 In the last class, we were looking at what is called as the half-carb model and then we reduced it to what is called as quarter-carb model, okay. We reduced this to what is called quarter-carb model. Now, let us understand this. There are, yes, there have been lot of equations on it. Let us understand what we are trying to do now with the quarter-carb model. As I told you in the last class that you can have a full-carb model, but to quickly understand the effect of the suspension system, okay and two of the very important frequencies that go with the system of suspension, sprung mass and unsprung mass as well as the tire of course, okay. We are looking at this quarter-carb model. In actuality, we have to consider the tire, sorry, the road which gives the input, how road is represented, how that goes as an input. All those things are important and we are going to see that from the next class onwards, okay. Next 4 to 5 classes, we will understand the road, the statistical nature of the road, how road is represented, how that affects the performance of the vehicle. All those things we will do that from next class. So, we will quickly finish some of the things that we want to do in the quarter-carb model and then we will go to half-carb model and finish that part as well. I hope we will be able to complete it or else we will, maybe it will spill over. So, what is the quarter-carb model? That is the, what is called as the M sprung mass, that is the unsprung mass, right. If I change anything, just tell me, I hope I use the same notation. If there is any difference in the notation, tell me, okay. This is one of problems like and so on, right. So, this is the road input R of t. This we called it as y and that we called it as z, right. I think that is why we left. Now, let us write down the equations for this. As I said, this is cut into two halves, the front and the rear and we had put M s to be the distributed mass between the front and the rear from the total sprung mass, which we called as B by L. This is in the front and for the rear, we said that it is M s into A by L, right. So, the system is the same. You have to now rewrite or replace M s by the corresponding front or rear, springs by front and rear and so on, okay. In other words, we are, in the quarter-carb model, we are delinking both, right. Sometimes we will analyze it with complete M s, okay, as one vehicle and the suspension and so on, you know, as this one whole vehicle as well, okay. Now, let me write down the two governing equations because why two? Because I have two degrees of freedom, which is y and z. So, M s into z double dot is the first one, okay, which is, we had already discussed this, C s into z dot minus y dot minus K s into z or z minus y. Same fashion, I can write down M u s into y double dot, okay. So, write down, yesterday we did that, write down the forces that are acting, okay. So, it will be minus C into, it will be in the opposite direction. So, y minus z minus K into y minus z. These are the two forces that are acting, okay and then the tire force minus K into y minus, right. These are the two forces that are acting. Now, we can do two things. As I told you, we are going to look at, that is the input of course, look at it as a linear time invariant system, okay. R can be expressed in terms of the sinusoid or an exponential function and so on and so forth, right. So, what we are going to do to follow this kind of system approach, we can find out the eigenvalues and the eigenvectors and then just express the result as a sum of the eigenvalues and the eigenvectors. So, that is what we are going to do. I think we did that already yesterday, yeah, we did that, right and did we write down z by r, you know that also we wrote down, okay, fine. I think I, is it by r and y by r, both of them we wrote down, sorry, yes. So, we wrote down both of them and so is equal to magnitude is equal to KT root of K squared plus CS squared omega squared divided by D squared plus CS squared omega squared E squared, okay and y by r is equal to KT into root of KS minus MS omega squared whole squared plus C squared omega squared by D squared plus C squared omega squared E squared, okay. So, this I think this is where we stopped, right. We had written that already, right. So, the other one, this is the displacements, the ratio of the displacements z and y due to r, right. The other one is actually what is called road holding, okay. Road holding is the force that the road exerts or the vehicle exerts onto the road. So, how do I express the road holding? What is road holding here? What is the force? KT into r minus y, okay. So, that is the force that is applied. Please note that all of them are perturbations about the static equilibrium, okay and since that it is the equilibrium due to the weights and so on. It is about that, that we are oscillating. So, that when I substitute it, rearrange it and so on. So, let me call that as how I called it H. So, let me call that as H into e power i omega t that is the force. So, that H divided by KT into r, okay is equal to omega