Techno Linear Motion Catalog30Technical Information1000 1.8 p 1T= 2 x 9.2 x ––––– x ––––––– x –––0.5 180 24T= 48.2 oz•in to accelerate inertiaB.Calculating torque required to accelerate and raisea weight using a drum and string. The total torquewhich the motor must supply includes the torquerequired to:a. accelerate the weightb. accelerate the drumc. accelerate the motor rotord. lift the weightThe rotational equivalent of the weight and the radiusof the drum is:I(eq)= wr2where:I(eq)= equivalent inertia (lb •in2)w = weight (lb)r= radius of drum (in)Example:Assume the following conditions:Weight = 5 lbs (80 oz)Drum = 3" O.D., 1.5" radiusVelocity = 15 ft per secondTime to Reach Velocity = 0.5 secondsMotor Rotor Inertia = 2.5 lb •in2Drum Inertia = 4.5 lb • in2 (for a 3" dia x 2" long steel drum)I(eq)= 5 x (1.5)2= 11.25 lb • in2I(drum)= 4.5 lb •in2I(rotor)= 2.5 lb • in2–––––––––––––––––––––––––––I(total)= 18.25 lb • in2since the velocity is 15 ft/sec using a 3" drum, the velocity in rev/sec can be calculated:15 x 12speed = –––––––– = 19.1 rev/sec3 pThe motor step angle is 1.8°, or 200 steps per revolution. Therefore:w'= 19.1 x 200 = 3820 steps per secondT= 2 x IOxxxT= 2 x 18.25 xxxT= 364 oz•in = torque required to accelerate the system. Torque required to lift weight equals:T= w r= 80 x 1.5 = 120 oz•inw–––tp q––––––1801–––243820––––––0.53.1416 x 1.8––––––––––––1801–––24WrMotor
