Question 9.8: Analyze the fastest turn performance for the fighter aircraf...

From Example 9.6, m = 16,770 kg, S = 38 m², b = 12.3 m, T_{\max} = 2 × 79.2 kN, e = 0.83, C_{L\max} = 1.4, C_{Do} = 0.022 (subsonic), C_{Do} = 0.032 (transonic); C_{Do} = 0.042 (supersonic), AR = 3.98, K = 0.096.

1. Sea level (ρ = 1.225 kg/m³)
We assume that the airspeed for fastest turn is subsonic, so a C_{Do} of 0.022 is used. If this assumption turns out to be false, it will be changed accordingly.

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

V_{ft}=\sqrt{\frac{2W}{\rho S}\sqrt{\frac{K}{C_{D_o}}}}=\sqrt{\frac{2\times 16,770\times 9.81}{1.225\times 38}\sqrt{\frac{0.096}{0.022}}}\Rightarrow V_{ft}=58.1\space m/s=112.9\space knot \quad \quad \quad \quad (9.94) \\ \space \\ V^{\ast}=\sqrt{\frac{2T_{\max}}{\rho_oS[KC_{L{\max}}^2+C_{D_o}]}}=\sqrt{\frac{2\times 127,000}{1.225\times 28.87\times[0.109\times 2^2+0.017]}} \quad \quad \quad \quad (9.90) \\ \space \\ \Rightarrow V^{\ast}=179.7\space m/s=349.3\space knot,

since V_{ft} < V^{\ast} , from Equation 9.96, we consider V_{ft} = V^{\ast} = 179.7.1 m/s = 349.3 knot. According to Table 9.3, the equations in the fifth column are used. Since the airspeed for fastest turn is subsonic, a C_{Do} of 0.022 is kept.
• Load factor

n_{ft}=\frac{\rho (V^{\ast})^2SC_{L_{\max}}}{2W}=\frac{1.225\times (179.7)^2\times 38\times 1.4}{2\times 16770\times 9.81}=6.4 \quad \quad \quad \quad (9.101)

• Turn rate

\omega_{ft}=\frac{g\sqrt{n_{ft}^2-1}}{V_{ft}}=\frac{9.81\times \sqrt{6.4^2-1}}{179.7}=0.345 \space rad/s=19.8 \space deg/s \quad \quad \quad \quad (9.103)

• Turn radius

R_{ft}=\frac{V_{ft}^2}{g\sqrt{n_{ft}^2-1}}=\frac{179.7^2}{9.81\times \sqrt{6.4^2-1}}=521 \space m \quad \quad \quad \quad (9.106)

• Bank angle

\phi_{ft}=\cos^{-1}\Big(\frac{1}{n_{ft}}\Big)=\cos^{-1}\Big(\frac{1}{6.4}\Big)=81^{\circ}\quad \quad \quad \quad (9.102)

• Time required to cover a half circle

t=\frac{\pi R}{V}=\frac{3.14\times 521}{179.7}=9.1 \space s\quad \quad \quad \quad (9.27)

2. 15,000 ft
At an altitude of 50,000 ft, the air density ratio is 0.153 and the air density is 0.188 kg/m³. The maximum engine thrust at this altitude is reduced to 24.2 kN (about an 85% reduction). The same calculations are repeated for this altitude. The results are shown in Table 9.4 for comparison.
It is observed that the fastest turn performance is extremely reduced at 50,000 ft altitude. Please note that the C_{Do} of the aircraft at 15,000 ft is slightly different from that for sea level. We intentionally ignored it for the sake of simplicity.

(9.94):        V_{ft}=V_{\omega_{\max}}=\sqrt{\frac{2W}{\rho S}\sqrt{\frac{K}{C_{Do}}}}=\sqrt{\frac{2W}{\rho S\sqrt{C_{Do}/K}}}

(9.90):        V^{\ast}=\Big[\frac{2P_{\max}\eta_P}{\rho S(C_{D_o}+KC_{L_{\max}}^2)}\Big]^{\frac{1}{3}}

(9.103):      \omega_{ft}=\frac{g\tan(\phi_{ft})}{V_{ft}}=\frac{g\sqrt{n_{ft}^2-1}}{V_{ft}}

(9.27):       t_{circle}=\frac{2\pi R}{V}

(9.96):       V_{ft}=\sqrt{\frac{2T_{\max}}{\rho S[KC_{L\max}^2+C_{D_o}]}}

Table 9.3 Summary of equations for the fastest turn parameters

Table 9.4 A comparison between fastest turn performances of F/A-18 at two altitudes