Question 9.9: A fighter aircraft has the following characteristics: mTO = ...

We first need to find two parameters:

AR=\frac{b^2}{S}=\frac{9^2}{40}=2.025 \quad \quad \quad \quad (3.9) \\ \space \\ K=\frac{1}{\pi eAR}=\frac{1}{3.14\times 0.7\times 2.025}=0.225 \quad \quad \quad \quad (3.8)

1. Application of the maximum engine thrust

V_{tt}=\sqrt{\frac{4KW^2}{\rho ST_{\max}}}=\sqrt{\frac{4\times 2.025\times (25,000\times 9.81)^2}{1.225\times 40\times 100,000}}=10.4.97 \space m/s=204\space knot \quad \quad \quad \quad (9.111) \\ \space \\  V^{\ast}=\sqrt{\frac{2T_{\max}}{\rho_o S[KC_{L\max}^2+C_{D_{o}}]}}=\sqrt{\frac{2\times 100,000}{1.225\times 40[0.0.225\times 1.3^2+0.014]}}=101.8\space m/s=198\space knot \quad \quad \quad \quad (9.90)

V_{tt} is slightly greater than the corner speed (V_{tt} ≥ V^{\ast}). So, we use the equations in the fourth column of Table 9.5.

n_{tt}=\sqrt{2-\frac{4KC_{Do}}{(T_{\max}/W)^2}}=\sqrt{2-\frac{4\times 0.225\times 0.014}{(100,000/25,000\times 9.81)^2}}=1.387 \quad \quad \quad \quad (9.113) \\ \space \\  \phi_{tt}=\cos^{-1}\Big(\frac{1}{n_{tt}}\Big)=\cos^{-1}\Big(\frac{1}{1.387}\Big)=43.87^{\circ} \quad \quad \quad \quad (9.116) \\ \space \\ R_{tt}=R_{\min}=\frac{V_{tt}^2}{g\sqrt{n_{tt}^2-1}}=\frac{104.97^2}{9.81\sqrt{1.387^2-1}}=1168.6 \space m\quad \quad \quad \quad (9.121)

Equation 9.119 yields the same value for the minimum radius.
2. Turn with a speed that is 90% of the corner speed but having the maximum lift coefficient. The new speed would be

V=0.9 V*=0.9× 101.85=91.66 m/s

When V < V*, but having the maximum lift coefficient, the load factor is

n=\frac{\rho V^2SC_{L_{\max}}}{2W}=\frac{1.225\times 91.66^2\times 40\times 1.3}{2\times 25,000\times 9.81}=1.091 \quad \quad \quad \quad (9.45)

The bank angle would be

\phi=\cos^{-1}\Big(\frac{1}{n}\Big)=\cos^{-1}\Big(\frac{1}{1.091}\Big)=23.6^{\circ} \quad \quad \quad \quad (9.12)

The turn radius would be

R=\frac{V^2}{g\sqrt{n^2-1}}=\frac{91.66^2}{9.81\sqrt{1.091^2-1}}=1958\space m \quad \quad \quad \quad (9.24)

This turn radius is 40.3% longer than the minimum turn radius.
A reduction in the airspeed requires a reduction in the engine thrust.

T=D=\frac{1}{2}\rho V^2S(C_{Do}+KC_{L\max}^2)=\frac{1}{2}\times 1.225\times 91.66^2\times 40\times (0.014+2.025\times 1.3^2) \quad \quad \quad \quad (9.38) \\ \space \\ T=81,000 \space N=81 \space kN

This thrust is 81% of the maximum engine thrust.

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

(9.111):      V_{tt}=V_{R_{\min}}=\sqrt{\frac{4KW^2}{\rho ST_{\max}}}

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

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

(9.119):       R_{tt}=R_{\min}=\frac{4Km^2}{g\rho ST_{\max}}

(9.45):        n_{\max}=\frac{\rho V^2SC_{L_{\max}}}{2W}    (when V < V*)

(9.12):         n=\frac{1}{\cos\phi}

(9.38):         D=\frac{1}{2}\rho V^2SC_D=\frac{1}{2}\rho V^2S(C_{Do}+KC_{L}^2)

Table 9.5 Summary of equations for the tightest turn parameters