Longitudinal Control and Trim Questions and Answers

Answer: C

Correct answer: 22 N per g

The tab's contribution adds to the surface's own:

\[ 8 + 14 = 22\ \mathrm{N\ per\ g} \]

The addition is the tab's whole purpose. A stabilator's bare stick force per g is low, because the surface is powerful and its hinge moments are small, and a pilot could reach a high load factor with almost no effort. The anti-servo tab deflects with the surface to oppose its motion, raising the force to a level certification requires - a deliberate degradation of an over-light control.

Answer: D

Correct answer: 0.08 degrees per knot

The gradient is the change in deflection over the change in speed:

\[ \frac{11 - 3}{200 - 100} = \frac{8}{100} = 0.08\ \mathrm{deg/kt} \]

The positive gradient - more up elevator as speed falls - is itself a stability requirement: certification demands that the stick move and the force build in the conventional sense with speed. The distractor 12.5 knots per degree inverts the ratio and is the same information expressed the other way round.

Answer: A

Correct answer: About 0.067 chord

Find the spare travel first, then convert it at the given rate:

\[ 22 - 14 = 8^\circ \]

\[ \frac{8}{1.2} = 6.67 \text{ increments of } 0.01\bar{c} = 0.067\bar{c} \]

This is in essence how a forward centre of gravity limit is established: the demanding trim case fixes the deflection already needed, and the limit is placed where the remaining travel still covers manoeuvre and gusts. Note that the calculation as posed leaves nothing in hand for those, so the certified limit would be more forward-restrictive still.

Answer: B

The difference of 4 degrees is what guarantees the foreplane reaches its limit first. Its stall then drops the nose automatically, giving the configuration an inherent protection against reaching the main wing's stall at all.

Answer: C

Correct answer: Both Statement I and Statement II are correct

Both are true and together they explain why the case is critical. At forward centre of gravity the tail must carry its largest download, and in the flare the aeroplane is at the lowest speed of the approach, where the elevator's moment per degree is least.

Ground effect compounds it: proximity to the ground reduces the downwash at the tailplane, raising its effective angle of attack, so it carries less download and the nose drops. Maximum demand coincides with minimum authority, which is precisely what a sizing case is.

Answer: D

Correct answer: Both Statement I and Statement II are correct

Both are true and the second follows from the first. Trimming adjusts a tab or the tailplane incidence so that the aerodynamic hinge moment is balanced and the pilot need hold no force - it does not centre the elevator.

At most trimmed conditions the elevator is deflected, often several degrees, because a download is needed to balance the wing's pitching moment. This is why a trim change with configuration or speed is felt as a force to be trimmed out rather than as the surface returning to neutral.

Answer: A

Correct answer: Both Statement I and Statement II are correct

Both are true and the second is the mechanism. As shock waves form on the upper surface at high subsonic Mach, the lift distribution moves rearward and the aerodynamic centre shifts aft - typically from about a quarter chord towards mid-chord.

The same lift then acts further behind the centre of gravity, producing a nose-down moment that steepens the dive and raises the Mach number further. A Mach trim system senses Mach and applies compensating tailplane or elevator movement automatically, since the divergence would otherwise be unacceptable.

Answer: B

Correct answer: Both Statement I and Statement II are correct

Both are true and the second is the reason for the first. Deflecting the whole surface changes its angle of attack directly rather than merely cambering it, so a stabilator generates far more moment per degree than an elevator of the same area.

Its hinge moments are correspondingly small, and the stick force per g would be too low to satisfy certification - a pilot could reach a high load factor with almost no effort. The anti-servo tab deflects with the surface to oppose its motion and restores an appropriate force.

Answer: C

Correct answer: Both Statement I and Statement II are correct

Both are true and the second explains the first. With a hydraulic actuator between the stick and the surface, the pilot moves a valve rather than the elevator, so none of the aerodynamic hinge moment reaches the hand.

Without feel there would be nothing to signal how much control was being applied, and gross overcontrol at high speed would be easy. Artificial feel supplies it: a q-feel unit stiffening the stick with dynamic pressure, springs for centring, and a bob weight adding force with normal acceleration.

Answer: D

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

The second statement is right - a tailplane in the wing's wake meets low-energy air and produces far less lift change per degree. But it refutes the assertion rather than supporting it.

A T-tail is the configuration most at risk, not immune: at extreme angle of attack the wake rises and can blanket a surface mounted at the fin tip, which is the deep stall mechanism. A low conventional tail is generally clear of it, which is one of the arguments for that layout.