 Hello everyone, I am Mr. Sachin Rathod working as assistant professor in mechanical engineering department from Wall Street Stop Technology, Sollapur. Today we are dealing with compound gear train. The learning outcome of this session is student will able to calculate the gear speed ratio of compound gear train. So first of all if you see the compound gear train, what is it called by compound gear train? Compound gear train in which on the intermediate shaft if more than one gear is mounted then it is called as a compound gear train. So this one is the input shaft and the output shaft. Input shaft is called as a driver and the output shaft is called as a driven. So in between these two shafts on which the two gears are mounted so it is called as an intermediate shaft. On the intermediate shaft the two gears are mounted so it is a case of your compound gear train. So what if suppose this is one gear, gear number one it is connected to another gear, gear number two. So on the gear number two one shaft is there, on the same shaft if another gear is mounted it is connected to another gear, this is gear number three, gear number four. So the gear number one is matching with the gear number two, this is the axis of the gear number one, it is connected to the gear number two, gear number three is connected to the gear number four. So this is your input shaft, gear number four is the output shaft. So this is a shaft for the intermediate. So on this one shaft two gears are mounted so it is a case of compound gear train. So in the case of the compound gear train we have to find out the speed ratio. Speed ratio, speed ratio is nothing but speed of driving gear. So input gear is gear number one divided by speed of the driven gear. So speed ratio is nothing but N1 by N4 so we have to find out this ratio. So gear number one is connected to the gear number two so we are getting the speed ratio is inversely proportional to the number of the teeth. So gear number one is connected to the gear number two so N1 by N2 is equal to T2 by T1, this is equation number one and gear number three is connected to the gear number four so we are getting N3 by N4 is equal to T4 by T3, this is equation number two. So just multiplying equation one and two we are getting N1 by N2 into N3 by N4 is equal to T2 by T1 into T4 by T3. As the gear number two and three are mounted on the same shaft so the speed of the gear number two and the gear number three are same so N2 and N3 will get cancelled therefore we are getting N1 by N4 is equal to T2 T4 divided by T1 T3. So this is nothing but the speed ratio. So speed ratio is equal to product of number of the teeth on the driven gear, this is the driven gear for the first gear, this is the driven gear for the third gear. So it is a product of number of the teeth on the driven gear divided by product of number of teeth on the driving gear. Speed ratio is equal to product of number of teeth on driven gear divided by product of number of teeth on the driving gear. So similarly we can observe for this compound gear train so we will see the mechanism gear number one is connected to the gear number two and the gear number two and three are mounted on the same shaft then gear number three is connected to the gear number four, four and five are mounted on the same shaft and the gear number five is connected to the gear number six. So we have to find out the speed ratio, speed ratio is equal to speed of the driving gear divided by the speed of the driven gear, this is the driving gear number one driven gear number six. So we are knowing the relation between N1 is connected to the N2 therefore, N1 by N2 is equal to T2 by T1, then gear number 3 is connected to the gear number 4, N3 by N4 is equal to T4 by T3, then gear number 5 is connected to the gear number 6, N5 by N6 is equal to T6 by T5, so this is gear number equation 1, 2, 3. So if you multiply equation 1, 2, 3 we are getting N1 by N2 into N3 by N4 into N5 by N6 is equal to T2 by T1 into T4 by T3 into T6 by T5, so here gear number 2, the speed of the gear number 2 and the speed of the gear number 3 will be the same because it is mounted on the same shaft, so it will get cancelled and the gear number 4 and 5 are mounted on the same shaft, so the speed will be the cancelled, but the number of the teeth are different, so we will keep as it is, so we are getting N1 by N6 is equal to T2, T4, T6 divided by T1, T3, T5, so this is nothing but speed ratio, okay the speed ratio is nothing but speed of the driving, driver divided by speed of the driven is equal to product of number of the teeth on the driven gear divided by product of number of teeth on the driving gear, like this way we can find out the speed ratio, so we can think about this, what is the application of the compound gear drain, generally in day to day life we are observing the compound gear drain in hand watches or the wall watches, we have taken one question or the problems regarding the compound gear drain, so we will see the question, now the gearing of machine tool shown in the figure, the motor shaft is connected to the gear number A, so this is the gear number 1, they are given and the rotates at 975 rpm, so they are given a speed of A is equal to 975 rpm, the gear wheels B, C, D and E are fixed to the parallel shaft rotating together, the final gear F is fixed on the output shaft, what is the speed of F, they are asked to find out the speed of F, the number of teeth on the each gears is given below, so these are the number of the teeth, so we have to calculate the speed of the F, so we will see the mechanism in this gear, gear number A is connected to the gear number B, B and C are the compound gear which is acting on the same shaft, so A is connected to the B then C is connected to the gear D, D and E are mounted on the same shaft, so what are the speed of the D, same speed we are getting at the E and the gear E is connected to the gear F, so already we have seen the speed ratio, speed ratio is equal to, the speed ratio is nothing but the speed of the driving gear to the speed of the driven gear and it is inversely proportional to the product of number of the teeth, so speed of the driving gear to the speed of the driven gear is equal to product of number of teeth on driven gear divided by product of number of teeth on driving gear, so here we are getting speed of the driving gear, driving gear gear is A and F, so we are getting N A by N F is equal to product of number of teeth on the driven gear, so here gear number B is a driven gear, D is a driven gear, F is a driven gear, so we are getting product of number of teeth, so T B into T D into T F divided by product of number of teeth on the driving gear, so E is a driving gear, C is a driving gear, E is a driving gear, so we are getting T A into T C into T E, so N A is the speed of the A they had given us, therefore speed of the A is 975 divided by N F is equal to product of number of the teeth 50 into 75 into 65 divided by 20 into 25 into 26, therefore N F is equal to 975 into 20 into 25 into 26 divided by 50 into 75 into 60, therefore we are getting the speed of N F output shaft is 52 rpm, so N F is equal to 52, so I have taken these references, thank you