 I am Sachin Rathod working as an assistant professor in mechanical engineering department from Walsh Institute of Technology, Sollapur. Today we are dealing with simple gear train. So the learning outcome of this session is student will be able to calculate speed ratio of simple gear train. So simple gear train, generally the gear train are used for reducing the gear ratio or the speed ratio. So in that simple gear train if on one shaft only one gear is mounted then we can transmit the speed or the power from one shaft to another shaft by using the different gears mounted on the shaft. So in this case the shaft are parallel to each other. The gear having the input shaft and this one is the output shaft. So on this input shaft and output shaft the input shaft is called as a driver and the output shaft is called as a driven or follower. So here we are giving the input speed and we are getting the output speed. So here we are occurring the speed ratio, we have to find out the speed ratio. So the speed ratio is nothing but the speed of driver divided by speed of driven. So in this case this is gear number one, gear number two. Gear number one having the tith t1, gear number two is having the tith t2, gear number one is having the speed of n1, gear number two is having the speed of n2. So we can find out the speed ratio is equal to speed of driving n1 by speed of driven n2. So in this case we can easily find out the speed ratio. So suppose we have taken two gears, gear number one and this is a gear number two. These are the PCD, pitch circle diameter. So from the gear number one to the gear number two we can transmit the power or the motion. Suppose the gear one is rotated in the clockwise direction, gear number two will rotate in the anticlockwise direction. So in this case this PCD, so if the two gears are mating with each other the module of the two gears should be the same. So the module m is given by the ratio of PCD divided by the number of the tith. So gear number one and gear number two having the same model. So we are getting d1 by t1 is equal to d2 by t2. So we are getting the relation d1 by d2 is equal to t1 by t. So we are having another relation the PCD of this gear that the two gears are mating with each other at that time the pitch line velocity should be the same. So we are getting pi d1 n1 is equal to pi d2 n2. Therefore we are getting d1 by d2 is equal to n2 by n1. So we are getting the relation from equation number one and two we are getting n1 by n2 is equal to t2 by t1 which is the equation number three. So we are getting the one relation between n and t that is n1 upon n2 is equal to t2 by t1 that is the ratio of the driving gear to the driven gear is equal to the number of tith on the driven gear to the number of tith on the driving gear. So the speed is inversely proportional to the number of the tith. So by using this relation we can easily find out the speed ratio. So in this first diagram A here the driver and driven so we can find out first one speed ratio is equal to speed of the driving gear that is n1 upon n2 is equal to now it is a inversely proportional to the number of the tith that is the t2 by t1. So if you observe the figure number b speed ratio is equal to speed of the driver this is a driven gear and this is called as the intermediate gear. Gear number two is called as the intermediate gear. Speed ratio is speed of the driver divided by speed of the driven n1 by n2 sorry n1 by n3 so we have to find out the n1 by n3. So in this case one is meshing with the gear number two that is n1 by n2 is equal to t2 by t1 and two gear mesh with the gear number three n2 by n3 is equal to t3 by t2. Therefore multiplying equation number one and equation number two we are getting n1 by n2 into n2 by n3 is equal to t2 by t1 into t3 by t2. So here t2 will get cancer and n2 will get cancer therefore we are getting n1 by n3 is equal to t3 by t1 therefore the speed ratio is equal to n1 by n3 is equal to t3 by t1. So here we are getting the speed ratio is equal to speed of the driver divided by speed of the driven is equal to number of the teeth on the driven gear divided by number of the teeth on the driving gear. So here we are not considering this number of the teeth on the intermediate gear. So similarly for figure number c we have to find out the speed ratio is equal to speed of the driver gear number one is a driver and gear number four is a driven therefore n1 by n4. So similarly we have to calculate the speed ratio of the n1 by n4. So for that purpose gear number one is matching with the gear number two so n1 by n2 is equal to t2 by t1 another n2 gear mesh with the gear number three n2 by n3 is equal to t3 by t2 another gear gear number three mesh with the gear number four n3 by n4 is equal to t4 by t3. So by multiplying equation over one two three we are getting n1 and 2 and 3 divided by n3 and n4. So here n2 will get answer n3 will get answer is equal to t2 t3 t4 divided by t1 t2 t3 t2 t3 will get answer is equal to t4 by t3. Therefore the speed ratio is equal to n1 by n4 is equal to t4 by t3. So here also we are getting the speed ratio is equal to the speed of the driver divided by the speed of the driven is equal to number of the teeth on the driven gear divided by number of teeth. So here we can think about this why intermediate gear is used. So here we can use this intermediate gear here the it is does not affect on the speed ratio. So we are using the intermediate gear for if the distance between the driver gear and the driven gear is more at that time we are connecting the driver gear and driven gear by using this intermediate gear and for giving the direction of the motion means if this gears two gears are there if the driver gear is rotated in the clockwise driven gear rotate in the anticlockwise. So if the driver gear is rotated clockwise and we require the driven gear rotated in the clockwise at that time we are using this intermediate gear. So there is a two use of this intermediate gear for changing the direction of the output gear and for increasing the distance between the two gear. So here one question is there the gearing machines as shown in the figure the motor motor shaft is connected to the gear one and rotates at 975 rpm the gear wheels 1 2 3 and 4 are fixed to the parallel shaft rotating together the final gear 4 is fixed on the output shaft what is the speed of the gear 4 the number of teeth on the gears are given below. So directly we are knowing speed ratio is equal to the speed ratio of this is a simple gear and because on the one shaft only one gear is mounted and it's called as a simple gear. So speed ratio is equal to n1 by n4 is equal to this is the speed of the driver the speed of the driven is equal to the number of the teeth on the driven gear developed by number of the teeth on the driving gear. So here in this question they are given us gear number one means the speed of the driver n1 is equal to 975 and they are asked n1 is equal to 975 975 divided by n4 is equal to t4 by t1. So number of the teeth on the 4 is 45 divided by number of teeth on the 40. So therefore we are getting the n4 is equal to 975 into 40 divided by 4866.66 rpm. So like this we are getting the speed of the outputs. So I have taken these two references. Thank you.