外文翻譯---行星齒輪變速箱的設計研究

1、<p><b>　　附錄 A </b></p><p><b>　　英文資料</b></p><p>　　Research on Design of Planetary Gearbox</p><p>　　Nanjing Artillery College　　Rao Zhenggang</p>&l

2、t;p>　?。跘bstract］The design and calculations of planetary gearbox are detailedly discussed in this thesis.The calculations of structure parameters,transmission ratio,torque of each shaft,brake torque,locking moment,and

3、 transmission power are explained here.Hence,this paper is of much importance to the design of planetary gearbox of caterpillar vehicle,tank,artillery and engineering mechanics.Key words: Planetary gearbox　　Structure pa

4、rameter　　Torque　　Power</p><p>　　c)Clutch L locks internal gear rign b and swiveling jib H(see figure 3,c)　　When clutch L combines gear ring b and swiveling jib H into one whole,nb=nH.Then torque TA of input

5、 shaft A can transmit through two ways:one is to transmit to output shaft B through central gear a,planet gear g and swiveling jib H;the other is to B by central gear a,planet gear g,internal gear ring b and clutch L.At

6、this moment,TA=Ta,TB=-TA and locking moment TL is equal to the torque of internal gear ring Tb,TL=Tb.Acco</p><p>　　TL=Tb=pTa　　(4-11)</p><p>　　From the formula above,when central gear a inputs an

7、d swiveling jib H outputs,the locking moment TL of the first scheme is the lowest and the size is smaller.So as to 2K-H(A) planetary gearbox,the first scheme is more reasonable.　　(1)Suppose internal gear ring b inputs,s

8、wiveling jib H outputs(see figure 4),TA=Tb,TB=TH;Let us discuss the locking moment T1 of friction clutch of three different locking pattern.　　a)Clutch L locks intermal gear ring b and centra gear a(see figure 4,a),that

9、is na=nb</p><p><b>　　(4-12)</b></p><p><b>　　So,</b></p><p>　　b)Clutch L locks central gear a and swiveling jib H(see figure 4,b),that is nH=na and TL=Ta.So T

10、L is</p><p><b>　　So,</b></p><p><b>　　(4-13)</b></p><p>　　c)Clutch L locks internal gear ring b and swiveling jib H(see figure 4,c),that is nH=nb and TL=Tb.Si

11、nce the clutch L links input shaft A and output shaft B,locking moment TL is equal to the torque transmitted by shaft A,namely TL=Tb=TA.So the locking moment TL is</p><p><b>　　TL=TA</b></p>

12、<p>　　From the formula above,when internal gear ring inputs and swiveling jib outputs,locking moment TL of the first scheme is the lowest.So it is the best to adopt the first locking scheme.　　After TL is calculat

13、ed,the necessary control units can be adopted to obtain the friction clutch with smaller size.But,in fact,the best locking scheme can't be always realized due to difficulty instructrual distribution.Under this situat

14、ion,the locking scheme should be adopted to get more compact structure of clu</p><p><b>　　(5-1)</b></p><p>　　Where,TA、nA—torque and rotational speed of input shaft A；TB、nB—torque an

15、d rotational speed of output shaft B.Ratio nA/nB=ip is kinematics transmission ratio;ratio-TB/TA=ip～is dynamics transmission ratio.Since TB has contains the friction loss of planetary gearbox,putting ip and ip～into form

16、ula(5-1),we can get the famous M.A.Kpe-Йюec formula:</p><p><b>　　(5-2)</b></p><p>　　So the efficiency of each step of multi-degree planetary gearbox ηp is equal to the ratio of its d

17、ynamics transmission ratio ip～and kinematics transmission ratio ip.　　As it is said above,its kinematics transmission ratio ip is function of characteristic parameter p of planet bar,that is:</p><p>　　ip=f(p

