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On 22nd Sept 2008, PM at ship in DR 60˚ 10’N 092˚ 27’ E, the sextant altitude of the star ARCTURUS was 25˚ 01.0’ when the chron (error 05m 01s SLOW) showed 00h 46m 31s. If IE was 0.2’ on the arc & HE was 17m, find the direction of the LOP and the intercept.
GMT 22 Sept 12h 51m 32s
GHA Ȣ(30d 12h) 181˚ 43.3’ Dec N 19˚ 08.3’
Incr. (51m 32s) 012˚ 55.1’ Lat 60˚ 10’ N
GHA Ȣ 194˚ 38.4’
SHA * (+) 145˚ 59.2’
GHA * 340˚ 37.6’
Long (E) (+)092˚ 27.0’
LHA * 073˚ 04.6’
P = LHA
= 73˚ 04.6’
We know that :
Cos CZD = ( Cos P × Cos Dec × Cos Lat) + ( Sin Lat × Sin Dec )
Cos CZD = ( Cos 73˚ 04.6’ × Cos 19˚ 08.3’ × Cos 60˚ 10’ ) + ( Sin 60˚ 10’ × Sin 19˚ 08.3’ )
CZD = 65˚ 05.3’
Sext Alt 25˚ 01.0’
IE (ON) (-) 00.2’
Observed Alt 25˚ 00.8
Dip (HE 17m) (-) 07.3’
App Alt 24˚ 53.5’
T Corrn. (-)02.1’
T Alt 24˚ 51.4’
TZD 65˚ 08.6’
TZD = 65˚ 08.6’
CZD = 65˚ 05.3’
Intercept = 03.3’ AWAY
We know that :
Azimuth = S 85.2˚ W
T Az = 265.2˚ (T)
LOP = 175.2˚ – 355.2˚
