Audited 27 Jul 2026·Last updated 28 Oct 2026·5 citations·Tier 1·0 uses

Torque Calculator

Calculate torque from force, lever arm and angle (τ = r F sin θ), with force in N, lbf or kgf and length in m, cm, in or ft; results in N·m, lb-ft and dyn·cm.

Torque Calculator

Force unit
Lever arm unit
Torque magnitude (τ)
45
The torque magnitude in newton-metres, r × F × sin θ worked out after converting the inputs to SI units.
Torque in pound-feet
33.1903
Torque in dyne-centimetres
450,000,000

Background.

Door handles sit on the edge farthest from the hinges for a reason. Push near the hinge and the door barely moves; push at the handle with the same effort and it swings open. What changed is the torque, the turning effect of a force, which grows with the distance between the pivot and the point where the force is applied, and with how squarely the force acts across that distance. A 150 N push at a handle 0.8 m from the hinges gives 120 N·m; the same push 0.1 m from the hinges gives only 15 N·m.

In symbols, τ = r × F × sin θ. The lever arm r runs from the pivot (a bolt's axis, a hinge, the centre of a shaft) to the point where the force is applied; F is how hard you push or pull; θ is the angle between the two. At 90° the whole force goes into turning. As the angle closes toward 0° or opens toward 180°, more of the force simply pushes along the arm, toward or away from the pivot, and contributes nothing to the rotation.

Force can be entered in newtons, pound-force or kilogram-force and the arm in metres, centimetres, inches or feet, in any combination. Everything is converted to SI first, and the torque comes back in newton-metres, pound-feet and dyne-centimetres at once, so a European specification in N·m and an American wrench scale in lb-ft can be compared directly.

What you get is the size of the torque from one force about one pivot. Direction of rotation is not shown, several forces are not combined, and there is no mode that solves backwards for the force you need, although rearranging the same equation does that in one line, as the formula section shows. For bolts, the figure is the torque you apply at the wrench, which is not the same thing as the clamping force the bolt ends up with.

What is torque calculator?

Torque, also called the moment of a force, measures how strongly a force tends to rotate an object about an axis. Its size equals the force multiplied by the perpendicular distance from the axis to the force's line of action, which is why r sin θ is called the moment arm. Formally torque is a vector, τ = r × F, pointing along the axis by the right-hand rule, and its direction tells you which way the rotation goes; the calculation here returns the magnitude only. The SI unit is the newton-metre. It has the same dimensions as the joule, but the SI Brochure notes that the joule is never used for torque, because a turning effect is not an amount of energy. Torque drives angular acceleration the way force drives linear acceleration, through τ = Iα, where I is the moment of inertia: a torque of 43.47 N·m applied to a wheel with I = 0.5 kg·m² speeds it up at about 86.93 rad/s² until friction or a load pushes back.

How to use this calculator.

  1. Enter the applied force and choose its unit. Kilogram-force suits a hanging mass or a person's weight: a 10 kg load pulls with 10 kgf.
  2. Enter the lever arm and its unit: the straight-line distance from the centre of the pivot to where the force is applied. On a wrench that is the bolt centre to the middle of your hand, not the overall length of the tool.
  3. Enter the angle between the lever arm and the force, anywhere from 0° to 180°. Use 90° when you push or pull straight across the handle.
  4. Read the torque in N·m, lb-ft and dyn·cm. For inch-pounds, the unit small-fastener specifications often use, multiply the pound-feet figure by 12.

The formula.

τ = r × F × sin θ

τ = r × F × sin θ, with r in metres, F in newtons and τ in newton-metres. Inputs are converted to SI first using exact definitions: 1 lbf = 4.4482216152605 N, 1 kgf = 9.80665 N (one kilogram under standard gravity), 1 in = 0.0254 m and 1 ft = 0.3048 m. The result is then given three ways: in N·m directly, in pound-feet by dividing by 1.3558179483314004, which is exact because it is simply 0.45359237 kg × 9.80665 m/s² × 0.3048 m, and in dyne-centimetres by multiplying by 10⁷, since a dyne is 10⁻⁵ N and a centimetre 10⁻² m.

The sine factor is the fraction of the force that acts across the arm. It is 1 at 90°, 0.5 at 30° or 150°, and 0 at 0° and 180°. Because sin θ = sin(180° − θ), an angle of 105° gives exactly the same magnitude as 75°; which side of square you pulled from does not change the size of the torque. Near 90° the sine curve is flat, so small errors in angle barely register: pulling 15° off square costs 3.4% of the torque and 10° off about 1.5%. Force and lever arm, by contrast, act in direct proportion, so doubling either doubles the torque.

An equivalent reading is force times the perpendicular distance from the pivot to the line along which the force acts. A 0.3 m arm pulled at 75° has a moment arm of 0.3 × sin 75° ≈ 0.2898 m, and that shorter effective distance is what the force works through.

Rearranged, the same equation gives the force needed for a target torque, F = τ ÷ (r sin θ), or the arm needed for a given force, r = τ ÷ (F sin θ).

At exactly 0° and 180° the result is 0 in every unit. A computer's value for sin 180° is a vanishing 10⁻¹⁶ rather than 0, so the results are rounded to 10 decimal places, which clears that residue; the same rounding means a real torque smaller than about 5 × 10⁻¹¹ N·m would also show as 0.

A worked example.

