The Force of Rear Axle and Front Axle Torque

That'll be torquing in this direction and that has to be equal to uh basically from Statics or Dynamics uh the force of the rear time B so this torquing Force multiplied by this torquing Force they have to be the same otherwise the car would be like bending and uh you know kind of destroying itself so that's just a basic equation um and so this gives us the force at the front is equal to the the force of the rear times B and then you just divide over that a so divided by a okay so now step five and things are going to start to get a little more tricky we're going to be substituting uh the equation that we have in four 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 MB^2r that makes sense okay now we're going to be uh basically you can see that f_r * uh plus f_r * B over a is the same thing as FFR * 1 + B over a and that's equal to mv^2 over R all right great that's simple now a + b equals L you can see that here L is just the wheelbase of the vehicle so if we take 1 + B over a that's the same thing as saying a plus b over a so a divid 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 * LL over A because 1 + B over a as we proved here is equal to lA so f_r * lL over a equal mv^2 and so now if we move LL over a to the other side of the equation you can see that f_r so you multiply that by a so ma over L multiplied v squared so we have our final equation from step five so FFR is equal to the mass times the distance a divided by L the wheelbase multiplied by the speed squared divided r.

Finding the Weight on the Rear Axle

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 * a is going to have to be equal to to uh W * P but if we sum the moments about it so we've got W * LWR here it's going to be providing this torque basically W the force times the length at which it's acting so that's WR * 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 wal MGA a a and then you can divide over L over l so W over G if you rearrange once again you divide this G over on this side W over g equals m a over L and this is useful because as you can see right here in this equation for the force of the rear ma over L is equal to WR over G so we're going to substitute that in uh for our equation number seven so FFR is going to equal W over G which is the same thing as ma over L * V^2 / r.

Substituting Equations

What we have here is we're going to substitute equation 7 into equation 1 so we know uh we'll do this for the rear so we know the slip angle at the rear will be equal to the weight uh at the rear divided by gravity time V^2 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 r of weight of the rear divided by the cornering stiffness at the rear multiplied by V^2 over gr and that's our final equation this gives us our slip angle uh and this is actually cool because you can use this now to calculate why your car would be under steering or over steering uh you can use it you know in determining where you should be placing weight in order to maximize uh the car's steering Dynamics and so it's a 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

