Each row uses τ = rF sin(θ). Counterclockwise is positive; clockwise is negative.
| Torque unit | Value |
|---|
Torque, or moment of force, is the turning effect of a force about a pivot. This calculator uses the magnitude relation |τ| = rF sin(θ), where r is the physical distance from pivot to application point, F is force, and θ is the angle between their vectors. The perpendicular moment arm is r⊥ = r sin(θ), so |τ| = Fr⊥.
| Angle | sin(θ) | Maximum torque |
|---|---|---|
| 0° / 180° | 0 | 0% |
| 15° / 165° | 0.2588 | 25.88% |
| 30° / 150° | 0.5 | 50% |
| 45° / 135° | 0.7071 | 70.71% |
| 60° / 120° | 0.8660 | 86.60% |
| 90° | 1 | 100% |
| Symbol | Meaning | Typical units |
|---|---|---|
| τ | Torque magnitude | N·m, lbf·ft |
| F | Force magnitude | N, kN, lbf |
| r | Pivot-to-force distance | m, cm, ft, in |
| θ | Angle between r and F | degrees, radians |
| r⊥ | Perpendicular moment arm | m, ft, in |
| Starting value | Equivalent |
|---|---|
| 1 N·m | 0.737562 lbf·ft |
| 1 N·m | 8.85075 lbf·in |
| 1 lbf·ft | 1.355818 N·m |
| 1 lbf·in | 0.112985 N·m |
| 1 kgf·cm | 0.0980665 N·m |
| 1 kgf·m | 9.80665 N·m |
Given F = 50 N, r = 30 cm, and θ = 90°. Convert r: 30 cm × 0.01 m/cm = 0.30 m.
|τ| = (0.30 m)(50 N)sin(90°) = (0.30)(50)(1) = 15 N·m.
Interpretation: the perpendicular push applies 15 N·m about the bolt.
Target τ = 85 lbf·ft, r = 18 in, and θ = 90°. Convert: 85 × 1.355817948 = 115.2445 N·m; 18 × 0.0254 = 0.4572 m.
F = 115.2445 / [(0.4572)sin(90°)] = 252.066 N. Convert: 252.066 / 4.448221615 = 56.67 lbf.
Interpretation: about 56.7 pounds-force at the end of the 18-inch bar reaches the target ideally.
Given τ = 60 N·cm, F = 12 N, and θ = 90°. Convert torque: 60 × 0.01 = 0.60 N·m.
r = 0.60 / [(12)sin(90°)] = 0.050 m. Convert: 0.050 × 100 = 5.0 cm.
Interpretation: apply the 12 N perpendicular force at least 5 cm from the pivot.
Given τ = 30 lbf·ft, F = 80 lbf, and r = 12 in. Convert r: 12 in ÷ 12 = 1 ft; the consistent ratio is 30 / [(1)(80)] = 0.375.
θ = sin−1(0.375) = 22.02°; the supplementary solution is 180° − 22.02° = 157.98°.
Interpretation: either direction has the same torque magnitude; the actual geometry determines which angle applies.
Calculation method: inputs are converted to N, m, N·m, and radians; the selected rearrangement of |τ| = rF sin(θ) is evaluated at full JavaScript precision; only the displayed answer is rounded. The angle solver rejects |τ|/(rF) > 1 as physically impossible for this model.
Rearrange the magnitude equation to theta = arcsin(|tau|/(rF)). The ratio must be between 0 and 1. Between 0 and 180 degrees, most nonzero ratios have a principal angle and a supplementary angle that produce the same torque magnitude.
They share the same base dimensions but describe different quantities. Torque is a vector-like moment of force and is written N·m; energy is a scalar and the special SI name joule, J, is used.
Enter the radius: the distance from the pivot axis to the force application point. If only a diameter is given, divide it by two before entering it.
A common planar convention treats counterclockwise torque as positive and clockwise torque as negative. The single-force solver reports magnitude; Advanced net torque applies that sign convention.
In torque specifications, lb-ft and ft-lb are commonly used aliases for pound-force foot, properly written lbf·ft. This calculator labels pound-force explicitly so it is not confused with pound-mass.
Torque is not power. Rotational power also requires angular speed: P = tau omega. You need rpm or another rotational-speed value before converting the resulting power to horsepower.
Arm length r runs from the pivot to the force application point. The perpendicular moment arm r_perp is the shortest distance from the pivot to the force line of action and equals r sin(theta).
