EXERCISE 1– PLANE AND PARALLEL SAILING (Numerical Solution)

In the following cases, find the position arrived:
Starting position
S No.
LATITUDE
LONGITUDE
COURSE
DISTANCE
1.         10˚ 20.0’ N060˚  20.0’E155˚300 M
2.         10˚ 12.0’ S120˚  11.0’W260˚ 458 M
3.         00˚ 10.0’ S179˚  40.0’W340˚510 M
4.         60˚ 11.2’ N120˚  18.6’E250˚312.4 M
5.         30˚ 14.0’ S168˚  12.0’W046˚ 12’426.8 M

SOLUTIONS:

  1. COURSE = 155˚ , T = S 25˚ E ,  Dist. = 300 MANS 1

We know that:
D’Lat =(Dist. × Cos Co.)
= 300 × Cos 25˚
= 271.9’
= 4˚ 31.9’ S

 

Lat left= 10˚ 20.0’N
D’LAT  = 04˚ 31.9’ S                             (as calculated above)
LAT ARR’D = 05˚ 48.1’ N

So, M’ LAT = 08˚04.0’N

Again, DEP. =  (D’LAT × Tan Co.)
= 126.79’

Now, D’LONG =( DEP./Cos M’LAT)
= 128.05’ E OR 002˚ 8’ E

LONG LEFT = 060˚ 20’ E
So, LONG ARR’D = 062˚ 28’ E

  1. COURSE = 260˚ ,  T = S 80˚ W ,  Dist. = 458 MANS 2

We know that:
D’Lat = (Dist. × Cos Co.)
= 458 × Cos 80˚
=  79.5’
= 1˚ 19’ S

Lat left            = 10˚ 12.0’ S
D’LAT              = 01˚ 19.0’ S
LAT ARR’D      = 11˚ 31.0’ S

So,  M’ LAT            = 10˚ 51.5’ S

DEP. = (D’LAT × Tan Co.)
= 450.86’

D’LONG =( DEP./Cos M’LAT)
= 459.08’ W OR 007˚ 39’ W

LONG LEFT                               = 120˚ 11.0’ W
LONG ARR’D                            = 127˚ 50.0’ W

  1. COURSE = 340˚, T= N 20˚ W , Dist. = 510 MANS 3

We know that:

D’Lat = (Dist. × Cos Co.)
= (510 × Cos 20˚)
= 479.24’
= 7˚ 59.2’ N

Lat left            = 00˚ 10.0’ S
D’LAT              = 07˚ 59.2’ N
LAT ARR’D      = 07˚ 49.2’ N

So, M’ LAT            = 03˚ 49.6’ N

DEP. =( D’LAT × Tan Co.)
= 174.42’

D’LONG = (DEP./Cos M’LAT)
= 174.42’ E OR 002˚ 54.4’ W

LONG LEFT                               = 179˚ 40.0’ W
LONG ARR’D                            = 177˚ 25.2’ E

  1. COURSE = 250˚,  T = S 70˚W,  Dist. = 312.4 MANS 4

We know that:
D’Lat =( Dist. × Cos Co.)
= 312.4 × Cos 70˚
= 106.8’
= 1˚ 46.8’ S

Lat left            = 60˚ 11.2’ N
D’LAT              = 01˚ 46.8’ S
LAT ARR’D      = 58˚ 24.4’ N

So, M’ LAT            = 59˚ 17.8’ N

DEP. = (D’LAT × Tan Co.)
= 293.55’

D’LONG = (DEP. / Cos M’LAT)
= 574.91’ W OR 009˚ 34.91’ W

LONG LEFT                               = 120˚ 18.60’ E
LONG ARR’D                            = 110˚ 43.70’ E

  1. COURSE = 046˚ 12’,  T = N 046˚ 12’ E,   Dist. = 426.8 MANS 5

We know that:

D’Lat = (Dist. × Cos Co.)
= (426.8 × Cos 46˚ 12’)
= 295.4’
= 4˚ 55.4’ N

Lat left            = 30˚ 14.0’ S
D’LAT              = 04˚ 55.4’ N
LAT ARR’D      = 25˚ 18.6’S
So, M’ LAT            = 27˚ 46.3’ S

DEP. = (D’LAT × Tan Co.)
= 308.04’

D’LONG =( DEP. / Cos M’LAT )
= 348.14’ E OR 005˚48.1’ E

LONG LEFT                               = 168˚ 12.0’ W
LONG ARR’D                            = 162˚ 23.9’ W

IN THE FOLLOWING CASES, FIND THE COURSE AND DISTANCE:
 