"WEBVTTKind: captionsLanguage: enhello 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 Jeffrey ball to come up with this derivation hopefully simplify it a little bit now these two equations here uh 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 in 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 uh you can close your eyes or plug your ears uh 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 okay so I've got three diagrams here which are going to be helping us out uh in explaining where this equation comes from the first one here lateral Force versus slip angle uh 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 uh this is a property of the tire multiplied by the slip flip 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 uh and so this represents the two front wheels this represents the two rear wheels basically everything's combined into one uh 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 the center of gravity in the middle so we have our first equation uh 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 the cornering 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 uh the force of the center of gravity of this vehicle going around a corner is equal to mv^2 r m 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 uh basically the force on that ball is going to be equal to mass time velocity s / R this is a basic equation from Dynamics I'm not going to derive it 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 uh we're looking at this bicycle model here uh if we sum the lateral forces than four small angles then we can come up with our next needed equation um basically you're going to have uh 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 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 uh because this is going to use the cosine of that angle the cosine of a small angle is going to be about one and so basically what I'm telling you is is uh 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 uh pretty simple equation there and if we substitute this in for the force of our center of gravity then we can see that the force on the rear plus the force on the front is equal to mv^2 over R okay moving on to the fourth equation let's sum the moment about the center of gravity so we're going to be summing the torque forces basically about this point right here so we have the force of the front time a that'll be torquing in this direction and that has to be equal to uh basically from Statics or Dynamics uh the force of the rear time B so this torquing Force multiplied by this torquing Force they have to be the same otherwise the car would be like bending and uh you know kind of destroying itself so that's just a basic equation um and so this gives us the force at the front is equal to the the force of the rear times B and then you just divide over that a so divided by a okay so now step five and things are going to start to get a little more tricky we're going to be substituting uh the equation that we have in four 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 MB ^2 r that makes sense okay now we're going to be uh basically you can see that f r * uh plus f r * B over a is the same thing as F FR * 1 + B over a and that's equal to mv^2 over R all right great that's simple now a + b equals L you can see that here L is just the wheelbase of the vehicle so if we take 1 + B over a that's the same thing as saying a plus b over a so a divid 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 * L A because 1 + B A as we proved here is equal to l a so f r * l l over a equal mv^2 and so now if we move L A to the other side of the equation you can see that f r so you multiply that by a so ma a you divide everything by L so m a over L multiplied v^ s r so we have our final equation from step five so F FR is equal to the mass times the distance a ID L the wheelbase multiplied by the speed ID squared divid 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 * a is going to have to be equal to to uh W * P but if we sum the moments about it so we've got W * L WR here it's going to be providing this torque basically W the force times the length at which it's acting so that's WR * L minus mg over a so mg coming down at a distance a mg being of course masstimes gravity so that would be you know the force coming down the weight of the car okay so rearranging this a bit you have wal MGA a 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 W over g equals m a over L and this is useful because as you can see right here in this equation for the force of the rear ma over L is equal to WR over G so we're going to substitute that in uh for our equation number seven so F FR is going to equal W over G which is the same thing as ma over L * V ^2 / r are okay we are almost there uh hopefully at least 5% of you are still with me and what we have here is we're going to substitute equation 7 into equation 1 so we know uh we'll do this for the rear so we know the slip angle at the rear will be equal to the weight uh at the rear divided by gravity time V ^2 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 r of weight of the rear divided by the cornering stiffness at the rear multiplied by V ^2 over gr and that's our final equation this gives us our slip angle uh and this is actually cool because you can use this now to calculate why your car would be under steering or over steering uh you can use it you know in determining where you should be placing weight in order to maximize uh the car's steering Dynamics and so it's a 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 watchinghello 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 Jeffrey ball to come up with this derivation hopefully simplify it a little bit now these two equations here uh 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 in 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 uh you can close your eyes or plug your ears uh 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 okay so I've got three diagrams here which are going to be helping us out uh in explaining where this equation comes from the first one here lateral Force versus slip angle uh 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 uh this is a property of the tire multiplied by the slip flip 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 uh and so this represents the two front wheels this represents the two rear wheels basically everything's combined into one uh 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 the center of gravity in the middle so we have our first equation uh 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 the cornering 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 uh the force of the center of gravity of this vehicle going around a corner is equal to mv^2 r m 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 uh basically the force on that ball is going to be equal to mass time velocity s / R this is a basic equation from Dynamics I'm not going to derive it 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 uh we're looking at this bicycle model here uh if we sum the lateral forces than four small angles then we can come up with our next needed equation um basically you're going to have uh 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 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 uh because this is going to use the cosine of that angle the cosine of a small angle is going to be about one and so basically what I'm telling you is is uh 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 uh pretty simple equation there and if we substitute this in for the force of our center of gravity then we can see that the force on the rear plus the force on the front is equal to mv^2 over R okay moving on to the fourth equation let's sum the moment about the center of gravity so we're going to be summing the torque forces basically about this point right here so we have the force of the front time a that'll be torquing in this direction and that has to be equal to uh basically from Statics or Dynamics uh the force of the rear time B so this torquing Force multiplied by this torquing Force they have to be the same otherwise the car would be like bending and uh you know kind of destroying itself so that's just a basic equation um and so this gives us the force at the front is equal to the the force of the rear times B and then you just divide over that a so divided by a okay so now step five and things are going to start to get a little more tricky we're going to be substituting uh the equation that we have in four 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 MB ^2 r that makes sense okay now we're going to be uh basically you can see that f r * uh plus f r * B over a is the same thing as F FR * 1 + B over a and that's equal to mv^2 over R all right great that's simple now a + b equals L you can see that here L is just the wheelbase of the vehicle so if we take 1 + B over a that's the same thing as saying a plus b over a so a divid 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 * L A because 1 + B A as we proved here is equal to l a so f r * l l over a equal mv^2 and so now if we move L A to the other side of the equation you can see that f r so you multiply that by a so ma a you divide everything by L so m a over L multiplied v^ s r so we have our final equation from step five so F FR is equal to the mass times the distance a ID L the wheelbase multiplied by the speed ID squared divid 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 * a is going to have to be equal to to uh W * P but if we sum the moments about it so we've got W * L WR here it's going to be providing this torque basically W the force times the length at which it's acting so that's WR * L minus mg over a so mg coming down at a distance a mg being of course masstimes gravity so that would be you know the force coming down the weight of the car okay so rearranging this a bit you have wal MGA a 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 W over g equals m a over L and this is useful because as you can see right here in this equation for the force of the rear ma over L is equal to WR over G so we're going to substitute that in uh for our equation number seven so F FR is going to equal W over G which is the same thing as ma over L * V ^2 / r are okay we are almost there uh hopefully at least 5% of you are still with me and what we have here is we're going to substitute equation 7 into equation 1 so we know uh we'll do this for the rear so we know the slip angle at the rear will be equal to the weight uh at the rear divided by gravity time V ^2 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 r of weight of the rear divided by the cornering stiffness at the rear multiplied by V ^2 over gr and that's our final equation this gives us our slip angle uh and this is actually cool because you can use this now to calculate why your car would be under steering or over steering uh you can use it you know in determining where you should be placing weight in order to maximize uh the car's steering Dynamics and so it's a 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\n"