On 17 June your 0815 zone time DR position...

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Question 1 of 2066636eea28f7522a1c51ab49

On 17 June your 0815 zone time DR position is LAT 25°27.0'N, LONG 47°16.0'W. At that time, you observe the Sun bearing 079.5°psc. The chronometer reads 11h 15m 03s, and the chronometer error is 01m 15s fast. The variation is 3°E. What is the deviation of the standard magnetic compass?

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Question 1 of 20

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🔍 Key Concepts

• Compass error = difference between true bearing and compass bearing • Steps to convert between bearings: psc ↔ per gyro ↔ per magnetic ↔ per true using deviation and variation • Use chronometer time + GHA/Dec of Sun (from Nautical Almanac) to find the Sun’s true azimuth at the DR position

💭 Think About

• How do you turn the observed Sun bearing of 079.5°psc into a true bearing? List each correction step in order, even if you don’t know the numeric values yet. • Once you compute the Sun’s true azimuth for that date, time, and DR position, compare it with your compass bearing. Which side is the compass pointing relative to true? • From the sign and size of the total compass error, how do you separate out variation (given) and deviation (unknown)?

✅ Before You Answer

• Be sure you applied the chronometer correction (chronometer fast) correctly to get the right UTC. • Confirm that you converted from zone time to GMT/UTC properly before entering the Nautical Almanac tables. • After finding total compass error, carefully apply: Compass error = Variation + Deviation (with proper E/W signs) before choosing the deviation.

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