The maths of the decathlon

While I'm quite steadfastly refusing to be swept along with either Olympics fever or anti-games grumbling, I stumbled across the Wikipedia page for the decathlon in a moment of nostalgia for my 1980s childhood tapping the keys of the ZX Spectrum - a system that was kind of like an XBOX, but obviously way better in every respect. And Daley Thompson's Decathlon was the finest game around.

I'd never really thought about how you scored the decathlon before - it's a two-day discipline that encompasses ten of the major athletics events (the 100 metres, long jump, shot put, high jump, and 400m on day 1, and the 110m hurdles, discus, pole vault, javelin and 1500m on the second), requiring good decathletes to be pretty good at everything. It turns out that there's a formula for working out the points in each event: roughly speaking, it's a power law based on how much better you do than a fairly rubbish benchmark.

More precisely, for the track events, it's $A(B - P)^C$, where $P$ is your score (e.g. 14.4 seconds) and $A$, $B$ and $C$ vary depending on the event. For the field events, it's similar: $A(P - B)^C$. The reason it's the other way around is that in the track events, a high number is a bad score (if you took 90 seconds to run the 100m, you wouldn't expect many points!) and in the field events, a low number (like a javelin throw of 10cm) is bad. Then, once you've worked out the score, you round it down.

Event A B C
100 m 25.4347 18 1.81
Long jump 0.14354 220 1.4
Shot put 51.39 1.5 1.05
High jump 0.8465 75 1.42
400 m 1.53775 82 1.81
110 m hurdles 5.74352 28.5 1.92
Discus throw 12.91 4 1.1
Pole vault 0.2797 100 1.35
Javelin throw 10.14 7 1.08
1500 m 0.03768 480 1.85

Table from Wikipedia under Creative Commons.

Finding the decathlon performance you need

This means you can work out the performance you need to get a particular number of points in an event. To break the world record, you'd need to score a total of more than 9026 points, so let's say we want to score 903 in the pole vault, always my favourite event because a) it's "how high can you jump if you have a stick to help you" and b) I like the name Sergei Bubka.

First thing is to set up the equation: it's a field event so we need $A(P - B)^C$. Looking in the table, A = 0.2797, B = 100 and C = 1.35. That gives us:

$$903 = 0.2797(P - 100)^{1.35}$$

We don't even need logs! Divide both sides by 0.2797: $3228.46 = (P - 100)^{1.35}$.

How do you get rid of a power of 1.35? Easy, just put both sides to the power of $\frac{1}{1.35}$ to cancel it out. That gives you $397.43 = P - 100$, so you'd need $P$ to be 497.43 or better. For jumping events, $P$ is measured in centimetres, so you'd need to vault about 5 metres to be on for your world record attempt.

* Edited 2016-07-24 to fix formatting.


Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.


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