Atlantic 57 Catamaran Capsized

[Just Another Sa]

Originally Posted by Cottontop

Humid air is less dense than dry air, no?

Yes, water vapor H2O only weights 18 grams/mole, while nitrogen N2 weights 28 grams/mole and oxygen O2 32 grams/mole.
But when you compare liquid water and gaseous air the difference is much larger, water being about 800 times heavier than air. Therefore just 0.1% of water by volume will multiply density by 1.82, from 1.22 kg/m^3 into 2.22 kg/m^3. Exact numbers of course depend on temperature and pressure.
That effect is very relevant on wind pressure when wind speed is 200 knots, but have hardly any effect at all when wind speed is less than 100 knots.

[Just Another Sa]

Originally Posted by Polux

That is not an article but a technical paper from someone that knows a lot about boat stability.

No, the paper is not wrong, it seems that what you know about the subject did not allow you to understand it.

One value is the product of the GM and displacement - 6.1 tonne meters

The other value is the maximum righting moment(4.8) that is the product of Max GZ and displacement.

By the way the author, Barry Deakin is a research engineer of the Southampton University, the one you were (rightly) referring to be a reference in what regards stability studies for monohull and multihull yachts and Naval Architecture related with them.

Ship & Boat Internatonal is a magazine, and your pic was taken from November 1994 issue page 35.

I did indeed miss the change from GZ into GM by no reason. GM at zero degrees is of course very close to zero for the trimaran, since its ama has hardly any waterplane area when there is no heeling angle. Since its use for such trimaran makes no sense (huge variation with very small changes of heel angle) I falsely assumed the article was about GZ all the time.

I assume the position of Barry Deakin has not stayed the same since 1994 and now, he may very well be an expert on his area now, even if he wasn't in 1994.

[Just Another Sa]

Originally Posted by Dockhead

You keep saying "everyone knows". Yet you don't provide anything with any substance to make your point. If it's so simple and obvious, can't you make a calculation or give us a citation, or give us a real life example? Every single other poster on this thread has called BS on this. I would be glad to find out that you are a lone genius with a unique understanding of how this works -- maybe we could all learn something -- but you don't give us anything to work with -- you give us only rhetoric and insults.


The paper I cited did not say that the heeling moment is negligible "compared to multihulls". It said that the heeling moment is negligible -- period. It's what all the rest of us have been saying -- knock a monohull down with a strong gust, and it will stay at 90 degrees (or 80, or 70, or whatever) until the gust passes, then it will pop back up. It will not go over in the absence of wave action. Because once the mast, rigging and superstructure is out of the wind, there is negligible heeling moment, and yet righting moment is still large.


Lever arm -- which you call "nonsense" -- is how you convert force to moment. The equation if M = F a, where M is moment, F is force, and a is arm. A certain amount of force centered up in the rigging will produce a lot of heeling moment, because there's a long lever arm. The same or even much more force, centered down at the heeled over hull, close to the center of buoyancy, produces proportionately less torque. That's why the wind can't capsize a typical ballasted keel mono, at least not in terrestrial wind ranges.

I have posted the calcs already, but here it is once again:
example values wind speed 50 m/s, density of air 1.22 kg/m^3, hull bottom area exposed to windage at 90 degrees of heel 20 m^2, cd=1.0, leverage for windage = 1.5 m. Weight of boat 4000 kg, ZM at 90 degs = 0.6 m g = 9.81 m/s^2.
Dynamic pressure = 0.5*rho*v^2 = 0.5*1.22*50*50 = 1525 Pa.
Windage (force) = Dpress * area * cd = 1525 * 20 *1 = 30500 N
Heeling moment = 30500 N * 1.5 m = 45750 Nm.
Righting moment at 90 degs = 4000 kg*9.91 m/s^2*0.6m = 29430 Nm.
HM / RM = 1.55 and the boat inverts if it's heeled to 90 degs by windage on mast & rigging, boom&lashed mainsail & hull side & ...

Of course exact numbers will vary, but there always exists a wind speed when this becomes a reality for such boat, and that windspeed can easily be exceeded by an extremely rare event that for example a microburst hits the boat directly side on. Even more so with any sails up.


