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:
- LAT P 24˚ 00.0’ N LONG P 074˚ 0’ W MP OF P 1474.54 NLAT 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
- LAT P 06˚ 00.0’ N LONG P 079˚ 0’ W MP OF P 0358.20 NLAT 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
- LAT P 40˚ 18.0’ N LONG P 100˚ 0’ W MP OF P 2631.10 NLAT 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
- LAT P 70˚ 20.0’ N LONG P 010˚ 0’ W MP OF P 6002.80 NLAT 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
- LAT P 32˚ 29.0’ S LONG P 064˚ 0’ E MP OF P 2050.10 NLAT 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 |
- Course = 241˚ = S 61˚ W
We know that:
D’Lat = (Distance × Cos Co.)
= (1897 × Cos 61˚)
= 919.68’ S
= 15˚ 19.7’ S
Lat Left = 36˚ 48.0’ N MP for Lat Left 2363.60 N
D’Lat = 15˚ 19.7’ S MP for Lat Arrived 1310.70 N
Lat Arrived = 21˚ 28.3’ N DMP 1052.90
D’Long = (DMP × Tan Co.)
D’Long = 1899.5’ or 31˚ 39.5’ W
Long Left 085 ˚ 53.0’ W
D’long 031 ˚ 39.5’ W
Long Arrived 117 ˚ 32.5’ W
POSITION ARRIVED :- LAT 21˚ 28.3’ N LONG 117˚ 32.5’ W
- Course = 333˚ = N27˚ W
We know that:
D’Lat = (Distance × Cos Co.)
= (4450 × Cos 27˚)
= 3965’ N
= 66˚ 05.0’ N
Lat Left = 06˚ 10.0’ S MP for Lat Left 0368.20
D’Lat = 66˚ 05.0’ N MP for Lat Arrived 4497.11 N
Lat Arrived = 59˚ 55.0’ N DMP 4865.31
D’Long = (DMP × Tan Co.)
D’Long = 2479’ or 41˚ 19’ W
Long Left 176 ˚ 47.0’ W
D’long 041˚ 19.0’ W
Long Arrived 141 ˚ 54.1’ E
POSITION ARRIVED :- LAT 59˚ 55.0’ N LONG 141˚ 54.1’ E
- Course = 124˚ = S56˚ E
We know that:
D’Lat = ( Distance × Cos Co.)
= (3256 × Cos 56˚)
= 1820.73’ S
= 30˚ 20.7’ S
Lat Left = 38˚ 18.0’ S MP for Lat Left 2476.64 S
D’Lat = 30˚ 20.7’ S MP for Lat Arrived 5712.16S
Lat Arrived = 68˚ 38.7’ S DMP 3235.52
D’Long = (DMP × Tan Co.)
D’Long = 4796.85’
= 79˚ 56.8’ E
Long Left 005˚ 11.0’ W
D’long 079˚ 56.8’ E
Long Arrived 074 ˚ 45.8’ E
POSITION ARRIVED :- LAT 68˚ 38.7’ S LONG 074˚ 45.8’ E
- Course = 056˚ = N56˚ E
We know that:
D’Lat = (Distance × Cos Co.)
= (2222× Cos 56˚)
= 1242.52’ N
= 20˚ 42.5’ N
Lat Left = 44˚ 11.0’ N MP for Lat Left 2944.81 N
D’Lat = 20˚ 42.5’ N MP for Lat Arrived 5141.25 N
Lat Arrived = 64˚ 53.5’ N DMP 2196.44
D’Long = (DMP × Tan Co.)
D’Long = 3256.3’
= 54˚ 16.3’ E
Long Left 140 ˚ 20.0’ W
D’long 054˚ 16.3’ E
Long Arrived 086 ˚ 03.7’ W
POSITION ARRIVED :- LAT 64˚ 53.5’ N LONG 086 ˚ 03.7’ W
- Course = 265˚ = S85˚ W
We know that :
D’Lat = (Distance × Cos Co.)
= (7126 × Cos 85˚)
= 621.07’ S
= 10˚ 21.0’ S
Lat Left = 18˚ 58.0’ N MP for Lat Left 1151.77 N
D’Lat = 10˚ 21.0’ S MP for Lat Arrived 0515.50 N
Lat Arrived = 08˚ 37.0’ N DMP 0636.27
D’Long = (DMP × Tan Co.)
D’Long = 7272.60
= 121˚ 12.6’ W
Long Left 072 ˚ 52.0’ E
D’long 121˚ 12.6’ W
Long Arrived 048 ˚ 20.6’ W
POSITION ARRIVED :- LAT 08˚ 37.0’ N LONG 048˚ 20.6’ W
