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Actually the idea is quite simple really, it consists of a floor above the hull with open drain plugs in the rear of the boat and above the water line. By this simple method as water comes in to the boat it will drain right out of the back. If too much weight is in the boat then water will simply come in through the drain holes. This is no worry to me because I can always just put some plugs in the drain holes when carrying a heavy load. I made the decision to build the aluminum skiff 17ft long and strong enough to hold an outboard motor with a mass of 141kg (approx. mass of 90Hp Mercury).
My Boat Dimensions
I made the decision to build the aluminum skiff 17ft long and strong enough to hold an outboard motor with a mass of 141kg (approx. mass of 90Hp Mercury). Knowing the amount of material I would need; I then went on to estimate the mass of the boat when finished to be 239kg. I also made the decision to build a fuel tank in the front of the boat holding 141kg of gasoline (≈ 38gal). The design of the boat permits the boat bottom to have an estimated area of submersion of 5.55m2 (L= 3.7m W=1.5m). With these measurements in hand I knew I needed a way to determine the height from the bottom of the boat at which I should build my second deck and drain plugs. I got a value from a veteran boat builder but surely there was an equation to help me out.
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.
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"Physics of Self Bailing Boats and How I Built One." 123HelpMe.com. 29 Feb 2020
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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:
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.
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.