# Why do we need a symmetrical wing profile

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### Symmetrical profiles

In this paragraph we will first discuss symmetrical profiles. In this case the profile has no curvature *dz*/*dx* = 0 and from Eq. (4.25) follows

To `` crack '' this integral equation, we first use a coordinate transformation

There *x* denotes a certain point on the profile chord and the range of values from encompasses the entire profile chord, there is exactly one to each *x* , namely .

It also follows from the transformation relationship

After substituting these transformations and taking into account the changed integration limits of at the front edge (where is) to at the rear edge (where is) we get

The solution to this integral equation is presented here without derivation. It can be found in standard mathematical works for integral equations and reads

This solution has a special feature on the rear edge of the profile, i.e. at . At this point we get an indefinite expression

Here the rule of L'Hospital helps us further and after we add the numerator and denominator to the variable, i.e. have derived that

I.e. the Kutta condition is also fulfilled!

We are now able to calculate the lift for a thin, symmetrical profile. The total circulation around the profile is

Using the transformation relationships Eq. (4.27) and (4.29) follow from this

and after substituting Eq. (4.31) yields this in the last relation

If we plug this relationship into Kutta-Joukowski's theorem, we get

The lift coefficient is defined as

in which

We set Eq. (4.37) and finally get for the lift coefficient

or for the increase in lift

The last two equations are again important results. They state that according to the theoretical derivation, *the lift of the profile is proportional to the angle of attack * and that the increase in lift is the same is. How well this result agrees with reality (wind tunnel tests) can be seen, for example, in Figure 4.17, which compares the development of lift and the aerodynamic moment, which we will then go into, with the theoretical predictions.

**Figure 4.17:** Comparison between experiment and theory for buoyancy and moment of the NACA0012 profile

The moment around the leading edge of the profile can be calculated as follows. We consider the vortex element again at the position , Figure 4.18. The circulation around this vortex element is . The lift generated by this element follows from Kutta-Joukowski's theorem . This lift in turn creates an infinitesimal moment around the leading edge . We then get the entire moment from the integration of all parts from the leading edge (*LE*) to the rear edge (*TE*) of the profile.

**Figure 4.18:** Calculation of the moment around the leading edge

Again we apply the coordinate transformation Eq. (4.27) on and after implementation of the integration we get without derivation

The moment coefficient is defined as

being again *S.* = *c*(1) is. Hence follows

But we already know that it is true

and finally get

For the *c*/ 4 - point applies

and after combining the last two equations we get

The pressure point is defined as the point around which the moments disappear. Another result of the theoretical investigation of slim, symmetrical profiles is that *In the case of symmetrical profiles, the pressure point is 25 percent of the profile depth*. With regard to the aerodynamic torque behavior, there is another excellent point and that is *aerodynamic center*. The aerodynamic center is defined as the point around which the aerodynamic moments are independent of the angle of attack are. From Eq. (4.48) it follows that this also applies to the 25 percent point. In other words, with symmetrical profiles, the pressure point and the aerodynamic center are together at the 25 percent point.

These results are also confirmed by the experimental investigations, Figure 4.17.

At the end of this paragraph we should summarize again. The basis of the theory for slim profiles is the determination of the vortex strength on the profile chord; in such a way that the skeleton line becomes the streamline and the Kutta condition is satisfied. We obtain the vortex strength by solving the integral equation Eq. (4.25) after transformation of in . The resulting vortex strength for a symmetrical profile is given in Eq. (4.31) and a substitution of this distribution in the Kutta-Joukowski theorem has given us the following important results

- Increase in lift
- With symmetrical profiles, the aerodynamic center and the pressure point are at the 25 percent point

**Next:**The arched profile

**Up:**Small perturbation theory:

**Previous:**Small perturbation theory:

*root*

Sun Dec 15 23:00:07 CET 1996

Sun Dec 15 23:00:07 CET 1996

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