The curves CD, a, and CM,CG for a light aircraft are shown in Fig. 13.9(a). The aircraft weight is 8,000 N, its wing area 14.5 m^2, and its mean chord 1.35 m. Determine the lift, drag, tail load, and forward inertia force for a symmetric maneuver corresponding to n = 4.5 and a speed of 60 m/s.

Question

The curves [latex]C_{ D }, \alpha[/latex], and [latex]C_{ M , CG }[/latex] for a light aircraft are shown in Fig. 13.9(a). The aircraft weight is 8,000 N, its wing area [latex]14.5 m ^{2}[/latex], and its mean chord 1.35 m. Determine the lift, drag, tail load, and forward inertia force for a symmetric maneuver corresponding to n = 4.5 and a speed of 60 m/s. Assume that engine-off conditions apply and that the air density is 1.223 [latex]kg / m ^{3}[/latex]. Figure 13.9(b) shows the relevant aircraft dimensions.

Accepted Answer

As a first approximation, we neglect the tail load P. Therefore, from Eq. (13.12), since T = 0, we have

[latex]L+P+T \sin \gamma-n W=0[/latex] (13.12)

[latex]L \approx n W[/latex] (i)

Hence,

[latex]C_{ L }=\frac{L}{\frac{1}{2} ho V^{2} S} \approx \frac{4.5 imes 8000}{\frac{1}{2} imes 1.223 imes 60^{2} imes 14.5}=1.113[/latex]

From Fig. 13.9(a), [latex]\alpha=13.75^{\circ}[/latex] and [latex]C_{ M , CG }=0.075[/latex]. The tail arm l, from Fig. 13.9(b), is

[latex]l=4.18 \cos (\alpha-2)+0.31 \sin (\alpha-2)[/latex] (ii)

Substituting this value of [latex]\alpha[/latex] gives [latex]l=4.123 m .[/latex] In Eq. (13.14), the terms [latex]L a-D b-M_{0}[/latex] are equivalent to the aircraft pitching moment [latex]M_{ CG }[/latex] about its CG. Equation (13.14) may therefore be written

[latex]L a-D b-T c-M_{0}-P l=0[/latex] (13.14)

[latex]M_{ CG }-P l=0[/latex]

or

[latex]P l=\frac{1}{2} ho V^{2} S c C_{ M , CG }[/latex] (iii)

where c = wing mean chord. Substituting P from Eq. (iii) into Eq. (13.12), we have

[latex]L+\frac{\frac{1}{2} ho V^{2} S c C_{ M , CG }}{l}=n W[/latex]

or dividing through by [latex]\frac{1}{2} ho V^{2} S[/latex],

[latex]C_{ L }+\frac{c}{l} C_{ M , CG }=\frac{n W}{\frac{1}{2} ho V^{2} S}[/latex] (iv)

We now obtain a more accurate value for [latex]C_{ L }[/latex] from Eq. (iv),

[latex]C_{ L }=1.113-\frac{1.35}{4.123} imes 0.075=1.088[/latex]

giving [latex]\alpha=13.3^{\circ}[/latex] and [latex]C_{ M , CG }=0.073[/latex].

Substituting this value of [latex]\alpha[/latex] into Eq. (ii) gives a second approximation for l, namely, l = 4.161 m. Equation (iv) now gives a third approximation for [latex]C_{ L }[/latex], that is, [latex]C_{ L }=1.099[/latex]. Since all three calculated values of [latex]C_{ L }[/latex] are extremely close, further approximations will not give values of [latex]C_{ L }[/latex] very much different to them. Therefore, we shall take [latex]C_{ L }=1.099[/latex]. From Fig. 13.9(a), [latex]C_{ D }=0.0875[/latex].

The values of lift, tail load, drag, and forward inertia force then follow:

[latex] ext { Lift } L=\frac{1}{2} ho V^{2} S C_{ L }=\frac{1}{2} imes 1.223 imes 60^{2} imes 14.5 imes 1.099=35,000 N[/latex]

[latex] ext { Tail load } P=n W-L=4.5 imes 8,000-35,000=1,000 N[/latex]

[latex] ext { Drag } D=\frac{1}{2} ho V^{2} S C_{ D }=\frac{1}{2} imes 1.223 imes 60^{2} imes 14.5 imes 0.0875=2,790 N[/latex]

Forward inertia force [latex]f W=D[/latex] (from Eq. (13.13)) = 2,790N

[latex]T \cos \gamma+f W-D=0[/latex] (13.13)

Question 13.4: The curves CD, a, and CM,CG for a light aircraft are shown i...

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