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Aerodynamic Derivatives Questions and Answers

Answer: A

A derivative is a partial rate of change: how much force or moment appears per unit change of one motion variable, with the others held at their trim values. That is what makes the linearised equations of motion possible, since each derivative is one coefficient in them. A total force is a trim quantity, not a derivative. The second time derivative of displacement is an acceleration, which the derivatives help to compute rather than being one. Lift-to-drag ratio is a performance parameter that appears in the phugoid damping estimate but is not itself a derivative.

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Answer: B

Linearisation is the whole basis of the method: the aerodynamic forces are expanded as a Taylor series about trim and truncated after the first-order terms, whose coefficients are the derivatives. That is why a set of derivatives is quoted for a stated speed, altitude and configuration, and why a different flight condition needs a different set. Tunnel capability has nothing to do with it. Aeroplanes leave trim constantly, which is precisely why the perturbation equations are needed. The units follow from the definition rather than from the choice of reference.

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3.

The non-dimensional derivative \( C_{m_\alpha} \) is obtained from a dimensional pitching moment derivative by dividing by

Answer: C

Pitching moment is non-dimensionalised by dynamic pressure, reference area and mean aerodynamic chord, since a moment carries an extra length compared with a force. Dividing by dynamic pressure and area alone would leave a length dimension, which is the correct treatment for a force coefficient rather than a moment. Density and speed alone omit the reference geometry entirely. Span is the reference length for the lateral moments, rolling and yawing, not for pitching, and mixing the two is a common source of factor errors.

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Answer: D

Static longitudinal stability requires \( C_{m_\alpha} < 0 \): an increase in incidence must produce a nose-down moment. Directional stability, by contrast, requires \( C_{n_\beta} > 0 \), and dihedral effect requires \( C_{l_\beta} < 0 \). Those three signs - negative, positive, negative - are worth memorising as a set, because two of them look inconsistent until you notice that a positive sideslip is from the right while a positive yawing moment is nose-right, so the aeroplane turns into the wind.

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5.

Match each derivative with the surface that supplies most of it.

List I

A. Pitching moment with pitch rate

B. Yawing moment with sideslip

C. Rolling moment with roll rate

List II

P. The wing

Q. The horizontal tail

R. The vertical tail

Select the correct match.

Answer: A

Pitch damping comes from the horizontal tail on its arm, directional stiffness from the fin, and roll damping from the wing across its span, so A-Q, B-R, C-P. Each pairing follows from where the relevant velocity change is largest: a pitch rate moves the tail vertically, a sideslip strikes the fin, and a roll rate changes incidence most at the wing tips. Getting these attributions right is what lets a designer predict which derivative a geometry change will move.

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6.

Derivatives with respect to a rate, such as pitch rate or roll rate, are made dimensionless using a factor built from the reference length and _____.

Answer: B

Rate derivatives use \( \bar c/2V \) in pitch and \( b/2V \) in roll and yaw, so the normalising factor is built from the reference length and twice the true airspeed. The factor of two is a convention, not a physical necessity, but it is universal enough that omitting it produces a derivative twice the intended size. The speed of sound appears in Mach corrections rather than in the normalisation. Dynamic pressure and area non-dimensionalise the force or moment itself, which is a separate step.

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7.

For \( \bar c = 3 \ \mathrm{m} \), \( V = 150 \ \mathrm{m/s} \) and a pitch rate of \( 0.20 \ \mathrm{rad/s} \), the non-dimensional pitch rate \( \hat q = q\bar c/2V \) is

Answer: C

Substituting into the standard normalisation:

\[ \hat q = \frac{0.20 \times 3}{2 \times 150} = 0.002 \]

so 0.002, a very small number, which is exactly the point: a rigid aeroplane manoeuvres at reduced frequencies far below unity, so quasi-steady aerodynamics is a good assumption. 0.004 forgets the factor of two in the denominator. 0.02 slips a decade. 0.0004 divides by the span instead of the chord in a way the definition does not permit.

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8.

Given \( C_{m_q} = -12 \), \( \bar q = 4000 \ \mathrm{Pa} \), \( S = 30 \ \mathrm{m^2} \), \( \bar c = 3 \ \mathrm{m} \), \( V = 150 \ \mathrm{m/s} \) and \( I_{yy} = 100\,000 \ \mathrm{kg\,m^2} \), the dimensional \( M_q = C_{m_q}\bar q S \bar c^{2}/(2VI_{yy}) \) is about

Answer: D

Working through the conversion:

\[ \begin{aligned} \bar q S \bar c^{2} &= 4000 \times 30 \times 9 = 1.08 \times 10^{6} \\ 2VI_{yy} &= 3.0 \times 10^{7} \\ M_q &= -12 \times 0.036 = -0.43 \ \mathrm{s^{-1}} \end{aligned} \]

so -0.43 per second. -0.86 omits the factor of two. -12 is the coefficient itself, which has no units of inverse time and cannot enter the characteristic equation. -0.14 divides by the chord once instead of squaring it, which is the commonest slip in this particular conversion.

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Answer: A

Quasi-steady means the flow adjusts so quickly that the forces depend on the current values of the motion variables alone, with no memory of how they were reached. It is a good assumption at the low reduced frequencies of rigid-body motion, and it fails for flutter and for gust penetration, where unsteady aerodynamics is needed. The downwash-lag derivative is the one deliberate exception retained in the rigid-body set. Trim, incompressibility and condition-independence are separate assumptions, and the last of them is simply false.

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10.

The lift curve slope of a thin two-dimensional aerofoil in incompressible flow is

Answer: B

Thin aerofoil theory gives \( 2\pi \) per radian, about 0.11 per degree, and it is the anchor value every finite-wing estimate is reduced from. Half that would be a very low slope no practical section shows. Four times \( \pi \) is the value sometimes quoted for a flat plate in supersonic flow at low Mach number in error, and \( 4/\sqrt{M^{2}-1} \) is the correct supersonic form. Unity per radian is far too small and would describe a very low aspect ratio surface.

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