 We were looking at how we started looking at vertical dynamics and that is what we are going to do for the next 5, 6 classes and we already noted that there are two important things in vertical dynamics, two important things that we have to look at in vertical dynamics. One is what we call as right comfort and the other is what is called as road holding. So, right comfort and road holding are the two important things that we are going to see in this in this part of the lectures. Now of course, road holding for a common car may not be an issue, may be more for a formula cars, racing cars this becomes very important because of certain differences between the usual car that you drive and that of a car which is driven for say formula 1 racing. So, what is more important for of these two, what is more important for us is what is called as this right comfort. As we had done earlier in this course where we had looked at a bicycle model and looked at handling from a very simple model in order to understand the concepts and then we of course, we did some modifications to this models. We looked at the assumptions that we had made and then slowly removed that assumptions and then brought out a complete picture. Like that we are going to look at this right comfort from a very similar perspective and we are going to first put down a very simple model and then draw some conclusions. The point is this that there are a lot of assumptions in the simple models, is it that it is not valid? One of the first things you have to notice is that when you have a very complex model that you may put up in one of the commercially available softwares, carsome, adams and so on, they need to first have a model in order that we put it inside this softwares. So, a first cut, first analysis is always good to do that to do before we go to bigger models and that is why these models are important. Let us go straight away to this model. There are a number of names to these models. We will know as we go along what are the names to these models. We will look at what is called as a half car and a quarter car model. This is a very standard technique. You would have noticed that in your control system course where we will go to replace the system with spring mass damper or maybe even in your vibration course. So, that is our simple model. You would notice that the tires have been replaced with springs which I would call as the spring stiffness, K front tire, K rear tires. Notice that we have not put a damper to the tire though it looks very attractive, but usually the damping effects are neglected, not very high and it is neglected. This is what is called as our unsprung mass. So, we will call by small m unsprung mass in the front and m unsprung mass in the rear. These are the stiffnesses and the damping characteristics of the suspension system. So, we will call that as K s f, C s f, K s rear, C s rear. That is the sprung mass and we will call that as m sprung mass and that we also include a J theta. Notice that we are going to have theta which is pitch can be considered as one of the degrees of freedom. So, we are also putting J theta in our analysis so that you know theta can be considered as in the analysis and we will follow the same thing as we had done before that is this distance A and that is the distance B from the front and the rear. The input to this is obviously the road input. So, we will call that as say road in the front and road at the rear. We assume here in this model, note this model what we have done is like we did you know compressing the bicycle model and so on. What we have done is to compress that and put it like this. So, in other words we are designing half the car. People call this sometimes as half car model because the other half is you know compressed and put into one thing. So, what is possible to have here is the pitch mode. We are not considering roll. In other words what we consider in this model or the assumptions that we make in the model is that the right input the right side of the tire input is the same as that of the left side of the tire both the inputs are the same or in other words the road is such that right and left side are the same here as well as here. So, of course this is a question of time that is the input into the model and we assume that there are displacements. Let us call this as Y1 and Y2 and let us just put the displacements here as Z1 and Z2 which can be expressed of course in terms of the what I will call as Z or ZS. In India we have a tendency to call Z as Z. So, you know why we are used to calling this as Z because Z is never used as Z you know if you get Z bra Z. So, if you notice that Z is always used as Z. So, the reason why usually people do not call this as Z, but they call it as Z. We are used to calling this as Z you know when I say see there are problems for some of you to follow what I mean. So, anyway so that is the Z1 and Z2. So, now what am I going to do very simple nothing great I am going to write down all the governing equations equilibrium equations which are going to be a second order differential equation that is all I am going to do. In fact you can write it down there is no issue. One what we are going to do of course I can write this in terms of ZS and theta or other words I can it is possible to relate both by means of this relationship or I can write it in terms of ZS and theta or in terms of Z1 and Z2. Now let me write down the equations the governing equations and you will understand what are the degrees of freedom. So, the first is so I said that again if I want I can change it that is all I want to know I want to state fine. Now let us consider each of these masses separately and write down the governing equations to it. And then of course these are the road inputs I just want to give you a small background to this road input. There is a tendency for people to just measure the road profile just measure the road profile measurement projects which I think couple of students have done before you may be 3-4 years ago. So, road profile measurements are done using laser sensors and so on. And there is a tendency to just input that straight away into this model is that correct is that right why it is not correct. No, that is correct, but