-
On 29th Nov 2008, AM at ship in DR 25˚ 30’S 107˚ 20’W, the sextant altitude of the star RIGEL was 35˚ 10.3’ when the GPS clock showed 11h 29m 20s. If IE was 2.8’ on the arc & HE was 12m, find the direction of the LOP and the intercept.
d h m s
GMT 29 11 29 20
LIT (W) (-) 07 09 20
LMT 29 04 20 00
GMT 29 Nov 11h 29m 20s
GHA Ȣ(29d 11h) 233˚ 42.3’ Dec S 08˚ 11.4’
Incr. (29m 20s) 007˚ 21.2’ Lat 25˚ 30’ S
GHA Ȣ 241˚ 03.5’
SHA * (+) 281˚ 15.0’
GHA * 162˚ 18.5’
Long (W) (-) 107˚ 20.0’
LHA * 054˚ 58.5’
P = LHA
= 54˚ 58.5’
Cos CZD = ( Cos P × Cos Dec × Cos Lat) + ( Sin Lat × Sin Dec )
Cos CZD = ( Cos 054˚ 58.5’ × Cos 08˚ 11.4’ × Cos 25˚ 30’ ) + ( Sin 25˚ 30’ × Sin 08˚ 11.4’ )
CZD = 54˚ 57.9’
Sext Alt 35˚ 10.3’
IE (ON) (-) 02.8’
Observed Alt 35˚ 07.5’
Dip (HE 12m) (-) 06.1’
App Alt 35˚ 01.4’
T Corrn. (-)01.4’
T Alt 35˚ 00.0’
TZD 55˚ 00.0’
TZD = 55˚ 00.0’
CZD = 54˚ 57.9’
Intercept = 2.1’ AWAY
Azimuth = N 81.9˚ W
T Az = 278.1˚ (T)
LOP = 008.1˚ – 188.1˚