18、1、p2、…pn)　　(5-3)</p><p>　　According to the relation of power equation,p is equal to ip multiplying (ηH)x then the function of p is obtained:</p><p><b>　　(5-4)</b></p><p>

19、;　　Where,ηH—transmission efficiency of planet bar in relative motion.As to 2K-H(A) planet bar,ηH=ηHag×ηHgb=0.97×0.99=0.96　　The equation of index xj is</p><p><b>　　(5-5)</b></p>

20、<p>　　j=1,2,…，n　　The mathematical meaning of sign is symbol.　　Index xj=1,its symbol is decided by (5-5)；if the value of sign is more than 0,sign is “+”，and xj=+1;otherwise,xj=-1。6　Sample of the design of planetar

21、y gearbox　　Suppose the transmission sketch of planetary gearbox of one certain caterpillar truck is shown as figure 5.Its degerrs of freedom is w=3,the number of steps is nd=4(three forwarding steps and one reverse step

22、).Try to calculate (1) number of control units m,number of planet bars </p><p>　　Put nd=4 into the formula above,</p><p><b>　　m2-m-8=0</b></p><p>　　It can be obtained fr

23、om the equation that:m≈3.4；Because C2m＞m＞W(wǎng)，the number of control units is:：m=W+1=4　　Then the number of friction clutches L can be got according to formula(2-8)</p><p>　　L=W-1=3-1=2</p><p>　　The

24、 number of brakes Z can be calculated throu-gh(2-9)</p><p>　　Z=m-L=4-2=2</p><p>　　The number of planet bars k is obtained by(2-12)</p><p>　　k=nd-(W-1)=4-(3-1)=2</p><p>

25、　　Finally,the number of basic motive components n0 can be calculated through formula(1-3).</p><p>　　n0=W+k=3+2=5</p><p>　　It is shown on figure 5 that its basic components are A、B、1、2 and 3.　　(

26、2)Transmission ratio ip of each step　　The module m of the gear in two planet bars of the said planetary gearbox is the same and m=4mm.The characteristic parameters are p1=p2=3.105.Because p1=Zb1/Za1 and p2=Zb2/Za2and no

27、rmally the number of teeth of central gear a Za＞17.we suppose Za1=Za2=19.Then the number of teeth of internal gear ring b is:</p><p>　　Zb1=p1za1=3.10519=59Zb2=p2za2=3.10519=59</p><p>　　The nu

28、mber of teeth of planet gear Zg is</p><p>　　The same as above,，Zg2=20　　Since the planetary gearbox has three forward steps and one reverse step,each step can be reached by combining two control components.T

29、he method to combine the control units is shown in table 6-1.　　The transmission ratios of each step of the planetary gearbox is calculated as followes:　　The transmission ratio of the first step(direct step):　　The firs

30、t step of planetary gearbox is obtained by locking two friction clutches L1 and L2.They combine central gear a1,a2 in</p><p>　　i=iAB=iH1b2=1</p><p>　　The transmission ratio of the second step:　

31、　The second step of planetary gearbox is obtained by locking brake Z2 and Clutch L2.The twoplanet bars participate in work.After input from input shaft A,the transmission power P distributed through swiveling jib in two

32、ways:one is through gear g1,a1,a2 and clutch L2 to output shaft B;the other is through gear g1,b1,swiveling jib H2,gear g2,a2 and clutch L2 to output shaft B.Now it is known that brake Z2 is locked,that is nb2=0.　　Accor

33、ding to the kine</p><p>　　Then through the connecting relations among units,it can be got that：na1=na2和nb1=nb2.From the equation above,the transmission ratio of the second step is:</p><p>　　The

34、transmission ratio of the third step:　　The third step of the planetary gearbox is obtained by locking brake Z1 and clutch L2.At this moment there is only one planet bar working.Transmission power P is input from shaft A

35、,then passes swiveling jib H,gear g1,a1 and a2,clutch L2 and reaches output shaft B.Now it is known that brake Z1 is locked,that is nb1=0.Then through the connecting relation,it can be got that na1=na2.Also through the k

