Aerodynamic Forces

(9.70) |

where the integral is taken over the surface of the airfoil, . Here, is an outward unit normal vector on , is an element of , and is the air pressure. From Bernoulli's theorem (in an irrotational fluid), we can write , where is a constant pressure. Because a constant pressure exerts no net force on a closed surface, we get

(9.71) |

where is the tangential air velocity just above the surface of the airfoil. Now,

(9.72) |

because on the surface. Hence,

(9.73) |

Making use of Equations (9.64) and (9.69), the previous expression can be written

(9.74) |

where

Here, , , and are the incident wind velocity, the velocity induced by the free vortices in the wake, and the velocity induced by the bound vortices covering the surface of the airfoil, respectively. The forces and are called the lift and the

Let us assume that

that is, that the bound vortices covering the surface of the airfoil run parallel to the -axis. This assumption is exactly correct for an airfoil of infinite wingspan and constant cross-section. Moreover, it is a good approximation for an airfoil of finite wingspan, provided the airfoil's length greatly exceeds its width (i.e., ). Now, the incident wind velocity is written . Moreover, , where is an element of length that runs parallel to the - plane whilst lying on the airfoil surface. Making use of the curl theorem, we can easily show that

(9.79) |

where the closed curve is the intersection of the airfoil surface with the plane , and is the air circulation about the airfoil in this plane. Thus, it follows from Equation (9.75) that

This expression is the generalization of Equation (9.31) for a three-dimensional airfoil of finite size. As before, the lift is at right-angles to the incident wind direction.

Let us make the further assumption--known as the *lifting line approximation* (because the lifting action of the wing is effectively concentrated
onto a line)--that

(9.81) |

throughout , where is the induced velocity due to the free vortices in , evaluated at the trailing edge of the airfoil. Here, the velocity is called the

Note that the induced drag is parallel to the incident wind direction. The origin of induced drag is as follows. It takes energy to constantly resupply free vortices to the wake, as they are swept downstream by the wind (note that a vortex filament possesses energy by virtue of the kinetic energy of its induced flow pattern), and this energy is supplied by the work done in opposing the induced drag. The drag acting on a well-designed airfoil (i.e., an airfoil with an aerodynamic shape that minimizes form drag) situated in a high Reynolds number wind (which implies that the friction drag is negligible) is generally dominated by induced drag.

According to Equations (9.62) and (9.77), the force is written

(9.83) |

We can interchange primed and unprimed variables without changing the value of the integral. Hence,

(9.84) |

Taking the half the sum of the previous two equations, we obtain

(9.85) |

However, . Thus, the previous expression yields

(9.86) |

But, the assumption (9.78) implies that

Consider a closed surface covering the small section of the airfoil lying between the parallel planes and . The flux of vorticity into the surface due to bound vortices at is . The flux of vorticity out of the surface due to bound vortices at is . Finally, the flux of vorticity out of the surface due to the free vortices in the part of the wake lying between and is . However, the net flux of vorticity out of a closed surface is zero, because vorticity is divergence free. Hence,

(9.87) |

which implies that

Finally, consider a semi-infinite straight vortex filament of vortex intensity
**
**
that terminates
at the origin,
, as shown in Figure 9.12. Let us calculate the flow velocity induced by this filament
at the point
. From the diagram
,
,
, and
**
**
. Hence, from Equation (9.61), the induced
velocity at
is
, where

(9.89) |

This result allows us to calculate the downwash velocity, , induced at the trailing edge of the airfoil by the semi-infinite free vortices in the wake. The vortex intensity in the small section of the wake lying between and is , so we obtain

where use has been made of Equation (9.88).