Steel-on-steel assumed (E = 210 GPa, ν = 0.3). Equivalent modulus E* ≈ 115.4 GPa. Typical gauges 1067 mm (narrow) / 1435 mm (standard). μ ≈ 0.30 dry, ≈ 0.10 wet, ≈ 0.05 with oil contamination.
Left = wheel/rail cross section / Right = top view of the contact ellipse (semi-major 2a × semi-minor 2b)
Hertz semi-ellipsoidal pressure distribution on the ellipse. p_max at the center, zero at the boundary. Contours mark 25/50/75 % of the peak.
The wheel-rail contact is an elliptical Hertz contact because the curvatures differ in two perpendicular directions. Equivalent curvatures A, B and curvature ratio k:
$$A = \tfrac{1}{2}\!\left(\tfrac{1}{R_{wx}}+\tfrac{1}{R_{rx}}\right),\ B = \tfrac{1}{2}\!\left(\tfrac{1}{R_{wy}}+\tfrac{1}{R_{ry}}\right),\ k=\tfrac{B}{A}$$Equivalent radius R_eq and equivalent modulus E* (steel-on-steel: E* ≈ 115.4 GPa):
$$R_{eq} = \tfrac{1}{2\sqrt{A B}},\quad \tfrac{1}{E^*} = \tfrac{1-\nu_1^2}{E_1}+\tfrac{1-\nu_2^2}{E_2}$$Semi-axes a, b of the contact ellipse (m, n depend on the curvature ratio k) and maximum contact pressure:
$$a = m\!\left(\tfrac{3F R_{eq}}{E^*}\right)^{1/3},\quad b = n\!\left(\tfrac{3F R_{eq}}{E^*}\right)^{1/3},\quad p_{max} = \tfrac{3F}{2\pi a b}$$Maximum tangential force at the adhesion limit (Coulomb friction):
$$F_{t,\max} = \mu\, P$$The values of m and n depend on the curvature ratio k. At k = 1 the contact is circular with m = n = 1; larger k gives a flatter ellipse. This tool linearly interpolates between the tabulated values at k = 1, 1.5, 2, 5 and 10.