Attitude and Heading Reference Systems Questions and Answers

Answer: A

Correct answer: 3 degrees

A constant bias integrates linearly with time. Five minutes is \( 300 \) seconds, so

\[ \Delta\theta = 0.01\ \mathrm{deg/s} \times 300\ \mathrm{s} = 3\ \mathrm{degrees} \]

The result shows why unaided integration is untenable: a bias small enough to be hard to measure produces an unacceptable attitude error within minutes. This is exactly the growth the accelerometer-derived vertical reference is there to bound, and why a Kalman filter estimates the bias as a state rather than trusting a factory calibration to hold.

Answer: B

Correct answer: 0.2 degrees

Angle random walk accumulates as the square root of elapsed time, and the coefficient is quoted in degrees per root hour:

\[ \sigma = 0.1 \times \sqrt{4} = 0.1 \times 2 = 0.2\ \mathrm{degrees} \]

The value \( 0.4 \) would follow from growth linear in time, which is the bias law, not the random walk law. The contrast is instructive: over four hours, random walk contributes only a fifth of a degree, whereas a bias of \( 0.5 \) degrees per hour would contribute two degrees. Bias, not noise, dominates the long-term error.

Answer: C

Correct answer: 11.5 degrees

The accelerometers sense the vector sum of the reaction to gravity and the specific force due to the aircraft's own acceleration. The apparent vertical therefore tilts by

\[ \theta = \arctan\left(\frac{a}{g}\right) = \arctan\left(\frac{2}{9.81}\right) = \arctan(0.2039) = 11.5\ \mathrm{degrees} \]

The option \( 2.0 \) mistakes the acceleration for the angle. This tilt is the acceleration-versus-gravity ambiguity in concrete form: an accelerometer alone would read a \( 11.5 \) degree pitch-up that is not there, which is why the vertical reference must be weighted down during manoeuvres.

Answer: D

Correct answer: 11.5 degrees per hour

The earth's rotation vector points along the polar axis, so at latitude \( \phi \) its horizontal (north) component is

\[ \Omega_H = 15 \cos\phi = 15 \cos 40^\circ = 15 \times 0.766 = 11.5\ \mathrm{deg/hr} \]

The vertical component is \( 15\sin\phi = 9.6 \) degrees per hour, which is the distractor offered. The horizontal component is what gyro compassing senses to find true north, which is also why the technique degrades towards the poles: \( \cos\phi \) shrinks while the sensor noise does not.

Answer: A

Correct answer: 2.5 ms

The update interval is the reciprocal of the rate:

\[ \Delta t = \frac{1}{400\ \mathrm{Hz}} = 0.0025\ \mathrm{s} = 2.5\ \mathrm{ms} \]

Rates of this order are chosen so the integration keeps up with airframe vibration and any coning motion; too slow a loop cannot follow the oscillation and accumulates coning error. In practice the fast attitude integration often runs at several hundred hertz while the filter that blends in the accelerometer and magnetometer references updates far more slowly.

Answer: B

Correct answer: 0.25 degrees

Bias produces error growing linearly with time, and thirty minutes is half an hour:

\[ \Delta\psi = 0.5\ \mathrm{deg/hr} \times 0.5\ \mathrm{hr} = 0.25\ \mathrm{degrees} \]

Keeping the units consistent is the whole of the problem - the distractor \( 15 \) comes from treating the thirty minutes as thirty hours. Note how much better this gyro is than the one in the earlier item: \( 0.5 \) degrees per hour is roughly \( 0.00014 \) degrees per second, so the heading drifts a quarter of a degree where the other drifted three degrees in five minutes.

Answer: C

Correct answer: Raise a comparator warning

The disagreement of \( 4.5 \) degrees exceeds the \( 3 \) degree threshold, so the comparator flags it. That is the entire function of the monitor: it detects that the two units are inconsistent to a degree that cannot be attributed to normal spread.

With only two units the system cannot tell which is at fault, so it must not select one or average them - averaging would produce a heading that is wrong by half the discrepancy and, worse, would conceal the fault. Ignoring a difference beyond the threshold would defeat the monitor. Identifying the faulty unit requires a third.

Answer: D

Correct answer: 271.2 degrees

Median voting sorts the three values and takes the middle one. Sorted, the readings are \( 271.0, 271.2, 275.8 \), so the median is \( 271.2 \) degrees.

The value \( 272.7 \) is the arithmetic mean, and it illustrates why the median is preferred: the outlier at \( 275.8 \) drags the mean more than a degree and a half away from where the two healthy units agree, whereas the median is unaffected by how far the outlier lies. The same property is what allows the deviant unit to be identified and excluded.

Answer: A

Correct answer: 14.0 degrees

The heading offset from magnetic north is the angle of the horizontal field vector:

\[ \theta = \arctan\left(\frac{B_E}{B_N}\right) = \arctan\left(\frac{5}{20}\right) = \arctan(0.25) = 14.0\ \mathrm{degrees} \]

The option \( 76.0 \) is the complement, from dividing the components the other way round - a common slip, avoided by remembering that the angle is measured from north, so the north component belongs in the denominator. Only the horizontal components are used; the vertical component carries no heading information, which is why the sensor must first be levelled using the attitude solution.

Answer: B

Correct answer: 1.8 m

A constant acceleration error integrates twice into a position error growing with the square of time:

\[ \Delta x = \tfrac{1}{2} a t^2 = \tfrac{1}{2} \times 0.001 \times 60^2 = \tfrac{1}{2} \times 0.001 \times 3600 = 1.8\ \mathrm{m} \]

The distractor \( 3.6 \) omits the factor of one half, and \( 0.06 \) integrates only once, giving the velocity error instead. This quadratic growth is precisely why an attitude and heading reference system does not attempt to output position: a bias too small to matter for attitude becomes tens of metres within minutes.