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 …
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…
Stability 1 – Chapter 5 Effect of Density Solutions > Exercise 5: Effect of density on draft and displacement – FWA ; DWA ; loadliness of ship. Solution: Displacement (W) = 16000 tTPC = 20 & SW draft = 8.0m We know that: Displacement when in SW = ( L X B x draft) x…
Stability 1 – Chapter 4 Solutions for Exercise 4: Some important terms – displacement; deadweight; form coefficient; reserve buoyancy; Tonnes per centimetre. Solution : Area of box shape vessel = (L x B)= 120m x 15mLight draft = 4m Light displacement = (u/w volume) x ( density)= (L x b x d )x (1.025)= (120…
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…
Solution : Area of the flat keel plate = (10 x 2) = 20m2 Pressure = (depth x density) = (8 x 1.025) = 8.2t /m2 Thrust = (pressure x area) …
SOLUTION: Given : L =16m , B = 15m, H = 6m, RD=0.7 =16x15x6 = 1440m3 We know that : Density = Mass /Volume0.78 =Mass/1440m3 Hence, Mass =1123.2 tonnes SOLUTION: Given: Diameter = 8m, Radius = (d/2)m=(8/2)m =4mHeight =10mMass = 400tRD =0.9 We know that: Volume of the cylindrical tank = π2h= 3.1416 x 4…