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Aerodynamic Derivatives Questions and Answers

31.

Assertion (A): Wind tunnel derivatives from a rigid model understate the pitch stiffness of a large flexible transport at high dynamic pressure.

Reason (R): Aeroelastic deformation of the wing and tail reduces their effective contribution to the restoring moment.

Select the correct answer from the options below.

Answer: C

Both are true and the second explains the first. A flexible aeroplane's surfaces twist under load in a direction that reduces the incidence change they actually achieve, so the real stiffness is lower than a rigid model measures, and the shortfall grows with dynamic pressure. Flexible-to-rigid correction factors are applied for exactly this reason, and the corrections are among the harder parts of a derivative set to get right before first flight.

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Answer: D

A control derivative has a control deflection as its independent variable, so pitching moment with elevator deflection is the one. The other three all differentiate with respect to a motion variable - incidence, sideslip and roll rate - and are therefore stability derivatives. The distinction matters in the equations of motion: stability derivatives populate the system matrix and set the modes, while control derivatives populate the input matrix and set the response to the pilot.

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33.

Which surface contributes most of an aeroplane's lift curve slope?

Answer: A

The wing supplies the overwhelming majority, typically ninety per cent or more of the whole-aeroplane value, because its area and aspect ratio dwarf everything else. The horizontal tail adds a modest amount in proportion to its area and to the downwash factor at its station. The fuselage carries a little lift and mostly adds a destabilising moment. The vertical tail is symmetric and contributes no lift in symmetric flight at all.

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Answer: B

A slender body develops its normal force forward, so the fuselage contribution is destabilising: it makes \( C_{m_\alpha} \) less negative and has to be offset by tail volume. Wetted area affects drag, not this sign. A body of revolution at incidence certainly does generate a moment, which is the entire point of the Munk correction. And the sign does not reverse with speed, though its magnitude changes with Mach number.

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Answer: C

A higher aspect ratio means weaker tip effects, so less of the section slope is eaten by induced downwash, and the finite-wing slope moves closer to the two-dimensional \( 2\pi \). The relation \( C_{L_\alpha} = a_0/(1 + a_0/\pi A) \) shows it directly. Aspect ratio is a shape parameter and does not by itself change area, which is held fixed in the comparison. Reynolds number depends on chord and speed, and the section thickness is unchanged.

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Answer: D

Sweep reduces the slope, since only the component of the flow normal to the quarter-chord line does the work, and simple theory scales the slope with the cosine of the sweep angle. That loss is accepted in exchange for a higher critical Mach number. Sweep lowers rather than raises the effective aspect ratio for this purpose. The section alone cannot determine a three-dimensional slope. And the reduction applies throughout the speed range, not only transonically.

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37.

Statement I: A T-tail places the horizontal surface out of the wing wake at low incidence.

Statement II: A T-tail therefore has a larger effective tail volume at low incidence than a conventional tail of the same area.

In the light of the above statements, choose the correct answer.

Answer: A

Both statements are correct. Mounting the tailplane at the top of the fin puts it in cleaner air, so its local dynamic pressure and effective lift curve slope are higher, giving more pitch stiffness and damping per unit area. The penalty comes at high incidence, where the surface can enter the wing wake and lose effectiveness abruptly, which is the deep stall problem. So the arrangement improves the derivative in the normal envelope and threatens it at the edge.

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38.

Which derivative most directly determines the elevator angle needed per g of normal acceleration?

Answer: B

The elevator angle per g is set by elevator control power, \( C_{m_{\delta e}} \), against the manoeuvre stability of the aeroplane. Lift with incidence determines how much incidence the g demands, which enters the calculation but does not by itself set the deflection. Pitch damping contributes through the manoeuvre margin at the pitch rate of a steady turn but is a smaller term. Drag with incidence plays no part in the pitching balance.

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39.

Consider the following pairs of derivative and the surface that dominates it.

Pitching moment with elevator - Horizontal tail

Yawing moment with rudder - Vertical tail

Rolling moment with aileron - Wing

Side force with sideslip - Horizontal tail

How many of the pairs given above are correctly matched?

Answer: C

The first three are right and the fourth is not, so three pairs match. Side force with sideslip comes from the vertical tail and the fuselage side area, not from the horizontal tail, whose contribution to a side force is negligible. Each control derivative belongs to the surface that carries the control, which is the easiest of the attributions, and the force derivatives follow the area presented to the relevant flow direction.

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Answer: D

The wing's downwash means the tail sees only a fraction \( (1 - \mathrm{d}\epsilon/\mathrm{d}\alpha) \) of the aeroplane's incidence change, so its contribution to pitch stiffness is reduced by that factor, typically to sixty or seventy per cent of the geometric value. It generally lowers the dynamic pressure at the tail slightly rather than raising it. Aspect ratio is geometric. And the wing's own aerodynamic centre is not moved by its downwash field.

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