How this instrument works
A spring constant folds an entire piece of hardware into one figure: newtons of push-back for every metre it moves, k = F ⁄ x. For a coiled compression or extension spring, that number is set almost entirely by the wire itself — k = Gd⁴ ⁄ (8D³n), where G is the wire material's shear modulus, d the wire diameter, D the mean coil diameter, and n the number of active coils. Because d is raised to the fourth power, a wire only 10 percent thicker yields a spring roughly 46 percent stiffer, which is why a spring designer treats wire diameter as the single most sensitive dimension on the drawing, well ahead of coil diameter or coil count.
The energy formula follows directly from the force formula, not from a separate law. Force is not constant as a spring moves — it grows from zero up to kx — so the work stored is the running total of force over distance, PE = ∫₀ˣ k·s ds = ½kx². That integral is the entire reason for the one-half: pushing a spring twice as far does not just double the force at the end, it doubles the force over a path that is also twice as long, and multiplying two doublings together quadruples the energy banked inside the coils.
Both formulas describe a straight-line relationship that real springs only honor over part of their travel. Compress a coil spring far enough and its turns close up against each other — solid height, in a spring designer's terms — well before which the wire may already have crossed its proportional limit and begun to yield permanently. Past that point k is no longer constant, F = kx and PE = ½kx² both stop matching what a force gauge reports, and the spring may return to a slightly different length than the one it started from.
- Set Spring constant, N/m to the value stamped on a datasheet, or measure it: hang a known load on the spring and note how far it settles, then divide that force by the settling distance.
- Enter how far the spring has moved from its unloaded length into Displacement; the field accepts millimetres, centimetres, or metres.
- Read Restoring force for the push-back in newtons, computed as F = kx.
- Read Stored elastic energy for the work banked inside the spring in joules, computed as PE = ½kx².
Worked example — a 500 N/m die spring compressed 20 mm
A die spring rated at 500 N/m sits inside a stamping tool and gets compressed 20 mm — 0.02 m — as the punch bottoms out on each stroke. Enter 500 into Spring constant, N/m and 20 mm into Displacement. The restoring force works out as F = kx = 500 × 0.02 = 10 N, the exact figure a toolmaker checks against the press's rated tonnage before running production parts.
The energy stored over that same 20 mm stroke is PE = ½kx² = 0.5 × 500 × 0.02² = 0.1 J, released back into the punch on its return stroke. Run the same die deeper, to 40 mm of compression, and the force only doubles to 20 N — but the stored energy quadruples to 0.4 J, because doubling the displacement doubles both the force reached and the distance it acted over.
Questions
What does the spring constant k actually describe?
It describes stiffness: how many newtons of push-back appear for every metre the spring is stretched or compressed, in N/m. A 500 N/m spring resists twice as hard as a 250 N/m spring at the same displacement. For a coil spring that number comes from the wire diameter, coil diameter, coil count, and material — not from how long the spring happens to be overall.
Why does doubling the displacement quadruple the stored energy instead of doubling it?
Because the energy formula carries displacement squared, not displacement itself. Force does rise in a straight line with x, but the work needed to get there accumulates over that whole path, and multiplying a doubled force by a doubled distance gives four times the total. Compress a 500 N/m spring 40 mm instead of 20 mm and stored energy climbs from 0.1 J to 0.4 J, even though the force only reaches 20 N instead of 10 N.
Is Hooke's law accurate for any amount of stretch or compression?
No — only up to the spring's elastic limit. Inside that range force and displacement track each other exactly, which is what makes F = kx exact rather than approximate. Push past the elastic limit and the wire begins to yield: the coils may close up solid, part of the deformation becomes permanent, and the spring's rated k no longer describes it until it is replaced.
Where do I measure displacement from — the spring's manufactured length, or wherever it currently sits?
From wherever the spring sits unloaded, at rest, with nothing pushing or pulling on it — not from any position it happens to be resting in under load already. A garage door torsion spring or a pre-tensioned extension spring is often already under some force before you add more, so measure only the additional travel you are applying on top of that resting position.
What's a typical spring constant, from something small to something industrial?
A kitchen drawer's soft-close spring sits around a few N/m, just enough to ease the drawer shut without a slam. A rail wagon's draft-gear spring, which cushions the jolt when couplers snap taut, runs into the tens of thousands of N/m, because it has to absorb a loaded freight car's momentum within a few centimetres of travel. The 500 N/m used in this page's example sits well below either, closer to a garage door extension spring or a heavy-duty drawer slide.
Why is F called a restoring force?
Because it always points back toward the spring's equilibrium position, opposite whichever way it was displaced. Compress the spring and F pushes outward against whatever is compressing it; stretch it and F pulls inward. This calculator reports magnitude only, so enter Displacement as a positive number whether you're stretching or compressing, and read Restoring force as the size of that push-back, not its direction.