Dynamic Longitudinal Modes Questions and Answers

Answer: C

With the stiffness term gone, the quadratic loses its constant term and factors into two real roots, one at the origin and one at the heave and pitch damping value. Physically the aeroplane has no preferred incidence to return to, so any disturbance in incidence simply subsides at the rate the lift and pitch damping allow, with no oscillation. An undamped oscillation would need the damping terms to vanish instead, which is the opposite case. A divergent oscillation needs negative damping. And the aeroplane certainly still moves; what it has lost is the tendency to come back.

Answer: D

Aft movement of the centre of gravity lowers the frequency, because it reduces pitch stiffness by shortening the arm to the aerodynamic centre; carried far enough it removes the stiffness altogether. The other three all raise frequency, since \( \omega_n \approx \sqrt{-C_{m_\alpha}\bar q S\bar c/I_{yy}} \) rises with dynamic pressure, with any change that makes \( C_{m_\alpha} \) more negative such as a larger tail, and with any reduction in pitching inertia. Reading that formula as a whole is the quickest route to the answer.

Answer: A

Spreading the same mass towards the ends of the aeroplane raises the pitching moment of inertia without touching the centre of gravity, and since \( \omega_n \propto 1/\sqrt{I_{yy}} \) the mode gets slower. The damping ratio suffers too, because the pitch damping term is also divided by inertia. Being unchanged would require dynamics to depend on the centre of gravity alone, which ignores the inertia in every moment equation. Frequency cannot rise, since inertia resists rotation rather than assisting it. Nothing here removes the stiffness, so the mode stays oscillatory.

Answer: B

A bigger tail adds both stiffness and damping, so A pairs with Q. A cleaner aeroplane has a higher lift-to-drag ratio and therefore, by \( \zeta_{ph} \approx C_D/(\sqrt{2}C_L) \), a worse damped phugoid, so B pairs with P. Rate feedback adds damping without stiffness, so C pairs with R. The correct set is A-Q, B-P, C-R. The alternatives swap the effect of a larger tail with that of the feedback, or attribute the phugoid penalty to the wrong change, which is worth guarding against because the aerodynamic clean-up looks like an unambiguous improvement until the phugoid is examined.

Answer: C

Short-period frequency rises with dynamic pressure while the phugoid's falls as \( \sqrt{2}g/u_0 \), so the two move apart. At 100 m/s the ratio might be twenty; at 250 m/s it can exceed a hundred. That widening separation is convenient, because it makes the two-degree-of-freedom approximations more accurate at high speed and least accurate at low speed, where a slow aeroplane's modes are closest together. The modes converging is the behaviour associated with an extremely aft centre of gravity, not with speed. The remaining options have one of the two trends inverted.

Answer: D

The first two are the standard results and the third is not. Period follows \( T = \sqrt{2}\pi u_0/g \), damping follows \( \zeta \approx C_D/(\sqrt{2}C_L) \), and neither expression contains the centre of gravity position at all. That insensitivity is one of the practical differences between the modes: a loading change that transforms the short period leaves the phugoid much as it was. Some second-order dependence does exist through the full quartic, but nothing that would justify calling it strong.

Answer: A

A damping ratio of 0.03 is positive but short of the 0.04 required, so the aeroplane achieves Level 2: acceptable, with an increase in pilot workload, but not satisfactory without improvement. Level 1 would need 0.04 or better, and the shortfall here is small enough that a modest drag increase or a simple autopilot altitude hold would recover it. Level 3 is where an unstable phugoid is tolerated provided the time to double amplitude is at least 55 seconds. Nothing in the numbers puts the aeroplane outside the levels altogether, which would need a faster divergence than that.

Answer: B

0.22 sits below the Level 1 lower bound of 0.35 but above the Level 3 minimum of 0.15, so the aeroplane is Level 2 on short-period damping: flyable, with compensation the pilot should not have to make in a manoeuvring task. Level 1 needs 0.35, and closing that gap is a textbook case for a pitch damper, since rate feedback raises damping without slowing the response. Level 3 is the floor rather than the verdict here. Unflyable would require the damping to fall below the Level 3 minimum or the mode to be divergent.

Answer: C

The parameter is \( \omega_n^{2}/(n/\alpha) = 4.0/15 = 0.267 \), which falls marginally below the Level 1 lower bound of 0.28. The verdict matters because a low value feels sluggish: too little initial pitch acceleration arrives for the load factor the aeroplane will eventually deliver, so the pilot cannot anticipate the response. Calling 0.267 comfortably inside the band misreads the boundary. 3.75 inverts the ratio. 0.13 halves the correct figure. A small increase in short-period frequency would fix it, since the parameter goes with the square of frequency.

Answer: D

The phugoid is the long-period mode, its period running from about thirty seconds on a light aeroplane to two minutes on a fast jet. The name was coined by Lanchester, and long period is simply the descriptive alternative. The short period is by definition the other one, lasting a second or two. The heave subsidence is not an oscillation at all but an aperiodic decay that appears when pitch stiffness vanishes. The tuck divergence is also aperiodic, and it is unstable rather than merely slow.