 Hello everyone, I am Mr. Sachin Rathod working as assistant professor in mechanical engineering department from Wall Street Institute of Technology, Swalapur. Today we are dealing with a reverted gear train. So, in this session the learning outcome is at the end of this session student will be able to calculate the speed ratio of So, the simple diagram is there for the reverted gear train. So, in this diagram, so reverted gear train is nothing but if the input axis and output axis of the shaft on which the gears are mounted are coaxial then it is called as a reverted gear train. So, in this gear, gear number one is a driver, consider gear number one is a driver and gear number four is a driven gear. So, we are getting the motion from one gear is connected to the second gear and on the second gear and the third gears are mounted on the same shaft. So, it is a kind of the compound gear train. So, second gear and third gear are mounted on the same gear, same shaft. So, from the gear number two to shaft and gear number three are rotated with the same speed. So, from the three it is transmitted to the gear number four which is the driven gear. So, we can transmit the motion from first gear to the second, second to the third and third to the fourth. And here this the shaft for the driver or all them it is called as the input shaft is having the coaxial with the output shaft. So, it is called as a reverted gear train. So, this kind of the reverted gear train we have to find out the speed ratio. Already we are knowing what is in my speed ratio. The speed ratio is nothing but speed ratio. The speed ratio is nothing but speed ratio is nothing but it is the ratio of speed of driver divided by speed of driver. So, this is a we can call driver gear divided by speed of driven gear. So, it is called as a speed ratio. So, the speed of the driver gear divided by speed of the driven gear. It is nothing but the speed of the driver is nothing but the one and driven is nothing but the four. It is equal to the N1 by N4. So, it is also kind of the compound reverted gear train. So, we can find out the speed ratio in terms of the number of the teeth is nothing but number of teeth. We can calculate it is equal to product of number of teeth on driven gear divided by product of number of teeth on driver gear. So, if you observe this mechanism. This is the driver gear for this is driven gear and this is a driven gear. So, this is a driver gear for this driven gear. So, one and three are the driver gear for 2 and four driven gear. So, we can call product of number of teeth on the driven gear. So, 2 and four are the driven gears. here. So, we can call T2 into T4 divided by the number of teeth on the driver here. So, 1 and 3 T1 into T3. So, it is nothing but N1 by N4. So, we can formula for finding the speed ratio. And second thing is that, if you observe this mechanism, the distance between the two shafts, these two shafts, suppose we can call the distance between the two shafts is x. So, we are getting, so this is the radius, suppose the gear number 1 is having the p series this much. So, the radius we are getting this much. So, this we can call for the gear number 1, this is r1, gear number 2, it is r2 that is the pitch circle radius, gear number 3, it is r3, gear number 4, it is r4. So, x is the distance between the two shafts is equal to r1 plus r2 is equal to r3 plus r4. So, just we require these two formula. By using these two formula, we can easily find out the speed ratio or any geometry from this reverted gear train. So, you can think about what are the applications of reverted gear train. So, this is the question for you. So, generally if you observe the lathe machine at the back gear box of the lathe machine, we are using the reverted gear train. So, the application main application in case of the back gear of lathe machine. So, consider this numerical. So, we have to solve this numerical. So, in this question, the speed ratio of the reverted gear train as shown in the figure is 12. So, speed ratio is nothing, but we are knowing speed of the driver divided by speed of the driven. So, here the NA and ND is the driver by driven. So, we are getting NA by ND is equal to 12. The module of the pitch of the gear A and B is 3.125. Therefore, MA is equal to MB is equal to 3.125 and C and D that is MC is equal to MD is equal to 2.5. So, calculate the suitable number of the tith on the gears. So, you have to calculate the number of the tith that is the TA, TB, TC, TD that you have to calculate. So, in this question, so the speed ratio they are given us. So, we can simplify this. So, the speed ratio is nothing, but NA by ND and in case of the reverted gear train we are knowing that the speed ratio that is the speed of the driver by speed of the driven is calculated by the product of number of the tith on the driven gear divided by product of number of tith on the driver gear. So, we are getting TB into TD divided by TA into TC. So, by using this formula we can calculate the speed ratio. So, here they are given as a speed ratio is equal to 12. So, we can simplify this we are getting TB by TA is equal to square root of 12 and TD by TC is equal to square root of 12. Because in these two gears the transmitted power should be the same that is why we are getting TB by TA is equal to if we make the product of this we are getting the speed ratio is equal to 12. So, we can easily calculate TB is equal to square root of 12 and TD is equal to square root of 12 TC. So, we are getting the relation between A and B. So, this is the first equation and the second equation. Next expression we are knowing the distance between a two shaft that is a x that given as 200 mm. Therefore, x is equal to RA plus RB is equal to RC plus RD. Therefore, we are getting x is equal to 200, 200 is equal to RA. We are knowing the relation between R and M. Here the R is equal to MT by 2. So, therefore, we are getting MA TA by 2 plus MB TB by 2 is equal to MC TC by 2 plus MD TD by 2. So, we are getting MA TB by 2. So, we are getting TA plus MB TB is equal to 400 and MC TC plus MD TD is equal to 400. Just we have taken a common 1 by 2 it will get 400, 200 into 2 we are getting the 400. So, we are getting the 2 equation. So, here the MA MB are the same 3.125 therefore, we are getting TA plus TB is equal to 400 divided by 3.125 and TC plus TD is equal to 400 divided by here the MC and MD are the same 2.5 it is coming to 2.5. Therefore, TA plus TB is equal to 400 by this TC plus TD is equal to 400 by this. So, we have to solve this. This is the equation number 3 equation number 4. So, by comparing the equation number 1, 2, 3 and 4 we have to calculate the value of TA TB, TC TD. So, for making the calculation we are getting here TA plus TB is equal to. So, for the calculation purpose 400 divided by 3.125 we are getting 128, 128 and TC plus TD is equal to 400 divided by 3.125 and TD is equal to 400 divided by 2.5 we are getting 160. So, this is equation number 5 equation number 6. So, here we are getting TB is equal to square root of 12, square root of 12, 3.464 3.464 TA and TD is equal to 3.464 TA and TD is equal to 3.464 TA and TD is equal to 3.464 TC. Therefore, by putting the value TB is equal to 3.464 we are getting TA plus 3.464 TA is equal to 128. Therefore, TA is equal to 128 divided by 4.35. So, we are getting 28.67. Similarly, TC and TD if you put TD is equal to 3.4 where therefore, we are getting TC plus 3.464 TD is equal to 160. Therefore, TC is equal to 160 divided by 4.464 is equal to 35.85. Therefore, if you consider is equal to 28 TA is equal to and TC is equal to 36. Therefore, if you put this value in this equation number 5 we are getting TB is equal to 128 minus 28 is equal to 100 and TD is equal to 160 minus 36. So, these are our answers. So, we are getting TA is equal to 28, TB is equal to 100, TC is equal to 36 and TD is equal to 30. So, like this we have to solve the equations. So, here are the calculation for this TD and TB calculation. So, these are my references. Thank you.