# Chapter 1: Earth (Numericals Solution)

**Solutions of Principle of Navigation: **Chapter 1: Earth (Numerical Solutions)

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# Chapter 1: Earth (Numericals Solution)

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**Solutions of Principle of Navigation: **Chapter 1: Earth (Numerical Solutions)

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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…

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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…

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Principle of Navigations Chapter 4: Sailing (Numericals Solution) _________ EXERCISE 4(A) Solutions EXERCISE 4(B) Solutions

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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…

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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…