TRANSVERSE VIBRATIONS OF A ROLLER CHAIN DRIVE WITH TENSIONER W. Coiand G. E. Johnson -resign Laboratory Mechanical Engineering and Applied Mechanics The University of Michigan Ann Arbor, MI 48109-2125

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TRANSVERSE VIBRATIONS OF A ROLLER CHAIN DRIVE WITH TENSIONER W. Choi and G. E. Johnson Design Laboratory Mechanical Engineering and Applied Mechanics The University of Michigan Ann Arbor, MI 48109-2125 Glen E. Johnson@um.cc.umich.edu Abstract A dynamic model for analysis of the performance of roller chain drive with a tensioners is presented. The model is based on the axially moving material model first given by Mote, but includes extensions that allow consideration of the effects of polygonal action, impact, and the periodic span length changes. Three equations of motion are obtained - one for the tensioner and one for each of the chain spans. The motion of the tensioner interacts with the motions of the chain spans through the chain tension, the contact angles, and the transverse displacement changes of the contact points. The effects of the periodic length change, tensioner stiffness, tensioner damping, tensioner position and external periodic load are investigated. Some of the results are compared to those obtained for a comparable chain drive without tensioner. Solutions are obtained by a finite difference method.

Nomenclature c: chain span velocity ci: contact damping coefficient in impact Ct: torsional damping coefficient of tensioner 12: moment of inertia for driven sprocket system It: effective moment of inertia of tensioner about its pivot point ki: contact stiffness coefficient in impact Kt: torsional stiffness coefficient of tensioner me: effective mass of tensioner meg: effective mass of tensioner for gravity effect mi: effective mass in impact ml: mass per unit length of chain Mt: moment due to torsional spring Mtd moment due to torsional damper Met: moment due to chain tension Mg moment due to gravity nl, n2: driving and driven sprocket tooth numbers rl, r2: radius of driving and driven sprockets rt: radius of roller PC: chain tension Tex: magnitude of external periodic loading Tp: tensioner preload torque Ts static tension u, v: transverse displacements of chain span Vre: relative velocity of an engaging roller to sprocket surface of: phase shift between engagement and disengagement T: angle between relative velocity of roller and the sprocket surface 0: tensioner arm angle 00: initial tensioner arm angle oI: angular velocity of driving sprocket c02: angular velocity of driven sprocket COex frequency of external periodic loading

1. INTRODUCTION Belt drives and chain drives are both commonly used methods of mechanical power transmission. Tensioners (or idlers) can be used on these two types of drives for several different reasons. In belt drives, tensioners are generally used to increase overall tension (to reduce slip). This has the additional benefit of increasing belt life. In roller chain drives, tensioners can improve drive performance by eliminating flop and sway on the slack side and reducing vibration and noise on the tight side [1-2]. In spite of the importance of the tensioner, there has not been much research reported in this area. Ulsoy, Whitesell and Hooven investigated a belt-tensioner system and reported mathieu type instabilities due to the belt tension variations caused by a periodic external load [3]. To date no theoretical investigations of chain drives with tensioners have been reported. In this study the chain drive system comprises two chain spans and a torsional tensioner. The modeling of each chain span is based on an axially moving material [4]. The modeling of polygonal action and impact is more completely described in reference [5]. Periodic length change for each chain span is also considered. The interaction between the tensioner and the chain spans and the performance of the chain drive with a tensioner are examined and the results are reported. 2. TRANSVERSE VIBRATION IN ROLLER CHAIN DRIVE WITH A TENSIONER The chain drive system with a tensioner includes the driving and driven sprockets, tensioner assembly, and chain spans (see figure 1). The tensioner assembly consists of a tensioner arm, a tensioner disk, a torsional spring, and a torsional damper. In order to simplify the analysis, a circular disk is adopted as the tensioner pulley and the contact line between the disk and the chain span is assumed to form a circular arc. The modeling of the tensioner and the chain span are done independently but they are coupled through the chain tension term, the inclination angles of the chain spans and the displacements of the contact points between the tensioner and the chain spans.

