Calculation of mass centre of cargo from Stern

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

  1. Ad Hoc
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    Ad Hoc Naval Architect

    Hi Robert,
    Sorry i forgot to follow up.
    Still a bit busy - again - however. Ive had a more detailed look, on the face of it - it is a simple straight fwd question, not overly difficult.
    Solving this algebraically is not so difficult using the first principals of each function and arriving at a common parameter. The problem comes is when reducing the problem down to an answer, there always appears to be 1 unknown.

    Are you 100% sure what you have copied - the question above - is correct?..since it appears on further inspection to be missing 1 key data. Since if i were to spend much more time on this - more than the 20mins or so that I have, suggests that either i) data is missing or ii) this is much more complex and requires a more detailed review of all the fundamentals, to test the student, of each term to isolate a single parameter; which suggests to me this is beyond the level of your course, perhaps (from your previous questions).

    Thus without spending more time than I have already, either a key data is missing or it is a cunning question to out fox you, the student, and those like me that haven't done these simple (but cunning) calc's for many years and i'd need to open up my old dusty books again to remind myself! Thus i would suggest that i) is more likely! It wouldn't be the first time - we had some errors in our Uni text books too, drove me nuts at the time - hahaha.
     
  2. Robert Wilkinson
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    Robert Wilkinson Junior Member


    Hi Adhoc

    Thanks for the reply.
    This is the quoted question from my Solving Problems in Fluid Mechanics Volume 1 by J F Douglas. I see what you mean sometimes on the odd occasion these textbooks can have one unknown.

    Can you please tell me what the unknown is to get the answer for this question.
    From the answer we know that Cargo load is 8.8m from stern, giving a trimming moment from centre of flotation.
    Maybe the unknown can be worked out from this and the question can be reworded to include this.

    When you first replied to my post this is what you mentioned.

    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.


    From looking through websites I have found that you can firstly use parallel sinkage at centre of flotation

    Change in draft aft from this will be the same as the centre of flotation and this = W / TPC
    W = Total mass of ship = 10,300 Tonnes and TPC = Tonnes per cm

    Change in draft aft from going from sea water to fresh, will be the difference in what's calculated for each

    Next you can calculate change in draft aft from moving cargo and possibly shift own mass of 10,000 tonnes to it's CG of 49m.

    Change in Draft aft = LCF / LBP * Change of Trim

    Change of Trim (COT) = Trimming Moment / MCTC

    With Trimming Moment what does it consist of ?
    Is it Cargo Load * 8.8 + Ship mass * Ship Lever
    and what is the Lever for the Ship, this is what I am confused by?

    MCTC = (W * GML)/ 100 * LBP where W = Total mass of ship = 10,300 Tonnes, GML = Metacentric Height = 91.5 m and LBP = unknown, but you don't need this because it cancels out later

    so Change in Draft Aft from this = LCF * [(100 * w * d)/(W * GML)]
    and the Change in draft aft going from sea water to fresh, will be the difference in what's calculated for fresh and sea water .

    Total change in draft aft going from sea water to fresh water = 30 cm = total from parallel sinkage change and mass movement changes


    I have also tried to do this another way on my post How to find Longtitudinal Centre of Gravity for Ship Cargo https://engineering.stackexchange.com/questions/29914/how-to-find-longtitudinal-centre-of-gravity-for-ship-cargo
    , but this takes into account Implied CG (due to change in volume) and I am not sure that this is the correct approach.

    Not sure how the BM (L) you mentioned can come into this because we can't use simpsons rule, half ordinates and parallel axis theorem to work it out.
    BM = area * distance^2 , but it seems that that is not enough to calculate BM

    I'd appreciate your reply on how you would calculate this if all data is known and also from what I have mentioned, because I think once we come to a final conclusion on all of this, this question can be ticked off.


    Cheers
    Robert
     
    Last edited: Sep 7, 2019
  3. Heimfried
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    Heimfried Senior Member

    Hi Robert,

    some thoughts, leading to a far more different solution.

    The question only makes sense if you compare the initial draft at stern before the additional load with the draft after loading and transition in fresh water. I think you calculations shown here are missing the first parallel sinking due to the additional load in sea water. For calculation it is common to imagine the load is first placed in the CoG of the ship, causing parallel sinking only and then moved to its real place causing trim (and heel, if so).

    The added load of 300 t placed in the CoG causes additional DISPV = 292.7 m³, divided by 1480 m² flotation area the increase in draft is 0.198 m (parallel sinking). The transition from sea water to fresh water generates additional 0.17 m increase in draft (10300 t / 1.000 t/m³ - 10300 t / 1.025 t/m³ = 251.2 m³; 251,2 m³ / 1480 m² = 0.17 m).

