How many foot pound in 1 erg? The answer is 7.3756214927727E-8. We assume you are converting between foot-pound force and erg. You can view more details on each measurement unit: foot pound or erg The SI derived unit for energy is the joule. 1 joule is equal to 0.73756214927727 foot pound, or 10000000 erg. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between foot pounds and ergs. Type in your own numbers in the form to convert the units!
1 foot pound to erg = 13558179.48331 erg
2 foot pound to erg = 27116358.96663 erg
3 foot pound to erg = 40674538.44994 erg
4 foot pound to erg = 54232717.93326 erg
5 foot pound to erg = 67790897.41657 erg
6 foot pound to erg = 81349076.89988 erg
7 foot pound to erg = 94907256.3832 erg
8 foot pound to erg = 108465435.86651 erg
9 foot pound to erg = 122023615.34983 erg
10 foot pound to erg = 135581794.83314 erg
You can do the reverse unit conversion from erg to foot pound, or enter any two units below:
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.
An erg is the unit of energy and mechanical work in the centimetre-gram-second (CGS) system of units, symbol "erg". Its name is derived from the Greek word meaning "work".
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