Gyroscopic Instruments Questions and Answers

Answer: A

Correct answer: 13.0 degrees per hour

Earth rate apparent drift is

\[ 15\sin\phi = 15 \times \sin 60^\circ = 15 \times 0.866 = 13.0\ \mathrm{deg/h} \]

the 15 being the earth's rotation in degrees per hour. The sine dependence means the drift vanishes at the equator, where the axis and the local vertical are perpendicular, and reaches its full 15 degrees per hour at the poles - which is why high-latitude operation demands more frequent realignment.

Answer: B

Correct answer: 0.04 rad/s

Precession rate is the applied torque divided by the angular momentum:

\[ \Omega = \frac{T}{I\omega} = \frac{0.02}{0.5} = 0.04\ \mathrm{rad/s} \]

The inverse dependence on angular momentum is what makes a fast, heavy rotor a good reference: the same disturbing torque produces less precession. It is also why a gyro whose rotor is slowing becomes progressively less reliable, drifting faster as its momentum falls.

Answer: C

Correct answer: 25 degrees

Apply the rule of thumb directly:

\[ \frac{180}{10} + 7 = 18 + 7 = 25^\circ \]

The approximation works because the bank needed for a given rate of turn rises with speed, and the linear form fits well over the range of light aircraft speeds. It shows why a standard rate turn becomes impractical at high speed: at 400 knots the formula gives 47 degrees, which is why airliners use half standard rate above about 250 knots.

Answer: D

Correct answer: 90 seconds

A standard rate turn is 3 degrees per second, so

\[ \frac{270}{3} = 90\ \mathrm{s} \]

The rate is chosen so that a full 360-degree turn takes two minutes, which makes mental timing straightforward - a quarter turn in 30 seconds, a half in 60. That convenience is the reason the standard rate is defined at all, and it underlies the timed procedure turns of instrument flying.

Answer: A

Correct answer: 3 degrees

Multiply the rate by the elapsed time in hours:

\[ 4 \times \frac{45}{60} = 4 \times 0.75 = 3^\circ \]

Real drift accumulates steadily and cannot be predicted from position, unlike apparent drift, so it can only be removed by realigning against the magnetic compass. Three degrees over three quarters of an hour is why the realignment interval is conventionally about fifteen minutes.

Answer: B

The sine of zero latitude is zero, so a horizontal-axis gyro sees no apparent drift in azimuth there. The effect grows with latitude and is greatest at the poles, where the local vertical coincides with the earth's axis.

Answer: C

Correct answer: \( 2.0 \ \mathrm{kg\,m^2/s} \)

Angular momentum is the product of moment of inertia and angular velocity:

\[ H = I\omega = 0.002 \times 1000 = 2.0\ \mathrm{kg\,m^2/s} \]

It is the quantity that determines both rigidity and precession rate: a larger \( H \) means the same disturbing torque produces proportionately less precession. Designers therefore spin the rotor as fast as the bearings allow and put the mass at the rim, where it contributes most to \( I \).

Answer: D

Correct answer: Half standard rate

Standard rate is 3 degrees per second, so 1.5 is exactly half.

The half rate is the normal choice for large aircraft: the bank angle for a standard rate turn grows with speed, and above about 250 knots it becomes uncomfortably steep. A half standard rate turn takes four minutes for the full circle rather than two, which is why holding patterns and procedure turns specify which rate is intended.

Answer: A

Correct answer: 240 degrees per hour

Scale the observed movement up to an hour:

\[ \frac{2^\circ}{30\ \mathrm{s}} \times 3600 = 240\ \mathrm{deg/h} \]

That is a very high rate for an instrument gyro, where real drift is measured in single degrees per hour, so it indicates a substantial applied torque rather than ordinary friction. The arithmetic is a units conversion, and the trap is dividing rather than multiplying by the number of seconds in an hour.

Answer: B

Correct answer: 4.2 degrees per hour

Transport wander is the rate of change of longitude multiplied by the sine of the latitude:

\[ 6 \times \sin 45^\circ = 6 \times 0.707 = 4.24\ \mathrm{deg/h} \]

It arises because moving east or west changes the direction of true north relative to a space-fixed axis, by an amount that depends on the convergence of the meridians. The same sine of latitude appears as in earth rate, which is why the two combine so simply.