Dynamic Longitudinal Modes Questions and Answers

Answer: A

The standard tool is a pair of two-degree-of-freedom models: incidence and pitch rate for the short period, speed and flight path angle for the phugoid. Each drops the states that barely move on its own timescale, and each reproduces the full quartic's roots to within a few per cent for a conventional aeroplane. A six-degree-of-freedom model is the full aeroplane including lateral motion, which is more than is needed and defeats the purpose. Quasi-steady describes an aerodynamic assumption rather than a modal one. A fully coupled model is what the approximations are an alternative to.

Answer: B

Speed enters both modes, and in opposite directions, so neither frequency is invariant. The phugoid frequency falls as \( \sqrt{2}g/u_0 \) while the short period rises with dynamic pressure, so as an aeroplane accelerates its slow mode gets slower and its fast mode gets faster and the two separate further. The phugoid period therefore changes, in direct proportion to speed. So does short-period frequency. The phugoid damping ratio depends on lift-to-drag ratio, which itself varies with speed through the drag polar, so it is not invariant either.

Answer: C

A nose-up pitch rate carries the tail downwards, so the tail meets the air at an increased local incidence of about \( q\,l_t/V \). The extra tail lift makes a nose-down moment that opposes the rotation, which is the damping. Because the increment is proportional to the arm and acts through the arm, the tail's contribution grows roughly with the square of the tail length. The wing's lift curve slope is a property of its geometry and does not depend on pitch rate. The centre of gravity is a mass property. Dynamic pressure at the tail is essentially unchanged by a small pitch rate.

Answer: D

The tail's vertical velocity from the rotation is \( q\,l_t \), so

\[ \begin{aligned} \Delta\alpha_t &= \frac{q\,l_t}{V} = \frac{0.10 \times 12}{120} \\ &= 0.010 \ \mathrm{rad} = 0.57^{\circ} \end{aligned} \]

a little over half a degree, and that small increment is the whole of the tail's damping contribution at this rate. 5.7 degrees is a factor of ten out, the commonest error being to forget that the result is in radians before converting. 0.06 degrees divides by the arm rather than multiplying. 1.2 degrees is \( q\,l_t \) itself, a velocity in metres per second, reported as an angle.

Answer: A

The wing's downwash takes about \( l_t/V \) seconds to reach the tail, so when incidence is changing the tail is immersed in the downwash belonging to an earlier incidence. That lag makes the tail's effective incidence depend on the rate of change as well as the value, which is exactly the derivative in question. It normally supplies a tenth to a third of total pitch damping. Boundary layer response is far faster and is not what the term models. Elevator inertia matters only in stick-free dynamics. Engine lag affects the phugoid through thrust, not this derivative.

Answer: B

The lag is simply the transport delay from wing to tail:

\[ \tau = \frac{l_t}{V} = \frac{15}{150} = 0.10 \ \mathrm{s} \]

so a tenth of a second. Compared with a short period of a second or two that is small but not negligible, which is why the derivative is retained. 1.0 s is a decade out and would be comparable with the whole mode. 0.01 s slips the other way. 2.25 s is the product of the two numbers scaled, which has the wrong dimensions entirely. Note that the lag shrinks as speed rises, so the term matters most at low speed.

Answer: C

An incidence increase raises lift, which curves the flight path upwards; the climbing flight path then reduces the very incidence that produced it. That heave, or plunge, damping is what \( Z_\alpha/u_0 \) measures, and it is the reason a high lift curve slope helps short-period damping. Drag with incidence is a much smaller effect and belongs mainly to the phugoid. Elevator lift is a control term rather than a stability derivative in this expression. Fuselage moments in sideslip are a lateral-directional matter and cannot appear in the longitudinal equations.

Answer: D

The first three are correctly paired and the fourth is not, so three match. Pitch damping and the downwash-lag term both damp the short period, and the axial force derivative with speed, dominated by drag, damps the phugoid. The normal force derivative with incidence appears as \( Z_\alpha/u_0 \) in the short-period damping, not the phugoid's, and it is one of the three terms in \( 2\zeta\omega_n \). Keeping the two force derivatives apart is the point of the question: one acts along the flight path and slowly, the other across it and fast.

Answer: A

Both statements are correct and the second is the reason for the first. With the pitch control surfaces only a short distance behind the centre of gravity, the effective arm is small, and since damping scales roughly with the square of that arm the available pitch damping is modest while the inertia of a long fuselage is not. The result is a lightly damped short period, which is one reason such aeroplanes are fitted with pitch dampers as a matter of course rather than as a refinement.

Answer: B

Every dimensional aerodynamic derivative carries a factor of dynamic pressure, and at constant Mach number dynamic pressure falls with density as the aeroplane climbs. Frequency goes as the square root of that, while the heave and pitch damping terms scale more nearly with it directly, so the damping ratio degrades with altitude. This is a familiar result: an aeroplane comfortable at low level can feel loose in pitch in the cruise, and the pitch damper earns its place there. Higher true airspeed does not rescue the damping. Fixing Mach number fixes neither density nor dynamic pressure. Nothing special happens exactly at the tropopause.