AOA, LIFT, DRAG, STALL, cL vs. α, MANEUVERING
Aerodynamic Force
 By Bernoulli’s Principle, the difference between static pressure on the
lower and upper wing surfaces creates a force known as the
aerodynamic force (AF)
  is the angle of attack (AOA), the angle between the wing chord line
and the relative wind
 The component of AF
o perpendicular to the flight path (relative wind) is lift (L)
o parallel to the flight path is drag (D)
Lift Equation—Airspeed in TAS
 L = (cL  V2 S) / (295.37) (#)
o cL is the coefficient of lift (dimensionless)
o  is density ratio (dimensionless)
o V is TAS (nm / hr)
o S is wing planform area (ft2)---constant for fixed configuration
Lift Equation—Airspeed in EAS
 V = TAS = EAS/  EAS2 =  V2
 L = (cL  V2 S) / (295.37) L = (cL EAS2 S) / (295.37)
 EAS (perfect IAS) and cL alone determine
o how much L an airplane develops
o how the airplane responds to control inputs
 However, low air density at high density altitudes causes an airplane to
be more sensitive to pitch and roll control input at high altitude than at
low altitude
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cL as a Function of AOA  / Stall
 cL increases linearly with AOA from low and moderate-to-high 
 at very high  (assuming G > 0)
o airflow over the upper wing surface begins to separate
o cL increases more slowly
 at the critical AOA CRIT, cL is maximum [cL)MAX]d
 at AOAs above CRIT
o cL decreases
o lift decreases
o however, the wing
 still produces lift
 does not “stop producing lift”
 CRIT is also called the stall angle of attack
 Since L = (cL EAS2 S) / (295.37), at CRIT
o L is maximum for fixed EAS
o the wing is NOT stalled
 definition of stall:  > CRIT
Drag
 Parasite drag DP
o Due largely to ram air pressure q =  V2 /2
o Consists mostly of pressure drag (profile drag) and skin-friction
drag
o Increases exponentially as q increases
 Induced drag DI
o Due to AOA
o Increases as AOA increases
 Total Drag DT = DP + DI
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Drag Equations
 D = (cD  V2 S) / (295.37)
 D = (cD EAS2 S) / (295.37)
 cD is the coefficient of drag—depends on EAS and AOA
Caveat about Lift and Drag Equations
 L = (cL EAS2 S) / (295.37)
 D = (cD EAS2 S) / (295.37)
 From an engineering /aircraft design point of view, these equations
o are simplifications
o suffice, nevertheless, for what pilots need to know to understand
how an airplane flies.
Unbalanced Lift Is Required to Maneuver
 Example—Steady-State Constant Altitude Banked Turn





Aircraft flying away from viewer
Right wing down
 is the bank angle
L cos  supports weight W
L sin  is the unbalanced force causing the
airplane to turn right (clockwise)
 If you can’t create unbalanced lift  you can’t maneuver
 Since L = (cL EAS2 S) / (295.37), to change L without changing aircraft
configuration, a pilot has two options and only two options:
o Change EAS
 Use throttle to change thrust
 Dive to increase airspeed or climb to reduce airspeed; i.e.
convert altitude to airspeed or airspeed to altitude
o Change cL
 Increase AOA to increase cL
 Decrease AOA to decrease cL
 Works only if cL ≤ (cL)MAX
o When AOA ≥ αCRIT (in any attitude at any airspeed)
 any attempt to increase L by increasing AOA will fail; cL
will decrease and thus L will decrease
 if AOA = αCRIT the only way to increase L without changing
aircraft configuration is to increase EAS
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G Force / Accelerated Stalls
 G = L / W (W is weight in #)
o W fixed at any flight moment
o Thus G is proportional to L being developed at that moment
 Accelerated stall
o a stall that occurs when G > 1.0
o “accelerated” refers to the increased G or acceleration force
imposed when an airplane is being maneuvered
o VS is the 1 G stall speed for fixed W
o VAS is the symbol we will use for accelerated stall speed
 Effect of G force on stall speed
o VAS = VS G
o An airplane’s stall speed is directly proportional to the square
root of the G force being developed
o Example: VS = 53 KEAS; (VAS)4G = VS G = 53 4 = 106 KEAS
Stall Recovery and Prevention
 At stall speed (regardless of G force)
o  = CRIT
o cL = (cL)MAX
o any attempt to increase G force will result in a stall
  will increase
 cL will decrease
 L will decrease
 If the stall is a high G accelerated stall, a departure from
controlled flight may result
 For fixed configuration, an airplane always stalls at the same angle of
attack, regardless of airspeed, altitude, or attitude
 The only way to recover from a stall is to reduce AOA to or below CRIT
 Stall prevention in any attitude at airspeeds below Vs
o The wing will not stall at 0 G, because no lift is being created
o When airspeed is low or rapidly decreasing—as in a very nose
high attitude
 going to 0 G will assure a stall cannot occur
 to achieve 0 G, AOA must be set so that cL = 0, resulting in
L=0
 to achieve 0 G, use forward yoke/stick until you feel
weightless in the seat
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EFFECT OF WEIGHT, BANK ANGLE,
CONFIGURATION, G, AND ALTITUDE CHANGES
ON VS, VBG (BEST GLIDE), VBR (BEST RANGE),
VBE (BEST ENDURANCE), &c.