squared root of m s m us omega squared minus Ks into m s plus m us whole squared plus c squared omega squared m s. The techniques are simple. The only thing is that the problem is you have huge, I would say equations are very long. That is because of the number of terms that are involved. So, apart from that they are not very difficult to understand. I suggest that you derive it because of the lack of time I am not going to do that. Please derive it and check how this equations are obtained, okay. So, most of the equations which I am going to express now I am not going to derive it because all of them are just algebra, okay. Bring it to the left hand side, bring it to the right hand side, put them together, divide it, you know those are the things. They are not very conceptually very, you know difficult to understand, right? Okay, yes. The damping, usually the damper effect is not very high for the tire, okay. Absolutely, absolutely, but those effects go to rolling resistance. Those effects are for rolling resistance, okay. It is a very difficult topic. See sometimes what we do in engineering is that when it is very difficult to handle, make an assumption that it does not exist, okay. See damping itself since you asked this question I want to comment on this. Damping is actually not a very easy topic to handle, to understand especially in the first course like this. Damping can be classified into what is called as proportional damping and non-proportional damping, okay. And the proportional damping in the sense that the damping, I am sure you had the background and vibration. So, you talked about mode shapes, mode superposition. You are going to see that again in the next course. So, where we had very nice matrices, okay, when I looked at the mode shapes, okay and they are orthogonal and so on, right. We actually, we did what we call as coordinate transformation and then we had those nice diagonal terms. We uncoupled the differential equations and so on, right. So, that is the beauty of a proportional damping where C is equal to say alpha times m plus beta times k. When I write like that, my C matrix behaves nicely and then I am able to get very similar expressions because C is now split into m and k, okay. I will go, I will take some of them to m and some of them to k and then I get a nice matrix, okay. If the damping is not proportional, in other words, if I am not able to write C is equal to alpha m plus beta k, which is the case actually with tires, non-proportional damping, then the way I have to handle becomes difficult, okay, becomes difficult. There is a way of handling it. It is, I mean it is quite an evolved ideas but not here in this course. So, people feel that the best way to handle this is, what difference does it make? If it is not much different then I will remove it. In fact, you would notice that in the next step I am going to do, I am going to look at the natural frequencies and even in that case I am going to remove the damping in order to calculate the natural frequencies of an undamped system, okay. Does it mean, especially, you know, the more important damping is this, the shock absorber. Does it mean that the shock absorber does not have any effect, okay? How, when I remove it, fortunately the difference between the natural frequencies of an undamped system and the damped system, the differences are very, very small, okay. Because though in the whole of mechanical engineering, this system is a very highly, one of the very highly damped, okay, within codes under damped system. It is a very highly damped system but still it is an under damped system, okay. So, the difference between, we are going to see that, the difference between the natural frequency of an undamped system and a damped system will be small. So, when I want to calculate the natural frequencies, okay, which looks like every, every mechanical engineer has such a good hold of it, okay, understands it. So, it is very easy to now calculate the natural frequency if I, if I do not have an undamped system. Fortunately, but I cannot say the same thing with respect to the motion or mode shapes. If I, for this motion rather, okay, the ratio of Z versus Y, I cannot say that because once I put this damping here, okay, the amount of motion, for example, in a typical car, if I do not have a damper, the ratio of motion will be 80 times for, say, for example, for omega 2. But if I put a damper, then the ratio of motion comes down to nearly 10 or in other words, there is a huge or nearly an order of magnitude difference in the ratio of that is Z by Y, okay, the ratio of the displacement or ratio of motion will be huge. So, if I want to calculate simple only natural frequencies, that is fine, I can do that. But if I do not want to calculate natural frequencies alone, I want to look at other things, then damping will have an effect, okay and so on. That is what we are going to see now. But in the case of tyre, okay, it has been a tradition to remove it. So, there have been, in fact, in fact, there have been very interesting papers which are now