The document discusses thin airfoil theory for calculating lift and moment coefficients. It covers the assumptions of thin airfoil theory, simplifying the governing equations using these assumptions, solving for the ideal flow solution using a vortex sheet, and determining the lift and moment coefficients algebraically in terms of the vortex sheet strength. Three examples are then provided to demonstrate calculating the aerodynamic characteristics of symmetric, parabolic, and NACA airfoils.

1. Thin Airfoil Theory

2. 2
• Lift coeffcient:
• Moment coeficient:
• Center of pressure:
• Aerodynamic Center:
Airfoil Characteristics
x
y
α
V∞
l
m
c
V
l
Cl
2
2
1
/ ∞
= ρ

3. 3
Thin Airfoil Theory – Setup
Non-penetration condition?
Kutta condition?
Bernoulli?
Assumptions:
1. Airfoil is thin η << c
2. Angles/slopes are small e.g.
sinα ≈ α, cosα ≈ 1, slope ≈ angle
3. Airfoil only slightly disturbs
free stream u', v' << V∞
α
V∞
ηu
ηt
ηc
ηl (<0)
u=V∞cosα+u'
v=V∞sinα+v'
x
Chord c
Camber
t
c
l
t
c
u
l
u
t
l
u
c
η
η
η
η
η
η
η
η
η
η
η
η
−
=
+
=
−
=
+
=
or
)
(
)
(
2
1
2
1
½ Thickness
y
Symbols:
(cusped TE)

4. 4
Thin Airfoil Theory - Simplifications
Bernoulli:
2
2
2
2
2
2
2
2
2
2
'
'
sin
'
2
cos
'
2
/
]
)
'
sin
(
)
'
cos
[(
1
/
)
(
1
∞
∞
∞
∞
∞
∞
∞
∞
−
−
−
−
=
+
+
+
−
=
+
−
=
V
v
V
u
V
v
V
u
V
v
V
u
V
V
v
u
Cp
α
α
α
α
Cp ≈
“Linearized Pressure
Coefficient”
Kutta Condition:
)
0
,
(
)
0
,
( −
+ = c
V
c
V
u=V∞cosα+u'
v=V∞sinα+v'
x
c
V(c,0+)
V(c,0-)
y
Assumptions:
1. Airfoil is thin η << c
2. Angles/slopes are small e.g.
sinα ≈ α, cosα ≈ 1, slope ≈ angle
3. Airfoil only slightly disturbs free
stream u', v' << V∞

5. 5
Thin Airfoil Theory - Simplifications
dx
d
dx
d
V
x
v
dx
d
dx
d
V
x
v
V
x
v
dx
d
dx
d
t
c
t
c
u
u
t
c
η
η
α
η
η
α
η
η
α
η
η
+
+
−
≈
+
+
−
≈
+
≈
+
∞
+
∞
∞
)
0
,
(
'
)
,
(
'
)
,
(
'
Small disturbances
and angles:
Airfoil thin:
Exact:
Likewise for
lower surface: dx
d
dx
d
V
x
v t
c η
η
α −
+
−
≈
∞
− )
0
,
(
'
α
V∞
ηu
ηt
ηc
ηl (<0)
u=V∞cosα+u'
v=V∞sinα+v'
x
t
c
l
t
c
u
η
η
η
η
η
η
−
=
+
=
y
α
V∞
x
y
Exact:
Linearized:
y=0+
y=0- u(c,0+)=u(c,0-)
≈
± )
0
,
(x
Cp
dx
d
dx
d
V
x
v t
c η
η
α ±
+
−
≈
∞
± )
0
,
(
'
Or:

6. 6
A Source Sheet
Jump in normal velocity component
A Vortex Sheet
Jump in tangential velocity component

