Dynamic Lateral Directional Modes Questions and Answers

Answer: C

Holding cross-controlled inputs pins bank angle and heading, which constrains every lateral degree of freedom and so suppresses all three modes for as long as the pilot holds them. That is why a steady sideslip is a static condition, analysed with control-power arguments rather than modal ones, and why the modes reappear the moment the controls are released at the flare. Suppressing only one mode would require the others to be independent of bank and heading, which they are not. And sideslip is precisely the state that excites the Dutch roll, so the last option is wrong twice over.

Answer: D

The first two are correct and the third is not. Rate feedback adds damping without adding stiffness, so the damping ratio rises and the frequency stays where the fin put it. But the fin is still needed: it supplies the directional stiffness that sets the frequency, it provides the surface the rudder hinges from, and an aeroplane must remain flyable with the damper failed. Augmentation reduces the aerodynamic surface required; it does not eliminate it, and certification for the failed case is what enforces that.

Answer: A

The two augmentation loops attack one mode each without the aerodynamic trade: the yaw damper raises Dutch roll damping and the wing leveller suppresses the spiral. Every purely aerodynamic pairing runs into the criterion \( L_\beta N_r - N_\beta L_r \), in which dihedral and fin volume pull opposite ways, so a larger fin with more dihedral improves each mode by an amount the other change partly undoes. That is the structural reason augmentation, rather than more surface, is the modern answer.

Answer: B

It is the roll subsidence, whose constant is the reciprocal of the roll damping and therefore short on any aeroplane with reasonable span and lift-curve slope. The Dutch roll is described by a frequency and a damping ratio rather than a single time constant, and its decay time runs to several seconds. The spiral is the slowest root by two orders of magnitude. The phugoid is longitudinal, and slower still than the spiral in period if not always in decay.

Answer: C

Both statements are correct. A lateral gust is a sideslip disturbance, and sideslip is exactly what the Dutch roll runs on, so every gust rings the mode and ride quality in the rear cabin depends on its damping. The spiral needs a sustained bank disturbance to develop and is far too slow to be driven by random turbulence; it is normally started by an unnoticed trim asymmetry or a control input instead. This is why turbulence complaints trace to Dutch roll damping and not to the spiral.

Answer: D

Decay is governed by the product \( \zeta\omega_n \), so a low frequency stretches the absolute time to decay even when the ratio itself looks acceptable. A damping ratio of 0.1 at 0.3 rad/s takes over twenty seconds to halve, which passengers experience as a persistent wallow. That is precisely why specifications constrain the product and the frequency separately rather than the ratio alone. Low frequency does not make a mode aperiodic, tells us nothing about the spiral, and does not affect roll control.

Answer: A

Weathercock stability is the stiffness term in the Dutch roll, so it sets the frequency through \( \omega_n \approx \sqrt{N_\beta} \). Damping comes from the rate derivatives instead, which is why a directionally stiff aeroplane can still have an objectionable Dutch roll. The two are plainly related, so the third option fails. And a negative yawing moment with sideslip would mean directional instability and a divergence in yaw, not a well-behaved oscillation, so the last option inverts the criterion.

Answer: B

There are three: the Dutch roll oscillation, the roll subsidence and the spiral. The quartic has four roots, but two of them form the single complex pair of the Dutch roll, so four roots give three named modes. Two would be the longitudinal count. Saying four confuses roots with modes, a distinction worth holding since a complex pair always describes one oscillation rather than two behaviours. One mode would leave three roots unaccounted for, and no aeroplane behaves that simply.

Answer: C

Only the Dutch roll oscillates; it is the single complex pair in the lateral quartic. The spiral and the roll subsidence are both real roots, so each is an aperiodic growth or decay with no period at all. That is the clearest structural difference from the longitudinal problem, which has two oscillations and no real roots on a conventional aeroplane. Calling all three oscillatory would need two more complex pairs than a quartic can hold.

Answer: D

The band is three to six seconds, set by the fin's directional stiffness against the yawing inertia. Thirty to sixty seconds is the phugoid's range, an order of magnitude slower. A tenth of a second belongs to structural or actuator dynamics. Minutes would be slower than the spiral. Carrying the three lateral timescales - seconds for the Dutch roll, tenths of a second for the roll mode and tens of seconds for the spiral - settles many questions without any calculation at all.