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A directional gyro is corrected for an apparent drift due to the rotation of the earth at latitude 30°S. During a flight at latitude 60°N, a drift rate of 15,5°/h to the right is observed. The apparent wander due to change of aircraft position is:
A) 2,5°/h to the left
B) 2,5°/h to the right
C) 5°/h to the right
D) 5°/h to the left
The earth rate experienced at 30°S = 15 x sin(30) = +7,5°/h (positive/to the left because it's the Southern Hemisphere)
The latitude nut corrected this earth rate so it is giving -7,5°/h (to the right/negative)
Now we are flying at 60°N where earth rate is = 15 x sin(60) = -13°/h (Northern Hemisphere so negative)
Latitude nut is still giving the -7,5°/h correction so total = (-13) + (-7,5) = -20,5
The Question states that we observed a drift of -15,5 , the difference between that and -20,5 = +5 (positive/to the left)
A) 2,5°/h to the left
B) 2,5°/h to the right
C) 5°/h to the right
D) 5°/h to the left
The earth rate experienced at 30°S = 15 x sin(30) = +7,5°/h (positive/to the left because it's the Southern Hemisphere)
The latitude nut corrected this earth rate so it is giving -7,5°/h (to the right/negative)
Now we are flying at 60°N where earth rate is = 15 x sin(60) = -13°/h (Northern Hemisphere so negative)
Latitude nut is still giving the -7,5°/h correction so total = (-13) + (-7,5) = -20,5
The Question states that we observed a drift of -15,5 , the difference between that and -20,5 = +5 (positive/to the left)
