Miscelleneous

EXERCISE 1– PLANE AND PARALLEL SAILING (Numerical Solution)

Starting position
DISTANCE
1.         10˚ 20.0’ N060˚  20.0’E155˚300 M
2.         10˚ 12.0’ S120˚  11.0’W260˚ 458 M
3.         00˚ 10.0’ S179˚  40.0’W340˚510 M
4.         60˚ 11.2’ N120˚  18.6’E250˚312.4 M
5.         30˚ 14.0’ S168˚  12.0’W046˚ 12’426.8 M

SOLUTIONS:

1. 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 5G Tech
1. 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 5G Tech
1. 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 Vikrant_sharma

• Niranjan nar Shandilya says:

Sir problem in stability chapter 3 ques no 4 is not correctely solve kindly correct all these types of error

• Krish says:

What is the mistake

• JeffReyes says:

I think some of your Dlong Computations are incorrect

• Ukeme Hillary Eboh says: