Maths on the Back of an Envelope: Clever Ways to (Roughly) Calculate Anything

by Rob Eastaway

  200 minutes per day × 5 likes per minute = 1,000 likes per day.

  1,000 × 365 days per year h 400,000 likes per year.

  That’s, like, nearly half a million!

  It wouldn’t surprise me if peak-likers utter the word ‘like’ well over a million times per year.

  Like, wow.

  ARE YOU DESCENDED FROM RICHARD III?

  In August 2012, a team of archaeologists digging in a city-centre car park in Leicester discovered a skeleton. DNA testing followed, which confirmed that they had found the long-lost body of Richard III, the king made into an infamous villain by Shakespeare as the hunchback monarch accused of murdering his young nephews in the Tower of London so that he could claim the crown. Richard of York (as he was also known) was defeated and killed by Henry Tudor at the Battle of Bosworth.

  The euphoria at the rediscovery of Richard’s bones was soon followed by an argument as to where they should be buried. The people of Leicester believed he should be buried in Leicester Cathedral, fittingly close to where he’d been found. But a small group, going by the name of the Plantagenet Alliance, declared that, as the descendants of Richard III, they should have the right to decide on the burial place of their ancestor. They wanted Richard to be buried in York.

  I was struck by this story because, after 500 years, I expected that there might be rather more than 15 descendants interested in this case. So I dived into Google, investigated Richard’s family, grabbed the back of an envelope and did some calculations.

  Here’s what I found.

  Although Richard III is known to have had three children, his only legitimate heir died in childhood, and his two others – both illegitimate – are not believed to have had any offspring. So there are no known direct descendants of Richard III. The so-called descendants lobbying on his behalf were actually descendants of his nephews and nieces.

  Richard had five siblings, and a number of nephews/nieces (though Shakespeare reckons he killed two of them, the famous Princes in the Tower). Richard’s eldest sister Anne had a daughter who had 11 children, and one of those grand-nieces herself had 11 children, so within a couple of generations there was a healthy stock of Richard III relatives to procreate.

  Let’s assume that each surviving descendant had two children who survived to child-rearing age, and that one generation is around 25 years. In the 500 or so years since Richard’s death, there have been about 20 generations, so if there was no interbreeding between descendants, there are around 220 descendants of his nieces and nephews today, which is about one million.

  But that’s a conservative estimate. If more than two children survive, e.g. 2.3 surviving children per generation (that’s a low estimate for wealthy families, who had higher survival rates because of better nutrition), the number of descendants leaps to an incredible 17 million – an example of the ‘sensitivity’ I discussed here.

  However, we know that marriage between distant cousins is inevitable. As you come down through the descendants of Richard’s family, there will be many who were marrying remote cousins. This will significantly reduce that figure of 17 million, but it’s reasonable to suppose that there are at least a million people who are descendants of Richard III’s siblings. To support this claim, Dr Andrew Millard of Durham University wrote a paper that makes a credible case for saying that if you have any English ancestry, you are almost certainly a descendant of King Edward III, Richard’s great-great-great-great grandfather. If it’s true for Edward III, then the odds for Richard III featuring in your ancestry must be quite high as well.

  My conclusion? Since there’s a chance that anyone with English ancestry is related to Richard III, the Plantagenet Alliance had no more right than the rest of us to decide on Richard’s burial place. That was also the view of the three High Court Judges, who dismissed the case when it went for judicial review. It was a victory for back-of-envelope maths (though an expensive one: it cost taxpayers over £200,00012).

  HOW MUCH FURTHER CAN YOU THROW A SHOT PUT WHEN YOU ARE IN MEXICO CITY?

  You might think that it makes no difference where you are when you throw a shot put. (Well, if distance is the objective, I suppose the top of the Eiffel Tower would be good – but I’m assuming we’re on a flat piece of ground here.)

  It turns out, however, that there are two variables that can have a significant effect on how far a projectile will travel: air density (which affects air resistance) and gravity.

  This was famously demonstrated in 1971, when the Apollo 14 Commander Alan Shepard pulled off a memorable stunt while standing on the moon. The moon has about one-sixth of earth’s gravity, and almost zero air resistance. Shepard had somehow managed to smuggle the head of a golf club and a couple of golf balls onto the spacecraft. On the moon, he improvised the handle of a golf club, and with a one-armed swing he managed to whack one of the balls – in his words – ‘miles and miles’. Later, he gave a more realistic assessment that the ball had gone ‘over 200 yards’. With a proper club and two-armed swing, it’s reckoned an astronaut on the moon could easily hit a ball over a mile.

  Back on earth we can’t avoid air resistance, but the density of the atmosphere does change with altitude. There is less air resistance at the top of Table Mountain than there is at the Dead Sea. So – other things being equal – a shot put released on top of a flat mountain will travel a little further horizontally than one released at the same speed at ground level. The difference, however, is small. Air resistance has a big effect in slowing a balloon, but on a dense lump of iron its impact is minimal. Even in a vacuum, a shot put would travel only a couple of centimetres further than it would in the open air.

