Empennage Sizing Questions and Answers

Answer: C

Correct answer: 6.0 m\(^2\)

The rudder covers 30 per cent of the chord over the full span, so it takes 30 per cent of the fin area:

\[ S_r = 0.30 \times 20 = 6.0\ \mathrm{m^2} \]

Chord ratios in that range are typical, and the reason is diminishing returns: control effectiveness rises with chord ratio but less than proportionately, while hinge moments and actuator loads rise steeply. Beyond roughly 40 per cent, an all-moving surface becomes the better arrangement than a larger flap on a fixed fin.

Answer: D

Correct answer: 1.33

With the same wing and the same tail arm, the tail volume coefficient is proportional to the tail area, so the areas are in the ratio of the coefficients:

\[ \frac{S_B}{S_A} = \frac{1.20}{0.90} = 1.33 \]

The distractor 0.75 inverts the ratio. A difference of this size between two designs usually reflects a difference in requirement rather than in quality - a wider centre of gravity range, or a more demanding trim case at the forward limit, buys tail area.

Answer: A

Correct answer: 1.48 m aft of the reference

The aft centre of gravity limit is the neutral point moved forward by the required static margin:

\[ x_{cg} = 0.42\bar{c} - 0.05\bar{c} = 0.37\bar{c} \]

\[ 0.37 \times 4.0 = 1.48\ \mathrm{m} \]

The distractor 1.68 m is the neutral point itself, which would leave no margin at all, and 1.88 m applies the margin in the wrong direction - aft of the neutral point, where the aircraft would be unstable. The margin must be subtracted, since stability requires the centre of gravity forward.

Answer: B

Correct answer: 50.4 kN downward

The load follows from dynamic pressure, area and lift coefficient:

\[ L = q\,S\,C_L = 12 \times 28 \times (-0.15) = -50.4\ \mathrm{kN} \]

The negative sign means the force acts downward, which is the usual sense for a conventional tail in trim. Keeping the units straight settles the magnitude: kilopascals times square metres gives kilonewtons directly, so no conversion is needed and the 504 kN distractor is a factor-of-ten slip.

Answer: C

Both are true and the second is exactly why the first holds. It is the single relation a designer leans on hardest when the fuselage length changes, since a stretch hands over tail volume without any change to the surfaces.

Answer: D

Correct answer: Both Statement I and Statement II are correct

Both are true and the second explains the first. Yawing and rolling moments are non-dimensionalised on the span, and pitching moments on the mean aerodynamic chord, so each tail volume coefficient uses the reference length appropriate to the moments its surface produces.

That is why \( V_H \) divides by \( S\bar{c} \) and \( V_V \) by \( Sb \), and why the two coefficients are numerically very different - typically around 1.0 and 0.08 - without the vertical tail being in any sense undersized.

Answer: A

Correct answer: Both Statement I and Statement II are correct

Both are true and the second is the reason for the first. On a T-tail the tailplane's lift, and the loads from elevator deflection and gusts, must be carried down the whole height of the fin as bending and torsion.

The fin therefore needs deeper spars and heavier skins than one carrying only its own side load, and it must be stiff enough to keep the tailplane's flutter speed acceptable. The penalty is usually put at 20 to 30 per cent of fin weight, which is why the configuration is adopted when the engine position demands it rather than for its own sake.

Answer: B

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

The second statement gives the definition correctly. But it refutes the first rather than supporting it: moving the centre of gravity aft brings it closer to the neutral point, so the static margin shrinks and stability decreases.

Taken far enough, the centre of gravity passes the neutral point and the aircraft becomes unstable, which is why an aft loading limit exists. The trade is real, since an aft centre of gravity reduces the tail download and hence trim drag - a gain modern transports pursue deliberately with fuel transfer to a tail tank.

Answer: C

The second is correct and it is exactly what makes the first false: a destabilising contribution is what a surface ahead of the centre of gravity provides. Canard stability comes from placing the centre of gravity well forward instead.

Answer: D

Correct answer: Both Statement I and Statement II are correct

Both are true and the second is the mechanism. Rudder side force is

\[ F = q\,S_V\,C_{L\delta}\,\delta \qquad q = \tfrac{1}{2}\rho V^{2} \]

so it falls with the square of airspeed. Halving the speed leaves a quarter of the force for the same deflection.

That is precisely why the minimum control speed exists: asymmetric thrust does not fall as the aircraft slows, while the rudder's opposing moment does, and below some speed the rudder simply runs out of authority.