Ellipsix Informatics: Blog

2009

Nov

Buoyancy, part 2

Posted by David on November 28, 2009 — Comments

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!

2009

Nov

Buoyancy, part 1

Posted by David on November 25, 2009 — Comments

Finally, time to get back to covering some old (by now) Mythbusters episodes. I'll start with the bonus episode aired a few weeks ago, "Ping-Pong Rescue" — an oldie but a goodie in which the Mythbusters try to raise a boat with ping-pong balls and lift a child off the ground with balloons.

This episode was all about buoyancy, the physical description of how stuff floats. Buoyancy goes all the way back to one of the scientific world's earliest experts, Archimedes. According to legend, he had been tasked with figuring out whether a crown, given as a gift to the king of Athens, was composed of pure gold or of other, less valuable materials with merely a gold coating. The straightforward way would have been to melt the crown down in order to make an accurate measurement of its volume and thus determine its density, but the king, for some reason, didn't want his crown damaged and so melting it was out of the question.

The bright idea that Archimedes eventually came up with was — we think — based on a principle that now bears his name (Archimedes' Principle): that the buoyant force on an object immersed in water is equal to the weight of the water displaced by the object. Back in Archimedes' time, this was a groundbreaking observation, but now we can consider it a consequence of the minimization of potential energy, and here's how: if some object of mass $m$ and volume $V$ is immersed in a fluid (like water) which has density $\rho$ , the total potential energy is equal to the sum of the potential energy of the fluid and the potential energy of the object,

$U = U_\text{fluid} + U_\text{object} = \iiint_\text{fluid} \rho g h\,\udc^3 V + m g h$

Well, great, but this isn't very practical to calculate — the integral is over a volume with an object-shaped hole in it. However, we can use the little trick of adding and subtracting a term that corresponds to that object-shaped hole,

$U = U_\text{fluid} + U_\text{flobj} - U_\text{flobj} + U_\text{object}$

If the space occupied by the object were filled with fluid instead, $U_\text{flobj}$ would be the potential energy of that amount of fluid. If that's not clear, look at the substitution:

$U = \iiint_\text{fluid} \rho g h\,\udc^3 V + \iiint_\text{object} \rho g h\,\udc^3 V - \iiint_\text{object} \rho g h\,\udc^3 V + m g h = \iiint_\text{all} \rho g h\,\udc^3 V + g h (m - \rho V)$

Now we've separated out the potential energy a different way: the first term represents the potential energy of the amount of fluid it would take to fill the entire volume, fluid and object, and the second term represents the potential energy of the mass defect, the difference in mass between the object and the amount of fluid that would be filling that space. (Yes I know mass defect has a technical meaning in nuclear physics) Basically it's how much extra potential energy the system has due to the object having replaced the equivalent volume of fluid. Anyway, now we can just calculate the net force on the object,

$F = -\pd{U}{h} = -m g + \rho g V$

and you can see both the normal gravitational force, $m g$ downwards, and the buoyant force, $\rho g V$ acting upwards.

Returning to the Mythbusters analysis, let's take a look at the idea of floating a child with helium balloons. It makes sense to proceed through the same reasoning that Scottie, Kari, and Tory did: first, how much weight does a single balloon lift? The critical point at which a balloon is able to lift a load occurs when the buoyant force on the balloon (and on the load, but I'm sure the buoyant force on paper clips is negligible) balances out the combined weight of the load and balloon, $-m_\text{load} - m_\text{balloon} + \rho V = 0$ . Time to estimate some numbers!

First, the density of air is about $\unit{1.2}{\frac{\kilo\gram}{\meter^3}}$ , and a normal helium balloon is approximately a sphere with a $\unit{15}{\centi\meter}$ radius, for a volume of $\unit{0.014}{\meter^3}$ ; multiplying these gives $\unit{17}{\gram}$ .
The balloon itself has a mass of about $\unit{2}{\gram}$ , and the helium inside it contributes another $\unit{2}{\gram}$ or so (multiply the density of helium, $\unit{0.16}{\frac{\kilo\gram}{\meter^3}}$ , by the volume $\unit{0.014}{\meter^3}$ ), leaving around $\unit{13}{\gram}$ to be occupied by the load.

Selected references (i.e. whatever came up on Google) suggest that one liter of helium can lift about one gram, roughly on par with what I've found here.

Given that result, in order to float a 44-pound child using party balloons, how many should we theoretically need? If $\unit{13}{\gram}$ can be lifted by one balloon, it's a simple calculation:

$\unit{44}{\pound}\times\frac{\unit{454}{\gram}}{\unit{1}{\pound}}\times\frac{\unit{1}{\text{balloon}}}{\unit{13}{\gram}} = \unit{1500}{\text{balloons}}$

And yet, clearly that wasn't nearly enough in the show — Kari, Tory, and Scottie wound up using nearly 3500 balloons, meaning that each balloon was actually holding up less than $\unit{6}{\gram}$ of Mattie's mass, less than half of what was predicted. You might quite literally say that they weren't pulling their weight.

Why not? Well, I have a couple of theories. First of all, this calculation doesn't account for the mass of the strings needed to hold the balloons, nor of the harness Mattie was riding in. We should have accounted for this extra weight when calculating the number of balloons by adding it on to the $\unit{44}{\pound}$ figure from above. Besides, it took the build team nearly an entire day to assemble the full set of 3500 balloons, and over that time, you're going to get some significant helium leakage from the first balloons inflated. So it's very likely that those early balloons had a reduced lifting power by the time they were actually able to get Mattie off the ground.

One more thing to consider is that the numbers involved in the calculation I did for one balloon were pretty small, meaning that even what seems like a minor inaccuracy could throw off the results quite a lot. Of course, science has a tool for dealing with exactly this sort of thing: it's called error analysis, and maybe someday, time permitting, I can revise these calculations to see just how (in)accurate that estimate of 1500 balloons really was. But this is getting long enough already.

Next up: part 2 takes a look at the role of buoyancy in raising the Mythtanic II with ping-pong balls.