 In the last class, we were looking at how the tires behave during longitudinal run and of course, we started lateral husband, but we will come to lateral in a minute or maybe after some time. We will look at the longitudinal force development through a very simple and very nice model called the brush model. The brush model is very nicely explained by Professor Pajekai in his classic text, Tire and Vehicle Dynamics, which has gone, I think it is the third edition. It is a very nice book, mathematically very rigorous. It would be very nice if you can have a look at many of the mathematical models that are developed in this book on tire mechanics, though we will not be able to go into those models in this course and we will be talking about very simple models, because we have to move into vehicle dynamics and discuss lot more things on vehicle dynamics. Tire mechanics being extremely important, I am going slow and we are looking at how actually the forces are developed and so on. There is a question just before the class whether we will have some mathematical model in this course. Yes, we are going to do a simple mathematical model, which is going to be called or which is called as the brush model. But there are very complex models that are available for the force development in tires, but we will not, of course we will talk about a very broad, I will give you a very broad picture in the next class about what are all the tire models that are available for modeling the longitudinal as well as the lateral forces, but this model brings out the physics of the tire behavior in the longitudinal direction. One of the things that we saw in the last class is that there is a pressure, I am just summarizing what we did. There is a contact pressure, there is a contact pressure under the tire and there is a longitudinal force. In other words, contact pressure can be converted into force and so on, but the total force F is it. So, there is a contact pressure and there is a longitudinal force due to which there is a shear. Now, we also saw the difference between braking and traction. So, we said that there is a region, there is a region where the tire is going to stick to the ground and there is a region after which when the shear stresses exceed the frictional mu into Fz, the normal force multiplied by mu or pressure multiplied by mu, then we said that there would be a slip. In other words, if this were to be the contact patch, then there is a region where the tire is going to stick to the ground, this is the region where it is going to stick to the ground and this is the region where it is going to slide. When I said slide, remember that these guys are going to just come in a bit till again equilibrium has reached and so on. So, in other words, the maximum force that could be withstood is obviously in a simple model of Coulomb's model is equal to mu into Fz. What is our goal? Our goal is to find out how Fx is developed mathematically, express it and what is it related to? What is it related to? In passing, we define what is called as slip and we will see how the slip actually enters into the mathematical model and what is the relationship between Fx, the force developed versus the slip. This is the whole idea. So, how do we do this? Simple, by means of what is called as a brushed model. These are the bristles which are sticking out say for example, from the carcass of the tire and they have a stiffness, let us say lateral stiffness, say lateral stiffness for unit length which we would call as Cpx which is the lateral stiffness for mm or unit length. So, what we are going to do in this derivation is that I am going to find out what is the deformation. So, the first step is what is the deformation of this bristle or this cone of bristle which I would call as u. So, once I know what is the deformation, the next thing I am going to do is to find out or multiply this to find out the force which is nothing but the Cpx into u. The third thing I am going to do is to assume a pressure distribution because one is left hand side and the other is right hand side of the equation. So, I will assume a pressure distribution which in this case we are assuming it to be parabolic. Assume a parabolic pressure distribution and call it as Qz. The next step I am going to write down the equations but I want to give a broader picture. The next step is obviously what is the next step? So, once I do it I have the left hand side of the equation, what is this force? The longitudinal force and this is the right hand side of the equation and I can find out mu into Qz. I can find out the left hand side and the right hand side and then what do I do? I find out point at which the slip starts or sliding starts. Be careful to use the word slip because we are going to give a very specific meaning for slip. So, in other words let us call that as Xs. Yes, this is a Cpx, the longitudinal stiffness. A good question. See please understand that this results have two stiffnesses. Stiffness is now you know what is stiffness. So, the stiffness is here in this direction