# How does one calculate how much a given hull will settle from the addition of weight?

Discussion in 'Boat Design' started by bntii, Mar 7, 2008.

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### bntiiSenior Member

If you add say 500 pounds of gear to a small vessel how much will she settle on her lines.

I know that the answer is by the volume of water needed to equal this additional weight. But how does one get a handle of how the specific hull form settles to displace this volume?

Thanks all

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### ted655Senior Member

First know the displacement of the boat. 12,000# displacement let's say.
That means that it will take that much weight to submerge the boat. 6,000# will leave the boat 1/2 way out of the surface of the water.
One of the 3 dimensions that were used to calculate displacement was height. You see the relationship?
Say the height of the boat was 24". Divide 24 into 12,000. The answer is the weight "per inch". 500# per inch, 250# per 1/2 inch, 125# per 1/4 inch, etc.
Every 500# will affect the freeboard of the boat, add or remove weight & the boat rises or submerges by the extrapolation of the inch to weight ratio.

You will get brainier solutions, but this is the redneck way to figure how many girls & beer you can get on your boat.

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### bntiiSenior Member

ok- a little thought and I have a good understanding of my problem

I need to know the area at the water line in as shown in a plan view. Which mean I suppose that I need this drawing on paper and a planimeter or software to give me the area of this shape.
As I don't have the plan view, is there a formula which could get close by using max breadth of water line and length etc?
I do have a table of offsets and a lines drawing for the boat. The most direct means I have is to set up a dwg. in Autocad from the offsets table and have the software give me the area.

Thanks all

Ted- thanks for the above. The hull is too complex to get a feel for the volume displaced via a box submerged kind of approach.

thanks all

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### tom28571Senior Member

It's not rocket science. You are right that you need to know the area of the waterplane with normal load. You can estimate this by using the offsets and interpolating the beam at several points along the waterline. Then use simple geometry, Simpson's rule or just averaging arithmetic to estimate the area.

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### PericlesSenior Member

The old PPI at DWL eh?

And then there is the Ex Perry Mental method. Weigh all your friends and have them climb aboard until boat sinks. This redefines the empirical method as "Use one less than that".

Good night, it's past midnight here.

Perry

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### alan whiteSenior Member

Calculating how much deeper is complicated. If you have a set of prints, it's not so difficult, but the crudest method, i.e., adding the weight, trimming the boat, and then measuring the difference in depth is by far the most accurate and foolproof. If your boat isn't any longer on paper but is in the water already, forget paper calculations and sink the boat with measured weights.

Alan

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### PericlesSenior Member

From Alan.

"If your boat isn't any longer on paper but is in the water already, forget paper calculations and sink the boat with measured weights."

Walk on, walk off. Don't go humping weights, in any sense of the word.

Pericles

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### alan whiteSenior Member

I meant body weight, or any weight. Walking IS humping weight. Step on a scale, it's a measured weight.

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### PericlesSenior Member

Alan,

Is it up late or up early to catch the tide?

Regards,

Perry

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### alan whiteSenior Member

Yes. Sometimes.

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### ted655Senior Member

.
Sorry, I should have seen your designation. I thought you were a newbe.

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### timothy22Junior Member

The old-ish way as mentioned in Skene's "Elements"

and others.
WPA in sq. feet = w/l length * w/l beam * 0.76

LBS/IN = WPA * 64 / 12

The factor of 0.76 was for "normal" (Cruising Club of America rule) hulls. The factor would be a little higher for a modern performance cruiser/racer a la Bruce Farr.

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### bntiiSenior Member

Ok- I should really look to my bookshelf first

I have Skene's............

Thanks- this is in fact a CCA type and looks like I have ~1000lbs/in