# Folding a piece of paper in half

A few weeks ago, I asked the following question:

“If you fold a (humungous) piece of paper in half 50 times, how thick would the pile be? 1cm, 1m, 1km, 1,000km, 1,000,000km…?”

Here I discuss the possibilities of folding paper in half 50, 25, 15 and then 9 times.

## 50 folds

### How thick?

With each fold, the thickness doubles, from 1 to 2 to 4 to 8 sheets thick and so on. After 50 folds, the pile would be 2^{50} folds thick. Now, knowing 2^{10} = 1024 ≈ 1000 = 10^{3}, and using a year 9 index rule *(a ^{b})^{c} = a^{bc}*, we get

2^{50} = (2^{10})^{5} ≈ (10^{3})^{5} = 10^{15}

Typical printer paper is 0.1mm thick, so the thickness of the pile would be

10^{15} x 0.1mm

= 10^{14}mm

= 10^{11} x 10^{3}mm

= 10^{11}m

= 10^{8} x 10^{3}km

= 10^{8}km

= 100,000,000km

This is about two thirds of the distance from the earth to the sun!

### How heavy?

Then I started wondering how heavy such a piece of paper would be.

Knowing the thickness of the paper, we then need its length, width and density.

I revisited a link to a report that in 2002, Britney Gallivan, a high school student in the US, had folded a single piece of paper in half 12 times. (Note added 21 March 2014: The record is now 13 folds. See here for details.)

She also derived equations for the minimum length of paper necessary to fold a piece of paper *n* times, either in alternate directions (N-S, then E-W, then N-S etc) or in a single direction (N-S, N-S etc) – as she did herself, using a (very long) toilet roll.

Let’s work out the weight of paper needed for single direction folding a piece of paper 50 times…

If *t* is the thickness of the paper, and *n* the number of folds, one of Britney’s formulas tells us that the length is

* L = **π t 4 ^{n}/6
*(I have simplified the latter formula a little, in a way that makes no difference for large

*n*. See the above article for details.)

For 50 folds, these equations mean that

* L **≈ 7×10 ^{25}m*

(This is more than 10% of the width of the universe!)

The pile should not be too narrow, otherwise it would buckle and crumple. Let’s say the width of the paper should be at least ¼ of the height, that is *2.5×10 ^{10}m*.

So the volume of paper is

* V = length x width x thickness
*

*≈ 7×10*×

^{25}m*2.5×10*

^{10}m x 10^{-4}m*≈ 2×10*

^{32}m^{3}Paper varies in density, but printer paper is typically 80gsm and (as noted above) is about 0.1mm thick, meaning a density of about 800 kg/m^{3}. So

* Mass = volume x density
*

*≈ 1.6×10*

^{35}kgThis is about 800 times the mass of our sun. The upper limit for stellar size is believed to be about 600 times the mass of the sun, and then only for one formed without elements heavier than helium.

Paper is just under 50% carbon, so the mass of carbon would be about 0.5x*1.6×10 ^{35} kg = *

*8×10*, still 300 times the mass of the sun. This piece of paper would be the heaviest known concentration of carbon in the entire universe. So this pile of paper would indeed crumple – it would collapse gravitationally under its own weight and become a supergiant sun, but be surrounded by a gas nebula uniquely rich in carbon.

^{34}kgSo, I should have asked whether it is possible, even in theory, to fold a piece of paper in half 50 times – clearly not.

Incidentally, the term ‘weight’ turns out to be a wildly inappropriate term here, as weight is normally used to mean the pull of the earth on the object. This object would have a mass about 25 billion times that of the earth. I should have said ‘mass’.

## 25 folds

Let’s step back a little, and make our calculation only totally unrealistic – rather than completely fantastical.

Repeating the above calculations for 25 folds produces

length = 5.9×10^{10} m = 59 million km

pile thickness = 3.6 km

width = 840 m (i.e. ¼ of the thickness)

mass = 4×10^{12} kg

In 2009, the global production of paper was 377 million tonnes = 3.77×10^{11}kg.

If all paper produced globally for a decade was made into one piece of paper for this magnificent obsession, then we could fold this piece of paper 25 times. Of course there would be some non-trivial issues still to deal with:

- The earth is 40075 km around at the equator. The paper would be so long it would wrap around the equator 1471 times.

“I want all 840 of you to stand a metre apart from each other. Now, move towards me, then keep going round the earth 736 times, and stop when you get back to me.”

“Great. Now, I’ll stay here, where the ends meet. You make sure the fold is neat, then pick the paper up there and go 378 times around the earth.” - The last fold would mean taking a pile of paper 1.8km thick and about 3.6 km long and lifting it up 1.8km over the other half of the paper.
- And it must not fall in the water.

I think it’s fair to say it’s not going to happen.

## 15 folds

Let’s try for something which just might be feasible.

Repeating the above calculations for 15 folds produces

length = 56 km

pile thickness = 3.3 m

width = 82 cm (i.e. ¼ of the thickness)

mass = 3.7 tonnes

A jumbo roll, as produced in a paper mill, weighs 15 tonnes, and is about 6 metres in width and over 30 km in length. So we would

- cut off a quarter of the roll to make a 1.64m strip 30 km in length, weighing 3.7 tonnes.
- fold it lengthways in half once
- fold it in half 14 more times end over end as normal

There would also be significant issues with allowing for suitable amounts of slack so the paper can fold without scrunching up or, worse, tearing.

It might be possible, but would be an industrial size undertaking.

## The humble toilet roll

So what could we do ourselves? Even we can debunk the supposed limit of seven folds.

A toilet roll may be 20 to 30 metres long, so you’ll need 10 to 15 meters to lay out the roll, assuming you immediately fold it double.

I was able to fold a normal toilet roll 9 times, in a scaled down version of what Britney did. Her formula says we need about 14 metres.

Do try this at home.

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