Linear Positioners Catalog_en-US_revA - Kollmorgen

Linear Sizing Calculations: Thrust CalculationPositioner MassIn applications where the acceleration force, (W t /g)a,is a significant part of the required thrust, the positionermass must be considered in the thrust calculation. After apositioner is chosen, the positioner weight (linear inertia),W positioner , is added to the weight of the load. W positioner canbe determined using the tables and equation in the positionerdata section. To illustrate, we will use the previous example.1. The first step is to pick a linear positioner with the abovethrust and speed capability. One such positioner is anEC3-AKM42-20-16B-300. This is an EC3 Electric Cylinderwith a AKM42 motor, a 2:1 gear reduction, a 16 mm leadballscrew, and a 300 mm stroke.2. The next step is to look up the effective Positioner LinearInertia in the tables located in the particular positionersection (do not include the “load” term in the equation). Anentry from this table can be seen in the table below. TheAKM42 motor inertia is 0.0013 in-lb-sec 2 . The effectivepositioner weight, calculated from the table is 297 lb.3. The final step is to add this weight to the weight of theload, W L , and recalculate the peak thrust required foreach section of the move profile (do not add this weight tothe gravity or friction terms):Acceleration Section:Ma = 447 lb/386 in/sec 2 × 11.25 in/sec 2= 13.03 lbF total = 169 lbSlew Section:Ma = 0 lb (since a=0)Deceleration Section:Ma = 447 lb/386 in/sec 2 × -11.25 in/sec 2 )= -13.03 lbF total = 193 lbWe can see from this calculation that the addition of this extra“acceleration weight” increases the thrust required duringacceleration, but reduces the peak thrust required duringdeceleration. The EC3-AKM42-20-16B-300 will work in theapplication.Vertical and Horizontal CasesIn a vertical system, θ is 90°, sin90 = 1, and F gravity is equal toW L . Since cos90 = 0, F friction = 0.W LF positioner = (W t /g)a + F applied + F gravityF positioner = (W t /g)a + F applied + W LIn a horizontal system, sinθ = 0, so gravity would play no part(F gravity = 0), and cosθ =1, so F friction would be equal to µW L ,or 50 lb.F positioner = (W t /g)a + F applied + F frictionF positioner = (W t /g)a + F applied + µW LRMS ThrustFor all Servomotor applications, the RMS Thrust needs to becalculated. This thrust must fall within the continuous dutyregion of the linear positioner. Use the following equationwhen calculating RMS Thrust:W LF T RMSF T ACC F T VEL F T DEC F T OFFF T RMS =t A t V t D t off(F T ACC ) 2 t a + (F T VEL ) 2 t v + (F T DEC ) 2 t d + (F T OFF ) 2 t offt a + t v + t d + t offThrust ProfileMove ProfileL I N E A R P O S I T I O N E R SEC Series Linear Positioner InertiaRotary Inertia (Reflected to Motor) = A + B* (Stroke, in) + C * (Load, lb)ModelScrew A B CRatio Reduction typeEC 3 Series Dia x Lead (mm) lb-in sec 2 lb-in sec 2 / in lb-in sec 2 / lbEC3-…-10-16B 1:11.188 E-03 1.176 E-05 2.604 E-05EC3-…-15-16B 1.5:1 Belt/pulley7.435 E-04 5.228 E-06 1.157 E-05EC3-…-20-16B 2:1 16 x 164.779 E-04 2.765 E-06 6.121 E-06EC3-…-50-16B 5:12.280 E-04 4.635 E-07 1.026 E-06Helical gearEC3-…-70-16B 7:1 1.975 E-04 2.401 E-07 5.314 E-07AKM42 Mechanical SpecificationsMotor Inertia(based on resolver)0.0013 lb-in-sec 2www.kollmorgen.com89