 In the last class we were looking at the governing equation for dynamics and we had in fact derived a very general expression for a body where the coordinates are centered at its say CG location for example and that we said this is the body centered coordinates. We looked at the inertial frame of reference and we also found out that there are some additional terms that are required especially when the body centered coordinates or in other words which is attached to the body of say for example vehicle and if it now starts moving the vehicle starts moving as well as it is rotating and so on. The body centered coordinates also rotates along with it and hence certain changes are needed. Some additional terms are required in order that we take into account this motion of the body. This is what we said and we derived a very general three dimensional expression for this body centered coordinates. You would have studied this in dynamics anyway that is a refresher. What I am going to do is that some of the things which you would have studied anyway we will go back and just refresh it a bit so that you will know that I am going to apply what you had studied before. I am going to do that for example you would have done a course on control systems I am going to do that again here in today's lecture. So ultimately as is usually the case in engineering you start with a grand you know equation and then slowly say that you know this I will neglect that I will neglect this I will neglect that I will neglect so that you know it ultimately becomes so to say handleable. So that you do not lose the physics at the same time the equations are simple. This is a usual fact of life in engineering so we will write that that way for what is called as a bicycle model. So that is the bicycle model which we had just started. So what I told I will please remember what I said in the last class that I am going to collapse the two wheels into one board in the front and the rear. Hence it looks like a bicycle so it is called a bicycle model nothing to do with actual bicycle stability of bicycle and so on. It is just called a bicycle model because it has two wheels collapse the two wheels into one wheel. The result of collapsing this is to reduce the degrees of freedom. So I am going to look at this with or in other words the equations that I am going to write down with consists of V and R where R is the yaw rate. So we already said that R is going to be the yaw rate. We are going to look at a system where U is a constant in other words steady state U is equal to a constant U we are not going to vary U and we want to write down all the equations in terms of V and R. In other words my differential equation which I am going to reduce that that huge guy 6 equations into simple 2 equations okay. So I am going to write down that equation into i z r dot what is i z r dot it is equal to tau it is the yaw which is perpendicular to the plane of the board okay. So for that I need to put down the forces so that is the force F alpha R let me okay F R or F alpha R why alpha R because as I told you before alpha R is the slip angle and F alpha R is the result of this alpha R. The same fashion we will put down a force that is acting in the front and we will call this as F alpha F. These angles are very small so I will make lot of again assumptions there. The whole idea most of the time is to bring down the equations to a linear range. The whole question in engineering is whether that assumption is valid whether or you are really operating the linear range and so on okay. Sometimes you would not be and that is where your judgment matters. Let us finish this expressions and then we will come back to we will take the questions okay. Now let us look at the front so this is the alpha R this is the F alpha R. Let us look at the front. The front you have hope this is not a very confusing picture may be I will use a chalk piece. So I have first look at that so that is what is the steering angle I have given okay. So that green one is the steering angle okay. I have this is not I am not steering the rear wheel it is possible to do that there are lot of interesting things that happen when you can steer the rear wheel as well. For example if you have huge vehicles for example vehicles which are used by space agencies where they want to transport the rockets which are to be launched. So you would see that there are 32 axles okay or 64 axles huge ones where you cannot how obviously drive them with only one driven axle or a steered axle. So it has to be more than one you know the problem becomes very interesting let us not get into all that we are only looking at now the front axle steered vehicle. So that is the alpha R and the alpha R okay sorry delta R which is the delta F rather or delta okay and that since I have to develop a force note that force requirement is that we I develop a slip angle which I call as alpha F right that is the slip angle. So with the result that I get the force which is F alpha F okay. So let us we can we will keep that as alpha F and alpha R if you want front Y and that will be easier this is F front in the Y direction and F front I mean sorry rear in the Y direction either way you can you can call this as alpha R or does not matter as long as you understand what we are talking about okay. No no no this is the turning please note that please understand the coordinate system we are going to take a turn okay this is some sort of a plan view okay you are looking at X and Y okay and that is X, Y and the Z is into the ground right that is the that is the coordinate. So you are you are looking at a plan view of the vehicle right. So obviously F is a centripetal force. So this is equal to the moment okay so what is that? A