    The sum of both sinkings is 0.37 m and already more than the allowed 0.3 m. That means the added load has to move forward not aft to reduce the draft at stern by 0.07 m.
    The distance from the CoG is e = DISPM / m * GM(l) * tan(theta) ; tan(theta)= 0.07 m / 55 m
    e = 10300 t / 300 t * 91.5 m * (0.07 m / 55 m) = 4.00 m
    This would lead to a distance from stern 49 m + 4 m = 53 m .

    Edit: corrections needed
     
    Last edited: Sep 15, 2019
  4. Robert Wilkinson
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    Robert Wilkinson Junior Member

    Hi Heimfried

    Yes I see what you mean, you have used difference in volume and trim angle and change in draft = change in volume / waterplane area.

    My best attempt to get close to the answer of 46.2 m is on this post How to find Longtitudinal Centre of Gravity for Ship Cargo https://engineering.stackexchange.com/questions/29914/how-to-find-longtitudinal-centre-of-gravity-for-ship-cargo
    I got the idea for this from Phil Sweet who was on my original post, but didn't reply when I wanted to clarify something.
    It has implied centre of gravity in it, is this the change in centre of gravity due to changes in volume moving from sea water to fresh water?
    It also involves putting weight at cg of 49 m.
    Do you understand this way of doing it? It gets close to the answer of 46.2 m

    I don't know if you have seen my recent reply to Ad Hoc

    That involves placing mass of 10,300 Tonnes at Centre of flotation to get parallel sinkage where draft changes evenly and then moving the ship mass of 10,000 Tonnes to 49 m cg and 300 Tonne mass to it's 46.2 m cg from stern.
    The problem is what would the levers be for the weight shifts to create trimming moment.
    This is what I am confused by.

    What I am thinking is where is the centre of buoyancy because doesn't that create the lever for the ship mass from cg of 49 m
    and is lever for the 300 Tonne cargo 8.8m from centre of flotation

    It seems like an OK question at first doesn't it, but when you go through the calcs it seems like there is an unknown and it is a trick question.

    I might have to ask Ad Hoc for a reply.
     
  5. Heimfried
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    Heimfried Senior Member

    This is not to happen. Because the center of flotation deviates from center of gravity (so there is a lever), putting load in the CoF creates a trimming moment: no parallel sinking.
    The textbook question demands to assume no change in waterplane area is caused by changing draft, so the ship is considered wall sided.
    In this case only loading in the CoG causes parallel sinking.

    Edit: Corrections needed.
     
    Last edited: Sep 15, 2019
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  6. Heimfried
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    Heimfried Senior Member

    Robert why do you think, a change in draft is causing a shift of CoG? The the location of the CoG is a property of the whole ship in a given state, not the property of the immersed part of the hull. Change in CoG will only occure, when any masses which are on the ship or part of the ship (loads, crew, gear, tank content, ...) are shifted or moving (relatively to the ships coordinate system).

    Edit: corrections needed.
     
    Last edited: Sep 15, 2019
  7. Robert Wilkinson
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    Robert Wilkinson Junior Member

    This was an idea that Phil Sweet had, I based calculation in Engineering Stack Exchange from that.
    I agree with you on centre of gravity, so it won't change if the ship goes from sea water to fresh water.
    The only time centre of gravity changes is in when the weight moves.
     
  8. Robert Wilkinson
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    Robert Wilkinson Junior Member

    I got the idea of weight placed at centre of flotation and moved from here LESSON TOPIC: https://fas.org/man/dod-101/navy/docs/swos/dca/stg4-04.html.
    Also doesn't the draft trim about centre of flotation and isn't the mean draft at centre of flotation, this is what has lead me to think this way about it.

    Is this question a different situation because centre of gravity is 6 m behind centre of flotation?

    I have found on this Quick Barge Stability and Trim http://hawaii-marine.com/templates/barge_trim_list_stability.htm that Trim Lever = LCG – LCB .
    This would be for total ship weight though.


    From what you have told me these are now my thoughts.

    So if both weights are put at 49 m cg, then the only lever involved will be from the 300 Tonne Cargo.
    If the answer of 46.2 m from stern is correct, then the lever will be 2.8 m and we can calculate lever from change of trim aft like I have shown before. This will come last though.

    Because the ship is already loaded in sea water, the parallel sinkage will be the difference between what's calculated for sea and fresh water.
    This will involve 10,300 Tonne and TPC at CG of 49 m.
    In sea water TPC = 10,330/ (1.025 * 1,480 * 1/100) = 678.97 cm
    In fresh water TPC = 10,330/ (1.0 * 1,480 * 1/100) = 695.95 cm
    So Parallel Sinkage difference = 16.98 cm and this will cause change in draft aft

    Now we have 30 - 16.98 = 13.02 cm remaining

    The rest of this is from change of Trim Aft from 300 Tonne Cargo and this where LCF comes into it

    Change of Trim = COT = (100 * LBP * w * d) / (W * GML)

    Change of Trim Aft = LCF / LBP * COT

    Change of Trim Aft = LCF * [(100 * w * d) / (W * GML)] = 55 * [(100 * 300 * d) / (10300 * 91.5)]

    Change of Trim Aft = 55 * [(30,000 d) / (10300 * 91.5)] = 1.75 d

    so 13.02 = 1.75 d, so d = 7.44 m

    so Cargo from stern = 49 - 7.44 = 41.56 m

    The only other way I can think of is just use 10,000 Tonne to do parallel sinkage calcs and then add 300 Tonne Cargo at LCF and calculate change of draft aft from this movement.
    From what you have told me maybe this is not possible.

