Data Pack G Iued March 2001 1502621407 Data Sheet Indutrial hock aborber (linear decelerator) Introduction Spring and buffer cannot atch the perforance of linear decelerator (the correct nae for indutrial hock aborber). A decelerator act jut like your hand doe when you catch a ball. The linear deceleration characteritic atch the peed and a of the oving object and bring it oothly to ret. The energy i releaed harlely, otly a heat. Spring and buffer act differently. They tore energy rather than diipate. So although they top the oving object, it bounce back. Dahpot when iued are not uch better. Their peak reitance coe at the beginning of the troke, and then fall away quickly. Thi caue greater reiting force than neceary. Figure 1 Dahpot inear Spring Principle of operation Virtually all anufacturing procee involve oveent of oe kind. In production achinery thi can involve linear tranfer, rotary index otion, fat feed, etc. At oe point thee otion change direction or coe to a top. Any oving object poee kinetic energy a a reult of it otion and if the object change direction or i brought to ret, the diipation of the kinetic energy can reult in detructive hock force within the tructural and operating eber of the achine or equipent. Kinetic energy increae with the quare of the peed and a operating peed rie under continual deand for higher productivity, hock daage to equipent becoe an increaingly eriou proble, reulting in cotly downtie and lo of production. An increae in production rate i therefore only poible by diipating kinetic energy and thereby eliinating the detructive deceleration force. Kinetic energy can be diipated by uing friction, a in a car whoe energy i changed into heat energy by friction at it brake urface. It can alo be diipated by echanically deforing another object uch a a pring or rubber buper. In practice uch device are unatifactory a they only tore energy and bounce it back into the yte cauing fatigue and preature tructural failure. Another device which i often iapplied in energy diipation i the dahpot or cylinder cuhion. The true function of a dahpot i to provide a contant velocity to a oving part and if ued to aborb energy it give rie to high force peak at the beginning of it troke. Mot coon energy aborbing device contribute to hock rather than reduce it becaue they do not diipate kinetic energy at a unifor rate. Due to thi non-linear deceleration the object being topped i ubjected to detructively high force level (hock) either at the beginning or the end of the deceleration troke. The ideal olution for thi proble i achieved when the energy of the object i linearly aborbed. That ean the required deceleration force i evenly ditributed over the entire troke length to give contant or linear deceleration. Shock aborber top oving object afely and effectively, without hock, by achieving controlled linear deceleration.
Figure 2 inear decelerator (hock aborber) are virtually aintenance free, elf contained hydraulic device with a erie of adjutable orifice. On ipact the piton rod i puhed into the hock aborber. The hydraulic fluid ituated in front of the piton (5) i initially diplaced through all the orifice (6) reulting in ooth pick up of the oving load. A the piton ove down it troke the orifice are progreively cloed off gradually lowing the oving object down to ret. Due to the pecific pacing of the orifice the preure generated in front of the piton reain contant throughout the entire troke a the velocity i reduced to zero. Conequently the reiting force reain