Fun with Math time.

My mom gave me a roll of toilet paper to put it in the bathroom, and looking at it I immediately wondered about this: is it possible, through very simple math, to calculate (with small error) the total paper length of a toilet roll?

Writing down some math, I came to this study, which I share with you because there are some questions I have in mind, and because as someone rightly said: for every problem there always are at least 3 solutions.

I started by outlining the problem in a geometrical way, namely looking only at the essential: the roll from above, identifying the salient parameters:

enter image description here


$r = $ radius of internal circle, namely the paper tube circle;

$R = $ radius of the whole paper roll;

$b = R - r = $ "partial" radius, namely the difference of two radii as stated.

First Point

I treated the whole problem in the discrete way. [See the end of this question for more details about what does it mean]


In a discrete way, the problem asks for the total length of the rolled paper, so the easiest way is to treat the problem by thinking about the length as the sum of the whole circumferences starting by radius $r$ and ending with radius $R$. But how many circumferences are there?

Here is one of the main points, and then I thought about introducing a new essential parameter, namely the thickness of a single sheet. Notice that it's important to have to do with measurable quantities.

Calling $h$ the thickness of a single sheet, and knowing $b$ we can give an estimate of how many sheets $N$ are rolled:

$$N = \frac{R - r}{h} = \frac{b}{h}$$

Having to compute a sum, the total length $L$ is then:

$$L = 2\pi r + 2\pi (r + h) + 2\pi (r + 2h) + \cdots + 2\pi R$$

or better:

$$L = 2\pi (r + 0h) + 2\pi (r + h) + 2\pi (r + 2h) + \cdots + 2\pi (r + Nh)$$

In which obviously $2\pi (r + 0h) = 2\pi r$ and $2\pi(r + Nh) = 2\pi R$. Writing it as a sum (and calculating it) we get:

$$ \begin{align} L = \sum_{k = 0}^N\ 2\pi(r + kh) & = 2\pi r + 2\pi R + \sum_{k = 1}^{N-1}\ 2\pi(r + kh) \\\\ & = 2\pi r + 2\pi R + 2\pi \sum_{k = 1}^{N-1} r + 2\pi h \sum_{k = 1}^{N-1} k \\\\ & = 2\pi r + 2\pi R + 2\pi r(N-1) + 2\pi h\left(\frac{1}{2}N(N-1)\right) \\\\ & = 2\pi r N + 2\pi R + \pi hN^2 - \pi h N \end{align} $$

Using now: $N = \frac{b}{h}$; $R = b - a$ and $a = R - b$ (because $R$ is easily measurable), we arrive after little algebra to

$$\boxed{L = 4\pi b + 2\pi R\left(\frac{b}{h} - 1\right) - \pi b\left(1 + \frac{b}{h}\right)}$$

Small Example:

$h = 0.1$ mm; $R = 75$ mm; $b = 50$ mm thence $L = 157$ meters

which might fit.

Final Questions:

1) Could it be a good approximation?

2) What about the $\gamma$ factor? Namely the paper compression factor?

3) Could exist a similar calculation via integration over a spiral path? Because actually it's what it is: a spiral.

Thank you so much for the time spent for this maybe tedious maybe boring maybe funny question!

  • 92
    $\begingroup$ Somewhere on the package, it will tell you how many sheets there are in a roll. Measure one sheet, and multiply by the number of sheets. $\endgroup$ Commented Jan 30, 2016 at 23:05
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    $\begingroup$ My wife used to work at Sears in the fabric department and she asked me once if you could estimate the amount of fabric left on a roll by counting the number of layers of fabric on the roll. After some time, I gave her a formular. She said "Yeah, that's the formula we use." She already knew the answer. She just wanted to see if I could derive it. $\endgroup$ Commented Feb 1, 2016 at 8:15
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    $\begingroup$ @Gerry get out of here with your practical solutions. We're trying to math here. $\endgroup$
    – corsiKa
    Commented Feb 1, 2016 at 15:34
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    $\begingroup$ @avid19 not only is this a wonderful question, it's quite applicable. My Dad is the Vice-President and primary programmer for Papersoft (See papersoft.com) . They use this calculation to determine how much paper is left on the big giant rolls in warehouses. Very cool stuff $\endgroup$
    – Daniel
    Commented Feb 1, 2016 at 20:09
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    $\begingroup$ Reminds me of a good label I saw in the supermarket once. Like many shops, Sainsbury’s price labels include a small breakdown by quantity, so e.g. “£0.65 per kg”, or for discrete items like e.g. dishwasher tablets, “£0.11 per tblt”. The toilet paper — on this occasion — was marked as (something like, I forget the exact number) “£0.33 per 100 shts”. $\endgroup$ Commented Feb 2, 2016 at 10:12

8 Answers 8


The assumption that the layers are all cylindrical is a good first approximation.

