Techno Linear Motion Catalog34Technical Informationwhere Kt is the motor torque constant (e.g., Nm/amp) and I is the drive current (amp). The choice of motorand drive must satisfy the following conditions:1. The product of Kt and peak drive current must give the required peak torque2. The product ofKt and continuous drive current must produce sufficient continuous torque.3. The maximum allowable motor current must be greater than the peak drive current.4. At maximum speed and peak current, the voltage developed across the motor must be less than80% of the drive supply voltage.The voltage across the motor is given by:E=KEw+R Iwhere KE is the motor voltage constant, w the speed, R the winding resistance (ohms) and I the peakcurrent (amperes). The speed units should be the same in each case; i.e., if the voltage constant is in voltsper radian per second, thenw should also be in radians per second.To make the most efficient use of the drive, the chosen solution should utilize most of the peak drive currentand most of the available voltage. Motor manufacturers usually offer alternative windings, and care shouldbe taken to select the most appropriate.Example:Leadscrew Length: 80 inLeadscrew Diameter: 1.5 inLeadscrew Pitch: 2.54 inTable Weight: 1000 lbLinear Table Speed Required: 472 inches / minAcceleration Time: 120 msD4LInertia of Leadscrew: J = –––––– = 11.25 lb • in236WInertia of Table: J = –––––– = 3.88 lb • in240 p2Total inertia = 15.13 lb • in2Maximum Speed = 472" / min = 1200 rpm (equivalent to 4000 full steps / sec)Acceleration Torque:J wT= –––––– = 660 oz•in (4.65 N•m)764tThis takes no account of motor inertia, so a suitable motor will be capable of producing around 1000 oz • intorque.Again, as with stepper selection, it is recommended to add a 20% factor of safety so that unexpecteddynamic loads are easily handled by the motor.Linear Table Driven by DC Motor
