# Ballast/Displacement Ratio-minimum for self-righting

Discussion in 'Sailboats' started by Doug Lord, Sep 17, 2009.

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### Doug LordFlight Ready

I've edited this question to more specifically explain what I am trying to achieve-instead of a new post. The reformulated question ignores the Ballast/Displacement Ratio.
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So I'll try again:
Assume the boat is lying with the mast more or less horizontal-maybe just the tip touching water.It is flat calm and the sails are dry(not touching the water):
1) the CG of the rig,incl. sails is 35' from the heeled CB of the hull and it all weighs 1250 lb.
2) 35X 1250=43750ft. lb.
3) the CG of the bulb is 12' from the heeled CB of the hull
4) to balance the boat in this position, the bulb would have to weigh 3646lb
(3645.834 X 12= 43750ft.lb.)
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My question is about judgement hence the desire to have an experienced designer/engineer answer:
We know it takes 3645.834Lb. to balance the boat in this position with no net RM or tendency to come upright. My question is: what multiple of this bulb weight would you use to bring the boat upright in flat water and no wind? Forget about sailing RM-it does not matter for the purposes of my
question. Forget about anything to do with the hull-it is largely* irrelevant for the purposes of this question.
* Assume the hull form contributes little to nothing to stability at any angle.
__________________

Last edited: Sep 18, 2009
2. ### Paul BPrevious Member

This is an idiotic question.

First, without any details of the hull shape no one can answer this. "Assume a carbon rig"? How tall, what's it weigh, etc? A 10' draft on a 60 footer?

Second, if you were doing "research" on something that has some preliminary design work done you would be able to calculate this yourself. Well, not you, but someone who knows how to design a sailboat.

Third, who wants to to stay knocked down until the wind stops?

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### fngJunior Member

Or have to pull down wet sails to get upright.

4. ### Paul KotzebuePrevious Member

Ballast / displacement ratio is not considered when evaluating stability. It's one of those ratios that are used to compare similar boats without drawing any kind of meaningful conclusion.

Every hull form has its own unique stability charactersitics. While there is no minimum ballast / displacement ratio for any given boat, there is a maximum vertical center of gravity (VCG). It is not uncommon for naval architects to calculate the maximum VCG at various drafts or displacements to meet a certain stability criteria.

Ballast does not work independently from the rest of the boat. As far as static stability goes ballast, boat, rig, etc. is all one object with a fixed center of gravity.

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### sorenfdkYacht Designer

There is no such thing as an idiotic question - only idiotic answers!

Paul B; it is clear that you know something about some things, but why do you so often have to act as an annoying know-it-all?

It is evident that we need some more information to answer the question. Please just ask for that information and leave it at that!

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### Doug LordFlight Ready

Please re-check the first post-I edited it to make the question reflect the judgement I am interested in more clearly.

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### fngJunior Member

At 60ft is this something that intends to go offshore ? If so you have catogory 0, 1, and 2 to comply with, and then you have to be built to iso 12215. In little old NZ vessels wanting to go offshore had to have a positive righting moment through to 115 degrees ( prior to the isaf/ iso changes )
this could be a good place to start

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### Doug LordFlight Ready

======================================

Thanks fng but this is not part of any design-it is a question of judgement on one specific issue and not related to any other issue.

9. ### Paul BPrevious Member

You are clearly wrong. Just last week I was involved in a discussion with some business colleagues regarding the sheer idiocy of the often quoted, "There are no stupid questions."

Every person in the group was able to quickly tell an anecdote showing this quote to be, well, stupid.

I'll simply note here you have not answered the question either.

You have been here long enough to realize this is simply a fishing expedition where The Lord of Non-Foiling is trying to get someone, anyone, to commit in writing a number that he can use in another argument about how adding lead ballast to his proposed 60 foot foiler is a bad idea.

So if you, Soren, give an answer then The Lord will be posting all over the internet that you, Soren F. on Boatdesign.net, have calculated the need for only X and that is insignificant in the whole scheme of his self-righting 60 foot monohull foiler.

He has done this before, citing other people by taking their answers out of context, even when they ask him repeatedly to stop quoting them.

So to make things easy, and to make you happy, I will answer The Lord's initial question: 37.

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### jehardimanSenior Member

[My emphasis in the above quote]

As pointed out, this is a contrived question. The real crux of the matter is that the hull is designed poorly, as has been stated.