I know full well what a leverage is, and have already explained 2 times what your mistake was.
There is 2 different equations one for HM and another for RM.
heeling moment HM = F1*a1, and Righting moment RM = F2*a2.
F1 is a force which is approximately horizontal in unheeled boat with uncanted mast, and tilts with heeling or canting. It's same magnitude as the component of aerodynamical resultant that is taken perpendicular to the boat centerplane. The leverage a1 has no relation what so ever with buoyancy or its center. a2 is very much related to center of buoyancy and is a horizontal dimension for too vertical forces, the other being buoyancy and another gravity. That is the last time I will explain that basic fact to you and mixing up a1 and a2 remains nonsense. You can not calculate leverage from center of effort of vertical force to the center of effort of horizontal force. Every time you do that you write pure nonsense.

"heeling moment is negligible -- period"
heeling moment is never negligible period, it depends on wind pressure, area the wind is acting on, aerodynamic coefficients and leverage. Increase wind pressure enough and any heeling moment that has in some other situation been negligible is no more such. When experts write on something they assume such basics are understood, and therefore assume their text is read with the context, not taken it out to totally different situation and taken as a religious person takes something written in the bible.

For a typical modern and wide 30 footer monohull, the leverage1*area1*drag coefficient1 of the hull at 90 degrees of heeling angle is the same magnitude than the leverage2*area2*drag coefficient2 of the bare mast of the same boat at zero degrees heel angle.
Typical values would be leverage1 = 1.5 m, area1 => 20 m^2
leverage2 = 8 m, area2 = 3 m^2.
A small difference in wind speed or density will easily make up any difference in drag coefficient*area*leverage.

The situation is very different for larger boats, because the increase of RM with size is much more rapid than increase of HM of windage on either hull or bare mast & rigging or on both.

[TeddyDiver]

Originally Posted by Just Another Sa

I have posted the calcs already, but here it is once again:
example values wind speed 50 m/s, density of air 1.22 kg/m^3, hull bottom area exposed to windage at 90 degrees of heel 20 m^2, cd=1.0, leverage for windage = 1.5 m. Weight of boat 4000 kg, ZM at 90 degs = 0.6 m g = 9.81 m/s^2.
Dynamic pressure = 0.5*rho*v^2 = 0.5*1.22*50*50 = 1525 Pa.
Windage (force) = Dpress * area * cd = 1525 * 20 *1 = 30500 N
Heeling moment = 30500 N * 1.5 m = 45750 Nm.
Righting moment at 90 degs = 4000 kg*9.91 m/s^2*0.6m = 29430 Nm.
HM / RM = 1.55 and the boat inverts if it's heeled to 90 degs by windage on mast & rigging, boom&lashed mainsail & hull side & ...

Of course exact numbers will vary, but there always exists a wind speed when this becomes a reality for such boat, and that windspeed can easily be exceeded by an extremely rare event that for example a microburst hits the boat directly side on. Even more so with any sails up.


I know full well what a leverage is, and have already explained 2 times what your mistake was.
There is 2 different equations one for HM and another for RM.
heeling moment HM = F1*a1, and Righting moment RM = F2*a2.
F1 is a force which is approximately horizontal in unheeled boat with uncanted mast, and tilts with heeling or canting. It's same magnitude as the component of aerodynamical resultant that is taken perpendicular to the boat centerplane. The leverage a1 has no relation what so ever with buoyancy or its center. a2 is very much related to center of buoyancy and is a horizontal dimension for too vertical forces, the other being buoyancy and another gravity. That is the last time I will explain that basic fact to you and mixing up a1 and a2 remains nonsense. You can not calculate leverage from center of effort of vertical force to the center of effort of horizontal force. Every time you do that you write pure nonsense.

"heeling moment is negligible -- period"
heeling moment is never negligible period, it depends on wind pressure, area the wind is acting on, aerodynamic coefficients and leverage. Increase wind pressure enough and any heeling moment that has in some other situation been negligible is no more such. When experts write on something they assume such basics are understood, and therefore assume their text is read with the context, not taken it out to totally different situation and taken as a religious person takes something written in the bible.

For a typical modern and wide 30 footer monohull, the leverage1*area1*drag coefficient1 of the hull at 90 degrees of heeling angle is the same magnitude than the leverage2*area2*drag coefficient2 of the bare mast of the same boat at zero degrees heel angle.
Typical values would be leverage1 = 1.5 m, area1 => 20 m^2
leverage2 = 8 m, area2 = 3 m^2.
A small difference in wind speed or density will easily make up any difference in drag coefficient*area*leverage.

