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Foot Pound to Joule Converter

Convert foot pound joule between units instantly. Includes conversion tables, common equivalents, and calculation formulas.

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Unit Conversion

Foot Pound to Joule Converter

Convert between foot-pounds, joules, BTU, calories, kilowatt-hours, and other energy units. Instant energy and work unit conversion.

Last updated: December 2025

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1 Foot-pounds
1.35582 J
Foot-pounds
1
Kilojoules
0.001356
BTU
0.001285

All Conversions

Joules (J)1.356
Kilojoules (kJ)0.001
Foot-pounds (ft-lbf)1
Calories (cal)0.324
Kilocalories (kcal)0
BTU0.001
Kilowatt-hours (kWh)0
Newton-meters (Nm)1.356
Ergs13,558,200
Your Result
1 Foot-pounds = 1.35582 J = 1 ft-lbf
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Formula

J = ft-lbf x 1.35582 | 1 BTU = 1055.06 J | 1 cal = 4.184 J

One foot-pound equals 1.35582 joules. The joule is the SI unit of energy defined as the work done by a one-newton force through one meter. Key relationships: 1 BTU = 1,055.06 J, 1 calorie = 4.184 J, 1 kWh = 3,600,000 J. The foot-pound measures work in Imperial units (one pound of force through one foot).

Last reviewed: December 2025

Worked Examples

Example 1: Engine Torque Conversion

An engine produces 350 ft-lbf of torque. Convert to joules and newton-meters.
Solution:
Joules: 350 x 1.35582 = 474.537 J Newton-meters: 474.537 Nm (same as joules for energy) Kilojoules: 474.537 / 1000 = 0.4745 kJ BTU: 474.537 / 1055.06 = 0.4498 BTU
Result: 350 ft-lbf = 474.54 J = 474.54 Nm

Example 2: Energy Comparison

Convert 5,000 joules to foot-pounds, calories, and BTU.
Solution:
Foot-pounds: 5,000 / 1.35582 = 3,688.10 ft-lbf Calories: 5,000 / 4.184 = 1,195.22 cal Kilocalories: 5,000 / 4,184 = 1.195 kcal BTU: 5,000 / 1,055.06 = 4.739 BTU
Result: 5,000 J = 3,688.1 ft-lbf = 1,195.2 cal = 4.74 BTU
Expert Insights

Background & Theory

The Foot Pound to Joule Converter applies the following established principles and formulas. Unit conversion is the process of expressing a quantity in a different unit of measurement while preserving its physical meaning. At the foundation of modern measurement lies the International System of Units (SI), which defines seven base units: the meter for length, kilogram for mass, second for time, ampere for electric current, kelvin for thermodynamic temperature, mole for amount of substance, and candela for luminous intensity. All other units, called derived units, are defined as algebraic combinations of these seven. Dimensional analysis is the principal method for performing unit conversions. By treating units as algebraic quantities that can be multiplied, divided, and cancelled, a conversion factor chain allows a value expressed in one unit to be rewritten in another without altering its physical magnitude. For example, to convert 60 miles per hour to meters per second, one multiplies by a chain of conversion factors each equal to one: (1609.34 m / 1 mile) ร— (1 hour / 3600 s). Metric prefixes enable compact expression of quantities across extreme ranges of magnitude. Standard prefixes span from nano (10^-9) through micro (10^-6) and milli (10^-3) up through kilo (10^3), mega (10^6), and giga (10^9), and beyond in both directions. These prefixes are strictly multiplicative and apply consistently to any SI base or derived unit. Temperature conversions require affine transformations rather than simple scaling. To convert Celsius to Fahrenheit the formula is ยฐF = (ยฐC ร— 9/5) + 32, while the conversion to the absolute Kelvin scale is K = ยฐC + 273.15. These formulas reflect the different zero points and degree-size conventions of each scale. Significant figures govern how precision is preserved through calculations. A result should not express more precision than the least precise input value permits. In digital storage, IEEE and IEC standards distinguish between decimal prefixes (kilobyte = 1000 bytes) and binary prefixes (kibibyte = 1024 bytes), a distinction that has practical consequences for how storage capacity is reported by manufacturers versus operating systems. Unit coherence โ€” ensuring that all quantities in an equation share a consistent unit system โ€” is essential for obtaining correct results.

History

The history behind the Foot Pound to Joule Converter traces back through the following developments. Human beings have been measuring and comparing quantities since before recorded history. The earliest known measurement units were body-based: the cubit (the distance from elbow to fingertip), the foot, the hand, and the digit. The furlong originated as the length of a furrow a team of oxen could plow without resting. These anthropomorphic standards were practical for local use but differed between regions and kingdoms, creating persistent difficulties in trade and construction. The ancient Egyptians standardized the royal cubit at approximately 52.4 centimeters and distributed calibrated granite rods to ensure consistency across building projects, including the pyramids. Roman engineers used the mile (mille passuum, one thousand double paces) and spread these standards throughout their empire via road networks. Despite these efforts, measurement diversity persisted across medieval Europe, hampering commerce. The French Revolution created political will for radical standardization. In 1795 France officially adopted the metric system, defining the meter as one ten-millionth of the distance from the equator to the North Pole along the Paris meridian. This gave the world its first fully decimal, rationally constructed measurement system. The Metre Convention of 1875 established the International Bureau of Weights and Measures (BIPM) in Sevres, France, creating a permanent international body to maintain physical artifact standards and coordinate global metrology. For over a century, the kilogram was defined by a platinum-iridium cylinder locked in a vault near Paris. In 1999, a stark demonstration of what unit inconsistency costs occurred when NASA's Mars Climate Orbiter was lost because one engineering team used pound-force seconds while another used newton seconds. The spacecraft entered the Martian atmosphere at the wrong angle and was destroyed, at a cost of 327 million dollars. In 2019 the SI underwent its most significant revision, redefining all seven base units in terms of fixed numerical values of fundamental physical constants such as the speed of light, Planck's constant, and the elementary charge. This eliminated any reliance on physical artifacts and made the measurement system permanently stable and universally reproducible.

