Foot-pounds Force to British Thermal Units Converter
1 foot-pound force ≈ 0.00128506746 British thermal units
Energy and torque share these dimensions. For fastener torque see the Torque category, where the same quantity is written lbf·ft.
Several BTU definitions exist. This converter uses the International Table BTU, the one used by HVAC equipment ratings.
1 foot-pound force equals 0.00128507 British thermal units
How to convert foot-pounds force to British thermal units
Calories here are thermochemical (4.184 J), while BTU uses the International Table definition. A kilowatt-hour is energy (3.6 MJ), not power; a food Calorie is a kilocalorie.
BTU = ft·lbf × 0.001285067463One foot-pound force is 0.001285067463 British thermal units, so a single multiplication converts any value.
0.0128507 BTU ≈ 10 × 0.001285067463Energy and torque share these dimensions. For fastener torque see the Torque category, where the same quantity is written lbf·ft.
Several BTU definitions exist. This converter uses the International Table BTU, the one used by HVAC equipment ratings.
Foot-pounds force to British thermal units conversion table
Common values in foot-pounds force, converted to British thermal units.
| Foot-pounds force (ft·lbf) | British thermal units (BTU) |
|---|---|
| 1 | 0.0012850675 |
| 2 | 0.0025701349 |
| 3 | 0.0038552024 |
| 4 | 0.0051402699 |
| 5 | 0.0064253373 |
| 10 | 0.012850675 |
| 20 | 0.025701349 |
| 25 | 0.032126687 |
| 50 | 0.064253373 |
| 100 | 0.12850675 |
| 250 | 0.32126687 |
| 500 | 0.64253373 |
| 1,000 | 1.2850675 |
About these units
foot-pound forceft·lbf
The work done by one pound-force acting through one foot.
British thermal unitBTU
The International Table BTU: 1,055.05585262 joules. The US heating and cooling unit.
Reference: NIST Special Publication 811.
Sources and precision
Table results are rounded to 8 significant digits; formula constants may be rounded to 10. Worked example answers use 6 significant digits. The converter starts at 6 significant digits and lets you change display precision. Calculations use the stored factors without intermediate rounding. Exact unit definitions do not make an input measurement more precise; conventional and measured factors follow the assumptions stated here.