick
rearward travel ... can't be done.