ME438 Aerodynamics is offered by Dr. Bilal Siddiqui to senior mechanical engineering undergraduates at DHA Suffa University. This lecture set is about prediction of lift on thin cambered airfoils.

1. Aerodynamics
ME-438
Spring’16
ME@DSU
Dr. Bilal A. Siddiqui

2. Thin Airfoil Theory – Recall 1
• For thin airfoils, we can basically replace the
airfoil with a single vortex sheet. For this case,
Prandtl found closed form analytic solutions.
• Looking at the airfoil from far, one can neglect
the thickness and consider the airfoil as just the
camber line. The airfoil camber is z(x).
• If we neglect the camber also, we can basically
place all the vortices on the chord line for the
same effect.

3. Thin Airfoil Theory- Recall 2
• From geometry
𝑉∞ 𝑛
= 𝑉∞ sin 𝛼 + tan−1 −
𝑑𝑧
𝑑𝑥
• For small angles sin 𝜃 ≅ 𝜃
• Both camber and angle of attack
are small, so
𝑉∞ 𝑛
≈ 𝑉∞ 𝛼 −
𝑑𝑧
𝑑𝑥

4. Thin Airfoil Theory - Recall 3
• Induced normal velocity at point x due to the vortex filament at 𝜉
𝑑𝑤 = −
𝛾 𝜉 𝑑𝜉
2𝜋 𝑥 − 𝜉
Total induced normal velocity at x is
𝑤 = −
1
2𝜋 0
𝑐
𝛾 𝜉
𝑥 − 𝜉
𝑑𝜉

5. Thin Airfoil Theory – Recall 4
• Therefore, the ‘no penetration’ (abstinence?) boundary condition is
𝑽∞ 𝜶 −
𝒅𝒛
𝒅𝒙
=
𝟏
𝟐𝝅 𝟎
𝒄
𝜸 𝝃
𝒙 − 𝝃
𝒅𝝃
• This is the fundamental equation of the thin airfoil theory.
• For a given airfoil, both 𝛼 and
𝑑𝑧
𝑑𝑥
are known.
• We need to find 𝛾 𝑥 which
• makes the camber line 𝑧(𝑥) a streamline of the flow
• and satisfied the Kutta condition 𝛾 𝑐 = 0

6. Thin Airfoil Theory – General case
• Let us again transform the variable 𝜉 to another variable 𝜃
𝜉 =
𝑐
2
1 − cos 𝜃 → 𝑑𝜉 =
𝑐
2
sin 𝜃𝑑𝜃
For any particular value of 𝜉 = 𝑥, there is a corresponding particular 𝜃 𝑥
𝑥 =
𝑐
2
1 − cos 𝜃 𝑥 [𝜃0 = 0 and 𝜃𝑐 = 𝜋]
• The fundamental equation can then be written equivalently as
𝑽∞ 𝜶 −
𝒅𝒛
𝒅𝒙
=
𝟏
𝟐𝝅 𝟎
𝝅
𝜸 𝜽 𝒔𝒊𝒏 𝜽
𝒄𝒐𝒔𝜽 − 𝒄𝒐𝒔 𝜽 𝒙
𝒅𝜽

7. Thin Airfoil Theory for Cambered Airfoils-2
• The solution to this integral equation can be shown to be
𝜸 𝜽 = 𝟐𝑽∞ 𝑨 𝟎
𝟏 + 𝐜𝐨𝐬 𝜽
𝐬𝐢𝐧 𝜽
+
𝒏=𝟏
∞
𝑨 𝒏 𝒔𝒊𝒏 𝒏𝜽
• This distribution has the 1st term similar to the symmetric airfoil distribution and
the 2nd term is the contribution due to camber. As expected!
• It can be shown that
𝐴0 = 𝛼 −
1
𝜋 0
𝜋 𝑑𝑧
𝑑𝑥
𝑑𝜃 𝑥, 𝐴 𝑛 =
2
𝜋 0
𝜋 𝑑𝑧
𝑑𝑥
cos 𝑛𝜃0 𝑑𝜃0
• Thus, the first term depends on the angle of attack as well as the camber, but the
second term only depends on the camber.
• It can be easily shown to satisfy the fundamental equation as well as the Kutta
condition 𝛾 𝑐 = 𝛾 𝜋 = 0.

