 Hello everyone, I am Mr. Sachin Rathod working as assistant professor in Mechanical Engineering department from Walsh University of Technology, Sallapur. Today we are dealing with epicyclic gear train. The learning outcome is at the end of this session student will be able to calculate the speed ratio of epicyclic gear train. So consider this diagram in which we have shown this is a supporting frame on which the gear A is mounted with an axis O1, this one is an arm which connects the gear B about the center O2. So in this mechanism if suppose the arm we have made the arm fixed and if we have driven the gear A so that the gear B will get rotated. So it is a kind of the simple gear train. If suppose we have fixed the gear A so the arm will forcefully will get rotates around the gear A and B also try to rotate up on the gear A. So the gear B is rotated up on and around the gear A that is why it is called as an epicyclic gear train. Epicyclic means around and epicyclic means up on. So up on and around the gear A gear B will get rotated so it is called as an epicyclic gear train. So in this case we have to find out the speed ratio or suppose the speed of the arm we are knowing or the speed of the gear A we are knowing. So we have to calculate the speed of the B. So for calculating this we should know what are the procedure for finding the speed. So first one so in this diagram if suppose we have to assume that the arm is fixed and gear A rotates through one revolution. Gear A rotates through plus one revolution you have to consider the anticlockwise direction as a plus so I can write that anticlockwise direction. So this is my first condition arm is fixed the gear A makes the plus one rotation in anticlockwise direction. So under this consideration if suppose in this diagram as the arm is fixed if the gear A makes the one rotation we can easily find out the rotation of the gear B by using the simple gear train. So we are knowing the relation between speed of the A divided by the speed of the B it is inversely proportional to the number of the teeth that is we are getting T B by T A therefore the speed of B is equal to T A by T B into N A as the A makes the one rotation it is a one therefore we are getting N B is equal to T A by T B. So as the arm is fixed the gear A makes the one rotation we are getting the speed of the B as T A by T B but if you observe that the gear A and gear B are meshes externally so if the gear A rotates in the anticlockwise direction so that the gear B will rotate in the anticlockwise direction. So we are getting the negative term. Then the next condition is that arm is fixed and gear A makes instead of the plus one here we are assumed that it rotates through the plus one but consider that the gear A makes the x rotation in anticlockwise direction. So we can easily find out as the gear A instead of the one it makes the x rotation therefore we are getting N B is equal to T A by T B into N A so here we are considering the A is equal to x therefore N B is equal to x into T A by T B and as the A rotates in the anticlockwise B rotates in the clockwise so we are getting the negative sign. So these are the two main basic condition of motion. So simply we will check this table directly on the table we are getting. So arm is fixed gear A rotates through the one rotation that is one revolution in anticlockwise direction. So first one is your the arm okay after that arm the gear A is there this is a gear A and gear A mesh with the gear B. So the arm is fixed so the revolution of the element arm is zero A makes the plus one rotation. So we are getting the motion of the B is minus T A by T B. Next condition arm is fixed so the speed of the arm is zero and instead of the one rotation it makes the x rotation so just for the x rotation we are getting minus x into T A by T B. Then the next step is that add y rotation to all element. So this is a y rotation if you add if suppose like this is arm we are rotating by the y rotation so each and every element is rotated by the y rotation so we have to add plus y revolution to each element so we have to make the total revolution. So here we have to make the addition of two and three we have to not consider the first condition because it is for the one rotation we are not knowing how much revolution of the gear A is there that's why don't consider this. So we are finding for the x revolution and also we are considering the motion of the arm that's why make the addition of condition number two and three so here we are getting the y x plus y y minus x into T A by T B. This is the table of the motion so by considering this if anyone turns we are knowing we can easily find out the speed of the other element. So one question is that what is the application of epicyclic gear train so generally in case of the differential gear box we are using the epicyclic means in that gear so like this arrangement is there so this is the arm so such kind of the gear is called as the epicyclic gear train means around this gear the other gears are rotated that is called as the epicyclic gear train otherwise in the back gear of the lathe machine the epicyclic gear train is there or in case of the Merton AV machines we are using the epicyclic gear train so lot of the applications are there of the epicyclic gear train. So the next one is that we have to solve this example so in this example and in an epicyclic gear train carries the two gears A and B having 36 and 45 teeth respectively so the given data sir so I will write down the given data ta is equal to 36, tb is equal to 45 if the arm rotates 150 rotation in the anti-clockwise means speed of the arm is equal to 150 in anti-clockwise that's why we have taken a plus sign for the anti-clockwise rotation we have to consider the plus sign and for the clockwise rotation we have to consider the negative sign, centers about the gear A the gear A which is fixed that means speed of the gear A is fixed the speed of gear A is fixed means the speed rotation is zero determine the speed of B we have to calculate the speed of B this is a question so under this question the arm is fixed gear A makes the one rotation that is the one rotation clockwise so these are the condition which is fixed just we have to formulate this table depending upon the motion so your the arm is fixed so this first one is the arm second one gear A gear A makes with the gear B so if suppose arm is fixed the rotation is zero gear A makes the plus one rotation so you have to find out the speed of the B we can easily find out the speed of the B that is the NA by NB is equal to tb by ta that's why we are getting the speed of the B is equal to ta by tb and though both gear are measures externally so we have to change the sign convention so this in this way we have to calculate the speed of the B so next one the arm is fixed gear A rotates through the plus x revolution just we have to multiply with the x the arm is fixed x into x into ta by tb next one add the y revolution to all element plus y plus y plus y so here we have to calculate the total motion y x plus y y minus x into ta by tb so this is a total motion so we have to check out the condition so here they are given a speed of the arm is equal to 150 so speed of the arm is nothing but your y therefore for the first condition y is equal to 150 rpm and if suppose gear A is fixed the gear A means x plus y is equal to fixed remain zero therefore x plus y is equal to zero if you put the equation number one in this case therefore we are getting y is equal to minus x therefore x is equal to minus 150 rpm okay so we are getting the value of the x is equal to minus 150 rpm right now we are knowing the value of y and x we have to calculate the speed of the B therefore speed of the B total speed of the B is y therefore speed of the B is equal to y minus x into ta by tb if you put the value y is equal to 150 150 minus x is equal to minus 150 so it will get plus 150 into ta by tb 36 divided by 45 so we can easily calculate this equation B is equal to 270 rpm in anti-clockwise direction okay so the next question again they are given us if the gear A instead of being fixed it makes the 300 rpm in clockwise direction what is the speed of the B so instead of this zero they are given us speed of the A is equal to 300 rpm in clockwise so that's why we have to consider clockwise means here the minus 300 what is the speed of B and B is equal to how much so just here they are given a speed of the A so here why we are doing speed of the A x plus y instead of the zero it will mix x plus y is equal to minus 300 just put the y is equal to 150 therefore x is equal to minus 300 minus y is equal to minus 300 minus 150 is equal to minus 450 so we are getting the value of the x is equal to minus 450 you have to find out the speed of the B speed of the B and B is equal to y minus x into ta by tb therefore y is equal to 150 minus x is minus 450 therefore plus 450 into ta by tb ta is 36 by 45 so the speed of the B is equal to 510 rpm so it is a plus sign we are getting in the anticlockwise direction so here the first answer is that 270 rpm in anticlockwise direction and for the second condition we are getting speed of the B is 510 rpm in direction so these are my references thank you