Techno Linear Motion Catalog31Technical InformationTotal torque required is, therefore:364 oz • in (accelerating torque)120 oz • in (lifting torque)––––––––––––––––––––––––––––––––––––––––––484 oz • in (total torque)C.Calculating the torque required to acceleratea mass moving horizontally and driven by a rackand pinion or similar device. The total torquewhich the motor must provide includes thetorque required to:a. accelerate the weight, including that ofthe rackb. accelerate the gearc. accelerate the motor rotord. overcome frictional forcesto calculate the rotational equivalent of theweight:I(eq)= w r2where:w = weight (lb)r= radius (in)Example:Assume that:Weight = 5 lbGear Pitch Diameter = 3 inGear Radius = 1.5 inVelocity = 15 ft per secondTime to Reach Velocity = 0.5 secondsPinion Inertia = 4.5 lb •in2 (assumed)Motor Rotor Inertia = 2.5 lb •in2I(eq)= wr2= 5 x (1.5)2= 11.25 lb×in2I(pinion)= 4.5 lb •in2I(rotor)= 2.5 lb •in2––––––––––––––––––––––––––––––––––––––––––I(total)= 18.25 lb • in2Velocity is 15 ft per second with a 3" pitch diameter gear. Therefore:15 x 12speed = –––––––– = 19.1 revolutions per second3 pThe motor step angle is 1.8° (200 steps per revolution). Therefore, the velocity in steps per second is:w'= 19.1 x 200 = 3820 steps per secondMotorrw
