**Compass Errors by Sun**

**At 1540 LMT, on 14th August 1991, the officer of the watch of a vessel in position 27°18’S 038°47’W observed the sun to check compass errors,**

**He obtained the following bearing: 321.5°C, 303° G. If variation is 23°W find each of the following: **

**a) The gyro compass error.**

**b) The error of the magnetic compass.**

**c) The deviation in the direction of the ship’s head.**

LMT = 15h 40m 00s

LIT (**038°47’W**) = 02h 35m 08s

GMT = 18h 15m 08s (14th august)

GHA of the sun for 18h = 088**°** 49.6′

Increment for 15m 08s = 003**°** 47.0′

GHA of the sun = 092**°** 36.6′

Long = (-) **038°47′** W

**LHA of the sun = 53° 49.6′**

Dec = 14**°** 21.2′ N

d corr. (-0.8)= (-) 0.2′

Dec = 14**°** 21′ N

Now,

A = tan lat / tan P = tan 27**°** 18′ / tan 53**°** 49.6′ = 0.37738 N

B = tan dec / sin P = tan 14**°** 21′ / sin 53**°** 49.6′ = 0.3169 N

C = A + B = 0.6943 N

tan AZ = 1 / c * cos lat = 1 / (0.6943 * cos 27**°** 18′)

AZ = N 58.3**°** W (**IF LHA 0° – 180°, body W**)

AZ = 301.7**°** (T)

We Know, C – D – M – V – T

321.5**°** – ? – ? – 23**°** – 301.7**°**

321.5**°** – ? – 324.7**°** – 23**°** W- 301.7**°**

321.5**°** – 3.2 E – 324.7**°** – 23**°**W – 301.7**°**

So, deviation is = 3.2**°** E

Compass error = 19.8 W

a) Gyro compass error = 303**°** – 301.7**°** = 1.3 (H)

b) Magnetic compass error = 19.8**°** W

c) Deviation = 3.2**°** E