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Wing and Airfoil Selection Questions and Answers

31.

For that flapped stalling speed of \( 47 \) m/s, the approach speed at \( 1.3 V_s \) is

Answer: C

Correct answer: 61 m/s

The approach speed is the required margin above the stall:

\[ V_{app} = 1.3 \times 47 \approx 61 \text{ m/s} \]

The factor of 1.3 is the certification requirement for the reference landing approach speed, giving a margin against gusts, manoeuvre and mishandling. Landing distance scales roughly with the square of this speed, which is why a small reduction in stalling speed is worth so much on the field length.

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32.

A wing with \( 25 \) degrees of sweep is changed to \( 35 \) degrees at the same cruise Mach number of \( 0.80 \). The normal Mach number falls from \( 0.725 \) to

Answer: D

Correct answer: 0.655

Applying the sweep relation at the new angle,

\[ M_n = 0.80 \times \cos 35^\circ = 0.80 \times 0.8192 \approx 0.655 \]

Ten more degrees of sweep buys another 0.07 of normal Mach number, which is a useful margin against drag rise. The cost is paid at low speed: more sweep means a lower maximum lift coefficient, a stronger tendency to tip stall, and a longer, heavier structural span for the same aerodynamic span.

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33.

A wing carries \( 2.5 \) degrees of washout with a root incidence of \( 3 \) degrees. The tip incidence is

Answer: A

Correct answer: 0.5 degrees

Washout is a reduction of incidence from root to tip:

\[ 3.0 - 2.5 = 0.5^\circ \]

The tip therefore flies at a lower angle of attack than the root and reaches its stalling angle later, so the root stalls first. That preserves aileron authority into the stall and gives a nose-down pitching moment rather than a wing drop, which is what makes the stall benign.

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34.

Aspect ratio rises from \( 8 \) to \( 10 \) at a lift coefficient of \( 0.6 \) and \( e = 0.85 \). The induced drag coefficient falls from \( 0.0168 \) to

Answer: B

Since \( C_{Di} = C_L^2/(\pi e AR) \), the value becomes \( 0.36/(\pi \times 0.85 \times 10) = 0.36/26.7 = 0.0135 \). A twenty per cent reduction in induced drag is worth having, and it is what the extra structural weight buys.

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35.

Statement I: A thicker aerofoil section generally offers a higher maximum lift coefficient.

Statement II: A thicker aerofoil section also raises the critical Mach number.

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

Answer: C

The first statement is correct: greater thickness usually improves maximum lift and provides structural depth. The second is the reverse of the truth, since a thicker section accelerates the flow more strongly and reaches sonic conditions at a lower free-stream Mach number.

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36.

Statement I: Wing sweep raises the drag-divergence Mach number.

Statement II: Wing sweep improves the wing's low-speed maximum lift coefficient.

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

Answer: D

Correct answer: Statement I is correct but Statement II is incorrect

Sweep does raise the drag-divergence Mach number, because only the component of velocity normal to the leading edge, \( M\cos\Lambda \), governs the section’s compressible behaviour. But it reduces the low-speed maximum lift coefficient for the same reason: the section sees a reduced effective velocity and the spanwise flow component thickens the boundary layer outboard, promoting tip stall. That penalty is exactly why swept-wing transports need such elaborate high-lift systems.

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37.

Statement I: A Fowler flap increases both the camber and the area of the wing.

Statement II: A plain flap increases the wing area when deflected.

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

Answer: A

Correct answer: Statement I is correct but Statement II is incorrect

The Fowler flap does add both camber and area, running aft on tracks as it deflects, and the area increase is much of the reason for its large lift increment. A plain flap simply hinges down about a fixed point: it adds camber and it certainly adds drag, but the planform area is unchanged. That difference in mechanism is the whole distinction between the two devices.

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38.

Statement I: A high wing position contributes an effective dihedral in sideslip.

Statement II: High-wing aircraft therefore often need less geometric dihedral than low-wing aircraft.

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

Answer: B

Both statements are correct and the second follows from the first. Flow around the fuselage in a sideslip adds a stabilising rolling moment for a high wing, so many high-wing transports use very little dihedral or even anhedral.

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39.

Statement I: Washout makes the wing tip stall before the root.

Statement II: Taper alone tends to raise the local lift coefficient towards the tip.

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

Answer: C

Correct answer: Statement I is incorrect but Statement II is correct

Statement II is correct and explains why statement I is the reverse of the truth. Taper reduces the chord faster than it reduces the local lift, so the local lift coefficient rises towards the tip and the tip would stall first. Washout is applied precisely to counteract that, setting the tip at a lower incidence so that the root stalls first and aileron authority is retained through the stall.

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40.

Statement I: The reference wing area includes the portion carried through the fuselage.

Statement II: Aerodynamic coefficients are meaningless unless the reference area is stated.

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

Answer: D

Both are correct. The trapezoidal reference extends the edges to the centreline, and since every coefficient is divided by that area, quoting a coefficient without its reference gives a number that cannot be compared with anything.

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