Help on calculations

Discussion in 'Stability' started by Grimorum, Apr 26, 2016.

  1. Grimorum
    Joined: Mar 2016
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    Grimorum Junior Member

    Hi All,

    I have encountered a questions on grounded ship and been unable to obtain the answer stated in the book.

    info: Displacement of ship = 5588 tonnef,
    Forward draughts: 4.57m,
    Aft draught: 5.03m,
    Forward marks from midship = 70.1m
    Aft marks from midship: 51.8m
    LBP: 146m
    TPC: .204MN/cm
    CF: 4.57 AFT OF MIDSHIP
    MCT BP: 1560 MN per metre

    The ship is grounded at a location 22.9m forward of midship, what is the force at the keel and what is the new draught at the draught marks when the tide falls 0.3m?

    I have done the calculations as follows:

    Assume the force at keel is P, the moment applied at LCF is 27.47P.

    parallel rise : P/20.4

    Change of draught at rock due to trimming: 27.47/146 * 27.47P/1560

    Total change of draught at rock = P/20.4 + 27.47/146 * 27.47P/1560 = 0.3

    The P I obtained is 5.73MN, but the answer in the book is 3.72MN.

    So I have no idea where I went wrong and I googled for answer but to no avail.

    Appreciate the guidance from all the esteemed architects here.

    Thanks,
     
  2. Rabah
    Joined: Mar 2014
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    Rabah Senior Member

    Hello Grimorum,
    It is a task № 18, page 88 from “Basic ship theory” of Rawson and Tupper, fifth edition, volume 1, 2001 /see the file/.
    But you abnormally quote it! The specified drafts are gauged not for FP and AP, but on the bow and the stern marks of drafts!
    But irrespective of it your calculation is exact. There is no error as you yet did not calculate the drafts. I too have received P = 5,734 MN.
    However I have found an error in condition of tasks:
    Two prescribed values contradict one another - first it is Displacement =5588 tonnef and second TPC = 0,204 MN/cm = 20,4 MN/m.
    One of them is false!
    1. If we shall take that Displacement =5588 tonnef is valid value, then
    D = 5588 tonnef = 54,8 MN
    Application of similarity of triangles easily we define values of drafts for initial WL up to a beaching:
    AP = 5,11m
    Aft Mark of draft = 5,03m /according to the task/
    Middle = 4,834m
    Fore Mark of draft =4,57m /according to the task/
    FP = 4,558m

    Or at T middle = 4,834m we have q = D / T middle = 54,8 MN / 4,834m = 11,336 MN/m
    I.e. q = 20,4 MN/m is false value.

    2. If to accept that D = 5588 tonnef is false value, and q = 20,4 MN/m is valid, then
    D = q * T middle = 20,4 * 4,834 = 98,6 MN> 54,8 MN
    If to accept q = 20,4 MN/m for valid value and generally to skip Displacement as it is made in solved Example 6/Fig 3.37/, page 81 from the same book then we can receive answer P = 3,72 MN only if MCT will not be 1560 MN*m per 1m trim, but it was equated on 162,7 MN*m. In this case are received Tfp = 4,043m and Tap = 5,222m for new WL when the tide has fallen 30 cm. Applying similarity of triangles it is defined and drafts at Marks.

    However as Naval Architect I would advise you to prolong calculations thus as I have made it:
    Has accepted value D = 5588 tonnef for the given and has prolonged with q = 11,336 MN/m and MCT = 1560 MN*m per 1m trim.
    In outcome has received the following:
    P = 3,28 MN
    Parallel rise = 3,28 / 11.336 = 0,289m due to P
    Trim = 0,0577 m from the moment of the force P
    Change of draft at FP from the trimming moment = 0,0018m
    Change of draft at AP from the trimming moment = 0,027m
    New draft at FP when the tide has fallen 30 cm:
    4,558 - 0,289 - 0,0018 = 4,267m
    New draft at AP when the tide has fallen 30 cm:
    5,11 - 0,289 + 0,027 = 4,848m
    Receiving similarity of triangles it is discovered and drafts at the Marks.
    We receive following Drafts for new WL when the tide has fallen 30 cm:
    AP = 4,848m
    Aft Mark of draft = 4,764m
    Middle = 4,557m
    Fore Mark of draft =4,278m
    FP = 4,267m

    ____________________
    NA Razmik Baharyan
     

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  3. Heimfried
    Joined: Apr 2015
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    Heimfried Senior Member

    Hello Razmik

    A given TPC is in my understanding related to the actual draft. Your calculation uses the TPC as an average value. It is an cargo ship and a typical hull shape will show at 1 cm draft ("just dipped in water" on even keel) a waterline area which is much lesser than the waterline area at 5 m draft.
     
