IS Code, weather criterion

Must be that I'm missing something pretty basic but can anyone explain?

If there is first a steady wind heeling lever and in turn a resulting angle of equilibrium/list angle. Wouldn't it be more critical that the ship is then:
1. assumed to roll owing to wave action leeward instead of windward

QTE
3.2.2 Recommended weather criterion
3.2.2.1 The ability of a ship to withstand the combined effects of beam wind and rolling should be demonstrated for each standard condition of loading, with reference to the figure as follows:

• .1 the ship is subjected to a steady wind pressure acting perpendicular to the ship's centreline which results in a steady wind heeling level (1w1).
  
  
• .2 from the resultant angle of equilibrium (θo), the ship is assumed to roll owing to wave action to an angle of roll (θ 1) to windward. Attention should be paid to the effect of steady wind so that excessive resultant angles of heel are avoided;footnote
  
  
• .3 the ship is then subjected to a gust wind pressure which results in a gust wind heeling lever (1w2);
  

UNQTE

 

Hi!

To see this, I think is really good to see the GZ curve:



So, the key consideration here is to account for both the wind lever (horizontal line) and the energy associated with rolling to the other side (area A). When a vessel rolls, it accumulates energy "against you" from its own GZ curve. The criteria are valid because when rolling windward, the righting lever "works against you", allowing you to accumulate energy for rolling to the opposite side. Additionally, on that side, you also experience a wind force, combining both effects.

To measure that you have enough energy to deal with it, IMO sets this:

• .4 under these circumstances, area b shall be equal to or greater than area a

Wich means that you must have enough GZ curve to resist the wind and the rolling combined.


Of course, To Wind and waves impacting to the same side, it's not desirable, but we would be talking about another type of criterion.

 

Hi Lopolito!

Thanks for joining the discussion!!

I do understand that the gz curve and the gust would work in the same direction resulting in a grater heeling angle leeward in the end. But if IMO assumes in point .2 that the waves induce a roll windward, wouldn't this same wave system then counter the combined effect of the gz curve and the gust when the vessel returns again to the leeward side?

 

Hi Jure,

The point is the method that is being used, here, IMO is using the Energy Balance Method, here you asume the vessel is just moving with a resonant roll due to the waves. So, it is assume this:

• You have a balance windward
• "Suddenly", when you are upright, you suffer a gust wind
• Then you have to sum the different areas (energies)

So, taken into account the energies, its just the same to have a wave windward or leeward, since if you are in beam waves, and you are measuring the balance of the ship, that would be the same regardless the side, your ship is symmetrical (Disclaimer: it is true that GZ curve in many cases is not symmetrical, but at low angles it use to be), so the energy to heel your ship to one side, would be the same on the other.

The point is to visualize the problem: we are balancing in beam waves, and suddenly, a gust of wind hits you when you are upright and you were about to balance to the other side. So its really pesimistic way to see it.

 

Hi Lopolito,

That makes it clear, thank you for taking the time!
I think I was misled by the Figure 2.3.1. It seemed to me that a wave that would roll the ship to θ 1 windward is quite significant and that it would help the gz curve when on leeward side by inducing a righting moment. But maybe θ 1 is not that big of an angle when you actually calculate it in comparison to how big it seems in the Figure.

 

You have to take in account something regarding this, under the effect of wind, the vessel equilibrium, is not at angle zero, but, in the intersection between GZ curve and wind lever. So in this case, the energy that goes "against" you, comes from this point (θ0) to θ1. So all this area "a" is acumulated energy that is going to make you heel more.

Thanks to you for raise this interesting question!

Greetings from Spain