The Joy of Archimedes Principle in Boat Building

Archimedes’s Principle for Floating Objects:
So hears the theory:

Archimedes principle says that the magnitude of the buoyant force always equals the weight of the fluid displaced by the object.  This buoyant force always acts upward through the point that was the center of gravity of the displaced fluid.  In the case of floating objects the buoyant force is equal to the force of gravity on the object.  Knowing that the change in pressure is equal to the Buoyant force per unit area (ΔP = B/A) we see that B = (ΔP)A and ΔP = ρgH where ρ is the density of the fluid g is the acceleration due to gravity and H is the height of the fluid displaced.  Therefore B= ρgH(A) = ρgV where V is the volume of the fluid displaced by the object.  If we look at the sum of forces acting in the vertical direction for a floating object we get the equation

                                                ΣF = B – Mg = 0                                                 
                                                       
where M is the mass of the object displacing the fluid.  By rewriting the equation to include what we know about the buoyant force and then solving for the height H we have what where looking for.

B = ρVg = Mg
ρgH(A) = Mg
                                                         

The equation gives us the height H of the fluid that is displaced by the object which is equivalent to the height from the bottom of our floating object to the water level on the object.  

    Because my boat is primarily used in saltwater I run the numbers (for dimensions click here) using the density of saltwater ρ = 1.03x103kg/m3.  I plug all my values in to the equation and I’ve got it.

H = Mg/ρgA = M/ρA = (141 + 141 + 239)kg / (1.03x103kg/m3 x 5.55m2)  = 9.11cm = 3.6 inches
 

By building the second deck greater than a height of 3.6 inches above the bottom of the boat I could ensure even with a full tank of gas the saltwater would not run back in to the boat.  The height I got from the local boat builder was a safe 5.75 inches.  So approximately how much extra mass can my boat hold before the plugs start letting in water?  By plugging the value of H into the equation we get the following:

5.75in=.0226m
ρgH(A) = Mg
M = ρHA = 1.03x103 x 0.226m x 5.55m2 = 129.19kg ≈ 283 lbs

 
My boat can hold approximately 283lbs of extra weight (in its center) before I need to put the plugs in the drain holes.  

Approximations:
The largest approximation I use in this is in assuming that the sides of my boat are actually straight up and down as well as the transom.  Because my sides are at 26o angles with the bottom of the boat the approximation will give a slightly larger area displaced and hence a lower height.