Dynamic Lateral Directional Modes Questions and Answers

Answer: C

The trade is real and specific: the dihedral effect and the fin stiffness appear in both the Dutch roll and the spiral criterion, with opposite effect, so pushing either derivative helps one mode and hurts the other. That is why augmentation is the normal answer rather than more aerodynamic surface. The modes are certainly not independent, being roots of one characteristic equation. Improvements are not universal, as the trade shows. And the spiral and the roll subsidence are two distinct real roots, separated by two orders of magnitude in speed.

Answer: D

Rolling raises the local incidence of the down-going wing, which gains lift and with it induced drag, so the nose is dragged away from the direction of roll. That is adverse yaw seen as a derivative rather than as an aileron effect, and it is one reason a roll input generates sideslip. The fin's sideslip angle is a consequence, not the cause. Dynamic pressure barely changes in a small-perturbation roll. Fuselage side force arises from sideslip, which is a different derivative altogether.

Answer: A

The fin and fuselage together produce a side force opposing any sideslip, and since the Dutch roll consists of sideslip reversing back and forth, that force opposes the motion and removes energy from it. It appears as \( Y_\beta/u_0 \) alongside the yaw damping in the damping term. It does not change the fin's effectiveness, which is a geometric property. Its rolling moment about the centre of gravity is a separate derivative. And inertia is a mass property that no aerodynamic force alters.

Answer: B

The first three are negative and the fourth is positive, so three pairs match. Directional stability requires a positive yawing moment with sideslip: sideslip from the right must produce a nose-right moment to turn the aeroplane into the relative wind. The other three are all restoring or damping terms and are negative by convention. Getting this one sign right matters, because it is the difference between weathercock stability and directional divergence.

Answer: C

A dorsal fin is a large-sideslip device: it generates a vortex that keeps flow attached over the fin and preserves the yawing moment with sideslip at angles where a bare fin would stall, which is what prevents rudder lock and improves spin resistance. At the small sideslip angles of a Dutch roll it does very little, since the fin is unstalled there anyway. Reducing effective dihedral is what a ventral fin does. The roll mode depends on wing roll damping, which a dorsal fin cannot touch.

Answer: D

The ventral surface's side force acts below the centre of gravity, so its rolling moment opposes that of the fin above and reduces the net roll produced by sideslip. Less roll content means a better damped Dutch roll and a smaller roll-to-yaw ratio, which is exactly why ventral fins appear on aeroplanes with plenty of sweep. Adding to the dihedral effect would make matters worse. It adds negligible inertia, and it increases rather than reduces directional stiffness.

Answer: A

Both statements are true and the second is the reason. Sweep produces a strong rolling moment with sideslip because the advancing wing meets the flow more nearly square on, and on a heavily swept wing that effect alone can exceed what is wanted. Geometric anhedral subtracts from it, trading a little spiral stability for a better damped Dutch roll. It is why so many swept-wing military transports carry visible anhedral while straight-wing light aeroplanes carry dihedral.

Answer: B

The sweep contribution to the rolling moment with sideslip is proportional to the lift coefficient, so it grows as the aeroplane slows and the lift coefficient rises. That is why a swept-wing aeroplane's Dutch roll is at its worst on approach, where the mode is both slow and roll-dominated, and why the yaw damper is required for landing on many types. Geometric dihedral is indeed a fixed contribution, but sweep is not, which is the point. Nothing here reverses the sign.

Answer: C

Frequency goes as \( 1/\sqrt{I_{zz}} \), since \( N_\beta = C_{n_\beta}\bar q S b/I_{zz} \) and \( \omega_n \approx \sqrt{N_\beta} \). The square root of 1.44 is 1.2, so the frequency falls by a factor of 1.2. A factor of 1.44 would follow if frequency went inversely with inertia rather than with its root, and 2.07 squares it instead. No change would require inertia to be absent from the derivative, which it is not. The same reasoning explains why a stretched fuselage variant wags more slowly than the model it was derived from.

Answer: D

A yaw rate changes the vertical tail's local sideslip angle in proportion to its arm, and the resulting side force acts through that same arm, so the fin supplies the bulk of the damping and its contribution grows with the square of the tail length. The wing tips contribute a little through differential drag but far less. Nacelles and the horizontal tail make minor contributions, the latter mainly through its own side area, which is small. The same arm-squared argument governs pitch damping from the horizontal tail.