11.07.2015 Views

Linear Positioners Catalog_en-US_revA - Kollmorgen

Linear Positioners Catalog_en-US_revA - Kollmorgen

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<strong>Linear</strong> Sizing Calculations: Move ProfileExample 1Calculate the peak acceleration and velocity for an object thatneeds to move 30 inches in 2.5 seconds. Assume a TrapezoidalProfile.Solutionv max2.5 sec.d = 30 in.totExample 3This is an example of a case wh<strong>en</strong> triangular and trapezoidalmove profiles are not adequate approximations. Assume amaximum positioner speed is 6 inches/sec. Sketch a moveprofile that will complete a 10 inch move in 2 seconds. What isthe minimum allowable acceleration rate in inches/sec 2 ?SolutionvTriangularTrapezoidalActualRequiredMoveL I N E A R P O S I T I O N E R Sv AVE =30 in2.5 sec= 12 in/sectv MAX = 1.5d totd tott tot= 18 in/secExample 2Calculate, in radians/sec, the peak acceleration and velocityfor an cylinder that needs to move 5 revolutions in 0.4 seconds.Assume a Triangular Profile.SolutionTriangularv AVE =10 in2 sec= 5 in/secv MAX = 2 × v AVE = 10 in/sec (v MAX > 6 in/sec – too fast)Trapezoidalv MAX = 1.5 × v AVE = 7.5 in/sec (v MAX > 6 in/sec – too fast)These are too fast, so we need to find t 1 as follows:a = 4.5 = 21.6 in/sec 2 in( t tot) 2 ( t tot – t 2)v max5 revsRequired Profile( t 1 + t 3)( 2 )d tot = v MAX + t 2d tot = 5 revs × 2π rad = 31.42 radrevv 31.42 rad in radAVE = = 78.550.4 secsecv radMAX = 2 v AVE = 157.1secda = 4 tot = 785.5 radT2 sec20.4 sec.d = + t 2 = t tot +22v MAX( )solving for t 2 ,( ) ( )td tot t tot2 sec2 = – × 2 10 inv = MAX 26 in/sec– 2× 2t 2 = 1.33 secNow assume t 1 = t 3 , sot 1 = (t tot - t 2 )/2 = 0.33 sec.Finally, calculate accelerationva = MAX = 6 in/sec = 18t 0.33 sec1sec2t 22www.kollmorg<strong>en</strong>.com87

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