Written by Colin+ in Uncategorized.

*This is based on a puzzle I heard from @colinthemathmo, who wrote it up here; he heard it from @DavidB52s, and there the trail goes cold.*

The Mathematical Ninja lay awake, toes itching. This generally meant that a mission was in the offing. Awake or dreaming? Unclear. But the thought had implanted in the Ninja brain: pull a rope taut around the equator, add six metres to one end and tie it to the other. Now lift the rope upwards until it’s taut. How high do you have to go?

Eyes closed. The rope will follow the equator most of the way around, then take off in a straight line. Let’s call the point it takes off from on one side T, the centre of the earth O and the point where the pole is P. We’ll put P’ at the top of the pole, and call angle TOP $\theta$.

The length of the straight bit of rope (on one side) is $R \tan(\theta)$, where $R$ is the radius of the earth. 6.4 million metres, the Mathematical Ninja knew somehow. Usually, something so obscenely… *tangible* wouldn’t stick.

But that length of rope would *also* be $R\theta + 3$, because if it laid ‘flat’ along the arc of a great circle, it’d reach three metres beyond $T$.

So (the Mathematical Ninja thought), we have our first equation: $R\tan(\theta) = R\theta + 3$, or $R(\tan(\theta) - \theta) = 3$.

P’, it turns out, is $R\sec(\theta)$ from O, which means the pole is $R(\sec(\theta)-1)$ tall.

If you’ve been keeping up with the new A-level spec (or, you know, just happen to know some useful maths approximations), you would know that $\tan(\theta) \approx \theta$.

This is entirely useless here: this would tell us that $0=3$, which is palpably untrue, and would give the Ninja the sort of nightmares **you** would never wake up from.

So we need to be cleverer. The first few derivatives of $f(\theta) = \tan(\theta)$ are $f'(\theta) = \sec^2(\theta)$, $f' '(\theta) = 2\sec^2(\theta)\tan(\theta)$ and $f ' ' '(\theta) = 2\sec^4(\theta) + 4\sec^2(\theta)\tan^2(\theta)$. Evaluated at 0, these work out as $f'(0) = 1$, $f''(0) = 0$ and $f'''(0) = 2$, so the Maclaurin series is $f(\theta) \approx \theta + \frac{1}{3}\theta^3$, ignoring $\theta^5$ terms and smaller.

That’s good! It tells us $\frac{1}{3}\theta^3 \approx \frac{3}{R}$, or $\theta^3 \approx \frac{9}{6,400,000}$. That’s an absurd thing to cube root in your head… isn’t it?

“Hold my ginger beer,” slurred the Mathematical Ninja. “It’s the cube root of 90, divided by 400. And you can *zzz* work out the cube root of 90 using logs. $\ln(9)$ is 2.2 or so. $\ln(10)$ is 2.3, so $\ln(90)$ is 4.5. Its cube root is $e^{1.5}$, which is 4.5. *snorfle*”

So we have $\theta \approx \frac{4.5}{400}$…

“I HEARD THAT”

… I mean $\theta \approx \frac{9}{800}$. Now what?

The height of the pole is $R(\sec(\theta) - 1)$, which could be a bit of a rabbit-hole - it’s tempting to think of it as $R\left( \frac{1 - \cos(\theta)}{\cos(\theta)}\right)$ and try to approximate from there.

“You’d be better to say that $\sec(\theta) \approx \frac{2}{2-\theta^2}$ and use a conjugate trick. *zzz*”

So you get $\frac{2(2+\theta^2)}{4 - \theta^4}$, which works out to $1 + \frac{1}{2}\theta^2$ - and the height of the pole is $\frac{1}{2}R\theta^2$.

“So that’s $0.5 \times 6.4 \times 10^6 \times \frac{81}{64 \times 10^4}$ *grunt*”

That’s right, sensei. Or $0.5 \times 10 \times 81$, which is a smidge more than 400 metres.

This is, it turns out, broadly parallel to @robjlow’s method mentioned in Colin’s post - but I thought it different enough (and first-principles enough) to write it up separately.

Now, if you’ll excuse me. The Mathematical Ninja is demanding coffee, a helicopter and a 400-metre pole. Be right back.

* Edited 2018-02-25 to fix some LaTeX and some labelling, and edit a link. Thanks to @colinthemathmo.