Page 212 - Aerodynamics for Engineering Students

Two-dimensional wing theory 195

Substituting Eqn (4.87) into Eqn (4.41) gives

-- [(2p - l)O, + sine,] + m [(2p - l)(n - 6,) - sin e,] (4.88)
m
-
TP2 -P)
Similarily from Eqn (4.42)

(4.89)

(4.90)

Example 4.2 The NACA 4412 wing section
For a NACA 4412 wing section rn = 0.04 and p = 0.4 so that

0, = cos-l(l - 2 x 0.4) = 78.46“ = 1.3694rad
making these substitutions into Eqns (4.88) to (4.90) gives

A0 = 0.0090, AI = 0.163 and A2 = 0.0228
Thus Eqns (4.43) and (4.47) give

~
CL = T(A~ - 2x40) + 2 ~ ~(0.163 - 2 x 0.009) + 2~a 0.456 + 6.28320 (4.91)
0
=
T T
CM,,, = --(AI - Az) = --(0.163 - 0.0228) = -0.110 (4.92)
4 4