Understeer or oversteer?

[Martin A]

Hi All

Imagine driving along a road at night on the outside of a bend of about 100m (330ft) radius reasonably close to the national speed limit cornering pretty enthusiastically at about 0.7G You suddenly see someone stagger onto the road. You brake hard also at about 0.7G, which I understand is just about possible if your tyres generate a maximum of 1G total grip. As you brake though the tyres continue to generate roughly the same forces if you keep the steering angle and the brake pressure constant. No ABS, ESP, etc.

Within a second you have scrubbed off about 15mph your speed is down to around 40mph. At this speed the cornering force will be bringing you round a corner of approximately 150ft radius. You are no longer on a collision course with the staggerer but you are heading across the centre of the road. This is scary.

Continue scared braking at the same rate and a second later you would be on a radius of about 50 ft. Can you picture this. The car will be rotating on a tighter and tighter radius. It will feel as if the back of the car is swinging out. It isn’t, but it feels like it. The sideways force you are experiencing hasn’t changed, the steering angle hasn’t changed, and you are bracing yourself so your perception of braking is skewed. After 3 seconds your turn radius is barely greater than the length of your car.

Trouble is by now you have probably pushed the brake even harder in panic or to brace yourself for the on coming crash. The grip of the tyres is now exceeded and so they begin to slide sideways and forwards, momentarily still rotating and providing some, but greatly lessened steering control. This control is gone pretty well instantly. As soon as the tyres slide the brakes will have greater friction and therefore stopping ability than the tyres have with the road and so the wheels will lock up and the car will continue to turn and slide.

This would seem to be oversteer at its worst however as the majority of braking is done at the front, the slip angles of the front tyres are likely to be greater than those of the rear as the tyres will be distorted more. If the slip angles are greater at the front than the rear the car must be understeering or is there something involving the tightening of the line and yaw that increases the rear slip angle so that the car does in fact oversteer?

Hoping for help and amusing comments

Best regards

Martin A

[rsv gone!]

Martin A said:

Hoping for help and amusing comments

Maybe get some advanced tuition?

Edited by rsv gone! on Thursday 26th March 20:37

[Mave]

You can't keep pulling 0.7g if you keep the steering angle constant whilst decelerating. You'll stay on the same radius turn pulling less and less g.

[Martin A]

Hi All

Mave said:

You can't keep pulling 0.7g if you keep the steering angle constant whilst decelerating. You'll stay on the same radius turn pulling less and less g.

Thanks for that, do you have a reference or the maths so I can see how the g deteriorates. I suppose that means that there's no real problem with firm braking and cornering then, which is nice, but contrary to what I understood.

Best Regards

Martin A

Edited by Martin A on Thursday 26th March 21:32

[RobM77]

You will indeed get oversteer just as you describe. The only way to try an emergency stop in a corner is to balance the braking force against the opposite lock and come to a halt that way (I've done it a few times quite successfully and it isn't that hard to do). It is possible to do as you suggest and brake so hard that you wash the front out (i.e. the opposite of accelerating so hard that the usual understeer you get with acceleration turns to oversteer), but of course unlike the drive from the engine in a RWD car, brakes operate on all four wheels, so unless you're accelerating at the same time to counter the rear braking, then what you describe with the front washing out won't generally happen because the brakes will affect the rear wheels too. Incidentally, this is how we arrive at the advanced spin correction technique of hitting both brake and accelerator - it washes the front out and brings the car back in line (see 1:35 into this clip: http://www.youtube.com/watch?v=QW4ef0E7DQs ).

It is possible in the wet (and without ABS) to get understeer in the manner you describe by hitting the brake pedal alone, but normally the weight transfer is so instant and dramatic that you'd gain yaw momentum and spin before you got a chance to get understeer.

The poster above who mentioned your high g values is quite correct. On a bumpy road this is dependent on the ride/handling compromise of your car (on a smooth track most road cars would just touch 1g). An Elise would probably just about manage 0.9 to 1g on a standard b road, but most cars with poorer damping would find abot 0.6g to 0.7g about as far as they could go. Nevertheless, I understand that your figures were merely illustrative, so I don't wish to dwell on that point.

HTH

Edited by RobM77 on Thursday 26th March 22:34

[Mave]

Martin A said:

Hi All

Mave said:

You can't keep pulling 0.7g if you keep the steering angle constant whilst decelerating. You'll stay on the same radius turn pulling less and less g.