contant and unifor or linear deceleration i achieved. An external adjuting ring (3) i ued to regulate the orifice ize enabling perfect deceleration to be achieved with varying load and peed and propelling force. The diplaced hydraulic fluid i tored in an internal accuulator containing cloed cell elatoer foa (8). A non-return valve (4) built into the piton head and the return pring (1) enable a quick reet ready for the next cycle. ACE linear decelerator (hock aborber) bring a oving load oothly to ret with the lowet poible force in the hortet poible tie. The aively contructed outer body, large area bearing urface and iniu eal coupled with paintaking quality control aure a long and trouble free working life. 1 2 3 4 5 6 7 8 Increae production rate Extend achine life Reduce contruction cot Reduce aintenance Reduce noie pollution. Security Indutrial hock aborber and autoobile braking yte have two crucial functional iilaritie. 1. Both hould bring a oving a quickly and afely to ret without any recoil or bounce back. 2. Both ut never uddenly fail without warning. If an autoobile braking yte ever failed uddenly without warning it would alot certainly reult in a eriou vehicle accident. If an indutrial hock aborber failed uddenly it could caue eriou daage to the achine it wa intalled on. In any cae thi could reult in a coplete production line being topped with reultant enorou downtie cot. Soe hock aborber anufacturer ue tube tock to ake their aborber bodie and inner preure tube. Thee tube are then ealed by an end plug held in with retaining ring or circlip and eated by elatoer O ring or iilar. Under overload condition the circlip or eal can fail or be extruded cauing udden and catatrophic failure of the hock aborber. In contrat RS offer ACE hock aborber bodie and inner preure chaber which are fully achined fro olid high tenile alloy teel. Thi give a copletely cloed end one piece preure chaber with no eal or circlip being neceary. The advantage of thi deign concept i that the ACE hock aborber i able to withtand uch higher internal preure or overload without daage giving a very high afety argin. The chance of a udden failure due to over- load etc. i effectively ruled out. Figure 3 Reaction force-energy capacitytopping tie coparion with other daping yte 1 3 4 Stopping troke 1. Hydraulic dahpot (high topping force at tart of the troke). With only one etering orifice the oving load i abruptly lowed down at the tart of the troke. The braking force rie to a very high peak at the tart of the troke (giving high hock load) and then fall away rapidly. The ajority of the energy i aborbed at the tart of the troke. 2 2
2. Spring and rubber buffer (high topping force at end of troke). Thee have an increaing force characteritic, becoing a olid top at full copreion. Alo they tore energy rather than diipating it, cauing the load to rebound back again. 3. Air buffer, pneuatic cylinder cuhion (high topping force at end of troke). Due to the copreibility of air thee have a harply riing force characteritic toward the end of the troke. The ajority of the energy i aborbed near the end of the troke. 