FROM
TO
 S N/O.
LATITUDE
LONGITUDE
LATITUDE
LONGITUDE
6.20˚ 30.0’ N179˚  36.0’ E16˚ 18.0’ N178˚  32.0’ W
7.     03˚ 12.0’ N004˚  11.3’ E02˚ 30.4’ S002˚  10.0’ W
8.     56˚ 12.5’ S046˚  12.5’ W50˚ 11.3’ S044˚  14.8’ W
9.     36˚ 11.7’ N075˚  12.6’ E40˚ 18.6’ N080˚  11.5’ E
10.     60˚ 11.6’ N076˚  44.3’ W55˚ 10.3’ N080˚  16.8 W

SOLUTIONS:

  1. FROM : LAT 20˚ 30.0’ N           LONG    179˚  0’ E
    TO: LAT   16˚ 18.0’ N     LONG    178˚  32.0’ W

D’LAT   04˚ 12.0’ S        D’LONG     001˚ 52.0’ E

D’LAT   04˚ 12.0’ S        D’LONG     001˚ 52.0’ E   and    M’LAT    18˚ 24.0’ S

We know that:

DEP. = (D’LONG × Cos M’LAT)
= (112 × Cos 18˚ 24.0’)
= 106.27 W

Tan Co. = (DEP. / D’LAT)
= 106.27 / 252
COURSE = S 22˚ 51.9’ E

DISTANCE = (D’LAT × Sec Co.)
= (252 × Sec 22˚ 51.9’)
DISTANCE = 273.5 M

  1. FROM : LAT 03˚ 12.0’ N           LONG    004˚  3’ E
    TO: LAT   02˚ 30.4’ S      LONG  002˚  10.0’ W

D’LAT   05˚ 42.4’ S        D’LONG     006˚ 21.3’ W

D’LAT   05˚ 42.4’ S        D’LONG     006˚ 21.3’ W   and    M’LAT  02˚ 51.2’ N

We know that:

DEP. =( D’LONG × Cos M’LAT)
= (381.3 × Cos 02˚ 51.2’)
= 380.82 W

Tan Co. = (DEP. / D’LAT)
= (380.82/342.4)
COURSE = S 48˚ 02.4’ W

DISTANCE = (D’LAT × Sec Co.)
= (342.4 × Sec 48˚ 02.4’)
DISTANCE = 512.1 M

  1. FROM : LAT 56˚ 12.5’ S           LONG    046˚  5’ W
    TO: LAT 50˚ 11.3’ S                   LONG 044˚  14.8’ W

D’LAT   06˚ 01.2’ N        D’LONG   001˚  57.7’ E

D’LAT   06˚ 01.2’ N        D’LONG     001˚ 57.7’ E   and    M’LAT  53˚ 11.9’ S

DEP. = (D’LONG × Cos M’LAT)
= (117.7× Cos 53˚ 11.9’)
= 70.50 W

TAN Co. = (DEP. / D’LAT)
= (70.50 / 361.2)
COURSE = N 11˚ 02.6’ E

DISTANCE = (D’LAT × Sec Co.)
= (361.2 × Sec 11˚ 02.6’
DISTANCE = 368.0 M

  1. FROM : LAT 36˚ 11.7’ N           LONG    075˚  6’ E
    TO: LAT 40˚ 18.6’ N      LONG 080˚  11.5’ E

D’LAT   04˚ 06.9’ N        D’LONG    004˚ 58.9’ E

D’LAT   04˚ 06.9’ N        D’LONG     004˚ 58.9’ E   and    M’LAT    38˚ 15.1’ N

We know that:

DEP. =( D’LONG × Cos M’LAT)
= (298.9 × Cos 38˚ 15.1’)
= 234.7 E

Tan Co. = (DEP. / D’LAT)
= (234.7/246.9)
COURSE = N 43˚ 32.9’ E

DISTANCE =( D’LAT × Sec Co.)
= (246.9 × Sec 43˚ 32.9’
DISTANCE = 340.6 M

  1. FROM : LAT 60˚ 11.6’ N           LONG    076˚  3’ W
    TO : LAT 55˚ 10.3’ N      LONG 080˚  16.8 W

D’LAT   05˚ 01.3’ S        D’LONG    003˚  32.5’ W

D’LAT   05˚ 01.3’ S        D’LONG     003˚ 32.5’ W   and    M’LAT  57˚ 40.9’ N

We know that:

DEP. = (D’LONG × Cos M’LAT)
= (212.5 × Cos 57˚ 40.9’)
=  113.6 W

Tan Co. = (DEP./D’LAT)
= (113.6/301.3)
COURSE = S 20˚ 39.5’ W

DISTANCE = (D’LAT × Sec Co.)
= (301.3 × Sec 20˚ 39.5’)
DISTANCE = 322 M

IN THE FOLLOWING CASES, FIND THE SET AND DRIFT OF CURRENT:
 
