Estimating Cable Length on a Reel

Question

I have been searching all areas of the internet to try and find a reliable formula for estimating cable length on a reel,

I'm trying to create a faster and more reliable way to estimate cable to enhance the speed of our Stock Take.

So far I have found one formula that makes sense to an extent but falls short on the layers calculation (step 6-8) this can be found here: http://www.ehow.co.uk/how_7464814_estimate-cable-reel.html

The reel sizes and cable diameter change per reel.

I used the guide above to measure a single wrap around of cat 5e, 4 layers thick and the results were out by at least 200mm, (77% inaccuracy)

Here are the measurements i recorded;

Center of the reel diameter : 57.5mm
Cable diameter : 4.3mm
Width of the Reel : 4.3mm (single layer of the cat 5e),
Flange Used : 17.2mm

Total Length : 756mm - Using the formula below

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Formula for working out the 4 layers is:
Using the Example Above
Layer 1 = (4.3*2)3.14
Layer 2 = (4.3*2+4.3)3.14
Layer 3 = (4.3*2+4.3+4.3*2)3.14
Layer 4 = (4.3*2+4.3+4.3*3)3.14

Can anyone see where i am going wrong?

I Manually counted the length and it was 980mm

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I noticed a trend in the increase of length per layer which was 30mm, is this relevant to factor in at any point in the guide?

I am by no means a Math Professional, so my mind is slightly boggled!

Any help at all would be appreciated.

Thank you

Answer 1

The classic way to make these estimates is by volume.

If you want to get really fancy you can get into some of the considerations in Calculating the length of the paper on a toilet paper roll, but the irregularities in winding cable are different from the irregularities in winding a sheet around a drum, so let's not go there this time.

I assume that "Center of the reel diameter" is the diameter of the cylindrical surface on which we start winding the cable, and that "flange used" is the distance from that cylindrical surface to the outer edge of the reel, where we stop winding cable. The "flange used" is therefore the total thickness in the radial direction of all the layers of cable wound on the reel.

So if $R$ is half the "Center of the reel diameter", $F$ is the "flange used", and $H$ is the width of the reel, the volume to be filled by the cable between the flanges is

$$ V = \pi ((R + F)^2 - R^2) H = \pi (2R + F) FH. $$

If the diameter of the cable is $d$, we can consider each unit length of cable to occupy the volume $d^2$. (The volume of actual cable is only $\frac\pi4 d^2$, but the cable takes up more volume than that because of the unfilled spaces between winds of cable.)

So the total length of cable would be

$$ L = \frac{\pi (2R + F) FH}{d^2}. $$

Plugging in $2R = 57.5$, $d = 4.3$, $H = 4.3$, $F = 17.2$,

$$ L = \frac{\pi (57.5 + 17.2) 17.2 \cdot 4.3}{4.3^2} \approx 938.7. $$

That's within $4.2\%$ of the measured length. Possibly the difference is accounted for by inaccuracies in measuring the input parameters, deformation of the cable allowing it to wind tighter, and other real-life sources of error.

Answer 2

1. Weigh the reel.
2. Weigh or estimate the weight of an empty reel (may be negligible).
3. Weigh a metre of cable.

$$Cable \space length = {[1] - [2] \over [3]}$$

Not mathematics.