Some days ago a pale from the ultimatecarpage.com forum asked for a quick explanation on the so called “liftoff oversteer”, mostly known for its effects on the Porsche 911.
After giving him an answer, I thought many others could be interested in this so to understand why it's a nice way to drive when you are in your Miata and a good way to kill yourself when driving dad's 935.
Long somehow technical rant after the jump.
So, you're driving pretty fast, no matter which car right now, and it's time to brake, and to brake hard.
This braking action acts towards the longitudinal direction of the car.
Given the front tires can provide enough grip, you already know that this will create a relative rotation between the car and the front axle, specifically the nose of the car will go down to kiss the road, and the tail will be up in the air like this:
What people could think is that this could be just a wonderful thing, because more weight is transfered to the front tires, which therefore will have more grip to count on for the next corner.
That's because what we generally call “grip” is the friction force, which is given by this relationship:
(or frictional coefficient)
First of all, no mass is transferred. We should clarify if we are talking about the mass or the force pushing something to the ground, the weight force, given by
That said, the mass doesn't change its position, and the gravitational acceleration isn't modified.
So what's happening? We'll see this later.
The point why this movement of the car, called yaw, pitch, or roll, isn't good at all because it will force the car to rely on mainly two wheels, so less stability, and because this is a rotation around the front axle, so the car is not “positioned” on the front tires, it's actually trying to rely on the noose, which isn't going to help you in any way.
The front of the car doesn't “weight” more, it's only the tail to weight less, and the front that is trying to dive into the ground. Just think about Formula 1 cars or ALMS prototypes, they don't have an evident roll right?
If you like, the car in this moment “weights” less, but it isn't correct, the weight is still the same, albeit not completely transferred to the ground Hence why the front tires aren't going to provide more grip than before. It's dynamic effect (more on this later, again).
That said, we left you while braking hard with the tail ready to take off into the sky, and we just understood, I hope, that this isn't going to help you shortening your braking distance.
Now, you're still braking, and a corner it's approaching, so while still braking (or even right after you released the brake pedal) you start turning the steering wheel in the direction of the corner.
Of course you do, it's called “inertia force”. Trying to make a corner, you're imposing a centripetal (toward the center of your trajectory) acceleration to your car, and in that moment you're on the car so you're feeling it as well. Now you may be asking, why I'm pushed on opposite direction?
That's the trick. Basically the inertia is the resistance to change your status, and it's proportional to the acceleration you just applied to the ground trough the steering wheel, and then the tires, and multiplied by the mass (NOT the weight).
It just acts in the opposite direction of the acceleration you imposed, so to fight the change the same original acceleration was trying to apply.
It's the same kind of force which created the relative rotation around the front axle when you started braking.
Explanation: when you brake, you're decreasing your speed, hence an acceleration towards the rear of the car is applied trough the brakes, and once again the tires. As we already said, at every acceleration (of a decent intensity) corresponds an inertia force.
Since the braking acceleration is directed to the back of the car, the inertia force is directed towards the front, right? Now, as you could know, every force can be considered as applied to the center of gravity if we are considering a rigid body (with this forces we can consider the car as a rigid body).
The center of gravity of the car is generally positioned higher than the axle (both rear and front obviously), hence this force directed forward and applied to a point higher than the axle around which the rotation is happening, provokes a clockwise rotation if you're looking the car from the right side, and anticlockwise rotation if you're looking at its left side.
Back to the corner, now you're braking, the tail is a bit suspended, and you're pushed towards the outer side of the corner. Even the car itself is pushed in the same “wrong” direction, and once again this force is applied to the center of gravity of the car, and transferred to the ground via the four tires (which will fight this force if enough grip is provided).
Now we could start distinguishing which kind of car we are talking about.
A Porsche 911 of the 997 generation has a weight distribution of 38:62, so the 38% of the weight is transferred to the ground trough the front tires, and the remaining 62% trough the rear tires. With older gens the situation should be even worst, but I don't have any reliable data to share.