Example

A 150 N push is applied at the end of a 0.3 m wrench, but the handle sits at 75° to the push instead of square to it. Square on, the torque would be 0.3 × 150 = 45 N·m. The angle scales that by sin 75° = 0.9659, so τ = 0.3 × 150 × 0.9659 = 43.4667 N·m. In the other units, 43.4667 ÷ 1.3558 = 32.0594 lb-ft, and 43.4667 × 10⁷ = 434,666,622 dyn·cm, or about 4.347 × 10⁸. The slant cost 45 − 43.4667 = 1.5333 N·m, 3.4% of the square-on figure. Reaching the full 45 N·m at the same 75° would take F = 45 ÷ (0.3 × 0.9659) = 155.29 N, about 5 N more push, which is why a pull a few degrees off square is rarely worth worrying about while a short wrench is.

angle75
force150
radius0.3

Frequently asked questions.

How do I convert between N·m, lb-ft and inch-pounds?
One pound-foot is 1.3558 N·m and one newton-metre is 0.7376 lb-ft. An inch-pound is a twelfth of a pound-foot, 0.1130 N·m, so 60 lb-ft is 720 in-lb. Specifications write the same unit as lb-ft, ft-lb or lbf·ft; for torque they all mean pound-force times feet. Check whether a small-engine or bicycle spec is in inch-pounds before setting a wrench, because mixing them up is a factor of 12.
How much does it matter if I don't pull the torque wrench at 90°?
Less than most people fear, as long as the error is modest. The torque falls with the sine of the angle, which barely changes near 90°: 10° off square loses about 1.5% and 15° off loses 3.4%. Hold a click-type wrench at the grip position its maker marks, because moving your hand along the handle changes the lever arm the scale was calibrated for.
I'm using an extension or crow's-foot on my torque wrench. What do I set it to?
If the adapter extends the wrench in a straight line, the effective arm grows from L to L + E, so set the wrench to the desired torque × L ÷ (L + E). With an 18-inch wrench and a 2-inch in-line crow's-foot, set 18 ÷ 20 = 0.9 of the target: 90 N·m for 100 N·m at the bolt. Turned 90° to the handle, the adapter adds almost no length and needs essentially no correction.
How is the torque on a bolt related to how tightly it clamps?
Only loosely, because most of the applied torque is spent overcoming friction in the threads and under the head or nut, and only a small fraction stretches the bolt. A common estimate from Shigley's machine-design text is T = K × F × d, with a nut factor K of about 0.2 for dry steel. A 10 mm bolt meant to clamp 20 kN needs about 0.2 × 20,000 × 0.010 = 40 N·m. Oil or grease lowers K, so the same torque then clamps harder.
What is the link between torque and horsepower?
Power is torque times rotational speed, P = τω, with ω in radians per second. An engine making 200 N·m at 3,000 rpm turns at 314.16 rad/s, so it delivers 62.83 kW, about 84.3 hp. In US units the same relation is hp = lb-ft × rpm ÷ 5,252, which is why torque and horsepower curves drawn on the same scale in those units cross at 5,252 rpm. Gears trade one for the other: a 3:1 reduction turns 43.47 N·m into about 130.4 N·m at the output shaft, at a third of the speed, before friction losses.
If torque and energy are both newton-metres, why aren't they the same thing?
Torque is a turning effect that exists even when nothing moves; work is energy transferred when something does move. They connect through the angle turned: work = torque × angle in radians. Holding 10 N·m on a stuck nut does no work at all, while turning a shaft one full revolution against 10 N·m takes 10 × 2π = 62.83 joules. That is why torque is written in N·m and energy in joules.
How do I use torque to balance a seesaw or a lever?
An object stays balanced when the torques on either side of the pivot are equal and opposite. A 30 kg child sitting 1.5 m from the pivot gives 30 × 1.5 = 45 kgf·m, so a 45 kg adult balances them sitting 1 m out on the other side. Enter each side separately, with the weight as kilogram-force and the angle at 90° for a level board, and the two results match. The board itself adds nothing when the pivot is at its middle, because its weight then acts through the pivot with a lever arm of zero.
Can I enter a mass in kilograms as the force?
Choose kilogram-force, which treats each kilogram as its weight under standard gravity, 9.80665 N. A kilogram is a mass, not a force, so the conversion assumes the load hangs or rests under ordinary gravity. A 10 kg weight hung from the end of a horizontal 0.5 m bar produces 10 × 9.80665 × 0.5 = 49.03 N·m about the bar's fixed end.
Can I add up the torques from several forces?
Yes, but one at a time. Work out each force's torque about the same pivot, then give it a sign by the way it turns the object, for example counter-clockwise positive and clockwise negative, because only magnitudes are shown. Adding the signed values gives the net torque: 43.47 N·m clockwise against 30 N·m counter-clockwise leaves 13.47 N·m clockwise. A net of zero means the forces balance and the object does not start to rotate.
Why does a longer handle loosen a nut that a short spanner will not?
Because torque rises in direct proportion to the lever arm while your strength stays the same. Pulling 40 lbf on a 9-inch spanner gives 30 lb-ft; the same pull on an 18-inch breaker bar gives 60 lb-ft, or 81.35 N·m. Do not slip a pipe over a torque wrench to get the same effect, since it changes the arm the scale was set for and can overload the mechanism.

References& sources.

  1. [1]Halliday, D., Resnick, R., and Walker, J. (2013). Fundamentals of Physics, 10th ed. John Wiley & Sons. ISBN 978-1-118-23072-5
  2. [2]Thomson, J. (1884). "On the Measurement of Electric Resistance." Phil. Mag. 18(5):316–320.
  3. [3]BIPM (2019). The International System of Units (SI Brochure), 9th ed.
  4. [4]Shigley, J.E. and Mischke, C.R. (1989). Mechanical Engineering Design, 5th ed. McGraw-Hill. ISBN 978-0-07-056899-0
  5. [5]Meriam, J.L. and Kraige, L.G. (2012). Engineering Mechanics: Statics, 7th ed. John Wiley & Sons. ISBN 978-0-470-61473-0

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τ = r × F × sin θ
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