EXERCISE 37- LONGITUDE BY CHRONOMETER STAR (Numerical Solution)

Obs. Long. = 040 ˚ 20.0’ W

Azimuth = N 64.7˚ W

T Az          = 295.3˚ (T)

Hence , LOP  = ( TAz  ±  90 )
LOP = 025.3˚ – 205.3˚

5. On 31^st Aug 2008, AM at ship in DR 40˚ 30’N 064˚ 56’E, the sextant altitude of the star DIPHDA was 21˚ 28.4’ when the chron (error 01m 06s FAST) showed 00h 21m 32s. If IE was 0.9’ off the arc & HE was 09m, find the direction of the LOP and the longitude where it cuts the DR latitude.

GMT     31 Aug 00h  20m  26s

GHA Ȣ(31d 00h)             339˚  32.7’                                           Dec         S  17˚  56.1’
Incr. (20m 26s)             005˚  07.3’                                           Lat            40˚  30’ N
GHA Ȣ                             344˚  40.0’
SHA *                        (+) 348˚  59.1’
GHA *                            333˚  39.1’

Sext Alt                            21˚ 28.4’
IE (OFF)                           (+) 00.9’
Observed Alt                 21˚ 29.3’
Dip (HE 09m)                  (-) 05.3’
App Alt                         21˚ 24.0’
T Corrn.                           (-)02.5’
T Alt                              21˚ 21.5’

We know that :

P = 38˚ 45.2’

Since, sight is AFTER mer. Pass. ,
So LHA = P

LHA          = 038˚ 45.2’
GHA         = 333˚ 39.1’
Long. E    = 065˚ 06.1’