and stiffness can also be in a direction which is perpendicular to plane of the board. So, which you would call as the lateral stiffness. So, which is Cpy. So, Cpy is the lateral stiffness. Oh sorry, sorry, sorry, sorry. Oh God! See longitudinal stiffness, longitudinal stiffness. So, Cpy is the lateral stiffness. I did not notice that sorry. That is the lateral stiffness. Though usually it is assumed that Cpx is equal to Cpy, it is not so there is a 50% difference between the 2 and so on. Good. I am sorry that I hope it is. So, the point at which sliding starts. Then as a last step what I am going to do is to find out the total force that is in the X direction which is the sum of the forces that is generated in the sticking region plus the sliding region. Sticking region plus F. So, F sticking plus F sliding. So, that is the total force. This is how our derivation is going to progress. Clear? The derivation is going to be exactly the same when we look at the lateral forces. We will come to that. The same similar model is going to be used for lateral forces, but I am not going to derive this in detail. People who are interested in this derivation which is going to be exactly the same can refer to a book by Prasekhar. Actually, he has derived the lateral force and said that is similar in longitudinal force. I am going to do the other way about so that you will have both the derivations with you. I am going to derive the longitudinal force and then say the lateral force is going to be very similar. Of course, we will discuss the physics of that a bit later. Just before we proceed with the word of a comment that though we are discussing longitudinal and lateral separately, there are conditions under which you will have both lateral as well as longitudinal forces developed. In other words, when you are cornering and breaking, for example, cornering and breaking, then there is a combined lateral and the longitudinal deformations and hence there is a force which is both fx and fy. We will just mention this as we go along and again we will not be able to derive that completely, but at least we will just see this like this and see what are the final equations. With this background, let us go ahead. It is always better that before we derive, better that we write down what are the steps we are going to do, then maths just follows the physics. Now, I am going to make some assumptions because I have told you this is a simple model. So, there are a number of assumptions which we are going to make to keep the maths within our bounds. Let us assume that this is the figure from Pasekha. We redrawn it and in order to understand how the deformations are going to take place. Let us assume that there is a point S. This is a point which lies on the rolling, this radius R which is such that it rolls on an imaginary circle without any slip. So, on an imaginary ground like that. So, in free rolling, S, the velocity of S in one direction, this direction which is omega into Re is equal to the vx. So, it does not slip, point number one. On the other hand, when there is breaking or when there is acceleration, the velocity of S is different from vx. We saw that. Hence, there will be a movement of vs or there will be a velocity of vs which we would call as vsx. We will see what this definition is in a minute, vsx depending upon the breaking or accelerating. It would be one direction or the other. We will use the same thing like the v which is in this direction for the tire can be assumed to be v in the opposite direction for the road and that it is rotating at a stationary position. Let us assume that there lies a point. There are slightly assumptions here because you may be confused as to about the Rs. We are going to make some assumptions. One of the assumptions we are going to make is that these radiuses are so large when compared to the lengths of these bristles and so on that we are going to write down this in terms of radiuses which we have defined. In other words, we are going to make some assumptions on the radiuses to make again things simple. Let us say that S prime is a physical point. It is a point in the tire which moves with the same velocity as that of S which moves the same velocity as that of S. We assume that these bristles which are sticking out, these bristles which are sticking out, though we have discrete bristles, actually you can assume that they are continuous but then continuous bristles do not give that physics and so on. You can do that and later when we integrate it, we will make it continuous. These bristles have a base, two things. One which is sticking to the, you can call as the tail which is sticking to the carcass and the other which is sticking to the ground, which is sticking to the ground. There is a difference between the two. What is sticking to the ground is going to move with the velocity of the ground and what is sticking to the base is going to move with the velocity of the tire, the tangential velocity of the tire. Since these two are going to be different in braking and that is what we are