36、inematics equation of planet x1(3-1):</p><p>　　na1-(1+p1)nH1=0</p><p>　　And the transmission ratio of the third step i3 is:</p><p>　　The transmission ratio of the reverse step：　　Th

37、e reverse step of planetary gearbox is obtained by locking brake Z1 and clutch L1.There are two planet bars participating in work.Transmission power P is input form shaft A,passes through swiveling jib H1,gear g1,a1 and

38、a2,g2,b2,clutch L1and reaches output shaft B.Now it is known that brake Z1 is locked,that is nb1=nH2=0，and planet bar x2 becomes semi-planet gear transmission(fixed-shaft transmission).So the transmission ratio of the re

39、verse step i</p><p>　　Where,Then i-1 is:</p><p>　　(3)Calcuations of touque T,locking moment TL and brake moment TZ of each unit　　Now suppose the torque of input shaft TA of the plantary gearbo

40、x is known,that is TA can be obtined through formula(4-1).Since the torque T,locking moment TL and brake moment TZ are different at each step,they should be calculated spearately according to their force bearing situatio

41、n.　　At the first step,i1=1(direct step).At this moment,clutch L1 and L2 lock the central gear a2,internal gear b2 and output shaft B.Tra</p><p>　　TL2=Ta2=TB=-TA</p><p>　　Accordign to formula(4-

42、3),Tb2=p2Ta2So,the locking moment TL1 is:</p><p>　　TL1=p2Ta2=-p2TA=-3.105TA</p><p>　　Since the brakes Z1 and Z2 are not locked,TZ1=TZ2=0?！　t the second step,i2=0.4278,At this time,brake Z2 lo

43、cks internal gear b2 and clutch L2 locks central gear a2 and output shaft B.Since TH1=TA,TL2=TB=-i2TA,locking moment TL2 is</p><p>　　TL2=TB=-i2TA=-0.4278TA</p><p>　　Since Ta2=Ta1，according to fo

44、rmulas(4—2) and (4—3)：</p><p>　　And since brake moment TZ2=Tb2，</p><p>　　TZ2=Tb2=-0.7564TA</p><p>　　At the third step,the transmission ratio i3=0.2463.Brake Z1 locks internal gear b

45、1 while clutch L2 locks central gear a2 and output shaft B.So TH1=TA，TL2=Ta2=Ta1=TB。According to formula(4-2)：</p><p>　　TA,So the locking moment is：</p><p>　　Since TZ1=Tb1，according to formula(4

46、-2) and (4-3)：　　,and the brake moment is:</p><p>　　At reverse step:i-1=-0.7564.At this time,brake Z1 locks internal gear b1 while clutch L1 locks internal gear b2 and output shaft B.So TH1=TA，TL1=Tb2 and TZ

47、1=Tb1。The same as above:</p><p>　　So the locking moment is</p><p>　　And the brake moment is</p><p><b>　　That is</b></p><p>　　(4)Calculation of transmission

48、efficiency of each step　　According to the formula(5-2),the transmission efficiency can be calculated.That is the claculation equation is</p><p>　　The transmission efficiency of the first step(direct step)　

49、　Since there isn't relative motions among the units,the kinematics transmission ratio is</p><p><b>　　ip=i1=1</b></p><p>　　The dynamics transmission ratio is</p><p>　

50、　So the transmission efficiency of the first step is(the friction loss is ignored)：η1=1　　The transmission efficiency of the second step：At the second step,the kinematics transmission ratio is</p><p>　　And i

51、ts dynamics transmission ratio is：</p><p><b>　　(a)</b></p><p>　　The symbol of the coefficient is decided as:</p><p>　　Put ηH and p1、p2 into equation (a),then</p>

52、<p>　　So the transmission efficiency of the second step can be obtained through formula(5-2)：</p><p>　　The transmission efficiency of the third step：　　At the third step,the kinematics transmission rati

