10. Fermi Estimates

The world is vast and complicated, and precise measurement of its systems can be hard, expensive, time-consuming, imprecise, or any or all of the four. Nonetheless, we sometimes find ourselves in need of - or even just itching with curiosity for - quick cheap estimates of magnitude or prevalence of some phenomenon that lives in the world. These can be as trivial as the number of piano tuners in New York City or as practical as the number of potential customers walking by a storefront on any given day or even as weighty as the explosive yield of a nuclear weapon. Lacking hard data, we cannot hope for precise answers, but might we work some magic to come up with a decent guess? The answer is yes, and it’s less of a miracle than you might think. A secret, a mystery: all knowledge is one, for the systems of the world are deeply and redundantly enmeshed with each other; out there in the world there is some true object or phenomenon, permanently inaccessible directly to the brain, but nevertheless there is a single correct answer as to its quantities and qualities, and any independently formulated way we might pick to try to measure it should give us the same answer, up to some independent error, as long as it’s a good and faithful mode of measurement that touches the world - this is called consilience, and when we find it to hold of these disparate good means of measuring, we should gain faith in both the measurements and the methods that produced them.

We are however finite creatures, so let us avoid holding ourselves to too high of a standard. If we can make a guess that’s within an order of magnitude of correct in either direction, we can justly declare victory. Anything better than that is just gravy. How many chicken eggs come in this box I saw last year? No, I won’t tell you what kind of box, only that it came from a normal-ish grocery store. “More than two and fewer than a hundred” is a good first answer, but try to tighten that too far and you’ll sometimes be wrong. “Somewhere between six and sixty” is a little tighter, but we can’t do much better than that; we should accept our limits and make that our guess and miss neither the half-dozen box nor the massive pallet of forty-eight.

Any description of how to do a Fermi estimate is necessarily incomplete, as there's many ways of doing it right, but the fundamental tactic is to start by understanding what the estimate requires you to know, including the units it’s naturally expressed in (including dimensionless!) and the kinds of connected (usually unitful) quantities that constrain or give connected information about it, and then recursively breaking down anything you don’t know into a product or ratio of things you do know, or can guess at. If you want a density, go looking for a mass and a volume you can estimate. If you then need the volume, try to remember the geometric formula that calculates it (or even a much simpler one!); keep going until you have simple lengths and times and masses and item counts that you can know or guess directly. You almost never want to add or subtract, unless you have a strong sense that one of those simple linear quantities has two major contributing factors. Three, tops. As a guiding principle, you can basically always simplify a little further and neglect constant factors that are between ⅓ and 3. Estimate π as 3; think of complicated shapes as cylinders, and cylinders as prisms if that’s too hard. Another is that your blind vaguely-reasonable guesses at those quantities - which can and should be taken from mental visualization and life experience! - are probably fine; you probably aren’t getting them wrong by multiple orders of magnitude. Guessing the world population as 1*10^9 and 5*10^10 would both be acceptable! If you want more precision, do a quick binary/higher-lower search, and always split the difference harmonically, not arithmetically: the average of 1 and 100 is 10, not 50. One final tip is to use metric, which makes unit conversions straightforward, especially between volume and cubic length; a helpful fact here is that the circumference of the earth is almost exactly 4*10^4 m as it was once defined to be, and another is that 1 L of water masses precisely 1 kg.

The key to why this method works is threefold: logarithmic reasoning, dimensional analysis, and the central limit theorem as applied to independent errors (remember consilience?). We start by noting that we always end up with a calculation that’s just multiplying a bunch of things together, where the things we multiply together are generally assorted reasonable-feeling guesses at numbers and half-recalled facts about the world and sometimes formulas using those guesses. We should expect those somewhat untrustworthy figures to be wrong, but crucially, wrong in independent and different directions, and wrong in ways that are best expressed in terms of orders of magnitude - a tenth or three times the right value; if we pile enough of those together, on average those wrongnesses will approximately cancel out, leaving us with a startlingly good answer. As a final sanity check of our work and answer, we should have a sense of the units that the answer will be in - anything from meters to molecules to number of cards, or even dimensionless - and the big multiplication problem that we do must necessarily work out to have the same units.

Here’s a first worked example thanks to SG (yes, the same one from the battle-short post): “How many micro SD cards can fit in a cargo 747? And how does that compare to the number produced each year?” You may want to stop reading here, think it over, and try this for yourself. Maybe I got it wrong - comment below with your reasoning and we can see whether we converge on numbers.