7. 7
Solving for the Flow
α
V∞
x
y
Linearized problem:
y=0+
y=0- u(c,0+)=u(c,0-)
∞
±
± −
≈ V
x
u
x
Cp /
)
0
,
(
2
)
0
,
(
dx
d
dx
d
V
x
v t
c η
η
α ±
+
−
≈
∞
± )
0
,
(
'
α
V∞
x
iy
Proposed Ideal Flow Solution:
y=0+
y=0-
x1 dx1
z
Source sheet
+ vortex sheet
∞
∞
∞
±
∞
±
±
−
=
⎭
⎬
⎫
⎩
⎨
⎧
−
= ∫ V
x
q
dx
x
x
x
V
V
x
W
V
x
v
c
2
)
(
)
(
)
(
2
1
)
0
,
(
'
Im
)
0
,
(
'
1
0 1
1
γ
π
∞
∞
∞
±
∞
±
∫ −
=
⎭
⎬
⎫
⎩
⎨
⎧
=
V
x
dx
x
x
x
q
V
V
x
W
V
x
u
c
2
)
(
)
(
)
(
2
1
)
0
,
(
'
Re
)
0
,
(
'
1
0 1
1 γ
π
m
So what?
Complex velocity at z
due to element dx1: 1
1
1
1
)
(
)
(
)
(
2
1
dx
x
z
x
i
x
q
−
− γ
π
Complex velocity at z
due to whole sheet: 1
0 1
1
1
)
(
)
(
)
(
2
1
)
(
' dx
x
iy
x
x
i
x
q
z
W
c
∫ −
+
−
=
γ
π
Complex velocity at
y=0± due to sheet:
K
1
0 1
1
1
)
(
)
(
)
(
2
1
)
0
,
(
' dx
x
x
x
i
x
q
x
W
c
∫ −
−
=
±
γ
π
N.B. Karamcheti defines the vortex
sheet strength with the opposite sign

8. 8
Solving for the Flow
α
V∞
x
y
Linearized problem:
y=0+
y=0- u(c,0+)=u(c,0-)
∞
±
± −
≈ V
x
u
x
Cp /
)
0
,
(
2
)
0
,
(
dx
d
dx
d
V
x
v t
c η
η
α ±
+
−
≈
∞
± )
0
,
(
'
α
V∞
x
iy
Proposed Ideal Flow Solution:
y=0+
y=0-
x1 dx1
Source sheet
+ vortex sheet
∞
∞
± ±
−
−
= ∫ V
x
dx
x
x
x
q
V
x
C
c
p
)
(
)
(
)
(
1
)
0
,
( 1
0 1
1 γ
π
∞
∞
∞
±
±
−
= ∫ V
x
q
dx
x
x
x
V
V
x
v
c
2
)
(
)
(
)
(
2
1
)
0
,
(
'
1
0 1
1
γ
π
∞
∞
∞
±
∫ −
=
V
x
dx
x
x
x
q
V
V
x
u
c
2
)
(
)
(
)
(
2
1
)
0
,
(
'
1
0 1
1 γ
π
m
Kutta
condition:
0
)
(
so
2
)
(
)
(
)
(
2
1
2
)
(
)
(
)
(
2
1
so
)
0
,
(
'
)
0
,
(
'
1
0 1
1
1
0 1
1
=
+
−
=
−
−
=
∞
∞
∞
∞
∞
−
∞
+
∫
∫ c
V
c
dx
x
c
x
q
V
V
c
dx
x
c
x
q
V
V
c
u
V
c
u
c
c
γ
γ
π
γ
π
Non-
penetration: ∞
∞
±
−
=
±
+
− ∫ V
x
q
dx
x
x
x
i
V
dx
d
dx
d
c
t
c
2
)
(
)
(
)
(
2
1
1
0 1
1
γ
π
η
η
α
Pressure:
Solution
order?
Need q(x)?
Pressure Difference:

9. 9
General Algebraic Solution
0
)
( =
c
γ
∞
∞
+
−
=
+
+
− ∫ V
x
q
dx
x
x
x
i
V
dx
d
dx
d
c
t
c
2
)
(
)
(
)
(
2
1
1
0 1
1
γ
π
η
η
α
How to solve for γ(x) w/o specifying ?
∞
∞
−
−
=
−
+
− ∫ V
x
q
dx
x
x
x
i
V
dx
d
dx
d
c
t
c
2
)
(
)
(
)
(
2
1
1
0 1
1
γ
π
η
η
α
add 1
0 1
1
)
(
)
(
2
1
)
( dx
x
x
x
i
V
x
dx
d
c
c
∫ −
=
+
−
∞
γ
π
η
α
+Kutta
)
(x
dx
d c
η
Non-penetration
α
V∞
x
0
θ
c
0
)
cos
1
(
/
)
cos
1
(
/
1
2
1
1
2
1
θ
θ
+
=
+
=
c
x
c
x
π
•Write γ as
•Solve integral for each term in
•Relate to for
dx
d c
η