DR
FIX
 
LATITUDE
LONGITUDE
LATITUDE
LONGITUDE
11.     46˚ 44.3’ N076˚  36.3’ E47˚ 00.6’ N076˚  10.4’ E
12.     30˚ 16.8’ S057˚  49.3’ E31˚ 00.7’ S058˚  20.4’ E
13.     00˚ 11.6’ N179˚  50.2’ W00˚ 40.3’ S178˚  40.1’ E
14.     60˚ 20.6’ S076˚  18.4’ W60˚ 01.7’ S175˚  54.9’ W
15.     50˚ 16.3’ N000˚  12.3’ E49˚ 50.4’ N000˚  20.1 W

SOLUTIONS :

  1. FROM : LAT 46˚ 44.3’ N           LONG    076˚  3’ E
    TO: LAT  47˚ 00.6’ N           LONG  076˚  10.4’ E

D’LAT   00˚ 16.3’ N        D’LONG     000˚ 25.9’ W

D’LAT   00˚ 16.3’ N        D’LONG     000˚ 25.9’ W   and    M’LAT    46˚ 52.4’ N

We know that:

DEP. =( D’LONG × Cos M’LAT)
= (25.9 × Cos 46˚ 52.4’)
= 017.70

Tan Co. = (DEP./D’LAT)
=(017.70/16.3)
COURSE = N 47˚ 21.5’ W

DISTANCE = (D’LAT × Sec Co.)
= (16.3 × Sec 47˚ 21.5’)
DISTANCE = 24.0 M

  1. FROM : LAT 30˚ 16.8’ S           LONG    057˚  3’ E
    TO: LAT   31˚ 00.7’ S           LONG  058˚  20.4’ E

D’LAT   00˚ 43.9’ S        D’LONG    000˚ 31.1’ E

D’LAT   00˚ 43.9’ S        D’LONG     000˚ 31.1’ E   and    M’LAT  30˚ 38.7’ S

We know that:

DEP. =( D’LONG × Cos M’LAT)
= (31.1 × Cos 30˚ 38.7’)
= 26.7

Tan Co. = (DEP. / D’LAT)
= (26.7 / 43.9)
COURSE = S 31˚ 21.7’ E

DISTANCE =( D’LAT × Sec Co.)
= (43.9 × Sec 31˚ 21.7’)
DISTANCE = 51.4 M

  1. FROM : LAT 00˚ 11.6’ N          LONG    179˚  2’ W
    TO: LAT 00˚ 40.3’ S         LONG 178˚  40.1’ E

D’LAT   00˚ 51.9’ S        D’LONG   001˚  29.7’ W

D’LAT   00˚ 51.9’ S        D’LONG     001˚ 29.7’ W   and    M’LAT  00˚ 26’ S

We know that:

DEP. = ( D’LONG × Cos M’LAT)
=  (89.7 × Cos 00˚ 26’) 
= 89.7

TAN Co. =(  DEP. / D’LAT)
= ( 89.7 / 51.9)
COURSE = S 59.9˚ W

DISTANCE = (D’LAT × Sec Co.)
= (51.9 × Sec 59.9˚)
DISTANCE = 103.5 M

  1. FROM : LAT 60˚ 20.6’ S           LONG    176˚  4’ W
    TO: LAT 60˚ 01.7’ S           LONG 175˚  54.9’ W

D’LAT   00˚ 18.9’ N        D’LONG   000˚ 23.5’ E

D’LAT   00˚ 18.9’ N        D’LONG     000˚ 23.5’ E   and    M’LAT    60˚ 11.1’ S

We know that:

DEP. = (D’LONG × Cos M’LAT)
= (23.5 × Cos 60˚ 11.1’)
= 11.7

Tan Co. = DEP. / D’LAT
= 11.7 / 18.9
COURSE = N 31˚ 45.6’ E

DISTANCE = (D’LAT × Sec Co.)
= (18.9 × Sec 31˚ 45.6 ’)
DISTANCE = 22.2 M

  1. FROM : LAT 50˚ 16.3’ N           LONG    000˚  3’ E
    TO : LAT 49˚ 50.4’ N      LONG 000˚  20.1’W

D’LAT   00˚ 25.9’ S        D’LONG    000˚  32.4’ W

D’LAT   00˚ 25.9’ S        D’LONG     000˚ 32.4’ W   and  M’LAT  50˚ 03.3’ N

We know that:

DEP. = (D’LONG × Cos M’LAT)
= (32.4 × Cos 50˚ 03.3’)
= 20.8

Tan Co. =( DEP. / D’LAT)
= 20.8 / 25.9
COURSE = S 38˚ 46.2’ W

DISTANCE = (D’LAT × Sec Co.)
= (25.9 × Sec 38˚ 46.2’)
DISTANCE = 33.2 M

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