On 21 February your 0823 zone time DR...

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Question 1 of 2066636eea28f7522a1c51ac8b

On 21 February your 0823 zone time DR position is LAT 21° 44.0' S, LONG 80° 14.0' E. At that time, you observe the Sun bearing 096° psc. The chronometer reads 03h 25m 19s, and the chronometer error is 01m 52s fast. The variation is 5° W. What is the deviation of the standard magnetic compass?

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🔍 Key Concepts

• Using an amplitude/azimuth of the Sun to find compass error • Time conversions with a chronometer error (fast/slow) to get GMT for Nautical Almanac data • The relationship T – V – D = C (True, Variation, Deviation, Compass) and correct sign conventions for E/W

💭 Think About

• How do you convert your chronometer reading and chronometer error (fast) into the correct GMT to enter the Nautical Almanac? • Once you have GHA and declination of the Sun, how do you compute the true bearing (true azimuth) of the Sun from your DR position? • You are given a compass bearing of the Sun (psc). How do you compare that to the true bearing using TVMDC to isolate the deviation?

✅ Before You Answer

• Make sure you have the correct GMT: apply the chronometer error in the right direction for a fast chronometer. • Check that you are using the right order: True → Magnetic → Compass (or reverse) and keeping W variation/deviation with the correct sign. • Verify that you apply the local variation (5° W) only once and do not confuse it with deviation, which you are solving for.

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