10. 10
General Algebraic Solution
Fourier Series Solution gives:
∑
∞
=
∞ +
−
−
−
=
1
)
sin(
2
sin
cos
1
)
2
(
/
)
(
n
n
o n
B
B
V θ
θ
θ
α
θ
γ
where: ∫
=
π
θ
θ
θ
η
π 0
)
cos(
)
(
2
d
n
dx
d
B c
n
and we have that ∞
=
Δ V
Cp /
)
(
2
)
( θ
γ
θ
∫
∫
∫ +
Δ
−
=
Δ
−
=
Δ
−
=
=
∞
∞
0
4
1
1
0
0
2
2
2
1
2
2
2
1
sin
)
cos
1
)(
(
)
(
1
π
θ
θ
θ
θ
ρ
ρ
d
C
c
x
d
x
C
c
x
pdx
x
c
V
c
V
m
C p
p
c
O
mO
Substitute for γ(θ) and evaluate: )
(
2 1
0 B
B
Cl +
−
= π
πα
Substitute for γ(θ) and evaluate:
)
(
)
2
(
2
1
4
1
4
1
2
1
0
4
1
2
1
B
B
C
B
B
B
C
l
mO
+
+
−
=
+
+
+
−
=
π
π
πα
α
V∞
x
0
θ
c
0
)
cos
1
(
/ 2
1
θ
+
=
c
x
π
p
C
Δ
mO
l
=
=
∞ c
V
l
Cl 2
2
1
ρ

11. 11
Transferring the moment - Conclusions
So,
• Lift varies linearly with α
• Lift curve slope is 2π
• Camber only acts to influence the zero lift angle of attack
• Thickness has no effect on lift and moment
• Lift acts…
• Aerodynamic center is…
•
• Center of pressure is…
=
mx
C
∫
=
π
θ
θ
θ
η
π 0
)
cos(
)
(
2
d
n
dx
d
B c
n
α
V∞
x
0
θ
c
0
)
cos
1
(
/ 2
1
θ
+
=
c
x
π
mO
l
mx
)
(
2 1
0 B
B
Cl +
−
= π
πα
)
( 2
1
4
1
4
1
B
B
C
C l
mO +
+
−
= π
Now:
( )( )
∑
∞
=
−
−
−
=
Δ
1
0
)
sin(
4
sin
cos
1
2
2
)
(
n
n
p n
B
B
C θ
θ
θ
α
θ

12. 12
Example 1 – Symmetric Foil
An airfoil has a straight camber line
defined as:
Determine the aerodynamic characteristics
and vortex sheet strength.
∫
=
π
θ
θ
θ
η
π 0
)
cos(
)
(
2
d
n
dx
d
B c
n
α
V∞
x
0
θ
c
0
)
cos
1
(
/ 2
1
θ
+
=
c
x
π
ηu
ηt
ηc
ηl
)
(
)
( 2
1
4
1
4
1
B
B
C
C l
c
x
mx +
+
−
= π
)
(
2 1
0 B
B
Cl +
−
= π
πα
0
/ =
c
c
η
∑
∞
=
∞ +
−
−
−
=
1
)
sin(
2
sin
cos
1
)
2
(
/
)
(
n
n
o n
B
B
V θ
θ
θ
α
θ
γ
2
2
)
/
(
/
2
cos
1
)
sin(
1
)
/
(
2
cos
c
x
c
x
c
x
−
=
−
=
−
=
θ
θ
θ

13. 13
Example 2 – Parabolic Foil
An airfoil has upper and lower surfaces
defined as:
Determine the aerodynamic characteristics
∫
=
π
θ
θ
θ
η
π 0
)
cos(
)
(
2
d
n
dx
d
B c
n
α
V∞
x
0
θ
c
0
)
cos
1
(
/ 2
1
θ
+
=
c
x
π
ηu
ηt
ηc
ηl
)
(
)
( 2
1
4
1
4
1
B
B
C
C l
c
x
mx +
+
−
= π
)
(
2 1
0 B
B
Cl +
−
= π
πα
)
/
1
)(
/
(
3
.
0
/
)
/
1
)(
/
(
5
.
0
/
c
x
c
x
c
c
x
c
x
c
l
u
−
=
−
=
η
η