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Frequently Asked Questions

To convert foot-pounds to joules, multiply by 1.35582. One foot-pound (ft-lbf) is the energy needed to exert a force of one pound through a distance of one foot. For example, 100 foot-pounds equals 135.582 joules. The foot-pound is commonly used in US engineering for torque specifications on engines and fasteners, while the joule is the SI standard energy unit used internationally.
A foot-pound of energy (ft-lbf) and a foot-pound of torque (lb-ft) have the same dimensions but different physical meanings. Energy foot-pounds measure work done when a force moves through a distance. Torque foot-pounds (often written lb-ft in automotive contexts) measure rotational force around an axis. Numerically they are equivalent, but conventionally energy uses ft-lbf while torque uses lb-ft to distinguish the concepts.
An erg is the CGS (centimeter-gram-second) unit of energy equal to exactly 0.0000001 joules (10^-7 J). One joule equals 10 million ergs. The erg is defined as the work done by a force of one dyne acting through one centimeter. While largely replaced by the joule in modern science, ergs still appear in astrophysics literature where extremely large energies are expressed in ergs. A supernova releases about 10^44 joules or 10^51 ergs of energy.
You may use the results for reference and educational purposes. For professional reports, academic papers, or critical decisions, we recommend verifying outputs against peer-reviewed sources or consulting a qualified expert in the relevant field.
All calculations use established mathematical formulas and are performed with high-precision arithmetic. Results are accurate to the precision shown. For critical decisions in finance, medicine, or engineering, always verify results with a qualified professional.
No. All calculations run entirely in your browser using JavaScript. No data you enter is ever transmitted to any server or stored anywhere. Your inputs remain completely private.
Educational Note: This calculator is provided for educational and informational purposes. Results are based on the formulas and inputs provided. Always verify important calculations independently. NovaCalculator processes calculator inputs client-side; optional analytics follow visitor consent settings. ยฉ 2024โ€“2026 NovaCalculator.

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Formula

J = ft-lbf x 1.35582 | 1 BTU = 1055.06 J | 1 cal = 4.184 J

One foot-pound equals 1.35582 joules. The joule is the SI unit of energy defined as the work done by a one-newton force through one meter. Key relationships: 1 BTU = 1,055.06 J, 1 calorie = 4.184 J, 1 kWh = 3,600,000 J. The foot-pound measures work in Imperial units (one pound of force through one foot).

Worked Examples

Example 1: Engine Torque Conversion

Problem: An engine produces 350 ft-lbf of torque. Convert to joules and newton-meters.

Solution: Joules: 350 x 1.35582 = 474.537 J\nNewton-meters: 474.537 Nm (same as joules for energy)\nKilojoules: 474.537 / 1000 = 0.4745 kJ\nBTU: 474.537 / 1055.06 = 0.4498 BTU

Result: 350 ft-lbf = 474.54 J = 474.54 Nm

Example 2: Energy Comparison

Problem: Convert 5,000 joules to foot-pounds, calories, and BTU.

Solution: Foot-pounds: 5,000 / 1.35582 = 3,688.10 ft-lbf\nCalories: 5,000 / 4.184 = 1,195.22 cal\nKilocalories: 5,000 / 4,184 = 1.195 kcal\nBTU: 5,000 / 1,055.06 = 4.739 BTU

Result: 5,000 J = 3,688.1 ft-lbf = 1,195.2 cal = 4.74 BTU

Frequently Asked Questions

How do I convert foot-pounds to joules?

To convert foot-pounds to joules, multiply by 1.35582. One foot-pound (ft-lbf) is the energy needed to exert a force of one pound through a distance of one foot. For example, 100 foot-pounds equals 135.582 joules. The foot-pound is commonly used in US engineering for torque specifications on engines and fasteners, while the joule is the SI standard energy unit used internationally.

What is the difference between a foot-pound of energy and a foot-pound of torque?

A foot-pound of energy (ft-lbf) and a foot-pound of torque (lb-ft) have the same dimensions but different physical meanings. Energy foot-pounds measure work done when a force moves through a distance. Torque foot-pounds (often written lb-ft in automotive contexts) measure rotational force around an axis. Numerically they are equivalent, but conventionally energy uses ft-lbf while torque uses lb-ft to distinguish the concepts.

What is an erg and how does it relate to a joule?

An erg is the CGS (centimeter-gram-second) unit of energy equal to exactly 0.0000001 joules (10^-7 J). One joule equals 10 million ergs. The erg is defined as the work done by a force of one dyne acting through one centimeter. While largely replaced by the joule in modern science, ergs still appear in astrophysics literature where extremely large energies are expressed in ergs. A supernova releases about 10^44 joules or 10^51 ergs of energy.

Is my data stored or sent to a server?

No. All calculations run entirely in your browser using JavaScript. No data you enter is ever transmitted to any server or stored anywhere. Your inputs remain completely private.

Can I use the results for professional or academic purposes?

You may use the results for reference and educational purposes. For professional reports, academic papers, or critical decisions, we recommend verifying outputs against peer-reviewed sources or consulting a qualified expert in the relevant field.

Does Foot Pound to Joule Converter work offline?

Once the page is loaded, the calculation logic runs entirely in your browser. If you have already opened the page, most calculators will continue to work even if your internet connection is lost, since no server requests are needed for computation.

References

Reviewed by Manoj Kumar, Mathematics Educator ยท Editorial policy