I am saying a particular road where you are doing measurements okay I measure this I can do that using okay that is very good we are going to consider the speed correct tire acts as a filter okay. Now because tire that is what that is what is called envelopment characteristics of the tire. So, the tire as it goes okay is not going to be like this, but it is going to be like this okay and it is not in other words it is not the true representation of the road what the car sees is not the true representation of the road. So, when I when I put this RF and RR they are not road inputs which is usually called in many textbooks not that just the road profile, but it is the profile which the car sees after this high frequency filter called tire which filters off not due to not due to its stiffness characteristics we are not talking about that I have already put it here, but more due to the envelopment characteristics. So that the what goes inside is what is felt if you go and sit in this tire here this region what is it that you feel is what we call by RF and RR yes. So, there are models you are correct there are models to understand and put that as an input okay this is a very important problem being tackled well in the entire models okay envelopment characteristics called if you go and look at Swift model okay then you will see that there is an envelopment characteristics is a part of the model, but that is very important. So, whatever model you put in it is very important that envelopment you understand what is envelopment envelopes okay that is what goes as an input that becomes very important. Let us let us now straight down straight away write down these equations now. So, ms into I would use that is equal to f front plus f rear which I have to be consistent that is why I have to refer to that nodes because I do not want to change my this let me call that as f1 and f2 okay. I have a tendency to change this you know I want to be careful if you do not understand anything okay. So, let us call that as f1 and f2 so f1 and f2 are the inputs from the front and the rear respectively. So, what is the second equation j theta m moment equation j theta into theta double dot is equal to moments f1 into a minus f2 into b okay. So, the next equations two equations I am going to write for the unsprung mass the front and the rear okay. So, m unsprung mass the front y1 double dot is equal to so write down the force balance equations okay m uf right. So, m uf into y1 double dot ma write down the force terms k front tire multiplied by now that is a spring. So, all of them are rf so rf minus y1 all of them are function of time. So, I am just removing this t plus a k suspension in the front. So, there are two things here so y1 and the other side is z1 okay. So, z1 is written in terms of of course I already written that plus right. So, similarly I can write down m ur into y2 double dot is equal to krt into rr minus yn plus ksf into minus y2 I am going to I am not going to write down or derive every step because this is algebra I said that I told you already that you have to work out some of the steps that you have yeah oh sir okay what is happening to me. So, good let me see if you say that I just put it so that I want to know whether all of you are awake okay. So, all of you are listening and you are understanding that is very good okay what is f1 now write down f1 minus kfs into z1 minus y1 minus cfs into z1 dot minus y1 dot okay and f2 is equal to minus kr spring into z2 minus y2 minus cr spring then what else have you written that is right okay. Now as I said I am not I am not going to do the rearrangement every step obviously I am going to write this down into this by the way what are the degrees of freedom how many degrees of freedom are there 4 what are they y1 y2 z1 z2 or y1 y2 zs theta okay so the 4d so this is basically a 4 degree of freedom model okay let me prompt what I am going to do the first thing I am going to do is to cut this and make it into what is called as a quarter car model okay I am going to compress that and put that as one quarter car model okay. In fact more interestingly what I am going to do is to cut this and consider the masses to be distributed to the front and the rear in the same ratio as we had done before and then consider this as two quarter car models one to the front and one to the rear okay. So, what I am essentially going to do is to rearrange this okay rearrange it I am I am going to write down the results of course u is equal to r which is the load of course m is quite straight forward okay so that is that is quite straight forward you can see all the 4 what am I doing I am very simple 4 equations I am just rewriting it in a matrix form okay you know why I am doing it it is quite straight forward. So, the next is the C matrix rearrangement gives us note that I am going to replace z1 z2 and all that so rearrangement gives me something like this CFS plus CRS sorry A minus CRS B minus CFS minus CRS one of the things which I had pushed in without elaborating is that the spring stiffness is a constant in other words I had pushed when I wrote down this equation I had pushed that the spring is a linear spring which is definitely not the case with most of the suspension system it is a hardening spring okay so the suspension spring is not going to be a linear spring but as I told you we want to get the first cut results what we call as first cut results the reason why we have made these assumptions and that this is you know the damping is characterized by a simple number and the spring stiffness is characterized by a simple number right and the K matrix we will write down the K matrix here the results are very interesting unfortunately we cannot jump to the results so we have to you would see notice that it is very similar to C matrix because there is no C there for the tire that is that is missing the stiffness of the front tire and the rear tire okay. We can solve this not a big issue okay you have MATLAB okay you know if I give you the functions okay you can solve it but I am not going to do that okay as I said you want immediate some results so this is a complete half car model okay in fact just for fun of it write down equations for a full car model okay write