Consider a small remote-controlled (RC) aircraft with the following characteristics:

m = 700 g, S = 0.2 m², b = 1 m; e = 0.8; $η_P$ = 0.7, $C_{Do}$ = 0.03, $C_{L\max}$ = 1.2

The airplane is employing a prop-driven electric motor where three cells of 2100 mAh, 12 V Li-Po (lithium polymer) batteries provide electric energy for the motor. Evaluate the tightest turn performance of this aircraft at the sea level.

Question

Consider a small remote-controlled (RC) aircraft with the following characteristics:

m = 700 g, S = 0.2 m², b = 1 m; e = 0.8; [latex]η_P[/latex] = 0.7, [latex]C_{Do}[/latex] = 0.03, [latex]C_{L\max}[/latex] = 1.2

The airplane is employing a prop-driven electric motor where three cells of 2100 mAh, 12 V Li-Po (lithium polymer) batteries provide electric energy for the motor. Evaluate the tightest turn performance of this aircraft at the sea level.

Accepted Answer

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

[latex]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 imes 0.8 imes 5}=0.08 \quad \quad \quad \quad (3.8)[/latex]

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

[latex]P_{\max}=IV=2.1 imes 3 imes 12=75.6 \space W \quad \quad \quad \quad (4.35)[/latex]

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

[latex]V_{ft}=\frac{8KW^2}{3 ho P\eta_PS}=\frac{8 imes 0.08 imes (0.7 imes 9.81)^2}{3 imes 1.225 imes 75.6 imes 0.7 imes 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}{ ho S(C_{D_o}+KC^2_{L_{\max}})}\Big]^{\frac{1}{3}}=\Big[\frac{2 imes 75.6 imes 0.7}{1.225 imes 0.2 imes (0.03+0.08 imes (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[/latex]

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

[latex]n_{tt}=n_{\max_C}=\frac{ ho (V^{\ast})^2SC_{L_{\max}}}{2W}=\frac{1.225 imes 14.4^2 imes 0.2 imes 1.2}{2 imes 0.7 imes 9.81}=4.44\quad \quad \quad \quad (9.143)[/latex]

The turn radius for the tightest turn ([latex]R_{tt}[/latex]) is

[latex]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)[/latex]

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

(4.35): P =IV

(9.139): [latex]V_{tt}=\frac{8KW^2}{3 ho P\eta_PS}[/latex]

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

Table 9.7 Summary of equations for the tightest turn parameters

No.	Tightest turn parameter	Symbol	If [latex]V_{tt}\geq V^{\ast}[/latex]	If [latex]V_{tt}\prec V^{\ast}[/latex]
1.	Airspeed corresponding to minimum turn radius	[latex]V_{tt}[/latex]	9.135	9.136
2.	Load factor corresponding to minimum turn radius	[latex]n_{tt}[/latex]	9.137 or 9.138	9.139
3.	Turn rate corresponding to minimum turn radius	[latex]\omega_{tt}[/latex]	9.146 or 9.145	9.145
4.	Minimum turn radius	[latex]R_{tt}[/latex]	9.143	9.142 or 9.144
5.	Bank angle corresponding to minimum turn radius	[latex]\phi_{tt}[/latex]	9.141	9.140

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

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