Longitudinal Control and Trim Questions and Answers

Answer: A

Correct answer: 15 degrees

Divide the moment wanted by the moment per degree:

\[ \delta_e = \frac{-0.18}{-0.012} = 15^\circ \]

The result should then be checked against the travel available, typically 20 to 25 degrees, since a demand near the stop leaves nothing for manoeuvre or gusts on top of the trim requirement. The distractor 0.0067 inverts the division, giving an answer far too small to be a physical deflection.

Answer: B

Correct answer: 10 degrees up

The speed change is 40 knots, which is four increments of 10:

\[ \frac{150 - 110}{10} \times 2.5 = 4 \times 2.5 = 10^\circ \]

The deflection is up, since slower flight needs a higher angle of attack and therefore more download at the tail. The relation is treated as linear here, which is a fair approximation over a modest speed range; across the whole envelope the elevator-against-speed curve is distinctly non-linear.

Answer: C

Correct answer: 4.5 degrees

The elevator angle per g gives the deflection for each unit of normal acceleration above the 1 g of level flight, so the increment is 1.5 g rather than 2.5:

\[ \Delta\delta_e = 3 \times (2.5 - 1) = 4.5^\circ \]

The distractor 7.5 degrees multiplies by the full 2.5 and forgets that the aeroplane is already trimmed at 1 g. It is the load factor change that the manoeuvre demands, which is why the gradient is defined against \( n - 1 \).

Answer: D

Correct answer: 4 degrees

Subtract both demands from the travel available:

\[ 25 - 15 - 6 = 4^\circ \]

Four degrees is a thin margin, and it is the quantity that fixes the forward centre of gravity limit: the limit is set where enough travel remains after trim and manoeuvre to cope with gusts and with dispersion in the estimates. Note that trim consumes the larger part here, which is typical of the flare at forward centre of gravity - the critical case for elevator power.

Answer: A

Correct answer: 2.5 degrees more up elevator

Divide the moment to be countered by the elevator's power:

\[ \frac{-0.03}{-0.012} = 2.5^\circ \]

The deflection is up, opposing the nose-down moment. Ground effect produces it by reducing the downwash at the tailplane: with less downwash the tail's effective angle of attack rises, it carries less download, and the nose drops. The demand arrives in the flare, at the lowest speed of the approach, where the elevator is least effective - which is what makes it a sizing case.

Answer: B

Multiplying gives \( 40 \times 0.8 = 32 \) kilonewton metres, and a thrust line below the centre of gravity pitches the nose up. Applying power therefore requires a trim change on any aeroplane with underslung engines.

Answer: C

Correct answer: 6.7 degrees more up elevator

Divide the moment change by the elevator's power:

\[ \frac{-0.08}{-0.012} = 6.7^\circ \]

The deflection is up, to oppose the nose-down moment the flaps have produced. Nearly seven degrees is a substantial share of the available travel, which is why flap-induced trim change is a design consideration rather than a nuisance - and why a variable incidence tailplane, which can absorb such a shift without using elevator travel at all, is fitted to large transports.

Answer: D

Correct answer: 12.5 N

A q-feel force is proportional to dynamic pressure, which goes as the square of speed:

\[ F_2 = F_1\left(\frac{V_2}{V_1}\right)^{2} = 50 \times \left(\frac{125}{250}\right)^{2} = 50 \times 0.25 = 12.5\ \mathrm{N} \]

The distractor 25 N assumes a linear relation. Reproducing the square law is the whole purpose of the system: on a fully powered control the pilot feels nothing of the real hinge moments, and feel that did not stiffen with speed would invite gross overcontrol at high speed.

Answer: A

Correct answer: 0.60 m aft

Multiply the fractional shift by the chord:

\[ (0.40 - 0.25) \times 4.0 = 0.15 \times 4.0 = 0.60\ \mathrm{m} \]

The movement is aft, and that is what produces Mach tuck: with the aerodynamic centre further behind the centre of gravity, the same lift generates a larger nose-down moment. The distractor 0.15 m omits the multiplication by the chord, taking the fraction for a distance.

Answer: B

Correct answer: 67 per cent

The improvement is measured against the original value:

\[ \frac{0.020 - 0.012}{0.012} \times 100 = \frac{0.008}{0.012} \times 100 = 66.7\% \]

The distractor 167 per cent gives the ratio of new to old rather than the increase, and 40 per cent divides by the new figure. The gain is real and is why stabilators are used on light aeroplanes and fast jets: deflecting the whole surface changes its angle of attack directly rather than merely cambering it.