Dynamic Lateral Directional Modes Questions and Answers

Answer: C

Frequency comes from directional stiffness, the yawing moment derivative with sideslip \( N_\beta \), which the fin supplies. The one-degree-of-freedom estimate is \( \omega_n \approx \sqrt{N_\beta} \), the yaw analogue of the pitch stiffness that sets short-period frequency. Roll damping sets the roll subsidence time constant instead. Lift-to-drag ratio damps the phugoid. The pitching moment derivative is longitudinal, and a candidate who reaches for it here has not separated the two problems.

Answer: D

The damping terms are \( N_r \), the yawing moment due to yaw rate, together with the side force term \( Y_\beta/u_0 \), both negative on a conventional aeroplane. The fin generates most of the first, since a yaw rate changes its local incidence in proportion to its arm. Rolling moment due to sideslip is the dihedral effect, which sets the roll content of the mode and generally makes the damping ratio worse rather than better. Ailerons and elevator are controls, not sources of damping.

Answer: A

The root of four is two, so 2.0 rad/s, a period of about three seconds. The estimate treats the aeroplane as a torsional spring in yaw whose stiffness is the fin's weathercock derivative, and it is useful for showing that a bigger fin raises the frequency. 4.0 rad/s is the derivative returned unrooted. 1.0 rad/s halves the answer. 16.0 squares it. The fuller two-degree-of-freedom treatment adds a roll term and gives a slightly different figure, but never a different order.

Answer: B

Dimensionalising the coefficient:

\[ \begin{aligned} N_\beta &= \frac{0.12 \times 5000 \times 25 \times 10}{150\,000} \\ &= \frac{150\,000}{150\,000} = 1.0 \ \mathrm{s^{-2}} \end{aligned} \]

so 1.0 per second squared and a Dutch roll frequency near 1 rad/s. The 0.12 option is the coefficient itself, which carries different units. 10 slips a decade. 0.5 halves the result. Note the structure: frequency rises with dynamic pressure and fin volume and falls with yawing inertia, exactly as the pitch case does with its own quantities.

Answer: C

Collecting the two negative terms:

\[ \begin{aligned} 2\zeta\omega_n &= 0.30 + 0.15 = 0.45 \\ \zeta &= \frac{0.45}{2 \times 1.0} = 0.225 \end{aligned} \]

so 0.225, which is respectable for an unaugmented aeroplane. 0.45 forgets to halve. 0.15 keeps only the side-force term and drops the yaw damping, which is the larger of the two. 0.90 doubles instead of halving. As with the short period, only the product \( \zeta\omega_n \) governs the decay, and it is that product the specifications constrain separately.

Answer: D

The damped period follows from the damped frequency:

\[ \begin{aligned} T &= \frac{2\pi}{\omega_n\sqrt{1-\zeta^{2}}} \\ &= \frac{6.283}{1.194} = 5.26 \ \mathrm{s} \]

so about 5.3 seconds. At this light damping the correction from \( \sqrt{1-\zeta^{2}} \) is only half a per cent, so the undamped period would serve; that is not true of a well-damped mode. The 8.7 s option would need a frequency near 0.72 rad/s, and 2.6 s halves the period, the error that follows from counting half cycles as whole ones on a trace.

Answer: A

Substituting, \( N_{1/2} = 0.110 \times 0.995/0.10 = 1.09 \), so a little over one cycle. That is the practical meaning of a damping ratio of a tenth: the aeroplane completes roughly one visible wag and rock before the motion is halved, which passengers notice but pilots can live with. 0.11 cycles is the formula's constant returned without the division and would describe a heavily damped mode. 11 cycles is a decade out. 2.2 cycles corresponds to a damping ratio near 0.05.

Answer: B

The complex pair is the Dutch roll and the large real root is the roll subsidence:

\[ \begin{aligned} \omega_n &= \sqrt{0.12^{2}+1.20^{2}} = 1.206 \\ \zeta &= \frac{0.12}{1.206} = 0.10 \\ \tau_R &= \frac{1}{2.50} = 0.40 \ \mathrm{s} \end{aligned} \]

so 0.10 and 0.40 s. The small root at \( -0.02 \) is the spiral, here just stable with a time constant of 50 seconds. The second option reports the real part as a damping ratio and the roll root as a time constant without inverting it, and the third takes the spiral root for the roll mode.

Answer: C

Both statements are true and the second explains the first. Sweep makes the rolling moment with sideslip strongly negative, because the advancing wing meets the flow more nearly square on, and that added roll response degrades Dutch roll damping at the same time as it helps the spiral. Combined with the fall in yaw damping at cruise altitude, the bare airframe cannot meet the requirement, so rate feedback to the rudder is provided as standard rather than as a refinement.

Answer: D

It feeds yaw rate to the rudder, producing a yawing moment that opposes the rate and so acting as an artificial increase in the yaw damping derivative. Roll rate to the ailerons is a roll damper, which addresses a different mode. Sideslip to the ailerons is a form of turn coordination or dihedral augmentation, not a yaw damper. Pitch rate to the elevator is the longitudinal equivalent and does nothing for the Dutch roll, since the two problems are decoupled.