The situation is very different for larger boats, because the increase of RM with size is much more rapid than increase of HM of windage on either hull or bare mast & rigging or on both.

Now you have a theory and a bit math to support. Now just go and make some tests to find out how this theory works. So far there's no evidence..

BR Teddy

[poiu]

Originally Posted by Just Another Sa

I have posted the calcs already, but here it is once again:
example values wind speed 50 m/s, density of air 1.22 kg/m^3, hull bottom ar

SNIP

, because the increase of RM with size is much more rapid than increase of HM of windage on either hull or bare mast & rigging or on both.

Great job Just Another Sa.

The explanation also needs to account for the dynamic balance of the hull in the horizontal plane as it will weathercock when heeled hard over, the bulk of the drag being to the stern, and that will massively reduce the heeling force.

There must be some real world examples of this, or modelling done.

Also I'll bet it can be nicely shown with a graph to show heel angle against wind speed under bare poles for each boat design. It would show I f there would be a sudden reversal after capsize as the keel hits the air flow. I suppose that would depend a lot on the keel and hull shape.

[TeddyDiver]

I can see two flaws instantly in JAS reasoning. Wind shear grows quite much with wind speed and the aerodynamics of a 90deg turned roundish hull side sticking out of water..

BR Teddy

[Polux]

Originally Posted by Factor

I have sailed multis for 30 or more years, know many boats, owners and designers, never not once ever have I seen or heard of or been told of a multi being rolled back level by a wave after being capsized.

Yes I agree. A very, very rare case, but this case seems to show that it is not impossible. As I have said if when inverted a hull fill with water much faster than the other and the boat is on a sea way with big waves, it doesn't luck difficult for me, even if joining all these conditions together would be very rare.

Regarding why a hull would fill with water much faster than the other that can easily happen if the seacocks on a hull are closed and open on the other one.

[poiu]

Originally Posted by TeddyDiver

I can see two flaws instantly in JAS reasoning. Wind shear grows quite much with wind speed and the aerodynamics of a 90deg turned roundish hull side sticking out of water..

BR Teddy

What would the wind be at say 1.5m height if measured at 20m height at 100kt?

[Dockhead]

Originally Posted by Just Another Sa

I have posted the calcs already, but here it is once again:
example values wind speed 50 m/s, density of air 1.22 kg/m^3, hull bottom area exposed to windage at 90 degrees of heel 20 m^2, cd=1.0, leverage for windage = 1.5 m. Weight of boat 4000 kg, ZM at 90 degs = 0.6 m g = 9.81 m/s^2.
Dynamic pressure = 0.5*rho*v^2 = 0.5*1.22*50*50 = 1525 Pa.
Windage (force) = Dpress * area * cd = 1525 * 20 *1 = 30500 N
Heeling moment = 30500 N * 1.5 m = 45750 Nm.
Righting moment at 90 degs = 4000 kg*9.91 m/s^2*0.6m = 29430 Nm.
HM / RM = 1.55 and the boat inverts if it's heeled to 90 degs by windage on mast & rigging, boom&lashed mainsail & hull side & ...

Of course exact numbers will vary, but there always exists a wind speed when this becomes a reality for such boat, and that windspeed can easily be exceeded by an extremely rare event that for example a microburst hits the boat directly side on. Even more so with any sails up.


I know full well what a leverage is, and have already explained 2 times what your mistake was.
There is 2 different equations one for HM and another for RM.
heeling moment HM = F1*a1, and Righting moment RM = F2*a2.
F1 is a force which is approximately horizontal in unheeled boat with uncanted mast, and tilts with heeling or canting. It's same magnitude as the component of aerodynamical resultant that is taken perpendicular to the boat centerplane. The leverage a1 has no relation what so ever with buoyancy or its center. a2 is very much related to center of buoyancy and is a horizontal dimension for too vertical forces, the other being buoyancy and another gravity. That is the last time I will explain that basic fact to you and mixing up a1 and a2 remains nonsense. You can not calculate leverage from center of effort of vertical force to the center of effort of horizontal force. Every time you do that you write pure nonsense.

