# Bouyancy

Discussion in 'Boat Design' started by Ted1785, Jul 30, 2002.

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### Ted1785Junior Member

I need to find out how bouyant my boat is. I'm building a boat. Dont know the volume of air in the pontoons yet. I need someone to tell me the equation or someway to find out how much bouyancy I have for whatever volume of air.

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### WillallisonSenior Member

Ted, perhaps you could give us a little more info to work on. From what you say, I gather the boat is supported by a couple of pontoons - are they cylinders?

Basically any boat - regardless of shape, building material etc will displace the same amount of water that is equivalent to its mass. So a boat weighing 1000 lbs will displace 1000lbs of water. To work out the volume, you need the density of water. Salt water weighs 64lbs per cubic foot, fresh water weighs 62.4.

You need to work out how heavy your boat is (including fuel, provisions, crew etc) and from that you can work out how much bouyancy is required to float it.

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### Ted1785Junior Member

The boats going to weigh about 200 tops with about 250 lbs between me and my gear onboard. And, I'll be just using it in freshwater lakes. And, the pontoons are going to be make out of just something that can hold air. Havent decided what to use or the size yet because I dont know how much bouyancy I need.

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### duluthboatsSenior Dreamer

Ted,
Lets round your displacement to 500lbs. You might catch a big fish or take some extra beer with you. 500lbs of fresh water will fill:
15ft of a 10inch tube,
12ft of a 12 inch tube,
4.5ft of a 18 inch tube.
These are rough numbers but should be near enough for your needs. If you double the length the water line will be half way up your tube. If I have this wrong, I’m sure it will be corrected shortly, many who lurk here are better than I at numbers.
Gary

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### james_rJunior Member

I get 10.2' for the 12" tube. The others, rounded off, seem correct.

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### duluthboatsSenior Dreamer

Thank you Checked myself, 10.19'
Gary

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### Ted1785Junior Member

is that 10.2 or 10.19' on each pontoon or on one for 12". And, I'm 17 yrs. old, Theres no way I'm going to be drinking on a lake.
Ted

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### duluthboatsSenior Dreamer

LOL!!! Your right of course, drinking and boating don’t mix. If you had 2, 12 inch diameter pontoons, 10.2 feet long, each pontoon would be half way under water. If the boat, gear, and captain added up to 500 pounds.
Gary

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### Ted1785Junior Member

thanks Gary. How bad would I be off if I have 2(8'x12") pontoons? Thats what I want to make them. That way it fits into my truck. And how do you know how to figure all this. I would just like to know so if I change what I'm going to use for pontoons that I can figure it myself.

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### james_rJunior Member

Time for a little math!

Using the original all-up weight of 500lbs in fresh water (62.4lbs/cu.ft.) your boat would displace 500/62.4=8.01cu.ft.

To calculate the volume of a cylinder, first you have to calculate the area of a circle of the same diameter (Pi*r^2) and then multiply that by the cylinder's length. In the case of a 12" cylinder first you divide by 2 to get the radius and then by 12 to convert to feet, which gives you a radius of 0.5'. The area in square feet is 3.1416*0.5^2=0.7854.

To get the length, divide 8.01 by 0.7854 and you get 10.2'. In other words, a 12" by 10.2' cylinder would float just below the surface if it weighed 500lbs. Two 10.2' cylinders would float halfway up. If you were to use two 8' long cylinders they would float deeper than halfway.

By increasing the diameter by just 2" to 14" your two pontoons would only need to be 7.5' in length each to float halfway. If you were to make them 8' long you'd be ahead of the game.

Use the following equation in Excel to make the calculations yourself: (B1/B2)/(PI()*(B3/24)^2) where B1 is the weight of the boat in lbs, B2 is 62.4 for fresh water or 64 for salt water and B3 is the outside diameter of the cylinder in inches.

You can also calculate the exact draft for any given diameter/length/ displacement combination but we'll have to get into a bit of trigonometry.

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### duluthboatsSenior Dreamer

Yah, what James said. Thanks James, if I had to explain it, Ted would have never figured it out. I’m not sure why it hadn’t dawned on me to use Excel, good tip.

Ted, I would also agree with James that 2, 12”X 8’ tubes would be marginal. I would use 3, or 2, 14” tubes.
Gary

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### Ted1785Junior Member

thanks for the help Gary and James, but I dont have excel....

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### lockhughesElectricGuy

wait a sec guys...
you're figuring on JUST ENOUGH bouyancy? D'you mean the craft will float, but be awash? Don't you think that'd make the newbie boaters anxious? How `bout a little freeboard eh?
Lock

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### james_rJunior Member

Ted:

Not a problem, Excel is overkill for this situation, a \$5 calculator can do it. If your calculator doesn't have Pi use 3.14159. If it doesn't square numbers, simply multiply the radius by itself. I only mentioned Excel because most computers will have a spreadsheet program of some kind which would allow you to quickly do a bunch of what-if scenarios. Or you can use pencil and paper, just like us BC (before calculator) guys used to do.

Lock:

Well, I thought Gary and I had been pretty clear about it. We're talking about using two tubes, not one, of the calculated length. As an example, one 14" by 7.49' cylinder weighing 500 lbs will be awash. Two of them, lashed together, will have a draft of 7".

Ted estimates that him and his gear will weigh 250 lbs and that the craft will weigh another 250 lbs. Personally I think that a couple of PVC pipes or aluminum tubes (or whatever other suitable material) of the required diameter lashed together with a few 2 by 4s will weigh less than that.

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