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 2017-06-03 

Big Ben Strikes Again

As a child, I was a huge fan of Captain Scarlet and the Mysterons, Gerry Anderson's puppet-starring sci-fi series. Set in 2068, the series follows Captain Scarlet and the other members of Spectrum as they attempt to protect Earth from the Mysterons. One of my favourite episodes of the series is the third: Big Ben Strikes Again.
In this episode, the Mysterons threaten to destroy London. They do this by hijacking a vehicle carrying a nuclear device, and driving it to a car park. In the car park, the driver of the vehicle wakes up and turns the radio on. Then something weird happens: Big Ben strikes thirteen!
The driver turning on the radio. Good to know that BBC Radio 4 will still broadcast at 92-95FM in 2068.
Following this, the driver is knocked out again and wakes up in a side road somewhere. After hearing his story, Captain Blue works out that the car park must be 1500 yards away from Big Ben. Using this information, Captains Blue and Scarlet manage to track down the nuclear device and save the day.
A map of London with a circle of radius 1500 yards drawn on it.
After rewatching the episode recently, I realised that it would be possible to recreate this scene and hear Big Ben striking thirteen.

Where Does Big Ben Strike Thirteen?

At the end of the episode, Captain Blue explains to Captain Scarlet that the effect was due to light travelling faster than sound: as the driver had the radio on, he could hear Ben's bongs both from the tower and through the radio. As radio waves travel faster than sound, the bongs over the radio can be heard earlier than the sound waves travelling through the air. Further from the tower, the gap between when the two bongs are heard is longer; and at just the right distance, the second bong on the radio will be heard at the same time as the first bong from the tower. This leads to the appearance of thirteen bongs: the first bong is just from the radio, the next eleven are both radio and from the tower, and the final bong is only from the tower.
Big Ben's bongs are approximately 4.2s apart, sound travels at 343m/s, and light travels at 3×108m/s (this is so fast that it could be assumed that the radio waves arrive instantly without changing the answer). Using these, we perform the following calculation:
$$\text{time difference} = \text{time for sound to arrive}-\text{time for light to arrive}$$ $$=\frac{\text{distance}}{\text{speed of sound}}-\frac{\text{distance}}{\text{speed of light}}$$ $$=\text{distance}\times\left(\frac1{\text{speed of sound}}-\frac1{\text{speed of light}}\right)$$ $$\text{distance}=\text{time difference}\div\left(\frac1{\text{speed of sound}}-\frac1{\text{speed of light}}\right)$$ $$=4.2\div\left(\frac1{343}-\frac1{3\times10^8}\right)$$ $$=1440\text{m}\text{ or }1574\text{ yards}$$
This is close to Captain Blue's calculation of 1500 yards (and to be fair to the Captain, he had to calculate it in his head in a few seconds). Plotting a circle of this radius centred at Big Ben gives the points where it may be possible to hear 13 bongs.
Again, the makers of Captain Scarlet got this right: their circle shown earlier is a very similar size to this one. To demonstrate that this does work (and with a little help from TD and her camera), I made the following video yesterday near Vauxhall station. I recommend using earphones to watch it as the later bongs are quite faint.

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Tube Map Platonic Solids, pt. 3
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 2017-06-04 
This is awesome and wonderful. I salute you.
Ben Sparks
 2017-06-03 
Wow! This has made my weekend.
Tony Mann
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 2017-03-27 

The End of Coins of Constant Width

Tomorrow, the new 12-sided one pound coin is released.
Although I'm excited about meeting this new coin, I am also a little sad, as its release ends the era in which all British coins are shapes of constant width.

Shapes of Constant Width

A shape of constant width is a shape that is the same width in every direction, so these shapes can roll without changing height. The most obvious such shape is a circle. But there are others, including the shape of the seven-sided 50p coin.
As shown below, each side of a 50p is part of a circle centred around the opposite corner. As a 50p rolls, its height is always the distance between one of the corners and the side opposite, or in other words the radius of this circle. As these circles are all the same size, the 50p is a shape of constant width.
Shapes of constant width can be created from any regular polygon with an odd number of sides, by replacing the sides by parts of circles centred at the opposite corner. The first few are shown below.
It's also possible to create shapes of constant width from irregular polygons with an odd number, but it's not possible to create them from polygons with an even number of sides. Therefore, the new 12-sided pound coin will be the first non-constant width British coin since the (also 12-sided) threepenny bit was phased out in 1971.
Back in 2014, I wrote to my MP in an attempt to find out why the new coin was not of a constant width. He forwarded my letter to the Treasury, but I never heard back from them.

