On 6 August your 1552 zone time DR position is LAT 24°26.0'S, LONG 73°19.0'E. At that time, you observe the Sun bearing 302°psc. The chronometer reads 10h 55m 07s, and the chronometer error is 02m 38s fast. The variation is 6°E. What is the deviation of the standard magnetic compass?

Question

On 6 August your 1552 zone time DR position is LAT 24&deg;26.0'S, LONG 73&deg;19.0'E. At that time, you observe the Sun bearing 302&deg;psc. The chronometer reads 10h 55m 07s, and the chronometer error is 02m 38s fast. The variation is 6&deg;E. What is the deviation of the standard magnetic compass?

Accepted Answer

Option D: 6.1&deg;W. Calculate the Sun's True Azimuth (Zn) using the DR position and GMT. Convert Zn to Magnetic Azimuth by applying Variation (6°E). Subtract the Magnetic Azimuth from the Compass Bearing (302°psc). The resulting deviation is 6.1°W, confirming the compass error relative to the magnetic meridian.