going to see now, there is going to be a deformation of these bristles. That is what we said first that we are going to look at the deformation of the bristles. This is the first thing first. So, let us look at what is V sx. So, V sx is, as I told you, is the difference between the velocity, that is we will write that as Vx, velocity of the ground minus omega into Re and that is where s is sitting, which we can call as r. So, let us say Vr, say Vr is equal to omega Re, say so that V sx is Vx minus Vr. Any questions? Slip, slip is a very technical quantity which we defined in the last class. We are going to define it again. English slip, English word slip means it is slipping. In tire mechanics, slip has a very specific definition. So, though you can say that slip and slide are synonymous, so you can use the words interchangeably, in tire mechanics slip is a definition. I am going to define that again in today's class. Sliding is the word which we use where when the force exceeds the mu into F, slides moves. So, do not get confused between the two. I know this is common confusion. So, I am going to define slip now. Just wait for a minute. Now, the first thing is that I want to know how this tail has moved. What is that? It is moved. So, in order to do that, let me introduce a coordinate system with this as the base, sorry, this is 0, x is equal to 0, positive in this direction. I know this is a very busy picture. We will understand this picture slowly. So, that is x is equal to 0 and that, this is the positive x direction and that this contact, look at that, this contact patch is plus a and minus a. So, that is again a. On either side of 0, this is plus a and minus a. Now, the first thing is that since these points here are moving, all these points here are moving with s prime which is moving with the velocity of v r, the motion of the base point, let us say to a point which is x from this point, which is x from here. In other words, that the base point travels a distance of a minus x at a time which is equal to delta t. The base point is travelling with the velocity v r and hence the base point moves with the velocity, sorry, with the at the time a minus x into v r. At this time, this is the base point moves. This is say for example, this is moved from here to here. At the same time, the tip of that bristle or the head of that bristle, that has moved, that has moved a distance of at the same time, the distance moved by the head is equal to v x into delta t. So, at the same time, the distance moved is v x into delta t. So, you can say that the distance moved from here to here, suppose I say that here to here, this distance or let me put it like this. This distance is v, this is a minus x and this distance is equal to v x into delta t. So, because it is simple, it is not even a figure, it is required that the base is moving with the velocity v r and hence that is the time that is required for it to move and the head is moving with the velocity v x and so the distance moved is equal to v x into delta t. Any questions? Simple. So, what is the deformation? The deformation is the, of this bristle, one at one hand it is moved this and the other hand that is the distance. So, the deformation is nothing but the difference between the two, so that you can write down the deformation u to be v r minus v x into delta t. It is the same as s, s prime velocity same and v s x. So, yes, the question may be, is it re? Is it sitting at re? What is that? Yeah. So, that is why I said initially that we are going to make some assumptions. We are going to say that r is so large that there is a small difference. It does not matter and that is how we are defining it. Substituting the expressions, so I can write that to be v s x, what is v s x? It is nothing but v r minus v x or v x minus v r because it is going to be, you know, just the difference. So, let me define that as v x minus v r or sorry, this can be v x minus v r. So, that is the deformation. That is actually the deformation. That is equal to v s x into delta t and delta t is a minus x divided by v r. So, into a minus x. So, we are considering breaking. I have to be careful correctly. We will write. So, v s x into v r a minus x. Since the deformation is in the, my coordinate system is positive along this direction. The deformation actually is in the negative direction. So, I have to write the deformation to be minus. I am going to define two slip quantities. One of the slip quantities called theoretical slip, sigma term given by Pasekhar that it is a theoretical slip is defined as this difference v s x by v r. So, that u can be written as, u can be written as minus a minus x into sigma x. We also define a quantity called practical slip kappa, which is used by the industry. Usually people in the entire industry talk about kappa, which is the, what we would call as the practical slip is given by minus v s x divided by, this also will put minus. So, that v r. We use the same quantity. There are plus, minuses. There are so many definitions. I will stick to that definition. So, that