THIN AIRFOIL THEORY

We will represent an airfoil using a vortex sheet, located along the camber line.

If the airfoil is represented by a vortex sheet along the camber line, then the camber line is a streamline of the flow.

The velocity induced at a point x by the entire vortex sheet is

where x (xi) is the distance along the chord line.

The “thin airfoil” assumption leads to the argument that

where w’ is the component of velocity normal to the vortex sheet at the camber line.

Boundary Condition

. So,

Note from Figure:

for small angle of attack and small surface slope.

This is called the “Fundamental Equation of Thin Airfoil Theory”. It says:

“Camber line is a streamline of the flow”.

Results for a Symmetric Airfoil

Symmetric airfoil: the camber line is straight.

. Use the transformation: . Thus,

This gives:

Consider some features of this function:

When you go close to the leading edge, the value goes shooting up towards infinity. As you go towards the trailing edge, it goes to zero. In between, there is is continuous variation. Actually, there is a “singularity” at the leading edge. In other words, we cannot determine what exactly happens at the leading edge from this function (we certainly don’t believe that there is a Vortex Sheet of Infinite Strength there!). This is one of the limitations of thin airfoil theory: it cannot tell you what really happens at the leading edge. This is unfortunate, because the leading edge is apparently where all the action is happening.

Total circulation around the airfoil is: or,

Therefore, .

Also, . But we know that . Thus,

. This beautifully simple result thus comes from thin airfoil theory. Also, differentiating this,

.

Pitching moment about the leading edge

. This reduces to:

. The moment coefficient is: . This can also be written as:

The negative value means that the pitching moment about the leading edge is nose-down. No wonder. Try holding the leading edge when the airfoil is producing lift (which acts through the center of pressure) and the tail will lift up and smack you.

The location of the center of pressure can be found as follows: The center of pressure is the point through which the resultant lift acts. Lets say that this point is located at x_cp from the leading edge. Thus the pitching moment about the leading edge is the moment due to the lift acting through this point. Thus,

Interesting: The resultant lift acts through the quarter-chord point. In other words, the moment due to the lift produced by the entire aft three-quarters of the airfoil is balanced by the moment produced by the lift in the frontmost one-quarter of the chord. Why is this? Let’s trace back where this came from:

The lift is obtained by multiplying the bound circulation by some constants. The bound circulation is obtained by integrating the vortex sheet strength along the entire chord.

Why the vortex sheet? The vortex sheet represents the difference between upper surface velocity and lower surface velocity at each chordwise location. If the velocity is different, the pressure must be different too, and the difference in pressure coefficients can be obtained using the difference in velocities. The difference in pressure between the upper and lower surfaces is what shows up as lift.

So, as we trace back through the derivation, we see that the variation of lift along the chord was determined by that function with the cosine and sine, describing the vortex sheet strength as a function of location along the chord. This function shot up towards infinity as we approached the leading edge. So this means:

MOST OF THE LIFT ON A SYMMETRIC AIRFOIL IN INCOMPRESSIBLE FLOW COMES FROM VERY NEAR THE LEADING EDGE!

This is indeed borne out by measurements: the suction peak occurs pretty close to the leading edge. Here comes another unfortunate revelation: you can’t use thin airfoil theory to get a reliable description of how the lift varies near the leading edge, but miraculously, the theory does indeed gives very good results for the integrated lift.

(Wish there were a way to construct a wing which was all leading edge only!)

Now since the center of pressure is at c/4 regardless of angle of attack (as long as the angle of attack is small, like below 10 degrees or so), the pitching moment about this point (c/4, or “quarter-chord”) must be zero, regardless of angle of attack.

The Aerodynamic Center

The aerodynamic center of an airfoil is the point located such that the pitching moment about that point is independent of angle of attack. The moment about c/4 is zero regardless of angle of attack, so this qualifies as the aerodynamic center.

Thus, for a symmetric airfoil, the center of pressure and the aerodynamic center are both at the quarter-chord in low-speed flow.

What about for other types of airfoils, or at other kinds of speeds?

If the airfoil is cambered, the center of pressure moves back from the quarter-chord. The aerodynamic center is still at the quarter-chord point. This can be seen as follows: We can consider the lift on the cambered airfoil to come from two things: the camber, and the angle of attack. The lift due to camber is independent of angle of attack, and acts through some point on the airfoil depending on the camber line shape. This is the lift when the angle of attack is zero. The part due to angle of attack still acts through the center of pressure. So if we take moment about the quarter-chord, we will get zero from the angle-of-attack part, regardless of angle of attack. We will get a constant value of moment due to the camber part, but this again will not change with angle of attack. So the pitching moment through the quarter-chord is still independent of angle of attack (though not zero). Thus the quarter-chord point is the aerodynamic center even for an cambered airfoil.

Now as speed increases into the regime where the flow density changes appreciably due to changes in velocity (the compressible regime), the center of pressure moves back over the airfoil, even for symmetric airfoils.

In supersonic flow, the center of pressure of a thin symmetric airfoil is at the midchord. This can be seen from supersonic thin airfoil theory.

Results for a Cambered Airfoil

(the derivation can be found in several textbooks, but is simply substitution in the relations derived before).

Assume that the camber line is given by z= z(x), with z =0 at x=0 and x=c.

The lift coefficient is:

Obviously, this consists of two parts: the first part is that due to angle of attack, and second due to camber. When the angle of attack is zero, the lift coefficient is just the second part. The angle of attack for zero lift is thus:

We have to leave these things as integrals because the precise expression will depend on the shape of the camber line for the given airfoil, i.e., the function (dz/dx).

. Here the coefficients A1 and A2 are seen from the following integrals:

The pitching moment about the quarter-chord is:

The location of the center of pressure is now:

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THE VORTEX PANEL NUMERICAL METHOD

To keep the solution simple, we will discuss a 2-dimensional problem. Extension to 3-D is straightforward, and is needed for swept wings and more complex geometries.

A Vortex Panel is a flat region with constant vortex sheet strength. Each panel has a “control point”. For steady-flow problems, this control point is usually the center point of the panel.

Potential at a point P due to the jth panel is:

where the integral is taken over the vortex sheet area on the jth panel.

Due to all N panels, the potential at P is:

. If P is the control point of the ith panel,

.

Boundary Conditions:

At the control points,

. i.e., =

i.e.,

There is one such equation for each panel: thus a total of N equations. There are N unknown panel vortex sheet strengths

In addition, we have the Kutta condition to impose:

At the trailing edge, . That is,

. We thus drop one of the control points and solve the remaining N-1 equations and the Kutta condition. This gives the unknown vortex sheet strengths. Then the total circulation is:

. Lift per unit span is:

To get the velocity at each point, we would use the Biot-Savart expression for induced velocity, summed up over the contributions of all panels. From this, using Bernoulli’s equation, we would find the local pressure and the pressure coefficient.