Miscelleneous

EXERCISE 3 — MERCATOR SAILING (Numerical Solution)

Find, by Mercator Sailing, the course and distance:-
 
FROM
TO
LATITUDE
LONGITUDE
LATITUDE
LONGITUDE
1.     24˚ 00.0’ N074˚  15.0’ W46˚ 00.0’ N053˚  45.0’ W
2.     06˚ 00.0’ N079˚  00.0’ W38˚ 00.0’ S179˚  00.0’ E
3.     40˚ 18.0’ N100˚  20.0’ W68˚ 00.0’ N140˚  10.0’ E
4.     70˚ 20.0’ N010˚  22.0’ W52˚ 50.0’ N009˚  45.0’ E
5.     32˚ 29.0’ S064˚  00.0’ E49˚ 50.0’ S005˚  15.0’ E
NOTE:
  • Where the distance exceeds 600 M,ex 3 q 1 it is generally recommended to use Mercator sailing so that results are more accurate results obtained.
  • To get DMP, apply same rule as for D’Lat i.e, same names subtract, different names add.
  • To get distance, use plane sailing formula.
  • In order to understand and solve such questions in a better and easy way, one must always draw the diagram.

SOLUTION:

  1. LAT P    24˚ 00.0’ N           LONG P   074˚  0’ W                   MP OF P        1474.54 N
    LAT Q   46˚ 00.0’ N         LONG Q     053˚  45.0’ W             MP OF Q      3098.70 N
D’LAT   22˚ 00.0’ N           D’LONG       020˚ 30.0’ W            DMP       1624.16

We know that:

TAN CO. = (D’LONG / DMP)
= (1230 / 1624.16)
COURSE = N 37˚ 08.23’ W

DISTANCE = (D’LAT × SEC CO.)
DIST. = 1655.81 M

COURSE = N 37˚ 08.23’ W   DIST. = 1655.81 M

  1. LAT P      06˚ 00.0’ N            LONG P      079˚  0’ W                MP OF P      0358.20 N
    LAT Q       38˚ 00.0’ S           LONG Q      179˚  00.0’ E            MP OF Q   2453.85 S
D’LAT       44˚ 00.0’ S             D’LONG      102˚ 00.0’ W        DMP         2812.05

We know that:
TAN CO. = (D’LONG / DMP)
= (6120 / 2812.05)
COURSE = S65˚ 19.31’ W  

DISTANCE =( D’LAT × SEC CO.)
DISTANCE  = 6323.03 M

COURSE = S65˚ 19.31’ W   DISTANCE  = 6323.03 M

  1. LAT P     40˚ 18.0’ N            LONG P    100˚  0’ W             MP OF P        2631.10 N
    LAT Q    68˚ 00.0’ N           LONG Q      140˚  10.0’ E     MP OF Q   5609.09 N
D’LAT       27˚ 42.0’ N             D’LONG      119˚ 30.0’ W        DMP        2978.00

We know that:
TAN CO. = (D’LONG / DMP)
= (7170 /2978)
COURSE = N 67˚ 26.7’ W

DISTANCE = (D’LAT × SEC CO.)
DIST. = 4332.97 M      

COURSE = N 67˚ 26.7’ W                   DIST. = 4332.97 M  

  1. LAT P       70˚ 20.0’ N         LONG P     010˚  0’ W             MP OF P     6002.80 N
    LAT Q     52˚ 50.0’ N           LONG Q      009˚  45.0’ E         MP OF Q    3728.51 N
D’LAT       17˚ 30.0’ S             D’LONG         020˚ 07.0’ E          DMP       2274.29

We know that:
TAN CO. = (D’LONG / DMP)
= (1207/2274.29)
COURSE = S27˚ 57.3’ E

DISTANCE = (D’LAT × SEC CO.)
DIST. = 1188.7 M

COURSE = S27˚ 57.3’ E         DIST. = 1188.7 M

  1. LAT P     32˚ 29.0’ S            LONG P  064˚  0’ E                  MP OF P        2050.10 N
    LAT Q    49˚ 50.0’ S           LONG Q      005˚  15.0’ E           MP OF Q       3441.05 N
D’LAT       17˚ 21.0’ S         D’LONG      058˚ 45.0’ W           DMP           1390.95

We know that:
TAN CO. = (D’LONG/DMP)
= (3525/1390.95)
COURSE = S 68˚ 28.00’ W 

DISTANCE = D’LAT × SEC CO
DIST. = 2836.18 M

COURSE = S68˚ 28.00’ W                                            DIST. = 2836.18 M

FIND, BY MERCATOR’S PRINCIPLE, THE POSITION ARRIVED:-
     STARTING POSITION
LATITUDE
LONGITUDE
COURSE
DISTANCE
6.     36˚ 48.0’ N085˚  53.0’ W241˚1897 M
7.     06˚ 10.0’ S176˚  47.0’ W333˚4450 M
8.     38˚ 18.0’ S005˚  11.0’ W124˚3256 M
9.     44˚ 11.0’ N140˚  20.0’ W056˚2222 M
10.     18˚ 58.0’ N072˚  52.0’ E265˚7126 M

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