 In the last class, we were talking about the braking distance. We had looked at how to calculate the braking distance. We had derived from a very simple dynamics relationship what should be the distance that the vehicle would travel when we go from velocity v1 to velocity v2 or v2 to v1. Now, what we are going to do is to in order to calculate the braking distance, we would now integrate it or rather v2 to v1 or I think that is what we wrote. We would like to look at what would be the braking distance when the velocity or the final velocity is equal to 0. Obviously, you know it. So, now you can write down the integration what we did and then you can calculate what is the distance that you would travel before we break the vehicle. Now, this is the formula which we wrote. If I remember right, this would be the I know whether I think that this is how we had given. So, that should be v1 to v2. I think this is how we wrote. vdv divided by fb plus frw, we wrote also cos theta s and you can make cos theta s equal to 1 plus or minus w sin theta s plus ra. We said ra is the aerodynamic forces and we just replace this with cae into v2 and we said that v2 being there that will have an effect on integration and you can integrate this straight forward. But there is nothing very difficult about it. Cg to cae ln of fb plus frw cos theta s plus or minus w sin theta s plus cae into v1 square divided by fb plus frw cos theta s sin theta s plus cae v2 square. Just quickly I am not going to spend time. You can do that. Then this is the distance traveled when the velocities vary by v1 to v2 or v1 goes to v2. Now, putting v2 is equal to 0 which is actually the stopping distance when the final velocity is equal to 0. This equation can you know you can write that equation to be. So, that gives you the stopping distance that you would get. Remember why we wrote that gamma that is for rotational inertia and other things and we also defined what is called as the efficiency of braking. Remember that we said the efficiency of braking is simply that all the f that is to be spent goes to braking the vehicle or in other words we said ma or m into the force that is the deceleration should be equal to w by or w by g. Let me write that as w by g into d should be equal to if you remember w that is the normal load that is acting that is the weight multiplied by mu. If you remember mu is equal to d by g and we said that that is the maximum efficiency. So, that is the reason that is one of the things that we wrote. You can also say that the maximum efficiency is when the braking force is such that it has already compensated for all the rotating inertia. In other words, gamma B is equal to 1. It has compensated for it and the rest of it is available for in order to break the linear case. So, in which case we can also write down the minimum braking distance. I will call this as minimum braking distance where we have fB replaced by mu into w that is the braking force. So, this can minimum braking force let me write it here. Let me s is equal to gamma B w divided by 2 g CAE natural log 1 plus CAE into mu and squared mu into w because that is the minimum all the forces go to plus fR w break the vehicle plus or minus w sorry sin theta s. So, that gives the minimum braking distance. I am just copying it. This is nothing very difficult. If there is any questions on this, all of them are just forces. I just did not want to leave any of the things and that is the reason why I am carefully writing it, but concept wise they are all very simple. So, that is the minimum braking distance. One of the things that is important well now a lot of research is going on is on the time lag between a driver applies the brake and the forces realized the braking forces realized. This becomes very important because there will always be a time lag. This time lag is due to right from application, we are talking about right from application to the point when the forces felt at the brakes and the brakes start acting. Delays are due to so many reasons here right from the pedal motion to the hydraulics that is involved or the fluid that is involved and so on. So, I just want to point out that that time lag need to be understood and taken into account if you are looking at braking distance. So, what we are talking here is the braking distance when the force is felt, but it is not from the time the vehicle or the driver applies the brake and then starts braking. So, usually that is a factor should not go into the details, I just want to point out that. So, we have covered quite a bit. We now know about acceleration braking in a passenger car. We will slightly deviate now, not that we are going to go away from the launch to the dynamics, but we have to learn lot more in launch to the dynamics. We will talk about tires, we will talk about aerodynamic forces and so on, but before we go there, let us complete the topic by looking at a tractor trailer. This is a very important vehicle today, especially in a country like India becoming very popular. Of course, the number of axles increase, when the number of axles increase, I would not be able to do the kind of calculation that I do here because the problem is not statically determinate. So, we are looking at it only from point of view of deriving a simple equation on forces acting. Hence, we would restrict ourselves to a tractor trailer where you will see that the trailer has one axle and that the tractor has two axles and that the tractor and the trailer are connected by means of what is called as the hitch point. There are a number of terminologies for this. Sometimes people call this as 50 or 50 or 50 or 50 or 50 or 50 or 50 or 50 or 50 or 50 or 50 or 50 or 50 and so on. So, we are, that is the hitch point. Let us understand what are the forces that can act on this. And as I said in the last class, the reference for this is the theory of ground vehicles by Wong. So, we will move away from this after this class to look at, let us say forces that are opposing, exactly the same thing. There is no difference in the