Convert foot-pound force to dekajoule -
Conversion of Measurement Units

›› Convert foot-pound force to dekajoule

›› More information from the unit converter

How many foot-pound force in 1 dekajoule? The answer is 7.37562149277.
We assume you are converting between foot-pound force and dekajoule.
You can view more details on each measurement unit:
foot-pound force or dekajoule
The SI derived unit for energy is the joule.
1 joule is equal to 0.737562149277 foot-pound force, or 0.1 dekajoule.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between foot pounds and dekajoules.
Type in your own numbers in the form to convert the units!

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›› Definition: Foot pound

In physics, a foot-pound force (symbol ft·lbf or ft·lbf), usually abbreviated to foot-pound, is an Imperial and U.S. customary unit of mechanical work, or energy, although in scientific fields one commonly uses the equivalent metric unit of the joule (J). There are approximately 1.356 J/ft·lbf, or exactly 1.355 817 948 331 400 4 J/ft·lbf.

›› Definition: Dekajoule

The SI prefix "deka" represents a factor of 10¹, or in scientific notation, 1E1.

So 1 dekajoule = 10¹ joules.

The definition of a joule is as follows:

The joule (symbol J, also called newton meter, watt second, or coulomb volt) is the SI unit of energy and work. The unit is pronounced to rhyme with "tool", and is named in honor of the physicist James Prescott Joule (1818-1889).

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