 This is Mr. Nagesh Arthambake, Assistant Professor, Department of Mechanical Engineering, Walshan Institute of Technology, Sholapur. Today we will discuss some problems on Ackerman Steering Systems. At the end of this session, students will be able to solve problems on Ackerman Steering System. This is the learning outcome of today's session. The Ackerman Steering Gear Mechanism in this, the entire steering gear mechanism is situated to the back of the front axle. Now here we can see, in this particular session, we will first look towards the turning circle radius. When a vehicle is taking a turn, that time, what is the turning circle radius of all the four wheels? Now here you can see this is the angle of outer turning wheel, outer front wheel, and this is the angle of inner front wheel that is theta. Now if we see, in this case, form a formula for a inner front wheel, outer front wheel, inner rear wheel and outer rear wheel. So first of all, I will calculate the turning circle radius for outer front wheel and that I will designate as ROF. Now ROF is what? Here i to this distance is nothing but that is ROF, turning circle radius of outer front wheel. Now first of all, I need to find out the distance from this end to this end, what this will be. Now here we can see, this is the triangle from which I can find out this particular length. Now here I can see and if I call this as ROF, so this becomes, this becomes V upon sin of phi, this becomes sin of phi, this much distance. So plus, now this much distance is remaining. So this is nothing but, now here A is nothing but the track length of the track length. Track length is there and C is nothing but the distance between the two pivot and B is nothing but the wheel base. Now to calculate this particular distance, we have to consider this track length and distance between the two pivot and this much distance is nothing but A minus C by 2. Therefore here I will write A minus C by 2 and from this particular formula, we will get the radius, turning circle radius of outer front wheel. Now we will discuss regarding turning circle radius of inner front wheel. Now again the same is there, here only the theta angle is there. We have to first calculate this particular distance from this I point to B point. From that particular distance if I calculate then that is B upon sin of theta and I need to go in a reverse way that is minus this much distance and that distance is again same that is A minus C divided by 2. Now we will come towards these two rear wheels. Now I will consider first outer rear wheel that I will designate as ROR. Now this ROR becomes. Now if I say I need to calculate I to this particular distance first. So for this purpose if I take tan, tan of phi then I will get tan of B upon tan phi that much distance ROR then if I then afterwards I need to calculate this much distance that I need to add that is again same that is A minus C by 2. This is the turning circle radius of outer rear wheel. Now I need to calculate last one that is turning circle radius of inner rear wheel that I will designate as ROR. Now this can be calculated as this much distance if I take then if I take the right angle of this particular theta then I will get B upon tan of theta B upon tan of theta that I will get. Now I need to calculate this much distance that is again the same that is A minus C by 2. So from these four particular formulas we can able to calculate the four different turning circle radius for different wheels. And we know that for correct steering for correct steering this equation must be satisfied by our steering mechanism that is cot phi minus cot of theta is nothing but C by B. So this equation if our steering system is satisfying then definitely that time we can call it as our steering system is correct and that time all the wheels are rolling perfectly. Now we will see the problem that is based on this particular equations. Now I will read the problem a motor car has a wheel base of 2.743 meter and towards center of 1.065 meter. Now here wheel base is given I will write the given data for it. So given data is nothing but that is B. B we have seen that is a wheel base distance between the two axles 2.743 meter and distance between the two pivots is given that is C 1.065 meter and next one that is track wheel the wheel track is given that is A 1.217 meter. Calculate the correct angle of outside lock and turning circle radius of outer front wheel and inner rear wheel when the angle of inside lock is 40 degree. Now here we need to calculate phi first and after that I need to calculate the turning circle radius of outer front wheel and turning circle radius of the inner rear wheel. So these three things I need to calculate for that purpose for correct steering for correct steering for correct steering what is our formula cot phi minus cot of theta is equal to C by B. Now here I know theta is what theta is 40 degree that is given that I need to write over here. Therefore cot of phi is equal to that is C is nothing but 1.065 divided by 2.743 and here plus this cot of 40 that is theta. Now I need to calculate this particular distance that is why I need to use a calculator. So if I find out this particular distance then what should I do? Now here 1.065 divided by 2.743 if I do this then I am getting 0.3882 so here I will write 0.3882 plus now this particular whatever the cot theta is there that is 1 upon tan. Therefore 1 divided by tan of 40 if I write then 1.19175 will be the answer. If I add this particular two figures then 3882 so this becomes 1.579 so this is cot of phi and now if I calculate this particular cot phi then I need to calculate 1 upon this particular tan. Therefore 1 upon answer is equal to if I do the tan inverse of this then I will get the angle that is phi is equal to 32.3 degree that answer I am getting. So phi I got first. Now second thing that I need to calculate is that is ROF. ROF how to calculate ROF? Now ROF becomes already that we have seen what is ROF. ROF is nothing but b upon sin of phi plus here a minus c divided by 2. Therefore b is nothing but that already we know 2.743 here and here sin of phi now we have calculated that is 32.3 plus a that we know 1.217 minus c we know that is 1.065 divided by 2. If I calculate this then I am getting ROF. So from this calculator we will calculate 2.743 divided by sin of 32.3 whatever the figure is there 5.133 I am getting 133 plus the bracketed quantity is 1.217 minus 1.065 this divided by 2 I am getting 0.076. If I add these two things then I am getting 5.209 means 5.209 meter this will be ROF this is my first answer. Now second thing that I need to calculate is the radius of inner rear wheel that formula is b upon tan of theta minus a minus c divided by 2. Now here again b that we know that is 2.743 the same thing tan of 40 over here because theta is 40 and a minus c that is 1.217 minus 1.065 divided by 2. Here I will calculate the first thing that is 2.743 divided by tan of 40. So whatever this quantity is there that is 3.268 minus 0.076 whatever this quantity is there that quantity is nothing but rir that is nothing but turning circle radius of inner rear wheel 0.076. So this is nothing but 3.2 meter that I am getting this is the rir. So in this way you can solve the problem regarding this Ackerman steering gear mechanism. These are the references thank you thank