Hi Peter

Thanks for the reply, your feedback is much appreciated.

I have tried to do this by the 3 stages you mentioned

For the 10,000 Tonne ship moving from sea water to fresh water, I have calculated the parallel sinkage at LCF, which is the same as the increase in Aft Draft.

Tonnes per centimetre (TPC) = Density of Sea Water * Area of Waterplane * 1/100

TPC = 1.025 * 1,480 * 1/100

TPC = 15.17 Tonnes / cm

Change in Draft aft = [(Sea water density - Fresh Water density) / Fresh Water Density] * Displacement of 10,000 Tonne ship / TPC

Change in Draft aft = [(1.025-1) / 1] * 10,000 / 15.17

Change in Draft aft = 16.48 cm from the 10,000 Tonne ship alone

For the 300 Tonne cargo load apply parallel sinkage as before, when moving from sea water to fresh water

Change in Draft aft = [(Sea water density - Fresh Water density) / Fresh Water Density] * Displacement of 300 Tonne Cargo Load / TPC

Change in Draft aft = [(1.025-1) / 1] * (300 / 15.17)

Change in Draft aft = 0.494 cm from the 300 Tonne Cargo Load alone

Remainder of Change in Aft Draft = 30 - 16.48 - 0.494 = 13.026 cm

The remainder 13.026 cm of this Change in Aft Draft is due to trim and Moment causing trim of 1 cm

and from this we can calculate the placement of the cargo load

Change of Trim (COT) = Trimming Moment (TM) / Moment causing Trim of 1cm (MCTC)

Trimming moment = w * d where w = cargo mass and d = lever from LCF

MCTC = W * GML / 100 * LBP where W = 10,300 Tonne ship and cargo, GML = Longtitudinal Metacentric Height and LBP = Length between perpendiculars

COT = (w * d) / [(W * GML) / (100 * LBP)] Multiply top and bottom by 100 * LBP

COT = (100 * LBP/ 100 * LBP) * {(w * d) / [(W * GML) / (100 * LBP)]}

so COT = (100 * LBP * w * d) / (W * GML)

Change in Draft Aft = LCF / LBP * COT where Longtitudinal Centre of Flotation (LCF) = 55m

Change in Draft Aft = LCF / LBP * [(100 * LBP * w * d)/(W * GML)]

LBP cancels out on top and bottom

Change in Draft Aft = LCF * [(100 * w * d)/(W * GML)]

Multiply both sides by (W * GML)

Change in Draft Aft * (W * GML) = LCF * (100 * w * d)

and rearranging

d = Change in Draft Aft * (W * GML) / (LCF * 100 * w)

From before Change in Draft Aft = 13.026 cm

so d = 13.026 * (10,300 * 91.5) / (55 * 100 * 300)

d = 7.440214364

d = 7.44m

So minimum allowable distance of mass centre of this extra load from stern = 55 - 7.44 = 47.56m

The text book answer is 46.2m , so I am out by 1.36m.

Is the answer in the text book correct or not?

Do you have any suggestions on something I may have missed.

I have noticed the centre of gravity of 49m from the stern for the ship has not been used in any of the calculations.

Yes my spreadsheet is messy, it's just a case of getting down ideas, it's not final

Thanks

Robert