# Garland

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:

$\rho(\theta)=\sqrt{x^2(\theta)+y^2(\theta)}$

$\rho(\theta)=\alpha\theta$

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

$\rho\to\rho+2\pi\alpha$

$z\to z+2\pi\beta$

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

$\displaystyle\tan\varphi'=\frac{2\pi\alpha}{2\pi\beta}$

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

$\displaystyle\frac{\alpha}{\sqrt{\alpha^2+\beta^2}}=\sin\varphi$

We 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.

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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

$dS=\sqrt{(dx)^2+(dy)^2+(dz)^2}$

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

$dS=\sqrt{(\alpha\cos\theta-\alpha\theta\sin\theta)^2+(\alpha\sin\theta+\alpha\theta\cos\theta)^2+(\beta)^2}$

$dS=\sqrt{\alpha^2+\beta^2+\alpha^2\theta^2}$

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

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

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

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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.