    Last edited: Apr 28, 2016
  4. devudegoa
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    devudegoa Junior Member

    Interesting observations by everyone above. Working backwards from the textbook answer (3.72), it seems the value of TPC should be in the range of 12.92 tf/cm (about 0.1267 MN/cm). This also gives a logical displacement which answers the query raised in the third comment above.
     
  5. jehardiman
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    jehardiman Senior Member

    Yes this is problem 18 from Chapter 3 of BST and yes the answer given in the text is correct. Before you plug and chug read the problem and draw a diagram showing all the points.

    Notice that the MCT is between PERPENDICULARS, not between draft marks.

    Soooooo….

    First calculate state before grounding
    draft at midships:
    16.5 –(16.5-15)*(170)/(230+170)= 15.8625 = 15’ 10.35”
    drafts at the perpendiculars:
    FWD: 15.8625-(16.5-15)*240/(230+170) = 14.9625 = 14’ 11.55”
    AFT: 15.8625+(16.5-15)*240/(230+170) = 16.7625 = 16’ 9.15”
    Trim between Perps = 1.8’ = 21.6”
    Draft at center of flotation = 15.8625+(16.5-15)*15/(230+170)= 15.91875 = 15’ 11.025”

    Now ground the ship
    Draft at grounding point: 15.8625-(16.5-15)*75/(230+170) = 15.58125 = 15’ 6.975”
    Total change of draft at the grounding point
    (P/52)+((75+15)/480)*((75+15)*P/1300) = 12”
    P(0.0192+0.0129)=12

    P=373.8 tons

    Parallel Rise in draft @ center of flotation = 373.8/52 = 7.188” = 0.599’
    New draft at center of flotation = 15.91875-0.599 =15.31975 = 15’ 3.875”
    Change in trim between perps = 373.8*(75+15)/1300 = 25.87”
    New trim between perps = 21.6 +25.87 = 47.47” = 3.95’

    New draft at
    Midships = 15.31975 - 15*3.95/480 = 15.196 = 15’ 2.35”
    FWD = 15.31975 – (15+230)*3.95/480 = 13.303 = 13’ 3.6”
    AFT = 15.31975 + (170-15)*3.95/480 = 16.595 = 16’ 7.14”

    And as a check
    Draft at grounding point: 15.31975 – (15+75)*3.95/480 = 14.579 = 14’ 6.94”
    So with rounding the difference is 0.035”

    Note, however the problem with this method. The change in draft at midships is 7.998" (15.8625-15.196 = 0.6665') and therefore implies a grounding force or decrease of displacement of 415.9 tons, not 373.8 tons. This is the issue with a normal set of D&O's when the ship is way off trim. Personally, as a salvor's agent, I'd do this with the Bonjeans or the digital lines that most ships now carry.
     
  6. Heimfried
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    Heimfried Senior Member

    Hello jehardiman

    The value of 1300, which schould be the MCT BP, but in what units?

    In the SI units version of BST the MCT BP is given with "1560 MN/m per m".

    (a Moment should be in MN*m)
     
