Calculation of GM

Discussion in 'Boatbuilding' started by Shidoran, Sep 21, 2005.

  1. Shidoran
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    Shidoran Junior Member

    Not sure if this is the right place to put a few questions, but I'm working on a project and am having trouble figuring out these calculations, or the method to do so. Any help is appreciated.

    1) This is an inclining experiment with the following recordings:

    Draft at each corner (with weights onboard) 0.5m
    Density of water = 1.0 t/m^3
    inclining weight used = 0.85 t
    Distance weight moved = 3.0m
    Angle of inclination = 2.5 degrees
    LCG of inclining weight = 6.0m from stern
    VCG of inclining weight = 0.5m above deck

    How would I go about calculating the GM? :?:

    2) A vessel with a displacement of 26 tonnes has a KG of 0.83m at the time on an inclining experiment. Assuming the inclining weights used weighed 0.85 tonnes, and that their vertical centre of gravity was 1.5m above the calculation baseline, how would I go about calculating the displacement and KG from the vertical datum with the inclining weights removed? (I also have a hydrostatic table to get the GM with the weights removed) :?:
     
  2. Tim B
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    Tim B Senior Member

    I'm not in the habit of doing people's course-work for them... but look at "Ship Hydrostatics and Stability" by Adrian Biran. It should be able to tell you all you need to know... failing that, look back at your notes.

    Tim B.
     
  3. Shidoran
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    Shidoran Junior Member

    Well...

    I can't see anything wrong with someone explaining what to do for the calculations instead of doing the calculations for me.... my lecturer sucks, the notes are crap.. there's nothing on this stuff. Thus why I'm asking. It's the theory I'm after (preferably online if there is anything) not someone to do it for me.
     
  4. CDBarry
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    CDBarry Senior Member

    You need displacement and KM to calculate GM from an inclining experiment. You get these from the hydrostatics tables for the vessel, based on the freeboard or draft.
     
  5. jehardiman
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    jehardiman Senior Member

    Draw a free-body diagram of the situation and everything will be clear. You've got way too much data so throw out the stuff you don't need out. And even if your lector and notes are crap, the basic texts aren't. You need to be able to "get" this concept or some of the later stuff (i.e. the whys of a BM calculation) is going to be really hard.
     
  6. Shidoran
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    Shidoran Junior Member

    Ok, so

    Alright, thanks for your input. I might sound like an idiot to you fellas, (as I'm making it rather obvious that I still know nothing about boats, - not that I haven't tried though) I have just a couple of clarification questions:

    The Vertical Centre of Gravity or "VCG" is the same thing as "KG"

    What does it mean if it said "draft at each corner (with weights on board) = 0.5m"?

    Thanks.
     
  7. jehardiman
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    jehardiman Senior Member

    VCG = KG if the VCG reference is the same as the KG reference, this is not always true, it depends on how the weights were done and the curves of form drawn.

    If the drafts are all the same at all four corners, then there is no trim/heel corrections to displacement, Ixx, Iyy, TPI, Awp, etc, and no need to calculate true midships draft.
     
  8. Shidoran
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    Shidoran Junior Member

    Ah. That makes sense. Well the drafts at the 4 corners are 0.5m, as shown in my first post (those are the calculations I'm trying to understand and work with).

    Could you tell me if I'm approaching this right?

    GM (of an inclining experiment) = Weight moved x distance moved
    ___________________________
    Displacement x (tan(angle of heel))

    Thus, to get the displacement (referring to the statistics in my first post)

    Draft = displacement (or immersed volume, which directly proportional) x density

    therefore

    Displacement = draft/density
    = (0.5m x 4)/1.0
    = 2.0 m^3

    Thus, GM = (0.85 x 3) / (2.0 x (tan 2.5))
    = 29.20 m

    If this is right, then I'm finally understanding it (yay for me! I'm getting somewhere!) if not, where would I be going wrong?

    Thanks for all your help... my lecturer is like impossible to get a hold of and I appreciate you taking a bit of time to help me out. I really do :)
     
  9. jehardiman
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    jehardiman Senior Member

    Not to be harsh or anything, but have you considered another field of study? Your posts show that you cannot or have not grasped the basic definitions and concepts and lack the problem solving ability necessary for simple hydrostatics.

    Again ...RTFB...RTFQ...start with the old greek.
     
  10. Shidoran
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    Shidoran Junior Member

    Alright, if it's gotta be that way...

