1. Lesson 13 Airfoils Part II
Aero Engineering 315
Lesson 13
Airfoils Part II

2. Where does the moment come from?
Fnet M
PL = Lower surface pressure distribution
PU = Upper surface pressure distribution
Note: M is negative for this example
In general:
M is < 0 for positive camber
M is = 0 for symmetric airfoils
M is > 0 for negative camber
Note: Shear stress also contributes to moment in the same manner…

3. Center of Pressure
M = 0

4. Center of Pressure Aerodynamic Force Lift Moment = 0 + Drag
Center of Pressure: the point on the airfoil where the total moment due to aerodynamic forces is zero (for a given  and V )

5. Aerodynamic Center a y Aerodynamic Force Mac x +
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.

6. Force and Moment Coefficients
Aerodynamic Force = FAERO = f( a , m , a , r , V , S )
Using Dimensional Analysis  FAERO = cf • q • S
Where: Cf = f (a, Rec, Mach) c
Lift:
Drag:
Moment:
Note: nondimensional coefficients!
Coefficients for NACA airfoils are found from charts
in the Supplemental Data package or Appendix B of text.

7. Lift-Curve Slope Terminology3 2 1 a
Sample NACA Data 4
[1] al=0 – Angle of attack () where the lift coefficient (cl) = 0,  no lift is produced; l=0 = 0 for a symmetric airfoil; al=0 < 0 for a positively cambered airfoil
[2] cla – 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

8. Changes to lift and drag curves due to Reynolds number
High Re
Changes to lift and drag curves due to Reynolds number c d l c l a
Low 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.

9. Changes to Lift Curves cl 1. Camber a Positive camber
Zero Camber (symmetric) cl
Negative camber a

10. Changes to Lift Curves c c a a 2. Flaps Without flaps With flaps
3. Boundary Layer Control (BLC) or increasing Reynolds Number c l a
Without BLC
With BLC

11. DATA SHOWN ON NACA CHARTS (2421)
Airfoil Shape
Data point symbols for various Reynolds numbers (R)
Location of aerodynamic center (a.c.)

12. DATA SHOWN ON NACA CHARTS (2421)
Lift Curve :
cl plotted against 

13. DATA SHOWN ON NACA CHARTS (2421)
Drag Polar:
cd plotted against cl

14. DATA SHOWN ON NACA CHARTS (2421)
Pitching moment coefficient at the quarter-chord
point (cmc/4) plotted against 

15. DATA SHOWN ON NACA CHARTS (2421)
Pitching moment coefficient at the aerodynamic
center (cmac) plotted against cl

16. Example Problem cma.c.= clmax = stall = l=0 = GIVEN: FIND: cl =
NACA 2421 airfoil cla = (cl / ) =
Reynolds number = 5.9x106 cd =
Angle of attack = 12° cmc/4 =
cma.c.=
clmax =
stall =
l=0 =

17. Example Problem (NACA 2421)
cl  1.3
Cla = ( )/(6°-0°) =
0.10/deg
Cm c/4 
Reynolds Number

18. Example Problem (NACA 2421)
cd  0.017
cl  1.3
Cm a.c. 
Reynolds Number

19. Example Problem (NACA 2421)
clmax  1.38
l=0  -2°
stall  15°
Reynolds Number

20. Example Problem
We just found: cl = 1.3 ; cd = ; cmac =
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( slug/ft3)(100 ft/sec)2 = lb/ft2
If c = 4 ft and S = 200 ft2, then:  
l = cl  q  S = ( )  ( lb/ft2)  (200 ft2) = lb
d = cd  q  S = ( )  ( lb/ft2)  (200 ft2) = lb
mac = cmac  q  S  c
= ( )  ( lb/ft2)  (200 ft2)  (4 ft) = ft-lb
1.3
3090.1
0.017
40.409
-0.045

21. 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