Buoyancy, part 2

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Following up on my calculation of the lifting power of helium balloons, it’s time to see how the same argument applies to ping-pong balls being used to raise a sunken ship.

Raising a ship with ping-pong balls is, in fact, nearly the same situation as raising a child with helium balloons. All you have to do is replace the air with water, the helium with air, the rubber balloons with plastic balls, and the child and harness with a boat (though preferably not in that order). The physical principle at work (Archimedes’ Principle) is exactly the same, and so the same equation I used last time is equally applicable here: the buoyant force on an object (ping-pong ball) immersed in a fluid (water) is equal to the weight of the water displaced by the fluid,

$$F = \rho g V$$

Let’s see what this says about how many ping-pong balls it would take to raise the Mythtanic II, which weighs about \(\unit{3500}{\pound}\) according to the show. We can start by figuring out how much mass it takes to balance out the buoyant force on a single ping-pong ball, using \(-m_\text{load} - m_\text{ball} + \rho V = 0\) as we did last time:

  • Water has a density of about \(\unit{1000}{\frac{\kilo\gram}{\meter^3}}\). What about the volume of a ping-pong ball? These balls come in two standard diameters, \(\unit{38}{\milli\meter}\) and \(\unit{40}{\milli\meter}\). The ones that Adam and Jamie used look like \(\unit{40}{\milli\meter}\) low-quality practice balls I’ve seen in stores, so let’s use \(\unit{40}{\milli\meter}\) as the diameter, giving a volume of \(\frac{4}{3}\pi r^3 = \snunit{3.35}{-5}{\meter^3}\). Multiplying that by the density of water gives \(\rho V = \unit{33.5}{\gram}\).
  • From that quantity, we need to subtract the mass of the ping-pong ball and the air inside it. Standard ping-pong balls weigh \(\unit{2.7}{\gram}\), easy enough. As for the air, I’m just going to assume that it’s at normal atmospheric pressure, so we can multiply the normal density of air \(\unit{1.2}{\frac{\kilo\gram}{\meter^3}}\) by the volume of the ball to get \(\unit{40.2}{\milli\gram}\). That’s a tiny amount of mass, about 2% of the mass of the ball itself, and our numbers just aren’t precise enough to bother caring about it. So the net lifting power of a single ping-pong ball in water is \(\unit{33.5}{\gram} - \unit{2.7}{\gram}\), or \(\unit{30.8}{\gram}\).

Given that figure, we can easily calculate how many balls it should take to lift \(\unit{3500}{\pound}\):

$$\unit{3500}{\pound}\times\frac{\unit{454}{\gram}}{\unit{1}{\pound}}\times\frac{\unit{1}{\text{ball}}}{\unit{30.8}{\gram}} = \unit{52000}{\text{balls}}$$

That’s a lot of balls.

But wait! We’re forgetting about the same thing that Adam and Jamie forgot about in their calculations — the buoyant force on the boat itself. It’s an easy mistake to make, because we’re used to dealing with objects in air, where there’s so little buoyant force that we usually just ignore it (except for things that blatantly “advertise” their buoyancy by floating away). Not so in the water, though; remember that buoyant force is proportional to the density of the fluid, and since water is so much denser than air, the buoyant forces are correspondingly larger.

Now, without knowing exactly what the boat is made of, there’s no way to tell just how much that force is, but let’s make a guess. Pretend the boat is made entirely of fiberglass. Now, the density of fiberglass) is about 1.5 to 2 times that of water, so if the boat is made of \(\unit{3500}{\pound}\) of the stuff, its volume would be

$$\frac{\unit{3500}{\pound}\times\frac{\unit{1}{\kilo\gram}}{\unit{2.2}{\pound}}}{\unit{2000}{\frac{\kilo\gram}{\meter^3}}} = \unit{0.79}{\meter^3}$$

(In case you’re wondering, yes a cubic meter is actually pretty sizable. Think about it.) With the Mythtanic underwater, all this volume of fiberglass is displacing water and contributing to the buoyant force along with the ping-pong balls. But of course, we know how to calculate how much mass this buoyant force can support: \(-m + \rho V = 0\), or in this case,

$$(\unit{1000}{\frac{\kilo\gram}{\meter^3}})(\unit{0.79}{\meter^3}) = \unit{790}{\kilo\gram}$$

and we also know how to calculate how many ping-pong balls it would take to lift that amount of mass in water:

$$\unit{790}{\kilo\gram}\times\frac{\unit{1}{\text{ball}}}{\unit{30.8}{\gram}} = \unit{26000}{\text{balls}}$$

That means that the boat itself provides the lifting power of about 26000 ping-pong balls, which in turn means that we only need an additional 26000 real ping-pong balls to achieve the target of 52000, which should bring the Mythtanic II back to the surface. And guess what? If you were watching the show, you would notice that it took Adam and Jamie just about 27000 balls to bring their boat up — practically right on target. Whaddya know, physics works!