EXERCISE 32 — LONG BY CHRONOMETER SUN (Numerical Solution)

  1. On 19th Jan 2008, PM at ship in DR 40˚ 16’S 175˚ 31’E, the sextant altitude of the sun’s LL was 43˚ 27.4’ when the GPS clock showed 03h 48m 00s. If IE was 1.5’ on the arc & HE was 22m, find the direction of the LOP and the longitude where it cuts the DR latitude.

                                d     h       m       s
GMT                       19   03     48    00
LIT (E)                    (+)    11     58   04
LMT                       19     15    46   04

      GMT     19 Jan 03h  48m  00s

GHA (19d 03h)            222˚  23.0’                                           Dec          S  20˚  28.9’
Incr. (48m 00s)          012˚  00.0’                                           d(-0.5)                 00.4’
GHA                             234˚  23.0’                                           Dec          S  20˚  28.5’
Lat                              40˚  16’ S

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’

dia32q2

NOTE:

In the following formula, if the LAT and DEC are of same name then sign is (-), If of contrary names then sign is (+).

e32q1p1

P = 49˚ 38.8’

Since, sight is after mer. Pass. ,so LHA = P

NOTE:

  • Before meridian passage (Mer. Pass.) , LHA wil be between 180 and 360. After meridian passage, LHA will be between 000 and 180.So, to calculate LHA,
  • If it given that the sight is taken AM at ship or meridian of east, then P = LHA.
  • If it is given that the sight is taken at PM at ship, then P = 360 – LHA
  • When P>90, the minus sign obtained for the value of A is to be ignored and is taken care by changing the name of A.

LHA          = 049˚ 38.8’
GHA         = 234˚ 23.0’
Long. W   = 184˚ 44.2’

Obs. Long. = 175˚ 15.8’ E

We know that:

e32q1p2

Azimuth = N 80.3˚ W
T Az= 279.7˚ (T)
LOP = 009.7˚ – 189.7˚

  1. On 31st Aug 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 direction of the LOP and the longitude where it cuts the DR latitude.
e32q2p21
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’
Lat                            10˚  15’ S

Sext Alt                              34˚ 54.0’
IE (on)                            (-)       01.5’
Observed Alt                    34˚ 52.5’
Dip (HE 22m)               (-)        07.3’
App Alt                            34˚ 45.2’
T Corrn. LL                  (+)       14.6’
T Alt                                34˚ 59.8’

dia32p1

NOTE:

 In the following formula,

  •  if the LAT and DEC are of same name then sign is (-),
  •  If of contrary names then sign is (+) .
e32q2p2
P = 46˚25.5’

Since, sight is after mer. Pass. ,so LHA = P

LHA          = 046˚ 25.5’
GHA         = 046˚ 30.5’
Long. W   = 000˚ 05.0’

Obs. Long. = 000˚ 05.0’ W

We know that :

e32q2p3

Azimuth = N 69.8˚ W
T Az= 301.8˚ (T)
LOP = 031.8˚ – 211.8˚

  1. 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 direction of the LOP and the longitude where it cuts the DR latitude.

                                   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’
Lat                                48˚  20’ N

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’

dia32q3
NOTE:

In the following formula,

  • If the LAT and DEC are of same name then sign is (-),
  • If of contrary names then sign is (+) .
e32q3p1
P = 59˚ 48.8’

Since, sight is after mer. Pass. ,so LHA = P

LHA          = 059˚ 48.8’
GHA         = 334˚ 15.9’
Long. E    = 085˚ 32.9’

Obs. Long. = 085˚ 32.9’ E

We know that :

e32q3p2

Azimuth = S 66.7˚ W
T Az          = 246.7˚ (T)
LOP = 156.7˚ – 336.7˚

  1. 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 direction of the LOP and the longitude where it cuts the DR latitude.e32q4p1
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’
Lat                                      38˚  11’ S

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’

dia32q4

NOTE:

 In the following formula,

  • If the LAT and DEC are of same name then sign is (-),
  •  If of contrary names then sign is (+) .

e32q4p2

P = 47˚ 52.9’

Since, sight is before mer. Pass. ,so LHA = 360 – P

LHA          = 312˚ 08.1’
GHA         = 161˚ 01.9’
Long. E    = 151˚ 06.2’

Obs. Long. = 151˚ 06.2’ E

e32q4p3

Azimuth = N 65.9˚ E
T Az          = 065.9˚ (T)
LOP = 155.9˚ – 335.9˚

  1. 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 01h 00m 52s. If IE was 3.1’ off the arc & HE was 20m, find the direction of the LOP and the longitude where it cuts the DR latitude.

                                   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’
Lat                                       00˚  20’ N 

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’

dia32q5

NOTE:

In the following formula,

  • If the LAT and DEC are of same name then sign is (-),
  • If of contrary names then sign is (+) 
e32q5p1
P = 44˚ 17.5’

Since, sight is before mer. Pass. ,so LHA = 360 – P

LHA          = 315˚ 42.5’
GHA         = 015˚ 55.7’
Long. W   = 060˚ 13.2’

Obs. Long. = 060˚ 13.2’ W

We know that :

e32q5p2

Azimuth = N 69.31˚ E
T Az          = 069.31˚ (T)
LOP = 159.3˚ – 339.3˚

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