AGMA 95FTM10-1995 Efficiency of High Contact Ratio Planetary Gear Trains《高重合度行星齿轮系的效率》.pdf

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1、 STD-ALMA 75FTMLO-ENGL 1995 Ob87575 0004757 354 H 95FTMlO Efficiency of High Contact Ratio Planetary Gear Trains by: John Colbourne, University of Alberta American TECHNICAL PAPER COPYRIGHT American Gear Manufacturers Association, Inc.Licensed by Information Handling ServicesSTD-AGMA 75FTMLO-ENGL 19

2、95 = Ob87575 00047b0 07b D EMlciency of High Contact Ratio Planetary Gear Trains John Colbourne, University of Alberta The statements and opinions contained herein are those of the auhor and should not be construed as an official , action or opinion of the American Gear Manufacturers Associti0n.1 Ab

3、stract A method is described for calculating the efficiency of pianetary gear Uains, and as an example the method is applied to the case of a fxed differential gear min. Copyright O 1995 American Gear Manufacturers Association 1500 King Street, Suite 201 Alexandria, Virginia, 22314 October, 1995 ISB

4、N: 1-555894594 COPYRIGHT American Gear Manufacturers Association, Inc.Licensed by Information Handling ServicesEFFICIENCY OF HIGH CONTACT RATIO PLANETARY GEAR TRAINS John R. Colbourne Department of Mechanical Engineering University of Alberta Edmonton, Alberta, Canada T6G 2G8 Abstract A method is de

5、scribed for calculating the efficiency of planetary gear trains, and as an example the method is applied to the case of a fixed differential gear train. Introduction Efficiency calculations for planetary gear trains fall into two parts. First, we find the tooth meshing efficiency of gear pairs with

6、fixed centers. Next, these efficiencies are used in the calculation of the overall efficiency of the planetary gear train. In this paper, a small improvement is suggested in the commonly used method for calculating the tooth meshing efficiency. An alternative method is described for gear pairs with

7、high contact ratios, which will be discussed later in the paper. Then a method is presented for calculating the efficiency of planetary gear trains, based on the dynamic equilibrium of the planet, and .taking into account the frictional tooth force components. Coefficient of Friction It is generally

8、 agreed that the effect of friction in meshing gears can be approximated by using a constant coefficient of friction. The value of the coefficient depends on the oil viscosity, the load intensity and the pitch line velocity. Expressions are given in AGMA 6023-A88 l for calculating the coefficient of

9、 friction, and these give values that range from 0.02 at low load intensity and high pitch line velocity, to 0.06 at high load intensity and low pitch line velocity. In Chapter 12 of Dudleys Gear Handbook 2, Shipley presents a table of coefficient of friction values measured by Ohlendori and Winter

10、3. The values range from 0.03 to 0.07, and Shipley uses the value 0.05 for his examples. Efficiency of Gear Pairs with Fixed Centers The method described by Shipley for calculating the efficiency of a gear pair with fixed centers is the same method used earlier by Buckingham 4 and Tuplin 5. The path

11、 of contact shown in Figure 1 extends from point T2 to T1. We use a coordinate s measured along the contact path from the pitch point, so that points above and below the line of centers have positive and negative coordinate values, and the length of the path of contact is then (sT-sT2). If we assume

12、 a constant tooth force W, the useful work done by one meshing tooth pair is equal to the force multiplied by the distance through which it acts, Work = W(sT1 -sT2) (1) The sliding velocity v, 6 is given by “, = Sb1 - 4 (2) 1 COPYRIGHT American Gear Manufacturers Association, Inc.Licensed by Informa

13、tion Handling Servicesthat the angular velocities are related by N, 01 N2% (8) and the loss factor is then given by 1 = p(l-)- Nl 1 (sT2)2 + (9) N2 Rbl (ST - ST2) Assumption of Constant Tooth Force Figure 1. The Path of Contact. We now examine the effect of the assumption made earlier, that the toot

14、h force remains constant throughout the mesh. The number of meshing tooth pairs passing the pitch point each second is equal to N,01/(2rc). We multiply this number by the work done per tooth pair, given by Equation (l), to obtain the power transmitted, where ai and 02 are the angular velocities, bot

15、h defined as positive when they are counterclockwise. This equation is valid for both external and internal gear pairs, but in an external gear pair the angular velocities have opposite signs, and are related by the expression, Power = w (sT1- sT2) N, w1 / (2 n) (10) Ni 01 + Nau2 = O (3) We replace

16、(s Ti- s T2 ) by mcpb, the contact ratio times the The velocity ds/dt 6) at which the contact point base pitch, express pb in terms of the base circle radius R, and then replace WRb, by WtRDi, where Wt is the moves along the contact path is given by transmitted force and Rp, is the pitch circle radi

