Aero Engineering 315
Lesson 13
Airfoils Part II
Where does the moment come from?
Aupper
Fnet 
Upper
surface
force
 PUdA
PU = Upper surface pressure distribution
M
Note: M is negative for this example
In general:
Alower
M is < 0 for positive camber
Lower distribution
surface force   PLdA
surfaceairfoils
pressure
L = Lower
M is = 0Pfor
symmetric
M is > 0 for negative camber
Note: Shear stress also contributes to moment in the same manner…
Center of Pressure
M=0
Center of Pressure
Lift
Aerodynamic Force
Moment = 0
+
V

Drag
Center of Pressure: the point on the airfoil where the
total moment due to aerodynamic forces is zero (for a
given  and V )
Aerodynamic Center
y
Aerodynamic Force
x
V
Mac
+

Aerodynamic Center: The point on the airfoil where the
moment is independent of angle of attack. Fixed for subsonic
flight  c/4. Fixed for supersonic flight  c/2.
The moment has a nonzero value for cambered airfoils
(negative for positively cambered airfoils). Moment is zero for
symmetric airfoils.
Force and Moment Coefficients
Aerodynamic Force = FAERO = f(  , m , a , r , V , S )
Using Dimensional Analysis  FAERO = cf • q • S
Where: Cf = f (, Rec, Mach)
Lift:
l  cl qS
Drag:
d  cd qS
Moment:
m  cmqS c
Note: nondimensional coefficients!
Coefficients for NACA airfoils are found from charts
in the Supplemental Data package or Appendix B of text.
Lift-Curve Slope Terminology
cl c
l
3
2
1
4

Sample NACA Data
[1] l=0
– Angle of attack () where the lift coefficient (cl) = 0,  no lift is
produced; l=0 = 0 for a symmetric airfoil; l=0 < 0 for a positively cambered airfoil
[2] cl
– Lift-curve slope (dcl /d); ‘rise’ of cl over ‘run’ of  for a linear
portion of the plot;  0.11/deg for a thin airfoil
[3] clmax
– Maximum cl the airfoil can produce prior to stall
[4] stall
– Stall angle of attack;  at clmax; maximum  prior to stall
Changes to lift and drag curves due to
Reynolds number
cd
cl

Low Re
cl
High Re
At higher Reynolds numbers the boundary layer transitions to
turbulent earlier, so it is more resistant to separation. Delayed
separation causes delayed stall and reduced pressure drag.
Changes to Lift Curves
1. Camber
Positive camber
c
Zero Camber (symmetric)
cl l
Negative camber

Changes to Lift Curves
2. Flaps
cl
Without flaps
With flaps

3. Boundary Layer Control (BLC) or increasing
Reynolds Number
cl
Without BLC
With BLC

DATA SHOWN ON NACA CHARTS (2421)
Airfoil Shape
Data point symbols for various Reynolds numbers (R)
Location of aerodynamic center (a.c.)
DATA SHOWN ON NACA CHARTS (2421)
Lift Curve :
cl plotted against 
DATA SHOWN ON NACA CHARTS (2421)
Drag Polar:
cd plotted against cl
DATA SHOWN ON NACA CHARTS (2421)
Pitching moment coefficient at the quarter-chord
point (cmc/4) plotted against 
DATA SHOWN ON NACA CHARTS (2421)
Pitching moment coefficient at the aerodynamic
center (cmac) plotted against cl
Example Problem
GIVEN:
FIND: cl =
NACA 2421 airfoil
cl = (cl / ) =
Reynolds number = 5.9x106
cd =
Angle of attack = 12°
cmc/4 =
cma.c.=
clmax =
stall =
l=0 =
Example Problem (NACA 2421)
cl  1.3
Cl = (.78-.20)/(6°-0°) = 0.10/deg
Cm c/4  -0.025
Reynolds Number
Example Problem (NACA 2421)
cd  0.017
cl  1.3
Cm a.c.  -0.045
Reynolds Number
Example Problem (NACA 2421)
clmax  1.38
l=0  -2°
stall  15°
Reynolds Number
Example Problem
We just found: cl = 1.3 ; cd = 0.017 ; cmac = -0.045
To calculate the lift, drag and pitching moment on the
airfoil we need to know the dynamic pressure, the chord,
and the planform area.
Given that we are at sea level on a standard day with V = 100 ft/sec,
q = ½ rV2 = ½(0.002377 slug/ft3)(100 ft/sec)2 = 11.885 lb/ft2
If c = 4 ft and S = 200 ft2, then:
l = cl  q  S = (1.3 )  (11.885 lb/ft2)  (200 ft2) = 3090.1 lb
d = cd  q  S = ( 0.017 )  (11.885 lb/ft2)  (200 ft2) = 40.409 lb
mac = cmac  q  S  c
= ( -0.045 )  (11.885 lb/ft2)  (200 ft2)  (4 ft) = -427.86 ft-lb
Next Lesson (T14)…


Don’t come to the classroom – go
straight to the Aero Lab
Prior to class


Read lab handout!
In Class

Gather wind tunnel data