We know that :
P = 55˚ 00.8’
Since, sight is AFTER mer. Pass. ,so LHA = P
LHA = 055˚ 00.8’
GHA = 162˚ 18.5’
Long. W = 107˚ 17.7’
Obs. Long. = 107˚ 17.7’ W
Azimuth = N 81.5˚ W
T Az = 278.5˚ (T)
Hence ,LOP = ( TAz ± 90 )
= 008.5˚ – 188.5˚
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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 longitude where it cuts the DR latitude.
GMT 22 Sept 12h 51m 32s
GHA Ȣ(30d 12h) 180˚ 44.2’ Dec N 19˚ 08.3’
Incr. (51m 32s) 012˚ 54.6’ Lat 60˚ 10’ N
GHA Ȣ 193˚ 38.8’
SHA * (+) 145˚ 59.2’
GHA * 339˚ 38.0’
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’
We know that :
