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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

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