2.1 Dynamic Load 2.1.1 Speed Tensioning, Impact and External Periodic Load The dynamic load due to speed tensioning is obtained in a way similar to the case of chain drive without a tensioner, i.e., D = Xmc2 (2-1) The pulley support constant X can be estimated as follows [6]. = kj(kb + k +kg) (2-2) where kb = (EA/L)(cos 21 + cos 2)2, ks = k/l2, kg = Ts(l + 1), Ci = d(cos fi)/dxs, EA = chain strength, L = overall span length. As for the impact, only the major impact between the engaging roller and the driving sprocket is considered. The dynamic load due to impact is determined in the same way as in the case of a chain drive without a tensioner [5] and it is given by Vrelsin 2fl Dip = 2 e-nt { (ki - ci n ) sin cdt + Ci Cod CS t (O<t< ) (2-3-a) d d Dip 0 (c-L <t<- ) (2-3-b) imp co 01 n where Vre, = oC Pc,, Pc: chain pitch, T1 = 35 + 240/nl + e deg., e <<1, Con = (k/mi)0-5, (Od = On (1-2)0.5, = Ci/(2 (mi ki)~'5). External dynamic loads are included by a single trigonometric term. Det = Tex cos ex t (2-4)

2.1.2 Polygonal Action The major parameter related to polygonal action is the fractional pitch and it is determined by the positions of the tensioner and sprockets. The lengths of span #1 and #2 and the length of the arc formed around the tensioner disk are shown in figure 1. The fractional pitch is calculated based on the geometric information and there are two different general tensioner positions. One is the case where the tensioner pivot point stays on the right side of the vertical line passing through the center of the tensioner disk (tensioner position #1) and the other is the case where the pivot point resides on the left side of the vertical line (tensioner position #2). L1 + L2 + (rt + rr)(e, + E2) - i*pitch P pitch where i: any integer such that 0 < f < 1, e = + + tan-l (PIY - 4Y) -tan-1 ( 1=2 p P4 r1 +rt+ rr PX 4X ~ = - + tan-1 (P2Y- 4Y_ tan-1 (r2 +k+. ) 2 2 P4- P2X r + r L1 = S - (r + rt + rr) S = (Px P4X) + (PY- 4Y) L2 = VS - (r2 + rt +rr)2 S2= (P4 2X) +(P4Y- P2Y) P4X = P3X - It cos (for tps #1) or P3x + It cos (for tps #2), PY = P3Y - It sin. The fractional pitch changes as time changes because E1 and 62 change. The fractional pitch was assumed to be constant in the case of a roller chain drive without a tensioner [5]. The phase angle (af) between engagement and disengagement is computed by multiplying one tooth angle of the driven sprocket by the fractional pitch. The angular velocity relationship between the driving and driven sprockets is formulated by equating the chain velocities from the driving sprocket side and the driven sprocket side (figure 1). Using the xl-yl coordinates and x2-y2 coordinates in the figure, the same equation used in the case of the chain drive without a tensioner is obtained. The

way to obtain the dynamic load due to the polygonal action for the rigid four bar model (applicable at low speeds) is the same as presented in reference [5] except that the constant fractional pitch is replaced with the time dependent fractional pitch given by equation (2-5). The angular motion of the driven sprocket and the dynamic load is given by 02 = 02int (02int + n (2-6-a) 2 2 0 =0 > (2int> nt (2-6-b) 02 = 2int i (0 2int 2 n2 (2-6-b) where 02int = cos' 1- (S 01 - cos 910) + cos 920o 9 = 91(0 ) = C — 2' 2 n 020 = 02(0) = 2 - - + af, af = 2 fp. 20 2(0) 2 i —c2 (2-7 Dpo = 2 = r2 sin (rl cos 01- r2 cos 02) (2-7) where 01=co)t+ -- 71L 2 n1 The way to obtain the dynamic load due to the polygonal action in case of the elastic four bar model (applicable at moderate and high speeds) is also the same as presented in reference [5] except that the time-dependent fractional pitch is used instead of the constant fractional pitch. The angular movement of the driven sprocket and the dynamic load are given by 02 =f02 dt 3= (q0 t + q1 sin pt + q2 sin 2pt + q3 sin 3pt + q4 sin 4pt - Pot - Pl cos pt - P2 cos 2pt - p3 cos 3pt - p4 cos 4pt + cin) (2-8) N3 a2 where c = (a + 2 2) + p + p2 + P3 + P4 - q0 T dint), inwhee Ci = n2r2 1 sin( —)

dint(x) rounds the value of x to the nearest integer towards zero. Dpl =k (x2 -x-L) = Np {(N a-11) sinpt+mlcos pt +0.5(Na2 -12) sin 2pt + 0.5 m2 cos 2pt- mi - 0.5 m2) (2-9) 2 ao 2 a0 where a = l2 a2 = 1- n2 1-4n2 The coefficients li, mi, pi, qi,are described in Appendix I. 2.2 Transverse Displacements of End Points and Periodic Length Change There are four end points - two for span #1 and the other two for span #2. This set includes the contact point between span #1 and the driving sprocket (end point #1), the contact point between span #1 and the tensioner disk (end point #2), the contact point between span #2 and the tensioner disk (end point #3) and the contact point between span #2 and the driven sprocket (end point #4). The vertical displacements of end point #1 and end point #4 are calculated as follows. u = r (1 - sin 0e) (2-10) o = r2 (1 - sin 02) (2-11) The angle of end point #4 is given by equation (2-6) at low speeds and by equation (2 -8) at medium and high speeds. The vertical displacements of end points #2 and #3 are determined by the motions of the contact points between the tensioner disk and the chain spans. The computation of the contact points is based upon two angles (0m and On in figure 1) formed by the chain spans and sprocket geometries. P5x = P4 + (rt + r) cos P = P4Y- (rt +r) sin m (2-12) P6X = P4x- (rt + r) cos P = P4Y - (rt + r) sin n (2-13) where m= l —e On= t2 -2 1' 2 2

