# Heeling and Righting moments calculation

Discussion in 'Hydrodynamics and Aerodynamics' started by ClaudioD, Sep 3, 2021.

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### ClaudioDJunior Member

Years ago a friend of mine passed a paper to indicate how to calculate the weight of the Bulb of a IOM sailing model as function of Wind Speed and Sail Area.
The most unclear parameter to me is the "C" Lift coefficient. Via many readings this Coefficient is changing values from 0.1 up to 1.5.
I'm a bit lost other than mathematics are not my cup of tea.
Thank you very much

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### DCockeySenior Member

The formula looks correct but it is not Bernoulli's equation.

Aerodynamic force coefficients are commonly defined as Coefficient = Force / (Reference area * 0.5 * fluid density * velocity ^ 2). The formula uses that definition of the coefficient "C".

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### TANSLSenior Member

The formula only measures the force exerted by the wind on the sail, without taking into account the force exerted on the hull, and with the added difficulty that the value of the coefficient "C" is not known exactly. That is why I would calculate the KN value at 30º to deduce the heeling / righting moment with it. As long as it is 30º the maximum heel that the boat will acquire. I think this is much more accurate than applying your "Benuilli formula".

Last edited: Sep 3, 2021
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### ClaudioDJunior Member

Hi David ,
Thank you !
If the formula is correct I will continue to use Lift Coefficient "C" = 1
Regards

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### ClaudioDJunior Member

TANSL
thanks a lot,
30° is just chosen as maximum acceptable Heel angle.

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### ClaudioDJunior Member

It is a sailing model of 1mt
Non hull stability expected
Hull surface against wind very low
Wind at 1 mt height totally different from wind at 3 mt height
This is why I do have some doubts including the Dellenbaugh angle formulation.

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### jehardimanSenior Member

It depends on what the formula is for. Basically, if the intent of this formula it to determine whether the IOM is knocked down (i.e. past 30 degrees or any other arbitrary angle up to 90) in a 12 knt wind when presented side on and fully sheeted in....then use C=2.0 and call it done.
Otherwise, the formula doesn't have much value unless tied to many other factors associated with sailing. For example, the Aspect Ratio of the sail is ~5:1, so at ~30 degrees apparent wind Coefficient of Lift is ~1.6 for a well shaped sail. But this lift force is perpendicular to the wind, not the hull. Additionally, there is a drag force in the direction of the wind. These two forces are attempting to overturn the hull and are resisted by the weight of the keel bulb as well as the shape of the hull. Furthermore, the keel itself is attempting to overturn the hull due to its buoyancy and lift.
If you want to know more, I would suggest the Aero-Hydrodynamics of Sailing by Marchaj.

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### ClaudioDJunior Member

More confused than before
Thanks

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### TANSLSenior Member

Mo misses me, I would be too.
Do you have a body lines plan to be able to work with it?

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### TANSLSenior Member

Nice model, very nice, but if you don't have a body lines plan, no estimate can be made. With it, it is much more accurate to calculate the righting moment for a 30º heel, given a certain displacement, than to calculate the heeling moment due to 12 knots of wind speed.

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### TANSLSenior Member

When the force exerted by the wind on the sails and on the hull (whatever this force is) causes the boat to list 30º and it is in equilibrium, it means that the heeling moment due to the wind (very difficult to calculate exactly) is balanced by a righting moment generated by the submerged part of the hull. Both moments are equal when the ship is in equilibrium. As this second moment can be calculated with great accuracy, the most practical thing would be to calculate it and forget if the coefficient "C" has a value of 2, 1 or 0.72356. Therefore, the best way to get the information you are looking for is to use the body lines plan.
I only mean to help, I do not mean to distract you from your thoughts or to disrupt this thread with nonsense.

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### ClaudioDJunior Member

all nice but you have numbers for the ballast ?

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### latestarterSenior Member

Getting an accurate ballast weight from theory is a problem. If your assumed wind speed is 20% out the overturning moment will be 44% out.
Are there similar successful model yachts that you could use as an estimate of what weight works.

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### TANSLSenior Member

That is precisely what needs to be calculated.

revintage likes this.

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### ErwanSenior Member

The maximum lead ballast should be driven by the "floatation line" of your design

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