Chapter 13: Great Circle Sailing (Numericals Solution)














Solution: Given:- Weight = 8800t,Original KG =6.2m,Cargo loaded = 200t,KG = 1.7m Ship’s weight KG VM 8800 t 6.2m 54560tm (+) 200 load 1.7m Final VM 54900 tm Final W – 9000 t Final VM 54900 tm So, we can calculate Final KG = (Final VM/ Final W)Final KG = (54900 / 9000)= 6.1m Here…
Solution : Volume of rectangular log = (L x b x h)= (8m x 2m x 2m)= 32m3 Weight = (U/W volume ) x (Density of displaced water)Weight = (8 x 2 x1.6) x(1)= 25. 6t RD = (mass / volume)= 25.6 / 32= 0.8t/m3 . Solution : Volume of rectangular log = (L x B…
NOTE: Also, we know that NOTES: Naming of Azimuth: The prefix N or S is the name same as C whereas suffix E or W depends on the value of LHA. If LHA is between 0 to 180, the body lies to the WEST and if it is between 180 and 360, the body lies to…
Solution: Ship’s weight KG VM 2000t (load) 4.2m 3400 tm 300t (load) 2.0m (+) 600 tm 200t (load) 3.2m (+) 640 tm 500t (load ) 1.0 m (+)500 tm Final W = 300t …
Principle of Navigations Chapter 4: Sailing (Numericals Solution) _________ EXERCISE 4(A) Solutions EXERCISE 4(B) Solutions