An aircraft of all-up weight 150,000 N has wings of area 60 m ^{2} and mean chord 2.6 m. The drag coefficient for the complete aircraft is given by C_{D}=0.02+0.04 C_{L}^{2} and the lift–curve slope of the wings is 4.5. The tailplane has an area of 10 m ^{2} and a lift–curve slope of 2.2, allowing for the effects of downwash. The pitching moment coefficient about the aerodynamic center (of the complete aircraft less tailplane) based on wing area is –0.03. Geometric data are given in Fig. 13.16. During a steady glide with zero thrust at 245 m/s EAS in which C_{ L }=0.09, the aircraft meets an upgust of equivalent sharp-edged speed 5 m/s. Calculate the tail load, the gust load factor, and the forward inertia force. Take \rho_{ o }=1.223 kg / m ^{3}.
Question 13.8: An aircraft of all-up weight 150,000 N has wings of area 60 ...
As a first approximation, suppose that the wing lift is equal to the aircraft weight. Then,
L=(1 / 2) \rho V^{2} S\left(\partial C_{ L } / \partial \alpha\right) \alpha_{ W }=150,000 N
so that
\alpha_{ W }=150,000 /\left[(1 / 2) \times 1.223 \times 245^{2} \times 60 \times 4.6\right]=0.015 rad =0.87^{\circ}
Also,
C_{ D }=0.02+0.04 \times 0.09^{2}=0.02
Taking moments about the CG and noting that \cos 0.87^{\circ}=1 approximately,
0.6 L-0.5 D+M_{0}=9.0 P
that is,
0.6(150,000-P)-0.5 \times(1 / 2) \rho V^{2} S C_{ D }+(1 / 2) \rho V^{2} c C_{ M , o }=9.0 P
Substituting the appropriate values gives
P=-10,813 N
Then,
L=W-P=150,000+10,813=160,813 N
The change, \Delta P, in the tailplane load due to the gust is, from Eqs. (13.29) and (13.30),
\Delta P=\frac{1}{2} \rho_{0} V_{ E }^{2} S_{ T } \Delta C_{ L , T } (13.29)
\Delta C_{ L , T }=\frac{\partial C_{ LT },}{\partial \alpha} \frac{u_{ E }}{V_{ E }} (13.30)
\Delta P=(1 / 2) \rho V^{2} S _{ T }\left(\partial L_{ L , T } / \partial \alpha\right)\left(u_{ E } / V_{ E }\right)
so that
\Delta P=(1 / 2) \times 1.223 \times 245^{2} \times 10 \times 2.2 \times(5 / 245)
that is,
\Delta P=16,480 N
The total load on the tail is then 16,480-10,813=5,667 N (upwards). The increase in wing lift, \Delta L, due to the gust is given by
\Delta L=(1/2)\rho V^{2} S\left(\partial C_{L}/\partial\alpha\right)\left(u_{E}/V_{E}\right)
Substituting the appropriate values gives
\Delta L=202,254 N
Then,
n=1+[(202,254+16,480) / 150,000)]=2.46
The forward inertia force, FW, is given by
F W=D=(1 / 2) \rho V^{2} S C_{ D }=(1 / 2) \times 1.223 \times 245^{2} \times 60 \times 0.02
that is,
F W=44,046 N