3K(Ⅱ)型行星齿轮传动设计
3K()型行星齿轮传动设计,3K()型行星齿轮传动设计,行星,齿轮,传动,设计
南京理工大学紫金学院毕业设计(论文)任务书系:机械工程系专 业:机械工程及自动化学 生 姓 名:学 号:设计(论文)题目:3K行星齿轮传动设计起 迄 日 期:2010年2月24日 2010年6月5 日设计(论文)地点:南京理工大学紫金学院指 导 教 师:专业负责人:发任务书日期: 2010 年2 月 23日任务书填写要求1毕业设计(论文)任务书由指导教师根据各课题的具体情况填写,经学生所在专业的负责人审查、系领导签字后生效。此任务书应在毕业设计(论文)开始前一周内填好并发给学生;2任务书内容必须用黑墨水笔工整书写或按教务处统一设计的电子文档标准格式(可从教务处网页上下载)打印,不得随便涂改或潦草书写,禁止打印在其它纸上后剪贴;3任务书内填写的内容,必须和学生毕业设计(论文)完成的情况相一致,若有变更,应当经过所在专业及系主管领导审批后方可重新填写;4任务书内有关“系”、“专业”等名称的填写,应写中文全称,不能写数字代码。学生的“学号”要写全号;5任务书内“主要参考文献”的填写,应按照国标GB 77142005文后参考文献著录规则的要求书写,不能有随意性;6有关年月日等日期的填写,应当按照国标GB/T 74082005数据元和交换格式、信息交换、日期和时间表示法规定的要求,一律用阿拉伯数字书写。如“2008年3月15日”或“2008-03-15”。毕 业 设 计(论 文)任 务 书1本毕业设计(论文)课题应达到的目的:1 熟悉机械传动系统的一般要求及设计准则2 熟悉传动机构的结构原理及总体设计思想3 熟练掌握机械传动系统的设计计算及参数选择4 掌握工程设计环节,完成总装配图及主要零部件图样设计2本毕业设计(论文)课题任务的内容和要求(包括原始数据、技术要求、工作要求等):1 设计指标:传动比:i2535功率:0.15kw输入转速:1450r/min效率:782 工作要求:了解国内外传动机构的发展现状,并进行创新性设计。毕 业 设 计(论 文)任 务 书3对本毕业设计(论文)课题成果的要求包括毕业设计论文、图表、实物样品等: 1完成总装配图设计2完成主要零件图3毕业论文说明书一份4主要参考文献:1 王华坤,范元勋. 机械设计基础(下)M. 北京:兵器工业出版社,2001.2 杨黎明. 机械设计M. 北京:兵器工业出版社,1998.3 饶振纲. 行星传动机械设计M. 北京:化学工业出版社,2003.4 孔恒,陈作模. 机械原理M. 北京:高等教育出版社,2001.5 徐灏. 机械设计手册(第3版)M. 北京:机械工业出版社,1991.6 卜炎. 机械传动装置设计手册(上、下)M. 北京:机械工业出版社,1999.7 唐嘉平. AutoCAD2006实用教程(第2版)M. 北京:清华大学出版社,2006.8 姚家娣,李明,黄兴元. 机械设计指导M. 北京:化学工业出版社,2003.9 饶振纲. 微型行星齿轮传动的设计研究J. 传动设计,2003,17(2):18-24.10 刘李梅. 行星齿轮减速器的设计和应用D. 无锡职业技术学院学报,2005.11 关岳编译. 微型机器超小型行星减速器Z. 世界发明,1993.12 饶振纲. 行星传动机构设计(第2版)M. 北京:国防工业出版社,1994.13 徐锡林微机械及其研究J中国机械工程,1993,(2):10-12.14 李占权, 李百宁, 战晓红. 行星齿轮减速器的设计J. 煤矿机械,2000,(11):12-13.15 饶振纲. 微型行星减速器的研究J. 机械制造与自动化, 1999,(2):10-15.毕 业 设 计(论 文)任 务 书5本毕业设计(论文)课题工作进度计划:起 迄 日 期工 作 内 容2010年1月20日3月19日3月20日4月30日5月1日5月15日5月16日5月20日5月21日5月25日6月1日6月5日查阅资料,完成外文翻译和开题报告基本总体方案设计、设计计算及论文分析完成总装配图、零件图设计完成论文初稿完成设计说明书的编写,并提出申请优秀论文优秀论文答辩论文答辩所在专业审查意见:负责人: 年 月 日系意见:系领导: 年 月 日南 京 理 工 大 学 紫 金 学 院毕业设计(论文)前期工作材料学生姓名:学 号:系:机械工程系专 业:机械工程及自动化设计(论文)题目:3K行星齿轮传动设计指导教师:讲师 (姓 名) (专业技术职务)材 料 目 录序号名 称数量备 注1毕业设计(论文)选题、审题表12毕业设计(论文)任务书13毕业设计(论文)开题报告含文献综述14毕业设计(论文)外文资料翻译含原文15毕业设计(论文)中期检查表12010 年 3 月注:毕业设计(论文)中期检查工作结束后,请将该封面与目录中各材料合订成册,并统一存放在学生“毕业设计(论文)资料袋”中(打印件一律用A4纸型)。 MMK TRITA-MMK 2005:01 ISSN 1400-1179 ISRN/KTH/MMK/R-05/01-SE Relations between size and gear ratio in spur and planetary gear trains by Fredrik Roos planetary gears are commonly known to be compact and to have low inertia. Keywords Spur Gears, Planetary Gears, Gearhead, Servo Drive, Optimization Contents 1 INTRODUCTION / BACKGROUND. 5 2 EQUIVALENT LOAD. 6 3 SPUR GEAR ANALYSIS. 8 3.1 GEOMETRY, MASS AND INERTIA OF SPUR GEARS. 8 3.1.1 Geometrical relationships . 8 3.1.2 Gear pair mass . 9 3.1.3 Inertia . 9 3.2 NECESSARY GEAR SIZE . 10 3.2.1 Hertzian pressure on the teeth flanks . 10 3.2.2 Bending stress in the teeth roots. 12 3.2.3 Maximum allowed stress and pressure. 13 3.3 RESULTS AND SIZING EXAMPLES. 14 3.3.1 Necessary Size/Volume. 14 3.3.2 Gear pair mass, geometry and inertia. 15 4 ANALYSIS OF THREE-WHEEL PLANETARY GEAR TRAINS . 19 4.1 GEAR RATIO, RING RADIUS, MASS, INERTIA AND PERIPHERAL FORCE. 19 4.1.1 Gear ratio and geometry . 19 4.1.2 Weight/Mass . 20 4.1.3 Inertia . 20 4.1.4 Forces and torques . 22 4.2 PLANETARY GEAR SIZING MODELS BASED ON SS1863 AND SS1871. 23 4.2.1 Sun planet gear pair . 23 4.2.2 Planet and ring gear pair . 24 4.2.3 Maximum allowed stress and pressure. 26 4.3 RESULTS AND SIZING EXAMPLES. 27 4.3.1 Necessary size/volumes . 27 4.3.2 Weight, radius and inertia. 28 5 COMPARISON BETWEEN PLANETARY AND SPUR GEAR TRAINS . 32 6 CONCLUSIONS. 34 7 REFERENCES . 