How much weight can my pontoons hold at half emersion?

hey ddrdan, netjaws was talking about the cylindrical pontoons. at the half immersed the pontoon will have the max waterplane area. Even a small additional weight may reduce the waterplane area, thus compromising on the stability of the vessel.

In your case the pontoon is square so the waterplane area wont change. But the overall reduction in the metacentric height will accompany the increase in the draft. i.e. stability reduces.

 

Thank you.

I'm just beginning to research metacentric height and center of buoyancy.

Found a good step by step calc assist site, and, a good layman's explanation site if anyone else is looking.

Engineering Stability and Metacentric Height

Metacentric Height in semi layman terms

I'd value an opinion if these sites aren't a good learning tool for my research.

Thanks again!

 

blah blah--ive built many a houseboat on pontoons- the two first equations posted got it right..its LxWXH X 2 pontoons (assuming 2 square ones) x cu ft weight of water....to bring them to waterlines at half depth of hulls divide by 2

use freshwater 62.2 lbs or salt water 64 lbs per cu ft...

if round pontoons -cmon folks remember preshool physics!
its dia x Pi x length x 2(pontoons) x waters weight per cu ft, divide again by two

...or 3/4 or 1/3 fro your required or preference for depth and safety, depending on how much you want to sink the pontoons....take a little off for the irregularites for the bow sections.

btw big diff between practical and theoretical bouyancy! --it is usually more bouyant..never go JUST by math figures.

 

ddran, the first one is good link to start with.. go on...

 

Hey! I think it's you that has forgotten your "preschool physics". The volume is Pi*dia*dia/4*L.
I thought this was the reason why your practical and theoretical bouyancy was different, but in my calcs it only becomes more bouyant when the diameter is over 4 feet.

 

Tugboat: Not to be a nit picker BUT....displacement, or enclosed volume, is calculated by finding the area of the section and multiplying by the length. To find the area of the round tube you can use one of two methods, both of which are mathematically equivalent. Area of a circle is found by multiplying the radius times the radius times pi. Another method is to multiply diameter times diameter times pi all divided by four. In mathese that is r^2 x pi = area and D^2 x pi/4 = Area.... In words...pi times the radius squared equals area.... or pi divided by four multiplied by diameter squared equals area. As I read your post, your area calculation is not quite correct.

There are folks here who will argue persuasively, even vehemently, about whether math is reliable. It is reliable in the absolute, if you are careful and accurate with the numbers and equations.

 

Good pickups there.

Can I also add that if the pontoon material is fairly uniform in thickness (say steel sheet) , its handy to also calcuate the surface area at the same time, multiply it by the weight of the hull material, and subtract it from the original bouyancy result.

This represents an even closer bouyancy estimate before doing the complete weight table.

Please check this methodology guys, its been a while
eg
Cylinder Surface (CS) = 2 x pi x radius x height
say 2 x 3.14 x .5 metre x 8 metres
= 25.12 square metres

Hull Weight (HW)= 1 kg per square metre
HW = 25.12 kilos

Original Displacement - Hull Weight = Closer Bouyancy Estimate.

 

Looks good to me RW.

The only thing I'd add is the submerged steel is only 70% as heavy due to buoyancy.

That would be nit-picking and to negate this effect would only add a margin of safety, however slight.

-Tom