questioned. For example, tyre noise, what is, what are the material parameters which have an effect on tyre noise? This is one of the, you know, say, issues. Stiffness has an effect, does G double prime has an effect and so on. Now people are re-looking at it. People said that initially in the 90s, there were papers which said that the materials do not have an effect, but now there are papers which say that the material has an effect on noise and so on, okay. So, in simple words, this is a very traditional approach, okay. In fact, tyre damping itself, this is our friend here, his research is on tyre damping and it is not a very easy thing to calculate. What is that? You are getting 2 percent, right. So, it is not, it is a PhD topic. He is looking at how tyre damping can be calculated, how tyre damping can be measured, you know. In fact, he is doing all that, okay. It is a topic which is of interest, which I would say is not at this level of vehicle dynamics, since an introductory level, okay, fine. Let me get back. So, we are now looking at, so that is the two things and now we will look at a situation where I have to now, given the sinusoidal input, I know how to do that. But I have a situation where I have to actually say tune the suspension. How do I optimize the suspension? So, in order to optimize the suspension, okay, I have to understand how the system behaves with respect to omega or in other words, excitation, okay. So, I have to know how the system behaves with respect to excitation. Why is it? Because the road consists of a number of frequencies. So, ultimately, when I use whatever be the technique that we will discuss it later, to split this up into a number of frequencies of excitation, I have to understand that the C value effects are the C value effects are not the same at every frequency. This you already know from elementary vibration classes, okay. I am going to exploit that and you will see right now that is not going to be a very straightforward technique. Before deriving, okay, let us look at the result. Yes, that is not the right way to do it, but this is just to motivate, okay. Now, why am I deriving it? That is why I am plotting this here. Suppose I now plot, I do the derivation of a free vibration system where I find out what would be the Z, okay, or transmissibility. Suppose I plot omega, the excitation frequency versus normalized Z, which I would let me plot it as, how do I plot it? Omega squared Z r, okay. That is a normalized plot, okay. When I plot it like this, then my plot is going to look like this for C is equal to 0, okay. At 2 places, you are going to have difficulties and those 2 places are the 2 natural frequencies which is omega 1 and omega 2, okay. Those are the 2 places where you are going to have trouble. Now, when I introduce damping, introduce damping, then this curve is going to change. Obviously, all of you know that this would not be infinity, okay, and it will come down, right. Now, so for example, at one level, it will come down like that and it will be like this, right. When I now increase damping, it may come out like this, okay, and then pass through the same point, it may increase, okay, pass through the same point and it can be like that. In other words, in other words, without damping I have 2 peaks and with various values of damping, I have curves where it will be bad, it will be good and so on. In other words, depending upon the frequency, my Z value would now vary or is actually the amplitude of the road excitation. Now, in other words, a good damper in one frequency, you cannot say that it is good or it will dampen out in another frequency, okay and so on. Higher damping, sorry, lower damping is good in one, okay, bad in another. Higher damping is good in one, bad in another and so on. Under these conditions, damping optimization becomes difficult. How do I, how do I dampen out, okay? Number one, that is why you have techniques where the damping varies with the frequency of excitation and that is in the realm of controls, okay. But every car does not have it, it is still, a lot of things are still in the, what you call as experimental stage. So, given a choice, what will you do then? You do not know, you do not, I mean, what frequencies are going to be excited, maybe all of them. So, what would be a choice? The best choice would be that I have C as flat as possible, okay. Is it a correct choice? No, but as a first cut, you can look at C as one which is almost flat, which is not that it is high at one place, low at other place and so on. So, in other words, optimizing the suspension, especially the shock absorber is not a task which is very obvious and trivial because of this kind of graph where, okay, its behavior is not going to be the same. One compromise results in an increase at other place vice versa, okay. In order to understand, this is clear. We are now going to see how we get this graph. That is going to be my derivation, right. So, with that background, let us quickly run through this. This stuff which you know already, so I am not going to spend a lot of