 Yeah, we said that we will start the lectures on lateral dynamics and we were talking about as a precursor when we started this in last class, we were talking about what we called as high site fall okay. We said that there is the graph and both of them are acting. I hope you are able to derive the expressions which we got we wrote in the last class. If there are any questions we will answer that in the tutorial class right and please derive that those expressions which we said or which we gave which we saw in terms of the combined slip okay the lateral and the longitudinal slip from the brush model right. Try it out or else we will do that sometime next week in the tutorial class right okay. So we were just talking about this as a precursor to the lateral dynamics. We said that suppose a motorcycle is taking a turn, you are taking a turn in the motorcycle and that you are applying a brake okay. When you apply the brake actually you what are you doing? You are getting both the longitudinal forces, you require a longitudinal forces for braking as well as due to cornering you also require a lateral force. Assume that you are at the edge of the friction circle or ellipse as you can call it or other words you are say at this position. Let us say that you are at that position. In other words you are making use of all the friction that is that is available right. Suppose you let go, you let go the brake that you have applied okay and you keep cornering. What would happen? That is the question. Suppose you let go the brake and then you continue your cornering. In other words what would happen now? The longitudinal force goes to 0 right. The longitudinal force goes to 0 and suddenly okay you are now shifting from this corner where you are at the boundary to the vertical axis okay with the same force, total force. You are now shifting from here to here. In other words the lateral force is more than what is required for the equilibrium of the vehicle. Yeah lateral force would remain the same but yeah because what happens is something like this I have plotted the slip curves. See suppose you are at this point okay you are at this point then if you now apply a brake the slip force remains the same. In other words you are moving from here. These are the slip curves, forces that are generated due to the slip angle. The slip angle remains the same okay but you are shifting from this corner on to the y axis. So in other words you are shifting from B to C. You are shifting from B to C position right. So because look at this curves how they are. This is the lateral force curve maybe this is equal to alpha is equal to 3 degrees and so on right. So you are shifting now to the y axis. In other words the forces the y axis force which was this much has suddenly increased from this point to that point right. So when the forces increase suddenly increase and is not it is more than what need to be compensated by the for the centripetal force then there is no equilibrium there is loss of equilibrium. So in other words you overturn clear okay. We will come back to this maybe towards the end of the end of this lateral dynamics where you understand it better but I am only telling you that there are practical applications of what we are going to study okay. Now we will get into the complete lateral dynamics derivations so that when I come back you would understand this better right. Let us look at how we are going to deal with lateral dynamics. Let us go into some fundamentals of dynamics right. Now I am going to deal with what is called as a body centered coordinate of a week. I am going to deal with what is called as the body centered coordinate of the week. In other words you can study the vehicle motion with respect to what is called as the inertial frame of reference which I would put in terms of capital X and capital Y or you can study this motion with a body centered coordinates which I would put as X and Y. It is a usual practice to study the motion of a vehicle in terms of body centered coordinates rather than the inertial frame of reference. You all have done engineering mechanics so you would understand what would happen when I have a body centered coordinate. You would see that you have to have additional terms to take care of the realignment of the body centered coordinates in order to take care of realignment of the body centered coordinates. In other words what are the body centered coordinates? Suppose this is the vehicle so I have a body centered coordinate in this vehicle so the vehicle takes a turn the body centered coordinate also takes a turn okay so they also move along with the vehicle. In other words in other words I have to take into account this motion of the vehicle and the body centered coordinates right. How do we take this into account? First thing is that is what we are going to see and then we are going to apply Newton Euler equation in order to understand the motion right. Let us say that the usual ijk is what defines the unit vectors along x, y and z right and let us say that the velocity of the vehicle in the three directions are u, v and w. These are the velocity of the vehicle in the three directions right and let us say that the rotation okay which we would call as p, q and r denotes the rotation of the vehicle or in other words the angular rotation velocity or angular velocity of the vehicle okay. So this is the angular velocity or rotational velocity and u, v, w is the linear velocity of the vehicle. In simple terms what is p? p is nothing but the roll, q is the pitch and r is the yaw of the vehicle okay so we call this as p, q and r. Now what are the peculiarities if I am going to deal with this kind of systems okay that is what we are going to see. You know like what are the additional terms that come into picture. Let