**The Airplane Aerodynamic Stall
Explained**

**In the aftermath of the downing by
Turkish F-16 fighter jets of the Russian Sukhoi tactical bomber Su-24 over
Syria, some have pointed out that Turkish claims the Russian jet was in Turkish
airspace for 17 seconds, but covered a distance of only 1.15 miles, would imply
the Russian jet would have had to be flying at only 243 mph, which some say
would be slower than the airplane’s stall speed and therefore impossible.**

**Others dispute this stall speed
figure and say an airplane of this type would have a stall speed closer to 150
mph. Who is right, and how can we find some objective answers? We can and we
will.**

**The most important point is that
there is no such thing as a single stall speed for any airplane, as Dr. Roberts
pointed out. In fact wing stall is not a function of airspeed, but of the
wing’s angle of incidence to the oncoming airflow, which is called the angle of
attack (AoA) or simply alpha (this angle is obviously a function of the
airplane’s “attitude” or pitch-up angle).**

**If wing angle of attack exceeds the
critical point, the airflow starts to separate from the top surface of the wing
(the suction side), and the wing will begin to lose lift and a stall recovery
maneuver is required.**

**The stalling angle of most
airplanes is on the order of 15 to 18 degrees of angle of attack, although some
combat aircraft have special design features, such as leading edge root
extensions (or LERX) that may increase critical AoA to about 30 degrees. Engine
power plays a big role too—a fighter aircraft with very powerful engines and
low airplane weight can sustain high angles of attack by using its engine power
to overcome the loss of lift due to wing stall.**

**It is also useful to know that wing
lift increases nearly linearly with an increase in the angle of attack. So if
we want to double lift, we must double the angle of attack.**

**For example when taking off, the
pilot will pull back on the control yoke and increase the pitch-up angle of the
airplane, thus increasing angle of attack and thereby total wing lift. The
extra lift launches the airplane into the air!**

**This maneuver is called takeoff
“rotation” which refers to the pilot in effect rotating the aircraft about its
pitch axis (which can be thought to run horizontally from wingtip to wingtip).**

**It is an old truism that every
student pilot is taught early on, that any airplane will stall at any speed, at
any attitude, and any bank angle. So when we say that an airplane has a certain
stall speed, it is simplifying things to the point of uselessness.**

**As Dr. Roberts pointed out, an
airplane that is maneuvering can have a higher stall speed than if it is simply
flying straight and level. When an airplane is in a banked turn the lift
created by the wing is also tilted, since lift is always perpendicular to the
wing. If the airplane is banked at 45 degrees, it means its lift vector will be
pointing 45 degrees from the vertical (or horizontal if you prefer), as seen in
the figure below.**

**The wing’s lift in a bank decreases
by the cosine of the bank angle. If the airplane is banked 45 degrees, the
cosine is 0.707 and the amount of vertical lift is only about 70 percent of the
total lift that is now pointing at 45 degrees.**

**In order to maintain the airplane’s
altitude in a turn, the pilot must then pull back on the yoke and increase the
angle of attack, so as to increase diagonal lift and thereby provide enough
vertical lift component to maintain the aircraft at the same altitude. ****That vertical lift component is shown in
the diagram.**

**It should be noted that the
airplane loses lift exponentially with increasing bank angle. In a 60 degree
bank the cosine is 0.5, (which is 0.707squared). At this bank angle the
vertical lift component is now only half of the total diagonal lift, and the
airplane now needs double the lift it would normally need at straight and level
flight. Again the pilot has to pull back the yoke and increase airplane pitch
attitude to an angle of attack about double what it would be in straight and
level flight.**

**In a 75 degree bank the cosine is
0.26 which means the airplane now needs almost four times as much lift.
Pitching the airplane into such an attitude could well exceed the critical
angle of attack and bring about wing stall—which could be dangerous at such a
steep bank angle.**

**Now it should be noted that all the
above examples of airplane bank are assumed for a level turn, where the
aircraft stays at the same altitude. This is the “standard” turn maneuver. If
the airplane were to simultaneously bank and descend, there need not be any
loss of lift (depending on bank angle and rate of descent), although the pilot
must now be careful that the descending airplane does not overspeed, or enter a
dangerous spiral dive.**

**Another technical point worth
noting in turning maneuvers is that (in a level turn), the airplane (and pilot)
weight increases due to centrifugal force, just as you may remember as a child
on those spin rides, where you felt heavy and unable to move your arms or legs.**

**In a level turn the g loading is
simply the inverse of our above bank-to-lift relation—ie. g = 1 / cosine bank
angle.**

**If the bank angle is 45 degrees the
cosine is 0.7 and the g-loading is therefore 1 / 0.7 = 1.41. In other words, if
the pilot weighs 100 pounds, she would now feel as if she weighed 141 lb. The
aircraft’s effective weight would be 1.41 times greater.**

**In a 60 degree turn g-loading is 2,
so the plane and pilot’s weight is effectively doubled. In a 75 degree turn the
g-loading is 3.9, which means that 100 lb pilot would feel like she weighs 390
lb, and the airplane is nearly four times as heavy (at least as far as the
unforgiving force of gravity is concerned). Again, this kind of steep bank
angle would likely result in a wing stall.**

**Another important point to consider
is altitude, which determines the air density. The higher the altitude, the
lower the density (due to less air pushing down from above) and the higher the
stall speed.**