down equation for full car model and it is possible to solve some very difficult and solve that as well but nevertheless you write it down just to get a feel of how we are going to deal with it in other words if you go to a full car model then you have a pitch then you have a roll right the other direction and then so for the body you have the z and then pitch and the roll right you can write it down now what I am going to do is to split this to get some very interesting results I am going to split this into two models or two quarter car models later you know I am going to consider a similar version of our half car model in order to understand okay more about this model. So the first thing is that this is what most of the companies do okay as a first cut why do they do it immediately if you want to look at suspension tuning if I want to look at optimized suspension okay then it becomes very simple to understand this right and so we use this for optimization of the suspension system so the first thing I am going to do is to look at what is called as the quarter car model as I told you before I am going to redistribute the mass right and put that front mass is equal to how do I put that B by L into this sprung mass and is equal to A by L into the sprung mass okay so now the equations are the same front and the rear so I am going to consider only one of them and put that M as M as a small M as and write down the equations okay. So how does this look like as I said I am going to cut it here so the front and the rear will have the same or similar or same equations only thing is you have to replace corresponding with the M as k front k rear and so on right so we will write down one common thing. So that is our sprung mass and then that is our suspension that is our unsprung mass that is my tire and the tire is you know on the road so that is my what is called as quarter car model so let me write down that is equal to as I told you this one we would call that as Z and that is Y minus C into Z dot minus Y dot I want you to check this is that correct of course that is KS that is C and that is KT is it positive is it correct or not and I know because I have few questions I mean last one of the earlier courses have been people are confused between say very easy when I put a force so when I have two things which are moving okay. So what happens when Z is more than Y that in other words this is positive okay what would happen let us just look at the spring Z is more than Y so what will happen pull down okay so the force will be in the negative direction so but my M is Z double dot is in the positive direction so I have to put a negative there right now your I know your question so any other question okay if you want write it like this if you want write it as MS is equal to force from the spring plus force from the damper okay and then write down right fine and so this is again note that we are looking at 2 degrees of freedom Z and Y of course R is the road input note that all of them are questions I mean are functions of time okay. Now I have to have a strategy to look at it why am I looking at this one I said that I want to optimize the suspension okay so in other words you can look at this simply from point of view of natural frequency determine the natural frequency okay and we will see we can optimize suspension and so on right in fact just want to state even before we go there so you will get okay in simple terms in terms of what you have studied you would see that there are two natural frequencies okay one is called as the wheel hop frequency and the other is the body frequency right but more important thing also is that I am going to write down the behavior or the ratio of the Y in terms of or Y to R and Z to R okay I am going to write that down. Now I am going to assume that this R is a complex exponential function which I would write that as R e power I say omega t okay as a complex exponential function this is you would have studied this already in controls okay so what is this system this is what is called as the linear time invariant system linear time invariant system nothing is going to change okay and that if your output is delayed input is delayed output is also correspondingly delayed and so on. The beauty of the system is that if I now give a complex exponential function like R e power I omega t the output is also a complex exponential with the same frequency but with a phase lag or phase difference with the phase right so let us call that as say for example this thing let us call that as Y e power I omega t plus phi okay which I would combine them together and write that as Y e power I omega t. So the Y is actually a complex function takes into account the phase as well right now why am I doing this or what is the use of this or what is how do I use this property called linear time invariant function that the complex exponential becomes a this is used how do we use it or why do we put it like this basically because whatever be the input can be split into in the time domain you can split it into complex exponentials or sum them up as complex exponentials okay. You had seen this for example in Fourier series right where you had a periodic function the periodic function was expressed in terms of sin n cos which are okay which can be expressed also in terms of complex exponentials okay that is for a periodic function you can also express say for example the road input as in terms of complex exponential in terms of a number of frequencies using another technique okay which is very well used called Fourier transforms okay and the way we are going to calculate that is what is what we are going to call that as fast Fourier transforms okay so a Fourier transform converts whatever be the input into a number of complex exponentials and hence if I now get a solution for one complex exponential okay and using this time superposition principle it is possible for me to get the output for a number of such complex exponentials so that is why I am interested in that complex exponential where y is also a complex function and so I write it like this alright with that background let us let me write down I am going to as I said quickly run through because we are running out of time there