Control Surface Hinge Moments Questions and Answers

Answer: C

Local shocks alter the pressure distribution ahead of and behind the hinge, so the hinge moment can rise sharply and even change character, which was the practical barrier that forced the move to powered controls on early jets. Gearing is mechanical and fixed. Effectiveness falls but does not vanish. And a trim tab does not reverse, though its effectiveness changes along with everything else on the surface.

Answer: D

A trim tab is deflected until the hinge moment is zero at the required control position, so the surface stays put with the stick released and the pilot holds no force. It does not extend the deflection range, which is set by the stops. It has a small drag effect either way and is not a drag device. And it has nothing to do with flutter, which is a mass and stiffness matter.

Answer: A

The tab moves down, which produces a nose-up hinge moment on the elevator and lets it float up with no stick force - and an elevator up gives a nose-up pitching moment on the aeroplane. The relationship is worth working through slowly, because tab, surface and aeroplane all move in a chain and the intuitive answer is often the reverse of the correct one. A tab at neutral is the untrimmed condition.

Answer: B

The Frise design lets the nose of the up-going aileron project below the wing, raising the drag on the wing that is going up and so counteracting adverse yaw, while also providing aerodynamic balance. The penalty is extra drag and sometimes a rougher feel. Retracting inside the wing describes a plain aileron at neutral. Sealing the gap is the internal balance idea. Mass balance is a weight rather than a shape.

Answer: C

Differential rigging gives the up-going aileron more travel than the down-going one, which raises the drag on the up-going wing and reduces adverse yaw, on the same principle as the Frise shape but achieved through the linkage. Equal deflection is the plain arrangement it improves upon. Only one moving at a time is not a rigging scheme. Symmetric droop with flap is flaperon behaviour, a different function altogether.

Answer: D

The first three are correct and the fourth is not, so three match. A trim tab lets a deflection be held without force; it does not make the surface any more effective about the aeroplane's centre of gravity. Keeping the purposes straight is the core of this topic, since four devices can all sit on one surface and each addresses a different complaint: force level, force gradient with g, flutter margin and trim.

Answer: A

The chain runs 1, 2, 3, 4, 5, and the crucial link is the second: it is the surface's inertia, not any aerodynamic effect, that produces the phase lag which lets the force do positive work on the structure. Mass balancing removes that lag and breaks the chain at its second step. Putting the aerodynamic force before the deflection inverts cause and effect, since the force exists because the surface has moved.

Answer: B

Multiplying out the definition:

\[ \begin{aligned} H &= -0.08 \times 6000 \times 1.5 \times 0.40 \\ &= -288 \ \mathrm{N\,m} \end{aligned} \]

so -288 newton metres. -720 omits the surface chord. -144 halves the answer, as though the area were 0.75 square metres. -28.8 slips a decade. The arithmetic is trivial and the trap is entirely in the reference geometry, which must be the surface's own area aft of the hinge and its own mean chord. Keeping the four devices and their purposes apart is most of the work in reading a control system drawing.

Answer: C

Stick force is proportional to the hinge moment, which is proportional to the derivative at a given deflection, so

\[ \frac{0.35}{0.60} = 0.58 \]

and the force falls to 58 per cent, a reduction of about two fifths. 35 per cent quotes the new derivative as though the old one were unity. 171 per cent inverts the ratio. 25 per cent would need the derivative to fall to 0.15. Note that control effectiveness about the aeroplane's centre of gravity is essentially untouched, which is what makes the change worthwhile.

Answer: D

The bobweight adds directly to the force per g:

\[ 30 + 20 = 50 \ \mathrm{N/g} \]

which is inside the 12 to 55 band, so the aeroplane remains compliant, though with little margin at the heavy end. That narrow margin is itself a design finding: a smaller bobweight would leave more room for the variation with centre of gravity and configuration. A bobweight cannot reduce the force per g, so the third option has the sign wrong, and it certainly does not leave it unchanged.