35 Relations between size and gear ratio in spur and planetary gear trains 4(35) Nomenclature Pressure angle rad Helix angel ( = 0 for spur gears) Transverse contact ratio Load angle rad Poissons number Mass density kg/m 3 F Root bending stress Pa Fmax Maximum allowed bending stress Pa H Hertzian flank pressure Pa Hmax Maximum allowed hertzian flank pressure Pa Angular velocity rad/s b Gear width m b c Carrier width m d Gear reference diameter m d a Gear tip diameter m d b Gear base diameter m E Module of elasticity Pa F Force N J Mass moment of inertia kgm 2 K F Factor describing the division of load between teeth K F Load distribution factor for bending K H Factor describing the division of load between teeth K H Load distribution factor for Hertzian pressure k ro Relation between outer and reference diameter of ring gear m Module m M Mass kg n Gear ratio of a complete gear train ( in / out ) p b Base pitch r Gear reference radius m S Safety factor T Torque Nm u Gear ratio of a single gear pair v Peripheral velocity m/s Y F Form factor for bending Y Helix angle factor for bending Y Contact ratio factor for bending z Number of teeth Z H Form factor for Hertzian pressure Z M Material factor for Hertzian pressure Z Contact ratio factor for Hertzian pressure Index 1 The small wheel (pinion) of a gear pair Index 2 The large wheel (gear wheel) of a gear pair Index r The ring gear in a planetary gear train Index s The sun gear in a planetary gear train Index p The planet gears in a planetary gear train Index c The planet carrier in a planetary gear train Relations between size and gear ratio in spur and planetary gear trains 5(35) 1 Introduction / Background This work was initiated within a research project about design and optimization methods for mechatronic systems. The goal with that research project is to derive methods for optimization of mechatronic actuation modules, with respect to weight, size and/or efficiency (Roos the heat generated in the motors winding is given by the RMS value of the motor current. Since the current is proportional to the motor torque, the RMS torque may be used for motor dimensioning. 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 1000 500 0 500 1000 time s Load profile, required torque and position Ang. Velocity rad/s Angle rad 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 40 20 0 20 40 60 RMS RMC Max time s torque Nm Figure 2. Example of an inertial load cycle, with RMS, RMC and max norms shown. Gear design is traditionally focused on strength of the gears. The load on the gear teeth is cyclic and therefore gear failure is most often a result of mechanical fatigue. The two classical limiting factors in gear design are surface fatigue and tooth root bending fatigue. The combination of cyclic loading of the gear teeth when in mesh and an applied load that varies with time makes it more difficult to find an expression for the equivalent load, than in the motor case. The exponent used in the torque norms used for gear sizing is not 2 as in the RMS norm, but ranges from 3 to 50 (Anthony 2003). These expressions for equivalent load are based on the so-called linear cumulative damage rule (the Palmgren-Minor rule). It assumes that the total life of a mechanical product can be estimated by adding up the percentage of life consumed by each stress cycle. The number of stress cycles on each tooth in a gear train can be huge during a lifetime. Anthony (2003) exemplifies this with a three-wheel planetary gear in which one sun gear tooth will be exposed to almost 3 million load cycles during an 8-hour period at 2000 rpm. Relations between size and gear ratio in spur and planetary gear trains 7(35) Area of unlimited load cycles Stres s l oad (l og S) Load Cycles (log N) E = 3 for bearings E = 8.7 for case hardening steel E = 17 for nitrided steel 10 6 10 9 10 3 10 12 E = 84 for carbo nitrided steel Figure 3. Whler curves for different steels. The exponent to use in the calculation of equivalent load depends on material type, heat treatments, and loading type (Antony 2003). It is however not obvious that the Palmgren-Minor rule can be used for infinite life design ( 10 6 load cycles), especially not in an application where the teeth will be subjected to the peak load more than 10 6 times. In fact only in applications where the total number of load cycles is below 210 6 is a higher load than the endurance limit load permissible (Antony 2003). This means that for infinite life dimensioning, the gears should be dimensioned with respect to the peak torque in the load cycle. Of course there are exceptions to this, for example load cycles where the peak load occurs while the gears are standing still. The calculated equivalent continuous torque, T cal is hence, for unlimited life design given by: max )(tTT cal = (1) This is the approach taken in this report; it is assumed that the teeth are subjected to the peak load more than 10 6 times, and therefore is the peak torque used for dimensioning. This research area is however very complex and are not investigated further in this report. By this approach, equation (1), is at least not a too low equivalent torque used. The sizing procedure gets even more complicated when the bearings are considered. For bearings, the Root Mean Cube (RMC) value of the load is often used as the equivalent continuous load (comp. Figure 3). This report will however only treat the actual dimensioning of the gears, not the bearings. But it should be noted that it may be the bearings that limit the maximum gear load. Relations between size and gear ratio in spur and planetary gear trains 8(35) 3 Spur gear analysis The analysis made here is mainly based on the formulas presented in the Swedish standard for calculation of load capacity of spur and helical gears, SS1871 and standard SS1863 for the spur gear geometry. Figure 4 shows a spur gear, to simplify the analysis only spur gears with no addendum modification are treated. Figure 4. Spur gear. 3.1 Geometry, mass and inertia of spur gears. 3.1.1 Geometrical relationships In order to simplify the rest of the analysis, it is useful to derive some simple geometrical relations. The gear ratio u is defined as: 2 1 1 2 1 2 = z z r r r r u in out (2) The center distance, a between the wheels is given by: 1221 rarrra =+= (3) Combining equations (2) and (3) gives: 1 1 r ra u = (4) Pinion Gear Wheel T in T out a r 2 r 1 Relations between size and gear ratio in spur and planetary gear trains 9(35) Finally, combining equations (3) and (4) gives the expressions for r 1 and r 2 1 1 + = u a r (5) 11 22 + = + = u au r u a ra (6) 3.1.2 Gear pair mass A gear wheel is here modeled as a cylinder, an approximation that is quite accurate. The mass, M of one gear is hence given by: 2 brM = (7) Where b is the face width, r is the reference radius and is the mass density of the wheel. The total mass of a gear pair can be expressed as: )( 2 2 2 121 rrbMMM tot +=+= (8) Finally by combining equation (2), (5) and (8) the following expression for the gear pair mass is obtained: 2 2 222 1 )1( 1 )1( u u baubrM tot + + =+= (9) 3.1.3 Inertia The inertia, J of a rotating cylinder is given by: 2 2 cyl cylcyl r MJ = (10) The inertia reflected on the pinion shaft (axis 1) of a gear pair is hence given by: 2 2 22 2 2 12 1 2 2 2 2 2 1 1 2 2 1 2 2 2 2 u r br r br u r M r M u J JJ tot +=+=+= (11) Figure 5. Gear mesh p F p F Relations between size and gear ratio in spur and planetary gear trains 10(35) Which, if combined with equations (5) and (6) result in the following expression of the gear pair