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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 longitude where it crosses the DR latitude.
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’
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’
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 :
P = 73˚ 11.2’
Since, sight is AFTER mer. Pass. ,so LHA = P
LHA = 073˚ 11.2’
GHA = 339˚ 38.0’
Long. E = 093˚ 33.2’
Obs. Long. = 093˚ 33.2’ E
Azimuth = N 81.5˚ W
T Az = 278.5˚ (T)
Hence ,LOP = ( TAz ± 90 )
= 008.5˚ – 188.5˚
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On 19th Jan 2008, at about 1900 at ship in DR 00˚ 02’N 170˚ 50’E, the sextant altitude of the star BETELGUESE was 43˚ 11.1’ at 07h 33m 44s by GPS clock. If IE was 1.3’ off the arc & HE was 18m, find the direction of the LOP and the longitude where it cuts the DR latitude.
d h m s
GMT 19 07 33 44
LIT (E) (+) 11 23 20
LMT 19 18 57 04
GMT 19 Jan 07h 33m 44s
GHA Ȣ(19d 07h) 223˚ 03.7’ Dec N 07˚ 24.6’
Incr. (33m 44s) 008˚ 27.4’ Lat 00˚ 02’ N
GHA Ȣ 231˚ 31.1’
SHA * (+) 271˚ 05.4’
GHA * 142˚ 36.5’
Sext Alt 43˚ 11.1’
IE (OFF) (+) 01.3’
Observed Alt 43˚ 12.4’
Dip (HE 18m) (-) 07.5’
App Alt 43˚ 04.9’
T Corrn. (-)01.0’
T Alt 43˚ 03.9’
P = 46˚ 30.0’
Since, sight is before mer. Pass. ,
So LHA = (360 – P)
LHA = 313˚ 30.0’
GHA = 142˚ 36.5’
Long. E = 172˚ 53.5’
Obs. Long. = 172˚ 53.5’ E
Azimuth = N 79.8˚ E
T Az = 079.8˚ (T)
LOP = 169.8˚ – 349.8˚
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On 30thApril 2008, PM at ship in DR 34˚ 18’S 040˚ 20’W, the sextant altitude of the star SIRIUS was 57˚ 51.9’ at 08h 53m 03schron time. If CE was 01m 40s FAST, IE was 1.2’ on the arc & HE was 21m, find the direction of the LOP and the longitude where it cuts the DR latitude.
GMT 30 April 20h 51m 23s
GHA Ȣ(30d 20h) 147˚ 06.9’ Dec S 16˚ 43.7’
Incr. (51m 23s) 012˚ 52.9’ Lat 34˚ 18’ S
GHA Ȣ 172˚ 00.7’
SHA * (+) 258˚ 37.3’
GHA * 070˚ 38.0’
Sext Alt 57˚ 51.9’
IE (ON) (-) 01.2’
Observed Alt 57˚ 50.7’
Dip (HE 21m) (-) 08.1’
App Alt 57˚ 42.6’
T Corrn. (-)00.6’
T Alt 57˚ 42.0’
We know that:
P = 30˚ 18.0’
Since, sight is AFTER mer. Pass. ,
So , LHA = P
LHA = 030˚ 18.0’
GHA = 070˚ 38.0’
Long. W = 040˚ 20.0’
Obs. Long. = 040 ˚ 20.0’ W
Azimuth = N 64.7˚ W
T Az = 295.3˚ (T)
Hence , LOP = ( TAz ± 90 )
LOP = 025.3˚ – 205.3˚
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On 31st Aug 2008, AM at ship in DR 40˚ 30’N 064˚ 56’E, the sextant altitude of the star DIPHDA was 21˚ 28.4’ when the chron (error 01m 06s FAST) showed 00h 21m 32s. If IE was 0.9’ off the arc & HE was 09m, find the direction of the LOP and the longitude where it cuts the DR latitude.
GMT 31 Aug 00h 20m 26s
GHA Ȣ(31d 00h) 339˚ 32.7’ Dec S 17˚ 56.1’
Incr. (20m 26s) 005˚ 07.3’ Lat 40˚ 30’ N
GHA Ȣ 344˚ 40.0’
SHA * (+) 348˚ 59.1’
GHA * 333˚ 39.1’
Sext Alt 21˚ 28.4’
IE (OFF) (+) 00.9’
Observed Alt 21˚ 29.3’
Dip (HE 09m) (-) 05.3’
App Alt 21˚ 24.0’
T Corrn. (-)02.5’
T Alt 21˚ 21.5’
We know that :
P = 38˚ 45.2’
Since, sight is AFTER mer. Pass. ,
So LHA = P
LHA = 038˚ 45.2’
GHA = 333˚ 39.1’
Long. E = 065˚ 06.1’
Obs. Long. = 065˚ 06.1’ E
We know that :
Azimuth = S 39.8˚ W
T Az = 219.8˚ (T)
Hence , LOP = ( TAz ± 90 )
LOP = 129.8˚ – 309.8˚