Suppose P5 has moved from [P5hs to [P5]i+l and P6 has moved from [P6]i to [P6]i+i during one time step. aa = [P5X]i+l- [PsX]i, a = [PY]i+l - [PY]i (2-14) = [6X]i+l - [P6X]i' b = [6Y]i+l- [P6Y]i (2-15) There are four different cases for each form of tensioner position (tensioner position #1 or tensioner position #2) in the computation of vertical displacements in terms of aa, oCb, Pa, and 1b. The related angles and the changes of vertical displacements are shown in figure 2 for tensioner position #1 and the expressions for the changes (ds and d6) are described in Appendix II. [u0]i+1 = [u0]i + [d5]i (2-16) [vn]i+1 = [V]i + [d6]i (2-17) where i, i + 1 represents time i*At and (i+l)*At respectively. The length changes occur for both span #1 and span #2 since all the end points are moving. In the case of the chain drive without tensioner, the span changes its length abruptly whenever there is engagement or disengagement between chain links and the sprockets. When the tensioner is present, each span changes its length, not only abruptly at the engagement and the disengagement of chain links, but also gradually at other times. The span lengths in a chain drive with a tensioner are determined by the end points. spll (length of span #1) = (Zlx - Psx) + (Zy - P)2 (2-18) spl2 (length of span #2) = (x - P6x)2 + Z2y P6y) (2-19) where Z1x = Plx - rl cos Ol Zly = Ply + r1 sin 0z 2x = P2x + r2 COs 0z2 2y = P2y + r2 sin z zl= 1-e1 e z2 = -2 02

2.3 Equations of Motion The tensioner assembly is modeled as a single degree of freedom system with torsional damper and torsional spring. The tensioner rotates about its pivot and the equation of motion is formulated by considering the moment equilibrium about the pivot point. It sc = M +tsM + Mg (2-20) There are four moments related to the pivot point of the tensioner and they are due to the torsional spring, torsional damping, chain span tension (see figure 3) and gravity. Mt = K (00 - 0) + Tp (2-21) M td= -C (2-22) Mbt = Pc lt sin (0 - el) - sin(0 + 2) ) (tsp #1) or Pc it sin (0 - e2) - sin(0 + e1) } (tsp #2) (2-23) where Pc(chain tension) = Ts + Ds + Dpl + Dimp + Dext M = m g t cos O g eg Substitution of the expressions for the moments into (2-20) leads to the final form of the equation of motion. me + + C +K, = Kt 00 +Pt +Pc Itsin (- el) - sin (0 + e2)) + meg g lt cosO for tensioner position #1 (2-24) me i e+CtO +Kt, =Kt0o +Pt+P ltsin ( -2) -sin (0 + e1)} + meg g it cos0 for tensioner position #2 (2-25) The motion of the tensioner was represented by one ordinary differential equation which is coupled with the equations of motion for the chain spans through the tension term

and inclination angles. Forward finite differencing was done about time to solve the equation. There are two chain spans in a chain drive with a tensioner. The chain span in contact with driving sprocket is denoted as span#1 and the other is span#2. Because of the presence of the tensioner, sag can be neglected in both chain spans. The equation of motion for a chain drive without a tensioner can be used for each chain span. The transverse vibration of chain span #1 is represented by u and the transverse vibration of chain span #2 is represented by v. Span #1: m1 u+ 2 m c ut +ml c2 u + cd(Ut + c ux) - (P + P) xx = 0 (2-26) Span #2: m vtt + 2 mi c vt + m1 c2 vx + cd(vt + c vx) - (P + P) v = 0 (2-27) where P = Ts + Dc, Pt = Dpol + Dimp + Dext The boundary conditions for span #1 are given by the equations (2-10) and (2-16) and the boundary conditions for span #2 are given by the equations (2-11) and (2-17). The finite difference formulation followed the finite differencing scheme used in the reference [5]. 3. COMPUTER SIMULATION 3.1 Time Step Control and Simulation Strategy A nonuniform time step scheme was utilized to execute computation efficiently. The basic idea is to use smaller time steps near the instant of the impact between engaging roller and sprocket tooth and larger time steps away from that instant so that the effect of the impact is included while the total computation time is minimized without causing numerical instabilities. To achieve this objective, an exponential function was utilized. The main objective of the simulation is to observe the change of the motions of the roller chain drive and the tensioner under different circumstances. In order to obtain the steady state response at an operating speed, the transient region during some amount of time from the beginning was ignored and the response after that was used to calculate the vibration amplitude. Next, the maximum amplitude from the equilibrium configuration of a chain span was chosen as a variable representing the vibration amplitude effectively over a wide range of operating speed. In order to represent the motion of the tensioner, two more

