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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
NOTE:
NAMING OF INTERCEPT: If the TZD is larger then intercept is named AWAY, if smaller than CZD then it is named TOWARDS.
We know that :
Azimuth = N 65.8˚ E
T Az= 065.8˚ (T)
LOP = 155.8˚ – 335.8˚
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On 22nd Sept 2008, PM at ship in DR 48˚ 20’N 085˚ 40’E, the sextant altitude of the sun’s UL was 20˚ 04.9’ when the GPS clock showed 10h 09m 38s. If IE was 2.2’ on the arc & HE was 25m, find the intercept & the direction of the LOP.
d h m s
GMT 22 10 09 38
LIT (E) (+) 05 42 40
LMT 22 15 52 18
GMT – 22 Sept 10h 09m 38s
GHA (22d 10h) 331˚ 51.4’ Dec N 00˚ 05.6’
Incr. (09m 38s) 002˚ 24.5’ d(-1.0)00.2’
GHA 334˚ 15.9’ Dec N 00˚ 05.4’
Long (E) (+) 085˚ 40.0’
LHA 059˚ 55.9’ Lat 48˚ 20’ N
P = LHA
= 59˚ 55.9’
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 = ( Cos59˚ 55.9’ × Cos 00˚ 05.4’ × Cos 48˚ 20’ ) + ( Sin 48˚ 20’ × Sin 00˚ 05.4’)
CZD = 70˚ 28.9’
Sext Alt 20˚ 04.9’
IE (on) (-) 02.2’
Observed Alt 20˚ 02.7’
Dip (HE 25m) (-) 08.8’
App Alt 19˚ 53.9’
T Corrn. UL (-)18.4’
T Alt 19˚ 35.5’
TZD 70˚ 24.5’
TZD = 70˚ 24.5’
CZD = 70˚ 28.9’
Intercept = 04.2’ TOWARDS
Azimuth = S 66.7˚ W
T Az= 246.7˚ (T)
LOP = 156.7˚ – 336.7˚
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On 19th Jan 2008, at about 1530 at ship in DR 40˚ 16’S 175˚ 31’E, the sextant altitude of the sun’s LL was 43˚ 27.4’ when the chron (error 02m 12s FAST) showed 03h 50m 12s. If IE was 1.5’ on the arc & HE was 22m, find the intercept & the direction of the LOP.
GMT 19 Jan 03h 48m 00s
GHA (19d 03h) 222˚ 23.0’ Dec S 20˚ 28.9’
Incr. (48m 00s) 012˚ 00.0’ d(-0.4) 00.0’
GHA 234˚ 23.0’ Dec S 20˚ 28.9’
Long (E) 175˚ 31.0’
LHA 049˚ 54.0’ Lat 40˚ 16’ S
P = LHA
= 49˚ 54.0’
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 (-).
Cos CZD = ( Cos P × Cos Dec × Cos Lat) + ( Sin Lat × Sin Dec )
Cos CZD = ( Cos49˚ 54.0’ × Cos 20˚ 28.9’ × Cos 40˚ 16’ S) + ( Sin 40˚ 16’ S × Sin 20˚ 28.9’)
CZD =46˚ 38.9’
Sext Alt 43˚ 27.4’
IE (on) (-) 01.5’
Observed Alt 43˚ 25.9’
Dip (HE 22m) (-) 08.3’
App Alt 43˚ 17.6’
T Corrn. LL (+)15.2’
T Alt 43˚ 32.8’
TZD 46˚ 27.2’
TZD = 46˚ 27.2’
CZD = 46˚ 38.9’
Intercept = 11.7’ TOWARDS
Azimuth = N 80.3˚ W
T Az= 279.7˚ (T)
LOP = 009.7˚ – 189.7˚
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On 30th April 2008, in DR 00˚ 20’N 060˚ 12’W, the sextant altitude of the sun’s UL East of the Meridian was 44˚ 13.4’ when the GPS clock showed 13h 00m 52s. If IE was 3.1’ off the arc & HE was 20m, find the intercept & the direction of the LOP.
d h m s
GMT 30 13 00 52
LIT (W) (-) 04 00 48
LMT 30 09 00 04
GMT 30 April 13h 00m 52s
GHA (30d 13h) 015˚ 42.7’ Dec N 14˚ 59.3’
Incr. (00m 52s) 000˚ 13.0’ d(+0.7) 00.0’
GHA 015˚ 55.7’ Dec N 14˚ 59.3’
Long (W) (-)060˚ 12.0’
LHA 315˚ 43.7’ Lat 00˚ 20’ N
P = (360˚ – LHA)
= ( 360˚ – 315˚ 43.7’)
= 44˚ 16.3’
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 = ( Cos44˚ 16.3’ × Cos 14˚ 59.3’ × Cos 00˚ 20’ N ) + ( Sin 00˚ 20’ N × Sin 14˚ 59.3’ )
CZD = 46˚ 07.1’
Sext Alt 44˚ 13.4’
IE (off) (+) 03.1’
Observed Alt 44˚ 16.5’
Dip (HE 20m) (-) 07.9’
App Alt 44˚ 08.6’
T Corrn. UL (-)16.8’
T Alt 43˚ 51.8’
TZD 46˚ 08.2’
TZD = 46˚ 08.2’
CZD = 46˚ 07.1’
Intercept = 01.1’ AWAY
We know that :
Azimuth = N 69.31˚ E
T Az= 069.31˚ (T)
LOP = 159.3˚ – 339.3˚
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On 31st August 2008, PM at ship in DR 10˚ 15’S 000˚ 00’, the sextant altitude of the sun’s LL was 34˚ 54.0’ when the chron (error 01m 20s FAST) showed 03h 11m 30s. If IE was 1.5’ on the arc & HE was 17m, find the intercept & the direction of the LOP.
GMT 31 August 15h 10m 10s
GHA (31d 15h) 044˚ 58.0’ Dec N 08˚ 21.4’
Incr. (10m 10s) 002˚ 32.5’ d(-0.9) 00.2’
GHA 046˚ 30.5’ Dec N 08˚ 21.2’
Long 000˚ 00.0’
LHA 046˚ 30.5’ Lat 10˚ 15’ S
P = LHA
= 46˚ 30.5’
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 = ( Cos46˚ 30.5’ × Cos 08˚ 21.2’ × Cos 10˚ 15’ )–( Sin 10˚ 15’ × Sin 08˚ 21.2’ )
CZD = 54˚ 04.3’
Sext Alt 34˚ 54.0’
IE (on) (-) 01.5’
Observed Alt 34˚ 52.5’
Dip (HE 17m) (-) 07.3’
App Alt 34˚ 45.2’
T Corrn. LL (+)14.6’
T Alt 34˚ 59.8’
TZD 54˚ 00.1’
TZD = 54˚ 00.1’
CZD = 54˚ 04.3’
Intercept = 04.2’ TOWARDS
We know that :
Azimuth = N 69.2˚ W
T Az= 301.8˚ (T)
LOP = 031.8˚ – 211.8˚
I guess in sum no. 2 the CZD is 70d 28.4m and not 70d 28.9m
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°