Calculation of mass centre of cargo from Stern

Discussion in 'Stability' started by Robert Wilkinson, Jan 31, 2018.

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Robert WilkinsonJunior Member

I am trying to workout how to arrive at the answer for the following question.

A ship displaces 10,000 metric tons and area of its plane of flotation is 1,480 m^2. 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/m^3.

Answer from book is given as 46.2m

May someone please guide through how I would arrive at the answer. Help would be much appreciated.

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DCockeySenior Member

Are you a student at a university or similar? Which book?

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IN order to work this out, you need to establish the MCT 1cm for the vessel.

So...what is the MCT 1cm of any vessel...what parameters define it?
Then how to obtain the GM(L)....and since you've been given the waterplane area of centroid, you can establish the BM(L).

Thus after ploughing through these you can finally calculate what the MCT 1cm for the vessel is, and thus work out the amount of trim and thus the moment which means the amount of weight x its lever...the lever being the distance you are required to find.

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Robert WilkinsonJunior Member

I'm an ex student. The text book is Solving Problems in Fluid Mechanics Volume 1 by J F Douglas.
Are you familiar with this type of problem?
I have tried to work out this question myself. I have taken into account the change of draft in sea water from parallel sinkage and trimming moment and when going from sea water to fresh water. This is what I have calculated
In Sea water:
parallel sinkage = w/TPC ; where w = cargo load of 300 Tonnes and TPC = Tonnes per centimetre immersion
Change of draft aft due trim = (Longtitudinal Centre of flotation/ Length between Perpendiculars) * Change of Trim
Change of Trim = Triming Moment/ Moment causing Trim of 1 cm ; where Trimming moment is moment from 300 Tonne cargo (w)
and MCTC = (W * GML)/ 100 * LBP
where W = 10,000 Tonne ship, GML Longtitudinal Metacentric Height
and LBP Length between perpendiculars

In fresh water :

Sinkage = Mean draft in Sea water * [(Sea water density / fresh water density) - 1 ]
So Final mean draft = Initial mean draft * (Sea water density/ fresh water density)

Since centre of buoyancy is in a vertical line with centre of gravity
I have calculated shift in centre of buoyancy = BB1 = Increment of volume * difference between LCB and LCF / (Underwater volume of vessel + Increment of volume)

From this Moment Changing Trim = W * BB1

So Change of Trim aft from change in density = Moment Changing Trim/ MCTC

and this is where I got the ugly equation of
= 0.1922LCF - (LCF * 100 * 7.31707d/ LBP)
and I have got stuck from here because d (centre of mass to centre of flotation) and LBP (length between perpendiculars) are unknown

Last edited: Feb 2, 2018
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Robert WilkinsonJunior Member

I have tried to work out this question myself. I have taken into account the change of draft in sea water from parallel sinkage and trimming moment and when going from sea water to fresh water. This is what I have calculated
In Sea water:
parallel sinkage = w/TPC ; where w = cargo load of 300 Tonnes and TPC = Tonnes per centimetre immersion
Change of draft aft due trim = (Longtitudinal Centre of flotation/ Length between Perpendiculars) * Change of Trim
Change of Trim = Triming Moment/ Moment causing Trim of 1 cm ; where Trimming moment is moment from 300 Tonnes cargo (w)
and MCTC = (W * GML)/ 100 * LBP
where W = 10,000 Tonne ship, GML Longtitudinal Metacentric Height
and LBP Length between perpendiculars

In fresh water :

Sinkage = Mean draft in Sea water * [(Sea water density / fresh water density) - 1 ]
So Final mean draft = Initial mean draft * (Sea water density/ fresh water density)

Since centre of buoyancy is in a vertical line with centre of gravity
I have calculated shift in centre of buoyancy = BB1 = Increment of volume * difference between LCB and LCF / (Underwater volume of vessel + Increment of volume)

From this Moment Changing Trim = W * BB1

So Change of Trim aft from change in density = Moment Changing Trim/ MCTC

and this is where I got the ugly equation of
= 0.1922LCF - (LCF * 100 * 7.31707d/ LBP)
and I have got stuck from here because d (centre of mass to centre of flotation) and LBP (length between perpendiculars) are unknown

Last edited: Feb 2, 2018
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HeimfriedSenior Member

Hi Robert,
is there a possibility, you didn't quote the question completely? Or, maybe the question in the book is not worded correctly.
Thoughts:
I don't see whether the ship is on even keel before the additional load is put on it.
The trimming moment induced by additional load occurres in sea water. But since the initial trim is unknown, the additional (or contrary) trim is also unknown.
The question is about the inceased draft caused by change from sea water to fresh water. This change would cause a parallel sinking without any change in trim, if the ship is wall sided.
After additional loading the ship has a DISPM = 10300 t, which is in sea water a DISPV(sw) = DISPM / (1.025 t/m^3) = 10049 m^3 in fresh water DISPV(fw) = DISPM / 1.000 t/m^3 = 10300 m^3 .
Additional Volume ist DiffV = DISPV(fw) - DISPV(sw) = 10300 m^3 - 10049 m^3 = 251 m^3 .
So the additional draft is 251 m^3 / 1480 m^2 = 0.170 m because the waterplane area will not change. No influence from any trimming moment.

