 Hello everyone, I am Mr. Sachin Rathod, working as assistant professor in mechanical engineering department from Oalchian Institute 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. Supporting frame on which the gear A is mounted with an axis O1, this one is the arm which connects the gear B about the center O2. So in this mechanism if suppose the R 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 get rotated around the gear A and B also try to rotate upon the gear A. So the gear B is rotated upon and around the gear A that's why it is called as an epicyclic one. So in this diagram if suppose we have to assume that the arm is fixed and gear A rotates through one revolution. We 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 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 the gear A rotates through the one rotation that is one revolution in anticlockwise direction. So first one is the arm after that arm the gear A is there this is the gear A and gear A makes 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 the 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 and is around this gear the other gears are rotated that is called as the epicyclic gear. 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. T A is equal to 36, T B 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. Center support the gear A. The gear A which is fixed means speed of the gear A is fixed. 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 the question. So under this question the arm is fixed gear A makes the one rotation. That is the one rotation in clockwise. So these are the condition which is fixed. Just we have to formulate this table depending upon the motion. So the arm is fixed. So the 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 meshes externally so we have to change the sign convention. So in this way we have to calculate the speed of the B. So next one the arm is fixed gear A rotates to the plus X revolution. Just we have to multiply with the X. So here 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 the total motion. So we have to check out the condition. So here they had 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. 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 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 anticlockwise direction. So the next question again they had 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 had given us speed of the A is equal to 300 rpm in clockwise. So that's why we have to consider clockwise means here minus 300 what is the speed of B and B is equal to how much. So just here they had given us speed of the A. So here Y we are knowing speed of the A X plus Y instead of the zero it will make 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. We 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. So these are my references.