variables were introduced - the average angle of the tensioner arm in vibration and the angular vibration amplitude of the arm about the average angle. 3.2 Simulation Results and Discussion The chain drive system used in this simulation consists of number 40 chain, two 24 tooth sprockets and a tensioner. The effect of the moment of inertia for a driven sprocket system is investigated for a chain drive with a tensioner and the results are shown in figure 4. First of all, the dual peaks corresponding to the short and long lengths for each chain span that were observed in [5] are no longer present. Instead, there is one peak corresponding to the average span length. This is because the span takes on the short span or long span length only instantaneously when a tensioner is present. The span length varies continuously between these two values during the remainder of the tooth period. The resonance of the tensioner is around 248 rpm and the angular vibration magnitude increases as the moment of inertia increases (especially around the resonances). This is because the increase in the moment of inertia induces an increase in the tension variation due to the polygonal action. The average angle of the tensioner decreases as the operating speed increases. This decrease is physically reasonable since the tension of the span increases as the speed increases due to speed tensioning and this tension increase pushes up the tensioner. The vibration amplitudes of span #1 and span #2 show similar trends. The vibration amplitude is not strongly affected by changes in the moment of inertia before the fourth resonance occurs, but the effect of changes in the ecmoment of inertia is clear around the fourth resonance. In the case of the largest moment of inertia of figure 4 (solid line), even instability can even occur due to the parametric excitation. The effects of tensioner stiffness are represented in figure 5. The location of the tensioner resonance increases as the stiffness increases and the amplitude at the resonance also increases as the stiffness increases since the excitation due to polygonal action increases as the speed increases. If the magnitude of the excitation remains the same at any speed, the amplitude will decrease as the stiffness increases. The amplitude of span #1 around the fourth resonance increases as the stiffness decreases while below the resonance the effect of the tensioner stiffness is insignificant. The amplitude of span #2 also shows the same trend. The effect of tensioner damping is shown in figure 6. Three cases (underdamped, critically damped, and overdamped) were considered. The effect of the damping is noticeable around the fourth resonance for span #1 and around the second and the fourth resonances for span #2. The position of the tensioner pivot point was changed in the vertical direction and the effects are shown in figure 7. One of the most interesting subjects for research in the area

of roller chain drives is optimal tensioner position. Vertical changes in the location of the tensioner pivot point make a difference in the fractional pitch of the total chain span. Three cases were investigated. One is the system with 0.5258 initial fractional pitch (dotted line), another is the system with 0 initial fractional pitch (dashed line) and the other is the system with 0.6721 initial fractional pitch (dot-dashed) at 500 rpm. The initial fractional pitch is calculated at the equilibrium configuration for a given operating speed. As far as these three cases are concerned, the angular vibration amplitude is largest for the system with 0.5258 fractional pitch and smallest for the system with 0 fractional pitch. The effect on the vibration amplitude becomes large as the operating speed increases and there are some differences among the resonance positions because each span length changes as the vertical position of the tensioner pivot point changes. The effect of the external periodic load on the case where the fractional pitch is 0 is also shown in figure 7. The effect is similar to that observed in the case of a roller chain drive without tensioner [5] and the addition of the external periodic load with the frequency oex = coi reduces the vibration amplitude around the resonances. The angular vibration amplitudes increase due to the external periodic load. Figure 8 shows the effect of the two different general tensioner position forms, corresponding to tensioner position #1 (P3 = (0.25, 0.1)) and tensioner position #2 (P3 = (.1702, 0.1). The vibration amplitude with tensioner position #2 is larger than the amplitude with tensioner position #1 around the second resonance for span #1 and around the fourth resonance for span #1. It is interesting to notice that tensioner position #1 is preferred to tensioner position #2 in recent automotive serpentine belt designs. The motions of the chain drive at medium operating speeds were also observed. Figure 9 shows the behavior of the chain drive when instabilities due to polygonal action in the elastic chain span occur. The system includes a relatively high inverse ratio of the tooth numbers (n2/nl = 50/15) and small moment of inertia for the driven sprocket system (12 =.0005 kg m2). As shown in the figure, the range of the vibration amplitude affected by the second instability is wider than the range affected by the first instability since there is a transverse resonance just above the second instability and the effect of the resonance is combined with the second instability. As the last simulation, a comparison between a chain drive with and without a tensioner was done for a system with the same center distance (30 pitches) between the driving and driven sprockets (figure 10). The tensioner position was set to let the initial fractional pitch at the operating speed of 500 rpm be zero. For the first and second cases (Ts = 25 N and Ts = 50 N) there are instabilities due to the combined effect of parametric (polygonal action) and external (curvature and transverse displacement changes at both ends)