us say that the force acting okay on the vehicle let me write down the forces as f x i plus f y j plus f z k is the three forces that are acting on the vehicle and the moment m or let me call that as tau okay the moment that is acting at when they are all vector quantities I am putting a squiggle at the bottom to indicate that they are vector quantities. So let me call that as L plus I will maintain the same thing so m what am I going to do I am going to first write down the linear momentum and the angular momentum and I am going to look at the rate of change of this momentum and the result is the Euler Newton Euler equation right that is all I am going to do. So let me write down first the linear momentum p m v and the angular momentum let me call that as h what is angular momentum I omega right now what is I now in this case the moment of inertia is it is actually a matrix right. So you actually have this as I okay omega what are omegas angular velocities how are we define this p q r okay so I omega and what is I is if you want we can replace that with a matrix notation okay I you would have studied this as one single quantity along xx and so on now we are just extending this to 3D so I am writing this as a matrix and so I xx so this is I xx I xy I xz so you all of you know the definition for I xx integral y squared plus z squared dm right so you know the definition for I xx throughout the volume of the body and you know also what is I xy minus of integral xy dm right okay from this we go to write the we are going to have some peculiarities we will come to that in a minute we go to write the Newton Euler equation right so what does the Newton Euler equation say so how do I if it is just f is equal to ma we call this as a Newton's equation so if it is a Newton Euler equation we also include the moment that is acting okay on the on the body so we write f is equal to where p is the linear momentum of the body similarly we write tau is equal to rate of change of angular momentum right so far it is very simple dynamics now we have to be careful in writing these terms okay we are going to write it with respect to the body centered coordinates I jk have introduced that for the body centered coordinates okay now I have to introduce them with body with respect to a body centered coordinates let us see how for a body centered coordinates a particular vector changes that is what is our goal now okay let us consider a vector a okay vector a some vector a we are we are we are following Kornop the reference for this is vehicle stability by Kornop k a r no p p Kornop vehicle stability is is the reference for this so let us say that I have a vector a which I am going to express in terms of I jk which are the unit vectors in the body centered coordinates I want to find out now the variation or rate of change of a so I have in other words if I now write da by dt I write this as a dot x I plus a dot y j a can be any vector it can be velocity we will see that a is a dot k plus because I is not the inertial frame of reference and I also changes with time so I have to rewrite that part so I will in other words all of you from a fundamental dynamics know how to write I dot how do I write I dot I dot j dot and k dot in other words di by dt you remember that in your earlier classes you would have done this suppose I have x and y along which I and j are defined and if it is rotating with say for example omega so the new coordinate due to a rotation of an angle theta gives you x prime and y prime right so the new value of I which let us say that I prime which is equal to I plus di by dt right is given by the changes okay here which is along j same way along I okay and so what is di by dt d theta by dt into j right so because this is a unit vector I is a unit vector r into theta gives me this right so this is r into theta it is along the direction of j so di sorry di by dt di is this di by dt is nothing but d theta by dt d theta by dt is omega omega into say j right so this you can do it for all three and write down this part right this we will keep that plus omega cross that is what happens omega in our case is pq and r clear so we will call this part we will give a new name to this and we will call this as dou a by dou t call this as relative so what is the physical meaning of the first term the physical meaning of the first term is simply that that is what an observer who is sitting in the body centered coordinate would see as the rate of change of a right so because I have to apply this my formulas in terms of inertial frame of reference now I know that when the body centered coordinates okay changes with time okay I take into account that as well by putting that omega cross a right okay what is omega cross a all of you know this where omega cross a how do I write that i j k what is omega as I said before pq r and a is a x a y a z so now d a by dt if you write that in terms of this write it in terms of that so what would happen for i a dot x plus pick that up from there plus i into q a z minus a y the whole thing is i right if I want to express this as a vector then I would remove that i and just write that as a vector because after all a that is a vector write down a dot y plus what are the terms that are there that is the that is for the j term r a x minus p into a z and a dot z plus p a y minus q a x yeah so that is d a by dt now I can apply my Newton Euler equation by assuming that a is are substituting for a the velocity I can write down the first of the equations which is d p by that is here that equation I can write it down of course m is a constant so m is you can take that out and so write down what is the first of the equations simple a f x f y f z is equal to the first term becomes so m into what is it u dot plus q into u v w is what