Air Data Instruments Questions and Answers

Answer: A

Correct answer: 250 knots

True airspeed follows from equivalent airspeed by dividing by the square root of the density ratio: \[ V_T = \frac{V_E}{\sqrt{\sigma}} = \frac{200}{\sqrt{0.64}} = \frac{200}{0.8} = 250\ \text{knots.} \] Multiplying instead of dividing gives 160 knots, which is offered as a distractor. True airspeed always exceeds equivalent airspeed at altitude, because the air is thinner, so the answer must be the larger number.

Answer: B

Correct answer: 6125 Pa

\[ q = \tfrac{1}{2}\rho V^2 = 0.5 \times 1.225 \times 100^2 = 0.5 \times 1.225 \times 10\,000 = 6125\ \text{Pa.} \] Omitting the factor of one half gives 12 250 Pa, the other plausible distractor. This is the quantity the airspeed indicator actually responds to, which is why its reading depends on density as well as on speed.

Answer: C

The local pressure is 10 hectopascals below standard, so the aircraft is about 270 feet lower than the standard setting suggests. Flying from high pressure to low pressure without resetting leaves the aircraft lower than indicated.

Answer: D

Correct answer: 306 m/s

\[ a = 20.05\sqrt{T} = 20.05 \times \sqrt{233} = 20.05 \times 15.26 = 306\ \text{m/s.} \] The speed of sound depends on temperature alone, not on pressure or density, which is why it falls with altitude through the troposphere and then holds constant through the isothermal layer above the tropopause. 340 m/s is the sea level standard value at 288 K.

Answer: A

Correct answer: 0.90

\[ M = \frac{V}{a} = \frac{270}{300} = 0.90. \] The inverse, 1.11, is the trap. A useful check is that a subsonic aircraft must give a value below one, so any answer above unity can be discarded immediately unless the question says the flight is supersonic.

Answer: B

Correct answer: 238 K

Rearranging the total temperature relation for the static value: \[ T_s = \frac{T_t}{1 + 0.2M^2} = \frac{268}{1 + 0.2(0.8)^2} = \frac{268}{1.128} = 237.6 \approx 238\ \text{K.} \] The static temperature must be the smaller of the two, since bringing the air to rest at the probe heats it - here by about 30 K, which is why the raw probe reading cannot be used as ambient temperature.

Answer: C

Correct answer: 750 ft/min down

\[ \frac{3000\ \text{ft}}{4\ \text{min}} = 750\ \text{ft/min.} \] The vertical speed indicator will settle at this value once the transient has passed, but it will lag behind the start of the descent by several seconds because it works by metering air through a calibrated restriction - which is why the attitude indicator, not the vertical speed indicator, is used to initiate a change.

Answer: D

Correct answer: 456 knots

\[ V_T = \frac{V_C}{\sqrt{\sigma}} = \frac{250}{\sqrt{0.30}} = \frac{250}{0.5477} = 456\ \text{knots.} \] A density ratio of 0.30 corresponds to roughly 30 000 ft, where the true airspeed is nearly twice the indicated value - a difference large enough that navigating on the indicated figure would put the aircraft badly out of position.

Answer: A

A lower sensed static pressure is interpreted as greater height, so the instrument over-reads by about 81 feet. The sign matters far more than the magnitude here, because it decides which side of an obstacle clearance the aircraft is on.

Answer: B

Correct answer: Increases as density falls

Equivalent airspeed is a measure of dynamic pressure, so holding it constant means holding \( \rho V^2 \) constant. As the aircraft climbs the density falls, so the true airspeed must rise to keep the product the same. This is why a climb schedule flown at constant indicated airspeed accelerates the aircraft in true terms all the way up.