Physics: Mechanics, Waves & Thermodynamics Rotation, Circular Motion & Gravitation Moment of a force, τ = r × F; SI and customary torque units per NIST SP 811

Torque Calculator

Torque is the turning effect of a force about an axis, and it depends on three things: how big the force is, how far from the axis it acts, and the angle between the force and the lever arm. Enter any two of torque, force and radius and this calculator solves for the third, reporting the answer in newton-metres, pound-feet and pound-inches. Add a rotation speed and it also gives the mechanical power that torque delivers, which is the number that decides what a motor or engine can actually do.

Calculator

This calculator runs in your browser. Enable JavaScript for live results — the inputs, formula and worked example below remain fully readable without it.

Inputs this calculator takes, with typical values
InputWhat to enterExample
What do you want to find?Choose the unknown; the calculator hides that field and solves for it.Torque
Applied forceMagnitude of the force applied at the end of the lever arm.250 N
Lever arm lengthDistance from the axis of rotation to the point where the force is applied.0.4 m
Angle between the force and the lever arm90° means the force is perpendicular to the arm, which is the most effective case. 0° or 180° means it points along the arm and produces no turning effect.90 °
Target torqueThe torque you need to reach. Used only when solving for the force or the lever arm.100 N·m
Rotation speedUsed only for the power figure. Power is torque times angular velocity, so torque alone does not tell you what a machine can do.1500 rpm

It returns

  • Torque — The moment of the force about the axis.
  • Torque
  • Torque
  • Force at the lever arm
  • Lever arm length
  • Mechanical power at this speed
  • Mechanical power

The formula

τ=Frsinθ
P=τω
τ=Iα

In plain text: τ = F · r · sin θ

  • τTorque, or moment, about the axis (N·m)
  • FMagnitude of the applied force (N)
  • rDistance from the axis to the point where the force acts (m)
  • θAngle between the force and the lever arm (°)
  • PMechanical power delivered, τ·ω (W)

Torque is properly the vector cross product τ = r × F; this scalar form gives its magnitude about the axis. The quantity r·sin θ is the moment arm — the perpendicular distance from the axis to the force's line of action.

Updated Category Rotation, Circular Motion & Gravitation Verified against published test cases Reading time 11 min

What torque is

Torque is the rotational equivalent of force: it measures how strongly a force tends to twist an object about an axis. Engineers often call it the moment of the force, and the two words mean the same thing. Its unit is the newton-metre, formed the same way as the joule but describing something entirely different — a turning effect, not an energy — which is why torque is written N·m and never J.

Three quantities set it. The size of the force is the obvious one. The distance from the axis is the reason a long breaker bar loosens a nut that a short spanner will not. And the angle matters because only the component of force perpendicular to the arm actually turns anything; a pull directed straight at the axis just tries to move the whole assembly sideways.

Combine those into τ = Fr sin θ and you have the whole subject. The product r sin θ is the moment arm: the perpendicular distance from the axis to the force's line of action. Sketch that perpendicular and you will never make a sign or geometry error again, because the moment arm is always shorter than the physical arm unless the force is exactly perpendicular.

Torque is what turns a screw, corners a car through the steering rack, holds a beam up at its supports, and defines what an engine can pull. It is also what limits a fastener: bolts are tightened to a torque figure because torque is what a person with a wrench can control, even though the quantity that actually matters is the tension in the bolt.

Why the sine is there, and why torque is not power

Torque is properly a vector cross product, τ = r × F, whose magnitude is Fr sin θ and whose direction is along the axis of rotation. The sine is what the cross product does: it keeps the perpendicular component and discards the parallel one. At 90° the sine is 1 and all the force turns the shaft; at 0° or 180° the sine is zero and none of it does, because the force's line of action passes straight through the axis.

Notice the symmetry: sin 30° and sin 150° are both 0.5, so a force pulling 30° off the arm and one pushing 30° the other way produce the same magnitude of torque. The calculator's angle sweep shows the whole curve, and it is the same shape whatever your force and radius.

The single most consequential thing to understand about torque is that it is not power. Power is torque multiplied by angular velocity, P = τω, so the same torque delivers ten times the power at ten times the speed. A stalled motor produces its full torque and zero power — every watt going in becomes heat. This is why gearing works: a reduction gearbox multiplies torque and divides speed in the same ratio, leaving power unchanged apart from losses. You never get anything for free; you trade speed for turning effort.

Torque also has its own version of Newton's second law: τ = Iα, where I is the moment of inertia about the axis and α is the angular acceleration in rad/s². It is the direct rotational analogue of F = ma, with every trap intact: the torque must be the net torque, and the moment of inertia must be taken about the axis actually being rotated.

