
 COMPARISON OF SHOCK ABSORBING OF DASH POTS, PU RUBBERS, SPRINGS AND CEC SHOCK ABSORBERS   FORCESTROKE TESTNG MACHINE  When shock absorber was affected by the force from outside, the main function of this testing machine is testing the relative curve between the pressure of inner tube undertake and the stroke of the shock absorber. That means through this curve, we can judge the efficiency of the shock absorber absorb the kinetic energy by the force it was affected.
  The springs and polyurethane rubbers are widespread to use in earlier period, but due to provide nonlinear deceleration and to result in strong resistance, all the kinetic energy of moving objects is not absorption and produce counter pressure, this is in low efficiency. If linear deceleration is necessary for a moving object, CEC Shock Absorbers is your best choice.    Examples of using CEC shock absorbers 
  OPERATING PRINCIPAL OF SHOCK ABSORBERS  Top   CEC Shock Absorbers main structure to combine with body, rod, bearing, inner tube, piston, fluid, spring. On impact the piston rod moves into the shock absorbers and the hydraulic fluid is pushed into the accumulator to produce resistant force, the pressure in the inner tube remains constant throughout the entire impact stroke. CEC Shock Absorbers providing a linear decelerationand brings the impacting object to stop smoothly and quietly. At the end of the impact stroke, the return spring pushes the piston to the original position for next cycle.   ROD  BEARING  PISTON  INNER TUBE  BODY    SELECTION OF HYRAULIC SHOCK ABSORBERS  Top   Three major factors to be considered 1.  How much energy to be dissipated each stroke?  2.  How much energy to be dissipated per hour?  3.  How much is your effective weight of your application?  After the three factors are calculated, go to the quick list to find the suitable shock absorbers.  Note: 1.  Shock absorbers series allow approximate 1 mm fixed stop before the end of the stroke, not to be driven into the final position under full load.  2.  Reuse is prohibited after disassembling; paint on the rod and threaded body is not allowed.  3.  Stop collar protects shock absorbers piston from bottoming out, Stop collar can be used in adjust the distance in stroke. Be sure the fixed board of the shock absorbers is strong enough.  4.  To assembling, the offcenter angle do not exceed 2.   
  ex. 1 Horizontal Impact without Propelling Force (Application) W = 20 kg V = 1 m / s C = 1000 / Hr   (Formulas and Calculation) E1 = 0.5 x W x V^{2} E2 = 0 E3 = E1 + E2 E4 = E3 x C We = W   E1 = 0.5 x 20 x 1^{2} = 10 Nm E2 = 0 E3 = 10 + 0 = 10 Nm / C E4 = 10 x 1000 = 10000 Nm / Hr We = 20 kg Model SC14151 is adequate  ex. 2 Horizontal Impact with Conveyor Driving (Application) W = 10 kg V = 1 m / s C = 600 / Hr S = 0.01 m = 0.25   (Formulas and Calculation) E1 = 0.5 x W x V^{2} E2 = W x x g x S E3 = E1 + E2 E4 = E3 x C We = 2 x E3 / V^{2}   E1 = 0.5 x 10 x 1^{2} = 5 Nm E2 = 10 x 0.25 x 9.81 x 0.01 = 0.25 Nm E3 = 5 + 0.25 = 5.25 Nm / C E4 = 5.25 x 600 = 3,150 Nm / Hr We = 2 x 