# Torque & Horsepower

## Torque and Horsepower Equivalents

 TORQUE: A turning moment or twisting effort that can be expressed in foot-pounds  or pound-feet. A foot-pound is the amount of energy expended  in lifting a one-pound mass a distance of one  foot against the pull of gravity.  FOOT-POUNDS INDICATES ENERGY A pound-foot is the movement created  by a force of one pound applied to the  end of a lever arm one foot long  POUND-FEET INDICATES TORQUE
 Torque (in Pound-Inches) =  63,025 x HP / RPM  = Force x Lever Arm (in inches)    Torque (in Pound-Feet) =  5,252 x HP / RPM  = Force x Lever Arm (in Feet)    Force = Working Load in Pounds  FPM = Feet Per Minute  RPM = Revolutions Per Minute  Lever Arm = Distance from the Force to the center or rotation           in Inches or Feet. EXAMPLE:  25 HP at 150 RPM = 875 Pound-Feet Torque  2.5 HP at 150 RPM = 1050.4 Pound-Inches Torque
 HORSEPOWER (HP): A common unit of mechanical power. One HP is the rate of  work required to raise 33,000 pounds one foot in one minute. HP = Force x FPM / 33,000 HP = Torque (in Pound-Inches) x RPM / 63,025 HP = Torque (in Pound-Feet) x RPM / 5,252
 OVERHUNG LOADS:   A bending force imposed on a shaft due to the torque  transmitted by V-drives, chain drives and other power transmission devices, other  than flexible couplings. Most motor and reducer manufacturers list the maximum values allowable for overhung  loads. It is desirable that these figures be compared with the load actually imposed by  the connected drive.
 O.H.L. =  63,000 x HP x F / N x R  HP = Transmitted HP x service factor  N = RPM of shaft  R = Radius of sprocket, pulley, etc.  F = Factor 1.00 for single chain drives             1.10 for Timing belt drives  F=      1.25 for spur or helical gear             1.25 for double chain drives             1.50 for V-belt drives             2.50 for flat belt drives
 Weights of the drive components are usually negligible. The formula is based on the assumption that the load is  applied at appoint equal to one shaft diameter from the bearing face. Factor F depends on the type of drive used.  EXAMPLE: Find the over hung load imposed on a reducer by a double chain drive transmission 7 HP @ 30 RPM.  The pitch diameter of the sprockets is 10’’, service factor is 1.3. O.H.L. =  (63,000) (7 x 1.3) (1.25) / 30 x 5