GM calculations with laterally shifted center of mass

[laukejas]

Hi,

A while ago I made another post asking about a discrepancy between two different approaches to calculating ship metacentric height (GM) dependent on the heel angle. Since then I have figured it out, but found a weird quirk with these calculations, and I would like a little bit of advice.

To reiterate, let's first look at these two methods of calculating GM.

1) When ϕ=0, the metacentric height is calculated by dividing displacement by the second moment of area of the waterplane, and adjusting it by the distance of the keel to center of buoyancy and keel to center of gravity:

GM = ∇/I + KB − KG

2) When ϕ≠0, the metacentric height is calculated by dividing the righting lever by the sin of heel angle:

GM = GZ / sin(ϕ)

Image for reference:

These equations, despite being very different, work really well together in normal scenarios, but I found a scenario where the results don't make sense anymore.

Let's see an example so I can explain.

Here is a simple slab drawn in CAD, 1 x 1 x 0.5 m. in size, volume 0.5 m^3, immersed in water with displacement of 0.1 m^3, center of mass assumed to be in the exact geometric center of the slab, heeled from 0° to 30°. The green shaded volume is above water, grey shaded volume is below water, blue square is center of buoyancy, green square is center of gravity.

Perspective of the slab immersed at 0°:

Front view of the slab immersed at 0°:

Front view of the slab immersed at 5°:

Front view of the slab immersed at 30°:

GM curve, calculate for every degree of heel, using #1 equation at 0° heel angle, and #2 equation for the rest of heel angles:

As you can see, there is no discrepancy between 0° and 1° degrees of heel, despite different equations used to calculate that GM. This also works fine with more complex models. This seems correct so far.

The issue:

If we shift the center of gravity just a little to the side, like 0.01m, the results look entirely different.

Front view of the slab immersed at 0° (notice the CoG shifted slightly right):

Front view of the slab immersed at 5°:

Front view of the slab immersed at 30°:

GM curve:

As you can see, now there is a massive discrepancy in GM - the results of the different equations at 0° and 1° or more don't line up anymore, there is this massive jump that cannot possibly be realistic. This, of course, is because the GZ≠0 when ϕ=0°.

I went over the equations in the software dozens of times, but they are correct, and since this shape is so primitive, it is easy to calculate it by hand - and manual calculations result in the same curve. Yet this cannot possibly be right.

You might ask, why does this matter? Well, in automated calculations based on CAD models, sometimes the GZ≠0 when ϕ=0° because center of gravity might be shifted a little bit, or the model might not be perfectly symmetrical. This is also true in real life. However, when trying to calculate GM, this completely messed up the results, and now it is unclear which equation is to be trusted, and how to bridge this gap.

Can anyone offer any insight?

 

[Ad Hoc]

.....

Front view of the slab immersed at 0° (notice the CoG shifted slightly right):

First of all, you need to check your condition.
As this cannot be the case, since this is no longer in equilibrium, it would rotate until it the CoB and CoG are in line.
It would be like this:

And as such the CoB has move to the right and vertically up, by a small amount, to maintain equilibrium.

AAhh..JEH did the other way +1

 

[jehardiman]

Again you are confused thinking that Initial Stability (GM) is somehow connected to Ultimate Stability (GZ). GM is meaningless when you have an unbalanced righting moment (i.e. GZ*displacement > zero which implies GZ not equal to zero). In all your cases the floating body would attempt to right itself but is prevent from doing so because you have applied an external moment (l, because it is along the x axis) such that l+GZ*disp = zero. The only reason you are able to calculate a positive GM is because the body is artificially held in the position you analyzed. Draw a FBD to satisfy yourself that this is the case.

This is not new or unexpected. Here is a classic example to be demonstrated to first year Naval Architecture students.

