Forces

The lift force is obtained by application of the Kutta-Joukowski lift theorem, where one accounts for the net resulting circulation

With the elimination of sin t in the first term and using orthogonality of the sines in the second, there remains only two terms

L’ = npU2c (A0 + Ai^ (3.67)

The lift coefficient reads

L’ 2Г / A1

° = Трит = Uc = n A0+t) (368)

The important result is that the lift coefficient only depends on the first two modes, A0 which depends on a and aadapt, and A1 which reflects the contribution to the camber of a parabolic plate, mode 1.

Fig. 3.19 Drag calculation

Application:

• the lift coefficient of a flat plate is C; = 2na

• the lift coefficient of a parabolic plate is Ci = 2n(a + 2 d)

Note, that at incidence of adaptation, the lift of the flat plate is zero, but it is 4n d for the parabolic plate.

The drag is zero (d’Alembert paradox). This seems quite surprising when con­sidering the case of a flat plate at incidence a, since the integration of pressure will produce a force F’ normal to the plate, hence a drag D’ = F’ sin a ~ F’a.

The proof of zero drag can be carried out in the general case as follows. A small element of a cambered plate is subject to a pressure difference, hence a small force dF’ perpendicular to the surface element

dF’ = 1 pU2 (C~(x) – C + (x}) ndl (3.69)

where ndl = (—dz, dx). See Fig. 3.19.

The projection of the force onto the x-axis gives the drag contribution

dD = dF’x = —1 pU2 (c— (x) – C + (x)^ dz

= -2pU2 (c— (x) — C+ (x)) dZdx (3.70)

Now using Cp = —2U, < u > = Г’ and = ™(Ц/0), one can write

dD’ = —pr'(x)w(x, 0)dx (3.71)

Upon integration and replacing w(x, 0) by its expression in terms of Г’ one gets

The last expression corresponds to the integration on a square domain [0, c] x [0, c] of an antisymmetric function, i. e. replacing x by £ and vice-versa changes the integrand in its opposite, as shown in Fig. 3.20 for M and M’. The result is D’ = 0 or Cd = 0 for all thin cambered plates.

Now the question remains concerning the result of the pressure integration along the flat plate and the apparent drag calculated to be

(3.73)

where w(x, 0) has been replaced by —aU. This apparent mystery is explained by the presence of a suction force acting at the leading edge of thin cambered plates in the direction of the tangent to the camberline, whenever the solution is singular at the leading edge. This suction force is not accounted for in the integration of pressure along the plate, but is found in a momentum balance on any contour that excludes the leading edge singularity (for example by using a small circle around it). The suction force FS balances exactly the “drag” from the pressure integration, i. e. F’s = —L’a, see Fig. 3.21.

Fig. 3.21 Suction force and z

resulting lift force ‘