EXERCISE 31 — INTERCEPT SUN (Numerical Solution)

  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 intercept & the direction of the LOP.

e31q1p1

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 (-).

dia 1

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 :

e31q1p2

Azimuth = N 65.8˚ E
T Az= 065.8˚ (T)
LOP = 155.8˚ – 335.8˚

  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 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 (-).

dia 2

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

e31q2p1

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

  1. 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.

e31q3p1

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 (-).

dia 3

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

e31q3p2

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

  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 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 (-).

dia 4

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 :

e31q4p2

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

  1. 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.

e31q5p1

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 (-).

dia 5

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 :

e31q5p2

Azimuth = N 69.2˚ W
T Az= 301.8˚ (T)
LOP = 031.8˚ – 211.8˚

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