Bouyancy formula

Discussion in 'Stability' started by Bruce1947, Aug 21, 2009.

1. apex1Guest

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FanieFanie

1.025 x 500

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conceptiaNaval Architect

the density of fresh water is 62.3 lb/cubic feet, while that of sea water is 64 lb/ cubic feet. so 500 X (64/62.3) will give you the weight capacity in sea water.

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conceptiaNaval Architect

Fanie, dont get confused over with the units. 1.025 is in SI and 500 is in US units.

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IkeSenior Member

not much more 500/60.4 = 8.278
8.278 X 62.4 = 516.556 (517 lb)

However small boats are generally rated for worst case which would be fresh water and that gives the lower figure.

Most of the answers above have gotten way to complex.

Scott and pamarine came closest.

For boats under 20 feet (6.96 meters) finding bouyancy and weight capacity is pretty simple. Actually the easiest way is if you have a lot of old weights laying around. For 500 lb (227 KG) you'll need 2500 to 300 pounds. Put them in the boat distributed evenly until water starts to come in at the lowest point. This takes a little juggling of weights.

The total weight is the displacement weight. You can compute the volume by dividing the weight by the wieght of a cubic foot of water or a cubic meter if that's your flavor.

But the maximum weight capacity is 1/5 of that for outboards and 1/7 of that for inboards.

displacement weight/5 or displacement weight/7

You can do this in reverse by calculating the volume of the boat below what is called the static float plane, that is, the waterline where water would start coming in. As was said if you have the boats lines in a computer program, many of them will do this for you, you just specify the waterline and it tells you the volume.

Maximum weight capacity minus engine weight gives person weight (for outboards)

For Inboards the engine weight is part of the boat weight so Maximum weight capacity can equal persons capacity but most builders use a smaller persons weight to leave a safety margin.

Persons weight + 32 /141 = persons. (yeah that's a screwy formula but it is the one that the USCG, ABYC, Canada and ISO all use to compute number of persons)

Last edited: May 12, 2010
6. Paul KotzebuePrevious Member

That is not correct. 1.025 is the specific gravity of sea water and has no units. It is applicable to both SI and US systems, so Fanie is right.

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conceptiaNaval Architect

do we get weight of an vessel by multiplying its volume with specific gravity?

8. Paul KotzebuePrevious Member

No. Weight = volume * density. And density of sea water = specific gravity of sea water * density of fresh water, regardless of units. So, weight of sea water displaced = specifc gravity of sea water * weight of fresh water displaced, regardless of units.

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Wynand NRetired Steelboatbuilder

Very elementary issue as Paul explained and one of the very basics one learns in boat design studies regardless of institution as a boat designer or by going a bit better, Naval Architect.

However, what mystifies my mind; you clearly state at your name "Naval Architect"...

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hoytedowCarbon Based Life Form

The weight of sea water displaced would equal the weight of fresh water displaced. the VOLUME of the sea water displaced would be less, so the boat would float higher.

11. Paul KotzebuePrevious Member

True for constant weight.

I stand by my statement, which is correct for constant volume. I was considering the question asked in post #15:

"So if I have a boat with a 500lb weight capacity in fresh water what is the weight capacity in salt?"

And the response in post #17:

"1.025 x 500"

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hoytedowCarbon Based Life Form

Fanie is a reliable source of information.

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PhotoBoatGuyRhino fanatic

Trevormlb:

If your boat will hold 500 lb in fresh water and you want to maintain the same waterline in saltwater, it will hold 500 * 1.028 = 514 lb.

Cliff W Estes
BaseLine Technology

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