ACE

12 12 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 6 Free fallingweight Example W =200 lbs D =15 in s =3 in (chosen) C =60 /hr 6.1Weight rolling/slidingdown incline Formula E 1 = 0.186 ·W · (√5.4 · D) 2 =0.186 ·W · V D 2 E 2 = (W · Sin(A)) · s E 3 =E 1 +E 2 E 4 =E 3 · C V D =√5.4 · D We=E 3 /(0.186 · V D 2 ) E 2 = (F–W · Sin(A)) · s E 2 = (F+W · Sin(A)) · s 6.1a propelling forceup incline 6.1b propelling forcedown incline 9 Swingingarmwithpropelling force (uniformweight distribu- tion) Example W =500 lbs V =6 ft/sec F =1,600 lbs s =1.91 in (chosen) R S =24 in R f =32 in L =48 in C =90 /hr E 1 =0.063 · 500 · 6 2 = 1,134 in-lbs T =1,600 · 32 = 51,200 lbs-in E 2 = (51,200 · 1.91)/24 = 4,075 in-lbs E 3 =1,134+4,075 = 5,209 in-lbs E 4 =5,209 · 90 =468,810 in-lbs/hr V D = (6 · 24)/48 = 3.0 ft/sec We=5,209/(0.186 · 3 2 ) = 3,112 lbs Chosen fromcapacitychart: ModelMC4550-4self-compensating 10 Weight loweredat controlled speed Example W =1,000 lbs V =3 ft/sec s =0.91 in (chosen) C =60 /hr E 1 =0.186 · 1,000 · 3 2 = 1,674 in-lbs F =1,000 lbs E 2 =1,000 · 0.91 = 910 in-lbs E 3 =E 1 +E 2 =1,674+910 = 2,584 in-lbs E 4 =2,584 · 60 =155,040 in-lbs/hr We=2,584/(0.186 · 3 2 ) = 1,543 lbs Chosen fromcapacitychart: ModelMC4525-3self-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 ) F =W Reaction forceQ [lbs] Q= (1.5 · E 3 )/s Stopping time t [s] t= (0.217 · s)/V D Deceleration ratea [ft/s 2 ] a= (9 · V D 2 )/s Formula E 1 =0.063 ·W · V 2 =6 · I · (ω · 0.01745) 2 E 2 = (T · s)/R S = (F · r · s)/R S E 3 =E 1 +E 2 E 4 =E 3 · C V D = (V · R S )/L=R S · ω/688 We=E 3 /(0.186 · V D 2 ) T =F · R f 8 Swingingarmwithpropelling torque (uniformweight distri- bution) Example I =3.895 lb-ft-sec 2 ω =70 °/sec T =15,000 lbs-in s =1 in (chosen) L =19 in R =12 in C =500 /hr E 1 =6 · 3.895 · (70 · 0.01745) 2 = 34.86 in-lbs E 2 =15,000 · 1/12 = 1,250 in-lbs E 3 =34.86+1,250 = 1,284.86 in-lbs E 4 =1,284.86 · 500 =642,430 in-lbs/hr V D =12 · 70/688 = 1.22 ft/sec We=1,284.86/(0.186 · 1.22 2 ) = 4,641.1 lbs Chosen fromcapacitychart:ModelMC4525-4 self-compensatingorMA4525adjustable Check the side load angle, tan α= s/R S , with regard to “Max.Side LoadAngle” in the capacity chart (see example 6.2) Formula E 1 =0.063 ·W · V 2 =6 · I · (ω · 0.01745) 2 E 2 = (T · s)/R E 3 =E 1 +E 2 E 4 =E 3 · C V D = (V · R)/L=R · ω/688 We=E 3 /(0.186 · V D 2 ) 7 Rotary index tablewith propelling torque Example W =195 lbs L =20 in V S =1.85 ft/sec T =1,700 lbs-in R =15 in C =60 /hr s =0.75 in (chosen) W a =195 · 20 2 /(2 · 15 2 ) = 173.3 lbs F =1,700/15 = 113.3 lbs E 1 =0.186 · 173.3 · 1.85 2 = 110.3 in-lbs E 2 =113.3 · 0.75 = 85 in-lbs E 3 =110.3+85 = 195.3 in-lbs E 4 =195.3 · 60 = 11,718 in-lbs/hr We=195.3/(0.186 · 1.85 2 ) = 306.8 lbs Chosen fromcapacitychart: Model SC300-4orMC225Hself-compensating Check the side load angle, tan α= s/R S , with regard to “Max.Side LoadAngle” in the capacity chart (see example 6.2) Formula E 1 =0.186 ·W a · V S 2 E 2 =F · s E 3 =E 1 +E 2 E 4 =E 3 · C V D = (V S · R)/L=ω · R We=E 3 /(0.186 · V 2 ) W a =W · L 2 /(2 · R 2) V =R · ω/688 F = T/R 6.2 Weight free fallingabout apivot point Check the side load angle, tan α= s/R S , with regard to “Max. Side LoadAngle” in the capacity chart tanα= s/R S Side load angle from shock absorber axis Formula V =√(5.4 · D) 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 ) F =W V =√(5.4 · 15) = 9 fps E 1 =0.186 · 200 · 9 2 = 3,013.2 in-lbs E 2 =200 · 3 = 600 in-lbs E 3 =3,013.2+600 = 3613.2 in-lbs E 4 =3,613.2 · 60 =216,792 in-lbs/hr We=3,613.2/(0.186 · 9 2 ) = 239.8 lbs Chosen fromcapacitychart: ModelMA4575adjustable Approximate values assuming correct adjustment. Add safetymargin if necessary. (Exact valueswill depend upon actual application data and can be provided on request.) FormulaandCalculations Calculation as per example6.1 except E 2 =0 E 1 =0.186 ·W · (√5.4 · D) 2 V D = (√5.4 · D) · R S /L