4. RS indutrial hock aborber (unifor topping force through the entire troke). The oving load i oothly and gently brought to ret by a contant reiting force throughout the entire hock aborber troke. The load i decelerated with the lowet poible force in the hortet poible tie eliinating daaging force peak and hock daage to achine and equipent. Thi i a linear deceleration force troke curve and i the curve provided by RS hock aborber. Reult: The peak topping force of a hydraulic dahpot occur at the beginning of the troke and i any tie higher than with a hock aborber that can aborb the ae aount of energy with le than half the force, achine wear and aintenance i thu dratically reduced. Figure 6 Stopping tie V t t Shock Hydraulic Figure 4 Energy capacity With the ae troke and the ae reaction force it i poible to aborb everal tie a uch energy with the hock aborber. (Energy capacity i repreented by the area under the curve). Reult: By intalling a hock aborber production rate can be ore than doubled without increaing deceleration force or load on the achine. Shock aborber Q Hydraulic Stopping troke Figure 5 Reaction force (topping force) Hydraulic Stopping tie Both device are topping the ae a at the ae ipact velocity and in the ae troke length. Therefore they are both aborbing the ae energy. Reult: The hock aborber top a oving load of the ae kinetic energy in a third of the tie of a dahpot or cylinder cuhion. A a reult cycle tie are reduced giving higher production rate. Adjutable hock aborber The adjutable hock aborber offer flexibility in application deign and election procedure. With the widet range of effective weight, one odel can cover any application. By iply tuning in another orifice when an effective weight change i neceary, the total orifice area change, providing the required contant internal preure. Self-copenating hock aborber In cae where non-adjutability i beneficial but the feature of an adjutable hock aborber are required, elfcopenating odel eet both need. With a wide range of effective weight, a elf-copenating hock aborber will provide acceptable deceleration depite changing energy condition. The orifice profile, deigned by a coputer that contantly arrange the ize and location of each orifice while inputting changing effective weight, neutralie the effect of changing fluid coefficient, weight, velocity, teperature and fluid copreibility. Shock aborber Q Stopping troke Both device are topping the ae a at the ae ipact velocity and in the ae troke length. Therefore they are both aborbing the ae energy (area under the curve). 3
A linear decelerator by definition decelerate a oving weight at a linear or contant rate of deceleration. The adjutable hock aborber i able to provide linear deceleration when operated within it energy capacity and effective weight range by dialling in the required orifice area. The reulting force-troke curve (figure 7) how the optiu (lowet) topping force Figure 7 Figure 10 how a faily of force-troke curve: a) Adjutable hock aborber properly tuned. b) Self-copenating hock aborber at the low end of it effective weight range. c) Self-copenating hock aborber at the high end of it effective weight. Figure 10 b c a Figure 8 how the force-troke curve of a elfcopenating hock aborber topping a weight at the low end of it effective weight range. Note how the reaction force are no longer contant but are till acceptable. The curve i kewed lightly higher at the beginning of the troke and dip lower at the end. Figure 8 A force-troke curve of the ae elf-copenating hock aborber,but at the high end of it effective weight range, i hown in figure 9. The energy curve i now kewed upward at the end of troke and till yield acceptable deceleration. Figure 9 Effective weight Effective weight i an iportant factor in electing hock aborber. A hock aborber ee the ipact of an object in ter of weight and velocity only; it doe not ee any propelling force. The effective