Physics

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Foot-Pounds of Energy Calculator.

Convert foot-pound-force energy to joules.

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Use Cases

Physics homework and lab work

Quickly convert energy values from foot-pounds to joules for assignments, experiments, or verifying calculations in mechanics and energy topics.

Example: Convert 25 ft·lbf to joules: 25 × 1.3558 = 33.9 J

Engineering and mechanical applications

Convert torque or energy specifications from imperial units to SI units for design, analysis, or comparing data across different systems.

Example: A torque spec of 50 ft·lbf equals 67.8 J of energy.

Frequently Asked Questions

What is a foot-pound of energy?: A foot-pound (ft·lbf) is a unit of energy or work in the imperial and US customary systems. It represents the energy transferred when a force of one pound-force is applied over a distance of one foot.
How do I convert foot-pounds to joules?: Multiply the energy value in foot-pounds by 1.3558179483 to get joules. For example, 10 ft·lbf equals 13.558179483 joules.
Why is the conversion factor 1.3558179483?: One foot-pound is defined as the work done by a one-pound force over one foot. Since 1 pound-force equals 4.4482216153 newtons and 1 foot equals 0.3048 meters, the product gives 1.3558179483 joules.

Tips & Common Mistakes

Tips

Enter the energy value in foot-pounds (ft·lbf) exactly as given; the calculator will output the equivalent in joules.
Remember that 1 ft·lbf is approximately 1.356 joules, so you can do quick mental estimates by multiplying by 1.36.
For precise engineering work, use the exact conversion factor 1.3558179483 to avoid rounding errors.
Double-check your input unit: this calculator only converts foot-pound-force energy, not pound-feet (torque) or other energy units.

Common Mistakes to Avoid

Confusing foot-pounds (energy) with pound-feet (torque) – they are different physical quantities, though the conversion factor is similar.
Using the wrong conversion factor, such as 1.356 instead of the more precise 1.3558179483, which can cause small errors in large values.
Forgetting to convert from foot-pounds to joules when the problem requires SI units, leading to incorrect results in physics calculations.

Last updated: August 13, 2026

Use Cases

Physics homework and lab work

Quickly convert energy values from foot-pounds to joules for assignments, experiments, or verifying calculations in mechanics and energy topics.

Example: Convert 25 ft·lbf to joules: 25 × 1.3558 = 33.9 J

Engineering and mechanical applications

Convert torque or energy specifications from imperial units to SI units for design, analysis, or comparing data across different systems.

Example: A torque spec of 50 ft·lbf equals 67.8 J of energy.

Frequently Asked Questions

What is a foot-pound of energy?

A foot-pound (ft·lbf) is a unit of energy or work in the imperial and US customary systems. It represents the energy transferred when a force of one pound-force is applied over a distance of one foot.

How do I convert foot-pounds to joules?

Multiply the energy value in foot-pounds by 1.3558179483 to get joules. For example, 10 ft·lbf equals 13.558179483 joules.

Why is the conversion factor 1.3558179483?

One foot-pound is defined as the work done by a one-pound force over one foot. Since 1 pound-force equals 4.4482216153 newtons and 1 foot equals 0.3048 meters, the product gives 1.3558179483 joules.

Tips & Common Mistakes

Tips

Enter the energy value in foot-pounds (ft·lbf) exactly as given; the calculator will output the equivalent in joules.
Remember that 1 ft·lbf is approximately 1.356 joules, so you can do quick mental estimates by multiplying by 1.36.
For precise engineering work, use the exact conversion factor 1.3558179483 to avoid rounding errors.
Double-check your input unit: this calculator only converts foot-pound-force energy, not pound-feet (torque) or other energy units.

Common Mistakes to Avoid

Confusing foot-pounds (energy) with pound-feet (torque) – they are different physical quantities, though the conversion factor is similar.
Using the wrong conversion factor, such as 1.356 instead of the more precise 1.3558179483, which can cause small errors in large values.
Forgetting to convert from foot-pounds to joules when the problem requires SI units, leading to incorrect results in physics calculations.

Use Cases

Physics homework and lab work

Quickly convert energy values from foot-pounds to joules for assignments, experiments, or verifying calculations in mechanics and energy topics.

Example: Convert 25 ft·lbf to joules: 25 × 1.3558 = 33.9 J

Engineering and mechanical applications

Convert torque or energy specifications from imperial units to SI units for design, analysis, or comparing data across different systems.

Example: A torque spec of 50 ft·lbf equals 67.8 J of energy.

Frequently Asked Questions

What is a foot-pound of energy?

How do I convert foot-pounds to joules?

Multiply the energy value in foot-pounds by 1.3558179483 to get joules. For example, 10 ft·lbf equals 13.558179483 joules.

Why is the conversion factor 1.3558179483?

One foot-pound is defined as the work done by a one-pound force over one foot. Since 1 pound-force equals 4.4482216153 newtons and 1 foot equals 0.3048 meters, the product gives 1.3558179483 joules.

Tips & Common Mistakes

Tips

Enter the energy value in foot-pounds (ft·lbf) exactly as given; the calculator will output the equivalent in joules.
Remember that 1 ft·lbf is approximately 1.356 joules, so you can do quick mental estimates by multiplying by 1.36.
For precise engineering work, use the exact conversion factor 1.3558179483 to avoid rounding errors.
Double-check your input unit: this calculator only converts foot-pound-force energy, not pound-feet (torque) or other energy units.

Common Mistakes to Avoid

Confusing foot-pounds (energy) with pound-feet (torque) – they are different physical quantities, though the conversion factor is similar.
Using the wrong conversion factor, such as 1.356 instead of the more precise 1.3558179483, which can cause small errors in large values.
Forgetting to convert from foot-pounds to joules when the problem requires SI units, leading to incorrect results in physics calculations.