Dynamic Longitudinal Modes Questions and Answers

Answer: C

Short-period damping comes from \( 2\zeta\omega_n = -(M_q + M_{\dot\alpha} + Z_\alpha/u_0) \), so the contributors are the pitch damping \( M_q \), the downwash-lag term \( M_{\dot\alpha} \) and the lift term \( Z_\alpha/u_0 \). All three are negative on a conventional aeroplane and all three therefore damp the motion. The speed derivatives of drag and thrust damp the phugoid, not this mode. Sideslip and yaw derivatives belong to the lateral problem entirely. Elevator effectiveness sets how hard the mode can be excited and where the aeroplane trims, but it does not appear in the damping term.

Answer: D

The ratio is \( 3.5 / 0.07 = 50 \). A separation of this order is typical, and it is the quantitative justification for treating the two modes independently: during one short-period cycle the phugoid variables have moved by about two per cent of their own cycle, so freezing them costs almost nothing. A ratio of 5 would be too close for the approximations to be safe, and is the kind of figure seen on an aeroplane whose short period has been slowed by a very aft centre of gravity. 500 and 25 both come from slipping a factor of ten or two in the division.

Answer: A

Substituting and converting frequency to period:

\[ \begin{aligned} \omega_n &= \frac{\sqrt{2} \times 9.81}{200} \\ &= 0.0694 \ \mathrm{rad/s} \\ T &= \frac{2\pi}{\omega_n} = 90.6 \ \mathrm{s} \end{aligned} \]

So about 91 seconds. The useful shortcut is \( T \approx 0.45\,u_0 \) with speed in metres per second, which gives 90 at a glance. 45 s is that result halved, 64 s divides by \( \sqrt{2} \) instead of multiplying, and 128 s multiplies by it. Note what the formula does not contain: mass, wing area and centre of gravity position are all absent, so the phugoid period depends on speed almost alone.

Answer: B

\( \omega_n = \sqrt{2}\,g/u_0 = 1.414 \times 9.81 / 250 = 0.0555 \ \mathrm{rad/s} \), so 0.055 rad/s, a period of about 113 seconds. The distractors are the three ways the \( \sqrt{2} \) can go wrong. 0.039 is \( g/u_0 \) with the root two left out altogether, 0.078 is \( 2g/u_0 \) from squaring it by mistake, and 0.111 doubles the correct answer. Remember that the frequency falls as trim speed rises, so a fast aeroplane has a slower phugoid, which is the opposite of what happens to the short period.

Answer: C

With \( C_D/C_L = 1/12 \), the damping ratio is \( 1/(\sqrt{2} \times 12) = 0.059 \). That is about six per cent of critical, which is why a phugoid takes so many cycles to disappear. 0.083 is \( 1/12 \) with the root two forgotten, 0.042 divides by two rather than by \( \sqrt{2} \), and 0.118 is twice the correct figure. The formula carries a design warning: damping is inversely proportional to lift-to-drag ratio, so the cleaner and more efficient the aeroplane, the more poorly damped its phugoid.

Answer: D

Since \( \zeta_{ph} \approx 1/(\sqrt{2}\,L/D) \), a large lift-to-drag ratio puts the damping ratio close to zero, so the sailplane has a poorly damped phugoid that can be seen wandering for minutes. It is one of the genuine penalties of aerodynamic efficiency: the only thing draining energy from the oscillation is drag, and an efficient aeroplane has very little of it. Heavy damping would need a high drag aeroplane. The period is set by speed, not by efficiency, and a sailplane's low speed makes its phugoid period short rather than long. The mode certainly still exists.

Answer: A

The envelope decays as \( e^{-\zeta\omega_n t} \), so

\[ \begin{aligned} t_{1/2} &= \frac{\ln 2}{\zeta \omega_n} \\ &= \frac{0.693}{0.003} \\ &= 231 \ \mathrm{s} \end{aligned} \]

which is nearly four minutes to lose merely half the amplitude. 116 s halves that answer, as though the exponent contained \( 2\zeta\omega_n \). 693 s drops the product \( \zeta\omega_n \) and divides by \( \omega_n \) alone times ten. 23 s is a decade adrift. The figure matters in flight test: at this damping the aeroplane must be left alone for several minutes before the record is clean.

Answer: B

Substituting \( \zeta = 0.04 \) gives \( N_{1/2} = 0.110 \times 0.9992 / 0.04 = 2.75 \), so between two and three full cycles. With a period of a minute or so that is several minutes of oscillation, which is exactly the pilot complaint the requirement on phugoid damping exists to control. 1.4 cycles would need a damping ratio nearer 0.08, twice what was given. 5.5 halves the damping instead. 0.11 cycles is the constant in the formula quoted back without the division, and would describe a very heavily damped mode.

Answer: C

Both are true and the second explains the first. Lanchester's derivation treats the aeroplane as a point mass trading height for speed with lift held proportional to the square of the speed at fixed incidence; no rotation about the pitch axis enters, so inertia in pitch cannot appear, and wing area cancels with the weight through the trim condition. What survives is \( \omega_n = \sqrt{2}\,g/u_0 \), a function of trim speed alone. It also explains the formula's limitation: having thrown away the loss mechanism, it says nothing about damping.

Answer: D

Inverting the shortcut gives \( u_0 \approx 68/0.45 = 151 \ \mathrm{m/s} \), so about 150 m/s, a plausible speed for a jet in a low-level cruise. The relation comes straight from \( T = \sqrt{2}\pi u_0/g \), which is worth carrying as \( T \approx 0.45\,u_0 \) in SI units. 68 m/s is the period read back as a speed. 31 m/s multiplies by 0.45 instead of dividing. 300 m/s would give a period near 135 s, twice what was measured.