Thanks for that, do you have a reference or the maths so I can see how the g deteriorates. I suppose that means that there's no real problem with firm braking and cornering then, which is nice, but contrary to what I understood.

Best Regards

Martin A

Edited by Martin A on Thursday 26th March 21:32

Well, you know that the forward speed is decelerating at 0.7g. So you can then work out your lateral g using the original radius and F = M w2 R

[DocJock]

http://www.carlimits.com/

Andy will take you through this scenario

[StressedDave]

Martin A said:

Imagine driving along a road at night on the outside of a bend of about 100m (330ft) radius reasonably close to the national speed limit cornering pretty enthusiastically at about 0.7G You suddenly see someone stagger onto the road. You brake hard also at about 0.7G, which I understand is just about possible if your tyres generate a maximum of 1G total grip. As you brake though the tyres continue to generate roughly the same forces if you keep the steering angle and the brake pressure constant. No ABS, ESP, etc.

No they don't - put simply the amount of slip angle required to generate 0.7g lateral and 0.7g longitudinal is significantly greater than that required to generate just 0.7g lateral. In fact one of the issues with brake initiated loss of control is that people suddenly jump off the brakes having applied a fair chunk more steering than they would normally and the car suddenly goes out of control - an 'oversteered' loss of control.

Best explanation of the theory of tyre behaviour is Pacejka http://www.amazon.com/Tire-Vehicle-Dynamics-Hans-P...

Martin A said:

Continue scared braking at the same rate and a second later you would be on a radius of about 50 ft. Can you picture this. The car will be rotating on a tighter and tighter radius. It will feel as if the back of the car is swinging out. It isn’t, but it feels like it. The sideways force you are experiencing hasn’t changed, the steering angle hasn’t changed, and you are bracing yourself so your perception of braking is skewed. After 3 seconds your turn radius is barely greater than the length of your car.

Doesn't happen like that - you need to investigate tyre yaw damping and the change in slip angle resulting from the change in speed:

Rear slip angle = (sideslip speed / forward speed) - (distance from c of g to rear axle x yaw rate / forward speed)
Front slip angle = (sideslip speed / forward speed) + (distance from c of g to front axle x yaw rate / forward speed) - steered angle

Milliken & Milliken http://www.millikenresearch.com/rcvd.html has about the best explanation of the dynamics involved

Martin A said:

Hoping for help and amusing comments

Best regards

Martin A

Can't think of anything amusing I'm afraid and '2/10, must try harder' is the second best way to start an argument on the Internet...

[7db]

I assume that recommending someone buy Milliken and Milliken and Pacejka is a close third...

[StressedDave]

Just be grateful I didn't include Blundell & Harty as well...

As a man with a pair - of mathematical qualifications - I'd assumed you'd have no trouble with either.

[Martin A]

StressedDave said:

Martin A said:

Imagine driving along a road at night on the outside of a bend of about 100m (330ft) radius reasonably close to the national speed limit cornering pretty enthusiastically at about 0.7G You suddenly see someone stagger onto the road. You brake hard also at about 0.7G, which I understand is just about possible if your tyres generate a maximum of 1G total grip. As you brake though the tyres continue to generate roughly the same forces if you keep the steering angle and the brake pressure constant. No ABS, ESP, etc.

No they don't - put simply the amount of slip angle required to generate 0.7g lateral and 0.7g longitudinal is significantly greater than that required to generate just 0.7g lateral. In fact one of the issues with brake initiated loss of control is that people suddenly jump off the brakes having applied a fair chunk more steering than they would normally and the car suddenly goes out of control - an 'oversteered' loss of control.

I was thinking that perhaps forward weight transfer would significantly increase the slip angle at the front allowing the car to slow as well as turn at 0.7g. I can see that coming off the brakes would cause oversteer if the tyres had exceeded the slip angle to where the tyre starts sliding but didn't know that would happen if they were still rolling. Just want to check we're talking about the same circumstances. Is that due to time delay of chassis and suspension?

StressedDave said:

Best explanation of the theory of tyre behaviour is Pacejka http://www.amazon.com/Tire-Vehicle-Dynamics-Hans-P...

I'm aware of Pacejka but sadly don't have the understanding of the maths to translate it into the real world. Is there a Pacejka-lite anywhere that gives general guidance on the ideas or is it explained in layman's terms?

