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Stability 1 – Chapter 6 Centre of Gravity
Solution: Given:Weight = 12000t,Initial KG =7.6m,Cargo shifted = 300t,Change of KG = (8 – 2)= 6m downwards Derrick Height = 20m,KG of the Weight = 8m (since the weight is on the UD) Case – 1 Ship’s weight KG VM 12000 t 7.6m 91200tm 300t (shift) 12m (+) 3600 tm Final W =12000…
Stability 1 Chapter 2 – Water Pressure
Solution : B = 14m, H =12m As we know that :Pressure = (depth x density)Thrust = (pressure x area) An right angle triangle PBC , PB =7mSince, Height divide AB equally into two parts.Triangle PBC and FEC are similarSo by law of similar angle triangle PB/ FE = PC / FC7 / FE =12…
EXERCISE 28 — AZIMUTH SUN Numerical Solutions
3. On Jan 19th 2008, in DR 40˚ 16’S 175˚ 31’ E, the azimuth of the sun was 267˚(C) at 1618 ship’s time. If the ships time difference was 11h 30m from GMT and variation was 2.3˚E, find the deviation for the ship’s head. d h m s Ship’s time 19 …