On 24 August in DR position LAT 26° 49.4' N,...

Question 1 / 2066636eea28f7522a1c51ad85

Question 1 of 2066636eea28f7522a1c51ad85

On 24 August in DR position LAT 26° 49.4' N, LONG 146° 19.4' E, you observe an amplitude of the Sun. The Sun's center is on the celestial horizon and bears 084° psc. The chronometer reads 07h 55m 06s and is 01m 11s fast. Variation in the area is 15° W. What is the deviation of the magnetic compass?

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🔍 Key Concepts

• Amplitude of the Sun to find true bearing of the Sun at the horizon • Relationship between true bearing, variation, deviation, and compass error • How to apply chronometer correction and longitude/Greenwich Hour Angle to get the correct true bearing at the observed time

💭 Think About

• From the DR position, date, and corrected time, think about how you find the Sun’s true azimuth (true bearing) at the instant its center is on the celestial horizon. • Compare the calculated true bearing of the Sun with the compass bearing (084° psc) to find the total compass error. • Once you know total compass error and you are given variation 15° W, reason out what the deviation must be, and whether it is east or west.

✅ Before You Answer

• Be sure you are using the correct sign for the chronometer correction (the chronometer is 01m 11s fast). • Confirm you are applying variation (15° W) with the right sign convention when moving between true and magnetic north. • Check whether the Sun’s true bearing you compute should be slightly more or less than 084° at that location and date; that will help you decide if the deviation is east or west.

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Question 1 / 2066636eea28f7522a1c51ad85

Question 1 of 2066636eea28f7522a1c51ad85

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Question 1 of 20

66636eea28f7522a1c51ad85

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🔍 Key Concepts

💭 Think About

✅ Before You Answer

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