time on this. You have done this already for say, a single degree of freedom system, right, where you know that there is a crossover point, okay, where the damping characteristics are different on either side of the point and so on. So same thing happens here, okay. Hence, we would run through this whole thing. So, for the, so we are looking at now and first let us look at, okay. We had finished this part. Let us first look at undamped quarter card model and it is very simple to arrive at what we call as the natural frequencies, okay. Follow the same thing. You have the equation, remove C, get the determinant, expand the determinant. All those things are very straightforward. You can do that and so you will get that ultimately. If you have any doubts, I will answer it but I think that is a known thing and hence, so that is a characteristic equation for the undamped system. Go back to that equation and then get this, right. If we solve this equation of course for omega and call that as omega 1 squared and omega 2 squared, okay and this results in Ks. That is the expression for omega 1 squared and omega 2 squared. I am not very happy with that expression. It is too long, okay. Let us do some simple tricks, immediately get the two natural frequencies without much. As I said, we keep on assuming, conscious of the fact that my assumption will not affect my results, okay. This is the standard technique, fine. But can someone suggest how else can we do it? One is to do a mathematical degree there. For example, the first thing we observe is that this prong mass is far, far greater than unsprong mass and the tire stiffness is far, far greater than the Ks. These two assumptions are not assumptions, they are facts of life. This should be Ks. I will just check that. I think that is Ks, okay. I will just check that. Next class, I will get this thing. So, these two are well known, right. Now, let us get some facts clearly. So, there are now two natural frequencies, okay. One, I would say corresponding to the body and the other corresponding to that of the unsprung mass, okay. Not that they actually participate but if you look at the ratios, it is the ratio of excitation, okay. We can assume as if that one is fixed and the other is exciting and so on, right. A typical frequencies, we will see in a minute, can be calculated by a very simple assumption here. So, what is the assumption? What do you think of this assumption? A very simple assumption here. The first thing, first is with respect to the body, okay. Since it is an undamped system, I am going to remove this and that, this mass is greater than this, okay. I can neglect it. So, I can actually look at it as if there is a mass, small mass here, okay, as if it is going to be like that and two springs in series with a mass ms and so that omega 1 squared is written as k1, k by m, okay. So, this is k spring, k tire, k tire into k spring divided by k tire plus k spring into ms, okay. So, this is a very straightforward way. In fact, you can achieve that by an approximate, say for example, you can do a Taylor series approximation here and you will get a very similar result, right. So, the other omega 2 is arrived at by assuming that the other end is fixed and you have this mass mus or m unsprung mass, okay. This is not in series because both of them are going to take the load so they are in parallel, okay. So, that omega 2 squared can be written in terms of k by m which is k is, in this case is kt plus ks divided by mus, right. So, that gives me the very simple one-minute answer for undamped natural frequency with all the approximations that are put in place, clear? Okay. Yes, we will look at some similar, I mean, typical results, okay, just to understand these factors. For example, for a typical car which you are taking from one of the textbooks is, say let us say that unsprung mass is 100 kg, the sprung mass is 1000, ks is equal to 70, kt is equal to say 560. If I calculate now omega 1 squared, so in terms of cycles per second hertz, so F1 happens to be 1.25 and F2 happens to be 12.64 and at F1 is equal to 1.25 and at F2, these are typical values. This omega 1 which is hertz of course, 1.25 is a typical value and usually is between 1 to 1.5 for all the cars. This is between 1 to 1.5 for all the vehicles. That is what we call as the body frequency, right. And omega 2 which results in a frequency of 12.64 hertz is a typical value and is called as the wheel hop frequency of the typical value for the unsprung mass. So, this is a typical wheel hop frequency, okay. So, look at that results. Yt for the second case, look at that 89 or 90 times Zt which means that the Z is almost, you know, negligible displacement and hence, in other words, you can assume as if this is stationary and this guy is jumping. But this is for an undamped system. That is what I was telling. If I damp the system, okay, this would come down to say minus 10 or minus 12 and so on. So, this may not vary much. This would vary minus 10 or minus 12. Sir, why is that minus sign? Yeah, why is there a minus sign? Because these are