1. 2009
Nov
28

## Buoyancy, part 2

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 …