Aerodynamic Derivatives Questions and Answers

Answer: A

Three are negative; the yawing moment with sideslip must be positive for directional stability, so that sideslip from the right yields a nose-right moment and the aeroplane turns into the relative wind. The side force, roll damping and yaw damping derivatives are all restoring or dissipative and therefore negative. This single positive sign among the lateral set is the one most often written down wrongly, and getting it wrong inverts the whole directional argument.

Answer: B

It is \( C_{l_{\delta a}} \), a rolling moment per unit deflection, one of the control derivatives rather than a stability derivative, since its independent variable is a control input and not a motion variable. Rolling moment per unit roll rate is roll damping, and the ratio of the two sets the steady roll rate. Yawing moment per unit sideslip is directional stiffness. Side force per unit rudder deflection is a rudder control derivative, useful for sideslip control and for engine-out trim.

Answer: C

Directional stiffness fixes the frequency, roll damping fixes the roll time constant, and the dihedral effect fixes how much rolling accompanies the sideslip, so A-Q, B-P, C-R. The third pairing is the one that decides whether a Dutch roll is felt as a wing rock or a nose wag, and it is also the derivative that trades against spiral stability. The alternatives all attach a stiffness derivative to a damping quantity or the reverse.

Answer: D

Assembling the conversion:

\[ \begin{aligned} \bar q S b^{2} &= 3000 \times 25 \times 144 \\ &= 1.08 \times 10^{7} \\ 2VI_{xx} &= 7.2 \times 10^{6} \\ L_p &= -0.45 \times 1.5 = -0.68 \ \mathrm{s^{-1}} \end{aligned} \]

so -0.68 per second, giving a roll time constant of about 1.5 seconds. -1.35 omits the factor of two in the normalisation, which doubles every rate derivative and is the classic error here. -0.45 is the coefficient itself, dimensionless and unusable in the equations of motion. -0.07 slips a decade.

Answer: A

Steady roll rate is \( p_{ss} = L_{\delta a}\delta_a/(-L_p) \), so raising aileron effectiveness raises the rate while raising roll damping would lower it. Dihedral effect and directional stiffness govern the slow lateral modes and do not enter the steady roll balance at all. The relation also explains why large ailerons and a stiff wing go together: effectiveness is worth little if the wing twists enough to undo it at high dynamic pressure.

Answer: B

DATCOM is a handbook of empirical charts and correlations keyed to geometric parameters, built from decades of tunnel and flight data, and it remains the quickest route to a first derivative set at the preliminary design stage. It solves no equations, so the second option describes computational fluid dynamics instead. It is built from measurements but is not itself a measurement method. And no exact potential-flow solution exists for a general configuration.

Answer: C

Only flight test identification measures the real aeroplane, so its derivatives contain the flexibility, the true Reynolds number, the propulsion effects and every geometric detail as built. That is why it is the final arbiter, and why a simulator model is tuned to match it. Thin aerofoil theory and vortex lattice methods are inviscid and rigid. A rigid tunnel model at low Reynolds number omits both of the effects named in the question.

Answer: D

The method adjusts the derivative set until the simulated response matches the recorded one for the same control input, typically by a maximum-likelihood or least-squares fit. Well-designed inputs, such as doublets and frequency sweeps, are what make the individual derivatives separable. Strain gauges measure loads rather than derivatives. The flight manual contains no derivatives. And differentiating a noisy record twice amplifies the noise without separating one derivative from another.

Answer: A

Dimensional derivatives carry dynamic pressure, and the non-dimensional ones still vary with Mach number, incidence, and the configuration of flaps, gear and stores. A set is therefore a table, not a single column, and modern simulation interpolates within it. Mass properties do change, and they matter, but they are inertias rather than derivatives. Tunnel limitations are irrelevant to why the physical quantity varies, and the requirement is engineering rather than administrative.

Answer: B

Substituting gives

\[ k = \frac{3 \times 3}{2 \times 150} = 0.03 \]

which is far below unity, so quasi-steady aerodynamics is justified for this short-period motion. That is the calculation behind the assumption rather than an assertion of it, and it is why flutter analysis, at much higher reduced frequency, cannot use the same derivative set. 0.06 omits the factor of two. 0.30 slips a decade. 0.003 divides by ten as well as by the factor.