How many foot pound in 1 GeV? The answer is 1.1817053329112E-10. We assume you are converting between foot-pound force and gigaelectronvolt. You can view more details on each measurement unit: foot pound or GeV The SI derived unit for energy is the joule. 1 joule is equal to 0.73756214927727 foot pound, or 6241506479.9632 GeV. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between foot pounds and gigaelectronvolts. Type in your own numbers in the form to convert the units!
1 foot pound to GeV = 8462346510.1609 GeV
2 foot pound to GeV = 16924693020.322 GeV
3 foot pound to GeV = 25387039530.483 GeV
4 foot pound to GeV = 33849386040.644 GeV
5 foot pound to GeV = 42311732550.804 GeV
6 foot pound to GeV = 50774079060.965 GeV
7 foot pound to GeV = 59236425571.126 GeV
8 foot pound to GeV = 67698772081.287 GeV
9 foot pound to GeV = 76161118591.448 GeV
10 foot pound to GeV = 84623465101.609 GeV
You can do the reverse unit conversion from GeV 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.
The SI prefix "giga" represents a factor of 109, or in exponential notation, 1E9.
So 1 gigaelectronvolt = 109 .
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