# Metacentric Height

Discussion in 'Stability' started by Paul Jones, May 1, 2009.

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### Paul JonesNew Member

Hi everyone,

I have a question I am thinking about for an experiment I did involving this topic. The questions I am thinking about are:

Does the position of the metacentre depend on the position of centre of gravity?

If yes, does the metacentric height vary with the angle of heel?

***EDIT Explained experiment a little further below.

Regards,
Paul

Last edited: May 1, 2009
2. ### Guest625101138Previous Member

The answer can be yes and no to the first question. If CofG changes directly vertically it will not alter the metacentric height. If the CofG moves laterally it can alter the metacentric height due to heeling, hopeful it increases to stiffen up. Likewise longitudinal shift in CofG will change trim and this can shift the metacentric height.

Rick W

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"..Does the position of the metacentre depend on the position of centre of gravity?.."

No, it is geometry related, from the buoyancy and the waterline.

"...If yes, does the metacentric height vary with the angle of heel..."

Yes, at small angle of heel, it is small enough to ignore. At large angles of heel the loci of the metacentre can be drawn to produce an "M curve"

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### Paul JonesNew Member

To put it into context it was from doing this experiment (image below) where the mass slider was moved from the left to the right (-40, -20, 10, 30, 60) and this was done for three different positions of the vertical sliding mass.

What would the answer be to this question (relating to this experiment)?

5. ### Guest625101138Previous Member

Moving the weight laterally will cause the hull to heel and this changes the transverse metacentric height at the new point of equilibrium.

Rick

6. ### Guest625101138Previous Member

I should add that the positioning of the weight on the rod will not alter the transverse metacentric height when the slider is centralised but once the hull heels, shifting the weight along the rod at any given lateral position will alter the angle of heel so moving the weight in this case will alter the transverse metacentric height.

Rick W

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### hbrJunior Member

calculation

starting from a hull with an ellips form breadth at deck equal 2*b and the depth of the hull equal to a in vertical position no heeling.
in stead of drawing all the different angles of heel, how can the stability curve be calculated at different angles of heel ?
Do you have a suggestion how to approach this in excel for instance ?

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### jehardimanSenior Member

Do not confuse the metacenter with the point the vessel rolls around or assume the metacenter controls the amount the vessel rolls.

The location of the metacenter (KM) is only a function of the vessels weight and form. The location of the CG (KG) only effects the GM (i.e. GM = KM-KG) not the metacenter. The GM, not the metacenter, is what effects the moment to heel some small angle (i.e. moment = W*GM*sin theta for small theta) and only if the wall sided assumption is correct. Many hulls never conform to the wall sided assumption, and therefore cannot be analyized by evaluation of GM.

See this discussion on GM...http://www.boatdesign.net/forums/boat-building/calculation-gm-8852.html#post60405 (and please ignore the scarcasm in it).

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Paul

Is that just a suspended weight, which has a slide function to allow it to move from one side to another, or is there some kind of buoyant shape under, which is immersed in water which then has the sliding function? It is not clear what the "box" of tricks is.

Also need to use correct terminology too. If the CoG is off-centre, then it will not cause an angle of heel. It will be 'list'. This should not be confused with the angle of 'loll'.

As I stated above the metacentre is a function of the geometry (which give the location of "B") and the waterplane (which gives the location of the intersection of the perpendicular from the new location of B to its previous location of B) to get M.

It has nothing to do with weight or CoG at all.

10. ### Guest625101138Previous Member

I imagine you are talking about a a hull having a constant elliptical section? Maybe a sketch would help.

In Z-Y plane the equation for an ellipse is:
Y = b * [1 - Z^2/a]^0.5

The waterplane can then be describe simply as a sloping line:
Y = hZ + c

h is the slope of the waterplane relative to the hull. c is determined by integrating the area between the waterplane line and the hull so that the area, At, is constant at any given angle of roll. This can be done numerically in Excel.

Once you have the value for c you can determine the intercepts of the waterplane and the hull. This gives the width of the waterplane. Lets call it Ywp

For a constant section hull the metacentric height can be calculated as:
BMT = {Ywp^3}/12/At

Sounds complex but if you set it up in Excel it is easy to do repeat calculations.

Rick W

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### jehardimanSenior Member

Now before we go too much further with this, it is important to remember that ultimate stability (i.e resistance to capsize) has nothing at all to do with initial stability (i.e the calculation of GM). For those that are interested, do the calculation of GM for a flat decked barge with a roller skate or marble on deck. You will find that the GM is negative and the barge will loll until the skate/marble hits the bulwark. It is a very enlightening exercise.

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### GuillermoIngeniero Naval

For a given floatation, of course.

This is more precise, but even for a given displacement the longitudinal and/or lateral position of CoG also influence the transversal KM by modifying floatation, don't you agree?.

Cheers.

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"...for a given displacement the longitudinal and/or lateral position of CoG also influence the transversal KM by modifying floatation, don't you agree?..."

CoG has nothing to do with the location or magnitude of KM.

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### jehardimanSenior Member

OK guys, Lets be very clear about the assumptions if we want to get into the more snarky aspects of KM.

The original poster wanted to know if GC effected KM, and the actual answer is No as defined in Axiom 1 below.

Axiom 1) For the static equlibrium condition where a body of arbitrary shape is free floating, the CG and CB are in line, and the location of KM is independent on the CG.

Now, we can contrive relative cases where we compare the KMs of a body of fixed shape and weight under various CG locations relative to some body fixed reference axes. However, each of these cases independently degenerate to Axiom 1 above. (For a really neat trick, try the above of a circular cylinder of various weight floating on it's side... )

FWIW, the calculation of KM is only of interest to find GM. And GM is only of interest to determine the initial instant stability, not the dynamic stability or the final stability. Whenever the CG is off centerline of a axisymetric free floating body, the body will "list" until CG and CB are in line. KM only effects the roll rate acceleration to reach this list, not the final static position.

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jehardiman

I'm not sure what you mean by "...get into the more snarky aspects of KM..."

Since you appear to be linking two separate technical issues, viz:
1) What and how to determine KM of "a floating body"
2) Stability and its link with roll period

As you noted in your axiom1), based upon the original Q "..Does the position of the metacentre depend on the position of centre of gravity..", no. Which answers 1).

"... Now, we can contrive relative cases where we compare the KMs of a body of fixed shape.." But the Q doesn't say/ask anything about roll period etc

So just wondering why you then deviated into another topic which just confuses many on here who are not naval architects, by going down the "stability" route?

However, it is still not clear to me, what Paul's 'little box of tricks' is doing, as there is no explanation to whether it is just a hanging weight or a floating body with a pendulum etc.

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