 In the last class we were looking at a very simple model which we called as bicycle model. We said that we will collapse the two types of the wheels into one in the front and the rear and define this model by means of two degrees of freedom is what we said so that if this is the coordinate system x and y and is that being so I thought we sorry we did this like this x and y is that being inside the ground this is the coordinate system which we saw and we said that there are two degrees of freedom u is not a degree of freedom we said that it is the yor rate and v the velocity along the y direction we said that the vehicle is travelling at a velocity u which means that you are giving that velocity so let me call that as u that is travelling there. In other words we are now looking at lateral dynamics when the vehicle is now taking a turn in simple terms or we are into a manual. A number of things I want to clarify because we have questions at the end of the class. One is that what are the assumptions that we have made here? It is quite obvious but let us put that very clearly. Now when the vehicle takes a turn obviously there is going to be a roll the vehicle rolls probably you would have heard about this term roll center roll axis and all that so there is a roll and that roll is not taken into account in this case obviously that is not a degree of freedom in this model so the roll is not accounted for. So in other words all that load transfer that comes due to roll is not there in this model. I am not considering the wheels individually so that the stiffness of the wheel rear wheel in this model is nothing but the addition of the stiffnesses of the rear and the front wheel clear that is the second thing that I am going to do. The third thing I am going to do especially in this model I can extend this definitely in this model is to consider a linear tire. So roll not considered so I am going to consider a linear tire model. I can remove that you know later of course that I am not considering the compliances of various bushes and other things that come into picture all those things I am neglecting and we will bring in all those things later. In other words what we are going to essentially do is to look at this model derive the equation understand the physics and then slowly include the effect of those aspects which I have neglected right now that is the whole idea. Obviously there is going to be a load or force rather that is acting which we call this as FIR and FIR and that is the delta which is the steer angle given and that is the alpha f. I mean these are all exaggerated so all angles are small just to show you from the center that is the delta that is given that is the delta f and that what was that V plus AR where R is this distance remember that is the distance and that distance is B. Then we said that that is beta so tan beta is equal to V by U which is approximately equal to beta and then we said that delta minus alpha is equal to V plus AR in actual terms as you look at it but this alpha f is actually negative in order to bring that into effect we define that to be negative right. So ultimately if you go back and look at the expression which we derive we said that delta is equal to L by R minus alpha f plus alpha r right this is what was defined. There is a small confusion in this because there are textbooks which give this equation for example if you look at Carnot vehicle stability he gives this equation. If you look at Jilles B he gives another equation the equation is slightly different so and there are other books many books which give the equation differently. So they write that as L by R plus alpha f minus alpha r so they give it like this. Now there is a confusion like which is correct both of them are correct there is no actually they are the same the only thing is this is the mod you know the actual value. Go back and look at the way we had written we now can write also I mean the variance of this is delta is equal to alpha f plus V plus AR by U I can write it like that. So that now alpha f would become the actual value of this angle and sin is not included but go and look at what are the way we had written or in other words in this case what happens is alpha f is equal to delta minus V plus AR by U right this is the actual alpha f actual alpha f. Now how is that how did we write it delta is equal to minus of sorry alpha is equal to minus of delta minus V plus AR by U that is how we wrote but on the other hand we wrote alpha f is equal to minus delta minus V plus AR by U okay simply because that minus sign is given because if I now consider alpha f it is outside okay it is like that which is actually negative and that is why this negative side. So you can use both it does not matter but be clear about the angle and be clear about the force that results from that angle okay. So science if you want to be consistent your the science are important follow the correct science or you know intuitively and so follow only the values. So in other words you can write this as if you are going to write that as like that and then you say that it is just this value in the same fashion if I see the force f y f okay it is alpha f is in this direction minus okay f y f is positive. So if you want to follow the science clearly then you can write that as f y f is equal to minus okay the constant which we called as c alpha f into alpha f that minus sign is again because the physics brings that sign one is negative direction gives a positive f y that is why it is right. If you do not want to follow this this is confusing okay fine then you can write that as like this but then this is the mod then correspondingly you should substitute. In fact if you put a minus here notice that ultimately the force nothing happens there is no no difference nothing happens because when I put minus alpha f actually that is the sign I am following sign conventions correctly when I put minus alpha f obviously it becomes plus so f y is in the plus direction okay. So this is what this there is no confusion in this but many books follow you know every book follows its own terminologies and the way it writes and so you have to be careful on that clear. Any other questions? Now there was one more question which I said I will answer it later let us not be in a hurry yes I let me I am going to answer that but then before the class I know some of you may have this doubt what is this you are talking about