migt have a big problem

gunship · Feb 10, 2010

This boat im theorycrafting (in maxsurf) was imported to Hydromax, and when i calculated everything, it turns out that my Cog is a little higher than my CoB. does that mean a naturallyunstable boat? i think so, but considering i have almost half the boats weight in the keel, and only the rig being any distance above the waterline, i find it hard to believe the CoG...

Crag Cay · Feb 10, 2010

Most 'normal' sailing boats of average proportions, have the C of G around or above the DWL. Obviously the C of B is always below this.

You need to look at how stability works (in particular the movement of the C of B with heeling) in order to understand how having the C of G above the C of B doesn't mean you will have an unstable boat.

gunship · Feb 10, 2010

i assume its also because as soon as the boat tils, CoB rapidly moves sideways while CoG is stationary (relativley) and thus you still get a rightening arm?

Crag Cay · Feb 10, 2010

That's right a positive righting arm has to develop with heal.
When completely upright (0° of heel) there is zero righting moment.

jehardiman · Feb 11, 2010

Crag is correct, CoG (KG) is generally above the CoB (KB), what keeps the boat upright is the positive GM which is linked to your idea of the CoB moving. Here is a explination of GM that I did for another thread.

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)
Click to expand...

gunship · Feb 11, 2010

Now that's covering the subject! thankyou people!

also, i realized my thread title is misleading, it should have "theoretical" whithin brackets. maybe a "solved" too.

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migt have a big problem

gunship Senior Member

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