concept, absolutely the same, but only thing is that when I take moments, there are not more forces that come into picture, okay. So let us just recapitulate what are the forces that are acting for the axle rating vehicle. One of our great interest is to find out what are the reactions that take place in these axles. What we would call as front axle, what I would call as rear axle, I have to be consistent. If there is any confusion on the terminology or on the notation, please let me know, I will be as consistent as possible W s. Now what are the forces that are acting? Of course, the aerodynamic forces, there are two aerodynamic forces that we take into account. One that is acting on the tractor. There is a tendency in the industry to call this also as a horse. They call this as a horse because it is something like a horse drawn carriage. So when they say horse driving a trailer, okay, do not be carried away by it because it is a loose term or a colloquial term that they use, okay, that is called a horse. So that is the RA1 is the aerodynamic forces that are acting, okay, on the tractor. The similar aerodynamic force acts here and that is RA2. The other forces that are acting, you know it, the famous rolling resistance force. So that acts everywhere on all the three, okay, the rolling resistance force. D'Alembert's force for acceleration, for deceleration it will just be opposite, same thing. The other force which is important are the hitch point forces, okay. Simple mechanics, the hitch point forces, directions you have to be careful, okay. You can look at this from two perspectives, one from the tractor perspective and the other from the trailer perspective. So these are all very simple mechanics. You can understand that. So you be careful in putting down the hitch point forces as they are called. Remember that we are considering the weights and then D'Alembert's forces, the tractor and the trailer separately. So that becomes the weight of the tractor and this becomes the weight of the trailer and the CG locations are written as H1, H2, right. We will make some simplifying assumptions. Not that it is necessary, so that I need not write the equation in a full equation on the board, I will make some assumption. Here again the concepts are simple, equations are going to be long, I am not going to spend much time, I am going to wind this up as soon as possible so that we will go to the next topic, okay. So any questions, ask me. So I am going to just write down the final equations, okay, telling you how I got it. You can derive it yourself straight forward, okay. So the first task for us is to find out what are the reaction forces that are acting onto the spake. Once we finish this, we will also go to the breaking which is very interesting, okay and that is the next step we are going to do. You can easily write down, why not you start writing down what would be the reaction forces that would result or that would be the result of this kind of arrangement. Note that I have put a point A, it is customary to take the moment about that point many times in order to find out the forces, okay. So let us do that, let us say that Ws into L2, I would take Ws into L2. Please note what I am going to do is to separate out, separate out, in your mind you can build this up, separate out the tractor and the trailer and then, you know, put the free body diagram. This is, this is just combined, I just put it like this. As I told you, this is important. Separate out the tractor and the trailer, when it is accelerating, of course, Fhi for the trailer will be in the other direction, okay, Whi won't change, okay and vice versa. So or in other words, there will be a support here for the trailer, okay. So note that and accordingly change it, okay. In other words, when I am considering the trailer part of it, this will be in the other direction, this will be in the opposite direction and vice versa. So well set, very simple. I am going to write down only the final equation for Ws, okay. This will be a good exercise for you. Verify it, okay. I am going to just make a small assumption so that I don't need to write down a lot of things. So I will write down H3 H2 L2 into W2 plus H2 W2. Note I am taking all the moment about the point A which I have defined, the hitch point forces, okay, that is the force which actually pulls the trailer. So that will be opposed by of course all the other forces so that the hitch point force can be written as, that is the, be very careful with this plus and minus. I don't want to every time give an explanation and confuse you but you would know how to write that, okay. So I am not going to do that but be very careful when you put a plus and when you put a minus. One of the easiest ways of doing it is that when it is going uphill, okay, accelerating, it is acting in the same direction as that of the RA toward the accelerating forces. When it is coming downhill, okay, the forces will be in the opposite direction. So the direction of forces is determined whether it is a plus or a minus, okay. Now using these two equations and substituting for whatever is in the brackets, here I can write down Ws to be W2 into D2 divided by L2 plus FR into H2. That is the reaction that happens in the trailer, the rear wheel of the trailer. Note that, note that these vehicles are rear wheel driven or in other words that is the wheel which has the drive and braking happens in all three wheels. So braking forces are applied to all three wheels. In other words, the driven wheel is this but the braking wheel is all three and so there is a distribution of the braking forces between these three wheels, okay. So the load at the hitch point, of course, all of you know how to calculate that. That is the W2 minus Ws because as I said the trailer is supported now by the rear as well as at the hitch point and so it can be written as W2 minus Ws and that is equal to 1 minus D2 divided by L2 plus FR into H2 into W2. So maybe you can call that, that equation is 1 and that equation is 2 and that equation is 3, that equation