are lot of things to cover okay so input and output are of the same form substituted what do I do put that in the governing equation I am going to get omega squared here and so on you do that I am going to express this final form you know the two equations do that of course r is real so from which I am solving this and writing down z by r and y by r okay so that is z by r the same thing as denominator let me write that down as something like this d plus ic omega e where d is equal to expanding that though this looks formidable there is nothing that is difficult in this because of the number of terms it becomes it looks very complex okay so we are going to as is usually the case I am going to simplify it later minus okay so of course if you want a solution look at that everything is known it is straight forward there is nothing very difficult right notice also that both y and z are complex okay we will do the interpretation in the next class okay but let us first understand what it is so in other words the very first thing is that given an input the road input as a complex exponential I can find out what would be my the displacement at the unsprung mass position what is that so let me call that as g let us say gz let us call this as gy you know why I am calling that as gz and gy okay so y is given as r multiplied by gy multiplied by e of i omega t same fashion you can write down okay how do I suppose I want I give you this and I want you to do a test how would you do practically I want to you to measure this before we go further sorry this is too much of match too much of equations okay tell me what are you trying to do what is that ultimately I am interested in okay right that is the first question I am giving you a car and I want you to measure first thing is I do not even know you talked about road development characteristics I do not even know how I get this road development characteristics and so on okay so I okay give me a car let me first do some experiments to understand what is happening first question so what are the what are those things that you have as an equipment is a sensor in order to do an experiment you have an accelerometer right okay very good all of you have used accelerometers so the first thing you would do is yes I have an accelerometer a data equation system right so what do I do I put an accelerometer here and I put an accelerometer here and then start recording the acceleration levels so now acceleration levels are obviously can be determined this by this y dot is equal to i r g y i omega t e power i omega t and i square that is in other words from here and it is minus of this quantity minus of that point right so once you measure this suppose I want to know how much of how much of road input has gone in I have measured it in other words I know what this quantity is right for a let us say that I do f of t I do all those things let us say let us take one frequency I know what is gone in here so I can easily find out what is the because I know I measured that I know the other quantities I can find out what is my actually r so this this is a for a particular tire okay that is a constant and so I can use that for any other vehicle I can use r into e power i omega t as an input for the road right so the first thing is that from here I am interested to determine the acceleration right and I am going to look at how my tire is is filtering right and also more importantly how my suspension is filtering all that input right why I am interested in acceleration because you would later see that acceleration is the deciding factor for right comfort you would see that it is at the tolerance level for a human being would be expressed in terms of accelerations okay so that you can do it by measurement as well as you can do it by a quarter car model we are going to give a small twist to this it is it exactly acceleration is it something else we will we will come to that we will give a small twist to it but nevertheless from here what we are interested in is the acceleration right if you want the magnitudes of z and r we will we will finish this class with that magnitudes of z and r of course you know how to get the magnitude get the real imaginary root of real plus imaginary so that is the two vertical magnitudes magnitude of acceleration for these two are given by omega squared multiplied by this omega squared multiplied by that would give you the two magnitude of acceleration levels clear any questions here wait wait wait you are you are right you are absolutely right you know the problem in this is that I have to derive it I have to show you the equation and then come to interpretation I know slightly impatient to why are you doing this right that is why I immediately jumped and talked about acceleration you are absolutely right that this is what we are talking about road profile is a spatial or in time domain okay it is sorry not spatial but is in the time domain is in the okay as a as a matter of distance spatial domain right it is a matter of distance actual profile right when I put this as an input I am converting the spatial domain into time domain so what is that linking factor speed so what is that you get you know we will with with the last statement we will finish this class okay so what is that you get you see that if I go in the same road very important point you brought out we will see go in the same road at different speeds and different speeds I am not going to get obviously the same output very simple you see that every day right at particular speeds you would see that suddenly your car starts jumping and after sometime it will come down okay so when is it going to jump that is an important question okay that is what we are going to see right so in other words for the same road speeds give you different okay inputs with respect to time absolutely correct how is it we will see that in the next class that does that answer your question come again so the speeds are important I have not yet tied tied this up okay speeds are important because speeds how put that down you know the velocity in terms of meters per second you know the spatial distribution terms of you know distances as meters connect them together and we will see that in the next class right.