excitations in a chain drive without tensioner, while there exist only relatively small vibrations in the chain drive with a tensioner. For the third and fourth cases (Ts = 100 N and Ts = 200 N), there are larger amplitudes at resonances but fewer number of resonances in the chain drive with a tensioner than in the chain drive without a tensioner since each chain span in the chain drive with a tensioner is smaller than the chain span in the chain drive without a tensioner. 4. SUMMARY AND CONCLUSION A model for the chain drive with a tensioner was developed including some important features, such as polygonal action, impact, and periodic length length of each chain span. The tensioner interacts with the chain spans in several ways including chain tension. The periodic length change for each chain span does not produce two distinct peaks around each order resonance as was the case for systems without tensioners. Instead, there is one peak at each resonance, corresponding to the average span length. This is because the tensioner causes the span length to vary continuously between the shortest and longest span lengths during one tooth period, rather than to abruptly take on either one value or the other. At low operating speeds, higher tensioner stiffness could reduce the vibration amplitude especially around high order resonances, but the location of the tensioner resonance increases. The vibration at the resonance also increases because the tension variation increases as the operating speed increases. Changes in the vertical location of the tensioner pivot point can be used to change the fractional pitch of the total chain span. The vibration amplitude can be reduced and sometimes instability can also be prevented by adjusting the vertical position to let the time dependent fractional pitch be approximately zero. At medium operating speeds, there is a danger of instabilities due to polygonal action. The possibility is generally low, but it increases as the inverse ratio of the tooth numbers increases and as the moment of inertia for the driven sprocket decreases. Compared to chain drives without tensioners, the tensioner can not only prevent instabilities for low static tension operations but can also reduce the number of resonances. 5. ACKNOWLEDGEMENTS The authors gratefully acknowledge the support of the National Science Foundation through Grants No. MSM-88-12957 and MSS-8996293. We also wish to thank N. C. Perkins for his thoughtful suggestions during the early stage of the model development.

BIBLIOGRAPHY 1. Les Gould, "Use an Idler to Solve Your Drive Problems", Modern Materials Handling, no. 45, Jun., 1990, p.75. 2. Kurt M. Marshek, "Standard Handbook of Machine Design - Chapter 32 Chain Drive", McGraw-Hill, 1986, pp.32.1 - 32.39. 3. A. G. Ulsoy, J. E. Whitesell, and M. D. Hooven, "Design of Belt-Tensioner Systems for Dynamic Stability", ASME Journal of Vibration, Stress, and Reliability in Design, vol.107, Jul. 1985, pp.282-290. 4. C. D. Mote, Jr., "Some Dynamic Characteristics of Band Saws," Forest Products Journal, January, 1965, pp 37 -41. 5. W. Choi and G. E. Johnson, "Vibration of Roller Chain Drives at Low, Medium and High Operating Speeds", Technical Report No. XXXXXXXXXX, Mechanical Engineering & Applied Mechanics Dept., University of Michigan, Ann Arbor, MI, 1992. 6. R. S. Beikman, N. C. Perkins and A. G. Ulsoy, "Equilibrium Analysis of Automotive Serpentine Belt Drive Systems Under Steady Operating Conditions", Proceedings of 22nd Midwestern Mechanics Conference, Oct. 1991, pp.533-534. APPENDIX I Coefficients in Equation (2-8) and (2-9) The values of 1,, ml, 12, and m2 are obtained from the following matrix equation. X = (AT A)-1AT F where X is a 4x1 matrix whose elements are unknown variables (li, ml, 12,m2). The elements of the matrices in equation (A-1) are as follows. c1 k co aokco b1 kco1 c k co1 a0k o1 b2 k co 11 2Np 12 2Np ' 3 4Np' 14 2p 4Np c2kco1 c212P 01 aokco, b kco1 bc2kco bb pp o, I2pI)1 a2 2Np 2 N3 a' 2 P P + + Np "2Np 2 a2 N3 a_ N2 '

