EXERCISE 3 — MERCATOR SAILING (Numerical Solution)

Find, by Mercator Sailing, the course and distance:-

	FROM		TO
	LATITUDE	LONGITUDE	LATITUDE	LONGITUDE
1.	24˚ 00.0’ N	074˚  15.0’ W	46˚ 00.0’ N	053˚  45.0’ W
2.	06˚ 00.0’ N	079˚  00.0’ W	38˚ 00.0’ S	179˚  00.0’ E
3.	40˚ 18.0’ N	100˚  20.0’ W	68˚ 00.0’ N	140˚  10.0’ E
4.	70˚ 20.0’ N	010˚  22.0’ W	52˚ 50.0’ N	009˚  45.0’ E
5.	32˚ 29.0’ S	064˚  00.0’ E	49˚ 50.0’ S	005˚  15.0’ E

NOTE:

• Where the distance exceeds 600 M, 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

2. 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

3. 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  

4. 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

5. 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’ N	085˚  53.0’ W	241˚	1897 M
7.	06˚ 10.0’ S	176˚  47.0’ W	333˚	4450 M
8.	38˚ 18.0’ S	005˚  11.0’ W	124˚	3256 M
9.	44˚ 11.0’ N	140˚  20.0’ W	056˚	2222 M
10.	18˚ 58.0’ N	072˚  52.0’ E	265˚	7126 M