# How suspension changes the handling balance of a car

You may have heard stiffening the rear suspension of a car causes more oversteer and vice verse but why?

Lets go right back to fundamentals.  Imagine a 100kg car with the weight in the middle of all four wheels.  Imagine this car cornering in such a way it just generates 1g of lateral acceleration.  Each axle must generate 50kg each to react the centripetal force.  If the weight of the car doubles to 200kg the force per axle is now 100kg each.  If the tyres can produce about the same grip in lateral force as the weight pushing down on them, they can generate this 1g of acceleration.  We could say this tyre has a co-efficient of friction of 1.0 which means the friction force divided by the weight is 1:

coefficient of friction (mu) µ = Friction Force / Weight     or

µ = Lateral Force / Normal Force   (normal is weight pushing down)

If we move the weight rearwards so that the rear tyres now have 150kg on them and the fronts only 50kg, the rear tyres now have to produce 150kg of lateral for the same acceleration but, as they have proportionally more weight on them and they produce the same amount of grip as force pushing down on them, that’s fine, they can cope.  The point is, putting more weight on one end of the car requires a larger lateral force to be generated by the tyres but it also gives the tyres a larger vertical load to create that force with.  However in this case, under steady state cornering, most tyres which were just on the limit with the weight in the middle would lead to the heavy end starting to slide.  That’s because tyres do not grip prortionally as well when a large load is placed on them as compared to when a lighter load is pushing them into the road.  They begin to saturate.

If we look at some real tyre data we can see the coefficient of friction for a rear ACB10 (the Formula Ford 1600 tyre) is higher with 84kg (the blue line) pressing down on it than when 306kg (the orange line) is pressing on it… considerably so too!  These are quite useful data to have as the weight of the rear of a FF1600 is about 300kg and the typical load on the inside tyre, when cornering fast, is about 80kg.  We’re only really interested in the ultimate value at about 9 degree of slip angle where the tyre begins to stall and can longer produce more lateral force despite any increase in slip angle.  Obviously the tyre creates a larger force at 300kg but proportionally for the weight pressing down on it, it is less.

What we want to achieve handling-wise is a car which slides a similar amount front and rear (if anything a touch more sliding at the rear but it’s up to the driver to control this with his right foot).  That means we need the coefficient of friction of the tyres to be about the same front and rear.  We know now that if we put a larger weight on one end of the car, the tyre will have a lower coefficient of friction at that end and will slide more.  So the only way around this is to use a fatter tyre at the heavy end which will saturate at a higher load.

As the rear of a FF1600 is about 300kg and the front 200kg we use a wider tyre at the rear which will give a higher coefficient of friction at 300kg than the narrower front tyre – it’s about 1.3 compared to 1.2 for the narrower tyre at 9 degrees of slip angle.  If you had the front tyre on both ends you’d get loads of oversteer as the rear is heavy.  The side force created by all of this weight would not be able to be reacted as well by the rear tyres despite the larger weight pressing them down just because the tyre would saturate earlier.  The wider tyre on the rear should give a FF1600 a reasonable balance before we start to tune it but a precision weapon like a FF1600 race car needs tuning and here’s how we do it…

We really don’t need to do any maths to understand this but we need to look at a basic diagram and have a think how we would calculate the forces on all the tyres.

The centre of gravity creates a force pushing the chassis down and also pushing the chassis sideways when cornering.  You would calculate all of the forces by working out all of the moments about a point like the circle around the bottom of the wheel.  But what is important is that we understand the total forces that point < that way equal the total forces pointing > that way and the same up and down.  The force pushing up on the tyres by the road is the same as the weight of the car.  The side force from the chassis is the same as the total side force created by BOTH tyres.  Because the centre of gravity (I’ll now call it ‘the weight’ for simplicity) is above the ground it has a tendency to lift the inside tyre off the ground as the car rotates about the circle.  The thing keeping it on the ground is the weight pushing it down rotating it the other way about that circle.

A real car obviously has two axles not just the one shown in the diagram.  The proportion of the total weight of the car on the front and rear axles is fixed by the location of the centre of gravity and nothing else (apart from forwards acceleration)!

We know the inside tyres have less weight on them compared with the outside tyres when you go around a corner, however, the total weight on the axle (i.e. the front inside and front outside tyres for example) never changes.  The front axle carries the same weight whether you’re going in a straight line or cornering that hard you’re on two wheels.  The only thing that can move weight from the front to the rear is accelerating/decelerating (roll bars can’t do this despite what idiots say!).

The cornering force increases the load on the outside tyres and decreases the load on the inside tyres.  Even though the total weight on an axle never changes we can change the proportion of weight on an axle’s inside tyre compared with the outside tyre on a car with suspension.  Imagine this…

Above is a car with a springy hinge one end and a rigid frame mounting the wheels at the other.  We know the total weight on each axle is fixed by the position of the centre of gravity and does not change.  However, imagine it cornering with enough cornering force causing the car to roll.  The inside wheel of the rigid end would instantly come off the ground as it’s so stiff whereas the end which can rotate will stay on the ground (unless it’s that fast it tips the whole car up – lets assume not!).  The whole inside of the car still weighs something and this load has to be carried by the inside tyre on the end which is free to rotate – the rigid end has no load on the inside because it’s rigid and any roll causes it to lift clear off the ground.  We know the total weight on the rigid end’s outside tyre must therefore be the total weight of the axle whereas the total weight of the hinged axle will be shared between the inside and outside tyres.

Now let’s say all four corners have the same tyre and the centre of gravity is bang in the middle of all 4 tyres like earlier.  Say each tyre has 25kg on it.  Up to now we’d say both axles would slide the same amount when going around a corner because the weight is equally shared, however, now we have a different situation.  Because of our basic suspension system we’ve invented, one tyre, the outside tyre on the rigid end has all of the weight of that axle on it (50kg) and the other could have almost the same weight on both the inside and outside tyre depending on how springy we make the hinge, position of CoG etc…  It doesn’t matter, what is important is that on the hinged (or soft) end the axle weight is shared between the inside and outside tyres and on the rigid (stiff) end the weight is all on the outside tyre.

To calculate the grip each axle has we need to know the coefficient of friction of the tyres on each axle.  What we do know is the tyre with the lowest coefficient of friction will be the tyre which is carries the entire weight of that axle on it’s own (the rigid axle’s outside tyre).  The next lowest will be the softer, hinged axle’s outside tyre followed by the hinged axle’s inside tyre.  The friction force from the rigid axle will be less than the other hinged/softer axle because the coefficient of friction of these wheels will be higher.

Because we’ve introduced a suspension system to the car which allows the chassis to roll we’ve created a situation where one axle can have more grip than the other because it can roll more than the other end.  The extreme case is where one end is completely rigid and the other end hinged but the principal applies to a stiffly sprung end and a softly sprung end.  One thing that is important is that the whole chassis which connects the ends together is very rigid so the angle each end rolls through is about the same!  If not the weight on the inside wheels may not be controlled by the other end so much.

We can make one axle stiffer than the other in the following ways:

• Stiffer/softer suspension springs – the stiffer end will slide more
• Antiroll bars – the more roll bar we use the less grip the axle will have
• Roll centre position – the lower the roll centre away from the Centre of Gravity the more leverage the CoG has on the roll centre hence more roll.  Roll centres are the point at which an axle is able to rotate
• Ride height adjustment – this mainly affects the axle’s roll centre height but if large enough the CoG will move closer to or further from the roll centre and change the leverage on it.

These will be discussed in seperate posts!

Michael