Use the Equation
TAS or EAS
V2  V1
W2
W1
cos 1
cos  2
W: weight changes
: level flight bank angle changes
CL: configuration changes
G: G force changes
: density altitude changes
cL 1
cL 2
TAS ONLY
G2  1
G1  2
These four parameters affect
both EAS and TAS changes
Affects TAS changes ONLY
NOTE!!!: Do NOT use the equation to calculate the effect of altitude change
on EAS, because EAS is independent of altitude changes.
How to use the Equation
1. To write the equation as you will use it, determine which parameters
have changed. Eliminate from the equation those terms where a
parameter has not changed. For example, if only weight and G force
have changed, eliminate the terms containing , cL, and ; keep the
terms for W and G. The result will be
V2  V1
W2
W1
G2
G1
2. Determine the original situation and the changed situation. Symbols
with subscript 1 are variables that define the initial flight condition.
Variables with subscript 2 are variables that define the changed flight
condition.
3. Substitute values for variables in the equation. For example, if the 1 G
stall speed is 95 KEAS and weight has changed from 2000# to 2500# and
G force has changed from 1 G to 3 Gs, then V1 = 95 KEAS, W1 = 2000 #,
G1= 1, W2 = 2500 #, and G2 = 3. The changed stall speed is given by
V2  V1
W2
W1
G2
2500 3
.
 95
G1
2000 1
4. Solve the equation: V2  V1
W2
W1
G2
2500 3
 95
 183.9667091 KEAS
G1
2000 1
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Example 1. VS = 140 KEAS (any altitude) @ 300,000# gross.
At 400,000# gross:
V1 = 140 KEAS; W1 = 300,000 #; W2 = 400,000 #
V2 = V1  (W2/W1) = 140 (400,000/300,000) = 161.6580753 KEAS stall
speed at 400,000 # (also KTAS at SL in SA, since TAS = EAS at SL in SA)
At 2.5 Gs and 300,000 #:
V1 = 140 KEAS; G1 = 1.0; G2 = 2.5
V2 = V1  (G2/G1) = 140 (2.5/1.0) = 221.3594362 KEAS accelerated stall
speed at 2.5 Gs (also KTAS at SL in SA, since TAS = EAS at SL in SA)
Note how G force drastically affects stall speed.
Example 2. VBG = 255 KTAS @ 820,000# gross, SL in SA
At 600,000 #:
V1 = 255 KTAS; W1 = 820,000 #; W2 = 600,000 #
V2 = 255 (600,000/820,000) = 218.1267253 KTAS best glide speed at
600,000# (also KEAS at SL in SA)
At FL300 and 820,000# in SA:
1 = SL = 1.0; 2 = FL300 = 0.37413:
V2 = V1  (1/2) = 255  (1.0/0.37413) = 356.6133625 KTAS best glide
speed (but 218.1267253 KEAS since EAS is invariant with altitude)
The TAS value is of little interest to a pilot, since the airspeed indicator
reads IAS (or EAS if IAS is corrected for pitot static system error and
compressibility error)
Note; VBG at FL300 of 218 KEAS corresponds to approximately 225
KCAS because of compressibility effect.
Example 3 (Light Airplane). VS = 53 KCAS dirty configuration, (CL)MAX = 8.2
dirty, (CL)MAX = 6.1 clean, SL in SA
Find stall speed VS clean in KEAS:
V1 = 53 KEAS since there is no compressibility error at SL in SA; (CL)1 =
8.2; (CL)2 = 6.1
V2 = V1  [(CL)1 / (CL)2] = 53 (8.2/6.1) = 61.44943278 KEAS (stall speed in
clean configuration (also KCAS in SA since compressibility correction at
SL is zero
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Example 4 .VS = 100 KEAS @ 14,000#.
Find stall speed @ 12,000# in a constant altitude 30O bank:
V1 = 100 KEAS; W1 = 14,000; W2 = 12000; 1 = 0O (because VS is 1G
wings-level stall speed); 2 = 30O:
V2 = V1 W2/W1) (cos 1/ cos  2) = 100 (12,000/14,000) (cos 0/cos 30) =
99.48584416 KEAS
Note:
V2 = V1  (W2/W1) = 100  (12,000/14,000 = 92.5820110 KEAS—effect of
weight change.
V3 = V2(cos 1 cos 2) = 92.5820110(cos 0/cos 30) = 99.48584416
KEAS—effect of bank angle change added to effect of weight change.
BUT DON’T ROUND THE INTERMEDIATE RESULT 92.5820110!!!
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