# Longtitudinal Centre of Gravity for Ship Cargo

I did a Civil Engineering course some years ago and this question from my textbook, I have found what seems impossible to get the correct answer of 46.2 m for. By putting this question up I am hoping that I will be able to come to a conclusion on how this should be solved.

A ship displaces 10,000 metric tonnes and area of its plane of flotation is 1,480 m^2. The centre of mass is 49 m and centre of area of plane of flotation is 55 m from the stern.

The metacentric height for pitching motion about transverse principal axis is 91.5 m. The ship is loaded in sea water with 300 metric tonnes of cargo. Find minimum allowable distance of mass centre of this extra load from the stern if, when ship passes from sea water into a fresh water canal, the stern draft must not increase by more than 0.3 m ?

Assume metacentric height and area of plane of flotation are not altered by the change in draft, density of sea water is 1,025 kg/m^3 and density of fresh water is 1,000 kg/m^3.

Just looking for some help on this from someone who understands ship stability well.

Here is my attempt at finding a Solution to this Question

The first step:

We will put the 300 Tonne Cargo Load at the ships centre of gravity of 49 m, so we don't have to correct ourselves.

Firstly we will treat this as if the cargo has a weight shift from centre of flotation to centre of gravity to calculate the aft added draft from the cargo.

We need to run the Change of Trim Formula

by using COT = (100 * LBP * w * d) / (W * GML) with d = 6 m because you want to have initial LCG AT 49m

COT = (100*LBP*300*6) / (10,300*91.5)

Aft Added Draft = (LCF / LBP) * COT = (55 / LBP) * [(100*LBP*300*6) / (10,300*91.5)]

and LBP cancels out,

so Aft Added Draft = (55*100*300*6) / (10,300*91.5) = 10.5 cm at original 49 m CG

The Second Step:

Now we need to take into account sinkage change due to density, for 10,000 Tonne Ship + 300 Tonne Cargo all at 49m CG

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

TPC = 1.025 * 1,480 * 1/100

TPC = 15.17 Tonnes per cm

Parallel Sinkage in Fresh Water = 10,300 / 15.17 = 678.97 cm

And because the Density Changes

at 49 m CG the Sinkage Change = [(1.025 - 1) / (1)] * (10,300 / 15.17) * (49/55) = 15.12 cm

The Third and Final Step:

The cargo will shift from the ships centre of gravity of 49m, to where it should be.

For this part we will firstly have to calculate the remainder draft.

The remainder draft = 30 - 10.5 - 15.12 = 4.38 cm

Like before we need to use Aft added draft formula, but this time we need to find d (distance moved)

So now find d in following formula:

Change in Draft Aft (from shifted load) = LCF [(100 * w * d) / (W * GML)] = 4.38 cm

Now solve for d

4.38 cm = 55 [(100 * 300 * d) / (10,300 * 91.5)] = 55 [30,000 d / 942450]

so d = 2.5m

and this gives my answer so far of 46.5 m from stern for Longtitudinal centre of gravity of cargo.

This is 0.3 m off the answer, so this is what I mean by it seems impossible to get the correct answer of 46.2 m. By putting my answer up along with the question, I am hoping that I will get feedback from someone who knows how this works and can point me in the right direction.

• You have answered your question, good. Apr 20 '20 at 6:39
• @SolarMike Thanks, I've tried my best to get this sorted and I've used steps to make it easy to follow. The reason I put this up, is that it's not quite the answer of 46.2 m, so I need to check if I have this correct. As you may have seen from my approach, I have initially placed cargo load at centre of gravity and to start with treated it like a cargo shift from Longtitudinal centre of flotation. From there I've allowed for sinkage change, going from sea to fresh water and finally worked out shift of cargo to where it should be placed. Is there something else that needs to be allowed for ? Apr 20 '20 at 12:06
• @SolarMike I've sent a message to you on my "Boundary Layer Thickness at end of Train Carriage question", to confirm some important things. Apr 20 '20 at 12:08
• So either you provided an answer or you have asked a question where an answer should be. If it is the second you need to correct that. Apr 20 '20 at 12:10
• @SolarMike i have put answer up with my question, as it shows that I have tried to work this out myself. Are you able to answer my question of is there something else that needs to be allowed for in my answer to this question ? Apr 20 '20 at 12:15

The first step:

We will put the 300 Tonne Cargo Load at the ships centre of gravity of 49 m, so we don't have to correct ourselves.

Firstly we will treat this as if the cargo has a weight shift from centre of flotation to centre of gravity to calculate the aft added draft from the cargo.

We need to run the Change of Trim Formula

by using COT = (100 * LBP * w * d) / (W * GML) with d = 6 m because you want to have initial LCG AT 49m

COT = (100*LBP*300*6) / (10,300*91.5)

Aft Added Draft = (LCF / LBP) * COT = (55 / LBP) * [(100*LBP*300*6) / (10,300*91.5)]

and LBP cancels out,

so Aft Added Draft = (55*100*300*6) / (10,300*91.5) = 10.5 cm at original 49 m CG

The Second Step:

Now we need to take into account sinkage change due to density, for 10,000 Tonne Ship + 300 Tonne Cargo all at 49m CG

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

TPC = 1.025 * 1,480 * 1/100

TPC = 15.17 Tonnes per cm

Parallel Sinkage in Fresh Water = 10,300 / 15.17 = 678.97 cm

And because the Density Changes

at 49 m CG the Sinkage Change = [(1.025 - 1) / (1)] * (10,300 / 15.17) * (49/55) = 15.12 cm

The Third and Final Step:

The cargo will shift from the ships centre of gravity of 49m, to where it should be.

For this part we will firstly have to calculate the remainder draft.

The remainder draft = 30 - 10.5 - 15.12 = 4.38 cm

Like before we need to use Aft added draft formula, but this time we need to find d (distance moved)

So now find d in following formula:

Change in Draft Aft (from shifted load) = LCF [(100 * w * d) / (W * GML)] = 4.38 cm

Now solve for d

4.38 cm = 55 [(100 * 300 * d) / (10,300 * 91.5)] = 55 [30,000 d / 942450]

so d = 2.5m

and this gives my answer so far of 46.5 m from stern for Longtitudinal centre of gravity of cargo.

This is 0.3 m off the answer,

so this is what I mean by it seems impossible to get the correct answer of 46.2 m. By putting my answer up along with the question, I am hoping that I will get feedback from someone who knows how this works and can point me in the right direction.