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Techno Linear Motion Catalog 29 Technical Information       W I = ––– (r12 + r22) for a cylinder        2 where W  = Weight in pounds r = Radius in inches Equivalent Inertia A motor must be able to: a.  overcome any frictional load in the system b.  start and stop all inertial loads including that of its own rotor The basic rotary relationship is:         Ia T   = ––––     24 where: T   = torque (oz•in) I = moment of inertia (lb •in2) a   = angular acceleration, in radians per square second (rad/sec2) Angular acceleration (a) is a function of the change in velocity (w) and the time required for the change.     w 2 – w1 a   = ––––––––          t or, if starting from zero,     w a   = ––      t where: w   = angular velocity (rad/sec) t    = time (sec)       steps per second since w = –––––––––––––––––– x 2p,     steps per revolution angular velocity and angular acceleration can also be expressed in steps per second (w') and steps per square second (a'), respectively. Sample Calculations A.  Calculating torque required to rotationally accelerate an inertia load:               w'       pq         1 T   = 2 x IO ––– x ––––– x –––              t       180      24 where: T   = torque required (oz•in) IO   = inertial load (lb •in2) p   = 3.1416 q    = step angle (degrees) w'  = step rate (steps/sec) Example: Assume the following conditions: Inertia = 9.2 lb • in2 Step Angle = 1.8° Acceleration = from 0 to 1000 steps per second in 0.5 seconds Motor Inertial Load