Dynamic Longitudinal Modes Questions and Answers

Answer: A

The real part is \( \sigma = \zeta\omega_n = 0.5 \times 4.0 = 2.0 \ \mathrm{s^{-1}} \), and it is given as unchanged, so

\[ \begin{aligned} \zeta_{2} &= \frac{\sigma}{\omega_{n2}} \\ &= \frac{2.0}{2.5} = 0.80 \end{aligned} \]

The damping ratio therefore rises to 0.80 as the frequency falls. This is the standard trade with aft loading, and it explains an apparent paradox in flight test: an aeroplane can feel better damped and worse controlled at the same time. 0.31 scales the wrong way. 0.50 assumes damping ratio is a fixed property, which it is not. 1.25 is the frequency ratio rather than the damping ratio.

Answer: B

The first three pairs are correct and the fourth is not, so three match. Pitch stiffness sets short-period frequency, pitch damping sets its damping ratio, and lift-to-drag ratio sets phugoid damping through \( \zeta_{ph} \approx C_D/(\sqrt{2}C_L) \). Pitching inertia, however, does not appear in the phugoid at all: Lanchester's period \( T = \sqrt{2}\pi u_0/g \) contains only trim speed and gravity, because the idealised phugoid involves no rotation about the pitch axis. Inertia belongs with the short period, where it divides every moment derivative.

Answer: C

A longer arm increases the tail's contribution to pitch stiffness roughly in proportion to the arm, and its contribution to pitch damping roughly with the square of the arm, since the tail's local incidence change from a pitch rate is itself proportional to the arm. So the change raises frequency and damping together, which is why a long-tailed aeroplane feels crisp and well damped. The second and third options each get one of the two trends backwards. Leaving both unchanged would require the tail to contribute nothing, which is only true of a tailless configuration relying on the wing alone.

Answer: D

Routh's necessary condition is that every coefficient shares one sign, conventionally taken as positive after dividing through by \( A \). A single sign change guarantees at least one root in the right half plane, so the aeroplane is unstable without any further calculation. It is necessary but not sufficient: an aeroplane can have five positive coefficients and still diverge in oscillation if the Routh discriminant is negative. Testing \( E \) alone catches the aperiodic divergence but misses that case. Allowing free signs elsewhere abandons the test. A zero sum would merely mean that \( \lambda = -1 \) is a root.

Answer: A

Taking the three products in turn:

\[ \begin{aligned} BCD &= 4.0 \times 9.0 \times 0.30 = 10.8 \\ AD^{2} &= 0.09 \\ B^{2}E &= 16 \times 0.10 = 1.6 \\ R &= 10.8 - 0.09 - 1.6 = +9.11 \end{aligned} \]

Positive, and with every coefficient positive as well, so the aeroplane is stable in both longitudinal modes. The second option is the answer for a different set of coefficients with a much smaller \( D \). The third stops after the first product and never subtracts the other two terms, which is the usual slip. Neutral stability would need \( R \) to vanish exactly.

Answer: B

All coefficients positive rules out a real root in the right half plane, so the instability has to be oscillatory: a complex pair with a positive real part. In practice it is the phugoid that goes unstable this way, because it holds so little damping to begin with, and the fault often traces to a small or negative \( D \). An aperiodic divergence would announce itself as a negative \( E \), which is not the case here. Light but positive damping would give a positive discriminant. A constant-amplitude oscillation is the boundary case \( R = 0 \), not \( R < 0 \).

Answer: C

Both statements are correct, and together they are the neatest link between static and dynamic stability. With \( A \) positive, the product of the four roots equals \( E/A \); a negative \( E \) therefore forces an odd number of negative-product combinations, which for a real system means a real root in the right half plane and a pure divergence with no oscillation. The classical result is that \( E > 0 \) is the dynamic statement of the static stability criterion, so an aeroplane that has just lost static stability shows itself dynamically as an aperiodic pitch divergence rather than as a growing oscillation.

Answer: D

Aft loading reduces pitch stiffness, so the short-period roots migrate towards the origin: the frequency falls, the damping ratio rises because the real part is roughly preserved, and at the neutral point the stiffness reaches zero. Past it, \( E \) changes sign and a root crosses into the right half plane, giving the aperiodic pitch divergence that defines the aft limit. Moving away from the origin is what a forward shift does. The phugoid is comparatively insensitive to centre of gravity, so it is the short period that moves, not the other way round. Damping ratio and frequency cannot change independently here, since both depend on the same stiffness.

Answer: A

Tuck appears when the pitching moment derivative with speed turns negative as the aerodynamic centre moves aft with Mach number. That drives \( E \) through zero and negative, producing a real positive root and a nose-down divergence with no oscillation, which is why a Mach trim system is fitted to sense the condition and apply compensating trim. Short-period frequency is set by stiffness in incidence and is not what changes here. The phugoid period is tied to trim speed and does not collapse. Purely imaginary roots would be the boundary of an oscillatory instability, a different case entirely.

Answer: B

The chain runs in the order written: 1, 2, 3, 4, 5. Compressibility moves the aerodynamic centre aft, which is what makes the moment derivative with speed negative; the nose-down moment then accelerates the aeroplane, and the higher Mach number deepens the moment that caused it. That feedback is what makes the motion divergent rather than merely a trim change. Putting the derivative before the aerodynamic centre shift inverts cause and effect, and the remaining orders break the loop that turns a trim shift into a divergence.