How many feet of rope to wrap a column

Question

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).

Answer 1

Some hints:

• What length of rope does it take to loop around the pipe once? (Call this $L_{loop}$.)
• How high up the pipe does this one loop cover? (Call this $h$.)
• How many loops will you need to cover the whole pipe? (Call this $N$. If $H$ is the height of the pipe, then $N = H/h.$)
• Then the length of rope is $L_{rope} = NL_{loop}.$

Answer 2

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

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

Answer 3

Wrapping a rope of diameter $d$ around a pipe of diameter $D$ is equivalent to wrapping a ribbon of width $d$ around a pipe of diameter $D+d$. For the latter problem, the length of ribbon, $L$, is whatever it takes to cover the total surface area of the pipe, hence

$$dL=\pi(D+d)h$$

where $h$ is the height of the pipe. Thus

$$L=\pi\left(1+{D\over d}\right)h$$

In the present problem, $D=2.55$, $d=.25$, and $h=105$ (all in inches), so $L\approx3694.5$ inches, or about $308$ feet.

(A caveat: You can't really cover all of the pipe with a helical rope or ribbon -- you'll have some uncovered potions to the two ends -- so you really can't get a mathematically exact answer.)

Answer 4

In 3 calculations, I got 3360 inches by assuming that the area of rope needed independent of wrapping or taking 16 (8” circumference/ 0.25” thick [diameter]) 8 foot pieces gluing them straight up the pole.

If I bought 10% more for shrinkage = 3396” !!!!!

Leo