Hooke's Law Calculator - Spring Force, Constant and Stretch

Spring force, spring constant and stretch from one law

A spring rated 200 N/m pulls back with 10 N when it is stretched 5 cm, and while it does, it holds 0.25 J of energy. That is Hooke's law, F = k × x, and this calculator runs it in whichever direction you need: the force from a known spring, the spring constant from a load you measured, or the stretch a given load will cause. Spring rates go in as N/m, N/mm or lbf/in, lengths in anything from millimeters to feet.

Quick start. Hang a known weight on the spring and measure how far it moves. Choose "Spring constant", type the weight with the unit set to kgf or lbf, type the stretch, and press Calculate. A 2 kg weight that pulls a spring down 4 cm gives 490.3 N/m, which is 2.8 lbf/in.

The order the form asks for things

What to find - force, spring constant or stretch. The field for the answer disappears, and so does its unit selector.
Spring constant and its unit - from a catalog or a label. Compression springs are usually sold in N/mm or lbf/in; physics problems use N/m.
Stretch and its unit - measured from the free length, the length of the spring with nothing on it, not the total length under load.
Load and its unit - newtons, kilonewtons, pound-force, or kgf when the load is a mass hanging from the spring.
Mass on the spring - optional. Add it and the result also shows the period of a bounce, the frequency and the sag at rest.
Read the result - the answer in two unit systems, the energy stored, a table of the same spring at other stretches, and the same load on springs arranged in series or in parallel.

Seven springs, from a catalog part to a coilover

Each case below was run through the calculator. Change one number and the rest of the result moves with it, which is the quickest way to get a feel for how stiff a given rate really is.

Case 1: a spring of unknown stiffness and a known weight

Situation: load 2 kgf, stretch 4 cm
Result: k = 490.3 N/m, or 2.8 lbf/in, with 0.3923 J stored.

Every centimeter of stretch on this spring costs 4.9 N, roughly the weight of half a kilogram.

Case 2: a hanging spring scale

Situation: full scale 50 lbf, hook travel 2 in
Result: k = 25 lbf/in, which is 4,378 N/m. At full scale the spring holds 4.167 ft·lbf, or 5.649 J.

A scale like this is only accurate while its spring stays in the linear range, so a reading past the printed maximum is not worth trusting.

Case 3: a small compression spring from a parts catalog

Situation: rate 0.5 N/mm, pressed 3 mm
Result: 1.5 N, or 0.3372 lbf, and only 2.25 mJ stored.

N/mm is the unit catalogs print. It is a thousand times the N/m figure, so 0.5 N/mm is 500 N/m.

Case 4: a coilover spring compressed two inches

Situation: rate 450 lbf/in, compressed 2 in
Result: 900 lbf, which is 4,003 N, with 75 ft·lbf (101.7 J) stored.

This is the rate of the spring itself. What the tire feels depends on where the spring mounts on the suspension arm, covered in the questions below.

Case 5: how far a load settles, and how fast it bounces

Situation: 900 lb resting on a 300 lbf/in spring
Result: it settles 3 in (7.62 cm) and, left to bounce, does so at 1.806 Hz, one cycle every 0.5539 s.

Type the mass into the optional field and use the stretch mode with the same load: the sag and the stretch come out equal, as they should.

Case 6: a lab mass on a lab spring

Situation: k 200 N/m, mass 0.5 kg
Result: sag 2.452 cm, period 0.3142 s, frequency 3.183 Hz.

Double the mass and the period grows by the square root of two, to 0.4443 s, not to twice as long.

Case 7: two identical springs, two ways of joining them

Situation: two 200 N/m springs, 10 N load
Result: one spring alone stretches 5 cm. End to end they stretch 10 cm; side by side, 2.5 cm.

The arrangement table in every result does this for your own spring and load.

Spring rate in every unit a catalog uses

The same stiffness appears under five names depending on who printed the label. The first column is the unit you have; read across for the others. The pound-force and kilogram-force factors follow from their definitions, so the table is exact to the digits shown.

One unit of	N/m	N/mm	lbf/in
N/m	1	0.001	0.00571
N/mm (same as kN/m)	1,000	1	5.710
lbf/in	175.1	0.1751	1
kgf/cm	980.7	0.9807	5.600
kgf/mm	9,807	9.807	56.00

Joining, stacking and cutting springs

Springs in series share the load and add their stretches, so the combination is softer. Springs in parallel share the stretch and add their forces, so it is stiffer. A coil spring's rate is inversely proportional to its number of active coils, which is why cutting one makes it stiffer rather than softer. The last two columns use a 200 N/m spring under 10 N.

Arrangement	Combined constant	Stretch under 10 N	Energy stored
One spring	k = 200 N/m	5 cm	0.25 J
Two in series	k / 2 = 100 N/m	10 cm	0.5 J
Two in parallel	2k = 400 N/m	2.5 cm	0.125 J
Cut to half its coils	2k = 400 N/m	2.5 cm	0.125 J
Cut to a third of its coils	3k = 600 N/m	1.667 cm	0.08333 J

Under one fixed load, the softer arrangement always stores more energy, because E = F × x / 2 and the same force travels further. Under one fixed stretch it is the other way round.

Stiffness, sag and bounce, answered briefly

What does a spring constant of 490 N/m actually mean?

It is the force per unit of stretch. At 490 N/m every meter would take 490 N, so every centimeter takes 4.9 N and every millimeter 0.49 N. A bigger number is a stiffer spring.

Is spring rate the same thing as the spring constant?

Yes. "Spring rate" is the word catalogs and the automotive trade use, usually in lbf/in or N/mm; "spring constant" is the physics name, usually in N/m. A 450 lbf/in spring is a 78,807 N/m spring.

Why does doubling the stretch give four times the energy?

Because the force grows while you stretch. The energy is the area under a straight line from zero to the final force, E = k × x² / 2. Twice the stretch means twice the final force over twice the distance, which is 4 times the area: 0.25 J at 5 cm becomes 1 J at 10 cm on a 200 N/m spring.

Does cutting a coil spring make it softer or stiffer?

Stiffer. For a helical spring k = G × d⁴ / (8 × D³ × n), where n is the number of active coils, so halving the coils doubles the rate. The shorter spring also runs out of travel sooner, so it can hit coil bind under a load the full spring handled.

What is the difference between spring rate and wheel rate on a car?

The spring usually sits partway along the suspension arm, so the wheel moves further than the spring does. The rate felt at the wheel is the spring rate multiplied by the square of that motion ratio. A 450 lbf/in spring at a ratio of 0.7 gives a wheel rate of about 220 lbf/in.

How do I measure k at home if the spring does not start moving right away?

Extension springs are often wound with initial tension, so the first bit of load moves nothing. Use two loads instead of one and divide the difference: k = (F₂ - F₁) / (x₂ - x₁). Hanging 1 kg and then 3 kg, with the hook moving 4 cm between them, gives 490.3 N/m whatever the preload was.

Why does the period only depend on how far the mass sags?

The sag at rest is m × g / k, so m / k equals sag / g, and the period T = 2π × √(m / k) becomes 2π × √(sag / g). A load that settles 2.452 cm bounces with a period of 0.3142 s, whether it is a small mass on a soft spring or a large one on a stiff spring.

When does Hooke's law stop working?

Past the elastic limit, where the wire starts to yield and the spring no longer returns to its free length, and near coil bind, where the coils touch and the rate shoots up. Datasheets give a maximum deflection; keep the stretch you type below it.

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