The first thing that I needed to know was that microSD cards are not the big standard kind, they’re the tiny kind that go in cell phones. Let’s tackle the first half first. I’ll round off a 747 cargo plane to being a cylinder. I know from having been in a 747 that there’s a fair bit of room between my head and the ceiling and that the passenger floor is at about the midpoint of the plane; since I’m about 1.7 m tall, I’ll put the radius of the plane at 2m. Also, there’s like 60 rows of seats and about 25% excess space between rows of seats - the emergency exits, the bathrooms, and such - and I’m pretty sure that lying down I’d stretch to cover about 2 rows, maybe a little more: 75 * 1.7/2 ~ 100 m. The volume of the cylinder is V_747 = πr^2h ~ 4 m^2 * 100 m * 3 = 1,200 m^3. On the other hand, microSD cards are roughly rectangular prisms and have an aspect ratio of something like 3:2 with a short side of about ¼ inch and they’re about half as thick as a quarter, giving me a guess at dimensions of 10 mm x 15 mm x 1 mm = 150 mm^3 = 1.5 * 10^-7 m^3. Dividing out, we get (1.2 * 10^3 m^3/airplane)/(1.5 * 10^-7 m^3/microSD) ~ 10^10 microSD cards/airplane. The estimate for yearly production is quicker and dirtier: lots of people (~half) have cellphones these days and go through about one phone every 18 months; also some people (~a quarter) and some industries use LOTS of them but not too many, so something like 1 card/person-year sounds about right? You could get a better estimate if you really tried but it would likely not be worth it; surely it’s more than 0.1 card/person-year and less than 10 cards/person-year, so we’ll stick with a rough estimate of 10^10 cards/year. They turn out to be very very close to each other!


Here’s another excellent one from KN: “How many tons of newspaper exist in the world?” Once again, you may want to stop reading here, think it over, and try this for yourself.





I’ll specifically operationalize this as newspaper and not newsprint: that is, nothing being used as wrapping paper or firestarters or tossed into a recycling bin or a gutter, no back stocks of newsprint sitting in printshops; however, old newspaper sitting as-yet unread on a table until the crossword can be done is fine, as are newspapers sitting as-yet unsold or unclaimed in vending boxes. Maybe my answer would increase by a factor of 3, if I count all the marginal cases? Anyway: probably a given newspaper lasts for more than 2 days and less than a month, so ~8 days; any given newspaper probably masses about 0.5 kg (mabetween local periodicals on one hand and Sunday editions on the other. For readership figures, probably about 10% of people these days use dead-tree newspapers, especially if we average out those periodical readers who never do, those who stick to Sunday editions, and those who order daily, and maybe half of all adults - a quarter of all people - read newspapers regularly at all - about 2 billion. Putting this all together, m_newspaper = t_life * (n_readers * prop_physical) * m_newspaper ~ 8 * 2*10^8 * 0.5 kg = 8*10^8 kg ~ 7*10^6 imperial tons, where this time I’ve elided some of the units like “days” and “people” because I’ve also elided obvious factors like “1 newspaper lifetime per newspaper” and “~1 newspaper per physical reader”. I think I even believe that there’s something like a ratio of 600 kg living human : 1 kg newspaper in the world?


One last one from OG: “How many trained acrobats are there in the US?” This resembles a classic: “how many piano tuners are there in New York City?” (Give that one a try if no other, if you want.)





To prove a point, I’ll first operationalize “trained acrobat” as anyone who has received substantial training in acrobatic feats and could safely perform any acrobatic trick. Probably ¼ of the US population is under 14 years old, and another ¼ is over 60. I’ll eliminate both. After this things get dicey: all I think I can really do here is go on a felt sense that there’s no way that 10% of people have been trained in acrobatics, but 0.1% feels somewhat too low. My very rough guess, then, will be 0.5%, and together with a figure of about 3.4*10^8 people in the US, we have a guess of 3.4*10^8 * 0.5 * 0.005 = 850,000.

Now I’ll operationalize “trained acrobat” as anyone who is or could be a professional acrobat. I’ll take a different approach: probably on any given day in the US, there’s 1 or 2 acrobatic acts going on, for about 500 per year, with an acrobatic troupe probably numbering around 10~20 and 2~50 at extremes, so I’ll guess 20; I expect that any given acrobat probably takes decent-sized breaks, only performs during an on-season, or in some cases recovers for a while, so that the 500 performances represent maybe 50 troupes’ worth of acrobats. So maybe about 1,000 professional acrobats, or maybe more like 3,000 if you throw in people who could be professional acrobats but have chosen not to. Do I believe a ratio of something like 1:500 for professionals to reasonably devoted amateurs? I think I do. Maybe that’s a little bit too extreme and there’s either fewer amateurs or more professionals.

The point here is threefold: first, that how exactly you operationalize a simple natural-language question matters a lot in your final estimate; second, that the more precise your operationalization, the more you have to go on in estimating; and finally that you can use even very rough guesses to anchor better ones on nearby topics.

Now go forth, pull untrustworthy numbers from your tail, and assemble them into uncanny insight!



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