# EXERCISE 34 – AZIMUTH STAR (Numerical Solution)

###### GMT     05 March 22h  35m  16s

GHA Ȣ(05d 22h)            134˚  01.0’                                           Dec         N  08˚  53.2’
Incr. (35m 16s)             008˚  50.4’
GHAȢ                              142˚  51.4’
SHA  *                          (+)062˚  12.4’
GHA *                            205˚  03.8’
Long (E)                     (+)083˚  46.0’
LHA  *                            288˚  49.8’

###### NOTE:

Working is very similar to that of azimuth- Sun, except that GHA Ȣ and SHA * have been used, which can be taken from the nautical almanac.

P = (360˚ – LHA)
= (360˚ – 288˚ 49.8’)
= 71˚  10.2’

We know that :

###### Azimuth = N 73.8˚ E

T Az          = 073.8˚ (T)
C Az          = 078.0˚ (C)
Error         = 4.2˚ W
Variation  = 3.0˚ W
Deviation = 1.2˚ W

1. ###### On 30th Nov 2008, PM at ship in DR 48˚ 57’ N 173˚ 18’ W, the azimuth of the star VEGA was 296˚(C) when the GPS clock showed 07h 39m 22s. If variation was 1˚E, calculate the deviation of the compass.

d       h      m      s
GMT                        01    07      39    22
LIT (W)                    (-)     11      33    12
LMT                         30    20    06     10

#### Vikrant_sharma

• Rohan singh says:

The direction of azimuth .. don’t you think should be N34.4W. it’s written east..in question no. 5 of azimuth star. Please let me know if it’s correct.

• Francis says:

Good day Sir,

Just a short question, how do we name every deviation obtained from Azimuth Sun sailing? How do we know its east or west? Thank you

• Pallav says:

I found a very usefull.
Keep it up
Sir!!

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