stability and you are you said that I am not right now not worried about Eigen values okay and I know that stability depends upon the Eigen values I know that it I have to calculate the Eigen values and that if the real part of the Eigen value is negative okay the system is stable but these are the things which I have studied you know you suddenly say that do not worry about that you know let us first look at stability and then look at this I thought there is a confusion I though I clarified it but let me put that for the sake of others yes of course but usually pedagogically you are absolutely like usually what we do is if it if the values are given if the all the values are known okay and if I give this as an equation when I substitute all the values then it is easy for me to calculate the Eigen values and it is easy for me to find out whether the system is stable or not but if I do not know the value I want to develop a condition for it I need not calculate the the Eigen values so I will follow a stability criteria like route stability criteria and so on and develop a condition okay which gives me stability this is not in variance with what you are defining okay and we will also see later a bit later why okay the stability condition what you learnt is also important and we will use that but right now I am not going into that because as I told you I do not have numbers I know I know only I now know the governing equation okay so I have to only look at that from that perspective I have to develop the stability criteria clear so that is why I made I said that let us first look at stability okay from a different perspective okay or in other words same principles are same from a different angle right so for that let me go back and see where I left I had written this in the state space form right I mean that is where we left and we wrote that as v dot or dot so that v and r is our the state equations and I mean so that this would become minus c alpha f plus c alpha r divided by m u minus a c alpha f plus b c alpha r divided by m u minus u minus a c alpha f minus b c alpha r divided by i u see i is nothing but the moment of inertia term i z term okay or i z z or i z z however you want to pronounce okay so that is that is what we had written in the last class right which we took off we wanted to define what is the stability criteria if an equation is given that is where we went about right and then we wrote down you remember that we wrote down the differential equation for nth order system right I think I think we did all that clear and then we took the what is that we did later we took the Laplace transform right and then we arrived at an equation Laplace in the Laplace domain and so on clear so ultimately we wrote down an equation for this as a naught a squared plus a 1 s is equal to 0 this is where we left right clear now for this system the first six second order system first and second order system you know route stability criteria I have done that in controls I am just going to not going to details of it I am just going to state restate it that the necessary and the sufficient condition for this system to be stable this is the system which we are dealing with now okay for system to be stable these quantities should be positive a 0 a 1 and a 2 should be positive okay now in other words if a matrix is a 11 if a matrix is written as a 11 a 12 a 21 a 22 remember that I have to do at s minus si minus a okay and so on right it is all done any questions so this is what we are going to use that is why we are not looking at right now whether eigenvalue the real part of the eigenvalue is negative or positive and so on that is we are not looking at that okay so in this system I am going to write down so I am going to write down a naught a 1 and a 2 okay go back and look at it and where a naught we had already written that see a a naught okay can be for example ultimately look that like this a 1 is equal to minus of a 11 plus a 22 and a 2 is equal to a 11 a 22 minus a 12 a 21 okay so that is where it actually boils down to now I am going to write that down you know from here we will take off we will look at that so a 11 is my first term so that is this term and so on a 12 a 21 a 22 so on so let me now write down a 1 a 1 is minus a 1 this is how you know I have normalized this is that a naught is equal to 1 and this is the equation okay go back and look at that so a 1 is equal to minus of a 11 plus a 22 and is equal to c alpha f plus c alpha r divided by m u okay plus a squared second term c alpha f plus b squared c alpha r divided by a u. Let us look at the second term a 2 or the third term rather so a 11 into a 22 minus and minus so that becomes plus so c alpha f plus c alpha r divided by m into u multiplied by a squared c alpha f plus b squared c alpha r divided by a u that is the first term minus what is a 12 minus a c alpha f plus b c alpha r divided by m u minus u multiplied by which term that is the term right so I would put that as plus because there is a minus there and so multiplied by a c alpha f minus b c alpha r divided by simplify this look at that expression and simplify it let us see what you are going to get take a minute simplify it so let us do that a squared c alpha f first term plus b squared c alpha f c alpha r plus a squared c alpha f c alpha r plus b squared c alpha r squared sorry that is right so that is the expression c alpha r squared divided by m i u squared we will do a small jugglery here let me put that as minus and write that as like this a small jugglery I will do why am I doing this becomes simple the first term now becomes minus a c alpha f minus b c alpha r whole squared divided by m i u squared that is because there is minus I had taken that out right then next term minus u because minus is there minus u into a c alpha f minus b c alpha r divided by i u clear simplify this expand this so you will get what is that you will get so this will be a squared c alpha f squared plus b squared c alpha r squared minus that is both of them are minus outside so what will happen this guy will go off and that guy will go off okay and you will get minus 2 a b alpha alpha f alpha r right clear and minus this minus is there there will be plus so you will have a squared so I will write that as a squared c alpha f c alpha r plus the second term b squared c alpha f c alpha r and the third term comes from here which will be plus 2 a b c alpha f c alpha r the