Dynamic Lateral Directional Modes Questions and Answers

Answer: C

The reciprocal of 2.5 is 0.40, so four tenths of a second: an aileron input reaches about 63 per cent of its steady roll rate in that time and is essentially complete in three time constants. 2.5 s is the derivative read back as a time. 1.73 s and 0.28 s come from folding in a logarithm of two that does not belong: the time to half amplitude is \( 0.693\tau_R = 0.28 \ \mathrm{s} \), which is a different measure and is worth keeping distinct from the time constant itself.

Answer: D

In the steady roll the aileron moment balances the damping moment:

\[ \begin{aligned} p_{ss} &= \frac{L_{\delta a}\delta_a}{-L_p} \\ &= \frac{0.60}{2.5} = 0.24 \ \mathrm{rad/s} \end{aligned} \]

which is 0.24 rad/s, about 14 degrees per second. 1.50 rad/s multiplies where it should divide. 0.42 rad/s inverts the ratio of the two numbers. 0.06 rad/s divides by ten as well. The relation shows why roll performance and roll damping pull against each other: the damping that makes the response crisp also limits the rate it settles at.

Answer: A

Substituting into the standard non-dimensional measure of roll performance:

\[ \begin{aligned} \frac{pb}{2V} &= \frac{0.24 \times 10}{200} \\ &= 0.012 \ \mathrm{rad} \end{aligned} \]

so 0.012 radian, or a little under a degree. The helix angle is the local incidence change at the tip caused by the roll, which is why it, rather than roll rate alone, is the quantity specifications constrain. 0.024 omits the factor of two in the denominator. 0.12 slips a decade. 0.0012 slips it the other way.

Answer: B

Rolling gives every spanwise station a vertical velocity, so the down-going wing meets the air at a higher incidence and the up-going wing at a lower one, and the resulting lift difference opposes the roll. The effect grows towards the tips, which is why span and lift-curve slope both matter. Sideslip at the fin is a yaw effect, not the source of roll damping. The centre of gravity is a mass property. Dynamic pressure changes very little in a small-perturbation roll.

Answer: C

The time constant is \( \tau_R = 1/0.80 = 1.25 \ \mathrm{s} \), which exceeds the one-second limit, so the aeroplane is non-compliant for Level 1 on this characteristic and feels sluggish in roll: the pilot rolls into a bank and waits for the rate to build. 0.80 s is the derivative read back as a time, the same slip as before but with the opposite verdict, which is why the reciprocal must be taken deliberately. A larger wing span or more effective outboard sections would raise the damping and shorten the constant.

Answer: D

Directional stiffness sets the frequency, roll damping sets the roll time constant, and yaw damping sets the Dutch roll damping, so A-Q, B-P, C-R. Sorting the lateral derivatives this way is the fastest route through most questions on the topic, and it mirrors the longitudinal pattern in which one derivative supplies stiffness and another supplies damping. The alternatives all swap a stiffness for a damping term, which is the error the mapping exists to prevent.

Answer: A

In a yaw, the outer wing has a higher local airspeed and therefore more lift, producing a rolling moment into the turn. That term appears in the spiral criterion multiplied by the fin's stiffness, and it is what makes a large fin spirally destabilising. It is not the largest derivative; roll damping and dihedral effect are bigger. The fin's incidence change with yaw rate produces the yaw damping instead. And the roll time constant comes from roll damping alone.

Answer: B

The first three are correct and the fourth is not, so three match. Dihedral helps the spiral, fin volume raises the Dutch roll frequency, and a yaw damper raises its damping. More roll damping, however, reduces the steady roll rate, since \( p_{ss} = L_{\delta a}\delta_a/(-L_p) \) has the damping in the denominator: it makes the response quicker to settle but slower in absolute rate. That tension between crispness and authority is worth carrying, because it is the reason roll requirements are written on both the time constant and the achieved rate.

Answer: C

A large fin gives directional stiffness, so the Dutch roll has ample frequency, while the spiral criterion suffers because the fin term is large and the dihedral term is supplied only by the wing's high position. High-wing effective dihedral is real but modest, so the spiral is marginal at best. The Dutch roll cannot be absent, since the coupling that produces it is always present. Roll subsidence is stable on any conventional wing, and longitudinal behaviour is a separate problem altogether.

Answer: D

The first two hold and the third does not. Dihedral improves the spiral and worsens the Dutch roll, and the roll subsidence, with a time constant of a few tenths of a second, is by far the fastest root. But certification does not demand spiral stability; it demands that a divergence be slow, and quotes a minimum time to double amplitude instead. Many aeroplanes in service are spirally divergent, which is why the requirement is written that way rather than as a sign condition.