 So, in the last class, we were looking at braking, may be stopped just sort of this or may be we did this derivation. Very simple derivation, there is nothing much in this. We looked at the moment equilibrium and then we derived this and we also made a comment that there is a load transfer in braking where the front forces are high or the front reactions are high or the load gets transferred to the front unlike the acceleration part where the load gets transferred to the rear. This is what we saw. We also said, we went to the next step and we said that we can dump the whole of the effect of tire onto the road by, dump the whole concept of this interaction into a factor called mu which is very similar to your well known friction condition, Coulomb's friction and said that W multiplied by mu will give me a limiting condition for the wheel to lock. In other words, the maximum force that it would take any tire, front or rear would be given by this condition. We can put that the braking force, the maximum braking force in the front is given by WAF and at the rear is given by WR. This is where approximately we stopped. We also said that the total force, similarly we can put for WR, the total force braking force is split into the front and the rear. Unlike we have cars which is front wheel driven, we have some cars which is rear wheel driven, we have cars which are or vehicles which are all four wheels being the driving wheels and so on, braking is distributed in the front and rear. In other words, obviously you know that you have front and the rear brakes. The total force, the braking force is now distributed between the front and the rear and let us call this fraction of the braking force which is distributed to the front as KBF. So the braking force in the front can be written in terms of into FB. Same fashion you can write it for KR, KBR and braking force in the rear is equal to KBR into FB. Obviously it is a fraction so KBF plus KBR is equal to 1. Now let us look at the physics. We have two wheels that are braking, that is two set of wheels rather front and the rear wheels which are braking. Now the front wheel brakes when the condition that the force generated in the front wheel is equal to mu into WF obviously and the rear wheel brakes in the same fashion. So assume for a moment that they are randomly distributed you know KBF and KBR are randomly distributed or FBF or FB are randomly distributed we do not know just one value. What would happen? One of the wheels will break first the front or the rear wheel. So what is this condition? The front wheel breaks which means that it starts skidding under one circumstances when you give more force to the braking force to the front. So look at this way, look at it this way that the rear wheel front wheel is now skidding the rear wheel has not locked so it still has some force that is unused for braking that is unused for braking. Hence optimum condition is one where you use the complete force FB, you use the complete force FB to break the brake. So when will this happen? When both the front and the rear wheel they lock together, only when they lock together you do not lose any braking force. So that is an ideal condition where you break both of them, you distribute the forces in such a fashion that both of them start skidding at the same time. When would this happen? When you distribute this braking force whether the front braking force or the rear braking force you will distribute it in such a fashion that it is proportional to when will it skip I mean sorry when will it lock when it reaches mu WF. So how would you now distribute both of them? Proportional to WF and WR. So in other words if you have the braking forces to be distributed or the maximum braking force develops at the same time means that they are distributed proportional to the front WF and the rear force or the rear reaction WR. So what is that condition? We simply say that KBF by KBR should be equal to both of them should be proportional to WF and WR. So this can be written as in a very simple fashion can be written as from this equation from this equation L2 plus H by L and I am going to replace FB the braking force by means of the deceleration quantities or you can say mu if you want in terms of mu into WF. So you can write this as plus H into mu plus FR and WR you can write that as L1 minus H into mu plus. So if the two forces are distributed proportional to WF which I have just rearranged it. So in this fashion then that is an ideal braking where both the front and the rear would brake at the same time. Typical values for this would vary from one position to the other or one vehicle to the other and even for the same vehicle it would vary. Why? See though we say that this is the ideal ratio it is not a constant even for a vehicle and definitely would vary from one vehicle to the other. Why would it vary from for the same vehicle for example? Because when you load the vehicle it is not necessary that L1 and L2 which are the distances from CG locations are the same. It may vary. For a truck it would vary a lot more than a car obviously. So when you load say for example a truck then definitely the CG locations can shift to the rear. Incidentally L1 in many of the textbooks and this we are following theory of ground vehicles from Ong this part. One of the books I think I forgot to mention by Ong. We will shift from one book to the other and our own experiences we will look at this course. So one of the things I just want to point out is that instead of L1 and L2 people also use A and B. A and B is very common. The notations A and B, A being the distance of the CG from the front, B being the distance of the CG from the rear. So I use capital L, small l interchangeably both of them means that it is the distance between the two wheels. So this is the ideal ratio. So why would it vary within a vehicle or for a vehicle? Obviously when the road conditions are different mu would be different, mu would not be the same. So mu would change. So for the same vehicle A and B or L1 and L2 are not a constant. I did not use A here because you may confuse it with acceleration and that is the reason why I had not used A and B here. I had