# Vertical COG relationship to Performance Displacement Yacht

Discussion in 'Stability' started by motorbike, Aug 9, 2022.

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### motorbikeSenior Member

Good evening, I would like to know if there is a way of determining the degree of performance advantage from changing the vertical COG on a 4 ton yacht in relation to overall weight, all things being equal.

Presuming that the ballast keel is approximately 1.5 tons and the COG is 900mm below W/L for one boat and the 750mm for another. Assume a 3m beam, how much stiffer would boat A be compared to boat B, is there a way of calculating the approximate difference in terms of weight on the rail and how much more power for a given sail area A has over B. I presume you will need a lot more data but I am initially interested in a broad overview and maybe what 50mm increments do. I am sure this has been done before but I dont have the language to describe the problem mathematically in sensible terms!

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### Robert BieglerSenior Member

How do you get the centre of gravity that deep? Rig heights are commonly about six times the beam, so roughly 12 metres. I don't know what a rig that size weighs. 200 kg seems vaguely reasonable, with a centre of gravity about 6 m above waterline. The centre of gravity of a 3 metre beam hull, assuming conventional proportions, might be about 0.4 metres above waterline, and you assume 2300 kg weight. The combined centre of gravity for hull and rig is then 0.848 m above the waterline, and therefore either 1.748 m or 1.598 m above the intended centre of gravity with keel. For a 1500 kg keel to balance the 2500 kg that are 1.748 m above the intended overall centre of gravity, the 1500 kg keel's centre of gravity of the 1500 kg keel therefore needs to be 1.748 m * 2500 kg/1500 kg further down. That is 2.91 m below the intended overall centre of gravity, which you intend to be 0.9 m deep. So the centre of gravity of the keel must be 3.81 m below the waterline. The ballast itself will need to be deeper than that to balance the structure above the keel's centre of gravity. Add another 0.5 m for 4.31 m? That is quite a deep draft. The equivalent calculation for the 0.75 m depth of the overall centre of gravity gives me a keel's centre of gravity that is 3.41 m below waterline.

You will need to decide on a hull shape to get an accurate answer to your question, and the answer will be a righting moment curve as a function of heel, rather than a single number. And that will need software that can calculate righting moment from hull shape, weight and heel angle. You also need to take into account that the deeper keel that you need for the lower centre of gravity has a lower centre of lateral resistance, therefore a greater lever arm relative to your centre of effort, so some of the gains in righting moment will be lost there.

The equivalent weight on the rail depends on that weights distance from the centre of buoyancy, which depends on hull shape and heel angle. Therefore that weight will be a function of heel angle, too.

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### jehardimanSenior Member

Pretty much.
A sailing vessel has two "stiffnesses" an apparent one and a real one. The apparent stiffness is a function of GM (which itself is a function of displacement, CG, and Iwp); this is how "tippy" a boat feels at any given moment due to weights moving on the decks. The real stiffness is a function of the BG couple (which itself is a function of displacement, CG, and hull shape); this is how much sail it can carry. For every angle of heel, the vessel has a GM and a BG couple that play off against the overturning moment of the sail. Sometimes GM dominates, sometimes BG dominates.
Really Motorbike, the question is a little to open. This is why you could have a 4t "plank on edge" and a 4t "skimming dish" both with maximum 3m beams and have very different boats. But, generally speaking....for similar hull forms (both above and below the waterline to give similar CB's and Iwp's at heel)...the boat with the lower CG (and yes, it is possible to have the CG too low)....will generally (again we will need specifics)....have more resistance to overturning.
NOTICE I DID NOT SAY "POWER" OR MENTION SAIL AREA! The ratios of driving power to overturning moment to sail area are a function of the rig and underbody shape, NOT the CG, BG, and GM.
Here is are some old posts I wrote about initial stability and final stability; please ignore the snark and latent anger.

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### AJBJunior Member

If we made a bunch of assumptions, say
1. Upwind heel 20 degrees
2. Crew weight contributes 20 of RM
3. Use VCG as minus 0.75...

A 50 mm change in VCG will change RM by around 1%

Which could speed you up by about 1/2 a tenth.....

Note that a 150 mm change in VCG is a very large amount.... usually not achievable without other performance or rating costs.

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### AJBJunior Member

Oops
Crew weight contributes 20% of RM

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### motorbikeSenior Member

Thanks you for your replies, I am not well versed in the right terminology- I should have said centre of buoyancy. However it may be better to describe the problem another way; Yacht A and B are class boat and identical, except one has 1.5tons of ballast keel and the other has 1.6 tons. The draft is 1.8m, beam is 3m, length is 10m, the measurer has no idea of the amount of ballast in yacht B's keel and can only determine it by an incline test. What is the best way to describe the difference/advantage in righting moment between the boats and if possible correcting it so performance is equalised

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### jehardimanSenior Member

There is no 'best' way as performance is never equal because each vessel sails in unique wind and water; and as I said before, righting moment has no close correlation to driving force. Choose between one and/or all of these methods:
1) Righting moment at some arbitrary angle of heel
2) Righting energy at some arbitrary angle of heel
3) Slope of the righting curve at some arbitrary angle of heel

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