StressedDave said:

Martin A said:

Continue scared braking at the same rate and a second later you would be on a radius of about 50 ft. Can you picture this. The car will be rotating on a tighter and tighter radius. It will feel as if the back of the car is swinging out. It isn’t, but it feels like it. The sideways force you are experiencing hasn’t changed, the steering angle hasn’t changed, and you are bracing yourself so your perception of braking is skewed. After 3 seconds your turn radius is barely greater than the length of your car.

Doesn't happen like that - you need to investigate tyre yaw damping and the change in slip angle resulting from the change in speed:

Rear slip angle = (sideslip speed / forward speed) - (distance from c of g to rear axle x yaw rate / forward speed)
Front slip angle = (sideslip speed / forward speed) + (distance from c of g to front axle x yaw rate / forward speed) - steered angle

Milliken & Milliken http://www.millikenresearch.com/rcvd.html has about the best explanation of the dynamics involved

Good, thanks for that I thought the yaw rate was involved. Does it change as the car slows down then, assuming the steering angle is constant, or does it remain the same? I mean without seeing some figures for different speeds I can't see that I can work it out from that as there is more than one variable and they may cancel each other out. Also does the maths explain the difference between an old 911 and Audi 100 for instance. Is the suggested scenario closer to one than the other? i.e. Tail heavy, nose heavy

StressedDave said:

Martin A said:

Hoping for help and amusing comments

Best regards

Martin A

Can't think of anything amusing I'm afraid and '2/10, must try harder' is the second best way to start an argument on the Internet...

Trying harder. 2.5/10?

Thanks

Best Regards

Martin A

[shoestring7]

Let Bosch sort it all out....

SS7

[StressedDave]

Martin A said:

I was thinking that perhaps forward weight transfer would significantly increase the slip angle at the front allowing the car to slow as well as turn at 0.7g. I can see that coming off the brakes would cause oversteer if the tyres had exceeded the slip angle to where the tyre starts sliding but didn't know that would happen if they were still rolling. Just want to check we're talking about the same circumstances. Is that due to time delay of chassis and suspension?

You're confusing two separate things... slip angle is a function of the equation I quoted - weight transfer doesn't come into it. Tyre force is a function of vertical load, slip angle and camber angle. You can get oversteer at any point - it's a function of the difference in yaw moment (side force x distance from c of g) between front and rear - you don't have to saturate a tyre to it's frictional limit to get it to occur.

Martin A said:

I'm aware of Pacejka but sadly don't have the understanding of the maths to translate it into the real world. Is there a Pacejka-lite anywhere that gives general guidance on the ideas or is it explained in layman's terms?

Not that I'm aware of...

Martin A said:

Good, thanks for that I thought the yaw rate was involved. Does it change as the car slows down then, assuming the steering angle is constant, or does it remain the same? I mean without seeing some figures for different speeds I can't see that I can work it out from that as there is more than one variable and they may cancel each other out. Also does the maths explain the difference between an old 911 and Audi 100 for instance. Is the suggested scenario closer to one than the other? i.e. Tail heavy, nose heavy

Seeing as yaw rate = speed / radius, I'll leave that for you to work out. It's far easier to do the maths with simulation, as effectively you have to do the calcs every 1/100th of a second or so. But once you do them, yes it does explain the difference between cars.

[new in today]

In the situation you describe you would need to declutch and lift the brake to bring the steering back and to stop the brakes from locking the rears . You said you had innitially slowed from 55 to 40 so at your crisis point you were then travelling at around 30mph?
Wether that would mean actual oversteer would depend on the type of car you were driving.
If you ever find your self heading where you don't want to be and locked up in extremis it is surprising just how quick the steering comes back (instantly) when you declutch and lift the brakes.
It may feel as if your car is suddenly going to go on two wheels but it won't. You will only roll/flip a car if the wheels catch a kerb or dig into soft ground.

[stefan1]

new in today said:

In the situation you describe you would need to declutch and lift the brake to bring the steering back and to stop the brakes from locking the rears . You said you had innitially slowed from 55 to 40 so at your crisis point you were then travelling at around 30mph?
Wether that would mean actual oversteer would depend on the type of car you were driving.
If you ever find your self heading where you don't want to be and locked up in extremis it is surprising just how quick the steering comes back (instantly) when you declutch and lift the brakes.
It may feel as if your car is suddenly going to go on two wheels but it won't. You will only roll/flip a car if the wheels catch a kerb or dig into soft ground.

I don't disagree, but a car is capable of rolling without the tyres hitting anything - at high speeds, with stiff suspension and very significant yaw, a roll is possible.

