 Hello everyone, I am Mr. Sachin Rathod working in mechanical engineering department from WIT Solapur. Today we will see the concept of law of gearing. So the learning outcome of this session is at the end of this session students will able to understand the law of gearing introduction. The law of gearing states that the normal that is the common normal at the point of contact between a pair of teeth must always pass through the pitch point. So you can think about this whether is it necessary to check the law of gearing. The answer is as the law of gearing states the condition which must be fulfilled by the gear tooth profile to maintain the constant angular velocity ratio between the two gears. For example, if you are going to design the gear that is the machine gear at that time after the designing if you are getting the failure between the velocity ratio we are not getting the constant velocity ratio between the two meeting gear then at the end result it will give the separation of the tooth will occur. So that is not that one we are not getting the proper alignment between the two gear that is the proper machine we are not getting. So for that purpose we require to check the law of gearing while the designing. So we will see what is in my law of gearing. So first thing that we will check so I will draw the two teeth of the gear from which we are getting whether we are satisfying the law of gearing or not consider this is a one tooth and it is mating with the another tooth of the gear. So like this one complete gear tooth profile is there and another gear tooth profile is like this so just I have taken a two tooth of the gear both are mating about this point. So consider this is a O1 center point and this is a O2 center point O1 is a center point of the gear number one O2 is a center point of the gear number two both are mating at point A. So this is a meeting point when we are drawing tangent line this will give you common tangent and perpendicular to this is common normal this is a meeting point A. So our ultimate aim about this law of gearing we have to get the angular velocity ratio between the two gear is constant. So suppose this is a gear number one which is rotated by angular velocity omega one so this gear number two will rotate by the angular velocity omega two. So suppose this is a O1A is nothing but the radius of meeting point at A and O2A is the radius of second gear about the A point. So the velocity is perpendicular to the radius for that purpose we have to draw the velocity perpendicular to the O1A1 this is rotated in the anticlockwise direction. So draw the perpendicular line so it will give the velocity of OA so for the simplification purpose you will give the naming as A1 and A2 A1 is a meeting point of your radius O1A1 and this is the O2A2 so this will give you the velocity of VA1 similarly the O2A2 radius we have drawn the another velocity as VA2. So draw the perpendicular line to common normal we will give name as M this one is a perpendicular line that is N the perpendicular line makes the angle with O1A1 that is radius is alpha and the perpendicular line with the radius it makes an angle beta so as per the geometrical consideration we are getting the angle alpha1 or alpha VA1 velocity of O1A1 radius we are getting the VA1 makes an angle alpha and VA2 makes an angle beta with the common normal because common normal is perpendicular to common tangent and we have drawn O1A1 is a perpendicular line and it makes an angle alpha so by considering this geometry we are getting the alpha and beta angle with the common normal so this is a diagram for satisfying the law of Gearing. Now we will check how to verify the law of Gearing so first of all the condition is that for two gear remains in contact the velocity component of the velocity that is a VA1 and VA2 along the common normal should be 0 the resultant of this velocity should be 0 otherwise if the resultant of this velocity is not going to 0 then the separation of these two tooth will occurs for that purpose VA1 cos alpha minus VA2 cos of beta should be equal to 0 so if this condition satisfy then and then only the two tooth remains in contact therefore we are knowing velocity is nothing but radius into the angular velocity so for the first gear the radius is O1A1 into the angular velocity omega 1 into cos of alpha is equal to VA2 that is nothing but the O2A2 into omega 2 into cos of beta then O1A1 into omega 1 into cos of alpha by considering this triangle cos of alpha we are getting this triangle O1M A cos alpha is equal to O1 O1M divided by O1A1 O1M divided by O1A1 is equal to O2A2 into omega 2 into cos of beta is equal to O2N divided by O2A2 so this will get cancelled so we are getting omega 1 into O1M is equal to omega 2 into O2N so by considering this similar triangle O1M so if I joining this two center so along the common or we are getting the p point that is called as a p-point so by considering this triangle O1Mp and O2Np triangle O1Mp and triangle O2Np by the similar triangle rule we are getting O1M divided by O2N is equal to O1P divided by O2P therefore if you put the value O1M divided by O2N is equal to O1P by therefore omega 1 by omega 2 is equal to O2N by O1M is equal to O2P divided by O1P therefore we are getting omega 1 by omega 2 is equal to O2P by O1P so by referring this equation the angular velocity between gear number 1 and 2 is equal to the ratio of the distance between a center point of O1 and O2 that is nothing but the ratio of omega 1 by omega 2 is equal to O2P by O1P nothing but the O2P by O1P therefore the p-point must be your the fixed point if you are satisfied the law of gearing so for that purpose we are knowing the law of gearing so by this equation omega 1 by omega 2 is equal to O2P by O1P the p-point must be the fixed point if you want to check the law of gearing then the ratio of the angular velocity between the two gear must be the 0 according that we have to design the distance O2P by O1P these are my reference thank you