Anyway, since we venture into the absurd, as long as the bulb weight is 1 mcg greater than the weight required to balance the rig on the fulcrum point as proposed, the vessel will always attemp to right. Of course the randomness of the environment will make it more or less successful, but in a perfectly benign environment as proposed it would always come to upright, it is just a matter of time. If you wanted it to come up in a given time you could calculate the mass moment of gyration and then work the additional weight needed (including resiting moments due to roll velocity on the laterial plane and the rig). Of course it may roll to the other side and then oscillate for a long time depending on damping, but it will eventually reach upright equlibrium. What is an acceptable length of time to reach upright stability is the question you are really asking.

Of course just having a microscopicly small righting moment will not give this vessel any abilty to carry sail. But that does not seem to be the issue here.

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### sharpii2Senior Member

Hi, Doug.

There are two kinds of stability.
1.) is form stability, based on the sectional shape of the hull, and
2.) is pendular stability based on the weight distribution of the craft as a whole.

Pendular stability is the type you describe. It is like a buoy or a fishing bobber, where the Center of Buoyancy (CB) is above the Center of Gravity (CG) when the craft is upright.

Form stability is usually when the CG is above the CB and, as the craft heels, the CB shifts to leeward faster than the CG. Obviously, there is a limit to how long this can go on. This is generally governed by the Beam of the craft. Eventually the CG shifts to leeward past the CB and then the craft has negative stability and capsizes.

Most seagoing monohulls use a combination of the two types of stability. The form stability is for sail carrying ability, because it is usually near its greatest while the craft is upright enough to make sail carrying practical, say 45 deg. heel or less (the pendular stability is helping here too, but to a lesser extent). Once the heel gets much beyond this point, say 60 deg.,the form stability starts to run out. At this point the pendular stability starts to kick in. But in all but the most extreme types, that starts to run out too. This is because the CG in these craft is still above the CB, but to a lesser extent.
Once that happens, the craft continues to capsize until completely upside down. But this condition doesn't usually last. Usually it was a wave that capsized the craft in the first place. That being so, it is all but certain other waves will hit the craft while upside down. This is where the magic comes in. If the wave that capsized it had to knock it down to 120 deg. heel, the wave that hits it while capsized has to only knock it past 60 deg to put it in a position to right itself. The righting process can start very slowly (assuming the wave that hit the craft had just enough energy to push it past 60 deg.), but as the boat becomes more and more righted, the righting moment increases until the craft reaches its maximum righting moment at about 60 deg. heel from upright. Then it starts to diminish until the craft is upright.

There are two extreme types of hulls where this is not true. The first are multihulls, which have only form stability, but an extreme amount of it. The second are plank-on-edge monohulls which have almost exclusively pendular stability.

The plank-on-edge monhulls had very narrow beam, comparable to hulls on a multhull. Their draft was often greater than their beam. The problem with these boats was their maximum righting moment was at 90 deg heel and diminished progressively after that. At about 60 deg heel from upright, they had already lost half their righting moment. And this was with the masts canted up only 30 deg from the waterline. The solution, of course, was to add more ballast (my guess would be at least 100% more). Then reasonable righting moment could be had to carry sail at a reasonable amount of heel, say 45 deg or less.

I only bring this up to give you an idea of what you're up against.

To answer your question, I propose the following formula:

((Righting Moment - Capsize Moment)/(Capsize moment))*((2pii*Capsize Arm)/32.2)

This will give you an approximation of how long it will take for your craft to right itself in seconds, under the following assumptions:

1.) Once righted, craft has no momentum left to heel it over to the opposite side, and
2.) There is no appendage friction of any kind.

Both of these assumptions are obviously false, but I think this formula could still be useful to help you get to where you want to go.

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### Doug LordFlight Ready

Thanks very much, Sharpii2! I'll work with this a bit and see what I find. I appreciate your effort.

13. ### Paul KotzebuePrevious Member

How did you derive your proposed formula? You did not specify units, but I'm assuming the moments are ft*lb, capsize arm is ft, and 32.2 is the acceleration of gravity in ft/sec^2. That being the case, the answer to your formula will be in seconds squared, not seconds.

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### boosterSenior Member

Hi!
Kotzebue and shapii2. Seems that the formula by sharpil2 needs a = to be a real formula. Perhaps something on the left side of the = helps it to be in seconds? For once I feel for the canting-keel-man who must be struggeling with the equation...
Regards,
Booster

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### Manie BSenior Member

sharpii2

thanks for a civil answer, nice to see gentlemen on the forum

a good read - and a good answer to what was intended as a relatively modest straight forward question, its nice to have an answer that can take one back to basics

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