Static Directional Stability Questions and Answers

Answer: C

Correct answer: 4 knots, with control speed the higher

The margin is the difference, and its sign is what the question tests:

\[ 92 - 88 = 4\ \mathrm{kt} \]

with \( V_{MCA} \) above \( V_S \). That ordering is required by certification, since an aeroplane that lost directional control before it stalled would be uncontrollable while still flying. It also explains why the take-off safety speed \( V_2 \) is set above both, and why on this aeroplane the controlling limit at low speed is the rudder rather than the wing - a small margin, so the engine-out case governs.

Answer: D

Correct answer: An increase of 11 per cent

Compare the totals rather than the individual surfaces:

\[ 2 \times 5 = 10\ \mathrm{m^2} \quad \text{against} \quad 9\ \mathrm{m^2} \]

\[ \frac{10 - 9}{9} \times 100 = 11.1\% \]

So twin fins here add area rather than merely dividing it. The comparison is not the whole story, though: two fins of half the height each have a lower aspect ratio and so a shallower lift-curve slope, which is why designers do not simply halve the area when going to a twin arrangement. The gain in placing them in cleaner air often outweighs that penalty.

Answer: A

The two requirements total 28 degrees against 25 available, so they cannot be combined at full value. Certification examines the critical combination rather than each case in isolation, which is what makes rudder sizing so demanding.

Answer: B

Correct answer: 29 per cent

The improvement is measured against the derivative before the addition:

\[ \frac{0.02}{0.07} \times 100 = 28.6\% \approx 29\% \]

The distractor 22 per cent divides by the new total of 0.09 instead of the original, and 350 per cent inverts the ratio. A ventral fin buys this improvement where it is most needed: mounted below the fuselage, it stays clear of the wing and fuselage wake at high angle of attack, so its contribution holds up in the very condition where the main fin's is falling away.

Answer: C

Correct answer: Both Statement I and Statement II are correct

Both are true and the second explains why the first is so. The asymmetric thrust moment is set by thrust and engine offset, neither of which falls with airspeed, while the rudder's opposing moment falls with dynamic pressure and hence with the square of speed.

So the binding condition occurs at the lowest speed at which flight must continue after the failure, and that is what fixes the rudder size. A crosswind landing is the other case examined, but on a multi-engine aeroplane the engine-out moment is usually the larger, so it governs the design.

Answer: D

Correct answer: Both Statement I and Statement II are correct

Both are true and the second is the common cause. Beyond the fin's stall angle the side force collapses, and two things follow: the restoring yawing moment is lost, so directional stability falls away in a large sideslip; and the separated flow reverses the rudder's hinge moment, so the surface floats hard over and cannot be recentred - rudder lock.

A dorsal fin addresses both at once because it addresses the stall itself, its swept leading edge shedding a vortex that keeps the fin's flow attached to a far larger sideslip angle.

Answer: A

Correct answer: Both Statement I and Statement II are correct

Both are true and the second is the reason for the first. A propeller in sideslip meets the flow obliquely, and the resulting side force acts at the propeller disc - ahead of the centre of gravity on a tractor installation.

A side force forward of the centre of gravity yaws the nose further from the relative wind, which is destabilising by definition. It is the same geometric argument that makes the fuselage destabilising, and the reason both must be overcome by the fin. A pusher propeller, acting behind the centre of gravity, contributes in the opposite and stabilising sense.

Answer: B

Correct answer: Both Statement I and Statement II are correct

Both are true and the second is the mechanism. At high angle of attack the wing and fuselage shed a broad wake of low-energy air, and a conventional centreline fin sits within it.

Reduced local dynamic pressure means less side force for the same sideslip, so the fin's contribution to \( C_{n\beta} \) falls and directional stability with it. The deterioration matters because it arrives near the stall and in spin entry, exactly where directional control is most needed - hence ventral fins below the wake, twin fins outboard of it, and strakes to re-energise the flow.

Answer: C

Correct answer: Statement I is incorrect but Statement II is correct

The second statement is sound and it refutes the first. A sideslip disturbance acts on the fin to produce a yawing moment and, through dihedral, sweep and the fin's height above the rolling axis, a rolling moment as well.

Because a single variable drives both, the equations are coupled and the two cannot be analysed independently. The lateral-directional modes exist only because of that coupling: Dutch roll alternates yaw and roll, and the spiral mode's stability turns on the relative strength of the weathercock and dihedral effects. Treating them separately would predict neither.

Answer: D

Correct answer: Both Statement I and Statement II are correct

Both are true and the second is the reason for the first. \( V_{MCG} \) is established using aerodynamic controls alone, without nosewheel steering, because the runway may be wet, icy or otherwise contaminated and the nosewheel's cornering force cannot be relied upon.

Certifying to the pessimistic case means the published speed remains valid on the day it is most needed. The same conservatism appears elsewhere in the definition - the critical engine is assumed to fail, the remaining engines are at take-off power, and a limited lateral deviation from the centreline is permitted.