u can now be written as, sorry, I will remove that. I will put a minus there. Can be written as, in terms of kappa as well, you can find out the relationship between sigma and kappa from this and can be written as a minus x kappa into 1 plus kappa. So the first step is over. So, we have found out the deformation. I hope my signs are correct because one more thing is that I am deriving, breaking. If there is any sign difference I would, hopefully it is correct and I will correct it later. My next step is, of course, finding out the force. We will come to that in a minute. My next step is to assume a distribution for the normal forces. I said that we will assume a parabolic distribution. So, that q z, very common distribution that is assumed all through is equal to. So, what it should be in terms of, total f z and in terms of a. So, we will write that as, 3 f z by 4 a into 1 minus, what should be the quantity there? It should be, you can take a guess, x by a whole square. So, that when it is x is equal to a or minus a, the term in the bracket goes to 0 and so q z goes to 0. So, it is maximum when x is equal to 0. Let me now get to q x. So, these two are fine. q x is quite simple. So, what is the way we write it? So, q x is what? I know the deformation multiplied by the stiffness C p x into a minus x into sigma x or kappa by kappa plus 1, 1 plus kappa. So, the point which I was trying to say is that this is what we define as slip. That is why I have to be careful when you use the word slip in time mechanics. That is fine. So, the next step for us is to find out when does the slip happen? Is it slip? When does the sliding happen? When does the sliding happen? When does the sliding happen? What do you think? Yes. So, mu n. Now, q z I know, q x per unit length. So, q z. So, it should be equal to q x should be when it is less than q mu q z. There would not be any slip, sliding, sorry, sliding and exceeds. So, I will write that condition is equal to mu into q z. Let us define a quantity called theta. Just to make things easier for us to write, let me define a quantity called theta, theta x, because you can define similar one for this is the, this one. That is the assumption I make as to how contact pressure is distributed. So, that you know the contact pressure is like this. Why? Basically, because when I integrate this expression, I should get F s n. Integrate it from minus a to plus a. I have to get the back this expression. So, that is the expression. Let us define theta x to be 2 by 3, let me just write that carefully, 2 by 3 C p x a square mu into F s n. Let me simplify this expression now. Let me simplify this expression. Can you do that? Let us simplify this expression. Write a minus a squared minus a is x squared into a plus x into a minus x, substitute for theta and get me a value of the relationship of a plus x into a plus x into a minus between theta and sigma x. So, the first thing let me, I mean nothing great. Let us say that x is the point at which, point at which sliding takes place, x is the point at which sliding takes place. Let me also call this as 2 a lambda following Paceka. It is a fraction of 2 a. It is a fraction of 2 a. 2 a is the total length of the contact patch. 2 a lambda is what we saw as x is. Now, let us take this expression and write down this expression, substitute for theta and write down what is x is. So, just do that and it is very simple, 3 by 4 mu F z divided by C p x a squared into, that is the a squared into a plus x s by, right. What I did was, just this is a plus x into a plus x into a minus x. My goal is to find out when total slip takes place. When is the total slip takes place? Lambda is equal to 1. That makes my job easy. Now, substitute for x s to be 2 a lambda and that is the point at which this lambda, I mean point from the, if this is the contact patch, that is what we called as 2 a delta. So, that lambda is equal to 1 is equal to 2 a, but in terms of our x, note this carefully. There is a difference, right. So, that because we are looking at x from that point, from the center, right. We are looking at x from the center. This is at x is equal to 0. So, in other words, this point is x is equal to plus a and that point is x is equal to minus a, right. So, let us substitute that in this expression. x s is a minus a minus 2 a lambda, that is what is x. In other words, that is x, that is a minus 2 a lambda, right. And hence, I would write sigma x to be 1 by 2 theta x into a plus a minus 2 a lambda divided by a, which I am going to simplify. I am not going to do every simplification. You can do that and that is why I am writing it. And hence, lambda, the point at which it starts is minus theta x sigma x. So, lambda is equal to, right, okay. I said that lambda is equal to 1 where the complete, you know, the sliding takes place. 