Pizza Cutting

When cutting a pizza into equal shaped pieces, the usual approach is to cut along a few diameters to make triangles. There are other ways to fairly share pizza, including the following (that has appeared here before as an answer to this puzzle):
The slices in this solution are closely related to a triangle of constant width. Solutions can be made using other shapes of constant width, including the following, made using a constant width pentagon and heptagon (50p):
There are many more ways to cut a pizza into equal pieces. You can find them in Infinite families of monohedral disk tilings by Joel Haddley and Stephen Worsley [1].
You can't use the shape of a new pound coin to cut a pizza though.
Edit: Speaking of new £1 coins, I made this stupid video with Adam "Frownsend" Townsend about them earlier today:

Infinite families of monohedral disk tilings by Joel Haddley and Stephen Worsley. December 2015. [link]

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New Machine Unfriendly £1 Coin, pt. 2
New Machine Unfriendly £1 Coin
The Importance of Estimation Error
Euro 2016 Stickers

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 2017-03-08 

Dragon Curves II

This post appeared in issue 05 of Chalkdust. I strongly recommend reading the rest of Chalkdust.
Take a long strip of paper. Fold it in half in the same direction a few times. Unfold it and look at the shape the edge of the paper makes. If you folded the paper \(n\) times, then the edge will make an order \(n\) dragon curve, so called because it faintly resembles a dragon. Each of the curves shown on the cover of issue 05 of Chalkdust is an order 10 dragon curve.
Top: Folding a strip of paper in half four times leads to an order four dragon curve (after rounding the corners). Bottom: A level 10 dragon curve resembling a dragon.
The dragon curves on the cover show that it is possible to tile the entire plane with copies of dragon curves of the same order. If any readers are looking for an excellent way to tile a bathroom, I recommend getting some dragon curve-shaped tiles made.
An order \(n\) dragon curve can be made by joining two order \(n-1\) dragon curves with a 90° angle between their tails. Therefore, by taking the cover's tiling of the plane with order 10 dragon curves, we may join them into pairs to get a tiling with order 11 dragon curves. We could repeat this to get tilings with order 12, 13, and so on... If we were to repeat this ad infinitum we would arrive at the conclusion that an order \(\infty\) dragon curve will cover the entire plane without crossing itself. In other words, an order \(\infty\) dragon curve is a space-filling curve.
Like so many other interesting bits of recreational maths, dragon curves were popularised by Martin Gardner in one of his Mathematical Games columns in Scientific American. In this column, it was noted that the endpoints of dragon curves of different orders (all starting at the same point) lie on a logarithmic spiral. This can be seen in the diagram below.
The endpoints of dragon curves of order 1 to 10 with a logarithmic spiral passing through them.
Although many of their properties have been known for a long time and are well studied, dragon curves continue to appear in new and interesting places. At last year's Maths Jam conference, Paul Taylor gave a talk about my favourite surprise occurrence of a dragon.
Normally when we write numbers, we write them in base ten, with the digits in the number representing (from right to left) ones, tens, hundreds, thousands, etc. Many readers will be familiar with binary numbers (base two), where the powers of two are used in the place of powers of ten, so the digits represent ones, twos, fours, eights, etc.
In his talk, Paul suggested looking at numbers in base -1+i (where i is the square root of -1; you can find more adventures of i here) using the digits 0 and 1. From right to left, the columns of numbers in this base have values 1, -1+i, -2i, 2+2i, -4, etc. The first 11 numbers in this base are shown below.
Number in base -1+iComplex number
00
11
10-1+i
11(-1+i)+(1)=i
100-2i
101(-2i)+(1)=1-2i
110(-2i)+(-1+i)=-1-i
111(-2i)+(-1+i)+(1)=-i
10002+2i
1001(2+2i)+(1)=3+2i
1010(2+2i)+(-1+i)=1+3i
Complex numbers are often drawn on an Argand diagram: the real part of the number is plotted on the horizontal axis and the imaginary part on the vertical axis. The diagram to the left shows the numbers of ten digits or less in base -1+i on an Argand diagram. The points form an order 10 dragon curve! In fact, plotting numbers of \(n\) digits or less will draw an order \(n\) dragon curve.
Numbers in base -1+i of ten digits or less plotted on an Argand diagram.
Brilliantly, we may now use known properties of dragon curves to discover properties of base -1+i. A level \(\infty\) dragon curve covers the entire plane without intersecting itself: therefore every Gaussian integer (a number of the form \(a+\text{i} b\) where \(a\) and \(b\) are integers) has a unique representation in base -1+i. The endpoints of dragon curves lie on a logarithmic spiral: therefore numbers of the form \((-1+\text{i})^n\), where \(n\) is an integer, lie on a logarithmic spiral in the complex plane.
If you'd like to play with some dragon curves, you can download the Python code used to make the pictures here.

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Dragon Curves
The Mathematical Games of Martin Gardner
MENACE
Is MEDUSA the New BODMAS?

