 In the last class, we were looking at rollover or in other words, we started discussions on rollover. So, this will be our last topic on lateral dynamics before we move over to vertical dynamics. One of the important results we got yesterday based on the static condition of rollover is a condition where we said that a critical threshold for A y or even A y by G, this T indicates threshold is equal to T by 2 H. So, what is T? T is the track and H which is the height and very intuitive the larger the height, you know that the A y will or the rollover propensity is high and that larger the track, larger this one is, smaller will be this ratio. So, larger will be the propensity, larger the track, larger will be this ratio and lesser will be the propensity for vehicle to rollover. So, this we arrived at with a simple, very simple equilibrium of moment which we have taken about one of the tires. And remember also that when we talked about rollover, we were just talking about the what is the condition? The tire just leaves the ground. In other words, we said that the inside tire had or will not have any reaction force, it is just of the ground. We are not interested whether it will rollover further or whether it will really rollover or come back and so on. But from a design point of view, we look at a condition where the wheel just is off the ground, okay. Actually, we are, we are not very justified in only analyzing it from a static perspective. We have to do this from a dynamic perspective. As I told you in the last class, basically the dynamic perspective is also to get controllers which would control this rollover, right. So, in other words, you can write these equations and the dynamic equations which are governing, okay. You can do that, maybe you can write it even in state space form and then you can design a controller for this. Unfortunately, this course, we would not have time to go into the dynamic case because we have to move over to vertical dynamics. We hardly have about 7, 8 classes remaining, so we need to go and look at vertical dynamics. So, we will only concentrate on the static case. We also said that there are various factors which would change this T and H and in fact, I would suggest, I would say that the first thing that any designer would look at is that ratio, okay, especially if you are looking at trucks as I told yesterday. So, the first thing that you have to look at is what are the parameters or what are the conditions under which both T and H are affected. One of the most important conditions under which this H factor H is affected is by the rolling of the vehicle, okay. Now, let us concentrate on that. That is a very important factor and in fact, you would see how that is going to have an effect, okay or in other words, what are the additional terms that is going to come. Let us understand this figure. Now, we have two rolls, okay. Here, the sprung mass, okay, which rolls about the roll axis has a roll angle of alpha 2 which is superposed on the alpha 1 of the axle, okay. So, we have alpha 1, this angle through which this axle rotates as alpha 1 and then we have alpha 2 as the angle through which this sprung mass rolls, okay. Now, what we are going to do is very simple, but the equations are going to be complex because we are going to bring right hand side to left hand side, left hand side to right hand side and so on, okay. That is the only condition under which this is going to be difficult. The rest of it is going to be very simple. Alpha 1 is the angle, okay. Let me explain this alpha 1. I thought I will do that later, but let me explain that. Remember that when it rolls, obviously there is going to be a difference in the load that is carried by on the say outer and inner or called as left hand right here or you can call it as outer and inner, okay. So, there is going to be a difference in the load. If there is a stiffness to the tire, let us call that stiffness to be CR, okay. Then the deformation of the tire is given by this being the stiffness of the tire or I can call that as KT if you want. I am, as I said, I am following George Rill. So, I am using the same thing. So, if there is this is the stiffness of the tire, the deflection of the tire, okay, is given by the load that is acting divided by this stiffness. Now, since these vehicle rolls and there is a distribution of loads, okay, the deformation of this tires are not going to be the same or in other words they are going to be different and that when there is a propensity to roll over, in other words the whole of this tire lifts off the ground and no reactions taking place, then all that load actually comes to the start. In other words, if FZ left is equal to FZR is equal to M into G by 2, okay, the weight, the total weight that is acting M into G by 2 is a usual case, this steady state running condition. Then if under the threshold condition, this whole thing is going to be M G, which means that the deformation here is zero of this tire because it