How many feet of rope to wrap a column

A heating pipe in my bathroom measures 105" in height. It is 8" in circumference (so about 2.55" diameter). I want to wrap it with a 1/4" thick rope. How many feet should I buy?

(All measurements in inches).

Solution 1:

The rope follows a helical path. Let's begin with the equation of a helix:

$$\begin{align*} x &= a\cos \theta, \\ y &= a\sin \theta, \\ z &= b\theta. \end{align*}$$

We now compute $a$ and $b$.

For $a$, we must consider the thickness of the rope in addition to the radius of the pipe! A rope is measured end-to-end, and when you wrap it, the inner radius compresses, while the outer radius stretches. Let's assume the neutral-stress point is at the center of the rope. Therefore, $a = r_{\textrm{pipe}} + r_{\textrm{rope}}$. Note that $r_{\textrm{rope}}$ is, of course, one-half the rope's thickness.

Now, in the equations above, the rise of the helix in one turn is given by $2\pi b$. Since we want to the rope to rise exactly one rope-thickness in a single turn, we'll compute this as: $$2\pi b = 2r_{\textrm{rope}} \implies b = \frac{r_{\textrm{rope}}}{\pi}.$$

Next, we wish to solve for the total number of turns we need, given this information. Let the length of the pipe be $H$; then, the number of turns is going to simply be given by $n=\frac{H}{2r_{\textrm{rope}}}$. Equivalently, this gives us $2\pi n = \pi \frac{H}{r_{\textrm{rope}}}$ radians.

Using the formula for the length of a parameterized curve, we will compute the total length of the rope:

$$L = \int_0^{\pi \frac{H}{r_{\textrm{rope}}}} \sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2+\left(\frac{dz}{d\theta}\right)^2}\ d\theta.$$

This simplifies:

$$L = \int_0^{\pi \frac{H}{r_{\textrm{rope}}}} \color{red}{\sqrt{a^2+\left(\frac{r_{\textrm{rope}}}{\pi}\right)^2}}\ d\theta.$$

The red term is constant, so we simply have

$$L = \sqrt{\left(r_{\textrm{pipe}}+r_{\textrm{rope}}\right)^2+\left(\frac{r_{\textrm{rope}}}{\pi}\right)^2}\left(\pi \frac{H}{r_{\textrm{rope}}}\right).$$