into F Y front or F YF let us easy to call that as F YF because many books the problem why I am writing like this is because if you follow a textbook then you would have given different terminology you understand that clearly F YF minus B into F YR okay Y in the rear direction right. So what is the second expression? It is go back and look at what we had done M V dot plus R U so it is not M V dot because of the body centred coordinates M V dot plus R U and that is equal to F YF plus F YF okay. So these are the two expressions we have. Now I am going to use again the dynamics along with geometry in order to write down deltas and so on. Now let me define a new term I hope you will not confused with that okay which I would call as beta. Obviously there is there is a V there is a U okay and hence vehicle actually moves along this angle beta okay that is the angle beta and beta is called as the vehicle slip angle called beta as the vehicle slip angle. Obviously from this diagram okay so you have U you have that angle beta and so that is V so tan beta we will do V by U. Now let us look at alpha. Now I have a R R is now what is your R we said that it is your R okay. So if I have to write down the velocity at the rear this is the velocity V and U are at the CG location okay hence if I have to write down the velocity at the rear what would be the velocity there would be look at this there would be a velocity which is due to R which is B into R B into R right that will be I will just put that here and then transfer it there. So that will be in this direction right B into R there will be a velocity which is V in this direction that will be a V in this direction okay. So B into R is due to R and V is the velocity so when you transfer the velocity it becomes V okay and Br since it is in the opposite direction okay I can write that as so in this direction I can write that as Br minus V okay Br minus V and tan alpha R because that is that is in the opposite direction to our coordinate system that becomes divided by U okay that is very simple look at that clear. Yeah of course non-student velocity U is constant we are not touching U so that is the same because the vehicle is moving in the same direction with the constant velocity U. So in other words please note this in other words when I have to look at U the variation of U I am looking at longitudinal dynamics the first equation in what we did in the last class we had once that we have done we have looked at the first equation in the last class so we had looked at the longitudinal dynamics. Now I am going to remove the longitudinal dynamics okay so I am saying now that I am going to only look at the lateral dynamics so that is what we are going to do now right okay yeah right we will come to that we have not we have not looked at tire yet I am going to have a very simple tire model okay Pleist air, cone city all those things will come everything we will have to then put the model just wait for some time let us finish this right just hold your questions let us go forward many of your questions will be answered as we go forward. I am going to make a assumption that tan alpha r is approximately equal to alpha r so alpha r is equal to V by U minus V into r by U right okay I am just going to so what is V by U we had already seen that V by U is beta tan beta which is again approximately equal to beta so that is equal to beta. Let us assume that this vehicle is taking a turn and the turning radius is equal to r capital R so r is the let us say that r is the turning radius I will finish this here and write it down there so now how do I write the second expression r is like omega use the tangential velocity so that is equal to radius of that is the r omega r r into r is equal to U so r by right r by U so that is equal to minus B by right so the first expression of interest to us I will write down here I will do all these things there so alpha r is equal to beta minus B by that is my first expression let us now look at alpha f is there any questions right I have same similar way I will look at alpha f which wait for no no this is the no no no no please understand that let us just wait for some time let us finish this expression okay we will come we will take the questions towards the end if this just hold your questions we will do that later so just say that see this is an r okay r is positive positive is right hand side so inside so you will see that this is the tangential velocity here due to r is actually in this direction so B into r is in that direction that is all we have said okay close that I mean closely watch that is why I put one B r like that and we like this clear now I have to write alpha f okay so what is alpha f look at that this whole thing is delta okay and that is alpha f so alpha f is equal to delta I am just giving a small gap there so here that is B r this is now a r right a into r a into r that is in the same direction as that of V so you will get delta minus delta minus that is a r plus B divided by U now now note that alpha f alpha f is in a direction which is in the negative direction of Z so I am going to put a minus here okay so I have to be careful because note that a negative direction alpha f produces a positive f y okay so that is why I put a negative so that I am clear so I do not have confusion so many books you would see that it is alpha f and when I when I develop f y f or f y it is just multiplied by c alpha multiplied by alpha okay there is a small confusion there is nothing wrong with this but you have to be very careful on the way you are writing it okay so I am going to write that you will see that I will put two minuses so that I will take care of that clearly the signs are clear okay now simplify that let us