is not going to be deformed due to the load and the deformation here is going to be FZ going to be M G with the result that if this is the ground, I am just exaggerating that figure. So, the tire would be like this, okay, because this tire would be compressed at the road being the same. So, there will be, so one tire has no deformation, the other tire has deformation, so the whole axle is going to be rotated and that is what we called by alpha 1. So, that is why I said that we are looking at static condition, we are not looking at dynamic case, okay. So, if you, so there is no time factor involved, it is the first thing I said, right, okay. Listen carefully to what we say. So, this is, there is no time factor involved, we are looking only at the static case. So, alpha 1 is that deformation. So, there are going to be lot of, in any of these derivations, there are going to be assumptions. So, these assumptions take shape as we go step by step in the derivation. So, it will be careful, you have to link that with the derivation. So, that is the alpha 1 which we are talking about. Of course, alpha 1 and alpha 2 are going to be small and we are going to use that as well, that both of them are small, okay. So, let us look at what is that we have written, F is at R, okay, into T by 2 that is we take the moment of the center into T by 2 is equal to write down the moment due to MA y and as well as due to MG, okay. Note that this is H1 and that is H2 and note that we are talking about small alpha 1 and alpha 2. So, we can write down that to be M into Ay into H1 plus H2, okay. This is in the anticlockwise direction. Then you have one more term which is again in the anticlockwise direction plus MG into H1 plus H2. You split this up because this alpha is up to this point alpha 1, so into alpha 1 plus the other one is H2, H2 into, right. So, as I said the threshold, when I put T, it is threshold. So, F is at threshold, we have put that left is equal to, okay. So, we will write this for the threshold condition which means that MG into T by 2 is equal to M into Ay threshold into H1 plus H2 plus MG into H1 plus H2 into alpha 1 threshold plus H2 into alpha 2 threshold. That means it is about to roll over, right. Remove this condition, I mean these things. We will rearrange it and we will write Ay threshold. So, this divided by G, okay, is equal to T divided by 2 into H1 plus H2. This is the first term, right. And bring it to the other side minus, since G has been taken over that G is out there. So, in other words, okay, minus H1 plus H2 is, I have divided that. So, that means that is gone minus H2 into H1 plus H2 into alpha 2 threshold, yes. Sir, H2 is height of roll centre, right. From the roll centre to the CG location. H2 is, this is the CG location. That is the roll centre, okay, about which it rolls. That is H2. In other words, I was just going to comment. In other words, H1 plus H2 is our H in the previous derivation. So, T divided by 2H is actually you have got, okay. And that is modified by two things. One is alpha 1 and the other is alpha 2, right. So, if I neglect this roll, I get back to the previous derivations, clear. Now, my job is now to find out what is alpha 1 threshold and alpha 2 threshold, okay. So, we already know what alpha 1 threshold is. So, we said that this is going to be a difference in deformation, okay. Let us call that as delta R because there is a radial deformation, okay. One is deforming, another is not deforming. So, what is delta R? Delta R is Fz. Please note that we are collapsing the front and the rear tyre by one tyre and so that is equal to Fz by the CR, okay. And that angle alpha can be written as alpha 1 threshold is equal to, what is this? This is T is equal to delta R by T and that is equal to, what is Fz? Now, mg because the whole load is taken there. So, mg divided by CR into, clear, okay. So, what is my next job? My next job is to find out what is alpha 2, okay. Alpha 2 is promoted by what is called as the roll stiffness, k phi which is the roll stiffness. We already know how to calculate the roll stiffness from the spring, okay. Half into ks into s square, right. In other words, this s is equal to T. So, you can, T and s are the same. You know, s is approximately the same. I should not say that they are the same because that s is actually that length. When do they become the same? When it is commercial vehicle with the suspension springs, okay, which are basically leaf springs and when they have rigid axle, then the distance between the springs, okay, it becomes s and if it happens to be an independent suspension, then s and p are the same, okay. Note that carefully and that we had already seen. So, we will now write this whole thing in terms of roll stiffness, clear, okay. I think let me, let me put on some numbers so that I will, I will remember these numbers and my most important equation is this equation. So, let me write that equation here and not remove that so that I will, when I want to substitute