whole thing divided by m i u squared okay that is the first term then I have the second term let me rewrite that term I will take care of all those minuses and say that plus I am adding that all that a squared plus b squared and all that you know this plus what is the second term b c alpha r minus second term ac alpha f upon i okay let me look at this carefully it is a squared plus b squared plus 2 a b so c alpha f alpha r I can take that out so I will write that as c alpha f c alpha r into a plus b whole squared which is l squared divided by m i u squared so that is the third term which I would which I called it as a 2 right okay now let us go back and look at the stability right I hope I have written correctly look at m i you know rest of it I know it is correct m i u squared I hope I have not left anything okay so now look at these three terms the Rhodes condition says that these three three things should be positive yes the second condition or the second one sorry not second condition second coefficient a 1 look at that coefficient can that be negative so that can be negative because we said that c alpha f alpha r positive m u everything is positive so that cannot be negative but let us get back to a 2 okay let us get back to a 2 first term no worry that will be positive but second term it is possible that the second term is negative okay so when ac alpha f is greater than b c alpha r obviously this is going to be negative right so the only possibility of the terms becoming negative is dictated by this guy here right so as long as in other words as long as c alpha r is greater than ac alpha f b c alpha r is greater than ac alpha f do I worry nothing to worry everything is positive on the other hand if I now if b c alpha f b sorry b c alpha r is less than ac alpha f then this becomes negative there is a cost of worry it is immediately it is not that the total sum becomes negative but there is a cost for worry that is all I would say right in other words this can be so high that I can overtake this guy and make a 2 a negative term all right not not that just because this is negative it has already you are done and doomed no clear so this condition we will later call this okay as an understeer condition why how this term all those things we will define we will define shortly what is called as understeer coefficient we will define all those things when b c r alpha r is less than ac alpha f okay then we have oversteer condition why not b c alpha f b c sorry b c alpha r is equal to ac alpha f yeah sure it is possible and that brings out a condition called neutral steer now why are we calling this as understeer oversteer you know what is this condition under which it will really become unstable these are the issues which are very interesting right and that is what we are going to see now now let us rearrange this term and quickly get into a position where it will really become unstable okay rearrange the term look at that and rearrange the term and let me know when it will become unstable so let me do that maybe I will write that as a 2 to be 1 by m i u squared plus b c alpha r minus ac alpha f divided by c alpha f c alpha r l squared multiplied by l squared c alpha f c alpha r let us let me do something like that okay let me take that I also out right let me bring another jugglery here I am I am I am just only manipulating this it is all the same because I am I am zeroing in on terms which which are going to be important so look at all these terms they all cannot be I want all these I mean this to be greater than zero so this terms are good guys they are all positive okay so I am not very worried about it so my condition is actually one let me do one more thing I will put this u squared here okay and then put u squared like that right so because u squared u squared is positive so no problem so my term actually reduces to m into b c alpha r minus ac alpha f divided by l squared c alpha f c alpha r into u squared okay that is how it reduces to and that should be greater than fine as long as this is greater than zero I have no issues but if this is less than zero then then this term is becomes negative right and at a point of time cross one okay and make this whole thing negative right so let me call this as say K3 why K3 I will tell you why I am calling that K3 a bit later let me call that as K3 okay so that I will write this as 1 plus K3 u squared okay to be greater than zero that is the condition and ultimately I read this reduces to a condition where K3 becomes negative okay so u squared yes so when u squared is equal to root of 1 minus minus of K3 okay that is where when K3 is negative this is the condition you would get okay when K3 when K3 is negative u squared is equal to root of minus 1 by K3 is the condition beyond which not u squared sorry u beyond which the vehicle is going to be unstable oh wow that is a very interesting conclusion clear so in other words what happens is that entry of the term u here in that expression right so what does it mean really let us physically understand it we will go into the equations a bit later okay so there is a K3 look at the K3 carefully okay K3 consists of m characteristics of the vehicle no problem it is going to be a constant okay A and B what are A and B the distances from the center of gravity to the front and the rear right alpha f and c alpha f and c alpha r are the result of assumption that the tire is linear clear linear so they are constants of the tire l squared c alpha c alpha r and so on so this that makes this whole thing K3 to be a vehicle parameter okay so the vehicle can be an understeer vehicle or an oversteered vehicle and the vehicle is an oversteered vehicle K3 becomes negative immediately does not become unstable K3 becomes negative okay I have not yet defined why I am calling this as understeer oversteer that we will come in a minute okay so that plus into K into u squared simply means that the oversteered vehicle has a velocity at which this third term goes to negative or becomes negative and that velocity is called as the critical velocity above which the vehicle is going to be unstable okay and what is this instability going to cause that also we will see in a minute okay so that is called as the critical velocity why I have put this as K3 because I am going to have two more definitions K1 K2 K3 there are books I am not sure I do not have the list but I know that some of the books okay call this as an