53、o is</p><p>　　And its dynamics transmission ratio is：</p><p>　　The symbol of x1 is deceded as：</p><p>　　Then the kinematics transmission ratio is:</p><p>　　So the trans

54、mission efficiency of the third step is：</p><p>　　The transmission efficiency of the reverse step：　　At the reverse step the kinematics transmission ratio is</p><p>　　And its dynamics transmissi

55、on ratio is：</p><p><b>　　(c)</b></p><p>　　The symbol of x1 and x2 is decided as：</p><p>　　＝Sign[-0.7564]=-1</p><p>　　Then the kinematics transmission ratio

56、 is:</p><p>　　So the transmission efficiency of the reverse step is:</p><p>　　After the calculations of the structural parameters,transmission ratio,moment and transmission efficiency of each st

57、ep，we can draw several concluusions:　　(a)The characteristic parameters of two planet bars x1 and X2 are p1=p2=3.105,which are within the reasonable range p=1.6～5.　　(b)Since the input unit of this planetary gearbox is s

58、wveling jib H1，the transmission ratio of each step is less than 1(excluding the driect step).It means the second,third and the reverse steps of the gearbox all have</p><p>　　附錄 B </p>

59、<p><b>　　中文翻譯</b></p><p>　　行星齒輪變速箱的設計研究</p><p><b>　　饒振綱</b></p><p>　　［摘要］本文較詳細地討論了行星齒輪變速箱的設計計算。文中闡述了行星齒輪變速箱的結(jié)構(gòu)參數(shù)計算，各檔的傳動比計算，各構(gòu)件的轉(zhuǎn)矩、制動轉(zhuǎn)矩和閉鎖轉(zhuǎn)矩計算，以及各檔的傳動效率

60、計算。同時，還附有具體的設計計算示例。因此，本文對于履帶車輛、坦克、自行火炮和工程機械等的行星齒輪變速箱的設計計算均具有較重要的指導意義。關鍵詞：行星齒輪變速箱　　結(jié)構(gòu)參數(shù)　　轉(zhuǎn)矩　　功率</p><p>　　c) 離合器L將內(nèi)齒圈b與轉(zhuǎn)臂H閉鎖(見圖3，c)　　當離合器L使內(nèi)齒圈b與轉(zhuǎn)臂H結(jié)合成一體時，即有nb=nH。此時，輸入軸A的轉(zhuǎn)矩TA可從兩條路徑傳遞：一路是經(jīng)過中心輪a、行星輪g和轉(zhuǎn)臂H傳到輸出軸B

61、；另一路是經(jīng)過中心輪a、行星輪g、內(nèi)齒圈b和離合器L傳到輸出軸B。此時，轉(zhuǎn)矩TA=Ta,TB=-TA。而離合器的閉鎖力矩TL等于內(nèi)齒圈b的轉(zhuǎn)矩Tb,即TL=Tb。據(jù)公式(4-3)可得閉鎖力矩TL的計算公式為</p><p>　　TL=Tb=pTa　　(4-11)</p><p>　　由上式可見，當中心輪a輸入，轉(zhuǎn)臂H輸出時，第一閉鎖方案(圖3，a)的閉鎖力矩TL為最小，且使其具有較小的結(jié)構(gòu)

62、尺寸。所以，對于2K-H(A)型行星排選取第一閉鎖方案較合理。　　(2)若內(nèi)齒圈b輸入，轉(zhuǎn)臂H輸出(見圖4)，即TA=Tb、TB=TH；討論在下列三種不同的2閉鎖方式下，其摩擦離合器的閉鎖力矩TL的計算公式?！　) 離合器L將內(nèi)齒圈b與中心輪a閉鎖(見圖4，a)，即有na=nb=nH;故該行星排變成為一個整體旋轉(zhuǎn)，則可得：輸出軸轉(zhuǎn)矩TB=-TA。而閉鎖力矩TL=Ta，據(jù)公式(4-2)可得TL的計算公式為：</p>&