Be aware that this is a static situation, once we start moving and an acceleration of some sort is applied, this isn't correct anymore. The problem becomes “dynamic” (see the point at the beginning).
That's explain why if the tail of the car is suspended it doesn't mean all the weight is relying on the front axle.
In the same way the inertia force caused by the braking creates a rotation of the car around the front axle, it also try to suspend the rear tires. So in this situation the rear axle is subjected, in the vertical direction, to:
INERTIAL EFFECT = [ (mass of the car * braking acceleration) * distance between the CoG and the front axle ] / distance between the front axle and the CoG of the rear.
Center of Gravity of the rear part of the car? What's that?!
Every object can be seen as a group of minor parts all rigidly connected to each other. Like, you have two arms, two legs and a chest, and all of these parts, if seen as single object, have their own CoG.
All of them is subjected to a gravity force, and if you sum all this forces, you're going to obtain the actual force pushing you against the chair. Nice, isn't it?
So right now we are thinking the car as split in two parts, one is the front of the car with it's own CoG, and one is the rear, with another CoG. We could have chosen different masses of substitution, the result wouldn't have changed, but it would have been a lot more complicated.
It's like seen the number 4 as made of 2+2, rather than 3+1 or 1+1+1+1, the result is always 4.
An important point is these two CoGs are rigidly connected, as the dimensions of the car aren't changing obviously.
Back on the car, the front of the car is basically fine for what concerns us right now, while the rear is in muddy waters.
Right now we have seen it is vertically subjected to two forces, with gravity directed towards the ground and the inertial effect towards the sky. This means the net force pushing the rear axle to the ground is lower now, and this also means the tires offer a lower grip too, as, even if the frictional coefficient is still the same, the acceleration pushing the car on the ground is now less if not even “negative” if the inertial effect (direct toward the sky) managed to overcome the gravitational acceleration (directed toward the ground)
So in the best situation, there still is a force, albeit minor than before, pushing the car to the ground, while in the worst picture the tail is completely suspended with “more” force pushing the tail in the air than to the ground.
Regardless of this and its effects in the distance necessary to slow down from a certain speed to another, the point is that with less force pushing the car to the ground or none at all the grip is going to be much less if not absent, obviously, and now we are considering the transversal direction.
This is when the differences between the Mazda Miata (aka MX-5) and the Porsche 911 come to the surface.
While the Miata is in this instable situation and you start turning, the inertial force pushing you on the outer side of the car is also pushing on the CoG of the rear part of the car, hence creating an oversteer which can help you turning quicker and therefore accelerate once again early, on the 911 the situation is way more delicate.
Since the rear of the car is much heavier, the inertial force acting on the rear CoG is much higher, because it is defined in this way:
As the 911's rear end is pretty heavy, the oversteer, the “liftoff” oversteer, is going to be much more evident, and difficult to manage.
Note that the inertial effect pushing toward the outer side of the corner is acting even on the front CoG, but it's contrasted by the grip offered by the tires, if everything goes well, otherwise, be ready to meet something hard and painful.
That's it, basically (it took three Word's pages, never mind).
With a lot of experience and bravery it can help you creating amazing trajectories on the road, but if you aren't much into high performance cars and track days, you better avoid old 911s for a while.
Modern 911s and modern cars in general are much easier to drive from this point of view, still, watch out!
A rear-end lighter car may suffer a minor inertial force acting transversally so easier to manage, but it's also easier to lift its tail while braking. This is indeed the only way to a see front wheel drive car having managing a bit of oversteering.
The same effect can obviously be obtained with the good old handbrake, which drastically reduce the grip at the rear axle.
Hope you liked and understood something out my rant, given I didn't mess with something in the middle. :)
Images sources: ducks, Lancia Stratos and amusement park via Flickr,
comparative, Ford Focus RS, Porsche 911 930,
Porsche Carrera GT (Hamilton's father accident), first and second illustration.