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Solving For Vessel's Trim

A vessel has a fwd and aft drafts of 28'-00" and the MTI is 1700 ft-tons. 250 tons of water is pumped from the forepeak into the after peak at a distance of 574 feet. What will be her trim after pumping is complete? (Tipping center is assumed to be at vessel's mid length).

ANALYSIS OF THE PROBLEM:
Our ship's trim is said to be "even keel" since we have the same draft forward and aft. However, a 250 tons of water from the forepeak is to be transferred into the after peak tank. So what will happen after pumping out is that there will be an increase in draft aft and a decrease in draft forward because the tipping center is at vessel's mid length.

SOLUTION:

 We have to use the formula for Change Of Trim (COT)
COT = (WEIGHT X DISTANCE)  ÷ MTI

Step 1. Multiply weight and distance.
              250 tons x 574 ft = 143500 ft-tons

Step 2. Divide the answer of step 1 by the MTI.
               143500  ÷ 1700 = 84.41176 inch is our change of trim.

Step 3. Divide COT by 2 to get the "half trim".
             84.4  ÷ 2 = 42.20588 inch

             We have to convert this into feet since the unit of ship's draft is in feet by dividing it by 12.
              42.20588 inch  ÷ 12 = 3.51715 ft or 3' 6.20" is our "half trim"

Step 4. The foreward draft is to be subtracted by "half trim" because it is  the forepeak tank that is being pump-outed.
               28' 00" - 3' 6.20" = 24' 05.8"
And the draft aft is to be added by "half trim" since it is the after peak tank where 250 tons of water is pumped into.
             28' 00" + 3' 6.20" = 31' 06.2"

So the answer is 24-05.8 FWD and 31-06.2 AFT.

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