we said q into w minus r into v right any questions no questions everything is crystal clear write down the rest of it write down the rest of it v dot plus r into u minus p w and w dot plus p v minus q so that is the first set of equations the second set is my tau is equal to d h by dt which is here tau is given by l m and n so that becomes my left hand side so let us let us write down that tau is equal to d h by dt again that is equal to dou h by dou t relative this is the first term that is the h dot term plus omega remember h is equal to remember h is equal to i into p q so omega is p q and r right omega is not p dot q dot and r dot it is not p q and r right this is a very standard terminology for example r small r is used in the literature in vehicle dynamics to indicate yaw okay this is a very standard terminology it is not usually written as omega x omega y and omega c right write down that that is going to be an interesting expression it is going to be a long expression do not differentiate i with respect of course t so the words of what is the first thing okay let me remove this so do you know that already l m n i x x p dot plus i x y q dot plus i x z r dot that is the first term in the vector plus write down omega cross h right okay I will write down the final things it is just nothing it is not very difficult so let me I will quickly write down that I hope I do not make any mistakes in this if there is anything just point out i y y q dot plus i y z r dot i z x p dot plus i z y q dot plus i z z r dot okay so that are the dou h by dou t terms so then I have to put the same way q into a z minus r into ay okay what is a z a z is h z h z is what remember that I had to multiply I had written already what is h so i x x i y y i y z multiply it and put that there and so there will be plus q into h z okay how many terms will be there and h z or h z how many terms there will be three terms right so write down those three terms so you can write down that as q into h z minus r into h y remember that h x h y h z is what i x x p plus i x y q plus i x z r so what is this term is the h x term and that term h y term h y and that term is the h z term right so that is what is going to come here and so on so the expression looks very formidable and long it is very simple to understand there is nothing much it is very simple to understand this expression okay do not worry I am not going to deal with this kind of huge expression I am going to make as usual some assumptions right what are the assumptions you think we can make what are the terms I do not like okay but I want I want all that you know let us u v w terms I need the terms usually I do not like are this i x y and so on right so how do I eliminate that look at I mean imagine that you have a car like I mean up to this is mundane there is nothing nothing great about it you know just the relationship between inertial frame of reference and body centered coordinates now what are the assumptions you think you can make symmetry fantastic so symmetry in a car is about what are the axis is that you can consider the symmetry what what is what do you think and we can consider so one is a longitudinal axis you can consider this as symmetry the other is that we are dealing with principal axis let us say that the I x x okay I y y and I z z and so on what you have chosen as x y and z for the car let us say that it is the principal axis or they may not but then in that case if I choose them to be the principal axis all others go out right okay so we make this kind of assumptions and we also say that the q the pitch is not is not of interest to us neglected and ultimately we will bring it down in a very simple handleable form right we are going to make all these assumptions and then we will bring it to handleable form so what is my first expression I am going to now write down right I mean is this this is clear any questions okay now with that assumptions I am going to rewrite my expression one say right here one and my expression to substitute that and write it going to be a huge expression and I am going to rewrite them in simple form handleable form which I am going to use so let us let us see write down what is what is the expression which I can use f x and f y let us write down f x and f y so f x that is the longitudinal force in the longitudinal direction which we had already seen okay but with respect to the body centred coordinate is m into u dot minus right so that is the body centred coordinate expression for longitudinal forces now write down for the lateral force f y f y is equal to m into v dot okay that is the first term our relative d a dou a by dou t so because of which I get a v dot term what is the next term I would get I am neglecting q and so on plus r q right and what is my l and m l and m from my expressions here so it will be the first term will be there i x x into p dot if you are not considering it just consider symmetry or not considering that you are under you are in the principal axis then i x z will be there so i x z into z into okay and so on right so you can write down for m r dot again plus the last is this term what we have is here the other terms because of symmetry goes off plus i z x into p so I am going to remove these two later and look at I would say the expressions for the lateral dynamics so this in other words this is the most general expression you can look at for the three dimensional case but I am going to from here I am going to make geometric assumptions so that ultimately the equations I am going to use become handleable right that is what I am going to do and from this handleable equations handleable in sense that you know I want to work it out in an analytical fashion I want to look at the results I want to interpret it and so on so from this I