ACE

13 13 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 EffectiveWeight (We) 19 Wagonagainst 2shockabsorbers Example W =2,000 lbs F =500 lbs V =5 ft/sec C =20 /hr s =1.91 in (chosen) E 1 =0.093 · 2,000 · 5 2 = 4,650 in-lbs E 2 =500 · 1.91 = 955 in-lbs E 3 =4,650+955 = 5,605 in-lbs E 4 =5,605 · 20 =112,100 in-lbs/hr V D =5 · 0.5 = 2.5 ft/sec We=5,605/(0.186 · 2.5 2 ) = 4,822 lbs Chosen fromcapacitychart: ModelMC4550-4self-compensating 20 Wagonagainstwagon 21 Wagonagainstwagon 2shockabsorbers Formula E 1 = 0.186 (W 1 ·W 2 )/ (W 1 +W 2 ) · (V 1 +V 2 ) 2 E 2 =F · s E 3 =E 1 +E 2 E 4 =E 3 · C V D =V 1 +V 2 We=E 3 /(0.186 · V D 2 ) Formula E 1 =0.093 ·W · V 2 E 2 =F · s E 3 =E 1 +E 2 E 4 =E 3 · C V D =V · 0.5 We=E 3 /(0.186 · V D 2 ) Example W 1 =4,000 lbs V 1 =2 ft/sec C =30 /hr W 2 =8,000 lbs V 2 =1.2 ft/sec F =1,000 lbs s =2.91 in (chosen) E 1 =0.186 (4,000 · 8,000)/ (4,000+8,000) · (2+1.2) 2 = 5,079 in-lbs E 2 =1,000 · 2.91 = 2,910 in-lbs E 3 =5,079+2,910 = 7,989 in-lbs E 4 =7,989 · 30 =239,670 in-lbs/hr V D =2+1.2 = 3.2 ft/sec We=7,989/(0.186 · 3.2 2 ) = 4,194 lbs Chosen fromcapacitychart: ModelMC4575-4self-compensating Note:Whenusingseveral shockabsorbers inparallel, the values (E 3 ), (E 4 ) and (We) aredividedaccording to thenumber of unitsused. A Weightwithout propelling force Formula We=W B Weightwithpropelling force Formula We=E 3 /(0.186 · V D 2 ) Example W =100 lbs F = 200 lbs V D = V=2 ft/sec s = 2 in (chosen) E 1 = 0.186 ·W · V 2 = 0.186 · 100 · 2 2 =74.4 in-lbs E 2 = F · s=200 · 2=400 in-lbs E 3 = E 1 +E 2 =74.4+400 =474.4 in-lbs We= E 3 /(0.186 · V D 2 ) = 474.4/(0.186 · 2 2 ) = 637.6 lbs C Weightwithout propelling force direct against shockabsorber D Weightwithout propelling force withmechanical advantage Theeffectiveweight (We) caneither be thesameas theactualweight (examplesAandC), or it canbean imaginaryweight representingacombinationof thepropelling forceor lever actionplus theactualweight (examplesBandD). FormulaandCalculations Formula E 1 = 0.093 (W 1 ·W 2 )/ (W 1 +W 2 ) · (V 1 +V 2 ) 2 E 2 =F · s E 3 =E 1 +E 2 E 4 =E 3 · C V = (V 1 +V 2 )/2 We=E 3 /(0.186 · V D 2 ) Example W 1 =4,000 lbs V 1 =2 ft/sec C =30 /hr W 2 =8,000 lbs V 2 =1.2 ft/sec F =1,000 lbs s =1.91 in (chosen) E 1 =0.093 (4,000 · 8,000)/ (4,000+8,000) · (2+1.2) 2 = 2,540 in-lbs E 2 =1,000 · 1.91 = 1,910 in-lbs E 3 =2,540+1,910 = 4,450 in-lbs E 4 =4,450 · 30 = 133,500 in-lbs/hr v D = (2+1.2)/2 = 1.6 ft/sec We=4,450/(0.186 · 1.6 2 ) = 9,346 lbs Chosen fromcapacitychart: ModelMC4550-4self-compensating Example W =100 lbs V D = V=2 ft/sec E 1 = 0.186 ·W · V 2 = 0.186 · 100 · 2 2 =74.4 in-lbs E 1 = E 3 =74.4 in-lbs We= E 3 /(0.186 · V D 2 ) = 74.4/(0.186 · 2 2 ) = 100 lbs Example W =50 lbs V D = V=2 ft/sec E 1 = 0.186 ·W · V 2 = 0.186 · 50 · 2 2 =37.2 in-lbs E 1 = E 3 =37.2 in-lbs We= E 3 /(0.186 · V D 2 ) = 37.2/(0.186 · 2 2 ) = 50 lbs Formula We=E 3 /(0.186 · V D 2 ) Formula We=W Example W =150 lbs V = 2 ft/sec V S = V D =0.5 ft/sec E 1 = 0.186 ·W · V 2 = 0.186 · 150 · 2 2 =111.6 in-lbs E 1 = E 3 =111.6 in-lbs We= E 3 /(0.186 · V D 2 ) = 111.6/(0.186 · 0.5 2 ) = 2,400 lbs