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Springback or Jump Ring Ahead

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Categories: Instruction and information; Jewelry

Word count/read time: 431 words; 2 minutes

Springback happens when tension on an object is released and it attempts to revert to its original shape. It is where elastic deformation stops and plastic deformation begins. For chainmaking, it's how tight (accurate) those jump rings are.

Wrapping wire around a mandrel makes a coil in that profile. Basic jump rings use round dowels and wire. The actual inner diameter compared to the mandrel diameter indicates the degree of springback. Every jewelry metal experiences springback except maybe dead-soft fine silver or gold.

Ideally, expected and observed diameters would be identical. However, the jump ring I.D. will be larger than the mandrel. Each metal is different so to get a perfect 1/4" inner diameter jump ring might require a 0.24" mandrel for annealed sterling silver compared to 0.22" for spring temper (hardened) stainless steel.

The I.D. is the dragon to chase. Aspect ratio is based on the I.D. measurement and the wire thickness. Sometimes an A.R. variation of 1% can make weaves fail. Other times it determines if a weave looks and behaves best.

There is no simple formula to compute springback. It depends on the metal, tension of the wire when winding, its diameter, temper, and mandrel diameter. Basically, it's trial and error. Reproducing the same conditions each time so that batch X is the same as batch Y, even if made years apart, is difficult.

The most direct way to compute springback is to measure the finished I.D. of the jump ring and divide it by the actual mandrel diameter. This requires a precise vernier (down to 0.05mm if possible) and an accurate measurement. Easier said than done sometimes.

Simple math might work here. Before the tension is released from the coiled wire, count the number of coils on the mandrel down to the smallest fraction possible (i.e., "3" wraps might actually be 3.1 or 3.15 wraps). When the tension is released, count again. Divide the first number by the second and subtract one to get the springback as a decimal.

Method one measured the inner diameter so there's no guesswork. The second method computes the I.D. increase. Thus, 0.015 springback (i.e. 1.5%) on a 0.500" mandrel yields a 0.5075" I.D. ring.

Example uses 1.6mm round wire and a 1.0cm mandrel with 1% springback:
Perfect: 1.0cm I.D. / 0.16cm wire diameter = 6.25 AR
Actual: 1.01cm I.D. / 0.16cm wire diameter = 6.31 AR
There are additional factors that could change the A.R. including how much the wire compresses, stretches, or squashes as it's being wound. The math changes as a result. Get it right before jumping all in.

Posted by M: March 12, 2023

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