63、lt;p><b>　　所以</b></p><p><b>　　(4-12)</b></p><p>　　b) 離合器L將中心輪a與轉(zhuǎn)臂H閉鎖(見圖4，b),即有nH=na和閉鎖力矩TL=Ta。仿上，按公式(4-3)可得閉鎖力矩TL的計算公式為</p><p><b>　　所以，</b><

64、/p><p><b>　　(4-13)</b></p><p>　　c) 離合器L將內(nèi)齒圈b與轉(zhuǎn)臂H閉鎖(見圖4，c),即有nH=nb和TL=Tb。由于離合器L將輸入軸A和輸出軸B相連接，所以，閉鎖力矩TL應等于輸入軸A所傳遞的轉(zhuǎn)矩TA,即有TL=Tb=TA。則得得閉鎖力矩TL的計算公式為</p><p>　　TL=TA　　(4-14)</p

65、><p>　　由上式可見，當內(nèi)齒圈b輸入，轉(zhuǎn)臂H輸出時，第一閉鎖方案的閉鎖力矩TL為最小值，故應選取第一閉鎖方案最為有利?！　榱双@得較小尺寸的摩擦離合器，利用以上各式求得所承受的閉鎖力矩TL后，便能夠合理地選取所需的控制元件。但是，實際上由于結(jié)構(gòu)布置上的困難，欲使最有利的行星排構(gòu)件閉鎖，并不是都能實現(xiàn)的。在此情況下，則應當從離合器的結(jié)構(gòu)緊湊和行星齒輪變速箱外形尺寸較小的觀點出發(fā)選取最合適的閉鎖方案。5　行星齒輪

66、變速箱的效率計算　　對于多級行星齒輪變速箱，其輸出功率PB與輸入功率PA的比值，則稱為該行星變速箱的傳動效率；即得：</p><p><b>　　(5-1)</b></p><p>　　式中，TA、nA—輸入軸A的轉(zhuǎn)矩和轉(zhuǎn)速；　　　TB、nB—輸出軸B的轉(zhuǎn)矩和轉(zhuǎn)速。其中，比值nA/nB=ip為其運動學傳動比；比值-TB/TA=p為其動力學傳動比。因為轉(zhuǎn)矩TB考慮

67、了行星齒輪變速箱的摩擦損失。將ip和p代入式(5-1)，則得到著名的克列依涅斯(M.A.Kpe-Йюec)公式：</p><p><b>　　(5-2)</b></p><p>　　由此可知，多級行星齒輪變速箱各檔的傳動效率ηp等于其動力學傳動比與其運動學傳動比ip之比值?！　∪缜八?，其運動學傳動比ip都是行星排特性參數(shù)p的函數(shù)，即</p><

68、p>　　ip=f(p1、p2、…pn)　　(5-3)</p><p>　　根據(jù)相對運動的功率方程式的關系，行星變速箱的動力學傳動比p應等于運動學傳動比ip關系式中的每個特性參數(shù)p值上乘以(ηH)X，則可得其動力學傳動比p的函數(shù)式為：</p><p><b>　　(5-4)</b></p><p>　　式中，ηH—在相對運動中，行星排的傳動

69、效率，對于2K-H(A)型行星排，且有：ηH=ηHag×ηHgb=0.97×0.99=0.96　　指數(shù)Xj的計算公式為：</p><p><b>　　(5-5)</b></p><p><b>　　j=1,2,…，n</b></p><p>　　sign的數(shù)學含義是表示符號。　　指數(shù)Xj=1,其符號

