Question 9.10: Consider the single-engine piston-prop acrobatic aircraft Ge...
Consider the single-engine piston-prop acrobatic aircraft General Avia F 22 Pinguino with the following features:
m_{TO} = 900 kg, S = 10.82 m², P = 130 kW, b = 8.5, V_s = 54 knot (with flap)
Assume: C_{Do} = 0.021, η_P = 0.8, e = 0.87.
Evaluate the fastest turn performance of this acrobatic aircraft at sea level.
We first need to find three parameters: K, AR, and C_{L\max}:
AR=\frac{b^2}{S}=\frac{8.5^2}{10.82}=5.92 \quad \quad \quad \quad (3.9) \\ \space \\ K=\frac{1}{\pi eAR}=\frac{1}{3.14\times 0.87\times 5.92}=0.06 \quad \quad \quad \quad (3.8) \\ \space \\ C_{L_{\max}}=\frac{2mg}{\rho SV_s^2}=\frac{2\times 900\times 9.81}{1.225\times 10.82\times (54\times 0.514)^2}=1.73 \quad \quad \quad \quad (2.49)
We need to compare the corner speed and the airspeed corresponding to the fastest turn:
V_{ft}=\frac{4KW^2}{P\eta_P\rho S}=\frac{4\times 0.06\times (900\times 9.81)^2}{130,000\times 0.8\times 1.225\times 10.81}=13.5\space m/s=26.3\space knot \quad \quad \quad \quad (9.126) \\ \space \\ V^{\ast}=\Big[\frac{2P_{\max}\eta_P}{\rho S(C_{D_o}+KC_{L_{\max}}^2)}\Big]^{\frac{1}{3}}=\Big[\frac{2\times 130,000\times 0.8}{1.225\times 10.82\times (0.021+0.06\times (1.73)^2)}\Big]^{\frac{1}{3}} \quad \quad \quad \quad (9.90) \\ \space \\ \Rightarrow V^{\ast}=42.9\space m/s=83.3\space knot
Since V_{ft} < V^{\ast}, the theoretical value for the fastest is not practical. From Equation 9.127, we consider V_{ft} = V^{\ast}= 42.9 m/s = 83.3 knot . According to Table 9.6, the equations in the last column are used.
• Load factor
n_{ft}=n_{\max_{C}}=\frac{\rho(V^{\ast})^2SC_{L_{\max}}}{2W}=\frac{1.225\times (42.9)^2\times 10.82\times 1.73}{2\times 900\times 9.81}=2.38 \quad \quad \quad \quad (9.130)
• Bank angle corresponding to the fastest turn
\phi_{ft}=\cos^{-1}\Big(\frac{1}{n_{ft}}\Big)=\cos^{-1}\Big(\frac{1}{2.38}\Big)=65.2^{\circ}\quad \quad \quad \quad (9.131)
• The maximum turn rate
\omega_{ft}=\frac{g\sqrt{n_{ft}^2-1}}{V_{ft}}=\frac{9.81\times \sqrt{2.38^2-1}}{42.9}=0.495 \space m/s=28.3 \space deg/s \quad \quad \quad \quad (9.133)
• Turn radius
R_{ft}=\frac{V_{ft}^2}{g\sqrt{n_{ft}^2-1}}=\frac{42.9^2}{9.81\times \sqrt{2.38^2-1}}=86.7\space m \quad \quad \quad \quad (9.135)
• Time required to cover a half circle
t=\frac{\pi R}{V}=\frac{3.14\times 86.7}{42.9}=6.3\space s\quad \quad \quad \quad (9.27)
(3.9): AR=\frac{b}{C}=\frac{bb}{Cb}=\frac{b^2}{S}
(2.49): V_s=\sqrt{\frac{2W}{\rho SC_{L_{\max}}}}
(9.127): V_{ft}=V^{\ast}=\Big[\frac{2P_{\max}\eta_P}{\rho S(C_{D_o}+KC_{L_{\max}}^2)}\Big]^{\frac{1}{3}}
(9.27): t_{circle}=\frac{2\pi R}{V}
Table 9.6 Summary of equations for the fastest turn parameters
| No. | Fastest turn parameter | Symbol | If V_{ft}\geq V^{\ast} | If V_{ft}\prec V^{\ast} |
| 1. | Airspeed corresponding to maximum turn rate | V_{ft} | 9.122 | 9.123 |
| 2. | Load factor corresponding to maximum turn rate | n_{ft} | 9.125 | 9.126 |
| 3. | Maximum turn rate | \omega_{ft} | 9.130 | 9.129 |
| 4. | Turn rate corresponding to maximum turn rate | R_{ft} | 9.131 | 9.131 |
| 5. | Bank angle corresponding to maximum turn rate | \phi_{ft} | 9.128 | 9.127 |