How to use this calculator
Enter the wire diameter, the mean coil diameter (outside diameter minus one wire diameter) and the number of active coils. Pick a wire material to fill in the shear modulus and a typical torsional yield strength, then enter the load.
Formulas
Worked example
Music wire, d = 2 mm, D = 16 mm, 8 active coils, 50 N (the default)
- C = 16 / 2 = 8, so Kw = 31/28 + 0.615/8 = 1.184.
- k = 81,700 × 2⁴ / (8 × 16³ × 8) = 4.99 N/mm.
- y = 50 / 4.99 = 10.0 mm.
- τ = 1.184 × 8 × 50 × 16 / (π × 2³) = 301 MPa, so n = 900/301 ≈ 3.0.
Assumptions and limits
- Round wire, closely spaced coils, and load along the spring axis.
- Linear elastic behavior; the spring does not reach solid height.
- Number of active coils excludes inactive ends (about total − 2 for squared and ground ends).
- Static load. For cycling loads, also check fatigue.
Frequently asked questions
Why does the Wahl factor matter?
The wire is curved, so stress on the inside of each coil is higher than the simple torsion formula predicts. Kw corrects for that curvature and for direct shear.
How do I make a spring stiffer?
Rate scales with d⁴, so a slightly thicker wire has a big effect. Fewer active coils or a smaller coil diameter also raise the rate.
What spring index should I aim for?
Between about 4 and 12. Lower values are hard to manufacture; higher values make springs that buckle and tangle.
For education and preliminary checks only. Have a qualified engineer verify any design before it is built.