  7. Rabah
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    Rabah Senior Member

    Hello jehardiman,
    Due to your calculations I have decided the puzzle!
    What is wrong in the fifth edition, 2001 of the book “Basic Ship Theory”?
    There is one zero more in the value MCT!
    In replacement MCT = 1560 MN*m per 1m trim it is necessary to be MCT = 156 MN*m per 1m trim
    I understood that you have made the calculation by the following input data:
    TPC = 52 t / inch
    MCT = 1300 t*feet / inch
    Or TPC = 52/2,54 t/cm = 20,47 t/cm = 0,20 MN/cm
    MCT = 1300*0,305/2,54 t*m per 1cm trim = 156,1 t*m/cm = 153 MN*m/1m trim
    Obviously in your edition of the book all measurement units specified in the American system as well that the editor and a corrector are seriously have made his work.
    I have tested - full coincidence of the outcome was received from us with you for drafts on initial WL up to the beaching / between foots and meters/.
    I have noted that you anywhere have not accepted in the calculation the value Displacement = 5588t. Obviously this value is erratic also excessive.
    If we shall make calculation with ultimately adopted input data, i.e. TPC = 0,204 MN/cm and MCT = 156 MN*m/1m trim, we shall receive P = 3,65MN = 372,38t. It means that the answer P = 3,72 MN too is not absolutely valid, but it is not so important.
    _____________________
    NA Razmik Baharyan
     
  8. Rabah
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    Rabah Senior Member

    Hi Heimfried,
    It is pleasant to see at the forum people which think!
    I congratulate you for good intuition! You have correctly guessed that TPC = 0,204 MN/cm is valid value!
    My calculation in which I am trusted on Displacement up to the beaching, but I do not accept TPC, will be exact only in the case when the vessel is fast-running.
    Then Cb = 0,55 till 0,64 for fast-running dry-cargo ships.
    If to accept Cb = 0,55 and B = 14m, in the sea it is received:
    D = γ *Cb*L*B*T = 1,025*0,55*146*14*4,834 = 5570t
    Existence of such vessel is quite probable, but for it will be smaller load-carrying capacity in comparison with the prototypes mentioned below, for the sake of smaller breadth and fine outlines.
    If the vessel is slow-speed the situation is different.
    Let's see in the books published in Ukraine in 2007 and 2008 for designing and construction of vessels of the mixed floating /river -sea/. The author - Professor G. V. Egorov, Doctor of Technical Sciences.
    We select the prototypes with close ship dimensions on L and maximum draft:
    Project 19610 "Volga" -Dry cargo carrier
    Project RSD12 - Dry cargo carrier
    Project 19612 - Tanker
    For them velocity is 10-11kt, B = 16,4 - 18m and Cb = 0,846 - 0,893
    We accept Cb = 0,847 and B = 16,4m
    It is received D = 1,025*0,847*146*16,4*4,834 = 10048,7t = 98,54 MN
    This calculation is very close with my calculation from the post №2 if TPC = 20,4MN/m is valid value:
    D = q * T = 20,4*4,834 = 98,6MN
    _______________________
    NA Razmik Baharyan
     
  9. devudegoa
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    devudegoa Junior Member

    @Rabah
    Thank you for that explanation Sir. As you mentioned, the answer is not important. What is important for a student of that textbook is to grasp the concept of how to solve that type of a problem. There is another similar problem a couple of pages before, which explains the concept very well without any error. As far as this problem is concerned, it is very difficult to say what the author had in mind while framing the question. Any way, it was a good learning exercise. :)
     
  10. jehardiman
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    jehardiman Senior Member

    The edition I used was the 2nd edition, 4th impression, 1981. LBP = 480 ft, TPI = 52 Tons/in, MCT1" BP = 1300 ft-tons. CF = 15 ft aft midships, Draft at marks 15.0 ft @ 230 ft fwd of midships, 16' 6" at 170 ft aft of midships.
     
  11. Rabah
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    Rabah Senior Member

    Hi devudegoa,
    We think equally with you, but for all that very much helps me wide experience because I am two times older than you. Always I test used formulas under several books and I compare to shun an error, especially if there is a variance in measuring systems. I am especially skeptical to computer programs - never I believe on made calculations only on one program. Always I test the account not less than on two programs. Life has learned!
    _____________________
    NA Razmik Baharyan
     
  12. Rabah
    Joined: Mar 2014
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    Rabah Senior Member

    Hello everybody,
    For those from you which are interested in computer calculation of the grounding I want to show the capabilities of the program Maxsurf Stability.
    See the file from the Maxsurf Stability Manual.
    ______________________
    NA Razmik Baharyan
     

    Attached Files:


  13. Niru
    Joined: Jul 2010
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    Niru Mr.

    Good day Rabah.

    Have to check this on my free time. never thought its possible on maxsurf stability.

    Thanks for the info.
     
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