Dynamic Lateral Directional Modes Questions and Answers

Answer: A

Dihedral effect drives spiral stability and fin volume drives Dutch roll frequency, so P is the spirally stable one and Q has the faster Dutch roll. The table is the design trade in two rows: neither aeroplane is better overall, and each pays for its strength with the other mode. The second and third options each award both virtues to one aeroplane, which the criterion \( L_\beta N_r - N_\beta L_r \) forbids, since the two derivatives enter it with opposite sign.

Answer: B

The period is \( 16/4 = 4.0 \ \mathrm{s} \), so

\[ \begin{aligned} \omega_n \approx \frac{2\pi}{T} &= \frac{6.283}{4.0} \\ &= 1.57 \ \mathrm{rad/s} \end{aligned} \]

giving about 1.57 rad/s. At light damping the damped and undamped frequencies differ by well under one per cent, so the approximation is safe here but would not be for a well-damped mode. 0.25 rad/s is the cycle rate in hertz rather than radians per second, a factor of \( 2\pi \) adrift. 4.0 rad/s is the period read back as a frequency, and 0.79 halves the correct answer.

Answer: C

The decay rate is the product itself, so

\[ \begin{aligned} t_{1/2} &= \frac{\ln 2}{\zeta\omega_n} \\ &= \frac{0.693}{0.10} \\ &= 6.9 \ \mathrm{s} \end{aligned} \]

which is about seven seconds, or roughly two cycles at a four-second period. 10 s is the reciprocal of the product without the logarithm, which is the time constant rather than the half-amplitude time. 0.69 s slips a decade. 14 s doubles the answer. This is the quantity the specifications constrain directly, precisely because it fixes how long a passenger feels the motion.

Answer: D

The two measures are different and both are used:

\[ \begin{aligned} \tau &= \frac{1}{0.020} = 50 \ \mathrm{s} \\ t_{2} &= \frac{\ln 2}{0.020} = 34.7 \ \mathrm{s} \end{aligned} \]

so 50 s and 35 s. The time constant is the time to grow by a factor of \( e \), the doubling time to grow by a factor of two, and since \( \ln 2 < 1 \) the doubling time is always the shorter. Reporting both as 50 s ignores the logarithm, and the third option swaps them. Requirements on the spiral are written in doubling time, so the distinction decides compliance.

Answer: A

For a first-order response the time to reach \( 1 - 1/e \), that is 63 per cent, of the final value is exactly one time constant, 0.50 seconds. Three time constants take it to 95 per cent, which is why a roll feels established almost at once on an aeroplane with a short constant. 0.32 s multiplies the constant by 0.63, confusing the fraction of the response with the fraction of the time. 1.50 s is three time constants, the 95 per cent figure. 0.15 s does not follow at all.

Answer: B

The damping sits in the denominator, so doubling it halves the steady roll rate while also halving the time constant, making the response quicker to settle at a lower rate. That is the trade behind roll requirements, which constrain both the constant and the achieved rate so that a designer cannot satisfy one by sacrificing the other. Doubling has the relationship inverted. Nothing here leaves the rate unchanged. The square-root factor belongs to frequency relationships, not to this first-order one.

Answer: C

The damping ratio of 0.30 is nearly four times the minimum, but the frequency of 0.35 rad/s falls short of 0.4, so the aeroplane fails on frequency alone. A low frequency with good damping describes a sluggish, over-damped directional response: the aeroplane is reluctant to point where it is put, which is why a minimum frequency is specified at all rather than damping alone. A larger fin, or sideslip feedback to the rudder, raises the frequency without spoiling the damping.

Answer: D

Rolling generates sideslip through the yawing moment due to roll rate and through aileron drag, and an interconnect applies proportional rudder to cancel that sideslip so the roll stays coordinated. The benefit is cleaner turn entries and less excitation of the Dutch roll during rolls. It does not add roll authority, so the steady rate is unchanged. The spiral is addressed by a bank-angle loop, and the roll time constant by roll damping, neither of which an interconnect alters.

Answer: A

On a configuration with most of its mass along the fuselage, the rolling inertia is far smaller than the pitching and yawing inertias, and a rapid roll then feeds the pitch and yaw equations through inertial terms strong enough to diverge. Several early jets departed controllably-flown rolls this way, and the cure was more fin and tail area plus roll rate limiting. Fin stall at large sideslip is rudder lock, a different phenomenon. Aileron reversal is aeroelastic. The spiral is a slow aerodynamic root and has nothing to do with it.

Answer: B

Minimum control speed is set by whether the rudder can generate enough yawing moment to balance the asymmetric thrust at that speed, with the bank angle limited by the certification rules. It is a control-power question rather than a modal one. Dutch roll damping, spiral divergence and the roll constant all describe the aeroplane's free motion, and while they matter for handling after the failure, none of them sets the speed below which directional control cannot be maintained.