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
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