The assumption that the layers form a logarithmic spiral is not a good assumption at all, because it supposes that the thickness of the paper at any point is proportional to its distance from the center. This seems to me to be quite absurd.

An alternative assumption is that the layers form an Archimedean spiral. This is slightly more realistic, since it says the paper has a uniform thickness from beginning to end. But this assumption is not a much more realistic than the assumption that all layers are cylindrical; in fact, in some ways it is less realistic.

Here's how a sheet of thickness $h$ actually wraps around a cylinder. First, we glue one side of the sheet (near the end of the sheet) to the surface of the cylinder. Then we start rotating the cylinder. As the cylinder rotates, it pulls the outstretched sheet around itself. Near the end of the first full rotation of the cylinder, the wrapping looks like this:

enter image description here

Notice that the sheet lies directly on the surface of the cylinder, that is, this part of the wrapped sheet is cylindrical.

At some angle of rotation, the glued end of the sheet hits the part of the sheet that is being wrapped. The point where the sheet is tangent to the cylinder at that time is the last point of contact with the cylinder; the sheet goes straight from that point to the point of contact with the glued end, and then proceeds to wrap in a cylindrical shape around the first layer of the wrapped sheet, like this:

enter image description here

As we continue rotating the cylinder, it takes up more and more layers of the sheet, each layer consisting of a cylindrical section going most of the way around the roll, followed by a flat section that joins this layer to the next layer. We end up with something like this:

enter image description here

Notice that I cut the sheet just at the point where it was about to enter another straight section. I claim (without proof) that this produces a local maximum in the ratio of the length of the wrapped sheet of paper to the greatest thickness of paper around the inner cylinder. The next local maximum (I claim) will occur at the corresponding point of the next wrap of the sheet.

The question now is what the thickness of each layer is. The inner surface of the cylindrical portion of each layer of the wrapped sheet has less area than the outer surface, but the portion of the original (unwrapped) sheet that was wound onto the roll to make this layer had equal area on both sides. So either the inner surface was somehow compressed, or the outer surface was stretched, or both.

I think the most realistic assumption is that both compression and stretching occurred. In reality, I would guess that the inner surface is compressed more than the outer surface is stretched, but I do not know what the most likely ratio of compression to stretching would be. It is simpler to assume that the two effects are equal. The length of the sheet used to make any part of one layer of the roll is therefore equal to the length of the surface midway between the inner and outer surfaces of that layer. For example, to wrap the first layer halfway around the central cylinder of radius $r$, we use a length $\pi\left(r + \frac h2\right)$ of the sheet of paper.

The reason this particularly simplifies our calculations is that the length of paper used in any part of the roll is simply the area of the cross-section of that part of the roll divided by the thickness of the paper.

The entire roll has inner radius $r$ and outer radius $R = r + nh$, where $n$ is the maximum number of layers at any point around the central cylinder. (In the figure, $n = 5$.) The blue lines are sides of a right triangle whose vertices are the center of the inner cylinder and the points where the first layer last touches the inner cylinder and first touches its own end. This triangle has hypotenuse $r + h$ and one leg is $r$, so the other leg (which is the length of the straight portion of the sheet) is $$ \sqrt{(r + h)^2 - r^2} = \sqrt{(2r + h)h}.$$ Each straight portion of each layer is connected to the next layer of paper by wrapping around either the point of contact with the glued end of the sheet (the first time) or around the shape made by wrapping the previous layer around this part of the layer below; this forms a segment of a cylinder between the red lines with center at the point of contact with the glued end. The angle between the red lines is the same as the angle of the blue triangle at the center of the cylinder, namely $$ \alpha = \arccos \frac{r}{r+h}.$$

Now let's add up all parts of the roll. We have an almost-complete hollow cylinder with inner radius $r$ and outer radius $R$, missing only a segment of angle $\alpha$. The cross-sectional area of this is $$ A_1 = \left(\pi - \frac{\alpha}{2} \right) (R^2 - r^2).$$ We have a rectangular prism whose cross-sectional area is the product of two of its sides, $$ A_2 = (R - r - h) \sqrt{(2r + h)h}.$$ Finally, we have a segment of a cylinder of radius $R - r - h$ (between the red lines) whose cross-sectional area is $$ A_3 = \frac{\alpha}{2} (R - r - h)^2.$$ Adding this up and dividing by $h$, the total length of the sheet comes to \begin{align} L &= \frac1h (A_1+A_2+A_3)\\ &= \frac1h \left(\pi - \frac{\alpha}{2} \right) (R^2 - r^2) + \frac1h (R - r - h) \sqrt{(2r + h)h} + \frac{\alpha}{2h} (R - r - h)^2. \end{align}