it, I would use that later. What essentially am I going to do in this? Very, very simple. I have already calculated this. I have to calculate this, substitute it, I mean rearrange it and get you one good picture. That is it. That is the most important equation which we will use it after we get this alpha 2 threshold, okay. Now, how am I going to get alpha 2 threshold? K phi into alpha 2, okay, that is the reaction force into alpha 2. Let me, is equal to, what are the things that go into this? M into A y, okay, into multiplied by h 2. Again, look at the assumption, I am making that they are small angles, cos of that angle is approximately equal to 1, sin of the angle is equal to that angle, all those things I am putting. So, be very careful in whether you are going to work in degrees or radians, you have to be very, very careful because when you make that approximation, okay, you put that in terms of radians. Plus M g into h 2 into alpha 1 plus alpha 2, okay. So, since we are looking at the threshold, let us call that as T. I have an expression for alpha 1 T already. So, I will just keep it, I will substitute it later, all right. Now, what are we going to do? We are going to rearrange it, okay. Let us see how we are going to rearrange it. So, I am just going to bring that alpha to the other side minus M g h 2 into alpha 2 into alpha 2 T is equal to M into A y into h 2 plus M g h 2 into alpha 1. In other words, alpha 2 T is equal to M into A y into h 2 divided by k 5 minus M g into h 2 plus M g into h 2 into alpha 1 divided by k 5 minus M g h 2. Which one? This one. What is this equation? Balance equation. So, this is the action-reaction equation about which point? Obviously, about this roll, okay. Well, there is a roll stiffness is k 5 into 5. We wrote this already. Remember that we wrote this equation already. In your previous class, go back and look at your notes. We had written this as k 5 into 5. Remember that we wrote that as k 5 into 5, the moment balance equation. That is exactly what we are writing now, okay. So, now what is my next step? I am going to substitute this into my, let me call that as star and I am just going to rearrange it. That is all, right. So, let me write that equation A y. That is a grovelling task of simplifying this expression is equal to T by 2 divided by h 1 plus h 2 minus alpha 1. I had already written there M g by C r into T. Correct me if I am wrong if I am going to make a mistake because there is going to be lot more minus h 2 into divided by h 1 plus, sorry, h 1 plus h 2 multiplied by the first term, put it like that, M into A y threshold into h 2 divided by k 5 minus M g h 2 plus M g into h 2 divided by k 5 minus M g h 2 multiplied by M g. So, what I will tell you? So, M a y and h 2 will have an alpha 1 plus m g into h 2 divided by k 5 minus M g h 2 multiplied by M g. No, this is, see the whole, this is attached to this body only. So, this is the role is not, that is why we had written alpha 2, the role is only with respect to this and then this whole body, you know, with respect to alpha 1 plus alpha 2, okay. We will discuss that later. We will, let us finish this first, okay. We will, let us finish this. Let us, let us, because this is going to be lot more things that we are going to do and then we will discuss it, okay. So, now, let us just rearrange this terms. Take a minute to rearrange it. What is that I am going to do? I am going to bring this to the left hand side and rearrange this term. So, let me, let me do that into 1 plus, what is that alpha A y? The first term, 1 plus h 2 divided by k 5 minus M g into h 2 divided by h 1 plus h 2 into M into h 2 divided by k 5 minus M g into h 2, right, which is the one? So, g is not in the room. So, I have to put a g there, right, okay. And then the rest of it in the right hand side. So, what are the terms? The first term is there, t by 2 into h 1 plus h 2 minus M g into C r into t, right, okay. Now, take a minute to rearrange this. By the time, let us look at the question which he had asked, okay and let us look at what, what is the question? Yes. Otherwise, M g would pass right through the point where we are going to go. Yes, of course. No, no, no, no, no, no, no. We already saw yesterday in the last class, M g is contributing by, how? Even if it does not rotate, okay, that is the restoring moment. You remember last time, we saw that how did we take the moment balance? No, no, whatever it is, it is not a question of taking the, it is participating, okay, this this and this. So, what I am what we are trying to say is that there is one moment which is due to the centri-fugal force in other words D'Alembert's force which is trying to rotate it in one direction and then this is going to rotate in the other direction. So the equilibrium is because of these two forces. It does not matter whether we took it along about this or not because that is how it is going to rotate. Now that one minute now that we have taken it about this okay. Of course, mg is the