am going to extract out the equations for lateral dynamics that is all I am going to do clear any questions yeah oh sorry n n l n okay for yam you can sorry n n n you are right and that would be okay that is the first that is this term and you can you can write this down I think you should have i z x into p yeah you are right good okay now we will move and be more specific and we will get into the expressions for the longitudinal dynamics so what are the two expressions of interest to us in longitudinal dynamics what are the two expressions which are of interest to us when you are talking about sorry lateral dynamics you are talking about lateral dynamics what is the force talking about f y right and what is the other thing that we are talking about is the moment due to lateral forces so the moment due to lateral forces will be along z direction right the moment due to the lateral forces will be along z direction so that is my interest so in other words just to quickly summarize so I have forces that are acting okay that is the vehicle then that forces create a moment okay which is perpendicular to the plane of the board which is the z so the two equations of interest to us are the lateral forces and the moment that is acting due to the lateral force which is this yes no yeah we will come to that okay so we are now going to look at a simple model which is called as the bicycle model okay so when I consider this that is why I have drawn it like this I am not going to consider the role okay so you are absolutely right I have to role is an important phenomena that happens you know during cornering but right now okay I am going to neglect the role I was expecting that question I am happy that you asked this so let us now look at these two expressions in other words what I am going to do is I am going to shrink these two wheels okay into one wheel like this the front and the rear wheels and look at the vehicle by means of these two wheels the front and the rear and call this as a bicycle model this is one of the most famous models in vehicle dynamics as he very correctly said we are not taking into account the role right now but then when after we study this we are going to introduce complexities one by one in order to understand what happens okay because of other factors which we have neglected what happens due to the other factors this is what we are going to study clear okay so we are into one of the most important models in the whole of vehicle dynamics which is called as the bicycle model right so I will remove this as I promised I will make it much easier so that is the one equation I will remove this I will remove this and these are the two equations that I am going to look at let me now expand this bicycle model now we will introduce after crossing the all these equations now we will introduce lot more again physics right okay so let us say that this is the rear wheel which I have collapsed which I have collapsed and let us say that this is the front wheel which is okay I am giving a turn to the front wheel let us say that that is just exaggerating this I am giving a steering input to the wheel okay of course that is the vehicle that is the center center of gravity location of the vehicle okay so that is the lambda which I had given which is the steering input now so I am taking a turn when I have when I take a turn I will of course need this Fy which we have been talking about right and how is Fy created or how is it generated we have already seen that this is due to the slip angle so a slip angle if this is the force that is acting okay the small angle differences which I am going to neglect right so that is the Fy in the front and this is the Fy at the rear so how is it generated by slip angle and what is the slip angle and that is the slip angle is alpha okay note out my coordinate system that is X, Y, Z is perpendicular to the plane of the board clear right now the very first point I want you to notice we will continue this in the next class is that is that the slip angle which we have seen it before but I am going to put some sign to it the slip angle is look at that is in this direction which is which is in the negative direction okay and the force is created in the positive direction of Y in the positive direction okay so this negative and positive directions we have to though we know this physically we know how Fy is created but this negative in the positive directions will be careful when we write down the equations right so there will be two alphas that are created because I want the front and the rear and let me call that as the rear alpha or alpha R the rear slip angle and let me call this as the front slip angle so note that this is a confusion I have said this again and again note that what is slip angle this alpha is not is nothing to do with this delta which is the steering angle clear I need a slip angle to generate the force now I have these two equations the n is equal to what is R that is R okay which is the yaw velocity and r dot is the rate of change of R right okay let me call this is again usually done like that we call this as L1 and L2 let me call that as A and B because in the lateral dynamics literature the distance from the of the CG location from the center of the tire is A and that is B right okay so that A into Fyf minus B into Fyr is the moment that is acting so n is equal to Fyf into A minus Fyr which is equal to Izzr dot clear Fy will come to this in a minute Fy is Fyf plus Fy we will continue this in the next class we will stop here and we will continue the bicycle model the this is probably the most important central piece of the whole lateral dynamics so we will look at this very carefully and then understand the vehicle behavior during cornering so we will continue in the next class.