Re: Pontoons

[protek@shreve.net]

Greg,
    Assume the pontoons are 24 inches in diameter, there is a hole in the
bottom and they are half way submerged (1 foot deep in water).  Also assume
that the water in the pontoon is 4 inches from the bottom, with none coming
in or going out.  In other words everything is in equilibrium.  The amount
of air pressure inside, in units of "Inches of Water" must be equal to the
difference between the level of water outside and the level of water
inside.  In this case it would be 12 inches - 4 inches or 8 inches of water.
    If the pontoon (ballast tank) were to be submerged totally, it would
take 24 inches of differential water pressure to blow the ballast or 2 feet
of water, or about .88 PSI.  Of course it is done with more pressure than
this but this would be in theory the lower limit.  If you were at a depth
where the outside pressure was 50 PSI then this .88 PSI would equate to a
requirement of 50.88 PSI.
    There are many other things to look at here.  It is possible to
calculate the rate that water leaves or enters an opening simply based on
differential pressure from the inside to the outside of the opening, and
the area of the opening.  This is an easier calculation for constant
differential pressure (blowing the tank) than for flooding the ballast tank.  
    Assume the pontoon scenerio above but this time we have a hole of known
area at the bottom.  We open the top to let air out with a valve of some
sort.  Initially the differential pressure of the air inside is 2 feet of
water or the .88 PSI.  But, when the water begins to fill the tank from the
bottom the pressure differential goes down, and thus the rate of inflow
goes down.  As the water level nearly reaches the top the differential
pressure is very small and the rate goes toward zero.  This kind of problem
is more complex to solve but gives some insight as to how large to make the
opening in the bottom of soft tanks.

Gary Boucher