# τ γ

z
Thin Airfoil Theory Summary
τ(x) = thickness
z(x) = camber line
x
c
•
Replace airfoil with camber line (assume small
τ
c
)
z
z(x) = camber line
x
c
•
Distribute vortices of strength γ (x) along chord line for 0 ≤ x ≤ c .
z
γ(x)dx
x
c
•
Determine γ (x) by satisfying flow tangency on camber line.
dZ 
γ (ξ )d ξ

V∞  α −
=0
−∫
dx  0 2π ( x − ξ )

c
•
The pressure coefficient can be simplified using Bernoulli &amp; assuming small
perturbation:
Thin Airfoil Theory Summary
p − p∞
1
ρV∞2
2
1
1
p + ρ {(V∞ + u ) 2 + V 2 } = p∞ + ρV∞2
2
2
2
2
(V + u ) + V
p − p∞
⇒
= 1− ∞
1
V∞2
ρV∞2
2
V 2 + 2V∞ u + u 2 + V 2
= 1− ∞
V∞2
u u 2 + V 2
= −2 −
V∞
V∞2
cp =
higher order
⇒ C p = −2
•
u
V∞
It can also be shown that
γ ( x ) = uupper ( x ) − ulower ( x )
⇒ ∆C p = C p lower − C p upper =
⇒ C p ( x) = 2
2
(uupper − ulower )
V∞
γ ( x)
V∞
Symmetric Airfoil Solution
For a symmetric airfoil (i.e.
γ (θ ) = 2αV∞
dz
= 0 ), the vortex strength is:
dx
1 + cos θ
sin θ
But, recall:
x=
c
(1 − cos θ )
2
16.100 2002
2
Thin Airfoil Theory Summary
x
c
⇒ cos θ = 1 − 2
sin θ = 1 − cos2 θ
x

= 1 − 1 − 2 
c

sin θ = 2
2
x
x
(1 − )
c
c
1−
⇒ γ ( x ) = 2αV∞
x
x
(1 − )
c
c
1−
γ ( x ) = 2αV∞
x
c
x
c
x
c
∆Cp
Thus, ∆C p = 4α
1−
x
c
x
c .
Some things to notice:
• At trailing edge ∆C p = 0 .
x
c
⇒ Kutta condition is enforced which requires pupper = plower
•
At leading edge, ∆C p → ∞ ! “Suction peak” required to turn flow around
leading edge which is infinitely thin.
The instance of a suction peak exists on true airfoils (i.e. not infinitely thin)
though ∆C p is finite (but large).
Suction peaks should be avoided as they can result in
2. low (very low) pressure at leading edge which must rise towards trailing edge
16.100 2002
3
Thin Airfoil Theory Summary
Cambered Airfoil Solutions
For a cambered airfoil, we can use a “Fourier series”–like approach for the vortex
strength distribution:




1 + cos θ ∞

⇒ γ (θ ) = 2V∞ Ao
+ ∑ An sin nθ 


sin θ
n =1
 
flat plate
cambered

contributions 
 contributions
Plugging this into the flow tangency condition for the camber line gives (after
some work):
π
1 dz
A0 = α − ∫ dθ 0
π 0 dx
π
An =
2 dz
cos nθ0dθ0
π ∫0 dx
After finding the An ’s, the following relationships can be used to find CA , Cmac , etc.
CA = 2π (α − α LO )
Cm c = Cm ac =
4
α LO
π
4
( A2 − A1 )
π
1 dz
1
= − ∫ (cos θ 0 − 1)dθ 0 = α − A0 − A1
π 0 dx
2
Note: in thin airfoil theory, the aerodynamic center is always at the quarter-chord
( c ), regardless of the airfoil shape or angle of attack.
4
16.100 2002
4