Toe out for turn in

[Smokes]

Can toe out be considered to be unstable but the caster angle stablises it which why it works on the front and not on the rear(has no caster).

[Jersey Tom]

Suffice to say, I do not agree with Carroll Smith's assessment, for reasons mentioned earlier.

If/when I retire and write a book, all will be explained in detail with data-driven examples.

[GSpeedR]

if the end result is some understeer, then the transient turn in behavior will feel more aggressive.

What's the mechanism that causes this? In traditional control theory if you increase the rate of response then the system becomes less stable, and vice versa... this would suggest that increasing the turn-in rate would lead to a more oversteery car...?

The response transfer functions derived from stability analysis (2DoF model) are all relative to a steady-state value. One response transfer function would be defined as a yaw rate output from a steer input (usually step-input). So while this analysis will show that an increase in understeer gradient will reduce yawrate 'response time' (and it will), we have to realize that response time is relative to a certain steady-state yaw rate...which is less for a vehicle with more understeer. To negotiate an actual corner, the driver will be required to input more steering. Some drivers relate response solely towards the total steer input required to achieve a desired yaw rate, and thus an oversteering car may "respond better", simply because the car turned more with less input. Interestingly, I remember that response will quicken with increasing understeer gradient and also with increasing front axle cornering stiffness (which always seems counter-intuitive).

Edit: it looks like Jersey Tom made a quick reference to the above on the previous page.

[munks]

With regard to compliances Road car rubber bushes are far from perfect, shore tolerances for production is +- 5. The also perish a different rates due to heat exposure age and wear and tear. I just replaced rear supension bushes in 9 year old M3 that has done 52000 miles. The bushes had distringrated, they are usally designed for 100,000 miles or 10 years for prodution road cars.

Don't mean to go off on a tangent here, but what do disintegrated bushings feel like when driving? I have a similar vehicle and am wondering if it's time for them to be replaced ...

[olefud]

A few comments to add to the confusion. Toe out would seem to be twitchy in a striaght line. The tire with the most weight controls steering. But with toe out, this induces turning and transfers weight to the other tire -particularly under braking. At best a driver could establish an instabile equilibtium.

Yet another thought on turn in. When a car turns in, it has to initiate both polar rotation and establish turning around a changing center(s) of rotation. The first is transient and disappears while the second requires continuing acceleration forces. Thus toe out, TOOT or Akerman can be useful if the beast is reluctant to turn.

Finally, NASCAR uses an opposite tactic with stagger, a larger OD outter rear tire than the inner tire to develop asymetric drive forces. Telementry show that turns are often entered with trailing brake. Given the locked diff, wouldn't stagger tend to resist turn in? And with only so much traction available from a tire, should it be used to rotate the already rotating car or for turning around the center of rotation?

The problem is too many variables which are constantly changing. It's more an art than science, though, when something works, it can be explained by science.

[munks]

A few comments to add to the confusion. Toe out would seem to be twitchy in a striaght line. The tire with the most weight controls steering. But with toe out, this induces turning and transfers weight to the other tire -particularly under braking. At best a driver could establish an instabile equilibtium.

In the linear range of the tire (small slip angles) the load is basically irrelevant. (EDIT: this is incorrect as noted in later comments.) Said another way, the cornering stiffness is typically constant over a wide range of loads. So in an almost straight-line I don't think it's true that the tire with the most weight controls the steering. In fact, looking at it from your weight perspective, I'd actually think toe-in would be the unstable one, because turning and adding more weight to the outside tire is a growing feedback loop.

But I believe in actuality your overall point here is correct for a different reason, which has to do with slip angles, not load.

[Jersey Tom]

Said another way, the cornering stiffness is typically constant over a wide range of loads.

Check RCVD Section 2.12 - "Goodyear Tire Data" and you will see this is very much not true.

[munks]

Of course you are right and I retract my comment. The first thing that goes is the memory, I don't remember the second thing.

[olefud]

MUNKS and I may be saying the same thing with tire load vis'-a-vis' slip angle. With the front tires a bit walleyed, the more lightly loaded tire will exhibit the greater slip angle while the tire with the greater load rolls straighter and largely determines the direction of travel. However, the more loaded tire will tend to veer towards its side, transfer weight to the other tire and swap slip angles. So, yes, slip angles contol, but they are a function of load.