"heeling moment is negligible -- period"
heeling moment is never negligible period, it depends on wind pressure, area the wind is acting on, aerodynamic coefficients and leverage. Increase wind pressure enough and any heeling moment that has in some other situation been negligible is no more such. When experts write on something they assume such basics are understood, and therefore assume their text is read with the context, not taken it out to totally different situation and taken as a religious person takes something written in the bible.

For a typical modern and wide 30 footer monohull, the leverage1*area1*drag coefficient1 of the hull at 90 degrees of heeling angle is the same magnitude than the leverage2*area2*drag coefficient2 of the bare mast of the same boat at zero degrees heel angle.
Typical values would be leverage1 = 1.5 m, area1 => 20 m^2
leverage2 = 8 m, area2 = 3 m^2.
A small difference in wind speed or density will easily make up any difference in drag coefficient*area*leverage.

The situation is very different for larger boats, because the increase of RM with size is much more rapid than increase of HM of windage on either hull or bare mast & rigging or on both.

At last some substance!! Wow. Now we see that you are not a troll after all, but a geek with knowledge but zero communications skills . This is good, and interesting, and there are some propositions in it which we will all have to agree with. In particular -- naturally, the heeling moment of the knocked-down boat in a strong wind is never going to be nothing. It needs to be calculated. An immediate quibble is that the CD of the hull is going to be a lot better than 1.0.

I will work with this when I have some time in the next couple of days; thanks.

[Polux]

Originally Posted by Just Another Sa

I have posted the calcs already, but here it is once again:
example values wind speed 50 m/s, density of air 1.22 kg/m^3, hull bottom area exposed to windage at 90 degrees of heel 20 m^2, cd=1.0, leverage for windage = 1.5 m. Weight of boat 4000 kg, ZM at 90 degs = 0.6 m g = 9.81 m/s^2.
Dynamic pressure = 0.5*rho*v^2 = 0.5*1.22*50*50 = 1525 Pa.
Windage (force) = Dpress * area * cd = 1525 * 20 *1 = 30500 N
Heeling moment = 30500 N * 1.5 m = 45750 Nm.
Righting moment at 90 degs = 4000 kg*9.91 m/s^2*0.6m = 29430 Nm.
HM / RM = 1.55 and the boat inverts if it's heeled to 90 degs by windage on mast & rigging, boom&lashed mainsail & hull side & ...
....

Very few Class A boats have a displacement of 4000kg. Typically a cruising boat as a much bigger displacement.

Than you talk about the boat righting moment of the boat at 90º and for determining it instead of multiplying the displacement by the GZ (arm) at 90º you make some mess of calculations that have no direct connection with determining the boat righting moment at 90º.

Than you talk about the windage of mast and rigging that is inexistente when the boat lay at 90º of heel.

And finally we are talking here about energies, the rotational energy that is needed to rotate the boat through the AVS point (that is far away from 90º) and on a class A boat with 4000kg will be at least 125º and the opposed energy that will not be given by the RM at 90º, as you wrongly suppose, but by energy resulting from all RM angles from 90º to 125º.

Putting it on a more correct way, the energy that corresponds to the positive area under the Right Moment curve between 90º and the AVS point, at 125º.

Finally you are considering that all energy that results from the contact from the wind with the surface of the boat is converted into rotational energy and that the hull is a vertical surface at 90º. That is wrong on the two points.

The hull has a rounded surface and the energy resultant from the contact of the wind with it will be very different from the one resultant with a vertical wall.

Not all energy that results from the contact of the wind against the boat hull will be converted in a rotational movement. A considerable part will be converted in a lateral sliding movement.

Yes there is a very rare situation where the wind can invert a class A monohull, so rare that it is till now a theoretical possibility (never happened) and that is a very strong wing coming from above. That kind of winds that typically happen on a microburts happen typically at altitude and have been responsible for bringing some airplanes down.

At the sea surface they had already turned in horizontal winds.

It seems you are a student at Southampton and I sincerely hope that you are still on the first year. Anyway your atitude is a very wrong one.

[Polux]

Originally Posted by Seaslug Caravan

Appologies if this has already been posted

Yes , the chamberlain 14 would make an interesting case study as 2 that I know of have capsized, (Big Wave Rider (discussed here recently), and Incinerator ( Devonhouse Recollections )

I am unsure of the exact number built. I know of only 3.

Regarding that perhaps it make sense to post here some considerations about cat design that I made on another thread replying to a cat sailor that has the opinion that my opinion about cats are largely correct.

