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Thin airfoil theory

Here we discuss thin airfoil in freestream of velocity V∞ under small angle
of attack . Camber and thickness are small in relation with chord length c.
In such case, airfoil can be described with a single vortex sheet distributed
over the camber line(Figure 4). Our goal is to calculate the variation of

(s), such that the chamber line becomes streamline and Kutta condition at
trailing edge,
(c) = 0, is satisfied.
3
Figure 4: Thin airfoil approximation. Vortex sheet is distributed over the
chamber line
The velocity at any point in the flow is the sum of the uniform freestream
velocity and velocity induced by the vortex sheet . In order the camber line
to be a streamline, the component of velocity normal to the camber line must
be zero at any point along the camber line.
w′(s) + V∞,n = 0 , (5)
where w′(s) is the component of velocity normal to the chamber line induced
by the vortex sheet and V∞,n the component of the freestrem velocity normal
to the camber line. Considering small angle of atack and defining (x) =
dz/dx as the slope of the chamber line, V∞,n can be written as (Figure 5)
V∞,n = V∞ −
dz
dx (6)
Because airfoil is very thin, we can make the approximation
w′(s) ≈ w(x) , (7)
where w(x) denotes the component of velocity normal to the chord line and
can be, using the Biot-Savart law, expressed as
w(x) = −Z c
0

()d
2(x − )
(8)
Substituting equations (6), (7) and (8) into (5) and considering Kutta con-
dition, we obtain
1
2 Z c
0

()d
x − 
= V∞ −
dz
dx

(c) = 0 (9)
fundamental equations of thin airfoil theory.
4
Figure 5: Determination of the component of freestrem velocity normal to
the chamber line
In order to satisfy this conditions , we first transform our variables x and
 into
 =
c
2
(1 − cos ) x =
c
2
(1 − cos 0) (10)
and equation (9) becomes
1
2 Z 
0

() sin d
cos  − cos 0
= V∞ −
dz
dx (11)
with a solution that satisfies Kutta condition
() = 0

() = 2V∞A0
1 + cos 
sin 
+

Xn=1
An sin(n) (12)
In order to find coefficients A0 and An, we substitute equation (12) into
equation (11) and use the following trigonometric relations
Z 
0
sin(n) sin d
cos  − cos 0
= − cos(n0) (13)
Z 
0
cos(n)d
cos  − cos 0
=
 sin(n0)
sin 0
(14)
and finnaly obtain
dz
dx
= ( − A0) +

Xn=1
An cos(n0) (15)
5
This equation is in form of a Fourier cosine series expansion for the function
dz/dx. Comparing it to the general form for the Fourier cosine expansion we
obtain
A0 = −
1
 Z 
0
dz
dx
d0 (16)
An =
2
 Z 
0
dz
dx
cos(n0)d0 (17)
The total circulation due to entire vortex sheet from leading to the trailing
edge is
􀀀 = Z c
0

()d =
c
2 Z c
0

() sin  d (18)
Substituting equation (12) for
() into equation (18) and carrying out the
integration, we obtain
􀀀 = cV∞A0 +

2
A1 (19)
hence the lift per unit span, given by Kutta-Joukowski is
L′ = ∞V∞􀀀 = c∞V 2
∞A0 +

2
A1 (20)
This equation leads to to the lift coefficient in form
cl = (2A0 + A1) = 2h +
1
 Z 
0
dz
dx
(cos(n0) − 1)d0i (21)
and lift slope
lS ≡
dcL
d
= 2 (22)
Last two results are important. We can see, that lift coefficient is func-
tion of the shape of the profile dz/dx and angle of attack , and that even
symmetrical wing produces lift, when set under an angle of attack. Lift slope
is constant, independently of the shape of the profile, while the zero lift angle
L=0 = −
1
 Z 
0
dz
dx
(cos(n0) − 1)d0 (23)
depends on the shape. The more highly chambered the airfoil, the larger is
L=0

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