Cl kicolcI2,pol bkol bkI2pol, c, kcol clI2p(o a23 4Np 2a2N3 a 4Np 2aNN3 3l 2Np a2N3 a = -C ko -cl Pl bP k b lkl 2k b12PO)^ 31 2 N p aN3 2Np Np aN3 2Np Np a~N3. a34=~a2o N3 ' b2 k ol b2kcop bIpo aa ko b o 2p 2Np a2 o a c32I Ik 3 c2 1 p 23o')i b2kco1 3b+ 2 2pc ck=, a31 c + 2 2Np p ' 2N2 Np aN3 43 4Np 2a N c2kcol 3c2I1p2ol b2kco1 3b2I2,po c k1o 3cI2po] a= 24Np 2 aN3 4Np aN3 =a5=O4 N p c] k coI 3 c] I2 p tol b1 k 3o1 3 b1 12 p co] a5 a43 a44 a,, = a,2 = 0, a43 4Np 2 N p 2 3 4 N p 2 4 N3 c2ko 2c2IP 01 b2kol 2b212 pco a- 4Np a2 N3 54 4Np a N3 aOkco b2 k o b2 I2 P C I2 P co c1 kcol c2 k o1 C2 I2 P W1 a =- P +2 N + nl P 2Np 2 a4 N3 ao N2 6 Np 2Np 2 aN3 bi k c bi ~po1, 3 c koclI2 p o b k co b 12 p Co +____ -, a 1 +I_ __I*I 63 4Np 2 a N3 a64 4Np 2a2N3 71 2Np apN3 c = c k o ] c 2kco l c2pco) ao ko 2I2pwo c2kco1 72=- 2Np Np a2 N3 'a73= aN2 2Np b2 kco 3b2I2P PO c2k _ol 3c22I2Po) baiko 3 b I+3 p a81= 2Np 2 ajN3 2Nk 21 32N3' 3 4Np 2aN3 a4c~kco 3c1~pow1 ~b ko 2bkolpco a84 = a9 =a92= 2=, a93 + 4Np 2 aN3 - a93- 4Np a2N3 c2ko. 2c2I2p0 a94 4Np a2N3 2=ml, x3=12, x4=m2, pf ^a kc alkw ^a2clko l alc2kpo clk2pco ' 2p 4p 4p 2p -,N2

f alclko) 20C212P0 f -)a2c1kco a1 c2 k col f5 a2 c2 k co - 2p a N2 4 4p ' 2p - 4p f6=a2blk ~alb2kol +aoalkNcw i b fapblkoIb1 alb2kco aoN2 8 al bl kol+ a0a2kNco1 2b212po1 f a2b2kcol _~ +,i~ 7 2p 2p a N2 9 4p The values of pi and qi are represented as follows. qO = -0.5 (b 11 + b2 12 + c m1 + c2 m2) + (ao N)2 ql = { —0.5 (b2 11 + b1 12 +c2 ml +c1 m2) - a n (b - 1)} q2 = 1 {-0.5 (b 11 - cl ml) - ao N (b2 -12)), 2 p q3 = 3{f-0.5 (b2 11 + bl 12 - c2 mi - cl m2), q4 = 4 {-0.5 (b2 12 - c2 m2)) pl = {(-0.5 (c211 - c 12- b2 ml + b m2)-ao N (c - ml) P2 = 2p-0.5 (cl 1, + bl ml) - ao N (c2 - m2)) 3 =3 {-0.5 (c2 11 +cl 12 +b2iml +bl m2)}, p4 = 1-{-0.5 (c2 12 +b22)) APPENDIX II Computation of Vertical Displacements of Contact Points Between Tensioner Disk and Chain Spans There ree four different cases to be considered in the computation of vertical displacements for each tensioner position (tensioner position #1 or tensioner position #2). For tensioner position #1, the related angles and the changes of vertical displacements are shown in figure 7 and computed as follows

d5 = o2 + 2 sin Ya ((aa > O) or -aca + 3 sin Ya (a(a < O) Case 2 (P3x < P5x): Ya = 2- m tan-1 (a) 2 + p+"2 sin Va2 21si d5 = V + a sin a (Ca, > 0) or - + 3 sin ya (aa < O) Case3( On>0 ): b =- +0n+ tan-1 (- ) -e -N (X2 + p2 - _V 0C2 +. d6 = +b2 + 3b sin Yb (ab > O) or - ol2 + 2 sin Yb (Ob < 0) Case 4 ( 0 < ): yb = j- On - tan'l (- ) d6 = - + b sin Yb (ab > O ) or a, + 32 sin Yb (ab < 0) In case 1 and case 2 the same expressions are derived but in case 3 and case 4 different expressions are presented. Finally the next vertical positions of end points #2 and #3 are obtained by adding the displacement changes calculated above to the present positions. For tensioner position #2, the procedure to determine d5 and d6 is exactly the same as for the tensioner position #1. Case 1 (P3x > P6x): b =2n tan- (-) d6 = - / + sin Yb ((b > 0) or + sin yb (ab < 0) Pb Case 2 (P3x < P6x): Yb = on + tanl (a") 2 m d6 = - N/a + p sin Yb (ab > O) or Ca + p sin Yb (Cb < 0) Case 3 ( em > 9): Y, = - 2 + 6 - tan^ (-) <

2 mca d = - 2 + a2 sinya (a > O) or Va2 + a sin, (a, < O) Case ( 6m <> ) y, = - tan(' (oc) 2 d5 = ~/a2 + 2 sin y (aa > O) or - /a2 + [~ sin ya (aa < 0)

P3 P tensioner position #2 \ tensioner position #1 Z2,/ 2 -AI,.~\, P2 Tensioner Disk r2 Pitch Circle Driven Sprocket Driving Sprocket Figure 1 Configuration of Chain Drive with a Tensioner

13 C D - - - - ^ ^ ^ ^ ^ - ) L i ('' C)i^^/^^v^^, ^^ l^ \0 ^r ^^,"**~^~~^^^ ' *o~~~~~~~~~~~~~~~~~~~~~~~~~~~~~rLP r^- \ B i Cr ~ K~~~~~~~~o t ^ ^ X. _____ ^ ^~~~~~~~~~~~~~~~~ ~p -I- >~ '~^ — / r~~~~~~~~~~~^.^< 5~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 0~

P4 I, /I\ I PI IP sin in el IE1 PC cos ~2 P6 5 Pc cos ~1 Figure 3 Calculation of Moment Applied to a Tensioner by Chain Tension

0.16 55.27 IL) 3 0.14- Dotted:12 = 0.0005 kg m2 d)^~ ^ \ ~~~~~Dashed:12 = 0.001 kg m2 ^ 55.265 0.12 - 5.6 01 ^^' ^ \ Solid: I =0.002 kg m2 I 0.1 \ - C f 0.08 - < 55.26 0~~~~~~~~~~~~~ 0.06 -/ ~.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~. 0.04 I K. — 55.255 7:1 0.02 -..... ^ o ^ ^ ^ ^ "._____________________________...'......... 55.25 250 ___________________________________ 5 055.25 50 100 150 200 250 300 350 400 450 500 50 100 150 200 250 300 350 400 450 500 Rotating Speed of Driving Sprocket (rpm) Rotating Speed of Driving Sprocket (rpm) 0.01 0.01 B E 0.008 - 0.008 I I o 0.006 -o 0.006 - Z 0.004 - 0.004 - ~ 0.002C 0.002 - 0.02-t.0 O t3 O 1-4~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 50 10 20I20.0 30 0 50 100 150 200 250 300 350 400 450 500 50 100 150 200 250 300 350 400 450 500 Rotating Speed of Driving Sprocket (rpm) Rotating Speed of Driving Sprocket (rpm) Figure 4 Effect of Moment of Inertia for Driven Sprocket System

_ 0.12.. 1 56.2 8 01L Dotted::k= 100 Nm: Dotted:l = 10 0 N~m /............................................ 0.1 Dashed:kt = 388 Nm \ - Solid: k, = 700 N-m ' 0.08 - \ 55.8 g 0.06 - / 55.6 - Rotating Speed of Drving Sprocket (rpm) Rotating Speed of Driving Sprocket (rpm) 0o.o0 '~ 0.01 0.008 055.4008 0.02 ~~06. ------- 7 55.2 006 M O l~________________ 558 0.01 0.01! ' g 0.008 - 00 0 0.006 - Rotating Speed of Driving Sprocket (rpm) Rotating Speed of Driving Sprocket (rpm) Figure 5 Effect of Tensioner Stiffness

0.1 55.27 3~tr~~ I~~ IDotted:cL =0.1 kg m/sec.. 0.08-. Dashed:ct =1 kg-m/sec::I |.' ~ |- Solid: c 10 kg.m/sec | 55.265 -0.06. ' 8 |:F:~~~~~~~~~. < 55.26 0.04 t ' o>,. ~ ~ 0 55.255, 0.02 - 0.02....... < 55.25 t 0.02 -.' ' ~I.......... 50 100 150 200 250 300 350 400 450 500 50 100 150 200 250 300 350 400 450 500 Rotating Speed of Driving Sprocket (rpm) Rotating Speed of Driving Sprocket (rpm) xlO-3 x10-3 4-E~~~~~~~~~ 4 -I. Io 44 3O - O Figure 6 Effect of Tensioner Damping 2E 2t )I ~l~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ F: C~~~~~~~~~~~~~~~ Figure 6 Effect of Tensioner Damping

