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FIGURING WITH FERMI
THE FERMI APPROACH
The man often credited as the guru of back-of-envelope calculations was Enrico Fermi. Fermi was a physicist who was involved in the creation of the very first nuclear reactor. Most famously, he was present at the detonation of the first nuclear bomb, the so-called Trinity Test in New Mexico, USA, in July 1945. At the time, scientists still weren’t sure how big the explosion would be. Some even feared it might be large enough to set off a chain reaction that would destroy the planet.
The story goes that Fermi and others were sheltering from the explosion in a bunker about six miles from ground zero. When the bomb went off, Fermi waited until the wind from the explosion reached the bunker. He stood up and released some confetti from his hand, and when it had landed, he paced out how far the confetti had travelled. He then used that information to make an estimate of the strength of the explosion. We don't know for certain how Fermi did this, but it probably involved him estimating the wind speed and working out how much energy was required to push out a ‘hemisphere’ of air from the centre of the explosion.
Fermi’s estimate of the bomb’s strength was 10 kilotons. Later, more rigorous calculations revealed that the real strength had been nearer to 18 kilotons, in other words Fermi’s answer was out by a factor of nearly two. Anyone submitting an answer that far out in a maths exam would probably get no marks, yet Fermi got huge credit for the accuracy of his back-of-envelope answer. The important thing was that his answer was in the right order of magnitude, and gave scientists a much better understanding of the potential impact of the weapon they were now dealing with.
What was so impressive about Fermi’s calculation was that he did it based on such crude data. His results also show that being in the right ballpark can sometimes mean being quite a long way from the exact right answer, and that an ‘inaccurate’ answer can still be useful.
Calculations that are done without access to much proper data have become known as ‘Fermi problems’. These calculations are typically done as an intellectual exercise: working things out for the sake of it.
There is, however, a practical benefit to honing your skills at solving Fermi problems, because this type of problem is notoriously used in interviews. I can still remember one of the questions I was asked at my university entrance interview: ‘What does an Egyptian pyramid weigh?’
One way to figure this out would be:
(1) estimate the dimensions of the pyramid and hence its volume;
(2) estimate the density of stone (in kilograms per cubic metre) and work out its volume;
(3) multiply the volume by the density to get a figure for the weight.
I doubt my interviewers were interested in the answer, not least because nobody has ever weighed an Egyptian pyramid. What the interviewers were looking for was the thinking process. I don’t remember what answer I gave, but it can’t have been too bad because I was offered a place.
For many employers it’s the same: companies like Google and Microsoft are famed for the (Fermi) questions they have posed to would-be employees, to see how they think on their feet.
Whether you’re practising for interviews, or are just exercising your brain for the fun of it, Fermi problems are an excellent work-out. In this section I have come up with a selection of Fermi problems that captured my imagination. In each case, I’ve shown the approach I would take when tackling them. You might well take a different approach. The only thing you and I can guarantee is that we are very unlikely to come up with the same answer, though with any luck we will at least both end up in the same ballpark.
COUNTING
When embarking on Fermi-style estimations, a good place to start is counting. Answering the question ‘How many …?’ is the most primitive mathematical challenge – though it often turns out to require a surprising level of skill, and is sometimes highly contentious too.
Who can forget Donald Trump’s anger when his own view that he had ‘probably the biggest crowds ever’ at his inauguration was contradicted by several sources who suggested the crowd was a lot smaller than his predecessor’s?
We’ve already seen examples of where even meticulous attempts to count exactly the right number often miss the mark (see vote-counting in elections here, for example). Fortunately, there are many situations where only a decent estimate is required. Here are some examples.
HOW MANY WORDS ARE THERE IN YOUR FAVOURITE BOOK?
Publishers like to do word counts – sometimes they even pay a rate per word. So how many words are there in a manuscript, or a book you pluck off the shelf?
There is, of course, in most word-processing software, a button that can be clicked to find the precise answer to this question: but that’s only for the electronic version of the book. For the paper version, you can either laboriously count each word or – more practically – make an estimate.
Word counting lends itself to the technique of taking a sample and then extrapolating that to estimate the whole book: word length and density tends to be pretty consistent through most books, so if you turn to a random page of full text somewhere in the middle of the book, that’s likely to be a good representation of every page.
How many words are there in, say, Pride and Prejudice?
The ultimate cavalier approach is to count the words in one line and base the entire estimate on that. Twelve words in the line, 38 lines on the page, 345 pages, using Zequals that’s
10 × 40 × 300 = 120,000 h* 100,000 words.
But it’s not much effort to take a bigger sample and come up with a more accurate estimate. Even counting three lines of text to get an average will improve the estimate considerably. If the word count over three lines is 34, that’s an average of about 11 words per line. And with the indents, sentences that end in the middle of the page, and half-pages at the end of the forty-odd chapters, we might say there’s the equivalent of about 36 full lines of text per page, and about 320 pages.
The approximate answer hasn’t changed: 11 × 36 × 320 h* 10 × 40 × 300 = 120,000 h* 100,000. But these better estimates of words per page and number of pages justify a more precise estimate: 11 × 36 ∼ 400, and 400 × 320 = 128,000.
Impressively, but perhaps fortuitously, it turns out that this estimate is remarkably close to the official word count of 122,000.
Working out the word count in the book you are reading now is a bit trickier – tables, numbers, diagrams and in-fill boxes make things a bit more complicated. But you can still make reasonable estimates to come up with a figure.
HOW MANY HAIRS ARE THERE ON AN ADULT HUMAN’S HEAD?
The number of hairs on a person’s head will, of course, vary hugely from person to person. Let’s try to figure out a number for somebody with a full head of hair, say. And let’s do it just using the imagination, and without looking at a scalp.
Picture (if you can) a square centimetre of scalp on a human head.
How far apart are the follicles? We can start by putting some upper and lower bounds on it. If hairs were 2 mm apart then the hair wouldn’t be dense at all, and you’d think we’d be able to see the scalp shining through. But if hairs are as little as 0.5 mm apart, the scalp would be so densely packed, it would be more like fur. So let’s go for 1 mm as a reasonable compromise.
That means that within each square centimetre of scalp, we’re looking at 10 × 10 = 100 hair follicles.
What’s the surface area of a scalp? Put your hands around your head, fingers at the top of your forehead, thumbs on your neck at the base of the scalp – call that a circle maybe up to 25 cm (10 inches) across.
The area of a circle with a diameter of 25 cm is: π × radius2, and the radius in this case is 12.5 cm. But this is all so rough and ready that Zequals is called for:
3.14 × 12.52 h* 3 × 102 = 300 cm2.
The surface area of the scalp isn’t a circle, it’s more like a hemisphere. For the same radius there will be more surface area on a hemisphere than on a circle, so let’s double it to 600 cm2.
The total hairs on the typical head is therefore, according to this estimate, 600 × 100 = 60,000.
That will vary a lot – the number probably going up to 100,000 at most, and down to 30,000 at least (if a receding hairline or thinning hasn’t set in).
The knowledge that a person has, at most, 100,000 hairs on their head means you can state with absolute certainty that there are at least two people in (say) Huddersfield who have exactly the same number of hairs on their head as each other.
The proof is found by first supposing that everyone in Huddersfield has a different number of hairs on their head. We know that the most hairs anyone has is going to be around 100,000. So let’s suppose that everyone else has fewer hairs, and that they all have a different number of hairs on their heads. Imagine now lining them up, starting with the person with no hair, then the one with one hair, two hairs etc.
We have 100,000 people with different numbers of hairs on their head. But what do we do with person 100,001? Or indeed with the other 50,000 or more Huddersfield...