Steel-on-steel contact (E = 210 GPa, ν = 0.3) is assumed, giving an equivalent modulus E* ≈ 115.4 GPa. Typical gauges are 1067 mm (conventional rail) and 1435 mm (Shinkansen). Typical friction coefficients are μ ≈ 0.30 when dry, μ ≈ 0.10 when wet, and μ ≈ 0.05 with oil contamination.
While paused, moving a slider updates the results immediately.
Left = wheel rolling on a rail and contact ellipse (Hertzian pressure) / Right = subsurface shear-stress field τ(z) and depth of τ_max. Changing the wheel load, wheel radius, or rail curvature updates the display every frame.
Left = wheel/rail cross section / right = top view of contact ellipse (major diameter 2a × minor diameter 2b)
Hertz semi-elliptical pressure distribution over the ellipse: p_max at the center and zero at the boundary. Contours show 25/50/75% of the maximum.
Wheel-rail contact forms an elliptical Hertzian contact because the curvatures differ in two orthogonal directions. Equivalent curvatures A and 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 of curvature R_eq and equivalent modulus E* (steel-on-steel: E* ≈ 115.4 GPa):
$$R_{eq} = \tfrac{1}{2\sqrt{A B}},\quad \tfrac{1}{E^*} = \tfrac{1-\nu_1^2}{E_1}+\tfrac{1-\nu_2^2}{E_2}$$Contact-ellipse semi-major axis a and semi-minor axis b (m and n are coefficients dependent on curvature ratio k), and maximum contact stress:
$$a = m\!\left(\tfrac{3F R_{eq}}{E^* }\right)^{1/3},\quad b = n\!\left(\tfrac{3F R_{eq}}{E^* }\right)^{1/3},\quad p_{max} = \tfrac{3F}{2\pi a b}$$Maximum tangential force at the adhesion limit (Coulomb's law of friction):
$$F_{t,\max} = \mu\, P$$The values of m and n depend on curvature ratio k. At k=1, m=n=1 (circular contact); as k increases, the ellipse becomes flatter. This tool evaluates k=1, 1.5, 2, 5, and 10 by linear interpolation of tabulated values.