    Alright, here's the deal. This is how it is:

    I don't get much of this hydrostatics stuff, and I'm asking for help. I am a little disappointed with the lack of effort people have put in to try and help me out, although I do realise if you do choose to help it would be off your own head, so thanks anyway, I appreciate any help that was given.

    "go read a book" kinda help isn't exactly what I'm looking for, there is no textbook for this unit, just a bunch of vague crappy notes, and it is only a half-unit that I have to do because there was nothing else. I'm looking for explanations on the principles that would be in a book, as often people explain it much better than books do.

    Also, because this is such a small unit the university doesn't have many books that are not way over my head (ie naval engineering level). I just want to get through this with the basics and I am struggling to find help anywhere.

    If you don't have the patience to help me then so be it, I'll have to keep looking or fail it... (which I don't want to do...) Obviously I'm trying to understand. Give me that much credit.

    In reply to your post (more specifically:)

    "Not to be harsh or anything, but have you considered another field of
    study?"

    Yes. I'm not going to study boats at career level, never planned to, and probably never will. I just want to get through this half-unit without failing.

    "Your posts show that you cannot or have not grasped the basic
    definitions and concepts"

    Without being offensive, but why else would I be asking for help?!

    "and lack the problem solving ability necessary for simple hydrostatics. "

    I don't appreciate this quirk. It means I don't understand the principles involved with hydrostatic calculations, not that I don't have the ability to solve them once I understand them.

    Well, Hope you have a nice day, and if you decide you'd like to help me out afterall, I'd love to hear from you. If not, enjoy your boats... I'll find someone more willing.
     
  11. CDBarry
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    CDBarry Senior Member

    Note that the concept of moment of inertia and how it resists bending moment in structures is exactly the same as how moment of inertia resists overturning moment in hydrostatics. If you look for the analogies in the two processes, this may help.

    Naval arch calcs are often confusing because theysometimes have very strange ways of approaching a simple subject. The peculiar ways are generally due to a need to do some specific calculation rapidly.
     
  12. jehardiman
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    jehardiman Senior Member

    Ok....even though this walks like a troll and quacks like a troll I'll bite. I do not see how you could have gotten into any university that would allow you to take an engineering "half unit" (WTF is that anyway?) and not understand basic secondary school physics.

    Anyway... I will try to use small words and you’re on your own for the math.

    The old greek defined buoyancy (B) as the upward force on a body in a fluid caused by the density of the fluid times the immersed volume(V, length units ^3).

    The immersed volume is identical with the displacement (W or delta in weight units) divided by the fluid density.

    The immersed volume is calculated by the integral of the instantaneous waterplane (Awp, length units^2) and the differential draft (T, length units).

    We will ignore pitch and trim for the moment and concern ourselves only with roll and heel for the rest of this discussion. Trim is similar to heel in all aspects of hydrostatics and the extension is elementary.

    We can plot the basic factors of hydrostatics in something called the curves of form. These are a set of curves, plotted against draft, which define the hydrostatic properties of any immersed body. They normally will include:
    Displacement (in force or weight units (W or delta) and length units^3, i.e. buoyant force, B=W=delta =V*density),
    TPI (“tons per inch” immersion; effectively a measure of waterplane area in weight units per some small draft change in length units using the wall sided assumption),
    Water plane inertia (Ixx, by convention in the x (roll) axis, a double integral over the length fore and aft and the beam of the waterplane differential area * distance squared off centerline (ff y^2 dy dx), the units are length^4) and
    KB ( the center of buoyancy (the integral over the draft of the waterplane times the draft then divided by V) measured from the baseline which is usually the bottom of the keel in length units(i.e. length^4/length^3=length) .

    On to GM

    For most floating bodies, the center of gravity (KG, measured from the baseline in length units) is above KB and the displacement is equal to the weight (W). The distance that KG is above KB is called the BG (i.e. KG-KB=BG). If life was perfect, the body could stay like this, similar to balancing a pencil on its point. But life is not perfect and the body will be perturbed and attempt to turn over to a more stable configuration. It is important to note that as the body just begins to roll, the moment that forces this is equal to W*roll angle (theta)*BG for very small theta. Draw a free body diagram to satisfy yourself.

