# Deriving Car Slip Angles – Tire Slip – WARNING: Maths

hello everyone and welcome in this video we’re going to be deriving the equation for slip angle and I’m going to be using this book automotive engineering fundamentals by Richard stone and Jeff people to come up with this derivation hopefully simplify it a little bit now these two equations here are for slip angle for the front and the rear of the vehicle are very important in understanding vehicle dynamics and I’m going to be referencing these equations and videos so I just want to make sure I’m thorough and so I’m going to derive this here now if you are terrified of math or you can’t stand math you can close your eyes or plug your ears you could quit watching the video but in all honesty you know when it comes to mechanical engineering this is pretty basic stuff so don’t be too afraid of watching ok so I’ve got three diagrams here which are going to be helping us out and explaining where this equation comes from the first one here lateral force versus slip angle and this as you notice has a linear curve for small slip angles so the lateral force will be equal to the slope which is the quartering stiffness this is a property of the tire multiplied by the slip angle that’s our first equation here this second diagram I’ve got here is basically what’s the bicycle model it’s a model of a car looking from top down and so this represents the two front wheels this represents the two rear wheels basically everything is combined into one and then this third model here is looking at the car from the side here you have the front tire the rear tire and you are looking at it from the side and there’s a center of gravity in the middle so we have our first equation which says the force the lateral force is equivalent to the cornering stiffness multiplied by the slip angle so in order to calculate the slip angle we can just divide the cornering for or the acquiring stiffness over so we get slip angle equals the lateral force divided by the cornering stiffness now our next equation this is a basic dynamic equation the force of the center of gravity of this vehicle going around a corner is equal to MV squared over R and being the mass V being the speed it’s going around and R being the radius so if you take a ball on a string and you swing it around a center point at a constant speed basically the force on that ball is going to be equal to mass times velocity squared divided by R this is a basic equation from dynamics I’m not going to but I will have a link in the video description that does derive it if you’re interested now if we sum the lateral forces we’re looking at this bicycle model here if we sum the lateral forces then for small angles then we can come up with our next needed equation and basically you’re going to have the force on the front tire the force on the rear tire and that’s going to be equal to the force of the center of gravity which is traveling the other way of course because your tires are holding your car on the road now I noted here for small angles and the reason that’s important is because actually this force here isn’t directly perpendicular to this force of the center of gravity it’s at an angle but for small angles because this is going to use the cosine of that angle the cosine of a small angle is going to be about 1 and so basically what I’m telling you is this is a very accurate representation for small angles where the force on the front tire and the force on the rear tire is equivalent to the force at the center of gravity pretty simple equation there and if we substitute this in for the force of our centre of gravity then we can see that the force on the rear plus the force on the front is equal to MV squared over r ok moving on to the fourth equation let’s sum the moment about the center of gravity so we’re going to be summing the torquing forces basically about this point right here so we have the force of the front times a that will be torquing in this direction and that has to be equal to basically from statics or dynamics the force of the rear times B so this torquing force multiplied by this torreón force they have to be the same otherwise the car would be like bending and you know kind of destroying itself so that’s just a basic equation and so this gives us the force at the front is equal to the force of the rear times B and then you just divide over that a so divided by a ok so now step five and things are going to start to get a little more tricky we’re going to be substituting the equation that we have in 4 into this equation that we have here in five so we can see that the force of the rear here plus the force of the front is equal to this so we can substitute that in and that equals MV squared over R that makes sense okay now we’re going to be basically you can see that F R times plus F R times B over a is the same thing as F R times one plus B over a and that’s equal to MV squared over R alright great that’s simple now a plus B equals L you can see that here L is just the wheel base of the vehicle so if we take 1 plus B over a that’s the same thing as saying a plus B over a so a divided by a would be 1 plus B over a so 1 plus B over a so we have a plus B over a that’s the same thing as saying L over a and the only reason we’re doing this is to substitute L into this equation so we can have now we can say F R times L over a because 1 plus B over a as we proved here is equal to L over a so fr times L over a equals MV squared and so now if we move L over a to the other side of the equation you can see that fr so you multiply that by a so MA you divide everything by L so MA over L multiplied by B squared over R so we have our final equation from Step five so fr is equal to the mass times the distance a divided by L the wheelbase multiplied by the speed divided it squared divided by R okay now we want to find the weight on the rear axle and in order to do this we’re going to be using this diagram and we’re going to be summing the moments about this front side right here so we know that basically mg times a is going to have to be equal to W R times B but if we sum the moments about it so we’ve got WR x LW R here it’s going to be providing this torque basically WR the force times the length at which it’s acting so that’s W R times L minus mg over a so mg coming down at a distance a mg being of course mass times gravity so that would be you know the force coming down the weight of the car okay so rearranging this a bit you have WR equals mg a and then you can divide over L over L so W R over G if you rearrange once again you divide this G over on this side WR over G equals M / L and this is useful because as you can see right here in this equation for the force of the rear ma / L is equal to WR / G so we’re going to substitute that in for our equation number seven so f R is going to equal WR / G which is the same thing as ma / L times V squared over R ok we are almost there hopefully at least five percent of you are still with me and what we have here is we’re going to substitute equation seven into equation one so we know we’ll do this for the rear so we know the slip angle at the rear will be equal to the weight at the rear / gravity times V squared over R and all of this is going to be divided by the cornering stiffness so once we divide that cornering stiffness we can put that in the equation the slip angle at the rear is equal to the rate of the weight of the rear divided by the cornering stiffness at the rear multiplied by V squared over G R and that’s our final equation this gives us our slip angle and this is actually cool because you can use this now to calculate why your car would be under steering or over steering you can use it you know in determining where you should be placing weight in order to maximize the car steering dynamics and so it’s very valuable equation and I will be using it in other videos so hopefully for those of you who are interested in where this equation came from that made sense and if that didn’t make any sense of course feel free to leave questions and comments below thank you for watching