Worked example: loosening a wheel nut

A wheel nut is specified at 110 N·m. You have a 380 mm wheel brace. How hard do you have to pull, and what changes if you cannot pull square to it?

  1. Perpendicular pull. F = τ ÷ (r sin θ) = 110 ÷ (0.380 × 1) = 289.5 N, about 29.5 kgf. That is a firm two-handed pull but well within reach.
  2. Pull at 60° instead. sin 60° = 0.86603, so the effective arm falls to 0.380 × 0.86603 = 0.32909 m and the force needed rises to 110 ÷ 0.32909 = 334.3 N — 15% more for the same result.
  3. Pull at 30°. sin 30° = 0.5, effective arm 0.190 m, force needed 110 ÷ 0.190 = 578.9 N. Half the effective arm, double the force. The geometry of how you stand matters as much as the length of the tool.
  4. Extend the bar. Slip a 600 mm tube over the brace, giving 0.98 m total, and the perpendicular force falls to 110 ÷ 0.98 = 112.2 N. That is the whole reason cheater bars exist — and also why they overtighten so easily, since the same comfortable pull now delivers 0.98 ÷ 0.380 = 2.6 times the torque.
  5. Check the units. 110 N·m × 0.7375621 = 81.1 lbf·ft, the figure a US-market service manual would quote.

The last step is worth doing every time. Mistaking lbf·ft for N·m under-tightens a joint by 26%; mistaking N·m for lbf·ft over-tightens it by 36%, which is enough to yield a small fastener.

Reading the result, and what torque does not tell you

Check the units before anything else. Newton-metres and pound-feet differ by a factor of 1.356, and pound-inches by a further factor of twelve, so a torque figure without a unit is worthless. Small fasteners and instruments are usually specified in pound-inches or newton-centimetres, vehicle fasteners in newton-metres or pound-feet, and industrial drives in newton-metres or kilonewton-metres.

Then check the effective arm. The calculator reports r sin θ separately for exactly this reason: it is the number that actually does the work, and if it is much smaller than your physical arm, your geometry is costing you.

Do not read a torque figure as a measure of how much work a machine can do. That is power. A motor rated at 200 N·m at 1,500 rpm delivers 31.4 kW; the same 200 N·m at 150 rpm delivers 3.14 kW. Comparing engines or motors on torque alone, without stating the speed, compares nothing.

Finally, on fasteners: torque is a proxy for what you actually want, which is the clamping tension in the bolt. Most of the torque you apply is spent overcoming friction under the head and in the threads, and only a small fraction becomes tension. The usual relation is T = K · D · Fpreload, with D the nominal thread diameter and K a nut factor around 0.2 for plain steel and lower when lubricated — figures that engineering handbooks give as rules of thumb, not as design data. Because K varies with lubrication, plating, surface finish and reuse, torque control alone is an imprecise way to set preload, which is why critical joints use angle control, bolt stretch measurement or direct tension indicators instead.

Torque unit conversions

All of these follow exactly from the defined values of the pound-force, the foot, the inch and standard gravity, as tabulated in NIST SP 811.
UnitIn newton-metres1 N·m equalsTypically used for
Newton-metre (N·m)11The SI unit — everything
Pound-foot (lbf·ft)1.35581790.7375621US vehicle and structural work
Pound-inch (lbf·in)0.11298488.8507458Small fasteners, instruments
Ounce-inch (oz·in)0.0070616141.6119Small motors, watchmaking
Kilogram-force metre (kgf·m)9.806650.1019716Older European and Japanese manuals
Kilogram-force centimetre (kgf·cm)0.098066510.19716Small assemblies, hobby servos
Kilonewton-metre (kN·m)1,0000.001Structural moments, large drives

A torque of 100 N·m is 73.76 lbf·ft, 885.07 lbf·in or 10.197 kgf·m. Always carry the unit — the numbers differ by more than a factor of eight between the two most commonly confused pairs.

Torque specifications come from the joint, not from a calculator

This page tells you the relationship between force, lever arm and torque. It cannot tell you what torque a particular fastener should be tightened to. That figure depends on the bolt grade, diameter, thread pitch, the material being clamped, the lubrication state and the joint's function, and it must come from the manufacturer's specification or the governing standard for the assembly. Applying a torque figure from a similar-looking joint is how threads get stripped and bolts get yielded.

