On 8 December in DR position LAT 21°56.1'S,...

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Question 1 of 2066636eea28f7522a1c51ae59

On 8 December in DR position LAT 21°56.1'S, LONG 17°21.6'E you observe an amplitude of the Sun. The Sun's center is on the celestial horizon and bears 240.5°psc. The chronometer reads 05h 27m 21s and is 00m 47s fast. Variation in the area is 3.3°E. What is the deviation of the standard magnetic compass?

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🔍 Key Concepts

• Amplitude of the Sun method to find compass error • Relationship between true bearing, variation, deviation, and compass bearing • Using date and DR latitude to get the Sun’s declination and then its true bearing at rising/setting

💭 Think About

• From the given date and DR position, determine the Sun’s declination from the Nautical Almanac. Is the Sun north or south of the equator on 8 December? • Use the amplitude formula (or almanac tables) with latitude and declination to find the Sun’s true bearing when its center is on the celestial horizon in the observed quadrant. • Once you have the true bearing, carefully step through the conversions: true → magnetic (apply variation) → compass (compare with the observed 240.5°psc) to isolate the deviation sign and magnitude.

✅ Before You Answer

• Be sure you are using apparent sunrise/sunset amplitude (center on celestial horizon), not an altitude sight method. • Check the sign convention: variation 3.3°E – does that make magnetic east or west of true in your working? • When you compare the computed magnetic (or true) bearing with the observed compass bearing 240.5°psc, confirm whether the compass is reading high or low to assign W or E to the deviation.

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