70、取決于公式(5-5)；若sign的值大于零，則“sign”為正號“+”，則得Xj=+1;否則，即得Xj=-1。6　行星齒輪變速箱的設計計算示例　　已知某軍用履帶車輛傳動裝置的行星齒輪變速箱的傳動簡圖如圖5所示。其自由度數(shù)W=3，檔位數(shù)nd=4(三個前進檔和一個倒檔)。試計算：(1)該行星變速箱所需的控制元件數(shù)m、行星排數(shù)k和基本構(gòu)件數(shù)n0；(2)各個檔位的傳動比ip；(3)各構(gòu)件的轉(zhuǎn)矩T、制動力矩TZ和閉鎖力矩TL；(4)各個檔位

71、的傳動效率ηp。</p><p>　　圖5　履帶車輛傳動裝置</p><p>　　Fig.5　The caterpilar vehicle transmission</p><p>　　解：(1) 控制元件數(shù)m、行星排數(shù)k和基本構(gòu)件數(shù)n0的計算?！　‘斪杂啥葦?shù)w=3時，其控制元件數(shù)m可按公式(2-6,a)計算，即</p><p>　　因nd

72、=4代入上式，則得：</p><p><b>　　m2-m-8=0</b></p><p>　　由上述方程可求解得：m≈3.4；又因C2m＞m＞W(wǎng)，則可得控制元件數(shù)為：m=W+1=4　　按式(2-8)可求得其摩擦離合器數(shù)L為</p><p>　　L=W-1=3-1=2</p><p>　　再按式(2-9)可得其制動器數(shù)

73、Z為</p><p>　　Z=m-L=4-2=2</p><p>　　按公式(2-12)可得其行星排數(shù)k為</p><p>　　k=nd-(W-1)=4-(3-1)=2</p><p>　　最后，可按其結(jié)構(gòu)公式(1-3)求得其運動基本構(gòu)件數(shù)n0為</p><p>　　n0=W+k=3+2=5</p>&l

74、t;p>　　由圖5中可見，其運動基本構(gòu)件為A、B、1、2和3共五個基本構(gòu)件?！　?2)各檔位的傳動比ip計算　　上述行星齒輪變速箱的兩個行星排中的各齒輪的模數(shù)m相同，且知，m=4mm，兩個行星排的特性參數(shù)為p1=p2=3.105。因p1=Zb1/Za1和p2=Zb2/Za2；一般，中心輪a的齒數(shù)Za＞17。現(xiàn)選?。篫a1=Za2=19,則得內(nèi)齒圈b的齒數(shù)為：</p><p>　　Zb1=p1za1=3.

75、105×19=59Zb2=p2za2=3.105×19=59</p><p>　　其行星輪的齒數(shù)Zg為</p><p>　　仿上，Zg2=20　　由于該行星齒輪變速箱具有三個前進檔和一個倒檔，每結(jié)合兩個控制元件便可獲得一個檔位。其各檔位結(jié)合控制元件的組合方法如表6-1所示。</p><p>　　表6-1　某行星變速箱各檔位結(jié)合控制元件的組合方

76、法Table 6-1　Combining method of control wnitsof one planetary gearbox</p><p>　　現(xiàn)計算上述行星齒輪變速器各檔的傳動比如下：　　一檔(直接檔)的傳動比：　　該行星齒輪變速箱的一檔是將兩個摩擦離合器L1、L2都閉鎖起來而得到的。它們將中心輪a1、a2和內(nèi)齒圈b2，以及輸出軸B結(jié)合成為一體，從而使得兩個行星排X1、X2連成為一個整體旋轉(zhuǎn)

77、；其各構(gòu)件之間均沒有相對運動，即轉(zhuǎn)速nA=nH1=nb1=na1=na2=nH2=nb2=nB。則得其傳動比為：</p><p>　　i=iAB=iH1b2=1</p><p>　　二檔的傳動比：　　該行星齒輪變速箱的二檔是將制動器Z2制動和離合器L2閉鎖而得到的；其兩個行星排均參與工作。傳遞功率P從輸入軸A輸入后，經(jīng)轉(zhuǎn)臂H1通過兩路分流：一路經(jīng)齒輪g1、a1和a2，離合器L2傳到輸出軸