is 4 and this equation is 5 and so on. We can write down this as a coefficient and call that as CHI, we follow Wong theory of ground vehicles and just write that down as CHI. So now we shift to the tractor. We consider tractor as a free body and draw the free body diagram for the tractor, do exactly the same thing and find out the reaction forces in the front and the rear, okay, again by taking moments and so on. So WR into, here it is, you need not take it about A. So WR into L1 is what I am going to take. So let me write the final result, W1 L1 plus RA1 into HA1 plus H1A W1 by G plus or minus W1 H1 sin theta S F HI H3 plus L1 minus D1 W HI1 HI divided by, okay. That is WR. Again you can write down, you can apply equation 1, let me call that as 6 and you can simplify this equation, okay. Now the technique is the same. So what is the next step? We are going to simplify this by looking at the longitudinal forces and hence the longitudinal force F, the balance of the longitudinal force. What is the longitudinal force here? Traction force, okay. So the longitudinal force which is a traction force has to now equilibrate all other forces that are acting which is again very straightforward. You can say that that is RA1, I am not going to write down that equation, it is very simple and then it is, that is the next force, okay. Then W1, let me write it like this, so that W1 that is the next force, okay plus of course the rolling resistance force, okay. So we can, yeah, plus, remember that we had something called a drawbar pull and we removed it but now it becomes very important and this is very similar to the drawbar pull which we had used in the last class, in one of the earlier classes. So that is the force that you have. You can substitute it and you can determine what would be the force of the, we can calculate what should be the maximum force, okay. Ranging into this expression, the previous expression, I can simplify this equation, okay, rearrange them and so on. I am going to skip that, rearranging, writing down and so on and I am going to only write this. The maximum force that can be supported by the tract, by the attractive effort or by the rear wheel is obviously equal to like what we had written, we are lumping all the effects of the tire into one number called mu into WR, so the maximum is given by f max is going to be, so mu into WR, substituting WR, simplifying all these equations, I can write f max to be, I am writing the final form, okay, will be equal to mu into L1W1 minus H1 into FR into W1 plus CHI, so that gives you, that gives you the maximum force that can be supported by a tractor, semi-trailer, technique is exactly the same, any questions? Yes, yeah, we are assuming all H is to be the same which is not correct, okay, I agree. This is just to write down laziness of writing down the equation that is all, okay, you can write down that more elaborately putting down HA2, obviously, obviously HA2 is not going to be equal to HA1, HA3 is going to be very different from H, sorry, H3 is going to be very different from H2, it is going to be different from H1 and so on, I agree. Just to write it down, this is not correct, okay, you can, you need not make that assumption, there is nothing in that assumption, you can substitute it and write down the final form. The equation will become long, concepts are simple, nothing will change, clear? You are absolutely right, that kind of assumption is not, is only for a classroom deriving the equations and in practice that would not be correct, okay. So that is for traction, we quickly finish breaking in this class so that we will move forward because it is longitudinal dynamics, we have to move, we have to accelerate. So we will look at breaking here. Any other questions, anything? Okay, again very simple dynamics is what we are going to use for breaking, okay. What we are going to do now is to divide this into three categories, we are going to look at tractor, we are going to look at the trailer and we are going to look at the tractor trailer together. So that is the first thing that we are going to do. We are going to get, in other words, we are going to get three sets of equations by considering these three systems or these three scenarios. That is the number one. Number two is that we are going to look at sigma fx, sigma fy in the moment about an axis which is perpendicular to the board. So in other words, for each of these scenarios we are going to get three sets of equations, okay and so we are going to get nine equations. Then we will see how we can simplify it so that we can find out what is the loads that are acting or the normal reactions at the three wheels. So this is the strategy. Why are we interested ultimately or what is our goal? Our goal is to find out what would be the reactions in the wheels. Our goal is to find out these three, right. Why are we finding this out? Because we want to make sure there is a locking sequence. So the most important thing in breaking is what happens when these wheels lock. So I have to find out when these wheels lock. So I have to find out what these are. So that is my first step. That is why I did all these things or I am going to do all those things. Please note that when I look at all these three Wf, sorry, three Ws, Wf, Wr and Ws, I will multiply this with Mu in order to look at the maximum breaking force that is possible before the wheel locks. So in other words, derivation is pretty simple. In other words, what we are ultimately interested in is whether the front wheel locks, the rear wheel locks or the trailer wheel locks. These are the three things, right. So in other words, what can be the sequence at which we can have locking would determine how we are going to distribute the breaking forces, clear? Now what happens when the front wheel locks? The situation is exactly similar to what we had before. We are going to lose directional stability or directional control. So we are going to lose directional control and the vehicle is going to go forward. What happens when, remember that scenario, it