For $n$ layers on a roll, using the formula $R = r + nh$, we have $R - r = nh$, $R + r = 2r + nh$, $R^2 - r^2 = (R+r)(R-r) = (2r + nh)nh$, and $R - r - h = (n - 1)h$. The length then is \begin{align} L &= \left(\pi - \frac{\alpha}{2} \right) (2r + nh)n + (n - 1) \sqrt{(2r + h)h} + \frac{\alpha h}{2} (n - 1)^2\\ &= 2n\pi r + n^2\pi h + (n-1) \sqrt{(2r + h)h} - \left( n(r + h) - \frac h2 \right) \arccos \frac{r}{r+h}\\ &= n (R + r) \pi + (n-1) \sqrt{(2r + h)h} - \left( n(r + h) - \frac h2 \right) \arccos \frac{r}{r+h}. \end{align}

One notable difference between this estimate and some others (including the original) is that I assume there can be at most $(R-r)/h$ layers of paper over any part of the central cylinder, not $1 + (R-r)/h$ layers. The total length is the number of layers times $2\pi$ times the average radius, $(R + r)/2$, adjusted by the amount that is missing in the section of the roll that is only $n - 1$ sheets thick.

Things are not too much worse if we assume a different but uniform ratio of inner-compression to outer-stretching, provided that we keep the same paper thickness regardless of curvature; we just have to make an adjustment to the inner and outer radii of any cylindrical segment of the roll, which I think I'll leave as "an exercise for the reader." But this involves a change in volume of the sheet of paper. If we also keep the volume constant, we find that the sheet gets thicker or thinner depending on the ratio of stretch to compression and the curvature of the sheet. With constant volume, the length of paper in the main part of the roll (everywhere where we get the the full number of layers) is the same as in the estimate above, but the total length of the parts of the sheet that connect one layer to the next might change slightly.

Update: Per request, here are the results of applying the formula above to the input values given as an example in the question: $h=0.1$, $R=75$, and $r=25$ (inferred from $R-r=b=50$), all measured in millimeters.

Since $n = (R-r)/h$, we have $n = 500$. For a first approximation of the total length of paper, let's consider just the first term of the formula. This gives us $$ L_1 = n (R + r) \pi = 500 \cdot 100 \pi \approx 157079.63267949, $$ or about $157$ meters, the same as in the example in the question. The remaining two terms yield \begin{align} L - L_1 &= (n-1)\sqrt{(2r + h)h} - \left( n(r + h) - \frac h2 \right) \arccos\frac{r}{r+h} \\ &= 499\sqrt{50.1 \cdot 0.1} - (500(25.1) - 0.05)\arccos\frac{25}{25.1} \\ &\approx -3.72246774. \end{align} This is a very small correction, less than $2.4\times 10^{-5} L_1$. In reality (as opposed to my idealized model of constant-thickness constant-volume toilet paper), this "correction" is surely insignificant compared to the uncertainties of estimating the average thickness of the paper in each layer of a roll (not to mention any non-uniformity in how it is rolled by the manufacturing machinery).

We can also compare $\lvert L - L_1 \rvert$ to the amount of paper that would be missing if the paper in the "flat" segment of the roll were instead $n - 1$ layers following the curve of the rest of the paper. The angle $\alpha$ is about $0.089294$ radians (about $5.1162$ degrees), so if the missing layer were the innermost layer, its length would be $25.05 \alpha \approx 2.24$, and if it were the outermost layer it would be $74.95 \alpha \approx 6.69$ (in millimeters).

Just for amusement, I also tried expanding $L - L_1$ as a power series around $h = 0$ (with a little help from Wolfram Alpha). (To make $L - L_1$ a function of one variable $h$ with constants $R$ and $r$, make the substitution $n = (R - r)/h$.) This turns out to be a series of powers of $\sqrt h$ whose leading term is $$ -\frac{(R + 2r)\sqrt2}{3\sqrt r} \sqrt h. $$ Plugging in the values from the example, this evaluates to approximately $-3.7267799625$. If you really wanted the length of the idealized toilet roll to the nearest millimeter, but could tolerate an error of a few $\mu\mathrm m$ (for typical dimensions of a toilet roll), a suitable approximation would be $$ L \approx \frac{\pi (R^2 - r^2)}{h} - \frac{(R + 2r)\sqrt2}{3\sqrt r} \sqrt h. $$

