A fighter aircraft has the following characteristics:

$m_{TO}$ = 25,000 kg, S = 40 m², $T_{\max}$ = 100 kN, $C_{Do}$ = 0.014; b = 9 m, $C_{L\max}$ = 1.3

Analyze the performance of the aircraft for the tightest turn at two flight conditions for sea level:
1. Theoretical solutions (i.e., application of the maximum engine thrust).
2. Turn with a speed that is 90% of the corner speed but having the maximum lift coefficient (i.e., point D in Figure 9.10).

Question

A fighter aircraft has the following characteristics:

[latex]m_{TO}[/latex] = 25,000 kg, S = 40 m², [latex]T_{\max}[/latex] = 100 kN, [latex]C_{Do}[/latex] = 0.014; b = 9 m, [latex]C_{L\max}[/latex] = 1.3

Analyze the performance of the aircraft for the tightest turn at two flight conditions for sea level:
1. Theoretical solutions (i.e., application of the maximum engine thrust).
2. Turn with a speed that is 90% of the corner speed but having the maximum lift coefficient (i.e., point D in Figure 9.10).

Accepted Answer

We first need to find two parameters:

[latex]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 imes 0.7 imes 2.025}=0.225 \quad \quad \quad \quad (3.8)[/latex]

1. Application of the maximum engine thrust

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

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

[latex]n_{tt}=\sqrt{2-\frac{4KC_{Do}}{(T_{\max}/W)^2}}=\sqrt{2-\frac{4 imes 0.225 imes 0.014}{(100,000/25,000 imes 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)[/latex]

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

[latex]n=\frac{ ho V^2SC_{L_{\max}}}{2W}=\frac{1.225 imes 91.66^2 imes 40 imes 1.3}{2 imes 25,000 imes 9.81}=1.091 \quad \quad \quad \quad (9.45)[/latex]

The bank angle would be

[latex]\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)[/latex]

The turn radius would be

[latex]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)[/latex]

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

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

This thrust is 81% of the maximum engine thrust.

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

(9.111): [latex] V_{tt}=V_{R_{\min}}=\sqrt{\frac{4KW^2}{ ho ST_{\max}}}[/latex]

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

(9.121): [latex]\omega_{tt}=\frac{g an(\phi_{tt})}{V_{tt}}=\frac{g\sqrt{n_{tt}^2-1}}{V_{tt}}[/latex]

(9.119): [latex]R_{tt}=R_{\min}=\frac{4Km^2}{g ho ST_{\max}}[/latex]

(9.45): [latex]n_{\max}=\frac{ ho V^2SC_{L_{\max}}}{2W}[/latex] (when V < V*)

(9.12): [latex]n=\frac{1}{\cos\phi}[/latex]

(9.38): [latex]D=\frac{1}{2} ho V^2SC_D=\frac{1}{2} ho V^2S(C_{Do}+KC_{L}^2)[/latex]

Table 9.5 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]	9.109	9.110
2.	Load factor corresponding to minimum turn radius	[latex]n_{tt}[/latex]	9.111	9.112
3.	Turn rate corresponding to minimum turn radius	[latex]\omega_{tt}[/latex]	9.117	9.117
4.	Minimum turn radius	[latex]R_{tt}[/latex]	9.114 or 9.115	9.114 or 9.116
5.	Bank angle corresponding to minimum turn radius	[latex]\phi_{tt}[/latex]	9.113	9.113

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

Related Answered Questions

Plot the flight envelope (combined V-n diagram) for the following GA acrobatic aircraft. Then, determine the maximum load factor. m = 2300 kg, S = 19.33 m², CLmax = 2, −CLmax = −1.2, AR = 7, CLα = a = 6.3 1/rad, Vc = 310 KEAS (at 10,000 ft). ...

Verified Answer:

Verified Answer:

Verified Answer:

Consider an acrobatic aircraft with a mass of 1500 kg diving with a velocity of 100 knot. a. If the aircraft pulls out of the dive with a radius of 150 m, determine the load factor. b. If the maximum allowable load factor is 6, determine the minimum radius for a pull-out of this dive. ...

Verified Answer:

Consider a small remote-controlled (RC) aircraft with the following characteristics: m = 700 g, S = 0.2 m², b = 1 m; e = 0.8; ηP = 0.7, CDo = 0.03, CLmax = 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 ...

Verified Answer:

Verified Answer:

Verified Answer:

Consider the single-engine turboprop utility aircraft Cessna 208 Caravan with the following features: mTO = 3995 kg, S = 25.9 m², P = 647 kW, b = 15.88, Vs = 61 knot, Vmax = 185 knot. Assume: CDo = 0.038, ηP = 0.85, e = 0.9. 1. Determine the corner speed of the aircraft. Then, determine the ...

Verified Answer:

Verified Answer:

Verified Answer: