Mechanical 'L/D' ratio of F1 tyres ?

[Tommy Cookers]

The tyres of an F1 car rounding Copse corner at 280 kph develop a lateral force equivalent to a centripetal acceleration of 4g
If the driver selected neutral, the car would start decelerate, in part due to an aerodynamic drag equivalent to about 0.7g

There would also be a mechanical 'drag' from the tyres (briefly) continuing 4g cornering (additional to the tyre's mechanical 'drag' ('rolling resistance') related to the car's weight)

What would be the value of this 'cornering mechanical drag' ?

From this what ratio of cornering forces would emerge to quantify the tyre's performance (equivalent to L/D ratio in aerodynamics) ?

[Jersey Tom]

Interesting way of thinking about it. Just a question of slip angle though.

If I had to guess and peg them at ~5 deg SA in a corner, and the car is what... ~1450-ish lb? At 4G lateral you'd have ~0.35G drag (~500 lbf) or a ratio of roughly 11.5:1.

[Greg Locock]

5 degrees is a bit on the low side for a heavily loaded F1 tire at max Fy, according to Wright Ferrari F1 2000.

[DaveW]

Would that depend on construction? In any event, PGW's "Formula 1 Technology" (figure 7.3, Page 108) suggests peak slip angle occurring at approx 3 degrees.

[silente]

Some of the F3 tyres i have worked on have peak forces at more than 7-9° (SA), i don´t know exactly where, because it was out of the measurement range of that test.

Some others where in the region of 4-6° depending on the load. I don´t know about F1 construction, but among F3 tyres you could really find important behaviour differences depending on the manufacturer (and so, i guess, on the construction).

[Jersey Tom]

1) The slip angle you end up at is going to be a function of the construction, compound, track condition, inflation pressure, camber, the corner you're taking, etc etc.

2) Slip angles you see in lab data (ie given by a tire company) do not necessarily equal what's going to happen on track.

So if you're looking for a very precise answer here, you're not going to get it. I'm using 5 as a really ballpark number here. There are times I'm sure it's more, and times it's less.

[silente]

JT, let´s say from lab you get 5 deg at a certain load. When in your opinion it could be more and when less on a real track (surface)?

[Jersey Tom]

JT, let´s say from lab you get 5 deg at a certain load. When in your opinion it could be more and when less on a real track (surface)?

No hard and fast rule. Very much a case by case basis.

There are times lab data will generally under predict slip angles, and there are times when it will over predict them. A lot comes down to the specifics of how the lab data is collected. Doesn't take much for it to be non-representative.

Then on top of that you have the question of who is fitting tire models to the lab data (presuming that's what most end users ultimately have access to). A typical Pacejka model these days is somewhere around 100 highly nonlinear coefficients to define the thing in its full form. Even with the best lab data, it's not difficult to create a bogus tire model from it - even among professionals.

Granted I'm spoiled in having had my hands in every step of the process, but if I just received a tire model from someone I'd tread lightly with it until I'd had some extensive conversation with the engineers involved in creating it.

[GSpeedR]

I think the answer would also depend on which coordinate system we are interested in for this 'L/D' ratio. If we are interested in the vehicle's coordinate system then steer angles become important (at the front at least) and the "drag" inducing element is 'sin(steer + slip_a)' relative to the chassis. However, if we are interested in the 'L/D' relative to each wheel (or tangential to the instantaneous turn center) then I think 'sin(slip_a)' would be used.