How much wire is needed to form eyes?

Estimate wire length needed to form eyes

Eyes refer to small loops used to connect components together. Forming eyes is one of the most common steps in any design. Use the tool to the right to estimate how much wire you need to form an eye. An equally weighted combination of circle and teardrop estimates was found to be most accurate.

Eyes refer to small loops used to connect components together. Forming eyes is one of the most common steps in any design. In the following experiment, I cut exact lengths of 18, 20, and 22ga wire and used different small diameter tools to form an eye at both ends of the wire. I then measured the the distance between the two eyes (the the wire ends make contact with the stem) to the nearest 0.25mm (I used the picture to the right magnified on my computer screen to help measure the distance). The original wire length minus this distance is the wire used to form two eyes. Divide this number by 2 to get the actual wire length used to create each eye.

An eye is formed by taking one rotation of the tool until the wire end touches itself. So the simplest way to estimate the length of wire needed is the circumference of the tool or a  = d𝝅 where d is the diameter of the tool. This method underestimates the actual wire length needed by 3.40mm on average. You can also see a pattern in the error correlating with wire gauge suggesting wire thickness should enter the equation.

As the wire curves around the tool, the inside edge compresses while the outside edge stretches, leaving the center of the wire the original length. So we really want to follow the path of the center of the wire. If we take wire thickness t into account, the new estimate is b  = (d+t)𝝅. This method reduces the average error to -0.64 mm and improves the accuracy. However, now you can see a pattern in the error correlating with tool diameter.

Upon closer inspection, the shape of the smaller eyes is more like a teardrop. The path travels only 3/4 the circumference (d+t)3/4𝝅. After adding the first and last part of the path, each traveling half of d+t, the new estimate simplifies to c = (d+t)(3/4𝝅+1). This method works somewhat better for smaller eyes but tends to overestimate larger eyes. Overall it overestimates by 0.54 mm on average and the accuracy isn't much better than estimation b.