mode shapes. So, obviously, okay, the way this, you know, when they oscillate, it is oscillating like this, okay. So, that is how this, the mode shapes are for this, okay. Fine. We had already derived the expression for the other cases and we had, that is approximately the omega 1 and omega 2 what we just now saw, okay. You can draw this graph from whatever we know. As I said, optimization of the suspension system is not a very straightforward problem. But one of the earliest attempts to do optimization in the 1950 is by just saying that, look at this graph for various C. Find out that C for which the curve, this is a, say, let us say this is a point A. All these graphs passes through point B, point C and so on, okay. It so happens that looking at the graph, people said that if the slope of the graph, the slope of this curve happens to be flat at A. In other words, slope is 0 or dy by dx is equal to 0 at that point. Then this curve is flatter almost throughout the range of omega. So, this is the first condition, you know, very early, okay, first condition that they had put. In other words, when they wrote dou of omega squared Z, of course, R is what is given as a constant, this is equal to 0, okay. This is the first cut optimized C. So, the result is that C optimized is equal to root of M s into K, K s divided by 2 into root of K t plus 2 K s divided by K t. Yeah, yeah. No, no, these are the natural frequencies. These are the natural frequencies and that frequencies for which the mode shapes are calculated, okay. These are the two mode shapes for that frequency. That frequency of oscillation, what is the motion, movement of the unsprung mass and the sprung mass, that is what we are expressing here. No, no, no, this is, we are talking about, I mean, 0, right hand side is equal to 0. This is the free vibration case, okay. If it were to vibrate in that frequency, okay, then what would be the relative motion? This is relative motion. Look at that carefully. This is the relative motion between the sprung mass and the unsprung mass. Clear? So, there is no input here. No? Yeah. So, this is, see, mode shapes are usually expressed as a vector, okay. So, you would say z, y or z, theta and so on, right, as a vector for a particular frequency say omega 1, right. I am expressing that as a ratio, z by y. Now, normalizing it and expressing it as a ratio, if z happens to be 1, y is equal to this and y happens to be, z is equal to 1, y is this and so on, right. So, it is just as a ratio I am expressing the, what we call as the mode shapes. Clear? So, this is an optimized one. I am not very happy with it. Hence, we have to move further away from this, okay. So, in other words, I have to look at the statistical nature of the road and see whether some optimization can be obtained from that perspective. As I said, first cut, this is fine. Now, I am going to shift gears. I am going to quickly run through a case where I am going to go back to my half car model, again make some assumptions and look at the, from the quarter car I will go over to the half car model. So, what are the lessons learned from in the quarter car model? The lessons learned is that two of the important frequencies, okay, which are not going to change. In fact, we have done extensive testing with many cars in India, okay. All these cars, the natural frequencies when I put an accelerometer at the axle position, the body position and all that, all these cars, you can very clearly see this Z and Y, okay, have the natural frequencies between 1.2 to 1.5, right. So, that is the natural frequencies which we have and the natural frequencies or the wheel hop frequencies is between 11 to about 13.514. You know, this is the range in which we have got this frequencies, okay. So, these are two important frequencies, wheel hop and the body frequencies, okay. We will see this importance again in the next class and let us look at what is called as the half car model. I am going to simplify this half car model, okay. I am not going to derive it again completely because I hardly have 4 or 5 classes. I want to shift to the statistical nature of the road. When in doubt, simplify. That is all, okay. I am going to look at the way I am going to look at the half car model. I am going to have, that is all is my model, okay. I am going to remove the unsprung mass, okay. This is going to give me some important inputs. Remove the unsprung mass, okay. Put this whole spring in the front, the tire as well as the suspension and call that as Kf and Kr, rather Kf and Kr, right and then just consider the displacement here as well as the theta, okay and call that as Jy. In other words, what essentially I have done is to compress some of them and express this in a very straightforward fashion, okay. Now what I am going to do is very standard. I am going to do how many equations I am going to get. I am going to get 2 differential equations, okay. So, we will call that as mass m, okay. We have removed or we have neglected the m, u, s. So, that is ms. So, we will call that as ms, Z double prime plus Kf into Z