EXERCISE 28 — AZIMUTH SUN (Numerical Solution)

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 :

e28q2p1

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

  1. On Jan 19th 2008, in DR 40˚ 16’S 175˚ 31’ E, the azimuth of the sun was 267˚(C) at 1618 ship’s time. If the ships time difference was 11h 30m from GMT and variation was 2.3˚E, find the deviation for the ship’s head.

    d      h     m     s

Ship’s time           19     16   18   00
Diff. (1130hr)      (-)11   30   00
GMT                      19    04    48   00

GMT     19THJan 04h 48m 00s

GHA (19d 04h)         237˚  22.8’                                        Dec         S  20˚  28.4’
Incr. (48m 00s)        012˚  00.0’                                      d(-0.5)                 00.4’
GHA                          249˚  22.8’                                       Dec         N  20˚  28.0’
Long (E)                (+)175˚  31.0’
LHA                         064˚  53.8’                                          Lat            40˚  16’ S

We know that:
P = LHA
= 064˚  53.8’

e28q3p1

Azimuth = S 89.3˚W
T Az          = 289.3˚ (T)
C Az          = 287.0˚ (C)
Error         = 2.3˚ E
Variation  = 2.3˚ E
Deviation = NIL

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