17、us. Power = WtmcRp, al (11) so the sliding velocity for an external gear pair becomes This equation is clearly incorrect, since it is known that the (5) vs = S(l+-)- Ni 1 ds N2 Rbi dt power is given by Power = W,Rp, o, (12) The power loss is equal to the friction force 1-W muitiplied by the absolute

18、 value of the sliding velocity, and this is integrated to give the energy lost during the meshing cycle of one tooth pair, and it is therefore necessary to discard the assumption made earlier. Nl 1 Lost Energy = pW (1 + -) - (s) + (sT1)T N2 2Rbl Finally, the lost energy is divided by the useful work

19、, to give the loss factor h, which is equal to (1-q), where q is the tooth meshing efficiency. P Figure 2. The Lowest and Highest points of Single-Tooth Contact. The derivation is the same for an internal gear pair, except 2 COPYRIGHT American Gear Manufacturers Association, Inc.Licensed by Informat

20、ion Handling Services STDeAGMA 75FTMLO-ENGL 1995 Ob87575 00047b3 885 We consider only gear pairs in which the contact ratio is between 1.0 and 2.0. Figure 2 shows the path of contact, and points L1 and H1 represent the lowest and highest points of single-tooth contact on gear 1. If W is the total to

21、oth force, equal to the applied torque on gear 1 divided by the base circle radius, then we will assume that the tooth force is W/2 between T2 and Ll , and between Hl and T1. This assumption is certainly closer to reality than the earlier assumption. The s coordinates of points L1 and H1 can be read

22、 from the diagram, T1 sL1 = s - pb (13) and the useful work done by one meshing tooth pair is as follows, This result is consistent with the expression for the transmitted power, given by Equation (1 2). To calculate the efficiency, we need to find the energy lost due to friction, which is equal to

23、the friction force times the sliding distance. We start by calculating the profile length of any involute from its starting point B on the base circle to a typical point A, as shown in Figure 3. The Figure 3. Involute Formed by a Point on a Bar Rolling on a Fixed Cylinder. involute can be represente

24、d as the locus of point A of a rigid bar, which rolls without slipping on a cylinder of radius Rb. If the profile length from B to A is h, then as the bar rolls the increase dh is given by dh = Rbe de (16) where E is the roll angle, .e. the angle between lines BC and CE shown in Figure 3. This equat

25、ion is integrated to give an expression for the profile length h, h = (R2 - RE)/(2Rb) (17) Referring to Figure 2, when the contact point is at a typical point s on the path of contact, the profile lengths h, and h, on gears 1 and 2 are given by h, = (Rbl tang + (2 Rbl) (18) The difference in the pro

26、file lengths between gears 1 and 2, as the contact point moves from T2 to L1 can be found from Equations (18 and 19), 1 Nl T22 Ah21 = - (1 +-) (S ) - (SL1)2 (20) 2Rbl N2 and we multiply by (pW/2) to obtain the energy lost to friction during this part of the meshing cycle. We repeat the process as th

27、e contact point moves from L1 to P, P to H1, and H1 to T1. The total energy lost during the meshing cycle is then given by Energy Lost = We divide by the useful work done to obtain the loss factor. The corresponding expression for an internal gear pair is found by replacing the plus sign in the firs

28、t bracket by a minus sign. 3 COPYRIGHT American Gear Manufacturers Association, Inc.Licensed by Information Handling Services STD-AGHA 95FTMLO-ENGL 1995 Ob87575 00047b4 711 = The same result could have been found by the method used earlier, where the friction force was multiplied by the sliding velo

29、city and integrated. The assumption of constant tooth force causes errors in the calculations of both the work done and the energy lost, and to some extent these errors cancel when we calculate the loss factor. Values given by Equation (22) for the loss factor are generally 10% to 15% lower than tho

30、se given by Equation (i). The reason for the difference is that near the ends of the contact path the sliding velocity is greatest, but the tooth force is halved. If we treat the tooth force as constant throughout the meshing cycle, we exaggerate the loss near the ends of the contact path. In Tupiin

31、s derivation 5, he stated that the complexity in assuming a variable tooth force is not justified, in view of the uncertainty about the value of the coefficient of friction. At the time he was writing, access to computers was very limited, and the simplicity of Equation (i) was an advantage. Today,

32、however, there is no difficulty in evaluating the expression in Equation (22). Efficiency of Planetary Gear Trains To calculate the efficiency of a planetary gear train, Shipley first finds the loss factor at each meshing gear pair, and then calculates the power loss at each mesh. We consider first

33、a sun gear S, a planet P and a planet carrier U. The energy lost as each tooth pair passes through mesh is given, as before, by Equation (21). The number of tooth pairs passing through the pitch point per second is now equal to the absolute value of Nc(%- mu)/(2). The power lost is equal to the ener

34、gy lost per tooth pair mutiplied by the number of tooth pairs per second, and the result is simplified by using Equation (22) to express the energy lost in terms of the loss factor, n T- Figure 4. Fixed Differential Gear Train. Power Lost = W, pb NS I(% - wu) I (2 a) (23) The base pitch is written i

35、n terms of the sun gear bast circle radius Fis, and the product WR, is replaced as before by WtRps, Power Lost = h W, RPs I (oc - wu) 1 (24) where W, is the transmitted force between the sun gear and the planet, and Re is the pitch circle radius of the sun gear. The efficiency of the gear train is f

36、ound by subtracting the power loss at each mesh from the input power, and dividing by the input power. The quantity WtReI(ws-wu)l in Equation (24) is referred to by Shipley as the equivalent power at the sun-planet mesh. It is difficult, however, to give a physical interpretation to this quantity, s

37、ince it may be larger than the input power, even in gear trains in which there is no circulating power. Example. Fixed Differential Gear Train. Figure 4 shows a fixed differential gear train with input eccentric U, compound planet P, fixed internal gear R, and output internal gear V. The center dist

38、ance (.e. the eccentricity) is C, the input torque Tu, and the toot1 numbers are N, N,.N, and N,. Figure 5 shows the planet pitch circle radii Rp12 and Q4. and the directions of the transmitted forces T, and TB, acting on the planet, for the case when Rp,2 is greater than R*,. The relative angular v

39、elocity (wp-w,) is equal to -N20U/N,. We find the transmitted forces by using the conditions of equilibrium for the force system in Figure 5, and we use Equation (24) to write down the power losses at each mesh, and hence the efficiency. Figure 5. Transmitted Tooth Forces Acting on the Planet. 4 COP

40、YRIGHT American Gear Manufacturers Association, Inc.Licensed by Information Handling ServicesSTD-AGMA 75FTMLO-ENGL 1775 Ob87575 000Li7b5 b58 Y12 = - Tu N3 (25) c (N1 - N3) Y34 = - Tu N1 (26) c (N, - N3) Power Input = Tuau (27) Since the two gear pairs have the same center distance C, the tooth numbe

41、rs would generally be chosen so that (N2-N,) is equal to (N4-N3), recutting in a simplification of Equation (30). Equilibrium Conditions There are two objections to the theory just described. First, the calculated efficiency of a gear train is the same, whether the gear train is used as a reducer or

42、 as a speed increaser. This is contrary to experience, since it is known that gear trains are less efficient when used as speed increasers, and are sometimes self-locking. Secondly, the conditions of equilibrium are not satisfied, when the friction forces are taken into account. Wt pl W W Figure 6.

43、Transmitted Force and Modified Transmitted Force. Figure 6a shows an external gear pair with fixed centers, similar to the one shown in Figure 1. Gear 1 is in dynamic equilibrium, so that the input torque T, is equal to RplW,. The transmitted force Wt also acts on gear 2, though in the opposite dire

44、ction, and the output torque T2 is therefore equal to R,W,. Since the pitch circle radii are in the ratio N,:N2, the output torque is equal to N2T,/N1. This result is only correct in the case when there is no friction. In the presence of friction, the output torque is equal to qN2T,/N,. To provide t

45、he correct ratio between input and output torques, the transmitted force W must act at a point Q, as shown in Figure 6b, which divides the line of centers in the ratio N, qN2. The distances from Q to the gear centers, which we will call the modified pitch circle radii Rp, and RIp2, are then given by

46、 Figure 7 shows the planet of the fixed differential gear train discussed earlier. The efficiency of the mesh between gears 1 and 2 is q12, and between gears 3 and 4 it is q34. Expressions for the modified pitch circle radii are shown on the diagram, and it is assumed that the input to the gear trai

47、n is the eccentric U. The planet is in dynamic equilibrium under the forces shown. We take moments about Q12 to find the transmitted force Wt34 in terms of the input torque Tu. We then multiply Wt34 by the modified pitch citcle radius of gear 4 to obtain the output torque i,. Wt12 Ql, Figure 7. Plan

48、et of the Reducer. Modified Transmitted Forces Acting on the 5 COPYRIGHT American Gear Manufacturers Association, Inc.Licensed by Information Handling ServicesIf there were no friction, the output torque would be given by Equation (33) when we set qlz and q34 equal to 1. The gear train efficiency is

49、 equal to the output torque with friction, divided by the output torque with no friction, The entire calculation can be repeated for the case when the input is through V. The planet and the modified pitch circle radii are shown in Figure 8, and the efficiency for this case is given by It can be seen from the expressions for the modified pitch circle radii shown in Figure 8 that, as the tooth meshing efficiencies q12 and q34 decrease, the radius gets smaller and R, gets larger. If the efficiencies are reduced sufficiently, and R