ACE

11 11 Issue7.2014 Specifications subject to change ACE Controls Inc. · Tel. 800-521-3320 · (248) 476-0213 · Fax (248) 476-2470 · E-mail: shocks@acecontrols.com · www.acecontrols.com ACEshockabsorbersprovide linear decelerationandare thereforesuperior toother kindsof dampingelements. It iseasy tocalculatearound90%of applicationsknowingonly the following5parameters: 1. Weight tobedecelerated W (lbs) 2. Impact velocityat shockabsorber V D (ft/sec) 3. Propelling force F (lbs) 4. Cyclesper hour C (/hr) 5. Number of absorbers inparallel n Key tosymbolsused E 1 Kinetic energy per cycle in-lbs E 2 Propelling forceenergy per cycle in-lbs E 3 Total energy per cycle (E 1 +E 2 ) in-lbs 1 E 4 Total energy per hour (E 3 · C) in-lbs/hr We Effectiveweight lbs W Weight tobedecelerated lbs n Number of shock absorbers (inparallel) 2 V Velocity at impact ft/sec 2 V D Impact velocity at shock absorber ft/sec ω Angular velocity at impact °/sec F Propelling force lbs C Cycles per hour /hr Hp Motor power hp 3 ST Stall torque factor (normally 2.5) 1 to3 T Propelling torque lbs-in I Moment of Inertia lb-ft-sec 2 g Accelerationdue togravity=32.2 ft/s 2 ft/s 2 D Dropheight excl. shock absorber stroke in s Shock absorber stroke in L/R/r Radius in Q Reaction force lbs � Coefficient of friction t Deceleration time sec a Deceleration ft/s 2 α Side load angle ° β Angleof incline ° 1 All mentioned values of E 4 in the capacity charts are only valid for room temperature. There are reduced values at higher temperature ranges. 2 V or V D is the final impact velocity of theweight.With acceleratingmotion the final impact velocity can be 1.5 to 2 times higher than the average. Please take this into account when calculating kinetic energy. 1 Weightwithout propelling force Formula E 1 =0.186 ·W · V 2 E 2 =F · s E 3 =E 1 +E 2 E 4 =E 3 · C We=E 3 /(0.186 · V 2) Example W =500 lbs V =3 ft/sec F P =0 lbs s =1 in (chosen) C =500 /hr E 1 =0.186 · 500 · 3 2 = 837 in-lbs E 2 =0 · 1 = 0 in-lbs E 3 =837+0 = 837 in-lbs E 4 =837 · 500 = 418,500 in-lbs/hr We=837/(0.186 · 3 2 ) = 500 lbs Chosen fromcapacitychart:ModelMC3325-3 self-compensatingorMA3325adjustable 2 Weightwithpropelling force Example W =14 lbs V =2.2 ft/sec F P =30 lbs s =0.4 in (chosen) C =100 /hr E 1 =0.186 · 14 · 2.2 2 = 12.6 in-lbs E 2 =30 · 0.4 = 12 in-lbs E 3 =12.6+12 = 24.6 in-lbs E 4 =24.6 · 100 = 2,460 in-lbs/hr We=24.6/(0.186 · 2.2 2) = 27.3 lbs Chosen fromcapacitychart: ModelMC75-3self-compensating 1 V is thefinal impact velocityof theweight:Withpneu- maticallypropelled systems this canbe1.5 to2 times theaverage velocity. Please take this intoaccountwhen calculatingenergy. 2.1 for verticalmotionupwards 2.2 for verticalmotiondownwards 3 Weightwithmotor drive Example W =2,100 lbs V =1 ft/sec Hp =2 hp ST =2.5 s =2 in (chosen) C =20 /hr F =550 · 2.5 · 2/1 = 2,750 lbs E 1 =0.186 · 2100 · 1 2 = 390.6 in-lbs E 2 =2,750 · 2 = 5,500 in-lbs E 3 =390.6+5,500 = 5,890.6 in-lbs E 4 =5,890.6 · 20 = 117,812 in-lbs/hr We=5,890.6/(0.186 · 1 2 ) = 31,670 lbs Chosen fromcapacitychart: ModelML6450orMC6450-4self-compensating Formula E 1 =0.186 ·W · V 2 E 2 =F · s E 3 =E 1 +E 2 E 4 =E 3 · C We=E 3 /(0.186 · V 2 ) E 2 = (F–W) · s E 2 = (F+W) · s 4 Weight ondriven rollers Example W =250 lbs V =2.5 ft/sec � =0.2 in s =1 in (chosen) C =180 /hr F =250 · .2 = 50 lbs E 1 =0.186 · 250 · 2.5 2 = 290.6 in-lbs E 2 =50 · 1 = 50 in-lbs E 3 =290.6+50 = 340.6 in-lbs E 4 =340.6 · 180 = 61,308 in-lbs/hr We=340.6/(0.186 · 2.5 2 ) = 293 lbs Chosen fromcapacitychart:ModelMA600 adjustableorSC650-3self-compensating Formula F =W · � E 1 =0.186 ·W · V 2 E 2 =F · s E 3 =E 1 +E 2 E 4 =E 3 · C We=E 3 /(0.186 · V 2 ) 5 Swingingweightwith propelling force Example W =20 lbs V =12 ft/sec T =50 in-lbs R =12 in L =16 in s =0.5 in (chosen) C =700 /hr E 1 =0.186 · 20 · 12 2 = 536 in-lbs E 2 =50 · 0.50/12 = 2.1 in-lbs E 3 =536+2.1 = 538.1 in-lbs E 4 =538.1 · 700 = 376,670 in-lbs/hr v D =12 · 12/16 = 9 ft/sec We=538.1/(0.186 · 9 2 ) = 35.7 lbs Chosen fromcapacitychart: ModelMC600self-compensating Check the side load angle, tan α= s/R, with regard to “Max.Side LoadAngle” in the capacity chart (see example 6.2) Formula E 1 =0.186 ·W · V 2 =0.186 · I · ω 2 E 2 = T · s/R E 3 =E 1 +E 2 E 4 =E 3 · C V D =V · R/L=ω · R We=E 3 /(0.186 · V D 2 ) 3 ST=^ relation between starting torque and running torque of themotor (depending on the design) Inall the followingexamples thechoiceof shockabsorbersmade from thecapacity chart isbasedupon the valuesof (E 3 ), (E 4 ), (We) and thedesiredshockabsorber stroke (s). Formula F =550 · ST · Hp/V E 1 =0.186 ·W · V 2 E 2 =F · s E 3 =E 1 +E 2 E 4 =E 3 · c We=E 3 /(0.186 · V 2 ) FormulaandCalculations Note:Donot forget to include the rotational energyofmotor, couplingandgearbox intocalculation forE 1 .