say that I am going to simplify that expression minus delta minus a r by U okay so that is equal to minus a by r minus beta right so that delta now can be written in terms of alpha bring the delta the other side and then rewrite that expression rewrite that expression okay and substitute it substitute my this expression for beta substitute that into this expression and you rewrite that expression so in other words I can say that when I when I take it delta the other side any questions note the difference between capital R and small r small r capital R is the radius of of this turning you know the turning radius actually let me give a bigger picture you are taking a turn okay so your vehicle is here and that radius that radius of of your turn is equal to capital R right so rearranging the terms this becomes l by r because a plus b is equal to l minus alpha f plus alpha r clear okay now we will come to the tire model there are a number of as I told you the number of tire models available we will go only for a simple model because I want to capture the physics of maneuvering so that I will write down f y to be minus c alpha into alpha so why am I writing this as minus because I know that a negative alpha produces negative alpha produces a positive f y okay and negative alpha produces a positive f y so in other words when I put alpha to be negative actually becomes f y becomes positive so that is why you put a minus there are books which treats this to be negative which is not correct okay c alpha stiffness is not correct so this is simple model what is this model this is a linear model so in other words we saw already that okay there is a region that is the suppose I plot alpha versus f y we saw that the curve looks something like this and I am going to take a linear model where that is the the slope is what I call a c alpha okay so c alpha into alpha gives me f y so this is a simple linear model from here when I put f y front I want I put c alpha f into alpha f and f y r I put minus c alpha r into alpha clear okay now so that f y now the total f y note that this is a general expression I am just going to remove it so that you do not get confused so now f y is equal to f y f plus f y r and that is equal to minus c alpha f into alpha f minus c alpha r into alpha r yeah of course right I will give you a small job write down alpha f and alpha r we had already derived it if I think it is there in that expression so just put that so minus is that so just substitute it minus of that so I am going to write down completely the final expression you can you can look at the final expression that is equal to minus a by u c alpha f plus b by u c alpha r into r minus c alpha f plus c alpha r into beta plus c alpha f into delta so that is going to be my f y the second term this is the second is minus a into c alpha f into alpha f okay plus b into c alpha r into alpha r substitute that so that m z is equal to minus a squared by u c alpha f minus b squared by u c alpha r okay so substituting that you get these two expressions now what is the next step what is the next step I have f y and m z go and substitute it back into my expression or my governing equation okay substitute that into governing equation so I get these are the two equations now what I have essentially done is to write down the right hand side okay so write down the right hand side I have written down the right hand side and so I have now okay substitute that back and write down the complete equation okay the left hand side and the right hand side right I am going to now what are the two variables or two degrees of freedom here r not u r and v r and v a by r which one by r right okay no I have put r outside out sorry I have put r outside so that is u here u is fine r is outside here okay r is outside and that is fine now I am going to rewrite this expression okay substitute that and rewrite this expression in a very familiar form okay so those two equations you know so those are the two expressions I mean so it is very simple you know only thing is you have to be careful in that in that algebra so the first expression now becomes I do not want rewrite it I z r dot is equal to m z that is the one and that is the second expression now I am going to write this down I leave a couple of steps to you to do it this is nothing in a form which is familiar to you I am going to write this as v what are the things that are available there or over my expressions so v dot okay that is the second expression I have and r dot okay those are the two things I am going to write down that as a readjust those terms v r right plus b into right essentially what is that I have done I have just put that rearrange the terms clear any questions nothing no no big calculations if I now put that form you know when I rearrange it I am going to I am going to write that final form v dot r dot is equal to minus c alpha f plus c alpha r divided by m u let me I think it is good to have that expression this we had I z r dot is equal to a c alpha f minus b c alpha r this word right yeah and m into v dot plus r u is equal to f y okay so r dot and v dot you know that is what we are we are writing there minus a c alpha f plus b c alpha r by m u minus u that is the second term and minus a c alpha f minus b c alpha r divided by i u minus a squared c alpha f plus b squared c alpha r divided by i u okay that would be my a and v r plus c alpha f by m a c alpha f by i okay any questions on this I skip two or three you know simple lines which I think you can fill up now this is a familiar form to you right this is a very familiar form to you what is this form what is this called as state space form okay so this is nothing but a state space form what are the state variables here v and