contribution from mg comes into picture because of roll. If I had taken this one about this point obviously obviously no no that is exactly what I am trying to say. This mg would have acted here and again I would have had mg into T by 2. So, that would not go anywhere. So, the question is whether you take it about this and then use this mg in order to compensate or in order to get back the system to equilibrium or you take it from here. It does not matter. Mg H2 into alpha 1 plus alpha 2. M a y into H2. Yeah, alpha m a y into H2. So, what would be the what would be the compensation? Is that the is that correct? Is that okay? Exactly. So, cos of please note this is exactly that is why listen to every question. Please note that what is actually you should multiply this with cos of cos of. So, this I exactly said this. This is sin of alpha 1 plus alpha 2. This would have been cos of alpha 2. So, I said that that is equal to 1 sin both of the angles are small. So, when I multiplied by cos see this actually what is that you do? You this horizontal distance is what you are going to take. The horizontal distance is cos of this angle cos of that angle is equal to 1 we take. So, actually this is multiplied by 1. Okay. So, so the angles themselves have no meaning. So, it is a sin of that in fact that what I should have written it written is to write this as cos of alpha 1 plus alpha 2 then sin of this then make that is equal to 1. This is equal to that is why I also made a comment that can you do that with degree? So, you have to do that with a radiance. No, no that is the whole assumption fundamental that when alpha is small, when alpha is small okay. What is the assumption that you usually make cos alpha is equal to 1 and sin alpha is equal to alpha okay. This is the assumption we have made here. So, that is why I said this derivation see there are two things in this derivation. One is I know this when you ask the question I know this I did not want to confuse it in the middle because I wanted to go through this and then come back because I had already made a comment that I am going to make all this alphas very good that you made this statement. So, that is why I said that there is a difference between concepts in every line and algebra okay of taking it to the right hand side dividing it okay rearranging it all those things as the high school stuff that is different but what is important is what we make assumptions here. Now the question that the more important question you should have asked is how do you know alpha 1 and alpha 2 are small okay that is the concept. How do you know alpha 1 and alpha 2 is small? Very important question. In fact this is the question first I asked you know we had one of the consulting work which we did this was a problem and I in fact interviewed a driver who had a big accident you know the guy escaped unheard in rollover okay. So, I wanted to have a feel you know because you can measure it of course there are things measure it but this was about a decade ago when we did not yet you know understand lot of things about rollover. So, I went and interviewed the driver he told me that this is in the the desert area Rajasthan. So, he said that he did not even feel the vehicle is going to rollover he said he was going very slowly according to him okay a turn a turn he didn't he said that if I had known I am rolling okay I would have stopped it but I didn't even know that there was a I was rolling and and the vehicle just turned over or rolled over. So, usually it has been measured also that these alphas are small and that is why we make that kind of assumption that cause alphas 1 and sign of that I am going to do one okay clear get back and rearrange this term see in every problem in vehicle then we are going to see that in the next derivation also this is going to be so much of assumptions basically that is also because when we work out with hand the first cut methods we cannot use very sophisticated you know methods of course when you design it in detail design comes you would do that in a much more sophisticated fashion okay rearrange it and let me write down that I am going to leave this to you because I know you are also bored I am also bored to rearrange that these terms and I will write down the final expression right. So, that is a much more interesting stuff then doing that t by 2 divided by h 1 plus h 2 plus you know from where it is coming okay h 2 divided by k 5 star minus 1 minus 1 divided by k 5 star or C r star just a second let me check that C r star work this out where we call this k 5 star to be k 5 divided by N g into h 2 and C r star is equal to this we use that C r right C r divided by N g okay rearrange this terms and you get this expression I am leaving this as an exercise for you to do that. So, that is the expression so when I neglect these terms okay I go back to my t by 2 divided by h 1 plus h 2 which is the height of the roll set from the ground okay. So, both of them are going to have an effect