On 27 June your 0734 zone time DR position...

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Question 1 of 2066636eea28f7522a1c51adb9

On 27 June your 0734 zone time DR position is LAT 22°14.0'N, LONG 53°52.0'W. At that time, you observe the Sun bearing 069.5°psc. The chronometer reads 11h 32m 51s and the chronometer error is 01m 26s slow. The variation is 5°E. What is the deviation of the standard magnetic compass?

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Question 1 of 20

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🔍 Key Concepts

• Azimuth of the Sun and how it is used to find compass error • Relationship between true bearing, variation, deviation, and compass (psc) bearing • Converting zone time to GMT (UTC) using the chronometer and its error

💭 Think About

• How do you go from the observed Sun bearing in per standard compass (psc) to a true bearing, and what corrections (variation and deviation) are involved? • What steps are needed to get the Sun’s true azimuth at the time of observation (think: correct chronometer time, then use the Nautical Almanac)? • Once you have the true azimuth and your observed compass bearing, how do you combine them to get total compass error, and then separate variation and deviation?

✅ Before You Answer

• Confirm you have correctly converted chronometer time to GMT, including applying the chronometer error with the correct sign (slow vs fast). • Verify you applied zone description (ZD) correctly when relating zone time to GMT. • Double-check the sign convention when going from true to magnetic (apply variation) and from magnetic to compass (apply deviation) so you don’t reverse East/West.

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