In the following cases, find the position arrived:
Starting position
S No. |
LATITUDE |
LONGITUDE |
COURSE |
DISTANCE |
| 1. | 10˚ 20.0’ N | 060˚ 20.0’E | 155˚ | 300 M |
| 2. | 10˚ 12.0’ S | 120˚ 11.0’W | 260˚ | 458 M |
| 3. | 00˚ 10.0’ S | 179˚ 40.0’W | 340˚ | 510 M |
| 4. | 60˚ 11.2’ N | 120˚ 18.6’E | 250˚ | 312.4 M |
| 5. | 30˚ 14.0’ S | 168˚ 12.0’W | 046˚ 12’ | 426.8 M |
SOLUTIONS:
-
COURSE = 155˚ , T = S 25˚ E , Dist. = 300 M

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
-
COURSE = 260˚ , T = S 80˚ W , Dist. = 458 M

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
-
COURSE = 340˚, T= N 20˚ W , Dist. = 510 M

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
-
COURSE = 250˚, T = S 70˚W, Dist. = 312.4 M

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
-
COURSE = 046˚ 12’, T = N 046˚ 12’ E, Dist. = 426.8 M

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’ N | 179˚ 36.0’ E | 16˚ 18.0’ N | 178˚ 32.0’ W |
| 7. | 03˚ 12.0’ N | 004˚ 11.3’ E | 02˚ 30.4’ S | 002˚ 10.0’ W |
| 8. | 56˚ 12.5’ S | 046˚ 12.5’ W | 50˚ 11.3’ S | 044˚ 14.8’ W |
| 9. | 36˚ 11.7’ N | 075˚ 12.6’ E | 40˚ 18.6’ N | 080˚ 11.5’ E |
| 10. | 60˚ 11.6’ N | 076˚ 44.3’ W | 55˚ 10.3’ N | 080˚ 16.8 W |
SOLUTIONS:
-
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
-
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
-
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
-
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
-
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’ N | 076˚ 36.3’ E | 47˚ 00.6’ N | 076˚ 10.4’ E | |
| 12. | 30˚ 16.8’ S | 057˚ 49.3’ E | 31˚ 00.7’ S | 058˚ 20.4’ E | |
| 13. | 00˚ 11.6’ N | 179˚ 50.2’ W | 00˚ 40.3’ S | 178˚ 40.1’ E | |
| 14. | 60˚ 20.6’ S | 076˚ 18.4’ W | 60˚ 01.7’ S | 175˚ 54.9’ W | |
| 15. | 50˚ 16.3’ N | 000˚ 12.3’ E | 49˚ 50.4’ N | 000˚ 20.1 W | |
SOLUTIONS :
-
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
-
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
-
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
-
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
-
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
Sir problem in stability chapter 3 ques no 4 is not correctely solve kindly correct all these types of error
What is the mistake
I think some of your Dlong Computations are incorrect
I enjoy learning navigation with your material. thanks