used L1 and L2 at this place. So this is a problem so you do not have an ideal. That is why you have so many other devices like ABS and other things. In order that we do not get into some conditions which we would discuss probably later in the course. Now that is the first thing. Then what is usually used? What is the ratio that you use? Various. Many of the truck companies today use 50 percent to 50 percent. Some of them from some of the car companies use 47, 53 and so on. So it is a very difficult ratio to arrive at because of the same vehicle there would be a change and hence we have this problem of fixing year ratio. Now why are we worried about locking? What happens during locking? Now let us understand that. Let us say that I have a vehicle, just a schematic diagram of vehicle which is travelling in this direction of course and let us say that one of the wheels lock. Let us say that rear wheels lock. What is meant by locking physically or from a physical principle? Locking simply means that all the mu that is available for that tire has been consumed, has been completely consumed. So no more that mu factor or the friction factor is available for any other force that acts in that wheel. Let us assume that the rear wheels have locked. All mu has been consumed. Now there is a small perturbation of the vehicle to the vehicle. There is a small perturbation. It is only in theory that the vehicle would run straight. There would be a camber of the road or there can be a cross wind or there can be a small undulation which you have to or a small bump whatever it is. There are perturbations in the roads and you cannot always drive straight and due to so many reasons your vehicle has to develop a lateral force. So it has to develop a lateral force. This guy here who is the rear, he has lost the ability to support any lateral force and so only the front wheel would develop the lateral force. So when the front wheel alone is capable of developing a lateral force, that is the situation. All of the lateral force that is required will be developed by the front wheel. So what happens? The vehicle starts spinning or in other words the vehicle loses what we would call as the directional stability. So it would spin. In fact it can spin so much depending upon the perturbations. It can just spin and even go 90 degrees or more. So when the rear wheel locks, the vehicle loses directional stability and would start rotating or yawing as it is called. On the other hand, let us consider what happens when the front wheel locks. When the front wheel locks, when the front wheel locks, obviously it loses the ability to take further forces and so all the force that is required to cater to the needs of the changing orientation of the vehicle, the lateral force is at the rear. So the vehicle would now lose directional control. A small change this side would not happen because the vehicle would now be yawed in the other direction. So it would now start going like this. So we call this as loss of directional control. Now if you are a driver, you have all the feel for the front wheel. You have all the feel for the front wheel because your steering would give you the feedback. So when the front wheel locks or front wheel is about to lock, you know that something is going wrong and it is possible, it is possible by an expert driver to get back that kind of loss of feel or in other words to steady the vehicle. On the other hand, if the rear wheel locks, even an expert driver cannot feel it and by the time he realizes everything is over, everything is over. So we need devices, special devices in order to avoid this kind of locking and loss of direction or directional stability and directional control. Is that clear? Now let us go into some details of this braking further. One of the key questions which we are going to ask is what is the deceleration that can be achieved by a given braking conditions? What is the maximum deceleration that I can achieve? Very important question, because ultimately brakes are also judged by deceleration and more importantly the braking distance. So in order to understand this, we need to understand what would be the deceleration level before the brakes stop. Start locking. This is what we are going to do now. I am going to take two equations and then just equate them and then get the, what is called as the maximum deceleration when the vehicle locks. Remember that our good old friend rolling resistance is going to help braking. So one of the conditions that we already wrote is, neglecting all other forces of course. Let us say that we neglect all other forces like aerodynamic force. You can add them, that can be an exercise to you. Just to bring out the physics, let us look at only these two forces. So this will be equal to, denote that we had replaced minus a by d and hence all the directions are, now so d is actually the deceleration. We also have, this is one equation, we also have an equation for W f and W r and that can be written as, how did we write it? We took W out, what is the first term for W f? L2 plus, what was the term which is inside? F b plus F r W. So what can you do from this equation? You can replace that, what was in the bracket in the previous case by this and can I not write this as the deceleration divided by g. So the same fashion I can write down W r W l minus d by g. What was the first term? In the bracket I just replaced it by the first equation. Let me write down F b from the first equation, call that as first equation for this second equation, third equation is equal to W by g d minus F r into W and I also know that front breaking force is equal to W by k b f into F b and the rear breaking force to be k b r into F b. So which is 1 minus k b f into W and so on. I also know that F b f will lock, that is the front force of maximum is equal to mu into, sorry, mu into W f. The maximum force of maximum is equal to before it locks is that and F b r mu into W r. So now what I am going to do is, so let me call this as 4, let me just rewrite it a bit. W divided by d by g minus F r is the, that and so this could become, this would be k b f w d by g minus F r. Now what I am going to do is, under the maximum condition, note