Slip angle and power

[Greg Locock]

I must admit I find it hard to explain the intuitive action of catching a RWD car's slide using throttle and steering wheel angle, by looking at friction circles. I am sure the answer is there tho. But it is an easy skill to learn so I doubt the reasoning is complex. The first thing you do is steer towards the sliding direction, then apply some throttle. then grin.

[Erunanethiel]

I must admit I find it hard to explain the intuitive action of catching a RWD car's slide using throttle and steering wheel angle, by looking at friction circles. I am sure the answer is there tho. But it is an easy skill to learn so I doubt the reasoning is complex. The first thing you do is steer towards the sliding direction, then apply some throttle. then grin.

Hmm, I hope there is somebody to explain

So if you slide a tire laterally, will you decrease its longitudional traction too?

[Tommy Cookers]

Maybe it will help to think of this way; a rear tire has a traction oval, usually with the lateral axis a bit larger. Thus if tire is at steady state maximum lateral traction with no power (forward thrust) and power is applied and torque is added to the tire, part of the traction budget will be taken by the torque and, available lateral traction will decrease. Since lateral traction was originally was at the max, all things being equal, the slip angle will increase big time. Of course all things won’t be equal since the torque will accelerate the vehicle thus adding wheel weight and changing the traction circle which, in turn, could either increase or decrease the slip angle.

I also found out that, even after the tire has past its best (most lateral force) slip angle, the tire doesnt lose much ability to take lateral force. Like, for examle, if the tire can produce its top g (3) at 8 degrees of slip angle, the tire also still makes 3 g at 12 degrees of slip angle. Then, why do we care about slip angle that much?

attempting to sustainably corner at 3g and 12 deg slip you will need more drive torque
(to develop the lateral force at 12 deg slip angle the tyre needs 50% more longitudinal force than for 3g 8 deg slip angle)
the required extra drive torque will not allow the tyre to give you the 3g 12 deg behaviour and cornering will degrade
and if you don't provide the extra drive torque your car will slow and cornering will degrade

[Erunanethiel]

Maybe it will help to think of this way; a rear tire has a traction oval, usually with the lateral axis a bit larger. Thus if tire is at steady state maximum lateral traction with no power (forward thrust) and power is applied and torque is added to the tire, part of the traction budget will be taken by the torque and, available lateral traction will decrease. Since lateral traction was originally was at the max, all things being equal, the slip angle will increase big time. Of course all things won’t be equal since the torque will accelerate the vehicle thus adding wheel weight and changing the traction circle which, in turn, could either increase or decrease the slip angle.

I also found out that, even after the tire has past its best (most lateral force) slip angle, the tire doesnt lose much ability to take lateral force. Like, for examle, if the tire can produce its top g (3) at 8 degrees of slip angle, the tire also still makes 3 g at 12 degrees of slip angle. Then, why do we care about slip angle that much?

attempting to sustainably corner at 3g and 12 deg slip you will need more drive torque
(to develop the lateral force at 12 deg slip angle the tyre needs 50% more longitudinal force than for 3g 8 deg slip angle)
the required extra drive torque will not allow the tyre to give you the 3g 12 deg behaviour and cornering will degrade
and if you don't provide the extra drive torque your car will slow and cornering will degrade

Is there specific equation for longitudinal force the tyre needs per sleep angle? It looks like it's what I need