Page 208 - Aerodynamics for Engineering Students

Two-dimensional wing theory 191

Differentiating Eqn (4.78) to satisfy (iii)
dY1 -
- 3a4 + 2a(b - 1)xl - ab = 0 when y1 = 1 (4.79)
-
dx1
and if xo corresponds to the value of x1 when yl = 1, i.e. at the point of maximum displace-
ment from the chord the two simultaneous equations are
1 = ax; + a(b - l)$ - abxo 1
0 = 34 + 2a(b - 1)XO - ab (4.80)
To satisfy (iv) above, AI and A2 must be found. dyl/dxl can be converted to expressions
suitable for comparison with Eqn (4.35) by writing
1
C
x=-(1 -case) or x1 =-(1 -case)
2 2
dYl
- L(i - 2cose + cosze) + a(b - 1) - a(b - 1) case - ab
=
dxl 4

(4.81)
Comparing Eqn (4.81) and (4.35) gives

3a s
A2 =--
8c
Thus to satisfy (iv) above, A1 = A2, i.e.
3s
-(;+ub):=ai- giving b=-- 7
8 (4.82)
The quadratic in Eqn (4.80) gives for xo on cancelling a,

-2(b - 1) f d22(b- 1)2 +4 x 3b (1 - b) &&z +b+ 1
-
x(j = -
6 3
7
From Eqn (4.82), b = - - gives
8
22.55
xo = - 7.45
or -
24 24
i.e. taking the smaller value since the larger only gives the point of reflexure near the trailing
edge:
y = 6 when x = 0.31 x chord
Substituting xo = 0.31 in the cubic of Eqn (4.80) gives

a=-- ' - 8.28
0.121