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 & 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
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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
1. leading edge separation
2. low (very low) pressure at leading edge which must rise towards trailing edge
⇒ adverse pressure gradients ⇒ boundary layer separation.
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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.
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