weight can be thought of a the weight that the hock aborber ee on ipact. Effective weight include the effect of the propelling force on the perforance of the hock aborber. Failing to conider the effective weight ay reult in iproper election and poor perforance of a hock aborber. Under extree condition an effective weight that i too low for the hock aborber ay reult in high force at the tart of troke (high on-et force). Converely an effective weight that i too high for the hock aborber ay caue high force at the end of troke (high et-down force). Conider the following exaple:- 1. A a of 3kg travelling at 6 etre per econd ha 54 Newton eter of kinetic energy (figure 11). On thi bai alone a Model A1/2 x 1 hock aborber would be elected. However, becaue there i no propelling force on the application the calculated effective weight i 3kg. Thi i below the iniu effective weight range of 5kg in to 450kg ax. If thi odel were ued the reult would be a high on-et force at the beginning of the take (figure 12). The olution in thi cae would be to ue a pecially orificed hock aborber to handle the low weight/high peed cobination or alternatively to ue the elf-copenating odel MC 1201 M-0. 2. A a of 50kg ha an ipact velocity of 0.3 etre per econd and i driven by a propelling force of 2000 Newton (figure 13). Total ipact energy i 53N and again a odel A 1/2 x 1 would be elected baed jut on the energy to be aborbed. However, the effective weight calculate to 1177kg which i well above the axiu for a tandard odel A 1/2 x 1. If thi hock aborber wa ued, high et-down force at the end of troke would reult (figure 14). The olution in thi cae i to ue a odel VA 1/2 x 1 which i deigned to work in low velocity, high effective weight application uch a thi. 4
Figure 11 ow Effective Weight 6 etre per econd The Ace effective weight concept provide a unique izing tool for Ace hock aborber and will enure that the hock aborber choen for your application will have the adjutent range to provide perfect linear deceleration a well a ufficient energy aborbing capacity. Further exaple of the calculation of effective weight follow: Figure 12 Figure 13 2000 N Figure 14 3kg ow Effective Weight High Effective Weight 0.3 etre per econd 50kg inear Deceleration Exaple 1: Orifice area i too all (high on-et) inear Deceleration Exaple 2: Orifice area i too large (high et-down) Forulae and calculation Ace hock aborber provide linear deceleration and are therefore uperior to other kind of daping eleent. It i eay to calculate around 90% of application knowing only the following 4 paraeter: 1. Ma to be decelerated (weight) (kg) 2. Ipact velocity v D (/) 3. Propelling force F (N) 4. Cycle per hour C (/hr) Key to ybol ued W 1 Kinetic energy per cycle (N) W 2 Propelling force energy per cycle (N) W 3 Total energy per cycle (W 1 + W 2 ) (N) W 4 Total energy per hour (W. 3 C) (N/hr) e Effective weight (kg) Ma to be decelerated (kg) *v Velocity of oving a (/) *v D Ipact velocity at hock aborber (/) ω Angular velocity (ω = 2 πθ 360 X) (rad/) F Propelling force (N) C Cycle per hour (/hr) P Motor power (kw) ST Stall torque factor (norally 2.5) 1 to 2.5 M Propelling torque (N) I Moent of inertia (kg 2 ) g Acceleration due to gravity = 9.81 (/ 2 ) h Drop height excl. hock aborber troke () Shock aborber troke () /R/r Radiu () Q Reaction force (N) µ Coefficient of friction t Deceleration tie (ec) g Deceleration rate (g ) α Side load angle ( ) ß Angle of incline ( ) θ Index degree ( ) X Index tie (ec) *v or v D i the final ipact velocity of the a. With accelerating otion the final ipact velocity can be 1.5 to 2 tie higher than the average velocity. Pleae take thi into account when calculating the kinetic energy. In the following exaple the choice of hock aborber ade fro the capacity chart i baed upon the value of (W 3 ), W 4 ), (e) and the deired hock aborber troke(). 5
