Another boat question. Assume l have a boat that displaces x tons of water. Then lets assume l never load the boat with more than y tons of cargo. To keep this simple, cargo includes everything put in and on the hull. How much of the boat hull do l have to fill with foam to make this un-sinkable even if it is all full of water and has cracks and holes (absolute worst case scenario). Please no dumb comments. l can not find the formula for calculating this buoyancy problem.

If the total weight of the vessel and its cargo, including the foam, is less than the volume of water displaced by the hull, it will float. You will have to calculate how much volume the hull has, and the weight, and how much boat you want to have sticking out of the water after the boat is holed (obviously if the boat is awash it wo not be very useful).

It is not the foam displacement> you would have to fill the complete hull to keep it from sinking> and the extra weight of the foam it self is adding to the weight> If the cargo is heavy than the foam would do nothing>It would only sink slower>

Will it help to know that 1 cubic foot of urethane flotation foam, weighs 2 pounds, & that it can support about 64 pounds. however as you've probably figured out, the weight of the boat plus the cargo must equal the floatation. If filled with water, it displaces noooo!!!!! water!!!! it is sunk! So! If u fill the boat completely full of foam, there is no room for water, but there is also no room fo cargo. So, if the cubic feet of displacement is x+y +the weight of the foam it must be equal to the weight of the boat+cargo ?????? Less the weight the foam can support. l am not sue it is possible. l hope this helps.

near as l can figure, it would actually require simultaneous equations to account for the weight of the foam, but here is a simplified version:

f = s - {(hw - e) - cv } / w}

where
h = maximum cubic feet of wetted hull
w = weight per cubic foot of water
c = weight per cubic foot of cargo
v = cubic feet of cargo
e = vessel zero-cargo weight (w/fuel, crew & provisions)
s = cubic feet of empty space that could fill with water or be filled with foam or cargo
and f is the cubic feet of foam required.

note that there is no contingency buffer built in here.

In this case, u have to reduce ur cargo weight by the weight of the foam.

the weight of seawater at the surface varies by density & temperature, but u could probably use 64 lbs per cubic foot if u did not need too much precision.

note that hw - e is the weight of the water it would take to sink the boat with no cargo on board, & (hw-e) / w is the cubic feet of water that would sink the boat.

I've specified ''cubic feet but should have said ''volume'' -- it wo not matter if u use metric instead of feet. Except for the nine-millimeter bullet, the metric system never really caught on in the states.

You will have to figure out the simultaneous equation thing yourself, or maybe ask the question over in the math section.

This is the best answer u can get, since u can not know how much foam u need until u know the fully-loaded weight of the vessel with cargo.