# Understanding a rope around the world

This is a slight break from the norm as it is something that has been flying around my brain for the past few days/decades.

The problem

Years ago (I must have been about 12 or 13) I was at a dinner party with the family and someone mentioned this mathematical problem.

For anyone not familiar it goes something along the lines of...

Take a length of rope and form it into a circle around a basketball.
Now add around 60cm to the rope and form it into a new circle.

Now...

Take a length of rope and wrap it around the equator of the earth. (assuming earth is a perfect sphere).
Now add the same 60cm to the rope and form it into a new circle.
Amazingly, mind-blowingly and converse to anything your brain might want to believe this new circle will now float 10cm above the earth all the way around.

All you've added is 60cm of length to a piece of rope that's about 40,000km long and it adds 10cm to the radius.

Year after year my brain has wandered back to this problem and always refuses to believe it. Even when I did the maths behind it and proved it to myself I still struggled to comprehend.

The maths...

Let $r$ be the radius of initial circle. $\displaystyle r=\frac{C}{2\pi}$

Let $r'$ be the radius of the new circle.
Let $x$ be the length added to the circumference. $\displaystyle r'=\frac{C+x}{2\pi} \\ r'=\frac{C}{2\pi}+\frac{x}{2\pi} \\ \therefore r' = r+\frac{x}{2\pi}$

In fact, plugging in the number you get an increase in radius of around 9.5493cm.

I did this years ago and still I couldn't wrap my brain around it.

Even if the original circle wraps around the edge of the universe it will still be 9.5493cm bigger in all directions!

Making my brain understand it

So, I started thinking about how to make it understandable enough for my brain to accept this obvious fact.

I wanted to see where each part of that 60cm of additional circumference length went. If I could picture this it would make the problem easier to see.

So I decided to explore what happens to a quarter of the circle when I add 15cm (one quarter of 60cm) in various different ways.

If I concentrate on the change in $h$ in the above diagram I might be able to understand what's going on.

So, initially I added all 15cm of (straight line) length at an angle of $0^{\circ}$ to the horizontal radius of the quarter circle.

Obviously in this we can see that the new height is r (the old height) plus the 15 cm we just added.

Now, let's do this again but instead I'll do it twice.

Now, how much height did we add? Well obviously at the bottom it's 7.5 cm. But at the half way mark? Let's enlarge this on its own. I have highlighted the added section in orange.

In this instance we let $\Theta = \frac{\pi}{4}rads$.

So, what is $\delta h$?

Well, we have a right angle triangle. We know the length of the hypotenuse, we know an angle and we want to get the length of the adjacent edge. So... $\displaystyle \cos\Theta = \frac{\delta h}{7.5}\\ 7.5\cos\Theta = \delta h$

Plugging in the values we get... $\delta h \approx 5.3cm$

So the total increase in height is... $\displaystyle 7.5\cos0 + 7.5\cos\left(\frac{\pi}{4}\right)\\ \to 7.5\left(\cos0 + \cos\left(\frac{\pi}{4}\right)\right)$
The first line can be through of as a triangle with angle 0.

So we get an increase of around 12.8033cm.

We can do this again...

We can see from this that, using the same geometry as we did previously, the increase in height is... $\displaystyle 5\left(\cos0 + \cos\left(\frac{\pi}{6}\right) + \cos\left(\frac{\pi}{3}\right)\right)$

To simplify this we could write it as a sum... $\displaystyle 5 \times \sum_{n=1}^{p=3}\cos\left((n-1)\frac{\pi}{2p}\right)$

To explain the above...

p is the number of sections added (3)
5 is the length of each section added to the line. In the is case 15 / 3 (the number of measurements).
n-1 because we want to start at zero radians.

We can now use this to work out the total change in height of around 11.83 cm.

Wolfram Alpha result

So we continue using this same method with more and more sections (triangles) and calculate the change in height each time. $p=1 \to \delta h=15\\ p=2 \to \delta h\approx 12.8033\\ p=3 \to \delta h\approx 11.8301\\ p=4 \to \delta h\approx 11.3013\\ p=5 \to \delta h\approx 10.9706\\ p=6 \to \delta h\approx 10.7447\\ p=7 \to \delta h\approx 10.5806$

In fact, we can plot this as a graph...

From here we can clearly see that the graph asymptotes towards a value of around 9.5cm... which if we look above in the post is exactly what we were expecting. The actual value of $\frac{60}{2\pi}$ which is around 9.5493cm.

Something things to consider

Yes, I was only looking at the change in height of quarter of the circle. But the problem is symmetrical. Not only is the change in height is translated to each of the other 3 quarters. But, dues to the symmetry of the cos and sin functions the increase in height is matched at every measurement by the "opposite" increase in width. i.e. for every triangle that adds a height of x there is an opposite triangle that adds a width of x and vice versa.

I have not made any assumptions about the size of the circle I have been adding sections to. The only thing I have used is the angle at which line segment is added. Therefore this works for a circle of any size.

Working through it this way I got to a point when I realised the obviousness of this whole problem.

Take this picture. I have done the first step of adding 15cm to each quarter circle. It is fairly obvious from this that by adding 4 sections of length 15 cm like this the point that is $90^{\circ}$ around from it will be 15cm further away. This can be comprehended even if the circle is a million light years in radius.

Lastly, I am pretty sure I'm just a couple of steps away from equating my summation using the circumference to the equation $C=2\pi r$ but I'll leave that for another day.

Conclusion

Anyway, I was quite pleased with how my calculations turned out and the little formula I came up with. It was nice to get my head into some proper analytical maths and for it to work so well after so many years off.

Who knows, I might make this a regular thing.

1. Madan
2. oliverfoggin