inertia: 4 2 4 4 24 4 4 )1( 1 2 )1()1( 2 u u ba u ua u ab J tot + + = + + + = (12) 3.2 Necessary gear size According to SS 1871, the necessary gear size is determined from the teeth flank Hertzian pressure and the teeth root stress. Neglecting losses such as friction the peripheral force, see figure 5, acting on a gear tooth is given by: au uT r T F out out out p )1( + = (13) For a given load the necessary gear size is determined as function of gear ratio and number of pinion teeth. Depending on material properties, gear ratio, number of teeth, etc., either the flank pressure or the root stress sets the limit on the gear size, in general both stress levels must be checked. 3.2.1 Hertzian pressure on the teeth flanks The Hertzian pressure on a teeth flank is given by (SS1871): ubd uKKF ZZZ HHcal MHH 1 )1( + = (14) For gears with no addendum modification the form factor Z H is given by: t b H Z 2sin cos2 = (15) As shown below, the transverse section pressure angle t is, for spur gears, the same as the normal section pressure angle, n . The pressure angle will therefore only be noted as from now on. = nt n t 0 cos tan tan (16) Since the helix angle is zero on the pitch cylinder () it will be zero on the base cylinder too ( b ) : 1cos0 cos coscos cos = bb (17) This leads to the following expression for Z H : 2sin 2 = H Z (18) The material factor Z M is given by (SS1857): + = 2 2 2 1 2 1 11 2 EE Z M (19) Relations between size and gear ratio in spur and planetary gear trains 11(35) Where E is the module of elasticity of the respective gear and is poissons number. For spur gears the contact factor, Z is according to SS 1871 given by: 3 4 =Z (20) Where is the contact ratio. For an external spur gear pair it is, according to SS1863, given by: + = ww baba b a dddd p sin 22 1 2 2 2 2 2 1 2 1 (21) Where a w is the center distance between the wheels, for gears with no addendum modification, aa w = . The base pitch, p b is given by: cosmp b = (22) Where m is the module, which is defined as: 12 2 1 1 )1( 2 zu a z d z d m + = (23) The tip and base diameters, d a and d b, are, for external spur gears, given by (SS1863): mdd a 2+= cosdd b = ) 44 (sin) 44 cos1( cos44cos)2( 2 22 2 2222 222222222 z z d z z ddd z d mddmmddmddd ba ba +=+= =+=+= (24) From equation (5) and (6) the following expressions for the gears diameters can be derived: 1 2 , 1 2 21 + = + = u au d u a d (25) By combining equations (24) and (25), and inserting them into equation (21), the following expression for the contact ratio is obtained: + + + + + = sin2) 44 (sin )1( 4 ) 44 (sin )1( 4 2 1 cos2 )1( 22 1 1 2 2 22 2 1 1 2 2 2 1 a uz uz u ua z z u a a zu += sin)1( 44 sin 44 sin cos2 22 1 1 2 2 1 1 21 u uz uz u z z z (26) Inserting equation (13) and (25) into equation (14) results in: 22 3 2222 2 )1( uba uT KKZZZ cal HHMHH + = This equation can be rewritten as: 2 max 2 3 2222 2 )1( H cal HHMH u uT KKZZZba + = (27) Relations between size and gear ratio in spur and planetary gear trains 12(35) Where Z H , Z M , Z , are given by equation (18), (19) and (20). K H and K H are factors describing the division of load between teeth and the load distribution on each tooth respectively. Generally K H can be set to 1. K H is more complicated since it only can be 1 in theory (if the gears are perfect). Here, for simplicity, it is set to 1.3, but if more exact data is available it should be used instead, see SS1871 for more information and guidelines about how to select this constant. Equation (27) gives the minimum size of the gear pair (with respect to Hertzian pressure) given a material, Hmax , E 1 , E 2 , 1 , 1 , a gear ratio, u, the number of pinion teeth, z 1 , the pressure angle and the calculated torque T cal . 