Longitudinal Stability Yuvaraj Adithya

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    By

    Yuvaraj R

    AdithyaManda

    LONGITUDINAL STABILITY

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    CONTENTSLongitudinal ta!ilityLongitudinal "#ta$#ntr#

    MCTC %

    Tri"Lot !uoyan$y "#thod

    Add#d ight "#thod

    'looda!l# l#ngth

    'a$tor o( u!diviion

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    Longitudinal StabilityLongitudinal Metacenter: Similar to the definition of the

    transverse meta center, when a ship is inclined longitudinallyat a small angle, A vertical line through the center of

    buoyancy intersects the vertical line through (before the

    ship is inclined) at .

    1B

    0B

    LM

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    The Location of the LongitudinalMetacenter

    For a small angle inclination, volumes of forward wedge immersedin water and bacward wedge emerged out of water are!

    10

    " 1 "0

    0 0

    (" ) ( tan ) where is the half breadth.

    (" ) ( tan ) , .

    #hus, " " , which indicates!

    moment of area forward of $ moment of area after .

    is the cente

    l

    L l

    l L l

    v y x dx y

    v y x dx v v v

    yxdx yxdx

    F F

    F

    =

    = = ==

    0 0

    0 0

    r of (mass) gravity of waterline , %

    is called of . #herefore, for e&ual

    volume longitudinal inclination the new waterline always

    passes through the ('.F

    W L

    W Lcenter of flotation

    center of flotation ).

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    Location of the LongitudinalMetacenter

    sing the same argument used in obtaining transverse metacenter.0 1 1 "

    0

    1 "0

    0" "

    0

    0 1 0

    ,

    (" ) ( tan ) " tan

    tan " " tan

    is the moment of inertia with respect to the

    transverse a*is passing the center of flotation.

    t

    l

    L l

    l

    FCL l

    FC

    B B vg g

    vg g y x xdx yx xdx

    yx dx yx dx I

    I

    B B B M

    =

    = +

    = + =

    =

    0

    0

    an , .

    .

    FC

    FC FC

    ML B B L B g

    IB M

    I IH B M Z Z GM Z Z

    =

    = + = + = +

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    Location of the LongitudinalMetacenter

    "

    0,

    0,

    sually Floatation 'enter ('.F) of a waterplane is not at the

    midship,

    ,

    where is the moment of inertia w.r.t. the transverse a*is

    at midship (or station +) and is the distance from F

    FC T

    T

    I I Ax

    I

    x

    =

    .'. to

    the midship.

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    Moment to Change Trim One cm(MCTC 1)

    #-! (moment to alter (change) the ships trim per inch) at

    each waterline (or draft) is an important &uantity. /e may

    use the longitudinal metacenter to predict #-

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    MCTC 1 ( a function of draft)ue to the movement of a weight, assume that the ship as 1 trim,

    and floats at waterline /.2., also e*pressed as MTIin inches

    ( )

    ( )

    0 1

    0 1

    0 1 0

    13 1tan , where is in feet.

    1"

    ue to the movement of the weight, moves to ,,

    tan tan

    tan tan

    1

    1"

    w

    L L G

    FCw L G w B G

    FCw B G

    LL L

    G GM w h G G

    G G G M HM Z

    IM HM Z Z Z

    IZ Z

    L

    = =

    = =

    = =

    = = + = +

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    MCTC 1 ( a function of draft)

    -f the longitudinal inclination is small, MCTC 1can be used to

    find out the longitudinalposition of gravity center ( ).

    4

    1 ,

    1" 1"

    56 lbft , 2ong #on $ ""60 lb, #- (ton7ft)6"0

    FC FC w FCG B w

    FCw

    I I IZ Z M

    L LI

    L

    >> = =

    = =

    1GX

    0 1

    0 0 0

    1 0 1

    #rimtan

    tan ,

    Since is in the same vertical line as . under ,

    F A

    FC FC FCB G G B

    FC

    G B B

    T T T

    L L L

    I I ITG G Z Z Z Z L

    G C B W L

    I TX X G G X

    L

    = = =

    = +

    = + +

    )

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    when a compartment is open to S

    0 0

    1 1

    /.2. before the damage

    /.2. after the damage

    W L

    W L

    -f W1L1 is higher at any point than the main deckat which the

    bulheads stop (the bulkhead deck) it is usually considered

    that the ship will belost (sink)because the pressure of water

    in the damaged compartments can force off the hatches and

    unrestricted flooding will occur all fore and aft.

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    (1) Lost buoyancy method

    0

    0 0

    (

    1) 'omputing the( )

    7 volume of lost buoyancy up to

    7 intact . . area8 7 area of the damage

    ") Find the (e*cluding damage

    aera ) area , '.F.,m y m

    vyA a

    v W L

    A W L a

    a A I

    =

    parallel sinkage

    midway !L!

    ) 0at draft 0.+

    ( 7 draft before the damage).

    cf d y

    d

    +

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    "

    "

    14) pdate the sinage .

    6) Find the trim , #rim (in)6"0

    distance between the '.F. of mean .

    % centroid of lost buoyancy of .

    (0.+ )+) raft aft #

    y

    y

    m

    mfc w

    d

    d

    m

    vyA

    I v x

    L

    x W L

    v

    v Ld

    A L

    +

    +

    =

    = =

    = +

    MTI MTI

    rim

    (0.+ ) raft forward #rim

    is the distance between '.F. % midship

    Aassuming that '.F. is aft of midship % the

    damage taes place of forward.

    m

    v Ld

    A L

    += + +

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    ! update the midway . .

    0.+ , then find new , and

    repeat the procedures given in the previous

    page.

    #he iteration may be stopped if the results are

    convergent, e.g. the mid

    m m

    W L

    d v A A+"econd iteration

    L

    way . . of the

    previous iteration % present iteration are

    different by an amount smaller than aprescribed tolerance error.

    W L

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    (0) (0), , wv x A a

    (())

    (0) (0)

    w

    vy

    A a=

    (0)

    1"

    (0) "yd d= +(0)

    mA(0)

    ma

    (1)(0) (0)

    m m

    vyA a

    =

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    (0) (1)y y

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    (2) dded !eight Method(considering the loss of buoyancyas added weight)

    also a #rial 9 error (iterative) method

    1) Find added weight vunder /020. #otal weight $ W + v

    ".) According to hydrostatic curve , determine W1L1 (or T) % trim (moment

    caused by the added weight % MCTC 1).

    4.) Since we have a largerT, and vwill be larger, go bac to step 1) re7compute

    v.

    #he iterative computation continues until the difference

    between two added weights vobtained from the two

    consecuti#e computationis smaller than a prescribed error

    tolerance.

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    $loodable length and its computation

    $loodable Length! #he $!L. at any point within the length ofthe ship is the ma*imum portion of the length, having its center

    at the point which can be symmetricallyflooded at the

    prescribed permeability, without immersing the margin line.

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    %ulkhead deck! #he dec tops the watertight bulhead

    Margin line! is a line :+ mm (or 4) below the bulheadat the side of a ship

    ithout loss of the ship: /hen the /.2. is tangent to

    the margin line.

    $loodable length(in short) #he length of (part of) the

    ship could be flooded without loss of the ship.

    &etermine $loodable length is essential to determine

    1. ;ow many watertight compartments (bulheads) needed

    '! $actor of subdi#ision(;ow many water compartments

    flooded without lost ship)

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    0 0 0 0

    1 1

    1 1 1 1

    1 0

    Steps for computing the F.2. given , , or

    1)

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    T*AN+

    YOU,,,