r okay the whole idea of writing this in a state space form is to look at the stability of the system that is the first idea later that we will use this in order to look at the response of the vehicle to dynamic inputs which is the steering input right so the first job here is to look at stability of the system now I will take questions any questions the only thing you have to be careful is c alpha f c alpha r minus plus whatever you want to use it it is fine as long as ultimately f y is in the positive direction right so that you have to be that is the only thing you have to be careful whatever you do ultimately you will get that expression only so minus to minus plus and all that you would get only that expression so now we are looking going to look at the stability of the system okay what do we yeah u is yes u is a particular velocity okay it does not change but it is an input right it does not change we are not going to calculate right okay please do please understand this a lateral dynamics so what do we mean by stability of the system how do we determine the stability of the system the immediate thing that comes to your mind when we talk about stability of the system is an eigenvalue problem right so you had studied that in your control system course and that there is no input the right hand side is equal to zero okay so and then you the left hand side which is the governing equation the right hand side which is the f okay is put to zero you go back and solve for the eigenvalues and so on and so forth this is what you have been doing okay here also you have to understand this carefully now we are going to do exactly the same thing we are going to put the inputs of to the system to be equal to zero in other words in other words I am going to put delta to be equal to zero this is the input to the system is equal to zero now there is a confusion if I put delta is equal to zero which means that actually I go straight now what is that I am doing when I go straight and I am putting delta is equal to zero and we are talking about stability in the lateral dynamics okay then what does it mean right that is the question so for that you have to understand what we mean by stability this is this usual thing because we are going to apply the concept of stability so we have to understand what is stability even if you are going straight okay we say that the system is stable in simple terms when there is a perturbation to the system the perturbation can be due to some suddenly wind that is blowing or due to a pot hole you take a small turn and so on you know a perturbation to the system you give the system returns back to its equilibrium position what is an equilibrium position the system continues to be in that position unless you it is forced to move out due to an input so a small perturbation which you give it gets gets back to its equilibrium position okay so that is why you put a very high school example of putting a ball in a well and when you distribute it sorry when you disturb it it comes back to this position because you are actually perturbing the ball about an equilibrium position okay and then it gets gets back to the so that is the philosophy here okay in other words in other words you cannot make a statement which I have seen many people making I am not going when I when I don't take a turn in a vehicle okay why am I worried about stability of the system no it is not true even if you go straight a small perturbation okay is enough to cause an unstable to to for the system to become unstable right so keep this in mind because this is going to be very useful for us to interpret the results later so we will get into the what is meant by stability how do you calculate stability and so on maybe you have done this before but we will do that again quickly okay now generally generally a system is expressed in terms of a differential equation okay which we will say that a dt power n yeah we will put that as is dt power n plus a n minus 1 d n minus 1 x dt n minus 1 okay this is the nth order system so the last term should be available plus a n into x is equal to 0 note that a n cannot be 0 if you are looking at an nth order system okay this is oh sorry a 0 into many times you write it the other way you can write this to be 0 1 and also as n okay either way you can you do that textbooks follow different procedures you can also write down okay this is a differential equation form the most immediate application if you look at this is in dynamics is your vibration equation you can write it as m x double dot plus c x dot plus k x okay you can also write this down in the state space state space form which we had written down okay which I can write down as x dot 1 okay in terms of what are called a state variables x dot 1 is equal to a 11 x 1 plus a 12 x 2 plus so on plus a 1 n x n and then x dot 2 is equal to a 21 x 1 plus a 22 x 2 plus and then so that x dot 1 x dot 2 x dot n is equal to a x okay is the is another way of writing it down there are two ways in which you can write down it does not matter in what way you write down okay if I want to look at the solution of this equation the solution of the equation can be written as x of t is equal to x e power s t okay that the solution of the equation can be written as x of t is equal to x e power s t this form of equation is valid only if we have certain characteristics for s I cannot put any s of course s has you know it can be real or imaginary and so on and so forth we will come to that in a minute so this x sorry this s okay has a characteristics which you have studied in your earlier courses and we will see what this characteristics of s has to be or what should this s be okay in order that this becomes a solution to the system right