3.2.2 Bending stress in the teeth roots The bending stress in a tooth, F can be calculated according to SS1871 as follows: bm KKF YYY FFcal FF = (28) The calculation of the form factor, Y F is somewhat complicated the way it is done in SS1871, therefore Y F is approximated with the following expression (Maskinelement handbok 2003): 14/ 1.32.2 z F eY + (29) Y F will always be larger for the small wheel (pinion) since it decreases with z. The helix angle factor Y is 1 for spur gears. Y is the so called contact ratio factor, and it is according to SS1871 calculated as follows: 1 =Y (30) Where the contact ratio, is calculated as before with equation (26). By combining equation (13), (23) and (28) the following is obtained: bua uzT KKYY cal FFFF 2 2 1 2 )1( + = (31) The expression above can be rewritten as follows: max 2 12 2 )1( F cal FFF u uzT KKYYba + = (32) Where Y F and Y is given by equation (29) and (30). K F and K F are factors describing the load division between teeth and the load distribution on each tooth respectively. If no other data is available K F can be set to 1, and K F to the same value as K H (SS1871). Equation (32) can be used to calculate the minimum size of a gear pair, with respect to bending endurance. Relations between size and gear ratio in spur and planetary gear trains 13(35) 3.2.3 Maximum allowed stress and pressure The maximum allowed bending stress Fmax and the maximum allowed hertzian pressure on the teeth, Hmax is of course largely dependent on material choice and safety factor. There are a lot of factors that can be included into the calculation of the stress limits, including the number of load cycles. Here the number of load cycles is assumed to be larger than the endurance limit and therefore that factor is disregarded. The maximum allowed stress and pressure is here simply calculated as follows: H H H S lim max = F F F S lim max = Where Hlim and Flim are material properties and S H and S F are the safety factors. For more advanced calculations see SS1871. Values of Flim and Hlim can for example be retrieved from Maskinelement Handbok (2003) or from the standards. It should be noted that if the safety factor for bending is doubled, the necessary gear size is doubled. But if the safety factor for the flank stress is doubled, it requires a four time larger gear pair, see equations (27) and (32) respectively. Relations between size and gear ratio in spur and planetary gear trains 14(35) 3.3 Results and sizing examples 3.3.1 Necessary Size/Volume In this section, equations (27) and (32) are applied on a (equivalent) load of 20 Nm. First, material data from an induction hardened steel with a Hertzian fatigue limit of 1200 MPa and a bending fatigue limit of 300 MPa is used. The Pressure angle is 20 degrees, and all other constants are set to the standard value (Table 1). Material Properties Gear Properties E 206 GPa 20 deg. 0.3 K H 1 7800 kg/m 3 K H 1.3 Load K F 1 T cal 20 Nm K F 1.3 Table 1. Values of material and gear parameters used in all examples. 0 5 10 15 10 15 20 25 30 35 40 0 0.5 1 1.5 2 2.5 3 3.5 4 x 10 5 Gear ratio Gear pair volumes (a 2 b). Hmax : 1200 MPa. Fmax : 300 Mpa T cal = 20 Nm No of teeth, wheel 1 (z 1 ) a 2 b m 3 0 5 10 15 10 15 20 25 30 35 40 0 0.2 0.4 0.6 0.8 1 1.2 x 10 4 Gear ratio Gear pair volumes (a 2 b). Hmax : 1200 MPa. Fmax : 300 Mpa T cal = 20 Nm No of teeth, wheel 1 (z 1 ) a 2 b m 3 root stress F1 root stress F2 flank stress H Figure 6. Gear pair volume as function of gear ratio and number of teeth. The safety factors SH and SF is 1 in the graph to the left and 2 in the graph to the right. The following plots are obtained for a non-hardened steel with a Hertzian fatigue limit of 500 MPa and a bending fatigue limit of 200 MPa (e.g. SIS1550). The load and all other parameters are the same as in the example above. 