  Gravity, however, is a different matter. Although we were all taught in school that gravity is a ‘constant’ on the surface of the earth, this is not strictly true. There are two things that affect it. The first is that gravity reduces the further you go from the centre of the earth. So (like air density), gravity is smaller on a mountain top than at sea level.

  The second is that while gravity is pulling us in to the centre of the earth, there is something else that is attempting to fling us outwards. Like a roundabout in the playground, the earth is spinning around, and without gravity we would be flung out into space. The faster you spin, the greater the force. Close to the North Pole, the speed at which you are spinning around the earth’s axis is close to zero, but at the equator you are hurtling around at about 1,000 miles per hour.

  These two factors – the lower gravity due to height, and the reduced gravity due to centrifugal force – mean that gravity is detectably lower at altitude on the equator (in Mexico City, for example) compared to sea level near the North Pole (Helsinki, say). In Mexico, gravitational acceleration (usually denoted as ‘g’) is about 9.77 m/s2, while in Helsinki it is 9.83 m/s2. These figures do vary, but that’s a difference approaching 1%.

  What impact does this have on the range of a shot put?

  There is a formula for the range of a projectile, based on Newtonian physics. I was going to show it here, but my editor warned me that it might immediately knock about 20% off the sales of this book. So I’ve buried the full thing in the Appendix and simplified it to this:

  OK, even that might not look ‘simple’. What it means is that as the value of g decreases, the distance that the shot put travels increases (by an amount proportional to 1 divided by the square root of g).

  A good shot-put throw travels about 20 metres. If we reduce gravity by 1%, we increase the range of the shot put by about 1/2%, or 10 cm. That’s not to be sniffed at when world records are at stake.

  WHERE ARE THE ALIENS?

  Aside from his general approach to back-of-envelope questions, Enrico Fermi is remembered for one calculation in particular.

  Sometime not long after the end of the Second World War, Fermi and some other scientists were having a conversation, and the subject turned to extraterrestrial beings. As the story goes, Fermi suddenly asked the question: ‘So, where is everybody?’ – by wh
ich he meant, given the billions of stars in the galaxy, one of which must surely have developed advanced life forms, why hadn’t we been invaded by aliens yet?

  His question became known as the Fermi Paradox.

  Several years later, the astrophysicist Frank Drake came up with an equation for expressing the number (N) of intelligent, communicating civilisations that are in our galaxy at any one time. His equation was this:

  where:

  R* is the average number of stars that form each year;

  np is the average number of planets per star;

  fL is the fraction of those planets that develop life;

  fi is the fraction of planets with life that develop intelligent life;

  fc is the fraction of civilisations that develop communication technology;

  L is the number of years that communicating civilisations survive.

  Although it looks sophisticated, this is really just common sense set down in mathematical form. The hard part – and the estimation – comes in putting values to each of the factors.

  For example, of the planets that form around any particular star, what proportion are capable of developing life? To even have a stab at coming up with a sensible figure requires an understanding of the chemicals and physical environment that are essential for life forms to be able to exist.

  Various scientists have attempted to come up with sensible figures for each of the factors.

  For the average number of stars that form each year, suggestions have ranged from 1 to 10. The number of planets per star that can develop life is reckoned to be in the range of 0.2 to 2.5, of which the percentage that form intelligent life is anywhere between 1% and 10%, and of these anything between 1% and 100% develop the technology to communicate. When they do communicate, different scientists have reckoned the civilisation will last anywhere between 100 and one billion years (with one scientist coming up with the suspiciously precise figure of 304).

  Picking middling values for each of the factors gives us:

  3 × 0.5 × 10% × 25% × 30% × 1000 h 10.

  So that suggests there might be 10 civilisations out there that are capable of communicating, and that might therefore be detectable.

  However, this number is extremely sensitive to the figures that you plug into the equation. The estimates for the number of civilisations out there communicating with the galaxy at any time could be anywhere between 1 × 10–10 (which is effectively zero) and 15 million. This must be the widest range of answers for a back-of-envelope question that has ever been produced, and it makes the forecasts for vCJD described here look like nanotechnology precision.

  The Drake equation is a fun intellectual exercise, but it’s also a good place to stop, because it demonstrates that, sometimes, attempts at estimation are bordering on the futile.

  LAST WORD

  WHEN THE ROBOTS TAKE OVER

  THE COUNTDOWN CONUNDRUM

  Before we finish, let’s briefly move away from estimation and back to the world of precise arithmetic.

  In a classic 1997 episode of Channel 4’s word and numbers game Countdown, the host, Carol Vorderman, picked the following six numbers from the table in front of her (four from the top, and two from the third row):

  The random number generator then produced a target of 952. The challenge, as always, was for the contestants to use some or all of the numbers on the cards no more than once, to get as close as possible to the target answer of 952.

  Perhaps you’d like to have a go now, to see how close you can get.

  To get close to the target answer, it’s a huge help to be able to play with numbers. And curiously, even though the task involves ‘exact’ calculation, rough estimates can be handy as a starting point: ‘952 … that’s going to be 9 × 100 and a bit … or 1,000 minus 50ish’.

