Rule of thumb · MechanicalNº 51 / 167

Double the wire, sixteen times the spring rate

A helical spring’s stiffness scales with wire diameter to the fourth power — doubling it multiplies the rate by 16.

Why it works

k = Gd⁴/(8D³n): d appears to the fourth power. A seemingly small jump in wire gauge changes a spring’s feel far more than instinct suggests.

When it fails

Only while the spring stays elastic and the coils stay apart. Compress it to coil bind and the rate is effectively infinite. And if you double the wire inside a fixed outside diameter you also shrink the mean coil diameter, which raises the rate by considerably more than 16×.

How wrong is it?

At 2 mm the rule says 3,399 N/m and the exact answer is 3,399 N/m — within half a percent. It holds to within 10% from 1.4 mm to 2.6 mm, and drifts outside that.

+10%0−10%12.54Wire diameter (mm)

The fourth power is exactly right — while the coil diameter holds still. In a real pocket the OUTSIDE diameter is what is fixed, so thicker wire also shrinks the coil, and rate goes as 1/D³ as well. Doubling the wire inside a 20 mm pocket multiplies the rate by far more than sixteen; the rule understates it by 30%.

The rule against the exact answer, computed across the range. Inside the shaded band the shortcut is close enough to use; outside it, reach for the calculator.

Do it exactly

Estimate with the rule, then check it against the calculator that models it properly.

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How does wire diameter affect spring rate?

A helical spring’s stiffness scales with wire diameter to the fourth power — doubling it multiplies the rate by 16. k = Gd⁴/(8D³n): d appears to the fourth power.

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