motorcycle cornering is aero-neutral
lean angles up to 64 deg are claimed (though presumably back-calculated from cornering data calculations)
some of the notional 64 deg toppling vector works against wheel gyroscopic reactions (continuous in steady-state conditions)
this 'some' was stated to be 2-4 deg (please feel free to calculate this, anyone)
Bear in mind that mu is load sensitive, so an f1 tire at 4g vertical load, ie a high speed braking event, will have a lower mu than it would at more moderate speeds.
Personally I think using the low speed longitudinal acceleration is fraught with difficulties, I use corners.
Typically at 0° Camber an F1 tire provides in the range of 1 kN to 3 kN Vertical Load a Coefficient of Friction of around 1.6 which then drops down to around 1.4 around 6 kN vertical load.
There had been friction coefficients of 2 for an F1 tyre being pedaled about on the forums here relatively recently which I found hard to believe.
I proposed another method to extract the coefficient of friction of the complete car and the downforce coefficient in another thread:
1. For a given circuit (or better still a number of circuits with similar aero setups) make note of the velocity and lateral acceleration through each corner. You can take this from an onboard lap or from one of the track diagrams that the various teams put out.
2. Put all of these datapoints on a scatter plot with speed as the horizontal axes and lateral acceleration as the vertical axes.
3. Fit a squared polynomial to the data in the form: LatAcc = a.Speed^2 + b
4. Then you have:
Tyre friction coefficient: mu = b/9.81
Downforce: SCz = (2 x mass x a)/(mu x AirDensity)
Obviously, as Greg pointed out you should take the data from corner which are traction limited only.
If i follow shooty81's method at the start of the race with downforce not considered as it is pretty much zero for the first few meters and
acceleration = 0,9G(1) and 1,1G(2)
mass(car+fuel) = 691kg + 95kg
wheelbase = 3,4m(1) and 3,5m(2)
COGh = 0,25m(2) and 0,3m(1)
wdr = weigth distribution rear = 54%(2) and 55%(1)
ft_corr = 1,01(1) and 1,03(2) (to account for the rotational energy of the front tyres and brake discs only, the rear axle is not accelerated by the rear tyre force)
accleration * mass * ft_corr = (mass * wdr *9,81N/kg + COGh/wheelbase * mass * acceleration) * friction_coeff
then (1) results in 1,444 and (2) gives 1,832 for the longitudinal friction coefficient.
Dear FIA, if you read this, please pm me for a redesign of the Technical Regulations to avoid finger nose shapes for 2016! :-)