is exactly similar when the rear wheel locks. When the rear wheel locks, remember that the whole force goes to the front and use yaw and so directional stability is lost. Now what happens here in this case when we have this wheel, it locks. Here in this case, what is going to happen is very interesting. The second one, so let me come back to this. When this locks, let us see what happens. Even we saw that as directional stability but here there is going to be small difference. Let us see what happens. I am going to break this and there is going to be, this guy is now moving forward. He is not, because if he has to be stopped, that has to be stopped and so on. So when this guy locks, more important effect is what is called as jack knifing because this kind of yawing which is interwheeler has to be looked at more closely because there are two forces that may be acting on either side. So may or may not happen, we do not know. But when this locks, there is going to be jack knifing, one going into the other, that would be the case and that would be a more important or dangerous situation. That would happen but we cannot talk about stability because that is for two-axle passenger car, it was very straightforward. So when this locks, there is going to be jack knifing. One, there will be a difference in the deceleration between the two and so one drives into the other. What happens when this locks? When this locks, there is going to be, please note that it is at the hitch point, this is something like supported here and this has a hinge there and so when this locks, this whole, this is going to sway and that is going to be dangerously swaying perpendicular to the boat. So the effects are such that each one of them has is important of course. One is more dangerous than the other two. Of course I said that locking in the front is the least of this danger. Why? Because the driver will be able to find out that he is going to lock or his wheel is going to lock. So he has a feel. So he will be able to correct it. So that is the least of the problems. Jack knifing is the worst and this is also very dangerous especially for vehicles which are coming or which are in the road, other vehicles because the whole trailer is going to sway. So this is, it is very important that these concepts are taking into account. So which locks first? So I would not allow this to lock. This would first lock, no issues. Why are we talking this? Because we want to distribute the total breaking forces. So this fine because I can find out what is happening and then this locks and then finally this locks would be a situation which would be useful in order to decide what should be the breaking forces. Of course this is only to understand the concepts. If you look at ABS which we will quickly run through, after understanding the behaviour of the tires you would appreciate the ABS in vehicles because this kind of situations, locking situations, how the ABS handles this kind of locking situation becomes important and the concepts will be very well understood at that point. So now what is my goal? I just want to write down the three sets of equations. I will do that right now and then write down what would be the W's that would happen in these three axes. So I am considering first the tractor. So let me consider the tractor first. The tractor WF plus WR, this is the, I said the sigma of forces equal to zero. This is the vertical forces that are acting. So WF plus WR, there are two vertical forces, the weight plus the hitch point. Then we will consider the breaking forces, the breaking force which acts in the, this would, we will consider that as the x as usual and the other one as the z direction and that as the y. So this is the x direction forces, the breaking forces. So CR into WF plus CR into WR, that is the coefficient of breaking is equal to as I said D by G into W1. Please note that in breaking F is in the opposite direction. So it should be plus into FHI right. So that is the second equation and the third equation that I am going to write is the moment equation in the y direction. So that is the set of equations for 1, 2 and 3 for the tractor. We have similar set of equations for the trailer, sometimes called as semi trailer because there is one wheel there. So anyway I will call this as trailer and again the same set or the same directions is what I am looking at, nothing difficult, the two directions that is the 4, 5. Then I said that I am going to take the tractor trailer combination and write down another three sets of equations. That is in the z direction, WF plus WR plus WS is equal to W1 plus W2 and then the total breaking force that is acting, those are what I would call as the breaking coefficients is equal to the, that is my F is equal to MA equation. Of course you know that D is the deceleration. So the signs are taken into account. Once I say that, that is the deceleration. So that is 7, 8 and 9. So these equations are now solved in order to get WF and WR. As I told you I am going to write down only the final expression. So the tractor, so I have to find out WF, WR and WS and I am going to just write down the final equation for this. You can derive that, the final equation for WF is a long equation multiplying, substituting and all that doing all this. I promise that there is going to be a long equation and it happens to be. Now we will write down WR, so obtained from these equations, WS, that means to be W2, that becomes 12. So we need to of course know CSE and CF and CR and so on in order to solve the equation. The key equation is this. Once you know this, you can go back to those 9 equations in order to find out WF and WR as well. So you need not go through this. You can as well from this equation, once you know WS, you can go to one of those 9 equations and you can determine WF and WR. So we will stop here just for you know assimilate this. We will quickly maybe spend 5, 10 minutes on the distribution of the braking force in the next class and then we will move to tire dynamics. So we will stop here for today.