On 16 June your 0430 zone time DR position...

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Question 1 of 2066636eea28f7522a1c51ab1a

On 16 June your 0430 zone time DR position is LAT 29° 24.0' S, LONG 36° 16.0' E. At that time, you observe Vega bearing 341.0° psc. The chronometer reads 02h 32m 06s, and the chronometer error is 01m 54s fast. The variation is 20.5° W. What is the deviation?

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🔍 Key Concepts

• Local hour angle (LHA) calculation using GHA, longitude, and chronometer time with error applied • Relationship between psc (per standard compass), true bearing, variation, and deviation: T–V–M–D–C (+W, –E) • Using an observed celestial bearing to solve for compass deviation at that heading

💭 Think About

• First, convert the chronometer reading and error into the correct UTC time of the observation. How does a chronometer that is "fast" affect the true time? • From the correct UTC, determine the GHA and declination of Vega, then compute the true azimuth (Zn). Compare: is this the true bearing or something else? • Once you know the true bearing of Vega and the observed psc bearing, carefully step through T–V–M–D–C to isolate the deviation on that heading. In what direction (E or W) must deviation act to turn the true bearing into the observed compass bearing?

✅ Before You Answer

• Be sure you apply the chronometer error with the correct sign (fast vs slow) when finding UTC • Keep track of east vs west longitude when forming LHA; a sign slip here will flip the azimuth • When using the T–V–M–D–C relationship, double-check that you are moving in the correct direction along the chain (from True to Compass, not the other way around) before deciding on E or W deviation

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