On 2 January your vessel's 1948 zone time position is LAT 21°42'S, LONG 39°12'W, when an amplitude of the Sun is observed. The Sun's center is on the celestial horizon and bears 260° per standard magnetic compass. Variation in the area is 19°W. The chronometer reads 10h 44m 36s and is 03m 24s slow. What is the deviation of the standard magnetic compass?
Official USCG Answer & Rule Citation
Correct Answer: Option A — 4.3°E
Calculate the Sun's declination and LHA to find the true amplitude. Compare the true amplitude to the compass bearing (260°psc) to find the total compass error. Subtract the 19°W variation from the compass error. The remaining value is the deviation of the standard magnetic compass, calculated as 4.3°E.
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