About the Coil Spring Rate Calculator
This coil spring rate calculator works out the stiffness of a helical compression spring from its wire diameter, outside diameter, number of active coils and wire material. It returns the spring rate in N/mm and lbf/in, the force at a chosen deflection, the corrected shear stress in the wire, the spring index, the solid height and how far the spring can travel before the coils touch.
It is meant for designers sizing a replacement spring, machinists and hobbyists measuring an unknown spring with calipers, and students checking the classic k = Gd⁴ / 8D³N formula. Measure the wire with a micrometer, the outside diameter across the coils, and count the coils that are free to move — the closed end coils do not count as active.
Results assume round wire, a cylindrical helix, small pitch angle and linear-elastic behaviour at room temperature. Shear modulus values are typical handbook figures; a spring maker’s tolerance on rate is often around ±10%, so treat the answer as a design estimate and confirm critical springs by testing.
With the default inputs, the spring rate (k) is 3.502 N/mm. Change any value above to recalculate instantly.
How to use the coil spring rate calculator
- 1Measure the wire diameter with a micrometer or calipers.
- 2Measure the outside diameter across the coils.
- 3Count the active coils — exclude closed end coils that touch their neighbour.
- 4Pick the wire material and the end type.
- 5Enter the free length and the deflection you expect in service.
- 6Read the rate, force and stress, and check travel to solid height.
Formula and method
A helical compression spring is a torsion bar wound into a coil, so its stiffness depends on the wire’s shear modulus G, the fourth power of the wire diameter and the inverse cube of the mean coil diameter. That is why a small change in wire size changes the rate dramatically: going from 2 mm to 2.2 mm wire raises the rate by about 46%. Only active coils store energy; doubling them halves the rate.
The wire stress uses the Wahl factor, which corrects the simple torsion stress for direct shear and coil curvature. Total coils, solid height and travel follow the standard end-type relations: closed-and-ground ends add two inactive coils and Ls = Nt·d; closed unground ends add one wire thickness; plain ends have Ls = (Na + 1)·d; plain ground ends add one coil with Ls = Nt·d. Dimensions are in mm and G in N/mm² (MPa), so the rate comes out in N/mm; 1 N/mm ≈ 5.71 lbf/in.
- k
- Spring rate (N/mm)
- G
- Shear modulus of the wire material (MPa)
- d
- Wire diameter (mm)
- D
- Mean coil diameter (mm)
- Na
- Number of active coils
- C
- Spring index D/d
- Kw
- Wahl stress correction factor
- τ
- Corrected shear stress in the wire (MPa)
Worked examples
Music wire spring: 2 mm wire, 20 mm OD, 8 active coils
The mean diameter is 20 − 2 = 18 mm, so k = 81,700 × 2⁴ ÷ (8 × 18³ × 8) ≈ 3.50 N/mm, almost exactly 20 lbf/in. Compressing it 10 mm takes about 35 N, giving a corrected wire stress of about 233 MPa. With closed, ground ends there are 10 total coils, a 20 mm solid height and 40 mm of travel from the 60 mm free length.
Stainless spring: 3 mm wire, 25 mm OD, 6 active coils, closed ends
D = 22 mm and G = 69,000 MPa give k = 69,000 × 3⁴ ÷ (8 × 22³ × 6) ≈ 10.94 N/mm. An 8 mm deflection needs about 87.5 N. Closed, unground ends give 8 total coils and a solid height of (8 + 1) × 3 = 27 mm, leaving 23 mm of travel.
Chrome-silicon valve-type spring, plain ground ends
With D = 28 mm and C = 7, k = 77,200 × 4⁴ ÷ (8 × 28³ × 5) ≈ 22.5 N/mm (about 128.5 lbf/in). At 12 mm deflection the load is about 270 N and the Wahl-corrected stress is about 365 MPa. Plain ground ends give 6 total coils and a 24 mm solid height.
Frequently asked questions
How do you calculate the spring rate of a coil spring?+
Use k = G·d⁴ ÷ (8·D³·Na), where G is the wire’s shear modulus, d the wire diameter, D the mean coil diameter (outside diameter minus wire diameter) and Na the number of active coils. With mm and MPa the answer is in N/mm.
What counts as an active coil?+
Active coils are the ones free to deflect under load. On a spring with closed (squared) ends, the end coil at each side touches its neighbour and is inactive, so active coils are usually total coils minus two.
How do I convert N/mm to lbf/in?+
Multiply by about 5.71. One newton is 0.2248 lbf and one inch is 25.4 mm, so 1 N/mm = 0.2248 × 25.4 ≈ 5.71 lbf/in. A 10 N/mm spring is therefore about 57.1 lbf/in.
Does cutting coils off a spring make it stiffer?+
Yes. Rate is inversely proportional to the number of active coils, so cutting a spring from 8 to 6 active coils raises the rate by a third. It also shortens travel and raises stress at a given load, which can cause the spring to take a set.
What is the spring index and why does it matter?+
The spring index C is the mean coil diameter divided by the wire diameter. Values from roughly 4 to 12 are easy to manufacture; lower values are hard to coil and highly stressed, higher values tend to tangle and buckle.
What is the Wahl factor?+
The Wahl factor corrects the basic torsional stress formula for the extra stress at the inside of the coil caused by curvature and direct shear. For a spring index of 9 it adds about 16% to the nominal stress.