Warren Buffett's Cube of Gold

In his 2011 annual letter to shareholders Mr. Warren Buffet, chairman of Berkshire Hathaway,  discussed the amazing rise in price of gold. Gold, at that time, was becoming extremely expensive (or valuable) and some shareholders wondered why Mr. Buffett did not buy gold to hold as the price increased.

Part of his response was:

"Today the world’s gold stock is about 170,000 metric tons. If all of this gold were melded together, it would form a cube of about 68 feet per side. (Picture it fitting comfortably within a baseball infield. [Bases are 90-feet apart.]) At $1,750 per ounce – gold’s price as I write this – its value would be $9.6 trillion. Call this cube pile A."

I remember reading that gold is extremely malleable and can be spread very thin. My thought was:

How much of our planet Earth could be covered by this hypothetical cube of gold?

To answer this question, I will use the data in the following table:

Item	Factor or Formula	Comment
Density of gold	19.32 g/cm^3	Compute mass of hypothetical cube
Convert feet to meters	3.2808 feet per meter
Ounces to kilograms	35.274 ounces per kg	Convert mass to another unit
Ounce of gold to area	1-ounce covers 100 square feet	Pounding gold to a large area
Radius of earth	6,353,000 meters	To find surface area of our planet
Surface area of a sphere (that is, Earth for our case)	\(4 \pi r^2 \)	Formula (r is the radius of the sphere)

Step 1: How much gold is in this cube?

For a cube of 68-feet per side there are \( 68 \times 68 \times 68 \; = \; 314, 432 \; \; ft^3 \) of gold. From the table above we have the density of gold, but not in units of feet. Let’s first convert the density from grams to kilograms:
\[ Density = 19.32 \;\; g \; /  \; cm^3 \times \left( 100 \; cm \; / \; m \right)^3 \\  = 19,320 \; \; kg \;  /  \; m^3 \]
And let’s convert the density from meters to units of feet:
\[ Density = 19,320 \; \; kg/m^3 \times \left( 1\; meter\; / \; 3.2808 feet \right)^3 \\ = 547.10 \; kg \;  per \; ft^3  \]
Now we can compute the amount of gold in this hypothetical cube as: \( 314, 432 \times 547.10 = 172,025,747 \) kg of gold.

 Step 2: How big an area will the gold cube cover?

This site notes an ounce of gold ("about the size of quarter") can be pounded into a 100 square foot sheet, which is pretty thin if you think about it. There are 35.274 ounces per kilogram so our cube has
\[172,025,747 kg \times \left(35.274 \; ounces\; / \; kg \right) \\ =  6,068,036,199   \; \; ounces \]
and for that many ounces of gold, we have a surface area of
\[ 6,068,036,199 \; ounces \;  \times \left( 100 ft^2 \; / \; ounce \right) \\  = 606,803,619,900 \; ft^2 \]
That cube covers over 606 billion square feet.

Step 3: What is the surface area of Earth?

The surface area of earth is:
\[ S_{earth} = 4 \times \pi \times r_{earth}^2 \] and we use the radius of Earth given in the table to find:
\[ S_{earth} = 4 \times \pi \times \left( 6353000^2 \right) \\ = 507,186,370,915,240 \; m^2 \]
which we can convert to square feet as:
\[  507,186,370,915,240 \; m^2 \times (3.2808 \; feet \; / \;  meter)^2 \\ = 5,459,175,891,528,360 \; ft^2 \]

Step 4: What is the ratio of the gold cube area to the surface area of Earth?

As a fraction, the gold cube covers 0.0111 percent of Earth:
\[ \frac{606,803,619,900}{5,459,175,891,528,360} \times 100  = 0.0111 \; percent \]
 which is a pretty small relative to the size of Earth.

What’s the point?

While this cube costs $9, 600 ,000 ,000 ,000 it only covers a bit more that one-hundredth of one percent of the planet. Of course, the whole idea of gold as valuable is because it’s so rare, and these calculations confirm that. But it was fun to do the math, don't you think?

Check out:

Gold fun facts from the American Museum of Natural History.