[Jersey Tom]

Slip angle is a geometric condition. It is entirely independent of load.

I believe there is a misconception that slip angle has something to do with tire stiffness or deflection or something like that. It does not.

[GSpeedR]

Slip angle is a geometric condition. It is entirely independent of load.

I believe there is a misconception that slip angle has something to do with tire stiffness or deflection or something like that. It does not.

What about a tire that has no vertical load?

[Jersey Tom]

Slip angle is a geometric condition. It is entirely independent of load.

I believe there is a misconception that slip angle has something to do with tire stiffness or deflection or something like that. It does not.

What about a tire that has no vertical load?

Makes no difference to me. All you're interested in is the relative angle between a velocity vector and orientation vector. Same way a Datron optical sensor works. It doesn't care if it's measuring sideslip of a wheel or car, on the ground or off the ground or whatever, just relative velocity.

I could pick a wheel up off the ground, with its axis of rotation pointed away from me, and throw it out in front of me. Could say that it's traveling at a sideslip angle of 90 degrees.

Or, could think of a plane in a crosswind landing:


The whole thing is at a sideslip angle... no different from the abbreviated "slip angle" concept of a tire - or more appropriately, of the beads of a tire or the rim shell itself.

[olefud]

I tend to think load does influence slip angle; at least in the instance of two tires constrained to run with toe-out. Since each tire wants to go in a different direction, the one with the greater load will prevail, though with a small slip angle. Unless it washes, the other, more lightly-loaded tire will assume a greater, opposite slip angle such that both tires travel in the same direction. It's much like changing the roll stiffness at one end of the car to change tire loading and over/under steer, i.e. slip angles. The load change between the four tires affects slip angles, or, put differently, the available turning force of each tire.

This is worth noting with regard to toe-out since the outer tire is thrusting in the wrong direction on turn-in and must transition to a new set before developing turning force. Toe-in, TOOT and Ackerman don't have this degree of delay.

[GSpeedR]

I agree with the definition of slip angle that you've presented, however I don't agree with the statement that "slip angle is entirely independent of load", though I make this claim more from a philosophical standpoint. Take caution, significant mental masturbation lies ahead with no clear objective.

Tires, being mechanical elements (albeit complicated elements), behave according to mechanical physics. Motion of a mechanical element is force-driven, by that I mean that forces are causal and motion is resultant...Newton's 3rd applies for both rigid body objects as well as finite-elements and continuous systems (object displacement and internal deflection). If this is true then it is true that springs deflect as a result of an applied force; there is no true mechanism for the reverse to occur (a deflection causing a force, for example). Force and motion are obviously directly coupled and when we deal with ideal elements (mass-less springs and rigid masses, for example) can apply laws in reverse to calculate various states...like Hooke's Law. We can often be tricked by machinery such as a shock dyno, which appears to be inputting a velocity and outputting a force, however, we'll hopefully realize that the dyno is still applying a force to the damper in order to achieve a desired velocity (output). I believe tire testing machines are responsible for the same trickery. The hub must input force to generate its desired slip angle.

A single tire, in isolation, when steered, will deform assuming the contact patch produces a frictional force. If the contact patch produces no friction then the tire will not deform. Strictly speaking, we would still have a "slip angle", which would be equal to our steering angle. The definition is what it is, whether I feel it is useful or not. That's not very interesting so let's take this the other direction and assume infinite friction at the contact patch (never slides, always in static friction), then the deformation of the tire is dependent upon the cornering stiffness of the tire system (carcass, tread, whatever) and the force applied. If the tire has an infinite cornering stiffness then this ideal tire will not provide a slip angle regardless of how much you steer it. Its heading will change instantaneously and its velocity vector will always be parallel to its orientation vector. A tire's heading (and thus slip angle) changes to suit the required force. A slip angle, just like a displacement, is just easier to measure than the corresponding force.

Now, just like everyone else, I create simulations and tire models using slip angle as an input and get tire force as an output. I input velocities into damper models to get force outputs. This is a convenience of using ideal elements. Everything Jersey Tom stated is accurate for making engineering decisions from data and simulation, but it's not how the real world works.

I apologize in advance.

[strad]

Now I'm getting confused...
JT...Doesn't slip angle boil down the difference between where I want to point the tires and where they actually go?