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 2017-02-25 

The Importance of Estimation Error

Recently, I've noticed a few great examples of misleading uses of numbers in news articles.
On 15 Feb, BBC News published a breaking news article with the headline "UK unemployment falls by 7,000 to 1.6m". This fall of 7,000 sounds big; but when compared to the total of 1.6m, it is insignificant. The change could more accurately be described as a fall from 1.6m to 1.6m.
But there is a greater problem with this figure. In the original Office of National Statistics (ONS) report, the fall of 7,000 was accompanied by a 95% confidence interval of ±80,000. When calculating figures about large populations (such as unemployment levels), it is impossible to ask every person in the UK whether they are employed or not. Instead, data is gathered from a sample and this is used to estimate the total number. The 95% confidence interval gives an idea of the accuracy of this estimation: 95% of the time, the true number will lie of the confidence interval. Therefore, we can think of the 95% confidence interval as being a range in which the figure lies (although this is not true, it is a helpful way to think about it).
Compared to the size of its confidence interval (±80,000), the fall of 7,000 is almost indistinguishable from zero. This means that it cannot be said with any confidence whether the unemployment level rose or fell. This is demonstrated in the following diagram.
A fall of 7,000 ± 80,000. The orange line shows no change.
To be fair to the BBC, the headline of the article changed to "UK wage growth outpaces inflation" once the article was upgraded from breaking news to a complete article, and a mention of the lack of confidence in the change was added.
On 23 Feb, I noticed another BBC News with misleading figures: Net migration to UK falls by 49,000. This 49,000 is the difference between 322,000 (net migration for the year ending 2015) and 273,000 (net migration for the year ending 2016). However both these figures are estimates: in the original ONS report, they were placed in 95% confidence intervals of ±37,000 and ±41,000 respectively. As can be seen in the diagram below, there is a significant portion where these intervals overlap, so it cannot be said with any confidence whether or not net immigration actually fell.
Net migration in 2014-15 and 2015-16.
Perhaps the blame for this questionable figure lies with the ONS, as it appeared prominently in their report while the discussion of its accuracy was fairly well hidden. Although I can't shift all blame from the journalists: they should really be investigating the quality of these figures, however well advertised their accuracy is.
Both articles criticised here appeared on BBC News. This is not due to the BBC being especially bad with figures, but simply due to the fact that I spend more time reading news on the BBC than in other places, so noticed these figures there. I quick Google search reveals that the unemployment figure was also reported, with little to no discussion of accuracy, by The Guardian, the Financial Times, and Sky News.

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The End of Coins of Constant Width
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 2017-01-13 

Is MEDUSA the New BODMAS?

I wrote this post with, and after much discussion with Adam Townsend. It also appeared on the Chalkdust Magazine blog.
Recently, Colin "IceCol" Beveridge blogged about something that's been irking him for a while: those annoying social media posts that tell you to work out a sum, such as \(3-3\times6+2\), and state that only $n$% of people will get it right (where \(n\) is quite small). Or as he calls it "fake maths".
A classic example of "fake maths".
This got me thinking about everyone's least favourite primary school acronym: BODMAS (sometimes known as BIDMAS, or PEMDAS if you're American). As I'm sure you've been trying to forget, BODMAS stands for "Brackets, (to the power) Of, Division, Multiplication, Addition, Subtraction" and tells you in which order the operations should be performed.
Now, I agree that we all need to do operations in the same order (just imagine trying to explain your working out to someone who uses BADSOM!) but BODMAS isn't the order mathematicians use. It's simply wrong. Take the sum \(4-3+1\) as an example. Anyone can tell you that the answer is 2. But BODMAS begs to differ: addition comes first, giving 0!
The problem here is that in reality, we treat addition and subtraction as equally important, so sums involving just these two operations are calculated from left-to-right. This caveat is quite a lot more to remember on top of BODMAS, but there's actually no need: Doing all the subtractions before additions will always give you the same answer as going from left-to-right. The same applies to division and multiplication, but luckily these two are in the correct order already in BODMAS (but no luck if you're using PEMDAS).
So instead of BODMAS, we should be using BODMSA. But that's unpronounceable, so instead we suggest that from now on you use MEDUSA. That's right, MEDUSA:
This is big news. MEDUSA vs BODMAS could be this year's pi vs tau... Although it's not actually the biggest issue when considering sums like \(3-3\times6+2\).
The real problem with \(3-3\times6+2\) is that it is written in a purposefully confusing and ambiguous order. Compare the following sums:
$$3-3\times6+2$$ $$3+2-3\times6$$ $$3+2-(3\times6)$$
In the latter two, it is much harder to make a mistake in the order of operations, because the correct order is much closer to normal left-to-right reading order, helping the reader to avoid common mistakes. Good mathematics is about good communication, not tricking people. This is why questions like this are "fake maths": real mathematicians would never ask them. If we take the time to write clearly, then I bet more than \(n\)% of people will be able get the correct answer.

Similar Posts

Dragon Curves II
The Mathematical Games of Martin Gardner
How to Kick a Conversion
How Much Will I Win on the New National Lottery?

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© Matthew Scroggs 2017