Traction Performance of Tires

Abstract

The traction performance of a tire is important to the safety of a vehicle. Traction models of a tire have been proposed for different road conditions, such as dry, wet, snowy, icy and muddy conditions. Simple analytical models for tire traction in braking and driving have been developed by extending the Fiala model of cornering performance. One is a model where the contact pressure distribution does not change under braking and driving forces and the sliding point between the adhesion and sliding regions can be determined using a simple equation. Another is a model where the contact pressure distribution changes under braking and driving forces and the sliding point between adhesion and sliding regions can be determined by iterative calculation. The hydroplaning phenomenon is analyzed employing the equilibrium of the hydrodynamic pressure and tire load, the two-dimensional Reynolds equation for squeezing water out of the tread block and computational fluid dynamics (CFD) simulation where a tire is modeled by FEA and water is modeled using the finite volume method (FVM). The snow reaction is analyzed using an analytical model where the shear strength of snow is determined from the density of snow and conducting CFD simulation in an approach similar to that adopted for hydroplaning. The traction on ice is analyzed using a brush model with a friction model on ice where the friction coefficient is a function of the sliding velocity and other parameters of thermodynamics, and another brush model where the shear force distribution of a freely rolling tire is included. The traction on mud is analyzed using an analytical model where the tire deformation is classified as being in a rigid mode or elastic mode according to the difference between tire rigidity and load retention properties of mud, and CFD simulation is conducted in a manner similar to that adopted for hydroplaning. FEA has recently become popular in the design of the tire pattern for traction on wet, snowy and muddy road conditions.

Notes

Note 12.1 Model of the Frictional Force Proposed by Moore [159]

Moore expressed a frictional force $\widehat{F}_{\mu }$:

$$\hat{F}_{\mu } = F_{\text{adhesion}} + F_{\text{hysteresis}} + F_{\text{plough}} ,$$

where F_adhesion is the adhesive force between rubber and the road, F_hysteresis is the force due to the hysteresis of rubber, and F_plough is a plowing force that includes the force due to the fracture of rubber. Furthermore, coefficients of friction between the rubber and hard ground, such as a paved road, are

$$\begin{aligned} \mu_{\text{adhesion}} & = C_{1} K \cdot E^{\prime } /p^{r} \cdot \tan \delta \\ \mu_{\text{hysteresis}} & = C_{2} \cdot p/E^{\prime } \cdot \tan \delta , \\ \end{aligned}$$

where μ_adhesion is the friction coefficient related to the adhesive force, μ_hysteresis is the friction coefficient related to the hysteresis force, C₁ and C₂ are constants, K is the shear stress of the interface, p is the contact pressure, tan δ is the loss tangent of rubber, E´ is the storage Young’s modulus of rubber, and r is a constant with a value less than 1 discussed in Sect. 7.4. Referring to Eq. (7.138), Hertz theory gives the relation r = 1/3.

Note 12.2 Eqs. (12.15) and (12.16)

In Eq. (11.52), parameters are changed as C_Fα → C_Fs, tan α → s_B and $\delta^{\prime } = 0$ is then used in the derivation for F_x. Differentiating F_x with respect to s_B, we obtain

$$\begin{aligned} \frac{{\text{d}F_{x} }}{{\text{d}s_{\text{B}} }} & = C_{Fs} s_{\text{B}} - \frac{{2C_{Fs}^{2} }}{{3\mu_{\text{s}} F_{z} }}\left( {2 - \frac{{\mu_{\text{d}} }}{{\mu_{\text{s}} }}} \right)s_{\text{B}} + \frac{{C_{Fs}^{3} }}{{9\mu_{\text{s}}^{2} F_{z}^{2} }}\left( {3 - 2\frac{{\mu_{\text{d}} }}{{\mu_{\text{s}} }}} \right)s_{\text{B}}^{2} = 0\\ &\to \left\{ {\frac{{C_{Fs} }}{{3\mu_{\text{s}} F_{z} }}\left( {3 - 2\frac{{\mu_{\text{d}} }}{{\mu_{\text{s}} }}} \right)s_{\text{B}} - 1} \right\}\left( {\frac{{C_{Fs} }}{{3\mu_{\text{s}} F_{z} }}s_{\text{B}} - 1} \right) = 0 \\ & \to s_{\text{B}} = \frac{{3\mu_{\text{s}} F_{z} }}{{C_{Fs} }}\frac{{\frac{{\mu_{\text{s}} }}{{\mu_{\text{d}} }}}}{{3\frac{{\mu_{\text{s}} }}{{\mu_{\text{d}} }} - 2}},\frac{{3\mu_{\text{s}} F_{z} }}{{C_{Fs} }}. \\ \end{aligned}$$

Note 12.3 Eqs. (12.51), (12.55) and (12.56)

Equation (12.51)

Note that Δx₁ is the cumulative relative displacement between the belt and road from the leading edge of the contact patch to the position x₁. Meanwhile, Δx₀ is the relative displacement between the belt and road at some time. Figure 12.21 shows that the tire is under braking. The slip ratio is then positive and is given by

$$s = \frac{{V_{\text{R}} - (V_{\text{B}} + \Delta V_{2} )}}{{V_{\text{R}} }} = \frac{{ - \Delta V - \Delta V_{2} }}{{V_{\text{R}} }}.$$

• Equations (12.55) and (12.56)

Referring to Fig. 12.21, we have

$$\begin{aligned} \Delta x_{2} & = \frac{{F_{x} }}{{R_{x} }} \\ \Delta x_{1} & = \frac{{F_{x} }}{{C_{x} wl}}. \\ \end{aligned}$$

Referring to Fig. 12.21, when torque is applied to the contact surface, we have

$$\begin{aligned} \Delta x & = \Delta x_{1} + \Delta x_{2} \\ G_{x} & = \frac{{F_{x} }}{\Delta x} = \frac{{F_{x} }}{{\frac{{F_{x} }}{{C_{x} bl}} + \frac{{F_{x} }}{{R_{x} }}}} = \frac{{R_{x} C_{x} bl}}{{R_{x} + C_{x} bl}}. \\ \end{aligned}$$

Referring to Eq. (12.9), C_Fs is given by C_Fs = C_xbl²/2.

Taking the limit ω/V_R → 0, we obtain

$$\begin{aligned} A & = 1 + \frac{{C_{x} b}}{{R_{x} }}\left( {l - \frac{{V_{\text{R}} }}{\omega }\sin \frac{\omega l}{{V_{\text{R}} }}} \right) = 1 + \frac{{C_{x} b}}{{R_{x} }}\left( {l - l\sin \frac{\omega l}{{V_{\text{R}} }}/\frac{\omega l}{{V_{\text{R}} }}} \right) \to 1 \\ B & = \frac{{C_{x} bV_{\text{R}} }}{{\omega R_{x} }}\left( {\cos \frac{\omega l}{{V_{\text{R}} }} - 1} \right) \to 0, \\ \end{aligned}$$

$$\begin{aligned}&\sqrt {\left( {\frac{{V_{\text{R}} }}{\omega }\sin \frac{\omega l}{{V_{\text{R}} }} - l} \right)^{2} + \frac{{V_{\text{R}}^{2} }}{{\omega^{2} }}\left( {\cos \frac{\omega l}{{V_{\text{R}} }} - 1} \right)^{2} } \\ &\quad \cong \sqrt {\left( {l\sin \frac{\omega l}{{V_{\text{R}} }}/\frac{\omega l}{{V_{\text{R}} }} - l} \right)^{2} + \frac{{V_{\text{R}}^{2} }}{{\omega^{2} }}\left( {\frac{{\omega^{2} l^{2} }}{{2V_{\text{R}}^{2} }}} \right)^{2} } \cong \frac{{\omega l^{2} }}{{2V_{\text{R}} }}, \end{aligned}$$

$$\begin{aligned} K_{\text{d}} &= \frac{{C_{x} bV}}{{\omega \sqrt {A^{2} + B^{2} } }}\sqrt {\left( {\frac{{V_{\text{R}} }}{\omega }\sin \frac{\omega l}{{V_{\text{R}} }} - l} \right)^{2} + \frac{{V_{\text{R}}^{2} }}{{\omega^{2} }}\left( {\cos \frac{\omega l}{{V_{\text{R}} }} - 1} \right)^{2} } \to \frac{{C_{x} bV_{\text{R}} }}{\omega }\frac{{\omega l^{2} }}{{2V_{\text{R}} }}\\ &= \frac{{C_{x} bl^{2} }}{2}. \end{aligned}$$

Taking the limit ω/V_R → ∞, we obtain

$$\begin{aligned} A & = 1 + \frac{{C_{x} b}}{{R_{x} }}\left( {l - \frac{{V_{\text{R}} }}{\omega }\sin \frac{\omega l}{{V_{\text{R}} }}} \right) \to 1 + \frac{{C_{x} bl}}{{R_{x} }} \\ B & = \frac{{C_{x} bV_{\text{R}} }}{{\omega R_{x} }}\left( {\cos \frac{\omega l}{{V_{\text{R}} }} - 1} \right) \to 0, \\ \end{aligned}$$

$$\begin{aligned} G_{\text{d}} &= \frac{{C_{x} b}}{{\sqrt {A^{2} + B^{2} } }}\sqrt {\left( {\frac{{V_{\text{R}} }}{\omega }\sin \frac{\omega l}{{V_{\text{R}} }} - l} \right)^{2} + \frac{{V_{\text{R}}^{2} }}{{\omega^{2} }}\left( {\cos \frac{\omega l}{{V_{\text{R}} }} - 1} \right)^{2} } \to \frac{{C_{x} bl}}{{1 + \frac{{C_{x} bl}}{{R_{x} }}}} \\ &= \frac{{C_{x} R_{x} bl}}{{R_{x} + C_{x} bl}}.\end{aligned}$$

Note 12.4 Eqs. (12.60), (12.63) and (12.87)

If the pressure distribution is expressed by a parabola, the maximum pressure p_m is equal to 1.5 times the average pressure p_av (p_m = 1.5p_av). Owing to the bending rigidity of tires, the average pressure p_av will be larger than the inflation pressure p_a (p_av > p).

• Equation (12.63)

The boundary conditions of real tires are expressed as u ≈ v at the tire shoulder and u = 0, v ≠ 0 at the tire center. Hence, Eq. (12.64) may not be satisfied in a tire.

• Equation (12.87) (Fig. 12.113)

  Fig. 12.113

  

  Average thickness of the water layer

$$\begin{aligned} h(x) & = j + \frac{{\bar{t}}}{w}x \\ h^{2} & = \frac{1}{w}\int\limits_{0}^{w} {h^{2} (x)} \text{d}x = \frac{1}{w}\int\limits_{0}^{w} {\left( {j + \frac{{\bar{t}}}{w}x} \right)^{2} } \text{d}x = j^{2} + j\bar{t} + \bar{t}^{2} /3 \\ \end{aligned}$$

Note 12.5 Eq. (12.99)

The Mohr–Coulomb failure criterion can be determined in shear tests for various normal stresses σ. The Mohr–Coulomb failure criterion is that when Mohr circle comes into contact with the line of Eq. (12.99), with plastic deformation occurring at this point. Suppose that the Mohr circle comes into contact with the Mohr–Coulomb failure criterion at point A of Fig. 12.114a and the angle between the σ-axis and the line connecting point A with the center of the Mohr circle is 2θ (where the counterclockwise direction is positive). A(σ, τ) in Fig. 12.114a expresses the stress field in the coordinate system at angle θ (clockwise direction) measured from the direction of the maximum stress σ₁. Hence, the shear stress τ is applied on plane A–A in Fig. 12.114b. Considering similarly at point A′ expressed by A′(σ, − τ) in Fig. 12.114a, the shear stress −τ is applied on plane A′–A′ with an angle π/2 − θ measured from the direction of the maximum stress σ₁ in Fig. 12.114b. According to internal friction theory, planes A–A and A′–A′ will become sliding planes.

Fig. 12.114

Physical meaning of the Mohr–Coulomb failure criterion

Note 12.6 Eqs. (12.102), (12.113) and (12.121) for the Heat Flow of a Semi-infinite Solid [160]

The governing equation for the heat conduction of a semi-infinite solid is

$$\frac{\partial T}{\partial t} = \alpha \frac{{\partial^{2} T}}{{\partial z^{2} }}\quad 0 \le z < \infty ,\quad t \ge 0,\quad T = T_{0} \quad 0 \le z < \infty ,\quad t = 0.$$

(12.222)

The solution to Eq. (12.222) is

$$T(z,t) = T_{0} \text{erf}\left( {\frac{z}{{2\sqrt {\alpha t} }}} \right),$$

(12.223)

where erf(z) is defined by

$$\text{erf}\left( z \right) = \frac{2}{\sqrt \pi }\int\limits_{0}^{z} {\text{e}^{{ - \xi^{2} }} \text{d}\xi } .$$

(12.224)

Suppose that initial condition and boundary condition are given by

$$T = T_{0} \quad 0 \le t \le l/V,\quad T = T_{\text{m}} \quad z = 0.$$

(12.225)

The solution to Eqs. (12.222) and (12.225) is

$$T_{\text{m}} - T = \left( {T_{\text{m}} - T_{0} } \right)\text{erf}\left( {\frac{z}{{2\sqrt {\alpha t} }}} \right).$$

(12.226)

