Longtitudinal Centre of Gravity for Ship Cargo

Question

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.

Answer 1

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.

Answer 2

I notice you've posted the same question on several pages including CFD Online, Boatdesign, DelfShip, Cruisers Forum, Make: Projects and Ships Nostalgia. This answer is in accordance with the procedure described by Peter Edmonds on the DelftShip forum, although I do make an additional assumption.

First, let me clarify that I do not get an answer similar to the answer provided in the textbook. The textbook you are referring to is Solving Problems in Fluid Mechanics, Volume 1, by John F. Douglas, which includes a lot of questions and single-number answers. I will not go as far as to claim the textbook is incorrect (and hence my answer the true solution), but considering the shear number of Q&As provided, I do not find it unlikely that an error is present in some of the questions and/or answers.

I hereof assume the terminology of ship stability as known, but for those not familiar, let me reference a similiar question of yours for an explanation; Ship Longtitudinal Centre of Gravity Position.

We assume the additional draft from adding the 300 tonne load to the ship is not considered.

Step 1: Parallel sinkage due to change in density In seawater, the ship displaces $\frac{10300 \,tonne}{1.025 \,tonne /m^3} \approx 10048\,m^3$. Moving into freshwater, the change in density gives a new displaced volume of $\frac{10300 \,tonne}{1.000 \,tonne /m^3} \approx 10300\,m^3$, giving a difference of $252 \,m^3$.

The force from extra buoyancy will attack in the center of flotation, meaning that we may consider parallel sinkage of the vessel. We assume the waterplane area to be constant in the vicinity of the current draft, giving an increase in draft of $\frac{252\, m^3}{1480 \, m^2} = 0.17 \,m$.

This leaves $0.13\,m$ before exceeding the limit of $0.3\,m$ additional draft at the stern.

Step 2: Change in trim Since we know the limiting draft left, we may use trigonometry to compute how far the load may be shifted. Recall that the center of flotation is the point about which rotation (trim) takes place, and hence where the trimming moment is taken about.

It follows that the trimming moment is given by $$M_{load} \,x = M_{vessel} \, GM_T \sin \theta$$ where $GM_T = 91.5\,m$, $M_{load} = 300\, tonnes, M_{vessel} = 10300\, tonnes$ and $\theta = \arctan \frac{0.13 \,m}{55 \,m} = 0.135\,rad$. If solving for $x$, the aftwards distance from the center of flotation, we obtain $x = 7.4\,m$. This gives as the final answer a distance of $\underline{\underline{47.6\,m}}$ from the stern.