8. Thin Airfoil Theory for Cambered Airfoils-3
• Total circulation is found by Γ = 0
𝑐
𝛾 𝜉 𝑑𝜉 =
𝑐
2 0
𝜋
𝛾 𝜃 sin 𝜃 𝑑𝜃
• This can be evaluation as
Γ = 𝑐𝑉∞ 𝐴0
0
𝜋
1 + cos 𝜃 𝑑𝜃 +
𝑛=1
∞
𝐴 𝑛
0
𝜋
sin 𝑛𝜃 sin 𝜃 𝑑𝜃
• Using trigonometric identities,
0
𝜋
1 + cos 𝜃 𝑑𝜃 = 𝜋
0
𝜋
sin 𝑛𝜃 sin 𝜃 𝑑𝜃 =
0, 𝑓𝑜𝑟 𝑛 ≠ 1
𝜋
2
, 𝑓𝑜𝑟 𝑛 = 1
• we can show
Γ = 𝑐𝑉∞ 𝜋𝐴0 +
𝜋
2
𝐴1

9. Thin Airfoil Theory for Cambered Airfoils-4
• The lift can now be calculated by the K-J theorem
𝐿′
= 𝜌∞ 𝑉∞Γ = 𝜌∞ 𝑉∞
2
𝑐 𝜋𝐴0 +
𝜋
2
𝐴1
Thus,
𝒄𝒍 = 𝟐𝝅 𝜶 +
𝟏
𝝅 𝟎
𝝅
𝒅𝒛
𝒅𝒙
𝒄𝒐𝒔𝜽 𝒙 − 𝟏 𝒅𝜽 𝒙
𝒄𝒍 𝜶
=
𝑑𝑐𝑙
𝑑𝛼
= 𝟐𝝅
This means the lift curve slope of a cambered airfoil is equivalent to the lift
curve slope of a symmetric airfoil. They only differ in the zero AoA lift.
Kia karain hai…Lift lift
hota hai

10. Thin Airfoil Theory for Cambered Airfoils-5
• The angle of zero lift is denoted by αL=0 and
is a negative value.
𝑐𝑙 = 𝑐𝑙 𝛼
𝛼 − 𝛼 𝐿=0 = 2𝜋 𝛼 − 𝛼 𝐿=0
• It is easy to see that
𝜶 𝑳=𝟎 = −
𝟏
𝝅 𝟎
𝝅
𝒅𝒛
𝒅𝒙
𝒄𝒐𝒔𝜽 𝒙 − 𝟏 𝒅𝜽 𝒙
The more highly cambered an airfoil, the
more negative its zero lift AoA

11. Thin Airfoil Theory for Cambered Airfoils-4
• For calculating moment about leading edge, the incremental lift 𝑑𝐿 =
𝜌∞ 𝑉∞ 𝑑Γ due to circulation 𝛾 𝜉 𝑑𝜉 caused by a small portion 𝑑𝜉 of
the vortex sheet is multiplied by its moments arm (𝑥 − 𝜉).
• The total moment is the integration of these small moments
𝑀′ 𝐿𝐸 = −𝜌∞ 𝑉∞
0
𝑐
𝜉𝛾(𝜉) 𝑥 − 𝜉 𝑑𝜉 = −
1
4
𝜌∞ 𝑉∞
2 𝑐2 𝜋𝛼 𝐴0 + 𝐴1 −
𝐴2
2
Moment coefficient then is 𝑐 𝑚 𝐿𝐸
= −
𝑐 𝑙
4
+
𝜋
4
𝐴1 − 𝐴2
Thus the moment too is that of symmetric airfoil, plus
a constant term due to camber.

12. Thin Airfoil Theory for Symmetric Airfoils-5
• Shifting this moment to the quarter chord point
𝑐 𝑚 𝑐/4 = 𝑐 𝑚
𝐿𝐸
+
𝑐𝑙
4
=
𝜋
4
𝐴2 − 𝐴1 ≠ 0
• Now recall that:
• Center of pressure is point on the chord about which there is zero moment
• Aerodynamic center is the point on the chord about which the moment about the
chord does not change with the angle of attack.
• A1 and A2 do not depend on AoA
• This means the quarter chord point on a cambered airfoil is the
aerodynamic center, but not the center of pressure!
• Center of pressure can be found by using the relation 𝑥 𝐶𝑃 = −
𝑐 𝑚 𝐿𝐸
.𝑐
𝑐 𝑙
𝑥 𝐶𝑃 =
𝑐
4
1 +
𝜋
𝑐𝑙
𝐴1 − 𝐴2
This is clearly dependent on angle of attack.

13. Experimental Validation of Thin Airfoil Theory
For Cambered Airfoils
• Consider the NACA 23012 airfoil. Its camber line is given as
• Calculate (a) the angle of attack at zero lift, (b) the lift coefficient
when α = 40, (c) the moment coefficient about the quarter chord, and
(d) the location of the center of pressure in terms of xcp/c, when α =
40. Compare the results with experimental data.

14. Experimental Validation