"Thanks, specially coming from a cat sailor. You know, I like all sailboats, including cats and the funniest thing is that even if I know that performance cats are much more prone to capsize than a condo cat and demand a much more careful sailing, personally I do like them more than condo cats, mostly due to a less windage and a better windward performance. I have to say that they look also a lot better to my eyes.

Another thing is to know how light a cruising cat can be, for a given size, to warrant a reasonably safety in what regards casual cruising sailing for average or even inexperienced sailors and that's a fact than lighter the cat the bigger it has to be, even if one does not need necessarily the space of a bigger boat. Bigger means also much more expensive.

When we see boats like the Chamberlain 14 (5T) with three boats made and two capsized and the TS 52 (7T) with two? boats made and one capsized, it seems to me that for that kind of Disp/Length 50ft cats don't not have a reasonably safe margin for cruising.

Unfortunately in what regards cruising cats the design of a cruising cats, with a weight that can make it safe with the size a couple needs for cruising (around 40ft), with low windage and daggerboards for a better upwind sailing and a reasonably price (similar to condo cats) has been completely abandoned due to the overwhelming number of cat clients that want them with the biggest possible interior space.

Regarding all that it seems contemporary mainstream cat design is compromising more and more sailability regarding older designs from the same brand. Fountain Pajot is a good example, with older models compromising less sailing, presenting less windage and less interior space.

Even saying all this it is obvious by the ARC results that modern condo cats are a good option in what regards passage making on the trade winds and that their performance, while cruising, on the trade winds with loaded boats is not very different from the one of performance cats. That is what show this ARC, as also other ARC editions."

[beiland]

Originally Posted by Polux

When we see boats like the Chamberlain 14 (5T) with three boats made and two capsized and the TS 52 (7T) with two? boats made and one capsized, it seems to me that for that kind of Disp/Length 50ft cats don't not have a reasonably safe margin for cruising.

You don't think the stability & safety margin would have anything to do with the size (and particularly the height) of the sailing rig do you
...as opposed to the weight of the vessel

[Just Another Sa]

Originally Posted by TeddyDiver

Now you have a theory and a bit math to support. Now just go and make some tests to find out how this theory works. So far there's no evidence..

BR Teddy

Sorry, but I can't take credit for well known physics theory:
https://en.wikipedia.org/wiki/Drag_(physics)
which is known to many highschool students.
There have already been tests made on windage of boats, nothing new there either. It's also well known fact that windage acting above water and lateral resistance acting below water produce heeling moment.
Just because some people in this forum don't know those theories doesn't mean they aren't widely known by others.

[Dockhead]

Originally Posted by Just Another Sa

Sorry, but I can't take credit for well known physics theory:
https://en.wikipedia.org/wiki/Drag_(physics)
which is known to many highschool students.
There have already been tests made on windage of boats, nothing new there either. It's also well known fact that windage acting above water and lateral resistance acting below water produce heeling moment.
Just because some people in this forum don't know those theories doesn't mean they aren't widely known by others.

There are many steps and a long way between understand how drag works, and understanding to what extent a ballasted keel monohull will be heeled by strong wind.

This is a kind of straw man argument -- you don't agree with my understanding Steps 13-39; therefore you are an idiot who doesn't understand Step 1 -- the theory of drag.

But it is not understanding drag which is the issue at all.

Everyone on here understands drag -- this is just an insult to their intelligence, and the reason why you evoke such negative reactions.

[Just Another Sa]

Originally Posted by TeddyDiver

I can see two flaws instantly in JAS reasoning. Wind shear grows quite much with wind speed and the aerodynamics of a 90deg turned roundish hull side sticking out of water..

BR Teddy

That depends on what causes the wind. Your assertion on wind shear applies when its the pressure gradients between lows and highs.
When the cause is a microburst there is practically any horizontal wind high up, and a lot of it closer to surface. Added fluid density from liquid water reverses the pressure shear, there is more pressure down low.
Hull sides are quite flat vertically these days, not much tumble home, and quite many boats have toerails to make the deck edge even less pointy. No roundness at all along the airflow.
There is also higher wind pressure below the keel than above it in those conditions, making the heeling moment a bit larger than that simple calculation. Some may call that lift, other use local flow direction and call it drag despite of being a vertical force.