  If you managed to get 950 (two away from the target), give yourself a bronze medal. This is the way that most people get there:

  100 × (3 + 6) + 50 = 950.

  If you managed to get to within one (953), give yourself a silver medal. To create the 3, you need to spot that:

  75 ÷ 25 = 3

  950 + 3 = 953.

  That would normally be enough to win you the round, but on the programme in question, you would have lost out to the other contestant, James Martin, a PhD maths student who managed to get the exact answer of 952.

  Here’s a slightly shortened version of his exchange with Carol Vorderman as he explained how he’d got his answer:

  JM: 100 + 6 = 106. Multiplied by 3 …

  CV: … is 318.

  JM: I’d like to multiply it by 75 …

  CV: Multiply 318 by 75? [Laughter] Good grief, I’m going to need my calculator for this one. [Eventually she does the calculation to get 23,850.]

  JM: Now take away 50.

  CV: [More laughter] 23,800.

  JM: And divide it by 25.

  CV: And divide THIS by 25? … [Can barely control hysterics as she writes out the division] Do you know – I think you’re right. That’s incredible!

  There are many calculation savants who could do all of the calculations above – and harder ones – in their heads. But James Martin was not a savant, he was just smart at manipulating numbers.

  Martin spotted that 106 × 9 = 954, which is 2 away from the target.

  There wasn’t a 9 available, but he could get to the same answer by multiplying by 3 twice: first by the 3 on the card, then by 75 ÷ 25, which is also 3. But where would he get the 2 that he needed to subtract from 954? What James spotted was that 50 ÷ 25 = 2, so he could use the 25 twice, by dividing it into 75 and 50.

  Written out in full, his solution was this:

  This might make it appear that he multiplied 318 × 75 (as you’d do on a calculator), but before multiplying, he divided 75 by 25 first to get 3 and 50 by 25 to get 2. This turned the calculation into:

  (106 × 9) – 2.

  Which is clever. But it is not genius.

  Back in 1997, when that solution was recorded, it’s unlikely that even somebody armed with a computer could have beaten James Martin to the score. But today, there are apps that can solve these Countdown problems instantly. It won’t be too long before a contestant could be wearing a pair of glasses that detects the numbers on the board and displays the solution on the lenses before the timer has even begun to count down.

  Which leads to an interesting Countdown conundrum.

  We are not far away from a world where we will all be equipped with Artificial Intelligence devices that can solve any numerical puzzles of this kind in an instant. When technology like this becomes readily available, not only will we not need calculators – some might question why we will need to learn any maths at all. When robots can work everything out, will games like Countdown continue to exist?

  There will of course be people who scoff at anybody who ‘wastes their time’ figuring out number puzzles when the solution is readily available to them. Just as there are many who scoff at the need to know how to do short division calculations manually, when it can be done by a calculator.

  But my prediction is that in 50 years’ time – even when computers can solve just about any numerical problem posed to them almost instantly – people will still have a huge interest in playing number games in their heads. And it won’t just be for a bit of TV amusement. We need to continue to be able to do back-of-envelope calculations without the aid of a calculator, or other artificial device.

  Why?

  Because we will always need to be equipped to challenge the information that is presented to us, whether it has come from a person or from a computer. If we leave every calculation and every decision to computers, we are in danger of becoming slaves to technology.

  And aside from all the practical benefits of being able to do maths on the back of an envelope, there is arguably another that is just as important: doing your own calculations keeps the brain stimulated, and gives it a valuable work-out. Some of us go further, and regard it as fun.

  APP
ENDIX

  SIGNIFICANT FIGURES

  The idea of rounding a number to one, two or three significant figures is a recurring theme in this book. Here’s a reminder of how it’s done.

  Let’s take the height of the Matterhorn in the Alps, which according to most sources is 4,478 metres. It’s likely that surveyors have readings that give them a height to the nearest centimetre, but given that mountains are always on the move, they have understandably rounded the Matterhorn’s height to a four-digit number that is correct to the nearest metre. The statistic therefore has four significant figures.

  To round the number to three, two, or one significant figure, lop off the number at the end and replace it with a zero – but with one proviso: if the digit being removed is 5 or more, the digit before it should be rounded up by one.

  Here’s what happens when we round the Matterhorn height:

  Rounded to …

  Three significant figures 4,480 (note that the 7 has been rounded up to 8)

  Two significant figures 4,500 (the second 4 is rounded up to 5)

  One significant figure 4,000 (note that 4,478 rounds down)

  The first significant figure of a number is always its first non-zero digit. So, for example, the first significant figure of 0.0063 is 6. It’s possible for a significant figure to be zero, including a number’s final digit. For example, if an athlete runs 100 metres in 10.28 seconds, that is 10.3 seconds to three significant figures, and 10 seconds to two significant figures.

  WHERE THE RULE OF 72 COMES FROM

  The ‘Rule of 72’ is found by working out how many iterations (years, for example) it takes for a number to double if it is growing at a fixed rate. To follow the derivation, you do need to be familiar with natural logarithms.

  Let’s call the annual interest rate R per cent. What we are looking for is the number of years, N, that it will take our starting pot of money, A, to double; i.e. after N years we will have an amount 2A: