Unifying small heel angle and large heel angle GM calculations

Hi,

I am currently improving the ship hydrostatics and stability simulation software that I wrote some years back, and I need some advice.

As you probably know, for small angles of heel (ϕ<10° or so, depending on the hull shape), metacentric height is calculated by formula GM = BM - BG, where BM=I/V (Moment of inertia of the waterplane/Volume of displacement) (source). This method assumes that B moves more or less in an arc, so BM doesn't change.

With larger angles of heel, with most hulls the B no longer moves in an arc, so the BM distance changes, and the first formula is no longer valid. Instead, the suggested method is to determine the actual position of B at specific angles of heel, and calculate metacentric height as GM = GZ / sin(ϕ) (source). However, this method doesn't work for ϕ=0, because it leads to division by zero. For very small angles of heel (say ϕ=0.01°) it may also lead to unreliable results due to insufficient numerical accuracy.

Now, the issue is that since these two methods approach calculating GM in very different ways, for certain hulls they might produce significantly different values. For instance, look at this GM vs ϕ curve, where GM at ϕ=0° was calculating using the first formula (based on moment of inertia), and GM at ϕ>0° was calculated using the second formula (based on GZ):



As you can see, the sudden change in GM from 0° to 1°, due to the changed method of calculating it, makes for some very confusing and suspicious results. If we were to switch the method at any other heel angle, say ϕ=10°, we would still get a sudden change in values.

So my question is, how can these two methods be unified to produce a method that works at any angle of heel, including 0°? Or is there another method I'm not aware of, that solves this issue? One idea is to use second method only, and for ϕ=0° calculate GM it at ϕ=1° and ϕ=2°, and then extrapolate to get the value at ϕ=0°. But this is an extra computation and feels like a cheat. Surely there must be a better way?

 

Why not simply use GM = KB + BM - KG with BM = I / V ?

 

Kind of yes...it depends.
The hull shape plays a part, and as such, one assumes, for small angles of inclination a - wall sided - formula. Which is about the 'wedge' of buoyancy either-side of the centreline created by the small angle of inclination.
Ship shapes do not follow 'perfect' arcs, in that sense, only a sphere would. But for very small angles, it is permissible.

Correct.
Since the initial meta centre, M, no longer remains in a constant position or even on the centreline - as is the case with small angle assumptions noted.
Additionally the waterline that are drawn at each angle of inclination, need not intersect on the CL of the vessel, as this then becomes very hull shape dependent. That could be the source of your error too.
So, what this means is that the value of the righting lever calculated by the small angle assumption, is no longer valid. So, the only way to obtain the stability is in terms of the righting lever, GZ using the method known as Atwood's formula.

This too is not with its own issues.
So the laborious way is to assume a fixed point and its perpendicular distance, much like the GZ, but as the SZ, a line of action owing to the buoyancy force.
So, GZ = SZ + SG.sin(angle)

 

Because that doesn't work for the large angles of heel, like I explained in my post...

That last formula you quoted - in my case, I already know GZ, it is very easy to measure with software, but I need the GM. The second formula that I quoted (GM = GZ / sin(ϕ)) was obtained by rearranging the traditional GZ = GM . sin(ϕ) formula. Perhaps I am misunderstanding what you meant there. I am not sure what the "fixed point" is, and how do we determine it (as well as SZ), I can't find any references to this on google. Can you please clarify?

Also, the issue remains: this formula won't work with heel angle of 0°, and what I am looking for is a way to calculate GM at ANY angle of heel, including 0° with a single method, rather than having to switch between different formulas, because that creates a discrepancy in values, as can be seen from my graph. Maybe I'm misunderstanding what you suggested, or maybe I inadequately explained myself in my post...

 

For large angle stability, you need to calculate the righting lever GZ first.

Basic Ship Theory Volume 1 - Rawson & Tupper, Chapter 4, has is all neatly explained.

See above.

 

Like I said, in my case, GZ is already calculated. What I need is the GM. And GM = GZ / sin(ϕ) doesn't work for ϕ=0° because that leads to division by zero (and also because GZ is typically zero too, since the ship is perfectly upright).

Thank you, I reviewed it, but the S point is really confusing there. It is first introduced in page 101 as a point from which a freely suspended weight is hung, but then seems to be used in an entirely different meaning in page 105, since there's nothing about any suspended weights there. I am not sure what S point is there.

But in any case, the issue I described above remains: all these methods rely on heel angle not being zero. And what I'm after is a consistent method for determining GM that works even when GZ and ϕ are zero.

EDIT: the more I read that book, the more it seems that for every method described there the various terms are used very loosely and inconsistently, with very little explanation of what is what. The point S and Z from which it is derived (in the context of page 105) is never explained. The same two points are referenced in page 112, but in an entirely different meaning, again, with no explanation how does one find that S. In any case, all these different methods seem to assume that GM is already known, whereas in my case GZ is known instead.

Ad Hoc, maybe I'm missing something obvious, but I don't see the answer I'm looking for in that book. If you could point me to the method that you had in mind for determining GM at any angle of heel (including 0°), it would be great.

 

Realistically, and traditionally, you just computed the B (and RM, etc.) at 2 degrees heel and that was where GM came from. In your case, given the very flat GZ curve, it should agree very well with the computation based on waterplane. Given the big divergence, there is an error in one of your computation methods. How are you computing the heeled volumes? Do you compute from scratch each time or do you rebalance the volume based on the previous condition? How are you computing static waterline moment?

 

I think that the concept you are missing is that the Metacenter, M, is fictious. It is a mathematical construct of the instantious Iwp/displacement for an infinately small angle of heel with a wall sided assumption. GM has no physical meaning except as an indicator of the present state of stability of the vessel. It does not effect the actual stability of the vessel, especially with respect to any angle other than the instant it is calculated. This is why the Cross Curves of Stability tables were developed; they allow stability at any angle to be calculated.

See this old thread and please ignore the snark in it.
Calculation of GM https://www.boatdesign.net/threads/calculation-of-gm.8852/#post-60405

 

If (Big If) is a light modern Sailboat there is a trick (which must be used with caution):

GZ = GM x sin (phi) x cos (phi)

GM = KB + BM - KG

KB = 0.64 T
BM = Bwl^2 / 10 T

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There is another way in JW Slooff Book

 

JW Slooff

(Deleted)

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I copied the formula incorrectly, and even after correcting it, I can't get it to work

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The formula above works well up to 20 degrees. At 30 degrees, it makes a very strange jump, and curiously, at 60 degrees, it regains its sanity.

 

The only confusion is R&T using the letter "S" for different scenario examples.
Don't conflate the two...

As JEH also notes:

Thus you are fixated on the metacentre, M, as an absolute, it is not.

You need to fully review - the section on Cross Curves of Stability, to understand why this is so.

 

I'm going to return to the question in the first post - why is the result of calculating the height of M significantly different at 0 degrees calculated using from the moment of inertia divided by the volumetric displacement compared to when calculated at small angles of heel using the center of buoyance and the sine of the heel angle.

The discrepancy in the example shown is around 0.055 m. In theory the small angle result should be very close to the zero angle result. Without knowing the size, proportions and shape of the vessel it is impossible to comment whether this discrepancy is signficant. I'm ignoring the shape of the curve shown because it appears to be the result of using a curve fit to data points at 0, 1, 2, 3, etc degrees with a step in the data between 0 and 1 degree followed by values at larger heel angles which vary slowly.

Similarly without any information about how the two calculations were made it is difficult to suggest specific reasons. I will suggest some general possibilities.

• Differences in the numerical algorithms.
• Differences in step size / mesh size and similar if numerical integration is used.
• Mistakes in the formulas used as the basis for the code.
• Mistakes in code. This can be something as simple as a sign error or an index error.
• Different reference points for determining the height.
• Discrepancy in the axis system for the heeled vessel compared to the zero heel vessel.

To diagnosis the source I'd start with very simple shape - a circular cylinder with the axis parallel to the water surface (CG in the middle of the cylinder) and the cylinder submerged something like 0.5 radius. For a circular cylinder with zero trim the metacenter will be on the axis of the cylinder for all heel angles. The waterplane will be rectangular so the moment of inertia will given by a simple equation. The submerged volume can be calculated by a simple formula. The center of buoyancy will be directly below the axis of the cylinder for all heel angles. Check whether the calculated values are correct. If they are not then find the cause(s) of the discrepancy.

If the circular cylinder results are correct then I would test a simple rectangular block.

Keep testing with increasingly more complicated geometry.

If you have control of the step size, mesh density/size or similar parameters for numerical integration steps then test how they influence the results.

If you have control of any convergence parameters then test how they influence the results.

 

JW Slooff

The Aero- and Hydromechanics of Keel Yachts https://link.springer.com/book/10.1007/978-3-319-13275-4

Up to 20 degrees it works very well.

(I don't know about further down the line, because maybe the sailboat I'm using for testing doesn't fit well with the formula)

Both formulas work very well for me up to 20 degrees.

 

After all, up to 20 degrees is the useful heel, even less in a modern, light sailboat.

It must be taken into account that beyond 15 degrees in a light sailboat, the centerboard root is very near the water surface

On the other hand, I've been reviewing what JW Slooff says about the rudder behind the centerboard, which I found a bit surprising:

The induced drag of the rudder is a function of the induced angle of the rudder + the induced angle of the centerboard.

Ignorance is a bad thing. The mind most often doesn't know, but rather recognizes or projects.

If we draw the problem, what JW Slooff says is geometrically evident.

So, a nice improvement is the solution we now see in the world of high-level racing: two rudders, and the windward rudder out of the water.

 

True Wind Speed @ 10 m H (TWS_10) = 30 Knots

On a small, light, 1-ton sailboat, the effect of loading 200 liters of seawater windward and aft (!) is monstrous.

We achieve four things:

1) We control the force of the sails that have a huuge lever arm and cause Pitch bow down, so we avoid nose diving

2) We move the sailboat's center of gravity even further aft of the Pitch axis, which passes through the center of Flotation, and so the boat rides the waves more smoothly

3) We increase Displacement, and

4) We move the center of gravity to Windward.