EXERCISE 28 — AZIMUTH SUN Numerical Solutions
2. On 22nd Sept 2008, PM at ship in DR 18˚ 20’ N 085˚ 40’ E, the azimuth of the sun was 265˚(C) when the GPS clock showed 10h 09m 38s. If variation was 2˚W, calculate the deviation of the compass.
d h m s
GMT 22 10 09 38
LIT (E) 05 42 40
LMT 22 15 52 18
GMT 22 Sept 10h 09m 38s
GHA (22d 10h) 331˚ 51.4’ Dec N 00˚ 05.6’
Incr. (09m 38s) 002˚ 24.5’ d(-1.0) 00.2’
GHA 334˚ 15.9’ Dec N 00˚ 05.4’
Long (E) (+)085˚ 40.0’
LHA 059˚ 55.9’ Lat 18˚ 20’ N
P = LHA
= 59˚ 55.9’
We know that :
Azimuth = S 79.7˚W
T Az = 259.8˚ (T)
C Az = 265.0˚ (C)
Error = 5.2˚ W
Variation = 2.0˚ W
Deviation = 3.2˚ W