0 5 10 15 10 15 20 25 30 35 40 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 x 10 4 Gear ratio Gear pair volumes (a 2 b). Hmax : 500 MPa. Fmax : 200 Mpa T cal = 20 Nm No of teeth, wheel 1 (z 1 ) a 2 b m 3 0 5 10 15 10 15 20 25 30 35 40 0 1 2 3 4 5 6 7 8 x 10 4 Gear ratio Gear pair volumes (a 2 b). Hmax : 500 MPa. Fmax : 200 Mpa T cal = 20 Nm No of teeth, wheel 1 (z 1 ) a 2 b m 3 root stress F1 root stress F2 flank stress H Figure 7. The same plots as in Figure 6, but for another steel. The safety factors are set to one in the plot to the left and 2 in the plot to the right. Relations between size and gear ratio in spur and planetary gear trains 15(35) From the plots above and a number of other examples not presented here, a couple of conclusions can be made. First of all, the Hertzian pressure is the limiting factor in the vast number of cases (it requires the largest gears). The root bending stress is the limiting factor only for steels with a large difference between the two stress limits, and only if the safety factor for Hertzian pressure is low. Of course it is possible to change a lot of constants to get a different result, but these are the results if the standard (SS1871) is applied with the constants set to the recommended values. Furthermore the surface representing the root stress of the pinion (wheel 1) is always higher than corresponding surface for the gear wheel (wheel 2). This is consistent with the previous conclusion that the root stress always is larger for the smaller wheel. Therefore the root stress is only calculated for the pinion in the rest of this report. The number of teeth has most influence on the root bending stress. Not surprisingly, the flank stress is in comparison almost independent of the number of teeth. By choosing a relatively small number of pinion teeth (wheel 1), the tooth flank stress will almost certainly be the limiting factor. 3.3.2 Gear pair mass, geometry and inertia By combining equation (9) with the results shown in the left part of Figure 6 (safety factors = 1, induction hardened steel) the following graph (Figure 8) of the gear pair mass is obtained: 0 5 10 15 15 20 25 30 35 40 0 0.2 0.4 0.6 0.8 1 Gear ratio Gearpair mass as function of number of teeth and gear ratio, T = 20 No of teeth z 1 Mass kg Figure 8. Gear pair weight as function of number of teeth of the pinion and gear ratio. To continue the analysis it will be necessary to lock one of the variables in Figure 8, at least if the results are to be visualized in 3D-plots. The number of teeth of wheel 1 (pinion) seems to be the most reasonable variable to lock. In the figure below (Figure 9) plots of two different pinion teeth numbers are shown, 28 and 17. The larger choice results in, depending on the gear ratio, that booth the flank and root stress are limiting. The choice of 17 teeth of the small wheel results in, as also seen earlier, that only the flank stress is limiting the volume of the gear pair, regardless of gear ratio. Relations between size and gear ratio in spur and planetary gear trains 16(35) 0 5 10 15 0.5 1 1.5 2 2.5 3 3.5 x 10 5 Cross section when z1 = const = 28, T = 20 Nm a 2 b m 3 Gear Ratio Flank Stress Root Stress limiting (max) 0 5 10 15 0.5 1 1.5 2 2.5 3 x 10 5 Cross section when z1 = const = 17, T = 20 Nm a 2 b m 3 Gear Ratio Flank Stress Root Stress limiting (max) Figure 9. Cross sections of the left part of Figure 6, at 28 and 17 teeth of wheel one. To eliminate unrealistic gear geometry t
收藏