The neutral point lies at 0.45 of the mean chord and the centre of gravity at 0.30. The static margin is
Answer: A
The margin is the difference \( 0.45 - 0.30 = 0.15 \) expressed as a fraction of the mean aerodynamic chord. Margins of ten to twenty per cent are typical for a transport aeroplane, giving adequate stability without excessive stick forces.
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A tailplane of \( 12 \ \mathrm{m^2} \) sits \( 10 \ \mathrm{m} \) behind the centre of gravity, on a wing of \( 60 \ \mathrm{m^2} \) and mean chord \( 2.5 \ \mathrm{m} \). The tail volume ratio is
Answer: B
Correct answer: 0.80
The tail volume ratio is the tail area and arm non-dimensionalised by the wing area and mean chord:
\[ V_H = \frac{S_t\,l_t}{S\,\bar{c}} = \frac{12 \times 10}{60 \times 2.5} = \frac{120}{150} = 0.80 \]A value near 0.8 is typical of a transport aeroplane. The 0.20 answer comes from the area ratio alone and 0.048 from dividing by the arm instead of multiplying by it.
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A centre of gravity moves from 0.25 to 0.35 chord while the neutral point stays at 0.42. The static margin falls from
Answer: C
Correct answer: 0.17 to 0.07 chord
The static margin is the distance from the centre of gravity to the neutral point:
\[ K_n = h_n - h \]Before the shift \( K_n = 0.42 - 0.25 = 0.17 \); after it \( K_n = 0.42 - 0.35 = 0.07 \). The neutral point is a property of the aerodynamic layout and does not move when the loading changes, so the whole of the 0.10 chord shift comes out of the margin. At 0.07 chord the aeroplane is still stable but only just, and the stick forces would be noticeably light.
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A pitching moment coefficient changes by \( -0.04 \) for a lift coefficient change of \( 0.50 \). The static margin is
Answer: D
The static margin is the negative of the slope, \( 0.04/0.50 = 0.08 \). Measuring the slope against lift coefficient rather than angle of attack gives the margin directly, which is why the curve is usually plotted that way.
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An aeroplane weighs \( 60 \ \mathrm{kN} \) and its tail carries a download of \( 3 \ \mathrm{kN} \) in trim. The wing lift required is
Answer: A
Correct answer: 63 kN
Vertical equilibrium takes the tail download as a load the wing must also carry:
\[ L_w = W + L_t = 60 + 3 = 63 \text{ kN} \]The wing therefore works five per cent harder than the weight alone would require, and the induced drag rises with the square of that. Subtracting the download, which gives 57 kN, would be right only for a canard, whose foreplane lifts upward.
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Downwash changes at \( 0.35 \) per unit change of wing angle of attack. For a two degree change in aeroplane incidence, the tail's incidence changes by
Answer: B
The tail sees \( 2 \times (1 - 0.35) = 1.3 \) degrees of the change. Downwash therefore reduces the tail's contribution to stability by roughly a third in this case, which is why the derivative appears explicitly in the analysis.
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A centre of gravity range runs from 0.18 to 0.35 chord on a mean chord of \( 3.2 \ \mathrm{m} \). The permitted travel is
Answer: C
Correct answer: 0.544 m
The range is expressed as a fraction of the mean aerodynamic chord, so the physical travel is
\[ (0.35 - 0.18) \times 3.2 = 0.17 \times 3.2 = 0.544 \text{ m} \]A little over half a metre is a generous range and typical of a transport aeroplane. The 0.17 m answer forgets to multiply by the chord, and 5.44 m misplaces the decimal point by an order of magnitude.
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Fuel of \( 4000 \ \mathrm{kg} \) burns from a tank \( 1.5 \ \mathrm{m} \) aft of the centre of gravity on an aeroplane of \( 40\,000 \ \mathrm{kg} \). The centre of gravity moves forward by approximately
Answer: D
Correct answer: 0.167 m
Removing a mass \( m \) at a distance \( d \) from the centre of gravity shifts it by \( m d \) divided by the mass that remains:
\[ \Delta x = \frac{4000 \times 1.5}{40\,000 - 4000} = \frac{6000}{36\,000} = 0.167 \text{ m} \]The fuel was aft, so removing it draws the centre of gravity forward and the static margin increases. Dividing by the original 40,000 kg rather than the final 36,000 kg gives 0.15 m, which is the usual slip: the moment must be taken about the mass that is actually left.
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A wing contributes \( +0.10 \) to the moment slope, the fuselage \( +0.04 \) and the tail \( -0.22 \). The total slope is
Answer: A
Correct answer: Minus 0.08, so the aeroplane is stable
The contributions add directly:
\[ \frac{\mathrm{d}C_m}{\mathrm{d}C_L} = +0.10 + 0.04 - 0.22 = -0.08 \]A negative slope means that an increase in lift coefficient produces a nose-down moment, which reduces the incidence that caused it, so the aeroplane is statically stable. The magnitude, 0.08, is the static margin in chords. Note how much of the tail\'s 0.22 is consumed simply in cancelling the wing and fuselage.
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An aeroplane has a neutral point at 0.40 chord. To achieve a static margin of 0.12 the centre of gravity must lie at
Answer: B
Correct answer: 0.28 chord
From \( K_n = h_n - h \),
\[ h = h_n - K_n = 0.40 - 0.12 = 0.28 \]The centre of gravity must be 0.28 chord aft of the leading edge of the mean aerodynamic chord, that is 0.12 chord ahead of the neutral point. Adding rather than subtracting gives 0.52, which would put the centre of gravity behind the neutral point and make the aeroplane unstable.
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