A balsa cube (density ~0.10) will float with a face up because GM is positive, an dense pine/ light oak cube (~0.55-0.6) will float with one corner up because it has negative initial stability (i.e. GM negative) so it will roll to a stable position, a teak/ebony cube (~0.9-0.95) will float with a face up because GM is once again positive.
In this demo KG and Iwp are unchanged, what changes is the height of KB and the volume. So for a unit cube trying to float face up...
KG 0.5
Iwp 0.083333333

Density KB BM GM Initial Stability
0.1 0.05 0.833333333 0.383333333 Yes
0.2 0.1 0.416666667 0.016666667 Yes
0.3 0.15 0.277777778 -0.072222222 NO
0.4 0.2 0.208333333 -0.091666667 NO
0.5 0.25 0.166666667 -0.083333333 NO
0.6 0.3 0.138888889 -0.061111111 NO
0.7 0.35 0.119047619 -0.030952381 NO
0.8 0.4 0.104166667 0.004166667 Yes
0.9 0.45 0.092592593 0.042592593 Yes

All will eventually have ultimate stability (i.e. B under G and GZ =0) and float, but won't float face up. What's great about this demo is you can just toss the cubes into water. You have to really try to get the balsa or ebony to float other than face up and you cannot get the oak to float face up unless you hold it there (i.e apply an external moment). It is an important lesson you need to review.

EDIT: X-post with Ad Hoc

 

[laukejas]

Thank you both very much for your replies. Yes, I am aware that GM doesn't make much sense if the cube is at non-equilibrium position (GZ > 0), but isn't that true for my first case too at any heel angle other than 0°? Meaning, that it is really useless to use the second equation (GM = GZ / sin(ϕ)) at all, because it only works when GZ > 0?

So perhaps it would be correct to use the first equation only, regardless of the heel angle, and if GZ>0, do not calculate GM at all, because it becomes meaningless?

 

[jehardiman]

Thank you both very much for your replies. Yes, I am aware that GM doesn't make much sense if the cube is at non-equilibrium position (GZ > 0), but isn't that true for my first case too at any heel angle other than 0°? Meaning, that it is really useless to use the second equation (GM = GZ / sin(ϕ)) at all, because it only works when GZ > 0?

So perhaps it would be correct to use the first equation only, regardless of the heel angle, and if GZ>0, do not calculate GM at all, because it becomes meaningless?

The fixing of GM is an important rules requirement because it is used to fix the initial angle of the actual stability requirement, the area under the righting curve (a function of GZ over ϕ) . Two of the hardest things to know in an actually constructed vessel are the true locations of KG and BG; they cannot be physically "measured", only inferred from the summation of weights and the lines plan. However, a shift in GZ can be measured for the completed vessel. This is the reason for the Inclining Test, where a known moment shift is correlated to a measurable heel angle. In this case we can use GM = GZ / sin(ϕ) because we know exactly much GZ is off from zero because we imposed a known external moment.

 

[laukejas]

The fixing of GM is an important rules requirement because it is used to fix the initial angle of the actual stability requirement, the area under the righting curve (a function of GZ over ϕ) . Two of the hardest things to know in an actually constructed vessel are the true locations of KG and BG; they cannot be physically "measured", only inferred from the summation of weights and the lines plan. However, a shift in GZ can be measured for the completed vessel. This is the reason for the Inclining Test, where a known moment shift is correlated to a measurable heel angle. In this case we can use GM = GZ / sin(ϕ) because we know exactly much GZ is off from zero because we imposed a known external moment.

So if I'm interpreting this correctly, it means that it's up to me to ensure that there is an equilibrium (GZ=0) at ϕ=0°, in order for the rest of GM curve to be correct, right? In other words, the way I'm doing these calculations is correct, but my scenario #2 doesn't make sense and shouldn't even be considered?

 

[fallguy]

So if I'm interpreting this correctly, it means that it's up to me to ensure that there is an equilibrium (GZ=0) at ϕ=0°, in order for the rest of GM curve to be correct, right? In other words, the way I'm doing these calculations is correct, but my scenario #2 doesn't make sense and shouldn't even be considered?

Based on my limited understanding; when the CoG moves, the CoB always must as well. It doesn’t mean the condition is desirable, but always occurs. If you think about it, this makes a lot of sense because if you load a vessel unevenly; it changes its attitude to attempt to continue to float (if you indulge me). I cannot speak to the failure conditions; that is for the NAs.

 

[TANSL]

When the initial equilibrium position has zero heel, the metacenter is calculated using the formula :

GM = Ix/Displacement​

where :

Ix = Moment of inertia of the waterline area about the X-axis
Displacement = Volume of the submerged hull and appendages.​

This is how the true metacenter is calculated.

