# 2-pontoon party boat

Discussion in 'Boat Design' started by Big Aussie, Mar 25, 2019.

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### Big AussieJunior Member

Hi everyone,. I am a new member and this is my first post.
I am building a 2-pontoon party boat. The sealed pontoons will have a flat bottom (jon boat system). The boat will be trailerable, so, 8 feet wide by 19 feet long. It will have walls and a roof 7 feet high. I propose to have a 4 foot drawbridge type deck/swimmout on each side, suspended by chain.
I have done some calculations re buoyancy but I am not sure exactly how many pounds it will take to completely submerge one pontoon as per "Figure 2"

Each pontoon will be 32 inches wide by 20 inches high by 19 feet long.

I would like to design for the worst possible scenario where all 10 big party goers (weighing 2500 pounds) suddenly decide to move to the edge of the deck at the same time! If required I can increase the width of the pontoons by a few more inches , or reduce the width of the decks by a few inches, or do both but I cannot increase the height of the pontoons.

I am enclosing 2 diagrams and would appreciate any comments/calculations that you may offer.
Thank you.

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• ###### jon boat pontoons.jpeg
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### DejaySenior Newbie

Hi Big Aussie, sounds like a cool project! Just out of curiosity, in what waters are you planning to party on the barge? And, more importantly, are we invited?

PolyCAD is free software that can perform hydrostatic analysis if you model the hull and calculate and input the center of gravity. Tutorial video. Of course you should know about metacentric height and all that jazz.

I'm a newbie so I can't really give you any advice other than that. But if you'd add flotation / buoyancy to the fold outs like with compartments or rigid foam, and they fold upwards when they hit the water, the lever would change in a way that isn't as easily modeled.

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### Mr EfficiencySenior Member

Is that individually, or total ? OK, jokes aside, those fold-out side decks look like an invitation to disaster. It is likely to tip over with virtually no warning, if it does go. One possibility might be to have some real depth to the side platforms so they offer buoyancy if the worst happens, but if they are non-buoyant it is a real risk of a sudden end to the party.

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### HeimfriedSenior Member

The fresh water density is 1 kg/dm³, so the bouyancy of one fully submerged pontoon is 8 dm * 5 dm * 58 dm * 1 kg/dm³= 2320 kg . Your first sketch shows pontoons immersed half of their depth, which means (zero trim) displacement is 2300 kg . Considering your party goers (1140 kg) are one half of this displacement: are you sure to achieve a build of a boat with 1150 kg including gear, load, crew? Half immersed pontoons are dangerous in operation.

The second sketch shows besides the fully submerged pontoon the other one just tilted, but still immersed with half of its volume. Is that meant the party goers bring the additional weight (mass) of 1140 kg? To calculate the force (weight) which would be allowed to cause the heeling angle of 10 deg. the displacement must be given first.

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### Big AussieJunior Member

Dejay. Inland waterways only.
Not sending out invites yet!

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### Big AussieJunior Member

Sorry, neither of the drawings were meant to reflect the exact level of immersion of the pontoons but were merely an illustration. The first drawing shows the weight of the various structures with furniture but no people. When calculations are made, the draft in fact would be much less than shown in the first drawing.
The second drawing makes it look as if it is going to tip over but will it? My calculations are as follows: The fully immersed pontoon is 32" X 20" X 228". That gives us a volume of just over 84 cubic feet. (The actual weight of the pontoon is 120 lbs but I have not factored that in). At 62 lbs per cubic foot, it would take 5208 lbs to submerge the pontoon. At 2500lbs, I am concluding that my 10 big party goers will stay high and dry. (and that's not considering the counter-balancing effect of the rest of the structure). Am I correct in my assumption?

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### Mr EfficiencySenior Member

Your boat can tip over in a mill-pond, if you put enough weight out on the side platform, and with a catamaran, there will be little warning before it goes. With the effects of wind and waves, it becomes more problematical. The ability to have substantial weights shifting to well outside the waterplane, is fraught, you only need to look at an 18-footer sail racer, if the crew stays out there when pressure on the sail reduces, they easily capsize..

Last edited: Mar 25, 2019
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### DejaySenior Newbie

I think another way to look at this is that the left pontoon will act as a fulcrum for all 2500 pounds of party goers stand on the left, and the rest of the weight of the boat will have it's center of gravity almost equally spaced (a bit further in). So your boat would have to weight 2500 pounds also to not tip over. Something like that?

Besides making the swimouts buoyant, another way might be to have the pontoons below the swimouts and fold them out from below the main platform. Just makes deploying it harder. Or fold them out from above and secure them with a beam.

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### BlueBell. . . . .

Cool project. What fun!

So your pontoons weigh 240 pounds, your friends 2500 pounds.
What does your pontoon attachment system weigh?
How much stuff will your friends bring on board, each?
What does the rest of the structure weigh?
Anchor weight? Chain? Docklines? Food and water?
What is the average max snowfall? Winds?
And how long might the hull sit to gather bottom growth and what will it weigh?
Fuel weight? Engine weight? Oil weight? Steerage weight?
And on, and on, and on.
Do you see what I'm getting at?
You're going to need a total weight and it's distribution.
You have much research to do before investing lots of money, time, effort, relationships, blood, etc.

What is the budget?

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### Big AussieJunior Member

Hi BlueBell, the example is hypothetical only. The final figures would be quite different. What I would appreciate help with is the scientific principles and calculations that will apply , based on the sizes and weights I quoted earlier and in a perfect scenario with no wind, no waves, no snow. Once I have grasped the scientific principles I will then make proper calculations, adjustments etc. looking at Figure 2, I would like to know how much weight on the deck will be required to submerge the pontoon closer to the deck, based only on the weights and sizes quoted. Thanks .

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### IkeSenior Member

I suggest you look at the standards for capacity for pontoon boats. I don't know if Australia publishes them on line but Canada does, and they are taken direct from American Boat and Yacht Council standards, and the same as ISO standards. The big plus is they are on line and free. Construction Standards for Small Vessels (2010) - TP 1332 E - Transport Canada https://www.tc.gc.ca/eng/marinesafety/tp-tp1332-menu-521.htm#wb85 If the link doesn't take you direct to the pontoon page it is section 4.5 Recommended Maximum Safety Limits For Pontoon Vessels. Of TP 1332 at Construction Standards for Small Vessels (2010) - TP 1332 E - Transport Canada https://www.tc.gc.ca/eng/marinesafety/tp-tp1332-menu-521.htm The calculations are pretty basic and you don't need a computer to do them.

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### Big AussieJunior Member

Thank you Ike

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### DejaySenior Newbie

The scientific principle is the same as in wind trying to push over a sailboat. Instead of the wind acting on the lever of the mast, the party goers act on the lever of the swimout. Your boat has a maximum righting moment and once that is exceeded by the weight of the party goers it's catapult time! (Somehow this plays in my head as a dance music video with the lyrics lamenting the loss of their favourite grill to the sea in a tragic boating accident)

With rectangular pontoons this should be relatively easy to compute. The wikipedia article has a link to the "second moment of area" and that has a link to typical area examples. Now you can calculate the metacentric height and the righting arm GZ and then the righting moment.

The torque the party goers create can be calculated by weight * g * distance to fulcrum (center of pontoon). Assuming one pontoon has enough buoyancy to carry the whole boat and passengers.

But since the pontoons are rectangular this could also be simplified using the scientific principles of the seesaw and the teeter-totter.

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