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1、Gear trains reviewn1 =-n3=120 rpmnH=600 rpmDetermine nHDifferentialsDirection of rotation Gear trains Differentialadding machinez3z1DifferentialsA differential is a device that allows a difference in velocity between in two elementsGear trains differentials1zzzz1-nnnni2132434113）（a) Driving power fo
2、rm Z5 z1 and z3 rotate uniformlyb) If resistance is met in z1 z3 will be faster than z1314nn2nGear trains DifferentialsGear trains DifferentialGear trains DifferentialsGear trains DifferentialsConstruction differentialsGear trains powertainCase 1: A=1（离合器工作） ,B=0nH directly output=?n3=0Case2 : A=0,
3、B=1n30, 2132H3H1zzzznnnnn3 Arm54653634zzzznnnn input n4 =n1 is given2132H3H1zzzznnnnn4=n1Gear trains powertrainGear trains powertain3-speed automobile transmission1st: z5,z6 meshing z3,z4 separation A,B separation2nd: z3,z4 meshing z5,z6 separation A,B separation3th: A,B meshing z3,z4 separation z5,
4、z6 separation4thBack z6,z8 meshing A,B separation z3,z4 separation z5,z6 separation Power inoutputTransmission in automobileBetter choice that spur gears are used in the caseGear trains powertrainGear trains powertrain18Consider: Speed ratio in the first case and the second case ?Cam Design Gears in
5、 machine toolPower transmission system in machine toolCam Design Gears in machine toolSpeed ratio diagram of power transmission system in machine toolCam Design Gears in machine toolGear trains APPLICATIONSCam Design EfficiencyThe general definition of efficiency is=output power/input power Efficien
6、cy of a conventional gear train is very high. power loss per gearset 12% 1) the external gearset ； = 0.982) An external-internal gearset: = 0.993）two stages gearset =1*24) Loss estimation Efficiency of gear trainsFor epicyclic trains, the efficiency is dependent on design)(.z21z113f2f is the frictio
7、n coefficient5) Helical gearset =9698% , cross helical gearset =5090% Cam Design efficiency1;ETTETT01H012;Efficiency of epicycle gear1) Torques balance T1+T2+TH=02) Power in =power out from shaft1 to shaft2 （=+1） E0T1 (1 -H )+T2( 2 -H )=0 HLHFH2H1H12R1iObtain the torques ratio through (1) (2) 1E1TT
8、1;ETT;ETT02H01H012Cam Design efficiency-1i1i1H1HHH1H2H1Let gear 2 is fixed,2=0, the from gear1 to arm HThe efficiency is that 1H1H11HHiTTTTbecauseand1ETT01Htherefore11EiTT01H1HSee case 5Cam Design efficiency 1;ETT01Hb) from ARM to shaft1（=-1） If 2=0 the gear 2 is fixed， power from H to 1H1H1HH11iTTT
9、T11ii -i -1H1H1H1HH1H2H1Because000E1)E1E1(See the case 6Cam Design efficiencyDetermine the efficiency form gear2 to arm1E1TT 02HBecause 1=01-i1i1-2H2HH2HH2H11EiTT02H2Htherefore1=-1case7Cam Design efficiency=+1Efficiency from gear1 to arm=-1Cam Design efficiencyH1H1HH11iTTTT12HZ1=99, Zg=25, Zg=25, z2
10、=100E0 =0.98Determine the overall efficiency199100ZgZ1Z2ZgiH12/solutionH1=-1000E1)E1E1(11ii -i -1H1H1H1HH1H2H1329099100980199100980.)(.Cam Design Gear trainExample in p.308Consider the same Fergusons paradox train as the following figure, which has the following tooth numbers and initial conditionsS
11、un gear #2 N2=100Sun gear #3 N3=99Sun gear #4 N4=101Planet gear N5=20Input to sun gear #2 0 prmInput to arm 100 prm counterclockwiseCam Design Gear trainCam Design Gear train-. 2=0solutionH2H3H2H3H32i=3Cam Design Gear train4SOLUTIONTwo teeth difference, but output will big difference 4Cam Design HOM
12、E WORK1) Give z1=z3=17, z2=z4=39, z5=18, z7=152, n1=1450 rpmA is loosing, B is braking (switch on)Determine rotation speed nH2) Give n1=3549 rpm,Z1=36, z2=60, z3=23, z4=49Z4=69, z5=31, z6=131, z7=94, z8=36, z9=167Determine output speed nHCam Design HELICAL GEAR3) Write calculation velocity formula i
13、n the epicycle train in Figure 9-35(l-k) Gear trains home work3-speed automobile transmission1st: z5,z6 meshing z3,z4 separation A,B separation2nd: z3,z4 meshing z5,z6 separation A,B separation3th: A,B meshing z3,z4 separation z5,z6 separation4thBack z6,z8 meshing A,B separation z3,z4 separation z5,z6 separation Power inoutput Write all speed ratio (1st4th) in the transmission4)Cam Design HOME WORK5)Give a pair of helical gear parameters z1=20, z2=40, mn=8, n=20, =15oDetermine the distance between gear and pinion, the diameter of addendum circle.