2 a, that is, this is completely the sliding. So, you can write that. So, theta x to be plus or minus depending upon. So, let us now find out the force f x. Now, let us now find out the force. As I told you before, the force f x is the force developed in the sticking region plus the force developed in the sliding region. So, I can write down x s from my previous expressions, whatever I have done, to be very simple. You can see that f z omega r e divided by v s x is equal to x into C p x. Very simple, whatever I have done, just substitute that. So, f x can be written as a to a minus 2 a lambda, sorry, lambda is equal to not 1, 0. I think you guys had not, okay. A lambda should be equal to 0 for this. That is why you get this. Just be awake, right. Lambda is equal to 0 is where complete sliding takes place. Not a lambda is equal to 1, right. So, that is why you get theta x sigma x equal to 1. No, good. That is the point, complete sliding, just moves like that, right. What is that I did? Simple, C p x sigma a minus x e d x e divided plus this one. And x s, I have got from these expressions. That is all, right. Integrate this and I am giving you a final expression. I am not going to do the integration. It is not a very difficult integration. So, you would notice that, this is the expression. You would notice that I can draw f x now in terms of sigma. So, now, if I now plot f x versus sigma, my graph would look something like this. This is the sigma versus f x. For very small slip, sigma you can assume it to be equal to kappa. First term gets into picture. So, you can, one of the quantities of interest to us, you will see that later is always the slope here, which can be approximated by, if you look at kappa, it is approximated to be 2 p, 2 c p to a square. And what is this quantity? That is the maximum force. And the maximum force would always, you need not even look at any of these things. Maximum force would always be equal to mu into f is it. So, this would be mu into f is it. So, essentially, this is the brush model, where we found out the deformations. We found out where the sliding takes place, the place where the sliding happens and we just found out the force due to just deformation and due to sliding and then we wrote down a complete final expression period. Look at that. This is slightly different from what is given in Pasekha's expression, but I think this is correct. Have a look at it. Now, practically this may not be a straight line and that is because we have assumed mu to be a constant, but there is going to be heating and there is going to be dynamic friction and so on. And usually the mu here as a constant may not be valid and you would have a drop here. We would talk about that again in the next class. But before that, let us now look at the lateral deformations. Any questions here? The slip condition at which the slip happens, that is what we said, the condition under which the slip happens is equal to, this is the condition where both of them are happening, is equal to the q x per unit length, force developed per unit length is equal to mu into q is it. This multiplied by dx, that in the q z is what I had written down there. So, that per unit length, so mu into q z is equal to q x developed and both of them are the same, then that would be slip. So, this may happen, say for example, at this point. If it is completely sticking, this is how it would happen. It would look like, when for example, here is where that excess where sliding is going to happen, then this guy would actually start slipping, sliding. He is going to slide. So, this is how it is going to slide. q z is a normal, see what I am using is nothing but mu into f or normal force, very well known is equal to, that is all I am using. Only thing is I am using it per unit length. So, whenever this is actually an inequality, this is not an equality, this is an inequality. So, whenever f is greater, it will not slide, that is all. Whenever it is equal, it will slide. No, sticking is a portion which is before. When it reaches, what we said is, when it reaches mu multiplied by this quantity q z, when it is equal to q z is equal to q x, it slides. That is the point by which we found out lambda or excess. Clear? So, they are when they are equal. So, in other words, if this is going to be the contact patch, sorry about my diagram, then we said that this is where it is sticking and this is what we defined as 2 a lambda or this is what we defined as excess, which is equal to a minus 2 a lambda, sorry, I should not say excess, 2 a lambda and that if this is the x is equal to 0 and then this distance is what is called as the excess, which is clear. So, we equated it to find out that point at which sliding takes place. So, at these points, we assumed that the force is given by mu q z per unit length dx multiplied force per an infinitesimal length dx, then we integrated it throughout. Actually, it is x into integral excess minus a, which is strictly speaking, I should write it like this, excess to minus a. Any questions? Clear? Now, let us look at quickly a lateral force development. I think we have to find get back to this figure again in the next class. Let us do that in the next class. Apurva has drawn it so beautifully, I feel bad that I have to rub this and we will do that in the