Calculating trim + where to put weights

Discussion in 'Stability' started by Chris Matthews, Aug 15, 2013.

  1. Chris Matthews
    Joined: Apr 2013
    Posts: 4
    Likes: 0, Points: 0, Legacy Rep: 10
    Location: Plymouth

    Chris Matthews New Member

    i have done A but i cannot see the logical step in finding the distance to put the mass to give an even keel ??

    A ship of displacement 15500 tonnes floats with drafts forward and aft of 7.3 m and 7.7 m. The following data from the vessel’s hydrostatic chart applies:
    LPP = 148 m
    LCF = 70.5 m
    LCB = 71.7 m
    TPCI = 22.8 t/cm
    MCTC = 188.2 t m/cm

    a. Find the arithmetic and hydrostatic mean drafts.
    ans(7.5 m, 7.509 m)
    b. 500 tonnes of cargo are to be loaded. Use LCF to find the longitudinal position required for this cargo in order to finish on a level keel. What is the final draft? (use hydrostatic draft for the calculation).
    ans(85.6 m from AP, 7.73 m)
    c. The LCB given above is from the hydrostatic chart and thus relates to a level keel. Before loading the 500 tonnes, the keel is not level, so LCG must be different to LCBLevel. Find the longitudinal distance from G to BLevel before loading (use MCTC and displacement for this calculation).
    ans(G aft of BLevel by 0.49 m)
  2. Ad Hoc
    Joined: Oct 2008
    Posts: 6,159
    Likes: 395, Points: 83, Legacy Rep: 2488
    Location: Japan

    Ad Hoc Naval Architect


    The way to approach this is by understanding moments.

    With a draft aft of 7.7m and fwd of 7.3m, this is a trim of 0.40m.

    So knowing the MCT, we can find out what moment is produced to create a trim of 0.40m. Or put another way a moment that is required to change from a trim of 0.40m back to zero trim.

    Firstly, you have first assume a parallel sinkage by adding the weight at the LCF (knowing the TPC) and then by similar triangles @ the LCF, knowing the distance moved to create that trim, you can work out where the weight needs to be from the LCF, since the change of trim = Moment/MCT.

    Think it through and you’ll understand the logic.
  3. Olav
    Joined: Dec 2003
    Posts: 296
    Likes: 30, Points: 38, Legacy Rep: 460
    Location: Filia pulchra Lubecæ

    Olav arch. nav.

    Okay, here we go:

    Given are the following particulars:

    Ta = 7.7 m, Tf = 7.3 m

    Trim = ΔT = Ta - Tf = 0.4 m = 40 cm

    We know that MCTC (Moment to Change Trim per cm) is 188.2 t*m / cm. Thus, we need a trim moment of Mtrim = 188.2 t*m/cm * 40 cm = 7528 t*m

    We need to create this moment with our cargo mass of 500 t. A moment is force times lever, so we need the inverse:

    7528 t*m / 500 t = 15.06 m of lever arm.

    To get to level keel, the trim moment is acting forward and we have to add our lever arm to LCF:

    15.06 m + 70.5 m = 85.56 m

    The vessel's mass is now

    15500 t + 500 t cargo = 16000 t

    We know that the vessel immerses one cm per every added 22.8 t, so

    ΔT = 500 t / 22.8 t/cm = 21.8 cm = 0.22 m

    Final draught is then

    Tfinal = T + ΔT = 7.509 m + 0.22 m = 7.73 m

    Basically the same thing as before:

    Trim is 40 cm, which requires a moment of 7528 t*m. The acting mass is the vessel's displacement of 15500 t (without the cargo), which means the required lever is

    7528 t*m / 15500 t = 0.486 m, rounded off to 0.49 m.


    Edit: Ad hoc was faster...
    1 person likes this.

  4. Chris Matthews
    Joined: Apr 2013
    Posts: 4
    Likes: 0, Points: 0, Legacy Rep: 10
    Location: Plymouth

    Chris Matthews New Member

    Thank you !!! such a help !!
Forum posts represent the experience, opinion, and view of individual users. Boat Design Net does not necessarily endorse nor share the view of each individual post.
When making potentially dangerous or financial decisions, always employ and consult appropriate professionals. Your circumstances or experience may be different.