Directional Gyro Indicator (DGI) — Page 169, Lesson 199

Let’s pick this up with the effect of latitude change on a compensated DGI. We already know the apparent drift rate due to the earth’s rotation varies with the sine of the latitude. Now, imagine an aircraft tracking due north, starting from the equator. At the equator, the initial apparent drift rate of an uncorrected gyro is zero. As the flight progresses, the reading of the DGI decreases. By the time the aircraft reaches 30°N, the DGI reading is decreasing at a rate of 7½° per hour. At 60°N, it is decreasing at 13° per hour. So, in flight due north or south from the equator at constant ground speed, the value of the apparent drift rate increases from zero to a value of 15 sin lat °/h at the pole. The rate of increase of drift rate is not linear — because of the sine function. The same applies if a compensated gyro is transported north or south of its latitude of correction. Now, errors due to unstable rotor rpm. Since the rate of precession of a gyro depends on rotor rpm, over which no precise control is maintained in a suction-driven DGI, the latitude nut compensation is only approximate. For instance, at high altitude with inadequate suction, the rotor rpm will be lower than the design value. This results in reduced gyroscopic rigidity, and the latitude nut produces too high a precession rate, so it over-corrects the apparent drift. Should rpm exceed the design figure — which is less likely to occur — the rigidity would increase, and the latitude nut would produce a lower rate of precession, so under-correcting the apparent drift. Next, transport wander. At any latitude other than the equator, meridians — which define local north — are not parallel. If the gyro is aligned to one meridian, then flown east to west, the new meridian will be inclined to the old by transport wander. Now, drift rate calculations. Example 1: an aircraft is stationary at 60°N. Calculate the hourly wander rate for an uncompensated gyro. Solution: apparent wander = -15 × sin 60° (decreasing) °/h = -12.99°/h. Finally, a quick check on the spin axis of a directional gyro. It is maintained in the yawing plane by means of air jets in an air-driven gyro, and by means of a torque motor in an electrically driven gyro. So the correct answer is option c: the yawing plane; air jets; torque motor.