14. 14
Example 3 – NACA 2412
A NACA 2412 airfoil has a camber line
given by the equations:
Determine the aerodynamic characteristics
∫
=
π
θ
θ
θ
η
π 0
)
cos(
)
(
2
d
n
dx
d
B c
n
α
V∞
x
0
θ
c
0
)
cos
1
(
/ 2
1
θ
+
=
c
x
π
ηu
ηt
ηc
ηl
)
(
)
( 2
1
4
1
4
1
B
B
C
C l
c
x
mx +
+
−
= π
)
(
2 1
0 B
B
Cl +
−
= π
πα
1
/
)
/
(
/
/
0
)
/
(
/
10
4
2
36
2
45
2
90
1
10
4
2
16
2
10
1
≤
≤
−
+
=
≤
≤
−
=
c
x
c
x
c
x
c
x
c
x
c
x
c
c
η
η
⎪
⎪
⎩
⎪
⎪
⎨
⎧
≤
≤
−
≤
≤
−
=
1
10
4
9
1
45
2
10
4
0
4
1
10
1
c
x
c
x
c
x
c
x
dx
d c
η
( )
( )
⎪
⎪
⎩
⎪
⎪
⎨
⎧
≥
≥
+
−
≥
≥
+
−
=
0
cos
1
18
1
45
2
cos
1
8
1
10
1
p θ
θ
θ
θ
θ
π
θ
η p
c
dx
d
°
=
=
+
=
102
7722
.
1
)
cos
1
(
4
.
0 2
1
rad
p
p
θ
θ

15. 15
∫
=
π
θ
θ
θ
η
π 0
)
cos(
)
(
2
d
n
dx
d
B c
n
α
V∞
x
0
θ
c
0
)
cos
1
(
/ 2
1
θ
+
=
c
x
π
ηu
ηt
ηc
ηl
)
(
)
( 2
1
4
1
4
1
B
B
C
C l
c
x
mx +
+
−
= π
)
(
2 1
0 B
B
Cl +
−
= π
πα
( ) ( )
( ) ( )
( ) ( ) 0138
.
0
2
cos
cos
1
2
1
10
1
2
2
cos
cos
1
18
1
45
2
2
0815
.
0
cos
cos
1
2
1
10
1
2
cos
cos
1
18
1
45
2
2
009
.
0
cos
1
2
1
10
1
2
cos
1
18
1
45
2
2
0
2
0
1
0
0
=
⎟
⎠
⎞
⎜
⎝
⎛
+
−
+
⎟
⎠
⎞
⎜
⎝
⎛
+
−
=
−
=
⎟
⎠
⎞
⎜
⎝
⎛
+
−
+
⎟
⎠
⎞
⎜
⎝
⎛
+
−
=
=
+
−
+
+
−
=
∫
∫
∫
∫
∫
∫
π
θ
θ
π
θ
θ
π
θ
θ
θ
θ
θ
π
θ
θ
θ
π
θ
θ
θ
π
θ
θ
θ
π
θ
θ
π
θ
θ
π
p
p
p
p
p
p
d
d
B
d
d
B
d
d
B
°
=
⇒
+
=
+
−
=
08
.
2
2279
.
0
2
)
(
2 1
0
ol
l B
B
C
α
πα
π
πα
0532
.
0
)
( 2
1
4
1
4
/
−
=
+
= B
B
Cmc π
Use Matlab!

16. 16
Comparison
with data
Cmc/4
Cl
http://naca.larc.nasa.gov/re
ports/1945/naca-report-824/
Summary of
airfoil data
Abbott, Ira H Von
Doenhoff, Albert E
Stivers, Louis, Jr

17. 17
Example 4 – Helicopter Rotor
The loading distribution ΔCp is measured
on a helicopter rotor airfoil section as a
function of angle of attack. Estimate the
change in loading produced by a 2 degree
change in angle of attack.
∫
=
π
θ
θ
θ
η
π 0
)
cos(
)
(
2
d
n
dx
d
B c
n
α
V∞
x
0
θ
c
0
)
cos
1
(
/ 2
1
θ
+
=
c
x
π
ηu
ηt
ηc
ηl
)
(
)
( 2
1
4
1
4
1
B
B
C
C l
c
x
mx +
+
−
= π
)
(
2 1
0 B
B
Cl +
−
= π
πα