1. Ma without propelling force Forulae Exaple 1 W 1 = 100 1.5 2 0.5 = 113N W 1 = v2 0.5 = 100kg W 2 = zero Figure 15 W 2 = 0 v = 1.5 / W 3 = 113 = 0 = 113N W 3 = W 1 + W 2 C = 500/h W 4 = 113 500 = 56500N/hr W 4 = W 3 C = 0.05 (choen) e = = 100kg v D = v e = 2. Ma with propelling force Forulae Exaple 2 W 1 = 36 1.5 2 0.5 = 41N W 1 = v 2 0.5 = 36kg W 2 = 400 0.025 = 10N Figure 16 W 2 = F *v = 1.5/ W 3 = 41 + 10 = 51N W 3 = W 1 + W 2 F = 400N W 4 = 540 1000 = 51000N/hr F W 4 = W 3 C C = 1000/hr e = 2 51 1.5 2 = 45kg v D = v = 0.025 (choen) 2 W3 e = v D2 *v i the final ipact velocity of the a: With pneuatically 2.1 for vertical otion upward W 2 = (F - g) propelled yte thi can be 1.5 to 2 tie the average 2.2 for vertical otion downward W 2 = (F + g) velocity. Pleae take thi into account when calculating energy. 3. Ma with otor drive Forulae Exaple 3 W 1 = 800 1.2 2 0,5 = 576N W 1 = v 2 0.5 = 800kg W 2 1000 4 2.5 0.102 =850N Figure 17 W 2 = 1000 P ST v = 1.2/ W 3 = 576 + 850 = 1426N v ST = 2.5 W 4 = 1426 100 = 142600N/hr P W3 = W 1 + W 2 P = 4kW e = 2 1426 1.2 2 = 1981kg W 4 + W 3 C C = 100/hr v D = v = 0.102 (choen) e = 2 W3 vd 2 Note: Do not forget to include the rotational energy of otor, coupling and gearbox into W 1. 4. Ma on driven roller Forulae Exaple 4 W 1 = 250 1.52 0.5 = 281N W 1 = v 2 0.5 = 250kg W 2 = 250 0.2 9.81 0.05 = 25N Figure 18 W 2 = µ g v = 1.5/ W 3 = 281 + 25 = 306N W 3 = W1 + W2 C = 180/hr W4 = 306 180 + 55080N/hr W 4 = W3 C µ = 0.2 (teel/teel) e = 2 306 1.52 = 272kg V D = v = 0.0.5 (choen) e = 2 W3 vd 2 5. Swinging a with Forulae Exaple 5 W 1 = 20 1 2 0.5 = 10N propelling force W 1 = v 2 0.5 = 0.5 1 ϖ = 20kg W 2 = 50 0.0125 0.5 = 1.3N W M 2 = v = 1/ W 3 = 10 + 1.3 = 11.3N Figure 19 V R M = 50N W 4 = 11.3 1500 = 16950nM/hr R v D M W 3 = W 1 + W 2 R = 0.5 v D = 1 0.5 0.8 = 0.63/ W 4 = W 3 C = 0.8 e = 2 11.3 0.63 2 = 57kg v R C = 1500/hr v D = = v R = 0.0125 (choen) e = 2 W3 (Check the ide load angle - ee exaple 6.2) vd 2 6. Free falling a Forulae Exaple 6 W 1 = 30 0.5 9.81 = 147N W 1 = g h = 30kg W 2 = 30 9.81 0.05 = 15N Figure 20 W 2 = g h = 0.5 W3 = 147 + 15 = 162N W 3 = W 1 W 2 C = 400/hr W 4 = 162 400 = 64800N/hr h W 4 = W 2 C = 0.05 (choen) v D = 2 9.81 0.5 = 3.13/ v D = 2 g h e = 2 162 = 33kg e = 2 W2 3.13 2 vd 6
6.1 Ma rolling/liding Forulae 6.2 Ma free falling about Check ide load angle fro down incline W 1 = g h = vd 2 0.5 a pivot point hock aborber axi Figure 21 W 2 = g inβ W 3 = W 1 + W Figure 22 2 R W 4 = W 3 C tan α = R h.g F ß vd = 2 g h 2 W 3 e = vd 2 a with propelling force up incline W2 = (F - g inβ) b with propelling force down incline W2 = (F + g inβ) Calculation a per exaple 6.1 7. Rotary index table with Forulae Exaple 7 W 1 = 1000 1.12 0.25 = 303N propelling torque W 1 = v 2 0.25 = 0.5 1 ϖ 2 = 1000kg W 2 = 1000 0.05 0.8 = 63N M v = 1.1/ W 3 = 303 + 63 = 366N Figure 23 W2 = R M = 1000N W 4 = 366 100 = 36600N/hr v W 3 = W 1 + W 2 = 0.05 (choen) vd = 1.1 0.8 1.25 = 0.7/ W 4 = W3 C = 1.25 e = 2 366 0.7 2 = 1494kg v R R = 0.8 R v D = = ϖ R M C = 100/hr Note: Forulae given are only correct for circular table with unifor weight vd ditribution 2 W 3 e = VD 2 (Check the ide load angle - ee exaple 