Empennage Sizing Questions and Answers

Answer: C

Correct answer: Aft hold only

Read the static margin column. Loading the aft hold without passengers puts the centre of gravity at \( 0.41\bar{c} \), only 0.03 forward of the neutral point - a margin of 3 per cent.

Transports generally require at least about 5 per cent, so this condition falls below the usual minimum while the next worst, at 0.06, just meets it. The loading limits exist precisely to exclude such a case: it arises from an unusual combination - freight aft with the cabin empty - which is exactly the sort of loading a published envelope prohibits.

Answer: D

Thirty kilonewtons falls between the 60 and 70 metres per second rows, nearer the middle, so about 65 metres per second. The quadratic growth of the force with speed is the reason certification of this case is always a low-speed problem.

Answer: A

Rear-fuselage engines leave no clean position low down, so the tailplane must climb the fin and the penalties are simply accepted, with a stick pusher covering the stall risk. It is why almost every rear-engined jet is a T-tail.

Answer: B

Correct answer: Both A and R are true, and R is the correct explanation of A

Both are true and the reason gives the mechanism. Since

\[ V_H = \frac{S_H\,l_H}{S\,\bar{c}} \]

holding \( V_H \) fixed makes the product \( S_H l_H \) constant, so area and arm trade inversely: double the arm and the area halves.

The trade is not free in either direction. A longer arm needs more rear fuselage structure and moves mass aft; a larger area adds wetted area and drag. Which is preferable depends on the fuselage length the aircraft has for other reasons.

Answer: C

The assertion is generally false: a twin loses half its thrust on one failure while a four-engine aircraft loses a quarter, so the asymmetric moment is typically smaller despite the longer arm. The reason as stated is true but supports the opposite conclusion.

Answer: D

Correct answer: Both A and R are true, and R is the correct explanation of A

Both are true and the second explains the first. The static margin is the distance from the centre of gravity to the neutral point, so moving the centre of gravity aft shortens that distance and the margin shrinks.

Carried far enough the centre of gravity passes the neutral point and the aircraft becomes unstable, which is what fixes the aft loading limit. The trade is worth noting: an aft centre of gravity reduces the tail download and hence trim drag, which is why modern transports move fuel aft in cruise within the permitted envelope.

Answer: A

Correct answer: Both A and R are true, and R is the correct explanation of A

Both are true and the reason is the hazard the pusher guards against. At extreme angle of attack the wing wake rises and, with rear-fuselage nacelles adding their own disturbance, can blanket a fin-mounted tailplane.

The elevator then lacks the authority to lower the nose, and the stall becomes self-sustaining. Because recovery may be impossible, certification for such configurations turns on prevention: a stick shaker to warn and a stick pusher to force the nose down before the critical angle is reached.

Answer: B

Correct answer: Both A and R are true, and R is the correct explanation of A

Both are true and the second is the reason for the first. A lifting surface ahead of the centre of gravity produces, in a disturbance, a moment that increases the angle of attack rather than reducing it - a destabilising contribution, which moves the neutral point forward.

Since stability still requires the centre of gravity forward of the neutral point, the whole permissible range shifts forward with it. Canard designs also usually arrange for the foreplane to stall first, so the nose drops rather than the wing dropping into a deep stall.

Answer: C

The quadratic dependence in the reason means the rudder is weakest precisely when the engine-out moment must still be balanced, which fixes the minimum control speed. Both statements hold and the second accounts for the first.

Answer: D

Correct answer: A is false but R is true

The reason is correct: a dorsal fin delays fin stall at large sideslip angles, by shedding a vortex from its swept low aspect ratio leading edge that keeps the flow attached.

The assertion misplaces the benefit. In cruise the sideslip is small, the fin is unstalled and working normally, and the dorsal fillet adds very little side force - its contribution is negligible until the angles become large. It is a fix for the extreme case, not a way of buying cruise directional stability, which comes from fin area and arm.