Static Directional Stability Questions and Answers

Answer: A

The ratio is \( (9 \times 11)/(70 \times 28) = 99/1960 \). Values around 0.04 to 0.07 are typical for transport aeroplanes, and the span is used as the reference length because yawing moments scale on the lateral dimension.

Answer: B

Correct answer: Plus 0.09, and therefore stable

Contributions to the yawing moment derivative simply add:

\[ C_{n\beta} = +0.14 - 0.05 = +0.09 \]

A positive \( C_{n\beta} \) means a sideslip to the right produces a nose-right yawing moment, turning the aeroplane back into the relative wind - so the aeroplane is directionally stable. The arithmetic also shows the fin doing exactly its job: the fuselage alone is destabilising at \( -0.05 \), and the fin's \( +0.14 \) more than cancels it. That surplus is the stability margin the designer is buying.

Answer: C

Correct answer: 20 degrees

Rudder power is the moment per degree, so the deflection needed follows by division:

\[ \delta_r = \frac{C_n}{C_{n\delta r}} = \frac{0.036}{0.0018} = 20^\circ \]

The result is worth checking against the physical limit. Rudder travel is typically restricted to about 25 to 30 degrees, so 20 degrees is a large but usable deflection - there is little margin left. That is the kind of margin a designer examines when sizing the rudder for the engine-out case, since a demand beyond the stop means the condition cannot be held at all.

Answer: D

Correct answer: 720 kN m

The yawing moment is the thrust times its perpendicular distance from the centre of gravity:

\[ N = T \times y = 90 \times 8 = 720\ \mathrm{kN\,m} \]

Two points make this the design case for the fin and rudder. The moment is independent of airspeed, since it comes from thrust rather than from dynamic pressure, while the rudder's opposing moment falls with the square of speed. And on a failure the drag of the dead engine adds to the asymmetry, so the real moment is somewhat larger than the thrust term alone.

Answer: A

Correct answer: 60 kN

Force follows from moment divided by arm:

\[ F = \frac{N}{l} = \frac{720}{12} = 60\ \mathrm{kN} \]

The fin area of 12 m\(^2\) is not needed for this step - it would be used next, to find the side force coefficient required and hence the sideslip or rudder angle to generate it. Noticing which data the question actually needs is half the work. The arm is what converts between force and moment, which is why lengthening the fuselage is as effective a way of improving directional stability as enlarging the fin.

Answer: B

Correct answer: 0.25

Sidewash is expressed as the fraction of the free-stream sideslip lost before the flow reaches the fin:

\[ \frac{d\sigma}{d\beta} = 1 - \frac{\beta_{\text{fin}}}{\beta} = 1 - \frac{6}{8} = 0.25 \]

So a quarter of the sideslip is washed out, and the fin's contribution is reduced by the factor \( (1 - 0.25) = 0.75 \). The distractor 0.75 is that retained fraction rather than the sidewash itself, and 1.33 is the ratio inverted. The effect is why a fin never delivers the stability its bare geometry would promise.

Answer: C

Correct answer: 1240 Pa

Dynamic pressure varies with the square of speed at constant density:

\[ q = \tfrac{1}{2}\rho V^{2} \qquad \frac{q_2}{q_1} = \left(\frac{V_2}{V_1}\right)^{2} \]

\[ q_2 = 2205 \times \left(\frac{45}{60}\right)^{2} = 2205 \times 0.5625 = 1240\ \mathrm{Pa} \]

The square law is why every aerodynamic control weakens so sharply as speed falls: dropping to three-quarters of the speed leaves only nine-sixteenths of the moment. That is precisely the difficulty at minimum control speed, where the asymmetric thrust to be balanced has not fallen at all.

Answer: D

Correct answer: 7.2 degrees

The drift angle is the angle whose sine is the crosswind component over the true airspeed:

\[ \sin\theta = \frac{15}{120} = 0.125 \qquad \theta = 7.18^\circ \approx 7.2^\circ \]

That is the crab angle the aeroplane must hold on approach to track the centreline, and therefore the yaw it must be swung through in the de-crab just before touchdown. The distractor 12.5 degrees is the ratio 15/120 read as a percentage rather than a sine, a mistake worth watching for since the two are numerically close at small angles.

Answer: A

Correct answer: 67 per cent

The increase is measured against the original value:

\[ \frac{25 - 15}{15} \times 100 = \frac{10}{15} \times 100 = 66.7\% \approx 67\% \]

The distractor 40 per cent divides by the new figure instead of the old, and 167 per cent gives the ratio of new to old rather than the increase. Physically, the dorsal fin works by generating a strong vortex along its swept leading edge that keeps the flow attached to the fin far beyond its natural stall angle - which is what preserves directional stability in a large sideslip.

Answer: B

Correct answer: 0.45 in the restoring sense

The derivative is a slope, so the coefficient is its product with the sideslip angle:

\[ C_n = C_{n\beta}\,\beta = 0.09 \times 5 = 0.45 \]

The sign carries the meaning. A positive \( C_{n\beta} \) gives a moment that yaws the nose towards the relative wind, reducing the sideslip that produced it - a restoring moment, which is precisely what directional stability means. Had the derivative been negative the same arithmetic would give a destabilising moment of the same size, which is why the sense must be stated with the number.