Group weights, water draft, air draft and density (Derrett Stability)
A ship’s draft is 6.40 meters forward and 6.60 meters aft. FWA = 180 mm. The density of the dock water is 1010 kg per cum. If the load mean draft in salt water is 6.7 meters, find the final drafts F and A in dock water if this ship is to be loaded down to her marks and trimmed 0.15 meters by the stern. (Centre of flotation is amidships.)
Here, FWA = 180 mm and density = 1010 kg per cubm
We know, DWA = FWA (1025 – D) / 25
= 180 (1025 – 1010) / 25 = 108 mm = .108 m
So, Dock water draft = salt water draft + DWA = 6.7 m + .108 m = 6.808 m
As the ship is .15 m by the astern and the center of flotation is amidships.
So, Fwd draft will be = (dock water mean draft – .15/2) = (6.808 – .15/2) = 6.773 m
Aft draft will be = (dock water mean draft + .15/2) = (6.808 + .15/2) = 6.883 m (Ans.)
Draft Density TPC FWA (Derrett Stability)
A ship floating in dock water of density 1005 kg per cum has the lower edge of her Summer load line in the waterline to starboard and 50 mm above the waterline to port. FWA = 175 mm and TPC = 12 tonnes. Find the amount of cargo that can yet be loaded in order to bring the ship to the load draft in salt water.
This ship is obviously listed to starboard side and if brought upright the lower edge of the summer load line on each side would be 25 mm above the waterline. Also, it is the upper edge of the load line mark which indicates the summer load draft by regulation and, since the load line mark is 25 mm thick, the ship’s draft must be increased by 25+25 = 50 mm to bring her to the summer load line in dock water. In addition there is a DWA by regulation.
DWA = FWA (1025 – D) / 25 = 175 (1025 – 1005) / 25 = 140 mm
Increase in draft = 50 mm + 140 mm = 190 mm = 19 cm
Cargo to load = increase in draft * TPC = 19 cm * 12 tonne = 228 tonne (Ans.)
Draft Density TPC FWA (Derrett Stability)
A ship’s light displacement is 3450 tonnes and she has on board 800 tonnes of bunkers. She loads 7250 tonnes of cargo, 250 tonnes of bunkers, and 125 tonnes of fresh water. The ship is then found to be 75 mm from the load draft. TPC = 12 tonnes. Find the ship’s deadweight and load displacement.
Total displacement = (3450 + 800 + 7250 + 250 + 125) = 11875 tonne
Increase in draft required = 75 mm = 7.5 cm
Cargo to load = increase in draft * TPC = 7.5 * 12 = 90 tonne
So, total load displacement = 11875 + 90 = 11965
Deadweight = Load displacement – Light displacement = 11965 – 3450 = 8515 tonne. (Ans.)
