On 16 April in DR position LAT 28° 07.0' N,...

Question 1 / 2066636eea28f7522a1c51ab06

Question 1 of 2066636eea28f7522a1c51ab06

On 16 April in DR position LAT 28° 07.0' N, LONG 81° 47.0' W, you observe an amplitude of the Sun. The Sun's center is on the visible horizon and bears 073.5° psc. The chronometer reads 10h 53m 41s and is 02m 23s slow. Variation in the area is 11° E. What is the deviation of the magnetic compass?

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🔍 Key Concepts

• Amplitude of the Sun and how it relates to compass error • Correcting the observed bearing: psc → per gyro/true using deviation, variation, and any corrections • Difference between deviation and variation and which one is local to your compass

💭 Think About

• Start from the observed compass bearing (073.5° psc). What corrections must you apply to reach a true bearing from an amplitude? • How do you use the date (16 April) and DR position to get the Sun’s true amplitude or true bearing at the time of observation? • Once you have both the true bearing and the compass bearing of the Sun, how do you find total compass error, and then separate variation from deviation?

✅ Before You Answer

• Be sure you are using the correct sextant/altitude correction tables for an amplitude on the visible horizon, not for a sight above the horizon. • Verify that the sign and order of applying variation and deviation are consistent (think: True ± Variation = Magnetic; Magnetic ± Deviation = Compass). • Confirm that you are using the correct west/east sign convention when extracting deviation from total compass error and known variation.

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Question 1 of 20

4.5° E

4.9° W

6.1° E

6.5° W

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Question 1 / 2066636eea28f7522a1c51ab06

Question 1 of 2066636eea28f7522a1c51ab06

Back

Question 1 of 20

66636eea28f7522a1c51ab06

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🔍 Key Concepts

💭 Think About

✅ Before You Answer

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