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## 2019/01/22 - Rope Around the Earth - Refined

As we previously mentioned, the Rope Around the Earth problem is a lovely one, and really, everyone should know it. But even if you do, there's still a nice added extra. In this post we'll have a look at that.

If you do know the problem then by all means skim forward to the section marked "Refining our answer", but for those who haven't met it, or who can't quite remember the details, let me state it carefully.

## The Rope Around the Earth

Someone discovers that the 40 million metres of rope either isn't 40 million metres, or that the Great Circle around which they stretched it isn't exactly 40 million metres (there's a surprise!), because when they stretched the rope around the world, the rope was six metres longer than needed.

Well, as in other puzzles involving 40 million metres of rope, they didn't want to cut it, because it was expensive, and tying a knot just seems too inelegant, so they fuse the ends together and think.

They come to the conclusion that making the circle bigger would fix the problem and take up the slack, so they make plans to prop up the rope, same height everywhere, (never mind that it would sag), and that would take up the excess.

So how high do the props need to be? There will be a lot of them, because they will need to be closely spaced, and we don't want to make them the wrong size.

So how high will the rope be?

• Atomic height?
• Paper thickness height?
• Hand thickness height?
• Could you crawl under it?
• Could you walk under it?

Spreading six metres of excess rope around 40 million metres circumference, just how much extra radius do we need?

If you've not met this before then I urge you most strongly to have a think about this, even if only for a minute or two. If you have met it before, make sure you can remember the answer, or work it out again. Remember, this version has six metres of extra rope, and is asking how high it needs to be propped.

## The solution

Whether you find this easy, hard, surprising, or obvious, depends a lot on your background, experience, and personality. I like it a lot, possibly because I met it a long time ago and regard it as an old friend. The simple answer is to say that the extra radius is $6/(2\pi)$, and since $\pi$ is roughly three, the answer is roughly one metre.

 I still find it simply wonderful that this result does not depend on $R$. The answer is the same whether you are encircling the Earth, the Moon, or a basketball. The algebra makes that clear, but for me, the calculus makes it even clearer. There we can see the $R$ vanish, leaving just the constant. Lovely.
You can get the answer with some simple algebra. Since $C=2\pi R$ we have that $C+6=2\pi(R+h)$, and solving that gives us our answer.

But there's an interesting bit of calculus here. Since we know that circumference is $2\pi$ times radius, so $C=2\pi R$, and we can differentiate with respect to $C$ to get

• $\frac{dR}{dC}=\frac{1}{2\pi}$.
While that seems overkill, it exactly says that the change in radius for every metre change in circumference is $\frac{1}{2\pi}$. The language of calculus gives us a way to think about this problem to make it clearer and a little less surprising.

There's a lovely refinement we can do on this estimated answer.

We got the answer of "about a metre" by assuming that $\pi$ is about $3$. But it's not, it's about $3.15$ (wait for it ...)

 This goes back to one of the results on the Elementary Estimates page where we observed that: $1/(1+x)\approx 1-x$
We can now observe that $3.15$ is bigger than $3$ by 5% (Exactly 5%, that's why I chose 3.15). So we actually want to divide by a number that's 5% bigger than $3$, and that means the answer will be about 5% smaller than 1 metre. In other words, assuming $\pi$ is roughly $3.15$ means our answer is about $0.95$ metres.

Let's get a calculator and do the actual divisions. We were dividing by $3.15$ and claiming it's roughly $0.95$, whereas the actual answer is:

• $3/3.15 = 0.95238...$

Well that's rather close.

If instead we want to divide by $3.14$ we can say that 3.14 is 0.01 less than 3.15, which is about $\frac{1}{3}$%, so we have to increase our answer by about 0.0033, giving an estimate of:

• $3/3.14\approx 0.9533...$

Using a calculator, the answer we get is:

• $3/3.14\approx 0.9554...$

But we're clearly into diminishing returns, and we might be better to approximate $\pi$ by $22/7$ and be done with it, or by $355/113$, which is even closer.

But using a calculator we get:

• $6/(2\pi)\approx 0.9549...$

I think we're close enough for Government work. Of course it's not exact, but for an estimate you can do in your head, getting two or three significant figures with almost no work is quite good.

Well, I liked it.

If you use $\pi\approx 22/7$ then we want to compute $3$ divided by that, so $21/22$.

Multiply numerator and denominator by 5 each to get $(105)/(110)$ and multiply again, this time by 9 to get $(945)/(990)$.

• $(945+9.45)/(999)$

or $954.45/999$

Make the approximation of adding 1 top and bottom, so we have $955.45/1000$, which is 0.95545.

Compare with $3/\pi$, which is $3.95493...$ and we're pretty close.

We can play the same game with $\pi\approx 355/113$ and compute (or approximate) $339/355$, but I think that's enough for now.

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