Raindrops
June 27, 2014
Our solution calculates the surface area of the planet then multiplies by the number of raindrops that fall at an average spot on the face of the planet.
Archimedes figured out long ago that the surface area of a sphere is 4 * pi * r^2. The earth has a circumference of 25,000 miles, or a diameter of 8000 miles, or a radius of 4000 miles, which is 12 * 5280 * 4000 = 250 * 10^6 = 1/4 * 10^9 inches. Thus the surface area of the Earth is 4 * pi * 1/4 * 1/4 * 10^18 square inches, the four constants (roughly) cancel each other, and the area is 10^18 square inches. Wolfram|Alpha agrees with our rough calculations; the exact answer is 7.9 * 10^17 square inches.
Figuring out the number of raindrops that fall per square inch is harder. We’ll assume that a raindrop is a cube 1/8 of an inch on a side, so there are 512 = 1/2 * 10^3 raindrops in a cubic inch. Some spots on the planet receive no rain (deserts) while other spots on the planet receive upwards of 120 inches (rain forest) of rain per year; we’ll arbitrarily say the average spot on Earth receives 20 inches of rain per year, which if it’s not right is at least within a factor of 2 or 3 from the right answer. Thus, each square inch of the Earth’s surface receives 1/2 * 10^3 * 20 = 10^4 raindrops per year, and the entire surface of the Earth receives 10^18 * 10^4 = 10^22 raindrops per year.
After solving the problem, I asked Google for other solutions. Oldguy makes similar calculations and comes to 7 * 10^18 raindrops per day, which is 2.5 * 10^20 raindrops per year, about 1/40th as much as our estimate. Doctor Ian estimates 10^19 cubic inches of rain fall per year, but counts only 16 raindrops per cubic inch, for a total of 1.6 * 10^20 raindrops per year, but his raindrops are much larger than ours, which fit 512 per cubic inch; if you make his raindrops and ours the same size, we would be within a factor of 2 of each other, which is comfortably close for this kind of estimate.
So the question comes down to how big is a raindrop, and how much is the average annual rainfall at any given spot on the planet.
Programming Praxis 
Let’s say that the earth is approximately a 10000km cube and that a cube has approximately 10 sides, that gives an area of 10^15 square metres, so with 100cm of rain per year, that’s 10^14 cubic metres of water. A raindrop is about a 5mm cube, so that’s 200^3 in a cubic metre, which is round about 10^7, so total drop count is 10^21. Or thereabouts…
Average raindrop size seems to be the big unknown here.
Apparently, average rainfall is more like a metre a year, and smaller raindrops are presumably the majority, so maybe 10^23 or even higher is probably closer.
back of the envelope calculation based on vaguely remembered factoids and only carrying one significant digit (using ~ to mean approximately)
earth radius ~ 4000 mi * 5000 ft/mi * 12 in/ft * 3 cm/in ~ 720e6 ~ 7e8 cm
earth surface area = 4 * pi * r^2 ~ 4 * 3 * 50e16 = 60e17 cm^2 = 6e18 cm^2
assume everywhere is same as rainiest place I know of (one of the Hawaiin islands), then scale at the end:
rainiest place ~ 500 in rain/yr * 3 cm / in = 1500 ~ 2e3 cm rain / year
total volume of rain = 2e3 cm / year * 6e18 cm^2 = 12e21 ~ 1e22 cm^3 / year
rain drop ~ 2.5 mm radius ( they split if much more than 5 mm) ~ .3 cm
volume of raindrop = (4/3) pi r^3 ~ (4/3)*3*(.3^3) ~ 4 * .027 ~ .1 = 1e-1 cm^3
number of raindrops = total volume of all rain / volume of one drop
~ (1e22 cm^3 / year) / (1e-1 cm^3) = 1e23 rain drops per year. Hmm, almost Avogadro’s number, a mole of raindrops per year ;-)
if the rainiest place gets 100 times as much rain as the world average, there would be around 1e21, or 1,000,000,000,000,000,000,000, raindrops per year.
Here’s a similar discussion about how many snowflakes it would take to get 6 feet of snow on earth: http://what-if.xkcd.com/104/
About the same as everyone else
Estimated earth circumference: 24,000mi (actual 24,901) error 3.6%
Estimated annual rainfall: 1meter (actual .99m precipitation) error 1%
Estimated volume of a raindrop: 0.2mL (No real data on this but from a few random places this seems in the ball park)
I forgot what the surface area of the sphere was but I remembered the volume was 4/3pi*r^3 and I knew the derivative of volume is surface area so I got to 4pi*r^2
After fixing a few zero issues, didn’t use a calculator, i ended with 1.9*10^21. Using the actual numbers and the 5 drops per mL you get 2.5*10^21 or an error of 25%.
Now the precipitation also includes snow and the problem just said raindrops so should probably take out 30%? from that.
Pretty incredible that a group of people can estimate such a figure that initially seems impossibly large and all get so close.