1. Clutch pedalFirst of all clutch pedal always have free-play area and also some non-functional space after clutch disengage. This is described in service manual for a car.
Usually in a real car free-play distance is something about 1 inch, working range is about 4 inch and disengagement position has 2 more inches until floor.
Thus total range of pedal movement should be divided in zones something like 15% - 60% - 25%.
2. Torque curveNext question is how transmitted torque is related to pedal displacement. I've found some articles about this, and most useful are:
http://dx.doi.org/10.1109/TMECH.2010.2047509http://dx.doi.org/10.1016/j.finel.2010.08.007Transmitted torque is related to load force by equation
T = n * μ * Req * F
T - torque;
n - number of pairs of contact surfaces (we have two because clutch disc has two sides, so n=2);
μ - friction coefficient;
Req - some equivalent radius of clutch disc;
F - normal component of load force applied to friction surfaces.
Load force F is dependent on clutch disc displacement, and controlled with the pedal.
Some typical curves look like these:
Displacement coordinates are measured relative to clutch disc, but we can assume they linearly dependent on pedal position.
I've come up with such empirical approximation for displacement-force dependence:
F(x):=Fmax*(1-(1-p*x)^γ)/(1-(1-p)^γ)
γ = 0.2
p = 0.95
Fmax = 5000 N
x range: 0.0 - 1.0
This is actually slightly modified gamma-correction curve from computer graphics. All parameters are very approximate.
I think it is a good starting point. It takes into account low rate at the beginning and has steep rate at the end of clutch working range.
Now we should define μ and Req coefficients. We could take these values:
Req = 89 mm
μ = 0.24
But the problem is friction coefficient μ is dependent on engine rotation speed.
Maximal torque will be transmitted only if motor is rotating fast enough to sustain “saturated” friction force at the clutch. If motor is rotating slower than some limit, friction force will be correspondingly smaller.
3. Slip speedFriction coefficient μ is dependent on angular frequency of motor rotation. To be more exact, friction coefficient depend on difference between rotation speed at the input and output of the clutch. If engine rotates with angular frequency ωf, and clutch output (transmitted to the gear box) rotates at the frequency ωc, then friction coefficient should be function of angular frequency
ωfc = |ωf-ωc|.
Note the absolute value is taken. I am not sure how the “direction” of applied torque is implemented in the game. If the road move faster than the engine rotates, the car will naturally decelerate.
Fenomenological model of dependence μ(ωfc) is given in the first cited article, and I repeat it here with some simplifications.
μ(ωfc) = μs + (μd-μs) * tanh[g*ωfc*Req]
ωfc = |ωf-ωc|, rad/s
μs = 0.15 - static friction coefficient
μd = 0.24 - dynamic friction coefficient
g = 1.5 s/m
Req = 89 mm = 0.089 m
Resulting torque depending on clutch displacement and including frequency looks like this:
We can notice, than friction coefficient μ(ωfc) saturate to its maximal value μd very fast. For rotation speed difference of 200 RPM it is almost at full value. So μ(ωfc) function is not actually relevant for the case we start moving the car from full stop, because idle speed is much higher (~800 RPM), and gear box is not rotating at all. But it became influential when speed of engine approaches the speed of road. And it has influence only if clutch pedal is depressed, because if the clutch is locked-up, the engine limits maximal torque, not the clutch.
We can conclude than most important part is to model pedal dead zones from section 1 and load force F(x) dependence from section 2. Dependence μ(ωfc) from section 3 is not so critical (can assume μ is constant) but could give more smooth transitions if implemented.