Dynamic Lateral Directional Modes Questions and Answers

Answer: A

The vertical tail does: its side force in sideslip acts on the tail arm and turns the nose into the relative wind, which is what makes the yawing moment with sideslip positive. The wing contributes very little directional stiffness and its main lateral role is the dihedral effect, a rolling moment. The horizontal tail's side area is small and close to the axis of symmetry. Nacelles and the forward fuselage are usually destabilising, since their side force acts ahead of the centre of gravity.

Answer: B

The yaw component gives a lateral displacement proportional to the distance from the centre of gravity, so the motion is worst at the extremities, and the rear cabin is the longest arm on most transports. That is why ride-quality complaints come from the back and why the flight deck can report the same oscillation as mild. At the centre of gravity the yaw contributes almost no lateral translation at all. The flight deck feels it, but less, and no aeroplane feels it equally throughout.

Answer: C

With the yawing moment derivative negative, a sideslip produces a moment that increases the sideslip, so the aeroplane diverges in yaw instead of oscillating: the stiffness that would have provided a Dutch roll frequency is working the wrong way. A well damped oscillation requires positive stiffness, so it cannot follow. The spiral criterion also fails when the fin term goes negative. And pitch behaviour belongs to the longitudinal problem, which is decoupled from this one.

Answer: D

It is the spiral, a slow bank and heading change in which the yaw rate is tiny and steady rather than oscillatory - and what little there is the washout filter deliberately blocks. Suppressing the spiral needs a bank-angle loop, which is what a wing leveller provides. The roll subsidence is a fast stable decay that needs no help, and the short period and phugoid are longitudinal modes that a rudder loop cannot reach at all.

Answer: A

The mode is first order with time constant \( \tau = 1/1.6 = 0.625 \ \mathrm{s} \), and the half-amplitude time is

\[ \begin{aligned} t_{1/2} &= 0.693\,\tau \\ &= 0.693 \times 0.625 = 0.43 \ \mathrm{s} \end{aligned} \]

so a little over four tenths of a second. The 0.63 s option is the time constant itself, which is the 63 per cent time rather than the 50 per cent one. 1.60 s is the derivative read back as a time. 0.87 s doubles the correct answer, which would follow from using \( 2\ln 2 \) in the numerator.

Answer: C

The undamped natural frequency is 2.165 rad/s and the damping ratio is 0.235.

For a 2 by 2 system matrix \(A\), the characteristic equation is \(s^{2} - (\text{tr}A)s + \det A = 0\). Comparing with \(s^{2} + 2\zeta\omega_{n}s + \omega_{n}^{2} = 0\):

\[ \begin{aligned} \det A &= (-0.26)(-0.76) - (-1)(4.49) \\ &= 0.1976 + 4.49 = 4.6876 \\ \omega_{n} &= \sqrt{4.6876} \approx 2.165\ \text{rad/s} \\ 2\zeta\omega_{n} &= -\text{tr}A = 0.26 + 0.76 = 1.02 \\ \zeta &= \frac{1.02}{2 \times 2.165} \approx 0.235 \end{aligned} \]

1.02 is \(2\zeta\omega_{n}\), not the damping ratio; 4.68 is close to \(\omega_{n}^{2} = 4.69\) and 4.49 is a single matrix entry, neither of them the frequency.