Dynamic Longitudinal Modes Questions and Answers

Answer: C

Multiplying the two squared frequencies:

\[ E = (0.07)^{2} \times (4.0)^{2} = 0.0049 \times 16 = 0.078 \]

so 0.078. The identity comes from factoring the quartic into two quadratics, whose constant terms multiply to give \( E \), and it is a quick sanity check on a computed quartic: a very small \( E \) is the fingerprint of a slow phugoid rather than of an error. 0.28 is the product of the frequencies themselves rather than of their squares. 0.0049 stops after the phugoid term. 16.0 keeps only the short period.

Answer: D

The phugoid is a swap between height and speed, that is between potential and kinetic energy, with total energy nearly conserved and drag removing a little on each cycle. That is why the mode is slow: the only restoring mechanism is the lift excess or deficit that follows a speed change at fixed incidence. Structural strain energy belongs to aeroelastic modes, which are orders of magnitude faster. Thrust and drag work balance in steady flight, but the phugoid persists with thrust fixed, so that is not the exchange in question. Rotational energy in pitch is negligible here, since the aeroplane barely rotates.

Answer: A

A ninety-second period can only be the phugoid; a damping ratio of 0.6 and a large incidence excursion both belong to the short period. So the pairing is A-Q, B-P, C-P. Reading the period as a short-period figure would require the mode to last a minute and a half, which contradicts its name. Attributing a damping ratio of 0.6 to the phugoid overstates its damping by more than a factor of ten. And a large incidence perturbation is precisely what the phugoid does not have, since it proceeds at nearly constant angle of attack.

Answer: B

In the short-period eigenvector the speed perturbation is very small while incidence and pitch rate are large. That is the mathematical statement of the physical argument that the aeroplane cannot change speed appreciably in a second or two, and it is what licenses setting the speed perturbation to zero in the approximation. Similar magnitudes would leave no basis for that simplification. Negligible pitch rate is impossible in a mode that consists of rotation. And there is no time to build a significant height change, which is the phugoid's signature rather than this mode's.

Answer: C

Both statements are true and the reason is the mechanism. With the column released the elevator becomes a body free to rotate about its hinge, driven by the hinge moment, so the elevator deflection becomes a state of the system and the characteristic equation gains order. The consequence is practical rather than academic: the stick-free short period can be less well damped than the stick-fixed one, and the elevator's own float can couple with the aeroplane's pitching. This is why handling requirements and flight tests are specified separately for the stick-fixed and stick-free cases.

Answer: D

A short, sharp doublet on the column, returned to trim, puts energy into the fast mode and leaves the slow one almost untouched, which is exactly what is wanted for a clean record. A slow throttle change is the standard way to excite the phugoid, because it perturbs speed rather than incidence. A steady pull establishes a new trimmed load factor and measures manoeuvre stability, not the transient. A rudder pulse excites the lateral modes, chiefly Dutch roll, and would only confuse a longitudinal record.

Answer: A

Several visible overshoots at a two-second period mean a genuine short period whose damping ratio is positive but too low: at a well-damped 0.5 or 0.6 the motion is over within a fraction of a cycle, so multiple overshoots point to something nearer 0.15 or 0.2. That is Level 3 territory in the requirements and usually calls for a pitch damper. Correct damping would not overshoot repeatedly. The phugoid is excluded by the period alone, which would be tens of seconds. A divergence would not oscillate at all, and would grow rather than settle.

Answer: B

A pilot closing the loop around an aeroplane whose short-period frequency lies near their own bandwidth, roughly one to three cycles per second once lag is included, can drive a sustained oscillation that neither aeroplane nor pilot would produce alone. Added control-system lag makes it worse, which is why digital flight controls are scrutinised for it. A long phugoid period is a nuisance to trim, not a coupling risk, because it is far too slow to interact with a pilot's reactions. A large static margin raises frequency and generally helps. Heavy hinge moments make an aeroplane tiring rather than oscillatory.

Answer: C

Deliberately moving the centre of gravity at or behind the neutral point makes \( C_{m_\alpha} \) zero or positive, so the bare airframe diverges in pitch and the flight control system must fly it continuously. The reward is a smaller tail, less trim drag and better manoeuvrability. The phugoid's period comes from trim speed and is not shortened by the change. Excessive damping is not the issue; the problem is missing stiffness. The quartic still has oscillatory roots, at least once the augmentation is in the loop, which is exactly what the control law is designed to restore.

Answer: D

Statements 1 and 3 hold. Static stability only guarantees an initial tendency to return, so an aeroplane can meet it and still oscillate divergently, which is the case of a positive \( E \) with a negative Routh discriminant. Statement 3 is the same coin's other face: without static stability the last coefficient goes negative and a real divergent root appears, so no convergent oscillation is possible. Statement 2 is therefore false for the bare airframe, and this is where care is needed: an augmented aeroplane appears to contradict it, but there the control system has supplied the missing stiffness and the effective airframe is no longer statically unstable.