Cockpit Display Systems Questions and Answers

Answer: A

Correct answer: 16.7 ms

The frame interval is the reciprocal of the refresh rate:

\[ \Delta t = \frac{1}{60\ \mathrm{Hz}} = 0.0167\ \mathrm{s} = 16.7\ \mathrm{ms} \]

Refresh rate matters for a flight display because the symbology must appear to move smoothly and, on a head-up display, must stay registered with the outside world. It also sets the floor on display latency: however fast the processing, information can be no fresher than the last frame drawn.

Answer: B

Correct answer: 57 ms

The total delay is the processing latency plus the worst-case wait for the next frame, which is one full frame interval:

\[ t = 40 + 16.7 = 56.7 \approx 57\ \mathrm{ms} \]

Latency of this order matters on a head-up display, where symbology must stay registered with the real world: at 150 knots the aircraft moves about 4 metres in 57 ms, so a conformal runway outline would lag visibly if the delay were much greater. It is why display latency is specified and tested rather than left to fall where it may.

Answer: C

Correct answer: Three

Each generator can drive three screens, so after one fails the surviving generator can drive three of the four - one display is lost.

This is the standard architecture question: the display units may be four, but the redundancy is limited by the generators behind them. It is also why reversion switching exists, allowing the crew to choose which three are driven, so that the primary flight displays are retained and the least critical content is the content lost.

Answer: D

Correct answer: 1.25

Aspect ratio is the horizontal extent divided by the vertical:

\[ \frac{30}{24} = 1.25 \]

The distractor \( 0.80 \) inverts the ratio. Head-up display fields of view are modest compared with the human visual field, which is why symbology must be placed carefully - the useful conformal area is small, and anything drawn near its edge is easily lost as the aircraft manoeuvres.

Answer: A

Correct answer: 0.0078 inch

Pixel pitch is the physical spacing between pixel centres:

\[ p = \frac{8\ \mathrm{inches}}{1024} = 0.0078\ \mathrm{inch} \]

This is about 128 pixels per inch. Pitch determines how fine a line or how small a character can be rendered legibly, and hence how much symbology a display of a given size can carry without the elements merging - a direct link between the hardware specification and the clutter problem.

Answer: B

Correct answer: One tenth of the time

Compare the two response times:

\[ \frac{3}{30} = \frac{1}{10} \]

A warning allows one tenth of the time a caution allows, which is precisely why the two categories are coded differently and treated differently. A warning is red, usually accompanied by an aural alert, and demands immediate action; a caution is amber and demands timely action. The colour coding exists so the crew knows which régime applies before reading the message text.

Answer: C

Correct answer: 93 metres

One minute of arc at the equator is one nautical mile, or 1852 metres, so one arc second is about 30.9 metres:

\[ 3 \times 30.9 \approx 93\ \mathrm{metres} \]

This is the resolution of the widely used 3 arc second terrain data. It is adequate for a terrain awareness and warning system, which needs to know the height of the ground ahead, but it is coarse relative to obstacles - a mast or a building falls between posts, which is why obstacle data is held separately as discrete points rather than in the grid.

Answer: D

Correct answer: 180 W

Multiply the per-unit demand by the number fitted:

\[ 45 \times 4 = 180\ \mathrm{W} \]

Display load matters for the electrical design and, more importantly, for what can be sustained on battery after a total generator failure. It is also a reason liquid crystal displays displaced cathode ray tubes: less power drawn means less heat to remove, and cooling in a flight deck panel is difficult and itself consumes power.

Answer: A

Correct answer: 35 feet

Convert the rate to feet per second and multiply by the response time:

\[ \frac{700}{60} = 11.7\ \mathrm{ft/s} \qquad 11.7 \times 3 = 35\ \mathrm{feet} \]

The point of the calculation is what a three second response allowance actually costs in height near the ground. On an approach, thirty-five feet is a substantial fraction of the remaining margin, which is why terrain warnings are timed to give the crew enough distance to arrest the descent, not merely enough to notice it.

Answer: B

Correct answer: 1060 feet per minute

Convert the speed to feet per minute and take the vertical component:

\[ 150\ \mathrm{kt} = 150 \times \frac{6076}{60} = 15\,190\ \mathrm{ft/min} \]

\[ ROD = 15\,190 \times \sin 4^\circ = 15\,190 \times 0.0698 = 1060\ \mathrm{ft/min} \]

This is why the flight path vector is so useful: its position relative to the horizon is the flight path angle, so the pilot reads the actual trajectory directly instead of inferring it from pitch attitude and vertical speed separately.