What is (BM*T)/B**2?

Discussion in 'Stability' started by Olav, Dec 16, 2007.

  1. Olav
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    Olav arch. nav.

    I wonder if anyone has an idea about the meaning of the term (BM*T)/B**2.

    I understand that BM is the distance between Centre of Bouyancy and the Metacentre, T the draught and B the beam of the vessel.

    The abovementioned term is from the Cross Curves of a fishing vessel where T/H (H = Height of hull from keel to deck) is plotted against (BM*T)/B**2.

    Thanks in advance for any input!
     
  2. Tim B
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    Tim B Senior Member

    I think it's a non-dimensionalisation of metacentric height with draught. Could you post the curves themselves?

    It is certainly an unusual way of expressing it though.

    Tim B.
     
  3. Olav
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    Olav arch. nav.

    Thanks for your reply! I have attached the graph in question below (it shows the curves for 3 different vessels).
     

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  4. Guillermo
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    Guillermo Ingeniero Naval

    I have been trying to find what they are, but I found nothing till now.
    Trying to work out something, I played around with formulas, like this:

    BM = I / V
    I = (1/12)*Cwp^2*L*B^3 (approx.)
    V = L*B*T*Cb

    then
    BM = 0.08*Cwp^2*B^2 / (Cb*T)

    from here
    T/B^2 = 0.08*Cwp^2/(Cb*BM)

    So
    (BM*T)/B^2 = 0.08*Cwp^2/Cp

    Now if admit the hull is an efficient one, we may write: Cwp = Cp ^ 2/3 (Schneekluth)
    so
    (BM*T)/B^2 = 0.08*Cp^1/3

    What this means or what is the interest of plotting it against T/H, I don't know. :confused:

    Cheers.
     
  5. Guillermo
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    Guillermo Ingeniero Naval

    But if we use another aproximation for BM, as this also common one:

    BM = 0.08*B^2/T

    then (BM*T)/B^2 = 0.08, which initially seems to make even less sense....but

    No! All the contrary! It begins to make sense....!

    Cheers.
     
  6. Guillermo
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    Guillermo Ingeniero Naval

    I think now those curves are intended to derive a more accurate factor k in formula:

    BM = k*B^2/T

    which from the graphs depends on the relation T/H and the type of vessel, probably because Cp -as per my first post- is within the formula and it varies with T/H.

    Most interesting.

    Could you tell us what kind of vessels are the ones in the study? I'd appreciate it a lot. Thanks in advance.

    Cheers.
     
  7. Guillermo
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    Guillermo Ingeniero Naval

    This is quite interesting as it can help us to better estimate KM for that kind of boats by using:

    KB = 1/3 (5T/2 - V/Afl) (Morris)
    BM = k (B^2/T), taking k from graphs posted by Olav.

    Cheers.
     
  8. Ekaiztea
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    Ekaiztea Seagoing Ships Master.

    Really good ¡¡¡

    Congratulations Guillermo ¡¡ Its quite nice to watch how your brain works and how you arrive to the "Eureka".

    Thanks a lot
     
  9. Olav
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    Olav arch. nav.

    Thanks Guillermo!

    Actually it wasn't exactly my problem but that of a friend of mine (who has announced that he's now going to register to boatdesign.net and say something about it himself) so all I know about the vessels is the fact that they are all relatively small fishing vessels of different size.

    The difference in size is the reason why the ordinate is labeled T/H (draft/side height) instead of just T in an attempt to "normalise" the draught.

    He's now come to the conclusion that (BM * T) / B**2 = BM * T/H * B^-1 where the middle term T/B can be understood as another form stability parameter in addition to BM. Admittledly I still don't get his explanations in full so I can just refer to his own yet to come comments.
     
  10. commodore
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    commodore New Member

    Hello at first
    Thanks for your efforts to solve my problem.
    As Olav stated this data are from near-shore fishing vessels. The draught to depth ratio may been chosen to compare different sized vessels.

    An useful additional information might have been, that for these vessels all data were known. So this figure was intended to compare stability characteristics of different ships.
    This is why I think it is not intended to estimate KM, though it surely can be used for that the way Guillermo stated.

    My latest idea is the following:
    BM is qualitatively I/V.
    I is qualitatively L*B³.
    So BM holds a B² and it increases for higher stability.
    T/B is a value that decreases for higher stability. (I guess it does, greater B means more stability, if KG is given greater T means less GM because of lower B. The emphasis which outbalances is an estimation.)
    1/B also decreases for less stability.
    Both, T/B and 1/B contain an influence on stability by breadth.
    So BM * T/B * 1/B is BM multiplied with two values decreasing for higer stability and somehow representing B.
    My conclusion was BM * T/B * 1/B is a stability attribute dispensed from a influence of the breadth.

    Hopefully I was able to let you get it and it is not to weired...
     

  11. Guillermo
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    Guillermo Ingeniero Naval

    Be very welcome to these forums, commodore. :)

    Let’s see:

    We have on one side

    BM = k*B^2/T

    Being "k" dependent on T/H (this is: on load condition) and type of boat, as per the curves Olav posted.

    On the other hand we know that:

    GM = (f *B/t)^2

    being
    - “t” the natural period of balance in the given load condition
    - “f” a factor that is obtained from rolling tests in calm waters. It depends also on load condition (T/H again!), taking a value around 0,76 for a fishing vessel with fuel, stores and equipment (0,78 or even 0,8 are usually taken as rough approximations for loaded condition)

    This method of estimating GM is provided to the Captains within the stability booklets in the form of a well known nomogram, or then tables.
    (Interested ones see: http://www.tc.gc.ca/marinesafety/tp/tp7301/part1.htm#STAB-2-APPENDIXB)

    I think there’s somekind of relation between all this, but I still do not know what it is. :confused:

    I’d greatly appreciate if you could tell me where did you get those curves from, and how can I get a copy of the work where they are published, be it a paper, book...whatever.

    Thanks in advance.
     

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