6.2) μ h v 8. Swinging ar with Forulae Exaple 8 W 1 = 0.5 56 1 2 = 28N propelling torque W 1 = v2 0.18 = 0.15 1 ϖ 2 I = 56kg 2 W 2 = 300 0.025 0.8 = 9N W2 = M ϖ = 1 rad/ W 3 = 28 + 9 = 37N Figure 24 R M = 300N W 4 = 37 1200 = 44400N/hr W 3 = W 1 + W 2 = 0.025 (choen) vd = 1 08 = 0.8/ W 4 = W 3 C = 1.5 e = 2 37 0.8 2 = 116kg R vd M v v D = v R R = 0.8 = ϖ R C = 1200/hr (Forulae are for ar with unifor weight ditribution) e = 2 W 3 VD 2 (Check the ide load angle - ee exaple 6.2) 9. Swinging ar with Forulae Exaple 9 W 1 = 1000 22 0.18 = 720N propelling force W 1 = v2 0.18 = 0.15 1 ϖ 2 = 1000kg W 2 = 7000 0.6 0.05 0.8 = 263N F r M v = 2/ W 3 = 720 + 263 = 983N Figure 25 W 2 = = R R F = 7000N W4 = 983 900 = 884700N/hr W 3 = W 1 + W 2 M = 4200N v D = 2 0.8 1.2 = 1.33/ W 4 = W 3 C = 0.05 (choen) e = 2 983 133 2 = 1111kg F V vd r R v D = v R r = 0.6 = ϖ R R = 0.8 (Forulae are for ar with unifor weight ditribution) 2 W 3 = 1.2 e = VD 2 C = 900/hr 10. Ma lowered at controlled Forulae Exaple 10 W 1 = 15000 1.5 2 0.5 = 16875N peed W 1 = v2 0.5 = 15000kg W 2-15000 9.81 0.152 = 22367N W 2 = g v = 1.5/ W 3 = 16875 = 22367 = 39242N Figure 26 W 3 = W1 + W2 = 0.152 (choen) W 4 = 39242 60 = 2354520N/hr W4 = W3 C C = 60/hr e = 2 39242 1.5 2 = 34882kg vd = v 2 W 3 e = VD 2 7
Reaction force Q (N) Stopping tie () Deceleration rate g Q = 1.2 W 3 t = 2.6 g = 0.6 vd2 vd g Approxiate value auing correct adjutent. Add afety argin if neceary. (Exact value will depend upon actual application data and can be provided on requet) Worked exaple Forulae W 1 = M. V 2. 0.5 W 3 = W 1 + W 2 W 2 = F. S W 4 = W. 3 C e = 2. W 3 V 2 Exaple M = 350kg V = 1/ F = 1500N C = 300/hr S = 0.05 (choen) W 1 = 175N W 2 = 75N W 3 = 250N W 4 = 75000N/hr e = 500kg By coparing W 3, W 4 and e with the value in the capacity chart the correct product ay be elected ie. in thi intance erie A type RS tock no. 834-308. Capacity chart - adjutable RS tock no. Shock aborber Mode Max energy capacity N Eff Weight e (kg) W3/cycle W4/hr in/ax 834-229 FA 1008 V-B 8 1.5 3600 0.6-10 834-235 FA 1210 M-B 10 3 6000 0.5-30 834-241 MA 225 M 19 25 45000 2.3-226 834-257 MA600 M 25 68 68000 5-1360 834-261 MA 900 M 40 100 90000 14-2040 831-272 A 1/2 x 1 25 115 85000 5-450 Capacity chart - elf copenating RS tock Shock aborber Max energy Eff Weight no. Mode capacity N e (kg) W3/cycle W4/hr in/ax 268-0535 MC 25 M 6 2.5 22600 0.7-2.3 268-0557 MC 25 M 6 2.5 22600 4.8-5.5 268-0563 MC25 MH 6 2.5 22600 3.6-13.6 268-0579 MC75 M 10 8.5 28200 0.2-15.4 268-0585 MC75 M 10 8.5 28200 0.9-16 268-0591 SC 190 HDM-1 16 20 34000 1.4-7 268-0614 SC 190 HDM- 2 16 20 34000 3.6-18 268-0620 SC 190 HDM-3 16 20 34000 9-45 268-0636 SC190 HDM-4 16 20 34000 23-102 268-0670 SC 300 HDM-1 19 33 45000 14-8 268-0664 SC 300 HDM-2 19 33 45000 4.5-27 268-0642 SC 300 HDM-3 19 33 45000 14-82 268-0658 SC 300 HDM-4 19 33 45000 32-204 268-0686 SC 650 HDM-1 25 73 68000 8-45 268-0692 SC 650 HDM-2 25 73 68000 23-136 268-0709 SC 650 HDM-3 25 73 68000 68-408 268-0715 SC 650 HDM 4 25 73 68000 204-1180 268-0721 SC 925 HDM-1 40 110 90000 14-90 268-0737 SC 925 HDM-2 40 110 90000 40-272 268-0743 SC 925 HDM-3 40 110 90000 113-726 268-0759 SC 925 HDM-4 40 110 90000 340-2088 RS Coponent hall not be liable for any liability or lo of any nature (howoever caued and whether or not due to RS Coponent negligence) which ay reult fro the ue of any inforation provided in RS technical literature. RS Coponent, PO Box 99, Corby, Northant, NN17 9RS Telephone: 01536 201234 An Electrocoponent Copany RS Coponent 1998