okay both the tire stiffness normalized tire stiffness I would call it and the normalized roll stiffness both of them okay has an effect on this roll. In fact in the book there is a small problem is very interesting problem you can maybe you can try it out okay heavy truck it is an heavy truck twin tire axle twin tire axle is loaded it is a 13 tonner what we call in India is a 13 tonner truck. So, m is equal to 13,000 okay the radial stiffness of the tire C r is 800,000 Newton per meter t is equal to the track is equal to 2 meters h 1 is equal to 0.8 and h 2 is equal to 1.0. So, now suppose now I calculate from these values the roll angles and as well as these values of A y and so on this is clear any questions very straight forward it is nothing very difficult about this you do that you what I suggest is this is a an exercise which I would like you to do just plot I am not I am not going to do that here say for example, if you plot this versus C or K phi versus the A y overturning limit how would the graph be and if I plot the K phi versus roll angle how this is going to be okay just plot it okay that will be an exercise for you. So, what happens when the roll stiffness increases calculate that so that is why you have is there an effect I mean the questions I am going to ask and you are going to answer them is if I have an anti roll bar put okay for example I increase the roll stiffness okay how do I is there is there going to be an effect okay, but I how do I realize whether that is going to stop my tendency to roll over this is a very important question because many people think for example many people who build this the truck I mean the car carriers they think that just keep on increasing the roll stiffness okay would stop them from or prevent the roll over of these trucks okay. So, I had this again from my experience I see that a lot of people think that just increase the roll stiffness it would not roll at all no in fact if you plot this graph you would notice that the graph would be yes it will be small there will be an effect after some time there would not be any effect of roll stiffness okay and the roll angle would actually come down like that and after some time there would not be any effect because of the contribution of these terms okay do that it is a good exercise to do and you will understand about the roll stiffness and so on okay. So, we will now switch gears I am we are running out of time we will move over to any questions good question what do you think will be an effect where see one of the things in handling is a good question one of the things in handling is that you should also look at the effects of other things okay this is roll over is one thing right what are the other factors what are the other things that will have an effect on rolls I mean effect due to variation of roll stiffness yeah so yes correct but exactly what what is the effect due to load transfer so in other words the roll stiffness is going to have an effect on roll load transfer so load transfer is going to have have an effect on slip angle or in other words it is going to have an effect on understeer gradient okay understeer gradient these the slip angle the delta is the result of variation of this understeer gradient which would have an effect on alpha f and so on. So, in other words okay we cannot de-link every aspect though for in order to study okay I say that handling you know it is over I am going to shift to roll roll roll over now I am going to say that this all over I am going to go to vertical dynamics does not mean that they can be looked at in compartments right fine. So, you should think about what happens it is a very good question I am not going to answer it now what happens when do I put it what happens when I put maybe I will answer it in the next class but I want you to think about it what happens when you put the front roll stiffness is higher for example roll bar in the front of a front wheel driven car okay when I put it and when I if I change it and put a roll you know stiffen on the rear what happens okay think about it we will answer that in the next class because I let us quickly get into what is called as vertical dynamics let us let us begin with vertical dynamics okay now vertical dynamics has a very important to roll to play in what is called as ride okay sometimes even people call this loosely as ride dynamics in fact there are two things that become important or we study vertical dynamics from two perspective one is to understand ride and the other is to understand what is called as road holding we want the tire to hug the road all the time okay obviously if the tire loses contact with the road okay we know that our grip is lost and stability is affected in that way actually vertical dynamics you cannot be totally uncoupled from the lateral dynamics but for for the purpose of studying this we are going to decouple it and we are going to study that vertical dynamics okay separately right so we are going to look