How much wire will fit on an industrial reel?

Question

I am trying to figure out how much cable would fit on an industrial reel. I have found a few calculators but nothing that will help me calculate this myself. If you have a reel with dimensions of 3m flange, 1.8m barrel and 1.5m traverse width, and a cable with a diameter of 12cm, how much cable can you spool on to it?

Answer 1

This is a packing problem. The reel has barrel diameter $b = 1.8\,{\rm m}$, flange height $f = 0.6\,{\rm m}$ and width $w = 1.5\,{\rm m}$. Each coil has diameter $d = 0.12 \,{\rm m}$

_{Note that flange height $f$ is given such that the outside diameter of the reel is 3 m.}

Above shown the general geometry with the position of an arbitrary wire indicated axially by $x$ and radially by $r$.

Each coil (circle in the diagram above) has length $\ell = 2 \pi r$ so if develop a relationship for each radius we can find the total length. Each stacked layer has a different radius $r$.

We also need to know how many coils fit along the width. Coils will tend to nestle in a hexagonal shape in layers. Each layer is offset by half a diameter from the previous one and two layers must fit in $w$.

The condition is $ (n + \tfrac{1}{2})\,d \leq w$ with solution $\boxed{n=12}$ coils side by side for each layer.

The total number of stakcs $m$ is whatever is needed to have the outer diameter of the top layer inside the flange diameter (unlike how I drew it above).

$$\begin{array}{r|c} \text{layer} & r \\ \hline 1 & r_1 = \tfrac{b+d}{2} \\ 2 & r_2 = \tfrac{b+d}{2} + d \tfrac{\sqrt{3}}{2} \\ 3 & r_{\rm i} = \tfrac{b+d}{2} + 2 d \tfrac{\sqrt{3}}{2} \\ \vdots & \\ i & r_{\rm i} = \tfrac{b+d}{2} + (i-1) d \tfrac{\sqrt{3}}{2} \\ \end{array}$$

The condition is $r_m + \tfrac{d}{2} \leq \tfrac{b}{2} + f$ with solution $\boxed{m = 5}$. Note that the outside diameter of the coils is ${\rm od} = 2 \left( \tfrac{b+d}{2} + (m-1) \frac{d \sqrt{3}}{2} \right) = 2.75\;{\rm m}$ the outside diameter of the flange is $b + 2 f = 3.0\;{\rm m}$ (per problem description)

The total length $L$ is thus:

$$ L = \sum_{i=1}^m 2\pi\,n \left( \tfrac{b+d}{2} + (i-1) d \tfrac{\sqrt{3}}{2} \right) = 2 \pi\, m\, n \left( \frac{b}{2} +d \frac{ (m-1) \sqrt{3} +2 }{4} \right) $$

$$ L = 2 \pi\, (5)\, (12) \left( \frac{1.8}{2} +0.12 \frac{ (5-1) \sqrt{3} +2 }{4} \right) $$

$$ \boxed{ L = 440.3\;{\rm m} } $$

Answer 2

Here is a quick approximation:

• We approximate every turn of the wire as a torus. The small diameter is the diameter of the wire. The large diameter is given by the position of the wire away from the axis of the reel.
• If the diameter of the wire is $d$ and the traverse width is $w$, on every layer you have $w/d$ tori.
• The length of each of these tori is $2\pi r$, with $r$ being the distance from the axis of the reel. So how much is $r$. In the first layer, $$r_1=r_b+\frac12 d$$Here $r_b$ is the radius of the barrel, or half of it's diameter.
• In the second layer the radius is increased by $d$, in the third layer by an additional $d$ and so on. So $$r_j=r_b+(j-\frac12)d$$
• So all we need to do is to sum up all these lengths. But how far do we need to sum? Until $r_max=r_f$, the radius of the flange.$$r_f=r_b+(j_{max}-\frac12)d$$or $$j_{max}=\frac{r_f-r_b}d+\frac12$$
• Total length of the wire is then $$L=\frac wd\sum_{j=1}^{j_{max}}(r_b+(j-\frac12)d)\\=\frac wd\left[j_{max}(r_b-\frac12d)+\frac12j_{max}(j_{max}+1)d\right]$$

Answer 3

Another approximate solution: The volume available to contain wire is $h\times(\pi R_2^2-\pi R_1^2)$, where $h$ is the traverse width and $R_2$ and $R_1$ are the flange and barrel radii, respectively. A cross section of the packed wire will look like a circle packing, so ideally the wire occupies $\frac\pi{2\sqrt3}\approx90\%$ of the available volume. (Realistically it will be less; John's answer gives a more conservative analysis.) The cable's volume is $L\times\pi r^2$, where $L$ is the length and $r$ is the cable radius.

So we have $L=\frac\pi{2\sqrt3} h(\pi R_2^2-\pi R_1^2)/(\pi r^2)\approx .9h\frac{R_2^2-R_1^2}{r^2}$.

Plugging in $h=1.5, R_1=0.9, R_2=1.5, r=.06$, we get $L=540$ meters.