Common mistakes

  • Confusing pound-feet with pound-inches. A factor of twelve, and the single most expensive unit error in the fastener world.
  • Writing torque in joules. Both are newton-metres dimensionally, but torque is a turning effect and energy is not. The SI convention keeps them apart deliberately.
  • Using the physical arm instead of the moment arm. If the force is not perpendicular, the effective arm is r·sin θ, and it can be far shorter than the tool you are holding.
  • Comparing motors on torque alone. Torque without a speed says nothing about capability. Power is torque times angular velocity, and it is the quantity that survives gearing.
  • Forgetting that a gearbox trades one for the other. A 10:1 reduction multiplies torque by roughly ten and divides speed by ten. Power out never exceeds power in.
  • Treating torque as a direct measure of bolt tension. Most of the applied torque goes into friction. Use angle control, stretch measurement or tension indicators where preload really matters.
  • Adding torques without regard to sense. Clockwise and anticlockwise moments have opposite signs. The net torque is the signed sum, which is exactly what τ = Iα requires.

Torque is one corner of a complete rotational mechanics that mirrors the linear one exactly. Force becomes torque, mass becomes moment of inertia, acceleration becomes angular acceleration, and Newton's second law becomes τ = Iα. Momentum becomes angular momentum L = Iω, with its own conservation law, paralleling the linear momentum calculator.

Energy transfers the same way. Work done by a torque is τθ with the angle in radians, and rotational kinetic energy is ½Iω². Those two are why a flywheel spun up by a given torque through a given angle ends at a predictable speed, and why the energy stored scales with the square of that speed.

In statics, torque appears as the moment of a force about a support, and the condition for equilibrium is that both the net force and the net moment vanish. That pair of conditions is the entire basis of beam analysis, of lever and pulley calculations, and of any free-body problem where a body is not free to translate. And in circular motion, a torque applied to a rotating body changes its speed, while the centripetal force merely holds it on its path without changing its speed at all — the clearest illustration of why forces along and perpendicular to the motion do such different things.

Frequently asked questions

How do I calculate torque?

Multiply the force by the distance from the axis and by the sine of the angle between them: τ = F·r·sin θ. A 250 N force applied perpendicular to a 0.4 m arm gives 250 × 0.4 × 1 = 100 N·m. When the force is perpendicular — the usual case with a spanner — the sine is 1 and the formula reduces to force times distance.

How do I convert newton-metres to foot-pounds?

Multiply by 0.7375621, since one pound-foot is exactly 1.3558179483 N·m. So 100 N·m is 73.76 lbf·ft, and 100 lbf·ft is 135.58 N·m. Be careful with pound-inches, which are a twelfth of pound-feet: 100 N·m is 885.07 lbf·in. Small fasteners and instruments are usually specified in pound-inches, so misreading the unit produces a twelve-fold error.

Why does a longer spanner make a nut easier to undo?

Because torque is force times lever arm, so doubling the arm halves the force you need for the same torque. A nut at 110 N·m needs 290 N on a 380 mm brace but only 112 N on a 980 mm bar. The trap is that the relationship works both ways: the same comfortable pull on the longer bar delivers far more torque, which is how over-length bars strip threads and snap studs.

What is the difference between torque and power?

Torque is turning effort; power is the rate at which that effort does work. They are linked by P = τω, with ω in radians per second, so the same torque delivers more power at higher speed. A stalled motor produces full torque and zero power. This is also why a gearbox can multiply torque without breaking any laws — it divides speed by the same factor, leaving power unchanged apart from losses.

Why does the angle of the force matter?

Because only the component perpendicular to the lever arm turns anything. A force pointing straight at the axis has no turning effect at all, however large it is, since its line of action passes through the pivot. The factor sin θ captures this, and the product r·sin θ — the moment arm — is the perpendicular distance from the axis to the force's line of action. Pulling 30° off square costs you half your effective arm.

Is torque the same as a moment?

Yes, in practice. Physicists tend to say torque and structural engineers tend to say moment, and both mean the turning effect of a force about a point or axis, computed the same way. Some texts reserve "torque" for a twist about an object's own axis and "moment" for bending about an external point, but the mathematics is identical and both are quoted in newton-metres.

How much torque does a motor need to accelerate a load?

Use the rotational form of Newton's second law: τ = Iα, where I is the moment of inertia of everything rotating and α is the angular acceleration in rad/s². Spinning a 0.99 kg·m² flywheel to 3,000 rpm (314.16 rad/s) in ten seconds needs α = 31.4 rad/s² and therefore about 31 N·m, on top of whatever torque friction and the working load demand.

Why is torque measured in N·m and energy in joules if they are the same units?

Because they are physically different quantities that happen to share dimensions. Energy is a force acting along a displacement, a scalar product; torque is a force acting at a distance from an axis, a cross product, and it is a vector along the rotation axis. SI keeps them typographically distinct — N·m for torque, J for energy — precisely so that the two are never confused. Work done by a torque, τθ, is in joules.

References