that this is valid only for maximum condition, I can equate this with this and rearrange the terms. So substituting from equation 2, I can write this as mu W by l l 2 plus d by g into h, that is equal to k b f W d by g minus F r. Let us, let us for a moment remove that, let us only look at this first. This is when, what is this condition? When we reach the maximum allowable before it starts locking. Rearrange the term and then I get this b b r mu into W r mu into W r mu into W r mu into W r mu d by g or in other words, the deceleration normalized by g, we will talk about it in a minute. Just rearrange it and tell me what can this be. Take it to the other side and let me know what should it be. So bring it to this side, not very difficult. Remove that W mu l 2 by l mu l 2 by l plus k b f into F r divided by, go to the other side, k b f that remains minus mu, this h is there h by l. So that is the maximum, maximum means what? When it locks. Please note that we have normalized with g and this is the language vehicle dynamics use. Whenever they talk about acceleration, they would not talk about meters per second squared and so on. They would usually normalize it with g and talk in terms of this normalized quantities. They would say that it is 0.3 g, 0.2 g, 1 g and so on. Hence, this expression normalized with g. Derive the expression for, this is, when is this condition? This is front wheel locks, the maximum deceleration that is possible when the front wheel locks. Now, derive what should this be when the rear wheel locks? What is the condition? Just put this, substitute it and rearrange it. See that you get this. This would be the condition. The same thing. I have two versions of the same equation. I am equating it, rearranging the terms and I am getting this. So, front lock, let me call that as that and this is the rear lock. You have to be very careful in interpreting this equation. For example, KBF is equal to 0. Let us say possible, I have all the braking to the rear wheel. KBF is equal to 0. What happens to this equation? It becomes minus. Is that correct? KBF is equal to 0. It becomes minus. Why should it become minus? The whole crux of this thing lies in this equation. Very simple. All of you know it. Sir, it is actually inequality. Exactly. So, it is very important that it is an inequality. So, it is an inequality. It is very dangerous to put, draw for example, for friction, to put this as W. Many books do that and then say that F and then put this as mu into W. Now, it can be, let us say that this is 100. Let us say that mu is equal to 0.3. So, if you put a diagram like this, what would happen? This would become say 30. So, I apply a force as 20. The equilibrium dictates that the actually the block moves in the opposite direction. So, you have to be careful in interpreting it because very correctly as I said it is an inequality. So, whenever I use equality which means that that is the condition under which we overcome the friction or else if you substitute it as you like, you would get into trouble and you will see that like here, that block would start moving in the opposite direction which is physically not correct. So, this is one word of caution. You know it, sometimes you get into, you stuck, you are stuck when you, when you derive it and that is very important that you understand this. Just now with that in mind, let us now interpret what happens when I now change, when I now change KBF with that in mind. So, it is a very important writer. Let us say that I plot KBF in this axis and I plot my deceleration. So, let us say that D by G in this axis. As I now keep increasing, increasing KBF, as I keep increasing KBF, if there is no more force in the front, so which will lock the rear only will lock. So, when I keep increasing KBF, more and more forces would go to the front. So, if I now draw, say let us say that that is about 0.5, if I now draw a limiting case for the rear wheel to lock, this would be the case. Or in other words, the deceleration at which the rear wheel would lock is this. So, that is the limiting values or the values at which the limiting values for the rear wheel to lock. Why is it increasing? Because the force of the rear wheel is now increasing. That small block which we drew, the force is small. Now, when I increase KBF, that means that I am giving force to the front. So, actually that would at some point that would start decreasing. So, that would be the curve at which the limiting value for the front wheel to lock, front wheel to lock. Look at that carefully. I have actually two curves, two curves. If you look at, let us call this point as A. If you look at a point which is to the left of A, who will lock first the rear wheels. To the right of A, it is the front wheel will lock. So, at point A, what would happen? Both wheels will lock. That is the ideal condition. I have used all my breaking force optimally when both wheels will lock. So, that would be the ratio which is ratio at which both wheels will lock. Unfortunately, this is not as I told you sometime ago, that this graph is not a constant. So, if there is a weight shift of L2 because this weights are higher or something like that, then it would take maybe something like this. So, this would vary depending upon the weight as well as depending upon the mu values. So, that is as far as deceleration is concerned. What is more important for us or what not more important, as important as this for us is the breaking distance. Though many companies put lot of importance on breaking distance, suppose you are a tire manufacturer, one of the things that would be required of you is to meet the breaking distance. We spend a few minutes on breaking distances. Though I said breaking distance is important, of course breaking distance depends upon deceleration. So, there is no question of removing this and talking only about the breaking distance. Let us look at an ideal condition. What is an ideal breaking? What is an ideal breaking? When I apply a force FB, I look at a car as one unit, just know what we do. Let us say that it has wheels and blah blah blah, so there is no issues about it. Let us say that mu is the factor. So, what would be the ideal