  • 49
    $\begingroup$ If I could, I would up vote you 10 times and more for all the effort you put into this! That's awesome! $\endgroup$
    – Enrico M.
    Commented Jan 31, 2016 at 23:17
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    $\begingroup$ This is more effort than I expected for an internet post about toilet paper. $\endgroup$
    – corsiKa
    Commented Feb 1, 2016 at 15:35
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    $\begingroup$ @corsiKa I suspect I wouldn't have put in half as much effort if this had been about something more "serious" than toilet paper. Compare the answers based on methods to track someone's production and/or inventory of rolled sheet metal, which give the simple formula $\pi(R^2 - r^2)/h$. $\endgroup$
    – David K
    Commented Feb 1, 2016 at 17:41
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    $\begingroup$ @corsiKa You obviously haven't been to travel.SE then... $\endgroup$ Commented Feb 2, 2016 at 9:02
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    $\begingroup$ @MarkB I have added a few calculations at the end. $\endgroup$
    – David K
    Commented Feb 3, 2016 at 4:31

Your approximation is an excellent one. It is easier to do $$\begin{align} L = \sum_{k = 0}^N\ 2\pi(r + kh) & = 2\pi r(N+1) +2\pi h\frac 12N(N+1) \\&=2\pi r(N+1)+\pi(R-r)(N+1) \\&=\pi(R+r)(N+1) \\&=\pi(R+r)\left(\frac {R-r}h+1\right) \end{align}$$

You don't have to worry about paper compression because $h$ is the distance between two layers as the paper is wound. This may be greater than the measured thickness of a sheet, but that is not an issue. As you say, you have modeled the paper as a series of concentric cylinders. To make it a spiral, it is not far wrong (and a bit better than what you have) to consider the radius increasing linearly from start to end. The paper then will be the hypotenuse of a right triangle with one leg the $L$ you have calculated and the other $R-r$, so the length becomes $$L'=\sqrt {L^2+(R-r)^2}\approx L\left(1+\frac {R-r}{2L^2}\right)$$ It is a small correction.

  • 3
    $\begingroup$ Firstly, let me thank you for the time you spent in reading and writing that. Secondly: great hint/help!! Thank you! $\endgroup$
    – Enrico M.
    Commented Jan 30, 2016 at 22:25
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    $\begingroup$ I was going to write an answer, saw it coincides with that by CiaPan.In practice winding tension T should decrease with higher radius so that inner layers do not get compressed, as per equilibrium $ T\cdot r = T_b\cdot b =$ constant. $\endgroup$
    – Narasimham
    Commented Feb 4, 2016 at 0:36

I suppose that you can take the paper layers are strictly circular, not spiral. The difference between a circle length and a corresponding spiral loop is negligible.

Then the area between circles with radii $R$ and $r$ is a side area of the paper tape, which is its length times thickness: $$\pi(R^2-r^2) = Lh$$ hence $$L=\pi\frac{R^2-r^2}h$$

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    $\begingroup$ Best answer, in fact because the paper so quite malleable it is probably more appropriate to use this model than the more complicated models above. $\endgroup$ Commented Feb 4, 2016 at 17:58

Lets do the spiral version. Using your notation, a spiral joining circles of radiuses $r$ and $R$ with $N$ twists has the form $S(t)=(r+\frac{tb}{2\pi N})e^{i t}$, where $t\in[0,2\pi N]$

The length $L$ of the spiral is

$$\begin{align} L & = \int_{0}^{2\pi N}|S'(t)|dt \\ & = \int_{0}^{2\pi N}\Big|\frac{b}{2\pi N}e^{it}+(r+\frac{tb}{2\pi N})ie^{it}\Big|dt \\ & = \int_{0}^{2\pi N}\sqrt{\Big(\frac{b}{2\pi N}\Big)^2+\Big(r+\frac{tb}{2\pi N}\Big)^2}dt \\ & = \frac{b}{2\pi N}\int_0^{2\pi N}\sqrt{1+\Big(\frac{2\pi Nr}{b}+t\Big)^2}dt \\ & = \frac{b}{2\pi N}\int_{2\pi Nr/b}^{2\pi N(r/b+1)}\sqrt{1+t^2}dt \\ & = \frac{b}{4\pi N}\big(t\sqrt{1+t^2}+\ln(t+\sqrt{1+t^2})\big)\Big|_{2\pi Nr/b}^{2\pi N(r/b+1)} \end{align}$$

  • $\begingroup$ Consider an arithmetic spiral in polar co-ordinates $r=k \theta$ with fixed $k>0>$. For the length $L$ along the spiral we have $dL/d\theta=k \theta.$ $\endgroup$ Commented Jan 31, 2016 at 7:04
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    $\begingroup$ Can you add a plot or just say how much of a difference this approach results in to the solution given in the question? Would be interesting to see $\endgroup$ Commented Jan 31, 2016 at 15:31

Use cross-sectional volume to calculate.