However, when the ship is heeled, the true metacenter is not calculated; instead, what is called the "false metacenter" is used. This false metacenter is the point "M", of intersection of the initial vertical line, passing through the center of the ship, with the vertical line passing through the new center of buoyancy. Once "M" is known, it is easy to calculate the GZ for each displacement and heel.

In stability calculations (GM, GZ, etc.), the CoG is considered fixed, and only the CoB moves, for whatever reason. The study of stability when there is a translation of weights or something similar is a different matter. But I believe that would be the second part of the lesson.

 

[laukejas]

Based on my limited understanding; when the CoG moves, the CoB always must as well. It doesn’t mean the condition is desirable, but always occurs. If you think about it, this makes a lot of sense because if you load a vessel unevenly; it changes its attitude to attempt to continue to float (if you indulge me). I cannot speak to the failure conditions; that is for the NAs.

Yes, that much is certain, I just figured that since we're artificially "holding" the hull at various heel angles (from 1° to 30°) in the #1 scenario, i.e. creating a non-zero GZ, then we might as well do that for 0° heel angle, as in the #2 scenario. But if I understand correctly what has been said so far, this just doesn't make sense as it wouldn't give any useful info; hence the GM curve looks weird as well. I hope I got that conclusion right.

When the initial equilibrium position has zero heel, the metacenter is calculated using the formula :

GM = Ix/Displacement​

where :

Ix = Moment of inertia of the waterline area about the X-axis
Displacement = Volume of the submerged hull and appendages.​

This is how the true metacenter is calculated.

However, when the ship is heeled, the true metacenter is not calculated; instead, what is called the "false metacenter" is used. This false metacenter is the point "M", of intersection of the initial vertical line, passing through the center of the ship, with the vertical line passing through the new center of buoyancy. Once "M" is known, it is easy to calculate the GZ for each displacement and heel.

Thanks for that clarification; in my case GZ is always known as it is very easy to measure in simulations, but I understand that it's the opposite way around in real life (like jehardiman described). Okay, so unless I'm still misunderstanding something, that probably clears it up. The final conclusion is that for these calculations to make sense, ship should be in equilibrium at initial heel angle (0°) in order to get a correct GM curve, so my #2 scenario is just meaningless.

 

[TANSL]

No, the initial equilibrium position doesn't necessarily have to occur with zero heel, although this situation is undesirable. The ship would have a permanent heel that must be avoided because, among other things, it will make any additional heel, caused by a wave or wind, for example, more dangerous.
You can calculate the initial metacenter for your initial equilibrium position (which has an alpha heel) but the GZ curve will not pass through the origin, it will not cut the x-axis (which represents the angles) at the origin, but will cut it at the point of x-axis = alpha.



There's an error in your reasoning because it's the GZ that we don't know for each degree of heel. The CoG, I repeat, is fixed (unless there's a weight shift), and if for whatever reason the ship heels by alpha degrees, the new position of the CoB is calculated. The GZ is then calculated from this.

 

[laukejas]

No, the initial equilibrium position doesn't necessarily have to occur with zero heel, although this situation is undesirable. The ship would have a permanent heel that must be avoided because, among other things, it will make any additional heel, caused by a wave or wind, for example, more dangerous.
You can calculate the initial metacenter for your initial equilibrium position (which has an alpha heel) but the GZ curve will not pass through the origin, it will not cut the x-axis (which represents the angles) at the origin, but will cut it at the point of x-axis = alpha.

View attachment 206349

There's an error in your reasoning because it's the GZ that we don't know for each degree of heel. The CoG, I repeat, is fixed (unless there's a weight shift), and if for whatever reason the ship heels by alpha degrees, the new position of the CoB is calculated. The GZ is then calculated from this.

Thanks for that clarification. This was what I was thinking too in my last post, but didn't explain myself properly. So basically it's important that the initial position is in equilibrium (regardless of whether heel is zero or not, although preferable to be zero in real life), and then the rest of the calculations will make sense.

Okay, I think that answers it all. Thank you all for your time!