On 13 October your vessel's 1722 zone time...

Question 1 / 2066636eea28f7522a1c51aa9a

Question 1 of 2066636eea28f7522a1c51aa9a

On 13 October your vessel's 1722 zone time DR position is LAT 27°36'S, LONG 136°16'E, when an amplitude of the Sun is observed. The Sun's center is on the celestial horizon and bears 266° per standard magnetic compass. Variation in the area is 2°W. The chronometer reads 08h 24m 19s and is 01m 43s fast. What is the deviation of the standard magnetic compass?

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🔍 Key Concepts

• Amplitude of the Sun formula and how it relates to latitude and declination • Converting between true bearing, magnetic bearing, and compass bearing using variation and deviation • How signs in southern latitude and south or north declination affect the computed amplitude direction (N or S of E/W)

💭 Think About

• From the DR latitude and the Sun’s declination for 13 October at the correct GMT, what is the Sun’s true amplitude (N or S of E/W)? • Once you know the true azimuth of the Sun at the time of amplitude, how do you step it through variation and then deviation to arrive at the observed compass bearing of 266°? • Does a westerly variation make magnetic north lie to the west or east of true north, and how does that affect the sign of deviation you must solve for?

✅ Before You Answer

• Make sure you compute correct GMT from the chronometer reading and its error before taking declination from the almanac • Check that you are using the correct amplitude formula: ( \sin A = \frac{\sin \text{dec}}{\cos \text{lat}} ), with proper signs for south latitude/declination • Carefully apply the relationship (True ± Variation = Magnetic; Magnetic ± Deviation = Compass) and track each sign step to see whether the resulting deviation is east or west

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Question 1 / 2066636eea28f7522a1c51aa9a

Question 1 of 2066636eea28f7522a1c51aa9a

Back

Question 1 of 20

66636eea28f7522a1c51aa9a

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🔍 Key Concepts

💭 Think About

✅ Before You Answer

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