# Placement of cargo from stern

DELFTship forum Hydrostatics and stability Placement of cargo from stern

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• #39123

I’d like to use the delft ship software to solve this following problem, but I’m finding it tricky trying to get the answer for this.
I have attached my calculations in an excel spreadsheet that I have done so far.
May someone please give me some tips on how to get the answer.
THE QUESTION IS AS FOLLOWS

A ship displaces 10,000 metric tons and area of its plane of flotation is 1,480 m2. The centre of mass is 49 m and centre of area of the 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 tons of extra cargo.

Find minimum allowable distance of mass centre of this extra load from stern if, when ship passes from sea water into a freshwater 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 change in draft and that density of sea water is 1025 kg/m3.

The answer listed in the text book is 46.2m

• #39147

Short answer, without trying it out.

There are 3 stages here. Do them one at a time.

Move the ship from seawater to fresh water, Treat this as a straight sinkage. Count this increase in draft at LCF as being increase in draft aft.

Add the 300 t of cargo at LCF position 55 m from stern. Again, count this as an increase in draft aft.

Calculate how much draft increase is left for the trim. This is 0.3 m – both these previous values.

Calculate trimming moment to generate this trim, using the displacement (now 10 300 t) and the longitudinal metacentric height. From this calculate the shift from LCF of 300 t to generate this moment.

Convert this position to distance from stern (= LCF position 55 m – the shift)

.I’ve just had a look at your spreadsheet. It looks messy relative to my 3 steps as above. DS software is irrelevant to the solution – I can’t see how I could use it.

Regards

Peter Edmonds
Naval Architect
Perth, Western Australia

• #39151

Hi,

It’s been a really long time since I practiced naval architecture (Nixon was president in the US) so I’m rusty but thought a quick Google search might help. I turned up this website which I think might help you.

http://hawaii-marine.com/templates/Trim_Article.htm

Andy

• #39153

Andy, Robert

I’ve had a quick look at Andy’s reference. It spells out the second step I outlined (adding with parallel sinkage), and addresses the third step, in reverse. The example derives draft change for defined shift; you need allowable shift for defined (derived by you) aft draft change.

You are on your own for the first element – the change of water density. This should be quite straight forward for you if you are tackling this kind of problem.

Please let me know if you can’t get to the nominated answer.

I used to teach (briefly) this kind of NA problem many years ago at Sydney Technical College.

Peter Edmonds
Naval Architect
Perth, Western Australia

• #39157

Hi Peter

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

• #43723

@edmondsausstep-com

I studied Civil Engineering some years ago and I had a Fluid Mechanics paper within it.
The textbook for it was Solving Problems in Fluid Mechanics Volume 1 by J F Douglas, which I still have kept.
I am really interested in solving this ship question in it and I have posted it up on delftship a few years ago, but I have really struggled to get the textbook answer of 46.2 m.
I have tried a lot of ways to solve it through forums, but never got a satisfactory result, so I feel that you are my only last hope and if you can help me, that will be much appreciated.

Just to bring back your memory of the question.

A ship displaces 10,000 metric tons and area of its plane of flotation is 1,480 m2. The centre of mass is 49 m and centre of area of the 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 tons of extra cargo.

Find minimum allowable distance of mass centre of this extra load from stern if, when ship passes from sea water into a freshwater canal, the stern draught must not increase by more than 0.3 m.

Assume metacentric height and area of plane of flotation are not altered by change in draft and that density of sea water is 1025 kg/m3.

From what you have told me before, you mentioned there was 3 stages

Move the ship from seawater to fresh water, Treat this as a straight sinkage. Count this increase in draft at LCF as being increase in draft aft.
For this I get TPC = Aw * Seawater Density = 1480* 1.25* 1/100 = 15.17 tpc
So straight sinkage = (1.025 – 1/1) *10300/15.17 = 16.97 cm

Add the 300 t of cargo at LCF position 55 m from stern. Again, count this as an increase in draft aft.
I assume you mean in sea water, so sinkage from cargo = 300/15.17 = 19.78 m

Calculate how much draft increase is left for the trim. This is 0.3 m – both these previous values.
But the problem is 16.97 + 19.78 = 36.75 cm, so this is over 30 cm.

Calculate trimming moment to generate this trim, using the displacement (now 10 300 t) and the longitudinal metacentric height. From this calculate the shift from LCF of 300 t to generate this moment.

I believe this is calculated from

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

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

LBP cancels out on top and bottom

Change in Draught Aft = LCF * [(100 * w * d)/(W * GML)]
from answer if d = 8.8 m,
then Change in Draught Aft = 55 * [(100 * 300 * 8.8)/(10300 * 91.5)] = 15.4 cm

Multiply both sides by (W * GML)

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

and rearranging

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

using what we calculated above d =  15.4 * (10300 * 91.5) / (55 * 100 * 300) = 8.8 m

So to get answer it seems that change in draught aft from trim has to be 15.4 cm, how is this possible ?