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Robert WilkinsonJunior Member

Are you able to show with a calculation how this works for this question?
With GM (L) do you mean the metacentric height for pitching motion about transverse principal axis (which is longtitudinal metacentric height) for this question?
Is BM (L) the centre of buoyancy to metacentre in the longtitudinal direction and we've been given centre of area of waterplane, but is this the same as centre of buoyancy for this question?

Last edited: Feb 3, 2018
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Robert WilkinsonJunior Member

Hi Heimfried

I've looked at the question and it was definitely the same as in the book. The book is Solving Problems in Fluid Mechanics Volume 1 by J F Douglas.
I think that with this question the ship is on an even keel before the cargo load is added.
From various websites I have searched I understand that there will be a parallel sinkage and trim from when the cargo is added. I listed those calculations before.
But, where I am having difficulty understanding is when it passes from sea water into fresh water.
It seems like you have to take into account change in centre of buoyancy due to the change in density.
The question mentions that the stern draft must not increase by more than 0.3m. Does this mean the change in stern draft only when it changes water density or throughout the whole process when ship gets loaded with 300 Tonnes of cargo and then passes from sea water to fresh water.

I don't know if you noticed adhoc reply to my question, Adhoc said
IN order to work this out, you need to establish the MCT 1cm for the vessel.

So...what is the MCT 1cm of any vessel...what parameters define it?
Then how to obtain the GM(L)....and since you've been given the waterplane area of centroid, you can establish the BM(L).

Thus after ploughing through these you can finally calculate what the MCT 1cm for the vessel is, and thus work out the amount of trim and thus the moment which means the amount of weight x its lever...the lever being the distance you are required to find.

This reply was listed above

This is my thoughts on that
MCT 1cm = Ships Displacement * Longtitudinal Metacentric Height/ 100 * Length between perpendiculars = W * GM (L)/ 100 * LBP
Change of trim = Moment changing trim/ MCT 1cm
Moment changing trim is the confusing bit, because is it the same from when it first gets loaded and then passes into fresh water
Change of trim aft = Centre of flotation/ LBP * Change of trim
Change of trim aft is the same as change of draft aft
In this question we don't know at the start LBP and distance cargo is from stern

I understand from parallel axis theorem that Second moment of area can involve area * distance squared from centroid for part of its calculation.
BM (L) I think means Metacentric radius longtitudinally and this is calculated from Second moment of area / Volume displaced.
GM (L) = 91.5m = BM(L) - BG (L), but how do you calculate BG (L)

Still an unsolved question for me, do you know of any naval architects who would know how to solve this?
Hopefully I can get this question solved some day.

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TANSLSenior Member

Imho, no one in this world, not even a naval architect, who is supposed to be a smart guy, could solve this problem if you do not provide more data.

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HeimfriedSenior Member

Hi Robert,
your trials to solve the problem are about the same as mine, when I tried it. In my opinion there is no way to solve this problem with the given data as Tansl says. (And I think, Ad Hoc was just sketching the usual way to solve such a problem, without doublechecking if the given data are sufficient.)

I think, the problems to solve in this book are worded a bit careless.

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Correct....i've not had time to look deeper into the question, just gave an outline of the procedure.

I may have spare time soon..

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Robert WilkinsonJunior Member

That's fine just when you have some spare time

I have actually put this question on delft ship forum as well called "Placement of cargo from stern"
this is the link DELFTship :: Topic: Placement of cargo from stern (1/1) http://delftship.net/DELFTship/index.php/forum/hydrostatics-and-stability/3865-placement-of-cargo-from-stern#5440

From the feedback I was told the following

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)

This was my attempt at using this method

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?
This is my best attempt at trying to get close to this answer.

I have noticed the centre of gravity of 49m from the stern for the ship has not been used in any of the calculations.
Do you have any suggestions on something I may have missed.
Your help so far is much appreciated.

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TANSLSenior Member

My advice, wait for Ad Hoc to have some spare time.

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Robert WilkinsonJunior Member

Just wondering if you managed to get some spare time to look over my question that I have not found the answer to.
This is from a textbook that have from a Civil Engineering course that I did some years ago.

A ship displaces 10,000 metric tons and area of its plane of flotation is 1,480 m^2. 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/m^3 and density of fresh water is 1,ooo kg/m^3.

Answer from book is given as 46.2m

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TANSLSenior Member

No, this is not correct.

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