condition, ideal deceleration? Ideal deceleration is mu into W, which is the breaking force. At note that I am not going to use, as I said inequality, I am just going to use equality, so you know what is the condition at which I am writing. So, this should be equal, this is the force that can be withstood by the tires, total, ideal condition. This should be equal to W by G into D. In other words, mu is equal to D by G. I said, we always normalize it with G. So, when D by G is equal to mu, that is the ideal condition which we can achieve. Any other condition is not, you cannot achieve it or you cannot overtake this condition. You can achieve only this. Hence, we define an efficiency of breaking. This is an ideal. I call this as an ideal condition or the best condition possible. So, obviously efficiency depends upon the best condition that you can achieve. So, efficiency of breaking, if I call this as an efficiency of breaking, so the efficiency of breaking is given by what you have achieved with a particular break and vehicle which is D by G, that is the normalized with G. Note that D is deceleration. I keep repeating so that you do not get confused divided by mu. So, that is the condition. It brings out all the simple facts you know that when mu is less, for example, you are driving in water or ice or whatever it is the mu dictates the breaking force, decelerations and so on. Now, we are looking at FB. FB also, the total force, FB also goes to decelerate or in other words when there is a deceleration. The deceleration also is for all the revolving components. In fact, this becomes very important in a motorcycle, not so very important or it is important not to that extent in a car and so on. So, in other words, since there are rotating components which also have to be controlled or as to come down, we need rotational inertia has to be taken into account. So, actually the whole breaking becomes quite complex if I have to write it. In order to avoid that complexity, what you usually do is to put a factor to the mass. So, you say that if W by G is the mass of the vehicle, then you say that I would multiply it by a factor which I would call as gamma B, gamma B. So, in other words, since I need to accommodate all those guys who are rotating, I would say that this is as if I am breaking a vehicle with a higher mass. So, gamma B is greater than 1 of course and it is typically more 1.05. Typically, no set rules, no set rules for some of the vehicles and for motorcycles and other things where you can apply similar things, it can go very high. It can be 18, 20 and so on, 20 percent and so on, 1.2, 1.15, 1.2 and so on. So, no set rules. I am just saying, I am just giving you a figure. Normally, it is about 1.05. There is a way to calculate this. I do not want to digress right now as to how this is done. But let us assume that I know it and I give you a figure and it is 1.05. So, my main idea is to calculate deceleration and I am going to use a very simple formula. Which all of you know, A D S is equal to B D V. This is high school physics. Now, I am going to write down this for A F by M which I would call this as F B plus sigma of all the resistances. Please remember that in the previous expressions I had left out certain resistances just to make it easier for you to understand. But in actuality, I have introduced all the other resistances as well which includes the gradient and which includes the aerodynamic forces. So, we need to include not only rolling resistance, we have to include aerodynamic forces. You need to include the gradient and so on. So, I am going to put all that as sigma r. We will expand that in a minute because we are going to get an interesting twist to this. And that is A divided by M. M I already said it is equal to W by G, gamma B into W by G and that multiplied by D S is equal to V D V. Now, this would have been very simple but for the fact that there is a term in R which depends upon V. So, in other words, what I am going to simply do is to take this to the right hand side, rearrange it and write it as gamma B, sorry, gamma B W by G divided by F B plus sigma r into V D V and then integrate the left hand side and the right hand side. It can be from, say, V 1 to V 2 and the corresponding distance we would call when it comes from velocity V 1 to V 2, we will call it as S, so that this would become S. As I told you, there is a small twist. What is the twist? The twist is because the resistance here involves aerodynamic forces and aerodynamic forces can be written by rho by 2, dense to the air, very important coefficient, drag coefficient C D multiplied by the projected area A into V square. So, since V square is there, I cannot take that out of the integral. So, I have to rewrite that integral to be S to be equal to, I can take down the, I mean, take out the first term, take out the first term and then write this as integral V D V divided by F B plus, write down all that you have for r. If you want to be very correct, then you will write that down into cos theta. Remember that this is what we talked about. We said we will make an assumption, remove that cos theta, but if you want to be correct, you can include that as well as F r w into cos theta plus or minus, I would say, y plus or minus, we said depending upon, we left it to you how we are going to write, plus or minus w cos theta S, whether uphill, downhill and so on. Plus r A, we will call this to be a factor called C. Let us say that I am dumping everything, because this is a constant, we can assume that this is a constant and write this as plus C into V square, oh, sin, sorry, sin into V square, sorry, sin into V square theta S, whether it is plus or minus is high school physics, I am not going to go deeper into that. You have this V square term. I have to now integrate this. So, regular integration D of V square, then 2 will be there. So, I will bring that 2 outside. So, D of V squared or you can say that this D of all this plus V squared also and then what would be the formula, what would happen, what is the, it is in terms of log to the basic, right. Integrate this from V1 to V2. We will continue this integration and further interpretation of this in the next class.