# How to determine how much energy recovered from regenerative braking



## brian_ (Feb 7, 2017)

It sounds like you could use some basic physics education. There should be lots of good material on websites.

The kinetic energy of a moving body is 
E = m * v²
where
_E_ is the energy (in joules if using SI units; a joule is a watt-second)
_m_ is the mass (in kilograms if using SI units)
_v_ is the velocity (in metres per second if using SI units)​
The energy which could possibly be recovered in regenerative braking is the difference between the kinetic energy of the moving vehicle at the beginning (v1), minus the kinetic energy of the moving vehicle at the end (v2), so
(m * v1²) - (m * v2²)
or
m (v1² - v2²)

You can also use your suggested power approach. The power dissipated by brakes or absorbed by regenerative braking is the force multiplied by the speed:
P = F * v
_P_ is the power (in watts if using SI units)
_v_ is the velocity (in metres per second if using SI units)​... but you need the force. To decelerate a mass, the force is the rate of deceleration multiplied by the mass
F = m * a
_F_ is the force (in newtons if using SI units)
_m_ is the mass (in kilograms if using SI units)
_a_ is the acceleration (in metres per second squared if using SI units)​
Example: if braking a one-ton (1000 kg or 2200 pound) mass at half "g" (half of the acceleration of gravity, so 5 m/s²), the force would be 1000 kg * 5 m/s = 5000 N (or 5 kN), and if the vehicle is moving at 20 m/s (72 km/h or 45 mph) the power would be 5000 N * 20 m/s = 100,000 N (or 100 kW). That's about as hard as anyone brakes on the street - normal would be less.

Any energy used by drag (rolling drag and aerodynamic drag) will not be available to be recovered, and of what is available there are the losses in the generator (motor) and battery.


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