Analyze the fastest turn performance for the fighter aircraft McDonnell Douglas (now Boeing) F/A-18 Hornet (Figure 8.12a) at sea level and 50,000 ft altitude (i.e., determine the maximum turn rate).

• What are the load factor, bank angle, airspeed, and turn radius for this turn?
• Determine the time required to cover a half circle (180°).

The characteristics of this aircraft are given in Example 9.6.

Question

Analyze the fastest turn performance for the fighter aircraft McDonnell Douglas (now Boeing) F/A-18 Hornet (Figure 8.12a) at sea level and 50,000 ft altitude (i.e., determine the maximum turn rate).

• What are the load factor, bank angle, airspeed, and turn radius for this turn?
• Determine the time required to cover a half circle (180°).

The characteristics of this aircraft are given in Example 9.6.

Accepted Answer

From Example 9.6, m = 16,770 kg, S = 38 m², b = 12.3 m, [latex]T_{\max}[/latex] = 2 × 79.2 kN, e = 0.83, [latex]C_{L\max}[/latex] = 1.4, [latex]C_{Do}[/latex] = 0.022 (subsonic), [latex]C_{Do}[/latex] = 0.032 (transonic); [latex]C_{Do}[/latex] = 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 [latex]C_{Do}[/latex] 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:

[latex]V_{ft}=\sqrt{\frac{2W}{ ho S}\sqrt{\frac{K}{C_{D_o}}}}=\sqrt{\frac{2 imes 16,770 imes 9.81}{1.225 imes 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}}{ ho_oS[KC_{L{\max}}^2+C_{D_o}]}}=\sqrt{\frac{2 imes 127,000}{1.225 imes 28.87 imes[0.109 imes 2^2+0.017]}} \quad \quad \quad \quad (9.90) \ \space \ \Rightarrow V^{\ast}=179.7\space m/s=349.3\space knot,[/latex]

since [latex]V_{ft} < V^{\ast}[/latex] , from Equation 9.96, we consider [latex]V_{ft} = V^{\ast}[/latex] = 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 [latex]C_{Do}[/latex] of 0.022 is kept.
• Load factor

[latex]n_{ft}=\frac{ ho (V^{\ast})^2SC_{L_{\max}}}{2W}=\frac{1.225 imes (179.7)^2 imes 38 imes 1.4}{2 imes 16770 imes 9.81}=6.4 \quad \quad \quad \quad (9.101)[/latex]

• Turn rate

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

• Turn radius

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

• Bank angle

[latex]\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)[/latex]

• Time required to cover a half circle

[latex]t=\frac{\pi R}{V}=\frac{3.14 imes 521}{179.7}=9.1 \space s\quad \quad \quad \quad (9.27)[/latex]

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 [latex]C_{Do}[/latex] 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): [latex]V_{ft}=V_{\omega_{\max}}=\sqrt{\frac{2W}{ ho S}\sqrt{\frac{K}{C_{Do}}}}=\sqrt{\frac{2W}{ ho S\sqrt{C_{Do}/K}}}[/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]

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

(9.27): [latex]t_{circle}=\frac{2\pi R}{V}[/latex]

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

Table 9.3 Summary of equations for the fastest turn parameters

No.	Fastest turn parameter	Symbol	If [latex]V_{ft}\geq V^{\ast}[/latex]	If [latex]V_{ft}\prec V^{\ast}[/latex]
1.	Airspeed corresponding to maximum turn rate	[latex]V_{ft}[/latex]	9.94	9.96
2.	Load factor corresponding to maximum turn rate	[latex]n_{ft}[/latex]	9.99	9.101
3.	Maximum turn rate	[latex]\omega_{ft}[/latex]	9.103 or 9.104	9.103 or 9.105
4.	Turn rate corresponding to maximum turn rate	[latex]R_{ft}[/latex]	9.106	9.106 or 9.107
5.	Bank angle corresponding to maximum turn rate	[latex]\phi_{ft}[/latex]	9.102	9.102

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

Altitude	V (knot)	n	ϕ (deg)	ω (deg/s)	R (m)	t (s)
Sea level	349.3	6.4	81	19.8	521	9.1
50,000 ft	371	1.1	25.2	1.39	7876	129.6

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

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