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. 
A pound-foot is the movement created 
by a force of one pound applied to the 
end of a lever arm one foot long 
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. 
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