Flight Data Recording Questions and Answers

Answer: A

Correct answer: 1 Hz

One appearance per one-second subframe is one sample per second, which is 1 Hz by definition.

The subframe is the basic unit of the recording format: parameters that change slowly appear once per frame of four subframes, at 0.25 Hz, while rapidly varying ones such as control positions appear several times within each subframe. Matching the rate to how fast the quantity can meaningfully change is what keeps the data stream within its bandwidth.

Answer: B

Correct answer: 0.25 Hz

A frame of four one-second subframes lasts four seconds, so one sample per frame is

\[ \frac{1}{4} = 0.25\ \mathrm{Hz} \]

- one sample every four seconds. That is the slowest rate the format offers, and it is reserved for parameters that cannot change meaningfully in the interval: fuel quantity, configuration discretes, and identifiers that hardly change at all. Sampling them faster would consume bandwidth needed by the fast parameters.

Answer: C

Correct answer: 3072 bits per second

Multiply the word rate by the word length:

\[ 256 \times 12 = 3072\ \mathrm{bit/s} \]

This is the standard ARINC 717 rate at 256 words per second, and it is modest by any modern measure - which is why solid-state memory removed the capacity constraint entirely. The format also runs at 64, 128, 512 and 1024 words per second, the higher rates carrying more parameters or the same ones more often.

Answer: D

Correct answer: 276 megabits

Multiply the rate by the number of seconds:

\[ 25 \times 3600 = 90\,000\ \mathrm{s} \]

\[ 3072 \times 90\,000 = 2.765 \times 10^{8}\ \mathrm{bits} \approx 276\ \mathrm{Mbit} \]

That is about 35 megabytes for 25 hours of flight data - a figure that made magnetic tape a genuine constraint and makes solid-state memory trivially adequate. The capacity limit that once shaped the parameter list has simply disappeared.

Answer: A

Correct answer: Already overwritten

A two-hour loop holds the most recent two hours and nothing before it, so material from two hours and ten minutes back has been written over by the ten minutes that followed.

This is exactly why the circuit breaker is pulled promptly after a serious incident: every minute the recorder keeps running destroys a minute of what may be the relevant evidence. The data recorder holds parameters rather than audio, so it cannot supply what the voice recorder has lost.

Answer: B

Correct answer: 18 days

The beacon runs 30 days from immersion and 12 have passed:

\[ 30 - 12 = 18\ \text{days} \]

The arithmetic is the whole of the operational problem. Locating a site in deep water can take longer than the transit alone, and once the battery expires the acoustic search ends and the far slower sonar mapping begins. That is the experience which drove the 90-day beacons now fitted to some installations.

Answer: C

Correct answer: 720 000

Multiply the rate by the total seconds:

\[ 8 \times 25 \times 3600 = 8 \times 90\,000 = 720\,000 \]

Three quarters of a million samples for one parameter, and a modern recorder holds hundreds of parameters - which is why the total volume is what matters rather than any single trace. The distractor 72 000 loses a factor of ten and 7 200 000 gains one; checking against the 90 000 seconds settles it.

Answer: D

Correct answer: 176 samples per second

Multiply the parameter count by the average rate:

\[ 88 \times 2 = 176\ \mathrm{samples/s} \]

The figure is what must fit within the format's word rate, so it is the constraint the frame layout is built to satisfy. Averaging conceals a wide spread: control positions may be sampled eight times a second while fuel quantity appears once every four, and the layout is a matter of fitting the fast parameters in without starving the slow ones.

Answer: A

Correct answer: One word, with one bit unused

An 11-bit parameter fits inside a 12-bit word, leaving one bit spare. Splitting it across two words would be possible but pointless, since it fits.

The spare bit is not necessarily wasted: formats commonly use such bits for sign, validity flags or a companion discrete. The word is the unit of allocation in the frame, so a parameter needing 13 bits would require two words - a step change in cost that shapes how resolution is chosen.

Answer: B

Swath width scales with range, so coverage falls as \( 1.5/2 = 0.75 \), a reduction of 25 per cent. If area rather than swath is considered the reduction is larger, which is why detection conditions matter so much to search planning.