[Martin A]

StressedDave said:

Martin A said:

I was thinking that perhaps forward weight transfer would significantly increase the slip angle at the front allowing the car to slow as well as turn at 0.7g. I can see that coming off the brakes would cause oversteer if the tyres had exceeded the slip angle to where the tyre starts sliding but didn't know that would happen if they were still rolling. Just want to check we're talking about the same circumstances. Is that due to time delay of chassis and suspension?

You're confusing two separate things... slip angle is a function of the equation I quoted - weight transfer doesn't come into it. Tyre force is a function of vertical load, slip angle and camber angle. You can get oversteer at any point - it's a function of the difference in yaw moment (side force x distance from c of g) between front and rear - you don't have to saturate a tyre to it's frictional limit to get it to occur.

Please be patient if I'm being stupid but if tyre force is a function of vertical load, slip angle and camber angle. Then if the tyre force remains constant doesn't the slip angle change if there is forward weight transfer during braking as the vertical load will change? Or am I missing something?


I understand that oversteer might occur at any point and maybe I should not have mentioned a wheel lock up. My main intended query was; keeping the same steer angle as the car slows down, would it naturally take a tighter and tighter line that would only appear to be oversteer but technically be understeer as the front slip angles would be exceeding those at the rear. Or would the car not behave like this and as soon as it slows down would it stay on the same radius with a diminished centripetal tyre force as already suggested in an early reply?

Or is the answer somewhere else?

Hope this explains my thoughts better.

Best regards

Martin A

[StressedDave]

Martin A said:

Please be patient if I'm being stupid but if tyre force is a function of vertical load, slip angle and camber angle. Then if the tyre force remains constant doesn't the slip angle change if there is forward weight transfer during braking as the vertical load will change? Or am I missing something?

You're being stupid... If you look back at the equation for slip angle I posted yesterday, you'll see that vertical load plays no part in the definition of slip angle, so irrespective of whether the vertical load changes, for a given combination of forward and sideslip speed, yaw rate and steer angle, the slip angle is defined. Of course the hard bit, and the reason why I suggested a mathematical simulation is far better at this than trying to work it out based on a series of vague assumptions. How the yaw rate and speed interplay is not something trivial to work out.

Martin A said:

I understand that oversteer might occur at any point and maybe I should not have mentioned a wheel lock up. My main intended query was; keeping the same steer angle as the car slows down, would it naturally take a tighter and tighter line that would only appear to be oversteer but technically be understeer as the front slip angles would be exceeding those at the rear. Or would the car not behave like this and as soon as it slows down would it stay on the same radius with a diminished centripetal tyre force as already suggested in an early reply?

IME - and it's something I do with clients a lot by getting them to fiddle with the brakes while holding fixed steering on a constant circle is that any such tendency tends to be engineered out using compliance steer because its a pretty undesirable behaviour.

Edited by StressedDave on Saturday 28th March 20:09

[new in today]

I see where you are going with this now...

You mean as the speed decreases , the slip angle becomes less as the steering forces needed are less and therefore by keeping the wheel the same you will find the car turning in on itself.

Yes this is quite true and as mentioned what you have to do if applying hard braking in a corner is anticipate this effect by adding a little stab of opposite lock as you start to break.
You will with experience actually apply the opposite lock at the same time as you brake in anticipation of the effect of your actions instead of waiting for the oversteer to start and build.

Its a bit like running over stepping stones you have to look and think ahead your actions being a flow of both expectectation and actual events and your subconscious doing a lot of the work.
If you have done a lot of karting then you will find you have to break a lot when you didn't plan to as other karts jossle for position at every corner. It becomes natural to apply a flick of opposite lock whenever you find you have to unexpectedly brake after the turn in.

[Martin A]

Hi All

Now that I know that it isn't slip angles alone (as seems to be suggested in many texts on the subject) that affect understeer and oversteer I can now see where my misunderstanding was. Tyre force also rarely seems to be mentioned in relation to vertical load and if regarded as 'grip' seems frequently to be 180 degrees out. Having revisited Pacejka (or the graphs rather than the maths which had frightened me off the graphs) I can now see how this works and understand why Stressed Dave is keen on being on the gas (which I would also teach) rather than on the brakes when entering a corner.

Thanks again for all you thoughts

Best regards

Martin A

[StressedDave]

Pacejka's formula is nothing more than a poncey curve fit to the graphs in any case... If you think of it as one-quarter of a sinewave stretched out to fit, then it's relatively easy to understand. The rest is just 50-odd coefficients to link camber, slip angle and vertical load so you get the right graph to fit to.