at both ride and road holding these are the two things that become important in vertical dynamics clear this is totally frequency driven right in other words this again a lot of confusion to look at what is the what is meant by vibration okay and what is noise because when I go on a road not only I there's a vibration that I feel but also a noise which I hear right both of them are there so we usually distinguish okay between in in ride dynamics we classify it into what is called as the primary ride and what is called as the secondary ride primary ride is from up to 5 hertz maybe 1 to 5 hertz and the secondary ride is from 5 to 30 some people put it as 25 but 5 to 30 hertz so when we talk about vibration of the vehicle our frequencies are in this region so frequency of vibration becomes very important not only because of the yeast with which we are going to study it okay but also ultimately our aim is to understand the effect of these on us humans who travel in the vehicle so in other words what is our tolerance level for these vibrations okay is there a relationship between the frequency and our tolerance level okay how are our fatigue you know we develop a fatigue in the physical fatigue and we drive and so on so all these things become important okay for example you would have heard about sea sickness so when you when you go in a ship okay people develop what is called as sea sickness why is it because the frequency of excitation okay of the ship is about 0.5 hertz which we cannot tolerate unless we get used to it we cannot tolerate but there's a very comfortable frequency okay where we do not feel sick at all so when I walk I can keep walking and I mean I feel tired but I won't I won't feel the nausea so there is again a frequency okay 1 hertz okay where we are comfortable and we walk so there are frequencies where we feel totally sick frequencies where we are comfortable where frequencies which are which when we are exposed to say some so many hours we feel comfortable and so on so in other words the frequencies are related to our well-being so we are going to study that the frequency becomes important okay so the other thing is is this we are going to do a very linear analysis one of the questions first question is that what are we what is that we are giving this as an input okay how are we going to study model this obviously our input is the road okay our input is a road okay how am I going to the road is random totally how am I going to characterize the road okay I know all of you know what is called as fast Fourier transform or FFT I know that you know there is a link between the time domain and the frequency domain and that probably a few of you at least are familiar with Fourier series where you had looked at say periodic signals and the sub and the connection between the time and the frequency domain so we are going to use a few concepts from FFT okay or in other words the results what FFT can give and we are also going to look at in this course this whole system of vehicle as what is called as the linear time invariant systems okay so in other words we are again look at that you know assumptions are very important listen to these assumptions carefully because these assumptions may not value b value okay under certain circumstances so one of the first assumptions I am going to make one of the first assumptions I am going to make is that the stiffnesses what we are considering are going to be linear okay in other words my superposition principle which I am going to apply okay by splitting this or by summing up using what is called complex exponential functions in simple words science and cost is valid only if I consider my system to be linear so that is the next assumption I am going to make right fine so with that as we go along look at all our assumptions with that let us start or let us see how we are going to attack this problem we will go into the full derivation in the next class what we are going to do is to consider what is called as the sprung mass okay which sit on two suspension systems okay which is we call that as the sprung mass and these two are the unsprung mass and that that is the tire okay and that is the road input we will consider this system we will go into the details okay first derive with what is called as four degree of freedom system then simplify it into two degrees of freedom system and understand these in other words simplify or split it into two systems okay come to what is called as a quarter car model okay and look at what we can get out of this quarter car model how do I now tune the suspension and that is going to be our major issue here we give we will give an introduction about the road its statistical characteristics but we won't be go we able to go beyond that we will look at these things the vibration and so on in a more detailed fashion in our next course and maybe a course on NVH would give lot more inputs into it so we will stop and we will continue with this model in the next class