# Buoyancy

### Mike Jensen

Posted Feb 28, 2011

A key part of many engineering projects is figuring out how nature works, and then using that knowledge to design useful things. I suppose that’s why physics is a common and required course of study in most university level engineering programs: physics helps us quantify how nature works, then we apply our specific engineering skills to complement or compensate for natural laws.  So with a new personal watercraft project floating around in my brain, I dusted off my physics memory engrams, grabbed one of my project notebooks, pulled my HP 32S calculator out of my desk drawer, and started calculating.

Elementary physics says that to keep myself from falling through a surface, that same surface must push up on me with at least the same force that I push down on it. So if F1 represents the surface’s force pushing up, and F2 is my force pushing down, I won’t fall through whatever I’m standing on if F1 is equal to or greater than F2. Each of these forces is represented by the well-known general equation:

F = m * a

Where “F” is the force, “m” is the object’s mass, and “a” is acceleration.

In my watercraft calculations I labeled the forces F(me) and F(water). Based on my reasoning above, minimum buoyancy requires the following relationship:

F(water) = F(me)

Substituting “F = m*a” into both sides of the equation:

m(water) * a = m(me) * a

Since “m * a” for an object is just its weight when “a” represents gravitational acceleration, this formula tells me I will float when the weight of water I displace equals my body weight. So whatever shape my watercraft takes, it must displace a volume of water whose weight is equal to or greater than my own body weight plus the weight of the craft. Easy enough.

Using English units, the weight of a gallon of water is approximately 8.33 pounds, and I tip the scales near the 300 pound mark (yeah…I’m a big guy – tall and stout). So dividing 300 pounds by 8.33 pounds per gallon tells me that my watercraft’s hull must displace at least 36.01 gallons of water in order for me to float. But I want to carry more than myself when water cruising, and I need to compensate for the weight of the craft itself, so I’m doubling the weight carrying requirements (me + boat weight + cargo) to 600 pounds. My hull design must therefore displace 72.02 gallons of water. So far, so good. Now it’s time for a few hull design calculations.

I’m a big fan of building projects, whenever possible, using readily available materials — things I can source locally. My watercraft project is no different. Since I’m building for casual cruising (speed is down the project specification ladder a wrung or two), hull hydrodynamics are a secondary consideration; I just need to float. So my choice of hull material is 12-inch PVC pipe — certainly not exotic, but durable, watertight, and available from my local building supply store. And equally important, the volume calculations are straightforward. With approximately 0.14 cubic feet in a gallon, displacing 72.02 gallons of water is equivalent to displacing roughly 10.08 cubic feet of water (72.02 * 0.14 = 10.08). So I need to choose a length of PVC pipe long enough to displace this volume of water. The volume of a cylinder is calculated as follows:

volume = pi * radius^2 * height

Since I want to know how long my hull needs to be, I solve this equation for the height and get:

height = volume / (pi * radius^2)

Substituting numbers for variables:

height = 10.08 / (3.14 * 0.5^2)

Crunching the numbers tells me if my 12-inch PVC pipe is at least 12.84 feet long, or if I use two pieces of pipe 6.42 feet long (which is the hull configuration I intend to use), my watercraft and its cargo will float.

So a little physics and a  bit of algebra help me understand the basics of buoyancy; many project calculations are just that simple. Stay tuned…as summer approaches I plan to post occasional watercraft project updates.