Page 196 - Aerodynamics for Engineering Students

Two-dimensional wing theory 179

Sine terms are not used here because practical camber lines must go to zero at the
leading and trailing edges. Thus yc is an odd function which implies that its derivative
is an even function.
Equation (4.22) now becomes

The solution for k as a function of 8 can be considered as comprising three parts so
that k = kl + kz + k3 where

(4.37)

(4.38)

(4.39)

The solutions for kl and k2 are identical to those given in Section 4.4.1 except that
U(a - Ao) replaces Ua in the case of kl. Thus it is only necessary to solve Eqn (4.39)
for k3. By comparing Eqn (4.26) with Eqn (4.39) it can be seen that the solution to
Eqn (4.39) is given by

0
k3(0) =2UxAnsinn8
n=l

Thus the complete solution is given by
cos8 c 00
k(8) = kl +k2 +k3 = 2U(a - 140)-+-+ 2U xAnsinnO
sin8 sin8 n= 1

The constant C has to be chosen so as to satisfy the Kutta condition (4.24) which
gives C = 2U(a - Ao). Thus the final solution is

(4.40)

To obtain the coefficients A0 and A, in terms of the camberline slope, the usual
procedures for Fourier series are followed. On integrating both sides of Eqn (4.35)
with respect to 8, the second term on the right-hand side vanishes leaving

l"gd8 = ~"Ao de = Aon