    What prevents the body from turning over is that as it rolls, it causes two wedges of changes to displacement to form. These are the emerged wedge on the high side and the immersed wedge on the low side. These to wedges have the effect of creating a buoyancy couple that rights the body, i.e. the emerged wedge decreases buoyancy on the high side and the immersed wedge increases buoyancy on the low side. For a deferential unit of length fore and aft and a very small roll angle, size of the wedges are .5*1/2 beam^2*tan theta and the centers are 2/3 the beam apart. This causes a small shift in the center of buoyancy off centerline and therefore a moment with respect to the center of gravity. How big is this shift? For small angles of roll, the restoring moment can be written as the integral over length for and aft of 1/12*beam^3*theta*density, notice that this is identical to the result for Ixx (the proof is left to the student)* theta*density. The magnitude of the shift is the change in moment/B. Rewriting, the moment shift is equal to Ixx*theta/V where density cancels. If this shift off centerline is greater than BG*theta, then the body is stable, i.e. the overturning moment W*theta*BG is less than righting moment (Ixx/V)*theta*B.

    Because people like graphical concepts, it was noticed that Ixx/V was used the same as the lever BG between B and G and effectively was the “pendulum point” that B swung around. It therefor was called the Metacenter (from greek for “above”),M. If you graphically lay it out it would appear to be on centerline the distance Ixx/V above KB , so therefor Ixx/V=BM; this shows the beauty of math because if you do a unit analysis (i.e. length^4/length^3=length) it is purity. Another thing to notice is that you really don’t have all that righting moment, it is reduced by the overturning moment. If we draw a freebody with G acting down through KG and B acting up through KM (KM=KB+BM) at some small angle of heel theta, there be a small distance between the two lines of action perpendicular from G. The point that this perpendicular hits the buoyancy line of action is labeled Z and this distance is called GZ. Notice that in the freebody this distance GZ is equal to the distance (KM-KG)*thetaor, rewritten GZ = GM*theta where GM = KB+BM-KG. This is the holy and mystic GM where the residual righting moment = GZ*B or GM*theta*B.

    NOTE: THE SHIP DOES NOT ROTATE ABOUT THE METACENTER AND GM IS NOT THE RIGHTING ARM. IT IS A CONCEPT FOR THE CONVIENCE OF CALCULATION ONLY!

    Now it is pretty easy to calculate KB and Ixx and so get KM, the hard part is to accurately determine KG. However this is where GM comes in. If the righting moment is always greater than the overturning moment, then the ship will always remain upright. In order to cause the ship to heel to a given angle, the residual righting moment must be exactly matched, i.e. if it takes a weight, w, to be moved an athwartships distance d, to produce 10 ft-tons to heel 1 degree, then GM*tan 1degree*displacement = 10 ft-tons. THIS ONLY WORKS FOR SMALL ANGLES OF HEEL! Therefore w*d = GM * tan theta*W or GM = w*d/( tan theta*W)

    This concludes today’s lecture. As I said before, you have everything you need to solve the original question. Until you grasp the above concepts it is useless to go on about free surface effects, lolling and other hydrostatic issues.
     
  13. Shidoran
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    Shidoran Junior Member

    Alright

    Thanks for the input, I appreciate the help. In reply to your comments:

    I didn't do physics in high school, I was never good at it and didn't like it. I am presently doing the equivelant of high school physics, (because I have to) and the course involves things like electricity and heat etc, and so far has been different to the boat course I am doing (and much easier for that matter). My boat lecturer has tried to cram way too much into such a small course, and most of the class is struggling to understand. Lucky for some, they have friends in the boatbuilding/naval engineering field and can seek help in that way, others like me are not quite as fortunate. I'm just trying to get help where I can.

    A "half unit" is considered to be half the workload of a "full unit" (although this isn't always the real situation, some half units are almost as much work as full units).

    In the university system here, to do "full time" study, you must have a workload of 100 "credits". Each "full unit" has the workload worth 25 "credits", each "half unit" only has 12.5 "credits". Thus, one can choose to do a "full unit" or two "half units" to get 25 "credits" towards a "fulltime workload" or "fulltime study". There ya go.

    Again, thanks for the info, I think its starting to click now. Heh, *grumble*abouttime*grumble ;)
     
  14. red_devil
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    red_devil New Member

    uni student lol

    wassup... i am also a student doing a project, and its also a half-unit. i think we are in the same class man. im at curtin studying engineering, n take this as an option as well.

    i was also having trouble with the assignment, cos the notes really are quite bad. so i came online to try and solve my problems.

    it was great to find this forum where someone actually posted the same problems that i am doing. its gonna help out alot.

    if your stuck with any other of the questions, or just wanna compare what you have and what i have, gimme an email, nward86@hotmail.com, n we could meet up at curtin to finish off the assignment, or just compare calculations ... if you havn't already

    cheers,

    Nick
     

  15. red_devil
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    red_devil New Member

    ****... didn't see the date of posts. you obviously did it last semester, 2005.
     
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