pound-foot | |
---|---|

Unit system | British Gravitational System, English Engineering Units |

Unit of | Torque |

Symbol | lbf⋅ft, lb-ft |

Conversions | |

1 lbf⋅ft in ... | ... is equal to ... |

SI units | ≈ 1.355818 N⋅m^{[1]} |

Gravitational metric system | ≈ 0.1382550 kgf⋅m |

A **pound-foot** (**lb⋅ft**), abbreviated from **pound-force foot (lbf · ft)**, is a unit of torque representing one pound of force acting at a perpendicular distance of one foot from a pivot point.^{[2]} Conversely one foot pound-force (ft · lbf) is the moment about an axis that applies one pound-force at a radius of one foot.

The value in SI units is given by multiplying the following exact factors:

- One pound (mass) = 0.45359237 kilograms
^{[3]}^{[4]}

- Standard gravity = 9.80665 m/s
^{2}^{[5]}

- One foot = 0.3048 m
^{[6]}

This gives the exact conversion factor:

- One pound-foot = 1.3558179483314004 newton metres.

The name "pound-foot", intended to minimize confusion with the foot-pound as a unit of work, was apparently first proposed by British physicist Arthur Mason Worthington.^{[7]}

Despite this, in practice torque units are commonly called the foot-pound (denoted as either lb-ft or ft-lb) or the inch-pound (denoted as in-lb).^{[8]}^{[9]} Practitioners depend on context and the hyphenated abbreviations to know that these refer to neither energy nor moment of mass (as the symbol ft-lb rather than lbf-ft would imply).

Similarly, an **inch-pound** (or *pound-inch*) is the torque of one pound of force applied to one inch of distance from the pivot, and is equal to 1⁄12 lbf⋅ft (0.1129848 N⋅m). It is commonly used on torque wrenches and torque screwdrivers for setting specific fastener tension.