78、B；另一路經(jīng)齒輪g1、b1，轉(zhuǎn)臂H2，齒輪g2、a2和離合器L2傳到輸出軸B。現(xiàn)已知：制動器Z2被制動，即轉(zhuǎn)速nb2=0?！　?jù)行星排的運動學方程式(3-1)可求得二檔的傳動比i2值。即得：</p><p>　　再由構(gòu)件之間的連接關系可求得：na1=na2和nb1=nH2。由上式經(jīng)整理后可求得其二檔的傳動比為：</p><p>　　三檔的傳動比：　　該行星齒輪變速箱的三檔是將特制動器Z

79、1制動和離合器L2閉鎖而得到的；此時只有第一個行星排X1參與工作。傳遞功率P從輸入軸A，經(jīng)轉(zhuǎn)臂H1、齒輪g1、a1和a2，離合器L2傳到輸出軸B?，F(xiàn)已知：制動器Z1被制動，即轉(zhuǎn)速nb1=0。再由連接關系得：na1=na2。據(jù)行星排x1的運動學方程式(3-1)得：</p><p>　　na1-(1+p1)nH1=0</p><p>　　則可求得其三檔的傳動比i3為：</p>&

80、lt;p>　　倒檔的傳動比：　　該行星齒輪變速箱的倒檔是將制動器Z1制動和離合器L1閉鎖而得到的；其兩個行星排均參與工作。傳遞功率P從輸入軸A，經(jīng)轉(zhuǎn)臂H1，齒輪g1、齒輪a1和a2、g2、b2，再經(jīng)離合器L1傳到輸出軸B?，F(xiàn)已知：制動器Z1被制動，即轉(zhuǎn)速nb1=nH2=0，行星排x2變成為準行星齒輪傳動(定軸齒輪傳動)。因此，其倒檔的傳動比i-1應按下列關系式求得：</p><p>　　式中，則可得其倒

81、檔的傳動比i-1為：</p><p>　　(3) 各構(gòu)件的轉(zhuǎn)矩T、閉鎖力矩TL和制動力矩TZ的計算　　現(xiàn)假設該行星齒輪變速箱輸入軸的轉(zhuǎn)矩TA為已知，即TA可由電動機的額定功率P和轉(zhuǎn)速n1按公式(4-1)求得。由于各構(gòu)件的轉(zhuǎn)矩T、閉鎖力矩TL和制動力矩TZ在各個檔位都是不一樣的。所以，它們應按各個檔位的受力情況分別計算。　　在第一檔位時，其傳動比i1=1(直接檔)，此時離合器L1和L2將中心輪a2、內(nèi)齒輪b2與

82、輸出軸B閉鎖。輸入功率P從輸入軸A，經(jīng)離合器L1和L2傳到輸出軸B。因為，在此考慮到TB=-TA，故離合器的閉鎖力矩TL1和TL2應從輸出軸B方面來確定。即有閉鎖力矩TL1=Tb2,TL2=Ta2。因轉(zhuǎn)矩TB=Ta2,所以，閉鎖力矩TL2為：</p><p>　　TL2=Ta2=TB=-TA</p><p>　　按公式(4-3)可得轉(zhuǎn)矩Tb2=p2Ta2所以，閉鎖力矩TL1為</p

83、><p>　　TL1=p2Ta2=-p2TA=-3.105TA</p><p>　　因制動器Z1和Z2均未制動，故其制動力矩為TZ1=TZ2=0。　　在第二檔位時，其傳動比i2=0.4278。此時，制動器Z2將內(nèi)齒輪b2制動，離合器L2將中心輪a2與輸出軸B閉鎖。因為，TH1=TA，TL2=TB=-i2TA。所以，閉鎖力矩TL2為</p><p>　　TL2=TB=-

