Static Longitudinal Stability Questions and Answers

Answer: A

Correct answer: The centre of gravity is moving forward, increasing the static margin

Fuel aft of the centre of gravity is being burnt off, so the remaining mass distribution shifts forward. The neutral point does not move, so the static margin \( h_n - h \) grows, and the stick force per g grows with it. The heavier feel is the expected consequence rather than a fault; what the crew must watch is that the forward limit is not passed before the approach, where the elevator power is least.

Answer: B

Without a tailplane there is no long moment arm to generate a strong restoring moment. The available static margin is inherently small, so a shift that a conventional aeroplane would absorb consumes most of what such a design has.

Answer: C

Correct answer: Static instability, remedied by moving the centre of gravity forward or enlarging the tail

A positive \( \mathrm{d}C_m/\mathrm{d}C_L \) means an increase in lift coefficient produces a nose-up moment, which raises the incidence further and diverges. The centre of gravity is behind the neutral point. Two remedies follow directly from \( K_n = h_n - h \): move the centre of gravity forward to reduce \( h \), or move the neutral point aft by enlarging the tail or lengthening its arm to raise \( V_H \). A positive slope is a perfectly measurable result, not an error - it is what an unstable layout gives.

Answer: D

A negative slope means an increase in incidence produces a nose down moment opposing it. The curve must also cross zero at a usable positive angle of attack, or the aeroplane would trim only at a condition it cannot actually fly.

Answer: A

Correct answer: neutral

The neutral point is the centre of gravity position at which \( \mathrm{d}C_m/\mathrm{d}C_L = 0 \). Ahead of it the aeroplane is statically stable, behind it unstable, and the distance between the two is the static margin. It is a property of the aerodynamic layout and does not move as the aeroplane is loaded - the centre of gravity moves towards or away from it. The manoeuvre point is a related but distinct quantity, lying somewhat aft of the neutral point.

Answer: B

Correct answer: Static

The static margin is \( K_n = h_n - h \), expressed as a fraction of the mean aerodynamic chord. It is the single number that says how stable the aeroplane is in pitch: it fixes the moment slope, it governs the stick force per g, and it is what the aft centre of gravity limit is set to preserve. The manoeuvre margin is its counterpart for pitching flight and is measured from the manoeuvre point instead.

Answer: C

Correct answer: volume

The tail volume ratio, \( V_H = S_t l_t/(S\bar{c}) \), is the dimensionless grouping in which the tailplane enters the stability estimate. The name comes from the product of an area and a length in the numerator having the dimensions of a volume, and from the wing area and chord in the denominator having the same. Two layouts with the same \( V_H \) make similar tail contributions even if their proportions differ.

Answer: D

Correct answer: Downwash

The wing leaves the air behind it deflected downward, and the tail flies in that deflected flow. When the aeroplane pitches up by \( \Delta\alpha \), the downwash angle grows too, so the tail sees only \( \Delta\alpha(1 - \mathrm{d}\varepsilon/\mathrm{d}\alpha) \). Since \( \mathrm{d}\varepsilon/\mathrm{d}\alpha \) is typically 0.3 to 0.5, a substantial part of the tail\'s potential contribution is lost, and the factor appears explicitly in the neutral point expression. Placing the tail high, as on a T-tail, is one way of moving it out of the strongest downwash.

Answer: A

Very large margins bring heavy controls, greater trim drag and a restrictive forward limit. The design aim is a margin that is sufficient for the handling qualities required rather than the largest that could be achieved.

Answer: B

Correct answer: It must overcome both the wing and the fuselage to make the aeroplane stable

The wing and fuselage together contribute \( +0.14 \), all of it destabilising. The tail supplies \( -0.22 \), and the sum is \( -0.08 \). So of the tail\'s contribution, nearly two thirds is spent simply cancelling the other two components, and only the remainder is available as actual stability margin. This is why a modest change to the fuselage or a shift in the centre of gravity can have such a large proportional effect on the total: the total is a small difference between larger quantities.