    God knows how you get the book answer of 46.2 m, to me this is a trick question or there is something missing.

    The only other thing is if they have a solutions manual online or from a library.
    My textbook is Solving Problems in Fluid Mechanics volume 1, Author is J F Douglas and Published by Longman Scientific & Technical in 1986.
     
    Last edited: Sep 15, 2019
  9. Heimfried
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    Heimfried Senior Member

    My statements above contained mistakes, sorry for that (source of this mistakes were papers from a nautic school). I need some time to rethink this.
     
    Last edited: Sep 16, 2019
  10. Heimfried
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    Heimfried Senior Member

    Following this link I was amazed because my knowledge was different (it was based on a faulty written lesson of a nautical school). Anyway - the double checking with a few books on ship theory convinced me, that parallel sinking is caused by a additional load placed at longitudinal center of flotation (lcf), not at longitud. center of gravity (lcg). Precondition (no heel) is the lcf remains at its location and the added load is small compared with vessels displacement. Next step was to understand why.

    Not that compliceted: the caused sinking creates a "slice" of displaced water, thickness of this slice equals amount of sinking its shape is the same as the waterline area. So the partial longitudinal center of buoyancy (lcb(slice)) is at lcf, the original part of lcb(orig) has a certain distance from it. The new lcb(2) is located between them vertically aligned with the new lcg.

    I'll try another attempt to calculate.
     
  11. Heimfried
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    Heimfried Senior Member

    Another try, another result.

    DISPM = 10,000 t
    POF = 1,480 m²
    LCG = 49 m
    LCF = 55 m
    MGL = 91.5 m
    m = 300 t
    T(diff,max) = 0.3 m
    rho (sea) = 1.025 t/m³

    rho (fresh) = 1.000 t/m³
    DISPV(sea) = 10,000 t / 1.025 t/m³ = 9,756 m³
    DISPM2 = DISPM + m = 10,300 t
    DISPV2(sea) = 10,300 t / 1.025 t / m³ = 10,049 m³
    DISPV2(fresh) = 10,300 t / 1.000 t/m³ = 10,300 m³
    LCB = LCG (at equilibrium)
    psi = trim angle

    1.
    The additional load m is placed at LCF to cause parallel sinking. The increase in displacement volume is DISPV2(sea) - DISPV(sea) = 293 m³.
    The sinking is equal to the thickness of a slice of Water with the area of POF: 293 m³ / 1480 m² = 0.2 m
    This parallel sinking causes a shift forward in LCG of (LCF - LCG) * m / DISPM = (55 m - 49 m) * 300 t / 10.000 t = 0.18 m
    New LCG is LCG2 = LCG + 0.18 m = 49.18 m ; LCF - LCG2 = 5.82 m

    2.
    The additional load is then shifted aft by a (yet unknown) distance d.
    This causes a further difference in LCG of m * d / DISPM2 = 300 t * d / 10,300 t = 0.0291 * d
    Because the load is shifted towards stern, the new LCG3 also does: LCG3 = LCG2 - 0.0291 * d = 49.18 m - 0.0291 * d ; LCF - LCG3 = 5.82 m + 0.0291 * d = e

    3.
    The ship now goes in fresh water canal. The displaced volume increases DISPV2(fresh) - DISPV2(sea) = 10,300 m³ - 10,049 m³ = 251 m³
    This causes a sinking at COF of 251 m² / 1,480 m² = 0.17 m .
    For the transition in fresh water the stern draft shall be of 0.3 m maximum. From this 0.3 m are 0.13 m left for additional trim.
    The trim is caused by change in LCB. A part of 251 t is moved by 5.82 m - 0.0291 * d
    tan(psi) = m * e / (DISPM2 * GML) ; added stern draft 0.13 m = LCF * (m * e) / (DISPM2 * GML);
    e = 0.13 m * DISPM2 * GML / (m * LCF) = 0.13 m * 10.300 t * 91.5 m / (300 t * 55 m) = 7.43 m
    d = (e - 5.82 m) / 0.0291 = (7.43 m - 5.82 m) / 0.0291 = 55.3 m

    This result is far away from textbook and is not probable.
    May be my calculation is wrong, may be the textbook question is wrong worded, may be both of it.

    I hope for questions/corrections/helpful remarks.
     

  12. Heimfried
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    Heimfried Senior Member

    Six weeks passed by, nobody interested to make a remark?
     
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