Static Longitudinal Stability Questions and Answers

Answer: C

Correct answer: 36 kN m

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

The long arm is what makes a modest tail load so effective: the same 4 kN acting at 1 m would produce only a ninth of the moment. This is also why lengthening the tail arm is such a cheap way of increasing stability, at no cost in tail area.

Answer: D

Correct answer: 37.5 N

Stick force per g is proportional to the manoeuvre margin, and with the margins taken as proportional here

Half as much margin again costs half as much force again. This is the direct link a pilot feels between loading and control feel, and it is why an aeroplane loaded well forward is described as heavy in pitch.

Answer: A

The contribution is the centre of gravity position less the aerodynamic centre position, \( 0.32 - 0.25 = 0.07 \), and a positive value is destabilising. Only the tailplane's negative contribution can then make the whole aeroplane stable.

Answer: B

Correct answer: 22 per cent

With \( S_t \), \( S \) and \( \bar{c} \) unchanged, \( V_H \) is proportional to the arm:

an increase of 22 per cent. The relation is linear in the arm, not quadratic, so 44 per cent is wrong; and 18 per cent is the change taken as a fraction of the new arm rather than the old, which is the wrong base.

Answer: C

Correct answer: Both Statement I and Statement II are correct

Both statements are true, and the second is why the first holds. Static stability asks only whether the initial moment following a disturbance acts to reduce it. It says nothing about what happens over the seconds and minutes that follow, which is the province of dynamic stability. An aeroplane with a positive static margin can still have a divergent phugoid or an unstable Dutch roll, so the initial tendency to return is necessary but not sufficient for the motion to actually die away.

Answer: D

Correct answer: Both Statement I and Statement II are correct

Both are true, and together they explain why the two limits have quite different characters. The aft limit is about stability: move the centre of gravity too far back and the static margin vanishes. The forward limit is about control authority: move it too far forward and the elevator, even at full deflection and at low approach speed where it is least effective, cannot generate the nose-up moment needed to flare or rotate. An aeroplane outside the forward limit may be very stable indeed and still unflyable.

Answer: A

Correct answer: Both Statement I and Statement II are correct

Both are true and the second explains the first. The wing deflects the air downward behind it, so the tail meets the flow at a reduced angle. When the aeroplane pitches up by \( \Delta\alpha \), the downwash angle also grows, and the tail sees only \( \Delta\alpha(1 - \mathrm{d}\varepsilon/\mathrm{d}\alpha) \). With \( \mathrm{d}\varepsilon/\mathrm{d}\alpha \) commonly 0.3 to 0.5, the tail\'s contribution is cut by a third or more, which is why the term appears explicitly in the neutral point expression.

Answer: B

Correct answer: Both Statement I and Statement II are correct

Both are true and the second follows from the first. With the centre of gravity ahead of the wing\'s aerodynamic centre, trim requires a tail download, and vertical equilibrium then gives \( L_w = W + L_t \). The wing carries more than the weight, and the extra induced drag is the trim drag. The penalty grows as the centre of gravity moves forward, which is one reason aircraft are loaded as far aft as the stability limits allow.

Answer: C

The second statement is correct but the first does not follow. A large margin brings heavy stick forces, greater trim drag and a restrictive forward centre of gravity limit, so the aim is an adequate margin rather than a maximal one.

Answer: D

Correct answer: Both Statement I and Statement II are correct

Both are true and the second is the reason for the first. An inclined slender body carries an unstable moment that acts to increase its incidence, so the fuselage adds a positive term to the moment slope. The effect is strongest for a long fuselage forward of the wing, which is why nose length and forebody shape enter the stability estimate and why a large fuselage may drive the tail size.