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On 5th March 2008, AM at ship in DR 38˚ 11’S 151˚ 10’E, the sextant altitude of the sun’s LL was 35˚ 59.1’ when the chron (error 00m 46s SLOW) showed 10h 54m 54s. If IE was 1.3’ off the arc & HE was 30m, find the intercept & the direction of the LOP.
GMT – 04 March 22h 55m 40s
GHA (04d 22h) 147˚ 06.9’ Dec S 06˚ 01.3’
Incr. (55m 40s) 013˚ 55.0’ d(-1.0) 00.9’
GHA 161˚ 01.9’ Dec S 06˚ 00.4’
Long (E) (+) 151˚ 10.0’
LHA 312˚ 11.9’ Lat 38˚ 11’ S
P = (360˚ – LHA)
= ( 360˚ – 312˚ 11.9’)
= 47˚ 48.1’
NOTE:
- While using below formula, one must be careful about + signs.
- If Lat and Dec are of same name then sign is (+).
- If of contrary names then it has to be (-).
We know that :
Cos CZD = ( Cos P × Cos Dec × Cos Lat) + ( Sin Lat × Sin Dec )
Cos CZD = ( Cos 47˚ 48.1’ × Cos 06˚ 00.4’ × Cos 38˚ 11’ ) + ( Sin 38˚ 11’ × Sin 06˚ 0.4’ )
CZD = 53˚ 51.7’
Sext Alt 35˚ 59.1’
IE (off) (+) 01.3’
Observed Alt 36˚ 00.4’
Dip (HE 30m) (-) 09.6’
App Alt 35˚ 50.8’
T Corrn. LL (+)14.9’
T Alt 36˚ 05.7’
TZD 53˚ 54.3’
TZD = 53˚ 54.3’
CZD = 53˚ 51.7’
Intercept = 02.6’ AWAY
I think your example 5 is incorrect. the LHA is 46 31.5 but in your ABC calculations you have used 47 31.5. also your Tan AZ of 2.705 is incorrect.
I believe the correct AZ is 290.2 degrees
I believe correct LHA is 47° 30.5′
My intercept is coming- 10.5° towards
True Az- 289.5°