Wheel Sinkage Depth I am trying to make a realistic car physics simulation for my game and the current thing that I am fighting to solve is rolling resistance. Although it's effect is minimal, this is important concept, because it is the force which prevents car from rolling away when it is standing still. So I lead to this article: https://en.wikipedia.org/wiki/Rolling_resistance where it's told that:
$$F_\mathrm{rr} = \frac{C_\mathrm{rr} W}{r}$$
There's no problem with that except for fact that it's told that $C_\mathrm{rr}$ is equal to square root of $z / d$ where:


*

*$z$ is sinkage depth

*$d$ is the diameter of the rigid wheel


So I want to know how to calculate the sinkage depth, what values do I need etc. And also I need to know if the diameter is the effective wheel diameter or it's diameter after the deformation(which actually causes this force). And if second version, than I wanna know how to calculate this.
 A: For a good reference, try Landau and Lifshitz book on Elasticity.
Suppose you have a wheel (approximated by an infinite cylinder) that is coming into contact with the flat ground. By symmetry, the area of contact will be a rectangle that will be very thin in one direction and very long along the length of the cylinder. Call the long length $a$ and the small width $b$ and the radius of the wheel $R$. Then $b$ and $R$ are related by (according to Landau/Lifshitz)
$$ \frac{1}{2 R} = \frac{N D}{\pi} \int_0^\infty \frac{d \xi}{(b^2+\xi)\sqrt{(a^2+\xi)(b^2+\xi)\xi}}, $$
where $N$ is the normal force (i.e. the weight of the load the wheel is carrying) and $D$ is a constant measuring the elasticity of the materials involved. Supposing that $a\gg b$, we have
$$ \frac{1}{2 R} = \frac{N D}{\pi} \int_0^\infty \frac{d \xi}{a (b^2+\xi)^{3/2}\sqrt{\xi}} =\frac{2 N D}{\pi a b^2}.$$
Now suppose that the wheel is perfectly rigid and only the ground deforms. Then the sinkage depth is related to the radius $R$ and the small width $b$ by 
$$z=R-\sqrt{R^2-b^2}\approx \frac{b^2}{2 R} \text{   (for $b\ll R$)}$$
It therefore follows that
$$z=\frac{2 ND }{ \pi a  }$$
Thus if you know the weight $N$, the width of the wheel $a$, and the value of $D$, then you are all set. And $D$ is given by
$$D=\frac{3}{4}\bigg(\frac{1-\sigma^2}{E}+\frac{1-{\sigma^\prime}^2}{E^\prime}\bigg),$$ 
where $\sigma,\sigma^\prime$ are the Poisson ratios of the ground and wheel (respectively) and $E,E^\prime$ are the Young's Moduli of the ground and wheel (respectively). Numerical values can be found for these online.
To address your last point, the diameter $d$ in the given equation on Wikipedia refers to the diameter without deformation because the equation (technically) requires that the wheel is perfectly rigid. Though in reality the size of the deformation of the wheel is so small compared to the diameter that it will make absolutely no difference for practical purposes. 