**Let’s first consider the “standard”
published stall speed, Vs, which is typically given for landing and takeoff
conditions at sea level standard day conditions, which means 15 C (59 F) and
atmospheric pressure of 101.3 kilopascals or 29.92 inches of mercury.**

**At high altitude and thinner air,
the stall speed can approach or even surpass the airplane cruise speed! A
friend has flown hundreds of USAF missions in the Lockheed U2 reconnaissance
airplane, considered the highest-flying jet in the world. The aircraft’s cruise
speed in that thin air at over 65,000 feet altitude is only several mph above
its stall speed, requiring constant pilot attention to keep the aircraft in its
very narrow safe operating speed range.**

**That is because the air molecules
are so few and far between that the pilot must fly at a high angle of attack
just to keep aloft. If he loses his concentration and the speed drops only
slightly, the airplane wing will stall, and a recovery under those conditions
might be very difficult.**

**In pilot circles this flight regime
is known as the coffin corner. There have been a number of deadly crashes of
high performance aircraft at high altitudes, prompting the FAA to issue an
advisory on high altitude flight.**

**Now there is a mathematical formula
that relates stall speed and air density:**

**Stall Speed = the square root of
wing loading (which is the aircraft’s weight divided by its wing surface area),
times 2, divided by air density, and divided again by the wing’s maximum lift
coefficient. In mathematical terms the formula looks like this:**

**Vstall = sqrt( (W/S) * 2 / rho /
CLmax)**

**Where W/S is the wing loading (or
airplane Weight divided by wing Surface), rho is the air density and CLmax is
the maximum lift coefficient.**

**So we can see just by looking at
that relation that the lower the air density, the higher the stall speed will
be.**

**Published data on the Su-24 stall
speed is difficult to find, but we can make an educated guess using a typical
estimate of maximum lift coefficient for combat aircraft that ranges between
1.2 and 1.8—in the so-called clean configuration, ie. no high lift devices like
landing flaps extended. In normal flight the aircraft would be flying clean to
reduce drag, but might deploy flaps in a tight turn to increase its lift and
help prevent stall.**

**If we assume the CLmax is between
those given values, say 1.5, then we can work out the rest since we know the
airplane’s wing loading is 133 lb / ft ^2 (wikipedia) and we know the airplane
was flying at about 6,000 meters altitude, about 19,700 ft. At this altitude
air density is only about half of what it is at sea level: in metric the air
density on a “standard” day at that altitude would be about 0.66 kg per cubic
meter, or in English units, 0.00128 slugs per cubic foot, where 1 slug equals
32.2 lb; (32.2 feet per second being the acceleration of gravity on earth).**

**So if we plug in the numbers we
get:**

**sqrt(133 * 2 / 0.00128 / 1.5) = 372
feet per second.**

**To convert to mph we multiply by 15
and divide by 22 and get 253 mph as our stall speed at straight and level
flight at 6000 meters altitude, about 19,700 ft. Now any turning or banking
maneuver would obviously increase stall speed further, so we can think of 253
mph as a minimum flying speed at that altitude for the Su-24 (again assuming
standard day conditions).**

**Now that wing loading assumes a
fully loaded aircraft, which probably was not the case. The Sukhoi would likely
not have been carrying full fuel of about 11 tons (22,000 lb), although it
might have been carrying its complete payload of over 3 tons of ordnance (6,000
lb). However reports of the sortie described the pair of bombers had already
made one bombing run, so full ordnance was probably not on board either.**

**We can take a stab at estimating
the operating weight of the aircraft at that time. The Su-24’s empty weight is
published as just under 50,000 lb (wikipedia). If we assume half fuel, that
would be about 11,000 lb, and if we assume half ordnance load, that would be
about 3,000 lb, giving a total aircraft weight of 64.000 lb.**

**The published wing area is 594
square feet, so the wing loading under this flying condition would be 64,000 /
594 = ~108 lb/ft^2.**

**If we now plug that into our
stalling equation that takes into account the air density, we get:**

**Vstall = sqrt(108 * 2 / 0.00128 /
1.5) = ~335 fps, which equals ~227 mph.**

**Again, this is an approximation of
the Su-24 stalling speed at part-weight and straight and level flight at 19,700
ft altitude. Any turning maneuver would increase stall speed as described
above. Also a warmer than standard temperature at that altitude would lower air
density (and increase the effective altitude). This is known as density
altitude and all pilots are trained to work only with density altitude, since
non-standard meteorological conditions are not uncommon.**

**Note also that the actual CLmax value
of the Su-24 may be something different than our guesstimate; a higher value
would decrease the stall speed, but a lower value would increase it. We are
making an educated guess. However, this puts us in the ballpark and hopefully
sheds some light on the technical issue under dispute.**

**On a practical level, it is outside
the realm of possibility that a bomber aircraft would be flying at the edge of
its stall at any time during a sortie, other than hard evasive maneuvers to
avoid enemy action. Since the rescued co-pilot reports that they did not even
see the missile coming, nor had time to take evasive maneuvers, it can be
assumed with strong certainty that the airplane would not have been flying near
the stall at any time.**

**More than likely the aircraft would
have been flying in the so-called “transonic” regime where airliners usually
operate, which is about 550 to 650 mph.**

**In light of this brief and inexact
analysis, there appears to be reasonable doubt as to the Turkish version of the
Sukhoi’s incursion into sovereign airspace under the conditions described.**

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