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I have looked all over and haven't seen any questions about this topic. This may be more of a mechanical question, but none the less, the dilemma on hand is how does twisting a set of equal length wires affect the end length?

Through some side work, customers have requested a certain twist per inch in their builds. For example, some customers have requested 2 twists of wire per inch of length such that 100" of wire will have 200 twists.

What I'm trying to get an idea of is some kind of equation that can be used to calculate how long of a wire I need to cut and how many twists will be needed to achieve the desired length of wire. I'm considered factors such as AWG, insulation thickness, and how many twists per inch (ranges from 1 twist to 4 twists).

Approaching it from a geometrical standpoint the helical length equation, \$L = \sqrt{H^2+\pi^2D^2} \$. Where L is the length of wire needing to be cut, H is the desired end length, D is the diameter from each wire core center. My trouble comes now trying to figure out how to go from the helical length equation and transforming it what I need... Any suggestions on where to start?

Don
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    I would calculate how much wire you need using your helical length equation, add a few percent, twist the wires and cut the excess to match the actual length. Unless you have a machine or technique that you've characterized well, you always run the risk of the wires having unequal twist radii. Worst case, one wire is completely straight and the other wire just wraps around it like the AM radio wire on a car antenna. This is especially true if the wires are of different gauges or otherwise have different stiffnesses. – vir Oct 18 '22 at 18:20
  • Used a vice and an electric drill - cheap twisting machine, just don't twist too much... – Solar Mike Oct 18 '22 at 19:46
  • your question is more suitable for https://engineering.stackexchange.com/questions – jsotola Oct 18 '22 at 19:55
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    I'd just twist some wire and measure the before and after. Freaking engineers think too much! – Kyle B Oct 18 '22 at 21:13
  • @vir I have a machine that can keep the twist count consistent! Thank you for the suggestion! – Don Oct 20 '22 at 13:49
  • @jsotola Thank you, first post on stack and wasn't too sure where it would have fell into! – Don Oct 20 '22 at 13:50
  • @KyleB So easy so simple! This is the way. – Don Oct 20 '22 at 13:51

2 Answers2

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One strategy is to have the twisted wire prepared using a wire twisting machine that works off long spools of the straight wire and then meter out the required length and cut it as needed.

Another strategy may be to approach this from an empirical standpoint and take apart already stranded wire (or prefab your own and then take that apart) and measure the length of the wires when smoothed out.

Lastly via some simple experiments you could determine a small percentage overage and when a customer askes for 200 feet sell them 210 feet and let them trim.

Michael Karas
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  • Thank you for the input! The empirical approach sounds like the most practical for my case. Time to go burn through a ton of wire! – Don Oct 20 '22 at 13:52
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http://inductor.thayerschool.org/papers/stranded.pdf, page 290, equation 4 shows a method of calculating the length factor for twisted wires.

$$ {l_{strand} \over l_{bundle}} = {1 + {{\pi^2 n d_s^2} \over {4 K_a p^2}} } $$

Where:
\$ l_{strand} \$ = actual length of the wire strand
\$ l_{bundle} \$ = length of the twisted bundle
\$ n \$ = number of strands
\$ d_s \$ = diameter of each strand
\$ p \$ = pitch length
\$ K_a \$ = packing factor = \$ A_e / A_b \$
\$ A_e \$ = sum of the cross-sectional areas of all the strands
\$ A_b \$ = overall bundle area

A test I ran on some 25/40 HPN litz wire, my measurements gave a ratio of around 1.0083 to 1.0125. The equation predicted 1.0075. Bundle diameter was taken from the MWS Wire "Litz wire" catalog.

qrk
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  • This is exactly something I was trying to find, thank you so much! Going to go test it out myself and see how the numbers check out, you're awesome! – Don Oct 20 '22 at 13:53