plus A theta. A, of course, you know, what is A? That is B plus Kr into Z minus B theta is equal to 0, okay. That is my first expression. My second expression is Jy into theta double dot plus Kf into A into Z plus A theta into these are the two equations. I am now looking at natural frequencies, undamped natural frequencies, okay. Because what happens is that for the proportional damping, this x is not that much affected. We will see that in a minute. So, I am removing the damping and we are looking at the undamped case. How far it is correct? Lot of questions, okay. First cut is fine, right. Now, we may not have time to complete this, but what are we trying to do here? We are trying to find out. So, there are two modes in this. Let us understand the physics equations follow. There are two modes to it. What are the two modes? One is what is called as the bounce mode and the other is the pitch mode, okay. A bounce mode and a pitch mode. So, those bounce and the pitch modes are given by these two things, okay. Now, because of the fact that look at that equation, because of the fact that the equations are coupled, unless I have a special, they are coupled. Rewrite it. You will see that both these equations are coupled equations, okay. So, what is meant by coupled equation? Simply means that is that we will have an effect on theta and theta will have an effect and so on. There is only one condition under which I can uncouple it. Yes, I uncouple it. Let us now look at, before we go into the details, look at the solutions for coupled and uncoupled equation. What do I mean physically by coupled and uncoupled equation? Suppose I have a method of uncoupling. The condition is very simple. We will see that in a minute, okay. So, in other words, you expand it. I have to make the first equation theta is equal to 0. That will be the condition. So, what will be the condition kf into a is equal to kr into b. It becomes an uncoupled equation. That is what you will see everywhere, right. If I am not, here also you will see the same thing, okay. I think minus b or maybe this is minus. Just check that. We will come to that in a minute. So, that both the couple, both the equations will become uncoupled. Yes, minus kf. What is meant by uncoupled equation? So, when there is a bounce mode, if this is the, my vehicle, when it is a bounce mode, it will, there will be a pure bounce at a particular frequency when it gets excited. It will be a pure bounce. And when it is a pitch mode, then it will be a pure pitch. It will be a pure pitch, okay. This is called as the uncoupled vibration. Sorry, uncoupled vibration. They are not coupled. On the other hand, when I have both of them coupled, how would it vibrate? I have ms, okay. It would vibrate in two modes, omega 1 and omega 2, okay. In one case, it will vibrate like this with respect to a particular point. It will vibrate. So, in other words, there will be a displacement as well as there is a rotation, which will take it to this. So, it will vibrate like this, okay. In another mode, it will vibrate like that, okay. So, both of them are now coupled. This is uncoupled and this is coupled, okay. Point number 1. Is this clear? Point number 2. What is the significance? Why are we doing this? Yes, I understand the assumptions, but why are we doing this? So, what is it that I want from this? I want minimum vibration, okay, or minimum disturbance when I sit in the sprung mass. So, where am I going to get the disturbance? When the vehicle goes, say, for example, a front wheel goes over a bump, okay. Now, what happens in this case? When the front wheel goes over a bump, wherever you are sitting, you are going to oscillate because that is the mode. So, rear wheel goes over a bump, again you are going to oscillate, okay. And when there is a bounce, whether you are sitting in the front or rear, you are going to oscillate. So, would you prefer this? No, you are going to oscillate all the time, okay. Let us look at this situation. Suppose this point happens to be in the front seat, let us say that you are front, okay, seat which is very close to the suspension and this happens to be at the rear, okay. Now, when I, one of them oscillate, okay, when it oscillates because at the place where you give the input, then you would see that the front bump won't affect, the rear rear bump won't affect the front and so on, okay. This is a much more optimum from point of view of the vibration characteristics when compared to this, right. So, in other words, how best can we get this kind of behavior or what are the conditions under which we get this behavior when compared to this is going to be the aim of the derivation, right. So, the first thing is very clear. I have to avoid coupling, right. The second thing is how do I place, what is this called as the node, okay. In the front, how do I place this node at the rear? This is my next question, okay. How am I going to answer this question? By looking at the mode shape for omega 1 and omega 2 and that is what we are going to derive from this. We will do that in the next class.