Question 9.11: Consider a small remote-controlled (RC) aircraft with the fo...

We first need to find two parameters, namely, AR and K:

AR=\frac{b^2}{S}=\frac{1^2}{0.2}=5 \quad \quad \quad \quad (3.9)\\ \space \\ K=\frac{1}{\pi eAR}=\frac{1}{3.14\times 0.8\times 5}=0.08 \quad \quad \quad \quad (3.8)

Based on the description of the battery, a maximum current of 2.1 A (i.e., 2100 milli-amps) at 12 V is provided for one hour (i.e., 3600 s). Thus, the engine power from these batteries is

P_{\max}=IV=2.1 \times 3\times 12=75.6 \space W \quad \quad \quad \quad (4.35)

We need to compare the corner speed and the airspeed corresponding to the tightest turn:

V_{ft}=\frac{8KW^2}{3\rho P\eta_PS}=\frac{8\times 0.08\times (0.7\times 9.81)^2}{3\times 1.225\times 75.6\times 0.7\times 0.2}=0.77\space m/s=1.5 \space knot \quad \quad \quad \quad (9.139) \\ \space \\ V^{\ast}=\Big[\frac{2P_{\max}\eta_P}{\rho S(C_{D_o}+KC^2_{L_{\max}})}\Big]^{\frac{1}{3}}=\Big[\frac{2\times 75.6\times 0.7}{1.225\times 0.2\times (0.03+0.08\times (1.2)^2)}\Big]^{\frac{1}{3}} \quad \quad \quad \quad (9.140) \\ \space \\ \Rightarrow V^{\ast}=14.4\space m/s=28\space knot

Since V_{tt} < V^{\ast}, the theoretical value for the tightest is not practical. From Equation 9.140, we consider V_{tt} = V^{\ast}= 14.4  m/s. According to Table 9.7, the equations in the last column are used. The load factor is

n_{tt}=n_{\max_C}=\frac{\rho (V^{\ast})^2SC_{L_{\max}}}{2W}=\frac{1.225\times 14.4^2\times 0.2\times 1.2}{2\times 0.7\times 9.81}=4.44\quad \quad \quad \quad (9.143)

The turn radius for the tightest turn (R_{tt}) is

R_{tt}=R_{\min}=\frac{V_{tt}^2}{g\sqrt{n_{tt}^2-1}}=\frac{14.4^2}{9.81\sqrt{4.44^2-1}}=4.89 \space m\quad \quad \quad \quad (9.146)

(3.9):       AR=\frac{b}{C}=\frac{bb}{Cb}=\frac{b^2}{S}

(4.35):     P =IV

(9.139):   V_{tt}=\frac{8KW^2}{3\rho P\eta_PS}

(9.140):   V_{tt}=V^{\ast}=\Big[\frac{2P_{\max}\eta_P}{\rho S(C_{D_o}+KC^2_{L_{\max}})}\Big]^{\frac{1}{3}}

Table 9.7 Summary of equations for the tightest turn parameters