understeer gradient call this as understeer call this as understeer gradient okay so that they would write this as 1 plus K3 u squared and K3 being an understeer gradient why am I being careful and write K3 because this definition varies from one book to the other okay the concepts do not change but the definition changes another book would just simply state L they will remove that L and say that is understeer gradient okay so we will see why the other books say like that so this is a simple way of defining this understeer gradient right okay so that is that is one thing the first thing that we have but why did we call this as understeer and oversteer okay in order to understand that let us get back to our original expression okay and treat a steady state condition for the body or for the vehicle okay let us get back to our original governing equation all of you remember the governing original governing equation how did we write okay can we remove this any questions clear we are good so remember that in the very first class I have to get back I think we called it as 1 and 2 I z r dot what was the expression which we wrote for this I z r dot you remember that is the first thing you wrote a into f y f minus b f y r right and what is the second expression we wrote m into v dot plus r u is equal to f y f plus f y r this is what we wrote let us now consider what is called as the steady state condition what is a steady state condition what do we mean by steady state condition what we mean by steady state condition is that that there is no time derivative terms okay so that I will write the first equation to be a f y f minus b f y r is equal to 0 and m r u is equal to f y f plus f y f right these are the steady state terms or sorry steady state equations let us now get back to my delta terms you know delta what I had written here write that down how we wrote it I hope there is no confusion on that I can write this in two ways as I said I can write that as l alpha f minus alpha r plus l by r now what I am essentially going to what is that I am going to do before we go further let us understand what we are going to do now I am going to look at the effect of delta effect on delta due to the new introduction or new term which we called as understeer gradient okay now I am going to look at how that delta is going to be affected okay how in a very in a steady state maneuver okay where we have simplified the equations and we are going to look at how that is going to have an effect this is what I am going to do right okay let us look at the first this equation m r u what is m r r what is r remember that it is the yaw velocity and so it is equal to u by capital R okay and so m u squared by r and that is equal to f y f note that I can also treat this as minus alpha f plus alpha r so I mean I have to be careful in that note that carefully right now let us look at the geometry part of the vehicle motion okay here it is so we define very clearly what is capital R and so on so obviously that is the radius that is the radius of the turn 1 by r is actually curvature right we know that okay so I have that angles what is this a and that is b approximately a lot of lot of it are approximation because every angle is small and all those assumptions we make so that that angle is what what is that angle okay and that angle is b by r clear okay and L is actually the total length of the vehicle so that is the geometry which is the you know the vehicle takes when it takes a turn right we have to go through a number of smaller things so that we understand it clearly one thing we know from our earlier classes how my normal forces distributed okay remember all this equation I am going to use them here and there okay maybe in the next class when you start we will write down all those things okay so have your notes ready so that we will when I put down an equation you will go back and refer to where from where we are getting now I know that w front which is the normal load that acts on the front is equal to in this in this way what is it b divided by L into w right and the w rear is equal to a divided by L into w okay now if f y is the total load that is acting obviously the same or how did how did I get that by taking a moment okay about the center or about one side and so on the same thing I can do it here as well because this is steady state I have no issues on taking the moments and hence f y f is going to be distributed in the same fashion as b by L into f y and f y are equal to a by L into f y where f y is actually the total force that is acting yeah any questions okay so its first thing is interesting to note that the lateral forces are distributed in the same fashion as that of the weights which I can write it as m u squared by r to a that is f y okay sorry into b by L that is m u squared by r is f y the total f y and that is equal to m u squared this is u squared by r into a by L right now I want to use this can I use that not very difficult this is nothing but mg and that is mg so I can substitute that into that expression and rearrange the terms in order to I mean and get f y f why am I doing it it will become clear in a minute okay so in other words the first thing is that the lateral forces are distributed in the same fashion as that of the normal loads okay rearrange it and tell me how I can write that in terms of w f m into b by L so w by g into u squared by r is that right same way f y r is equal to w r by g into u squared by r okay I took some time to get back to these two things okay now we will go fast we will start from here and then we will substitute that into delta we will see what happens and so on right okay there are the only thing is that there are a number of equations please go through them okay if you have any doubts ask me in the next class I am going to use them now okay to bring out the physics so you should be clear with every equation that we had written right we stop here and we will continue the next class yeah yeah right so that is why I said that it is an exaggerated figure and that it is all those angles are neglected and then that the forces are distributed something like like this okay that is all yeah of course you is what no we are we are not looking at longitudinal forces this is a lateral force alone is what we are looking at okay so we are not looking at longitudinal forces the three equations if you remember we wrote and fx equation we are not considered okay.