On 13 October your vessel's 1722 zone time DR position is LAT 27°36'S, LONG 136°16'E, when an amplitude of the Sun is observed. The Sun's center is on the celestial horizon and bears 266° per standard magnetic compass. Variation in the area is 2°W. The chronometer reads 08h 24m 19s and is 01m 43s fast. What is the deviation of the standard magnetic compass?
Official USCG Answer & Rule Citation
Correct Answer: Option A — 2.8°W
Sun amplitude computation yields the true bearing by applying latitude and calculated declination. Comparing the resulting true azimuth with the standard magnetic compass bearing of 266° and applying 2°W variation derives the compass deviation of 2.8°W per Bowditch methods.
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