Since the initial problem is measured as diameters and a spiral will not have a true final "diameter", we clearly have to use approximation.

We know that the volume of the cross-section of toilet paper is $V = \pi * (R^2 - r^2)$

If we laid out the toilet paper to its full length, the cross-section would be $V=L * T$

Setting these two equal to each other, we get $L*T = \pi * (R^2 - r^2)$

Reducing this to length, we get $L = \pi * (R^2 - r^2) / T$

I work in the metals industry, and this is how we calculate the length of a coil of metal, based on knowing its thickness, inner diameter and outer diameter. Your challenge with toilet paper is measuring the actual thickness. Unlike metal, toilet paper is often "quilted", which makes its thickness hard to measure.

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    $\begingroup$ This is the same as CiaPan’s answer, which was posted almost a day earlier than yours. $\endgroup$ Commented Feb 2, 2016 at 6:09
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    $\begingroup$ Yes, but if you just scanned the answers to see if anyone had mentioned using the volume of paper on the roll to come up with a solution, you wouldn't have realised that that what was CiaPan was doing. $\endgroup$
    – Steve Ives
    Commented Feb 2, 2016 at 14:30
  • 1
    $\begingroup$ Well, if the only think this answer strives to do is explain that, when CiaPan says "side area", he (she?) really means "volume", that should have been done with a comment — Keeta has the privilege to do that. $\endgroup$ Commented Feb 2, 2016 at 19:08

If you have access to precision scale take let's say 5 sheets of your toilet paper roll and weight that. Then measure the weight of whole roll. This way you will know how many sheets is in the roll. Take length of one sheet and multiply it to number of sheets in the roll.

I guess it goes to Physics but after reading previous answers I thought that it doesn't matter that much...

  • 4
    $\begingroup$ You would have to deduct the weight of the cardboard core before determining the number of sheets. $\endgroup$
    – TripeHound
    Commented Feb 2, 2016 at 10:47
  • $\begingroup$ @TripeHound oh, I forgot about that... I guess if you want want to be precise with my method you will have to find one :) $\endgroup$
    – lowtech
    Commented Feb 2, 2016 at 16:18
  • 2
    $\begingroup$ And here, ladies and gentlemen, @BrockAdams provides a real-life example of equating film to toilet paper and deriving a meaningful result. Let this be a lesson to all of you unenlightened physicists and mathematicians. $\endgroup$ Commented Feb 3, 2016 at 23:54
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    $\begingroup$ @TripeHound Ah, but you already have an empty cardboard core to weigh to determine the tare weight. Recall that OP was replacing a roll. Thus, there would be an empty core from the previous roll. $\endgroup$ Commented Feb 4, 2016 at 8:18
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    $\begingroup$ @DavidConrad I wasn't doubting the availability of an empty core (providing you've bought the same brand...), only pointing out that you would need to deduct its weight from that of the whole roll (+its core) before dividing by the weight of the five sheets. $\endgroup$
    – TripeHound
    Commented Feb 4, 2016 at 8:50

The answers above appear to be rather exact, and equally difficult to calculate.

If you want a more simple, and possibly less accurate formula, what about

\begin{align} L &= (\pi * (R^2 – r^2) ) / T &\text{where}\\ L &= \text{length}\\ R &= \text{outside radius}\\ r &= \text{inside radius}\\ T &= \text{average thickness} \end{align}

$T$ can be computed as \begin{align} T &= (R - r)/W &\text{where}\\ W &= \text{wraps, turns, or layers on the roll} \end{align}

So, $$L = (\pi * (R^2 – r^2) ) / ((R - r) / W)$$

That could probably be reduced some, but at the moment I'm too tired to work that part out.

Parts of this came from here: http://www.koolshiz.com/engineering/roll-of-material-calculation.html


If there are $N$ layers, the thickness is $h=(R-r)/N$.

For cylindrical layers, the lengths increase linearly with layer number, so we can take the average circumference $2\pi\bar r$, where $r= (r+R)/2$, times $N$: $$ L = 2\pi \bar r N $$

If we have an Archimedean spiral, the radius increases linearly with azimuthal angle. Thus, we can take the average radius $\bar r$ multiplied by the total angle $2\pi N$, and again: $$ L = 2\pi N \bar r . $$


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