5.25 / 1^{2} = 10.5 kg Model SC12102 is adequate  ex. 3 Horizontal Impact with Propelling Force (Application) W = 50 kg V = 1 m / s F = 1000 N C = 500 / Hr S = 0.04 m   (Formulas and Calculation) E1 = 0.5 x W x V^{2} E2 = F x S E3 = E1 + E2 E4 = E3 x C We = 2 x E3 / V^{2}   E1 = 0.5 x 50 x 1^{2} = 25 Nm E2 = 1000 x 0.04 = 40 Nm E3 = 25 + 40 = 65 Nm / C E4 = 65 x 500 = 32500 Nm / Hr We = 2 x 65 / 1^{2} = 130 kg Model FC2540 is adequate  ex. 4 Vertical Impact with Force from Top to Bottom (Application) W = 100 kg V = 1 m / s F = 1200 N C = 400 / Hr S = 0.025 m   (Formulas and Calculation) E1 = 0.5 x W x V^{2} E2 = ( F + W x g ) x s E3 = E1 + E2 E4 = E3 x C We = 2 x E3 / V^{2}   E1 = 0.5 x 100 x 1^{2} = 50 Nm E2 = ( 1200 + 100 x 9.81 ) x 0.25= 54.5 Nm E3 = 50 + 54.5 = 104.5 Nm / C E4 = 104.5 x 400 = 41800 Nm / Hr We = 2 x 104.5 / 1^{2} = 209 kg Model FC3625 is adequate  ex. 5 Vertical Impact with Force from Bottom to Top (Application) W = 200 kg V = 0.5 m / s F = 3000 N C = 500 / Hr S = 0.05 m   (Formulas and Calculation) E1 = 0.5 x W x V^{2} E2 = ( F  W x g ) x s E3 = E1 + E2 E4 = E3 x C We = 2 x E3 / V^{2}   E1 = 0.5 x 200 x 0.5^{2} = 25 Nm E2 = ( 3000  200 x 9.81 ) x 0.05 = 51.9 Nm E3 = 25 + 51.9 = 76.9 Nm / C E4 = 76.9 x 500 = 38450 Nm / Hr We = 2 x 76.9 / 0.5^{2} = 615.2 kg Model FC3650 is adequate  ex. 6 Horizontal Impact with Motor Driving (Application) W = 50 kg V = 1.5 m / s ST = 2.5 HP = 2 KW C = 100 / Hr S = 0.06 m   (Formulas and Calculation) E1 = 0.5 x W x V^{2} E2 = 1000 x HP x ST x S / V E3 = E1 + E2 E4 = E3 x C We = 2 x E3 / V^{2}   E1 = 0.5 x 50 x 1.5^{2} = 56.25 Nm E2 = 1000 x 2 x 2.5 x 0.06 / 1.5 = 200 Nm E3 = 56.25 + 200 = 256.25 Nm / C E4 = 256.25 x 100 = 25625 Nm / Hr We = 2 x 256.25 / 1.5^{2} = 227 kg Model SC36602 is adequate  ex. 7 Free Fall Impact (Application) W = 300 kg h = 0.5 m C = 300 / Hr S = 0.08 m   (Formulas and Calculation) E1 = W x g x h E2 = W x g x s E3 = E1 + E2 E4 = E3 x C We = 2 x E3 / V^{2}   E1 = 30 x 9.81 x 0.5= 147 Nm E2 = 30 x 9.81 x 0.08 = 23.5 Nm E3 = 147 + 23.5 = 170.5 Nm / C E4 = 170.5 x 300 = 51150 Nm / Hr We = 2 x 170.5 / 3.1^{2} = 35.5 kg Model SC25801 is adequate  ex. 8 Free Moving Load Down an Inclined Plane (Application) W = 30 kg L = 1 = 30 S = 0.04 C = 250 / Hr   (Formulas and Calculation) E1 = 0.5 x W x V^{2} E2 = W x S x Sin E3 = E1 + E2 E4 = E3 x C We = 2 x E3 / V^{2}   E1 = 0.5 x 30 x 2.2^{2} = 72.6 Nm E2 = 30 x 0.04 x 9.81 x 0.5 = 5.9 Nm E3 = 72.6 + 5.9 = 78.5 Nm / C E4 = 78.5 x 250 = 19625 Nm / Hr We = 2 x 78.5 / 2.2^{2} = 32 kg Model SC25401 is adequate  ex. 9 Rorary with Propelling Force (Application) W = 100 kg V = 1.1 m / s T = 2000 Nm S = 0.06 m RT = 1.25 m RS = 0.8 m C = 100 / Hr   (Formulas and Calculation) E1 = 0.25 x W x V^{2} E2 = ( T x S ) / RS E3 = E1 + E2 E4 = E3 x C Vs=( VT x RS ) / RT We = 2 x E3 / Vs^{2}   E1 = 0.25 x 100 x 1.1^{2} = 30.3 Nm E2 = 2000 x 0.06 / 0.8 = 150 Nm E3 = 30.3 + 150 = 180.3 Nm / C E4 = 180.3 x 100 = 18030 Nm / Hr Vs = 1.1 x 0.8 / 1.25 = 0.7 m / s We = 2 x 180.3 / 0.7^{2} = 736 kg Model SC36603 is adequate     