0.1 55.65 '~ 55.6 0.08 ' ' 5 a^~~~~.' ~~ ^Co~~~~~~~C 55.55 - 1 0.06 - 55.5 I ~ ~ '.^'\ ~< 55.45 --- —-- 0.04 6 I.4 ~~ -r~ 'r ~. - 55. > 55.35 -^ 0.02.'. -. ' "-~ - ~.....o 1 55.3 ".. 55.25 50 100 150 200 250 300 350 400 450 500 50 100 150 200 250 300 350 400 450 500 Rotating Speed of Driving Sprocket (rpm) Rotating Speed of Driving Sprocket (rpm) x10-3 xl 0-3 6. 6 Cl4 5~. ~... i cI 4- c.' 4.4 -o,o., 0 0, 3 -' '3 -' 1 ' 0 3..,. ~' T' ~I I: HV!.,' '' 0 2I,;-,.~i i.' ','. -~~~~.~~ ~-. x. C,;> 0;> 50 100 150 200 250 300 350 400 450 500 50 100 150 200 250 300 350 400 450 500 Rotating Speed of Driving Sprocket (rpm) Rotating Speed of Driving Sprocket (rpm) Figure 7 Effect of Tensioner Position and External Load Dotted:P3Y= 0.1 (fp =0.5258 at coi = 500 rpm) Dot-Dashed:P3 = 0.1 15 (fp =0.6721 at o1 = 500 rpm) Dashcd:P3 =0.1303(f =Oatoj=500rpm) Solid: P3Y = 0.1303 w/ext. load

_ 0.1 55.28 Solid: tensioner position #1 525..~~. (P3=(0.25, 0.1)) 55.275-........ ^ 0.08- Dotted: tensioner position #2 ~ (P3 (0.1702, 0.1)) c 55.27 -0I06 - 5 5.265 - 0.06 ^ 55.26 -0.04 -'1 55.255 - > > 5^g 5.255 - ^ t 0.02 - 55.25 bo ~~~~~~~~~~~~~~~~~~~~~~~55.245 -01 ^^ ^" ^-55.24 50 100 150 200 250 300 350 400 450 500 50 100 150 200 250 300 350 400 450 500 Rotating Speed of Driving Sprocket (rpm) Rotating Speed of Driving Sprocket (rpm) - 3.5^____________________________ 0.01_________________________- ____ 0.008 - 2.5 -0 ~ ^ ^ 0.006 - -II 2 1.5 0.0 0.004 - 0 C, 0.5 >0 0 50 100 150 200 250 300 350 400 450 500 50 100 150 200 250 300 350 400 450 500 Rotating Speed of Driving Sprocket (rpm) Rotating Speed of Driving Sprocket (rpm) Figure 8 Effect of Different Tensioner Positions

1.2 55.8 55.7 ) I\ 0.8 - 55.6 a) 0.6 55.5 - 0.4 - 55.4 0.2 55.3 5~~~~~~~~~~~~~~~~~~~~~~~~~~5. 505254565860500 520 540 560 580 600 0 520 540 560 580 Rotating Speed of Driving Sprocket (rpm) Rotating Speed of Driving Sprocket (rpm) 0.01 - 0.01 0.008 - 0.008 - " 0.006 -- 0.006 - ' 0.004 -\ 0.004 0.002 - 0.002 - > 0 >0 500 520 540 560 580 600 500 520 540 560 580 600 Rotating Speed of Driving Sprocket (rpm) Rotating Speed of Driving Sprocket (rpm) Figure 9 Behavior of Chain Drive with a Tensioner at Medium Operating Speeds (n2/n = 50/15, 12 = 0.0005 kg.m2)

5 x10-3 Solid: chain drive w/o tensioner ' 4- Doted:span# I of chain drive w/ tensioner -'3 3 Dashed: span#2 of chain drive w/ tensioner 2 2 o 10 4 I *..50 100 150 200 250 300 350 400 450 500 Rotating Speed of Driving Sprocket (rpm) (a) T, = 25 N xl0-3 4 - 50 100 150 200 250 300 350 400 450 500 3.2 2 50 100 150 200 250 '300 350 400 450 500 Rotating Speed of Driving Sprocket (rpm) (b) T, = 50 N x10-3 1 4 S, 13- 2. - _. 50 100 150 200 250 300 350 400 450 500 Rotating Speed of Driving Sprocket (rpm) (d) T, = 200 N Figure 10 Comparison between Chain Drives with and without a Tensioner -, / 50 100 150 200 250 300 350 400 450 500 Rotating Speed of Driving Sprocket (rpm) (d) T, = 200 N

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