since x is equal to this thing x of x dot t is equal to s x e power s t and x double dot t is equal to s squared x e power s t and so on substituting that into governing equation we will do that for this case up in a minute right so substituting that in that expression what is that you get a naught s power n plus a 1 s power n minus 1 plus so on plus a n e power s t x e power s t is equal to 0 the next step is obvious but then this what is that for the non-trivial solution to exist okay this has to be equal to 0 okay what is the trivial solution x is equal to 0 because once this is not equal to 0 then x has to be equal to 0 and that is a trivial solution so the trivial solution or non-trivial solution to exist plus a 1 s power n minus 1 plus a a n is equal to 0 for non-trivial solution so there are s is nothing but the roots of this equation there are n roots and so the solution of the equation can be expressed as the sum of these roots and so on right okay. Now that is known now let us now look at the state space form please understand that this is one way there is no unique way of writing this one way of writing the state space it is not necessary that you will have or you need to follow only a set called x dot 1 x dot 2 x dot and so on for state space you can have another form as the state variables okay and express this equation in another form it does not matter you will ultimately see that the stability of the system obviously will not be affected by the way you write down the state space equation clear okay so let us now look at how to write this this is this clear it is exactly now extended here so how do I write this instead of x I will write that as x e s t okay so that x dot now becomes s x e s t substitute that into this expression okay x dot is equal to a x s x e s t okay is equal to a x e s t put an i here so that which is the identity matrix right and so I will write this down as s i minus a into x e s t is equal to 0 okay this is just a revision I know you you guys know this once we finish it we will go and apply it to this expression that is what we are going to do so for a non-trivial solution for this expression to exist what is it what is determinant of this that is the characteristic that is what is called as the characteristic equation and so the same condition for non-trivial solution to exist so the determinant of s of i minus a should be equal to 0 or else okay that is simple or else if it happens to be non-singular then obviously you will get end up with the only the trivial solution hence this has to be equal to 0 and that throws up values of s yes no s 1 s 2 no that is that is what that is what we are going to get now s 1 and and s 2 we are not we are we are looking still at stability okay we will come back to other quantities a bit later okay we are still we have not solved for s we have not found out the eigenvalues yet we are still looking at the equation for stability what is the condition for stability right I have not solved it yet for s 1 s 2 and so on right for this that is what I am going to do now first I let me look at the stability of this equation yes that is what I am saying yes okay I have not I have not yet solved it I have just writing down I said that there is s 1 and s 2 that is all what are they I have not yet determined them yeah so we will s 1 and s 2 will there will be two roots you know just wait for a minute let us finish this we will we will look at that you know that is what we did here I wrote this if it is a second order system it will be s squared plus you know that will be that that is what I said that there are n roots just wait just wait for some time okay let us first finish it and then we will go to the stability so now that is the determinant of that system is to be should be equal to 0 okay now what I am going to do now is to express I know what is A that is a that is a right what I am going to do is to substitute that a here and expand it and write down and write down some quantity of s squared plus some quantity s plus some quantity is equal to 0 what are those things we will do that in the next class just quickly so that you do not have a confusion as to what I am doing right so I am going to write this down like this right then look at stability of the system that is what first I am doing so look at stability of the system the stability of the system for for which an expression governing the characteristic equations like this is given by a very simple criteria roots criteria which states that what is here should be positive if I call this as a0 this is a1 and this is a2 the necessary and sufficient condition is that a0 a1 and a2 should be positive okay so that is the first thing I am going to do we are going to do problems later okay to look at the transient analysis and so on then we will see we will rework this out and see how I am going to write down expression of e power at you know matrix called e power at and so on we will do that later right okay so the my first step is to find out a0 a1 and a2 and see under what condition a0 a1 and a2 are going to be positive so the question which we are going to answer in the next class is is it possible for a system to be unstable for a vehicle to be unstable if so under what condition the vehicle is going to be unstable okay we will finish this class by saying that note what I mean by unstable if you are going straight okay it is possible that if the system is unstable or vehicle is unstable a disturbance that comes out or comes about in your system may not die down and may not bring you back to your equilibrium position right so that is the unstable case right we will now our job is to substitute a here write down that expression we will do that in the next class okay