How many poker chips will fit inside a Boeing 747-8?

Question

Assuming the cargo hold of a Boeing 747-8 is 854.4 cubic meters, and a typical poker chip has a thickness of 3.3mm and a diameter of 39mm. How would you go about working out how many chips will fit inside the 747?

Answer 1

If you want a realistic answer to this, you would of course need to take into account that you can't pack the Boeing 'perfectly'; there will be lots of gaps and spaces depending on how the cargo bay is shaped. Even if you had a perfect rectangular box you would of course still have gaps between the poker chips, since they are round.

But for a rough approximation, maybe do exactly that: consider the Boeing cargo bay to be a rectangular box, and consider stacking the chips as little rectangular 'boxes' (a hexagonal packing would be more efficient, but someone else can do the math for that ...) of 39mm by 39mm. So then the volume of a single chip is $39 \cdot 39 \cdot 3.3 \ =5019.3 mm^3$

And the Boeing has a volume of $854.4 \ m^3 = 854.4 \cdot 10^9 \ mm^3$

So that would be $\frac{854.4 \cdot 10^9}{5019.3} \approx 0.17 \cdot 10^9$, which is about $170$ million poker chips

Answer 2

Technically, the answer depends on the shape of the cargo hold. For example, if the hold is a tube of diameter 3cm and a length of $302182$ meters, then no chip will fit inside.

(yes I know the Boeing 747-8 is not 302 kilometers long, I am just making a point about why shape is important).

But for an approximate answer to the question, you need to find the maximum number $N\in\mathbb N$ such that $$N\cdot V_{chip} <V_{cargo hold}$$

Answer 3

As others have pointed out, there are a lot of uncertainties in this problem, mostly related to the exact shape of the plane and the chips. In such a case, perhaps Fermi estimation might be a more reasonable plan of attack?

The basic idea is to work at the level of orders of magnitude. That is, round everything off to the nearest power of 10, and work in a realm where the mental arithmetic can be done more easily.

Each poker chip has a volume of approximately $40\cdot 40 \cdot 3 \approx 5000 \text{ mm}^3 = 5\times 10^3 \text{ mm}^3$. The plane has a volume of approximately $900 = 9\times 10^2 \text{ m}^3$. Finally, recall that there are $1000=10^3 \text{ mm}$ per meter. Thus we should be able to fit something on the order of $$ \frac{9\times 10^2\text{ m}^3}{5\times 10^3 \text{ mm}^3} \cdot \frac{10^9\text{ mm}^3}{\text{m}^3} \approx \frac{9}{5} \times 10^8 \approx 2\times 10^9$$ poker chips in the plane. That is, the plane will hold about 200 million chips. I'll note that this answer is pretty close to that given by Bram28, but can be computed quickly without the aid of a calculator, computer, or slide rule.