84、i2TA=-0.4278TA</p><p>　　因Ta2-Ta1，按公式(4—2)和(4—3)可得：</p><p>　　因制動力矩TZ2=Tb2，所以，</p><p>　　TZ2=Tb2=-0.7564TA</p><p>　　第三檔位時，其傳動比i3=0.2436。此時，制動器Z1將內(nèi)齒輪b1制動，離合器L2將中心輪a2與輸出軸B閉鎖

85、。因此可得：TH1=TA，TL2=Ta2=Ta1=TB。由公式(4—2)得：。所以，其閉鎖力矩為：</p><p>　　因制動力矩TZ1=Tb1，由公式(4-2)和(4-3)可得：,所以，其制動力矩為：</p><p>　　在倒檔傳動時，其傳動比i-1=-0.7564。此時，制動器Z1將內(nèi)齒圈b1制動，離合器L1將內(nèi)齒輪b2與輸出軸B閉鎖。因此可得：TH1=TA，TL1=Tb2和TZ1=T

86、b1。仿上，可得：</p><p>　　所以，其閉鎖力矩為：</p><p>　　仿上，可得其制動力矩為</p><p><b>　　即得：</b></p><p>　　(4) 各檔位的傳動效率計算　　按照前蘇聯(lián)學者M.A.克列依涅斯提出的公式(5-2)可計算各檔的傳動效率。即傳動效率的計算公式為：</p>

87、<p>　　一檔(直接檔)的傳動效率：</p><p>　　在一檔時，由于各構(gòu)件間均沒有相對運動，故其運動學傳動比為：</p><p><b>　　ip=i1=1</b></p><p><b>　　其動力學傳動比為：</b></p><p>　　所以，其一檔的傳動效率(忽略軸承的摩擦

88、功率損失)為：η1=1　　二檔的傳動效率：在二檔時，其運動學傳動比為：</p><p><b>　　其動力學傳動比為：</b></p><p><b>　　(a)</b></p><p>　　先確定系數(shù)x1、x2的符號：</p><p>　　將已知的ηH和p1、p2值代入(a)式可得其動力學傳動比

89、為：</p><p>　　所以，由公式(5-2)可得二檔的傳動效率為：</p><p>　　三檔的傳動效率：　　在三檔時，其運動學傳動比為：</p><p><b>　　其動力學傳動比為：</b></p><p><b>　　(b)</b></p><p>　　確定系數(shù)x1

90、的符號：</p><p>　　據(jù)公式(b)可得其動力傳動比為：</p><p>　　所以，由公式(10)可得三檔的傳動效率為：</p><p>　　倒檔的傳動效率：　　在倒檔時，其運動學傳動比為：</p><p><b>　　其動力學傳動比為：</b></p><p><b>　　(c

91、)</b></p><p>　　先確定系數(shù)x1、x2的符號：</p><p>　　據(jù)公式(c)式可得其動力學傳動比為：</p><p>　　所以，由公式(5-2)可得倒檔的傳動效率為：</p><p>　　通過對上述行星齒輪變速箱的結(jié)構(gòu)參數(shù)、各檔傳動比、力矩和各檔傳動效率的計算，經(jīng)研究分析后可得到如下幾點結(jié)論：　　(a) 該行星

92、齒輪變速箱兩個行星排X1、X2的特性參數(shù)p1=p2=3.105,此值是在行星排的特性參數(shù)的合理范圍p=1.6～5之內(nèi)?！　?b) 由于該行星齒輪變速箱的輸入件為轉(zhuǎn)臂H1，故其各檔的傳動比(除直接檔的傳動比i1=1外)均小于1；即表示該行星變速箱在二檔、三檔和倒檔時均起著增速器的作用?！　?c) 該行星齒輪變速箱各檔的傳動效率都較高，經(jīng)比較可知：η1＞η2＞η3＞η-1；且知η2=0.9736與η3=0.9696較接近，倒檔的傳動效率