How much tinsel do you need to decorate a Christmas tree?

Our idealised tree will be a cone with opening angle 2\varphi. The tinsel lies on the surface of the cone, and I’ll state that the curve described by the tinsel can be expressed parametrically by the equations

\bold{r}(\theta)=\begin{pmatrix}\alpha\theta\cos\theta \\ \alpha\theta\sin\theta \\ \beta\theta\end{pmatrix},\quad\,\alpha,\beta\in\left[0,\infty\right)

where the z axis coincides with the tree’s axis of rotational symmetry, and \bold{r}(0) corresponds to the tip of the tree. This vector describes the locus of a conical spiral, for which the distance between successive loops is constant.

To make sure this spiral lies on the surface of the cone, we have to make sure the constants \alpha and \beta are chosen appropriately. To do so, it’s useful to define the tinsel’s radial distance \rho from the tree’s axis of symmetry:



Then if \theta\to\theta+2\pi,


z\to z+2\pi\beta

So after wrapping around the tree once, the tinsel has moved outwards by an amount 2\pi\alpha and downwards by an amount 2\pi\beta. I hope you’ll agree that this means the angle made by the tinsel’s spiral path with the z axis satisfies


Hence for the tinsel to lie on the surface of the tree, we require that \alpha and \beta satisfy

\displaystyle \frac{\alpha}{\beta}=\tan\varphi

which we can also express as


tannenbaumWe can also introduce a second property of the tinsel, defined by

\displaystyle \lambda=2\pi\sqrt{\alpha^2+\beta^2}

This is the distance between successive loops, as measured along the surface of the tree. It’s a bit like the tinsel’s ‘wavelength’, which is why I’ve called the quantity \lambda.

For now we’ll continue to work with the parameters \alpha and \beta, and switch to the physical quantities \varphi and \lambda when it’s appropriate.


To measure the length of the tinsel, we use an integral of small line elements:

\displaystyle S=\int dS

The length of each element can be written in terms of the differentials dx, dy, dz using the equation


which can conveniently be expressed using the parameter \theta:

\displaystyle dS=\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2+\left(\frac{dz}{d\theta}\right)^2}d\theta

Using the equation for \bold{r}(\theta), we can work out the necessary derivatives:

\displaystyle \frac{dx}{d\theta}=\frac{d}{d\theta}(\alpha\theta\cos\theta)=\alpha\cos\theta-\alpha\theta\sin\theta

\displaystyle \frac{dy}{d\theta}=\frac{d}{d\theta}(\alpha\theta\sin\theta)=\alpha\sin\theta+\alpha\theta\cos\theta

\displaystyle \frac{dz}{d\theta}=\frac{d}{d\theta}(\beta\theta)=\beta

Substituting these in gives



\displaystyle dS=\sqrt{\alpha^2+\beta^2}\sqrt{1+\frac{\alpha^2}{\alpha^2+\beta^2}\theta^2}

Now would be a good time to change to the parameters we introduced before:

\displaystyle dS=\frac{1}{2\pi}\lambda\sqrt{1+\sin^2\varphi\,\theta^2}

So we need to solve the integral

\displaystyle S=\frac{1}{2\pi}\lambda\int\sqrt{1+\sin^2\varphi\,\theta^2}\,d\theta

By exploiting the identity

\cosh^2 y-\sinh^2 y=1

the integrand can be simplified using the substitution

\sinh y=\sin\varphi\,\theta

\cosh y\,dy=\sin\varphi\,d\theta

This gives

\displaystyle S=\frac{1}{2\pi}\lambda\csc\varphi\int\sqrt{1+\sinh^2y}\cosh y\,dy

\displaystyle S=\frac{1}{2\pi}\lambda\csc\varphi\int\cosh^2 y\,dy

The next few steps require the ‘double-angle’ formulae

\cosh 2y=2\cosh^2 y -1

\sinh 2y=2\sinh y\cosh y

allowing us to say

\displaystyle S=\frac{1}{4\pi}\lambda\csc\varphi\int \left(\cosh 2y+1\right)\,dy

\displaystyle S=\frac{1}{4\pi}\lambda\csc\varphi\left(\frac{1}{2}\sinh 2y+y\right)

\displaystyle S=\frac{1}{4\pi}\lambda\csc\varphi\left(\sinh y\cosh y+y\right)

\displaystyle S=\frac{1}{4\pi}\lambda\csc\varphi\left(\sin\varphi\,\theta\sqrt{1+\sin^2\varphi\,\theta^2}+\text{arcsinh}\left(\sin\varphi\,\theta\right)\right)

which is equivalent to

\displaystyle S(\theta|\lambda,\varphi)=\frac{1}{4\pi}\lambda\csc\varphi\left(\sin\varphi\,\theta\sqrt{1+\sin^2\varphi\,\theta^2}+\ln\left(\sin\varphi\,\theta+\sqrt{1+\sin^2\varphi\,\theta^2}\right)\right)


Let’s try a few values. For a Christmas tree with opening angle 35 degrees adorned with 5 turns of tinsel separated by a length of 40 cm, the total length of tinsel needed is

S(10\pi|0.4,17.5^\circ)\approx 10\,\text{m}

That’s just for a modest tree! Try some for yourself. For very long lengths of tinsel, the expression above can be approximated as

\displaystyle S(\theta|\lambda,\varphi)\approx\frac{1}{4\pi}\lambda\sin\varphi\,\theta^2

since the logarithmic term grows much more slowly than the quadratic term.

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