Dynamic Longitudinal Modes Questions and Answers

Answer: A

The band is roughly thirty seconds to two minutes, and the reason is \( T = \sqrt{2}\pi u_0/g \): at cruise speeds between about 70 and 250 metres per second the formula returns 32 to 113 seconds. One to five seconds is the short period's range. Five to fifteen seconds falls in the gap between the two modes, where a conventional aeroplane has nothing. Tens of minutes is far too slow for any aerodynamic mode, and would put the oscillation on the timescale of a fuel burn trend rather than a dynamic response.

Answer: B

The mode is over in a second or two: with a natural frequency near \( 3 \ \mathrm{rad/s} \) the damped period is about two seconds, and a damping ratio around 0.5 means the amplitude halves within a fraction of one cycle. Half a minute would put it in the phugoid's range. Minutes are longer still, slower even than most phugoids. A hundredth of a second belongs to structural or control-surface dynamics, three orders of magnitude faster than any rigid-body mode of the whole aeroplane.

Answer: C

The characteristic equation is a quartic, and for a conventional aeroplane it factors into two complex conjugate pairs, hence two oscillatory modes: the phugoid and the short period. One would leave two roots unaccounted for. Three or four oscillatory modes are impossible in a fourth-order system, since each oscillation consumes a conjugate pair and only two pairs are available. Unusual configurations can replace one pair with two real roots, giving fewer oscillations rather than more.

Answer: D

Lanchester named the phugoid in 1908, taking the word from a Greek root he believed meant flight. It is the slow trade of potential for kinetic energy at nearly constant incidence, and his idealised analysis still gives the period to within a few per cent as \( T = \sqrt{2}\pi u_0/g \). The short period is the fast longitudinal mode and was not his. Dutch roll and the spiral mode are both lateral-directional, so neither can be an exchange of height for speed in symmetric flight.

Answer: A

Airspeed and altitude are the phugoid's own variables, and both swing slowly and visibly while the mode runs, which is exactly why it is noticed on the instruments rather than felt in the seat. Angle of attack and pitch rate barely move in this mode; they carry the short period instead. Bank and heading belong to the lateral modes and stay put in a symmetric longitudinal oscillation. Engine parameters do not change at all, since the mode proceeds with the thrust setting untouched.

Answer: B

The short period is normally finished within a second or two, faster than a deliberate control input can be planned and applied, so a well-damped one registers as nothing more than a firm response to a gust. That is also why its damping is specified so carefully: the pilot cannot fix it in the loop, so the airframe must have it. The phugoid is the opposite case, slow enough to be corrected with ease but persistent enough to be tiresome if left alone. Saying both are equally noticeable ignores a frequency separation of fifty to one.

Answer: C

Height is one of the two variables the phugoid swaps, so a loop that holds altitude removes the freedom the mode needs and the oscillation is suppressed. This is why a poorly damped phugoid is much less of an operational problem in an autopilot-equipped aeroplane than the bare-airframe numbers suggest, and why the requirements still specify it: the aeroplane has to be flyable with the autopilot off. Saying the airframe property cannot be altered confuses the open-loop mode with the closed-loop response. Nothing here changes the period systematically, and no feedback turns one mode into the other.

Answer: D

The decrement is the natural logarithm of the ratio of successive peaks, and the damping ratio follows from it:

\[ \begin{aligned} \delta &= \ln(10.0/8.0) = 0.223 \\ \zeta &= \frac{\delta}{\sqrt{4\pi^{2}+\delta^{2}}} = \frac{0.223}{6.29} = 0.036 \end{aligned} \]

So 0.223 and 0.036, a lightly damped phugoid as expected. The second option reports the decrement twice, forgetting that the two quantities differ by a factor of about \( 2\pi \). 0.800 is the peak ratio itself rather than its logarithm. 0.125 is the fractional decrease in amplitude, which approximates the decrement only for very small decrements.

Answer: A

Rearranging, \( \zeta \approx 0.110/3 = 0.037 \), so under four per cent of critical. That is a typical figure for an unaugmented transport and just short of the 0.04 wanted for Level 1. The approximation drops the \( \sqrt{1-\zeta^{2}} \) factor, which at this damping changes the answer by less than a tenth of one per cent. 0.330 multiplies where it should divide. 0.110 is the constant returned unused. 0.003 is a decade adrift and would describe a mode that oscillates for the better part of an hour.

Answer: B

The period is \( 450/5 = 90 \ \mathrm{s} \), and inverting \( T \approx 0.45\,u_0 \) gives \( u_0 \approx 90/0.45 = 200 \ \mathrm{m/s} \), so 90 s and 200 m/s. Timing several cycles and dividing is the standard flight-test technique, since it averages away the difficulty of picking a single peak on a noisy trace. The second option forgets to divide by the number of cycles. The third reads the period as a speed. The fourth halves the period, which would follow from counting half cycles as whole ones.