Calculate total kinetic energy that brakes must absorb during a stop and the resulting energy per unit rotor mass.
Input Parameters
Results
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Kinetic Energy
ft-lbs
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Kinetic Energy
kJ
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Energy Per Rotor
BTU
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Specific Energy
BTU/lb
Note: A 3500 lb vehicle at 70 mph has approximately 750 kJ of kinetic energy. Brake rotors must absorb and dissipate this energy as heat during each stop.
Reference
KE = 0.5 x (W/g) x (V1² - V2²)
where V in ft/s, W in lbs, g = 32.174 ft/s²
KE (kJ) = KE (ft-lbs) x 1.35582 / 1000
KE (BTU) = KE (ft-lbs) x 0.001285
Specific Energy = Total BTU / Rotor Weight