Wrapping Up Wrapping Up The Earth

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(Forgive the odd links, I've put this in a temporary position for people to have a look at, prior to moving it to its final resting place. As a result, the auto-linking is, well, "odd".)

We're coming to the end of this series. Our friend has purchased too much rope, too little rope, too much miraculous wrapping sheet, and now, finally, we come to the question of too little. But first, let's look at the different "rope" scenarios, and see what happens in each of the analogous "sheet" cases.

The Rope Around the Earth problem

We have 6 metres too much rope, and having joined it in a circle we proceed to prop it up the same height everywhere. How high will that be? It turns out to be about a metre, regardless of the size of the Earth!

That's a very surprising result for most people, but the sums are irrefutable. We also perform some "back of the envelope" calculations to refine the answer here:

The analogous version for wrapping the Earth and discovering that you have a million square metres left over gives us a height of just 6.25mm, and that does depend on the size of the Earth.

Sums: WrappingTheEarth

The Other Rope Around the Earth problem

Now instead of propping it up everywhere we prop it up in just one place with a tent pole. Turns out, that's quite a tall tent pole. The sums are here:

That does depend on the size of the Earth, but perhaps the fun part about this is that we can again do this on the back of an envelope, once you approach it the right way.

The analogous version with the wrapping is a bit more difficult, but it yields before too long, and gives a surprisingly elegant result that once again, just like the original Rope Around the Earth problem doesn't rely on the size of the Earth:

• TheOtherWrappingTheEarthProblem

The $(Other)^2$ Rope Around the Earth problem

OK, this one is silly, where the rope turns into a rigid ring that encircles the Earth, and we ask what happens. You can read about that here:

• TheRingOfSteel
The analogous "Wrapping" version has the Earth floating inside a rigid shell, and we ask what happens. That, however, is answered by The Shell Theorem, which we discuss in TheRingOfSteel, so we won't spend any more time on that.

The $(Other)^3$ Rope Around the Earth problem

The rope is now too short, so we use the ends and throw the loop over the Earth like a lasso or quoit, and ask how close it gets to the equator. The actual calculations are here:

• OtherOtherOtherRopeAroundTheEarth
We can ask for an equivalent with the wrapping problem, and the closest I can think of is to leave a gap around the equator and ask how wide it is. But that's genuinely trivial. We need an area of a million square metres, and the circumference is 40 million metres, so the answer is about an inch (in old money) or 25mm.

That doesn't seem very interesting.

The $(Other)^4$ Rope Around the Earth problem

But now we come to the remaining case, where for the rope version a hole is drilled through the Earth as a short cut to save the 6 metres, and we ask how deep that will be. You can see that here:

• RoundingUpTheRopes
So what is the equivalent for the wrapping problem? Surely now we cut off a cap, perhaps at the North Pole, and ask how deep that will have to be.

So let's do that.

Cutting off the cap

We are removing the cap, here shown at depth $d$ and radius $r$, and replacing it with the circle of radius $r$. So we need to know the area of the cap, the area of the circle, and the difference has to be our material deficit of a million square metres.

But we know from the ArchimedesHatBoxTheorem that the area of the cap will be $\pi(d^2+r^2)$, and the circle is just a circle, so its area is simply $\pi r^2$.

So the difference is just $\pi d^2$.

That seems a bit easy.

 Let's approximate $\frac{1}{\sqrt{pi}}$ by hand. Firstly, $\pi$ is 5% bigger than 3, and $1/\sqrt{3}$ is about 0.5774, so the answer is 2.5% smaller than 0.5774. That's one fortieth, so we divide 0.5774 by 40 to get 0.01443, and subtract, giving us 0.563... Alternatively, $\pi$ is close to $\sqrt{10}$. Now $\frac{1}{\sqrt{10}}=\frac{\sqrt{10}}{10}$, so $\frac{1}{\sqrt{10}}\approx 0.316$. Hence $1/\pi\approx 0.316$, although $\pi$ is a little smaller than $\sqrt{10}$, so maybe it's closer to 0.318. Now think about the square root of 0.32. We know that $(0.55)^2=0.3025$, so looking at the ratio $0.3025 : 0.318 : 0.36$ we can guess that $\sqrt{0.318}$ is between one and two fifths of the way from 0.55 to 0.6, perhaps 0.56 or 0.57. So our answer would be that $\frac{1}{\sqrt{pi}}\approx 0.565$.
But it's true, it works, and we get the answer that the depth of the cap cut has to be:

• $d = \sqrt{10^6/\pi}$
So $d$ is roughly 564 metres, which is the same as the answer we had for our previous answer. But even more, this time it really doesn't depend on the size of the Earth. The previous result only worked when the Earth was big enough to make angle $a$ small enough, but this is exactly true.

Which I find oddly satisfying.

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