The rate of heat flow is

$$\dot{Q}_{\text{c}} = - k\left. {\frac{{\text{d}T}}{{\text{d}z}}} \right|_{z = 0} = k\left( {T_{\text{m}} - T_{0} } \right)\frac{1}{{\sqrt {\pi \alpha t} }}.$$

(12.227)

Expressing T_m − T₀ = T_b and transforming from the temporal domain to the spatial domain using t = x/V yields

$$\dot{Q}_{\text{c}} = - k\left. {\frac{{\text{d}T}}{{\text{d}z}}} \right|_{z = 0} = kT_{\text{b}} \left( {\frac{V}{\pi \alpha x}} \right)^{{\frac{1}{2}}} .$$

(12.228)

The total heat that flows through a unit area from t = 0 to t = l/V is therefore

$$Q = \int\limits_{0}^{l/V} {\dot{Q}_{\text{c}} \text{d}t} = 2k\left( {T_{\text{m}} - T_{0} } \right)\left( {\frac{l}{\pi \alpha V}} \right)^{{\frac{1}{2}}} .$$

(12.229)

Note 12.7 Eq. (12.104) for heat flow of a semi-infinite solid [160]

Consider the problem that the flux of heat at x = 0 is a prescribed function of time, the initial temperature is 0 °C, and the flux $\dot{Q}$ per unit time per unit area is constant. $\dot{Q}$ is expressed by

$$\dot{Q} = - k\frac{\partial T}{\partial x}.$$

(12.230)

$\dot{Q}$ satisfies the differential equation at T:

$$\frac{{\partial \dot{Q}}}{\partial t} = - \alpha \frac{{\partial^{2} \dot{Q}}}{{\partial x^{2} }}\quad x > 0,\;t > 0,$$

(12.231)

where α is the diffusivity constant. The solution to Eq. (12.231) with

$$\dot{Q} = \dot{Q}_{0} \quad x = 0,\;t > 0$$

(12.232)

is

$$\dot{Q} = \dot{Q}_{0}\text{erfc}\left( {\frac{x}{{2\sqrt {\alpha t} }}} \right).$$

(12.233)

erfc(x) is defined by

$$\begin{aligned}\text{erfc}\left( x \right) & = 1 - \text{erf}\left( x \right) = \frac{2}{\sqrt \pi }\int\limits_{x}^{\infty } {\text{e}^{{ - \xi^{2} }} \text{d}\xi } \\ \text{erf}\left( x \right) & = \frac{2}{\sqrt \pi }\int\limits_{0}^{x} {\text{e}^{{ - \xi^{2} }} \text{d}\xi } . \\ \end{aligned}$$

(12.234)

Equations (12.230) and (12.232) yield

$$T = \frac{{\dot{Q}_{0} }}{k}\int\limits_{x}^{\infty } {\text{erfc}\left( {\frac{x}{{2\sqrt {\alpha t} }}} \right)\text{d}x} = \frac{{2\dot{Q}_{0} \sqrt {\alpha t} }}{k}\text{ierfc}\left( {\frac{x}{{2\sqrt {\alpha t} }}} \right) = \frac{{2\dot{Q}_{0} }}{k}\left\{ {\left( {\frac{\alpha t}{\pi }} \right)^{{\frac{1}{2}}} \text{e}^{{ - \frac{{x^{2} }}{4\alpha t}}} - \frac{x}{2}\text{erfc}\left( {\frac{x}{{2\sqrt {\alpha t} }}} \right)} \right\}.$$

(12.235)

In the derivation of Eq. (12.235), applying integration by parts, we obtain

$$\begin{aligned} \text{ierf}(x) & = \frac{2}{\sqrt \pi }\int\limits_{x}^{\infty } {\int\limits_{\eta }^{\infty } {\text{e}^{{ - \xi^{2} }} \text{d}\xi } \text{d}\eta } = \frac{2}{\sqrt \pi }\left[ {\eta \int\limits_{\eta }^{\infty } {\text{e}^{{ - \xi^{2} }} \text{d}\xi } } \right]_{x}^{\infty } - \frac{2}{\sqrt \pi }\int\limits_{x}^{\infty } {\eta \text{e}^{{ - \xi^{2} }} \text{d}\eta } \\ & = - \frac{2}{\sqrt \pi }x\int\limits_{x}^{\infty } {\text{e}^{{ - \xi^{2} }} \text{d}\xi } + \frac{1}{\sqrt \pi }\text{e}^{{ - x^{2} }} = \frac{1}{\sqrt \pi }\text{e}^{{ - x^{2} }} - x\text{erf}(x), \\ \end{aligned}$$

(12.236)

where

$$\text{ierfc}\left( x \right) = \int\limits_{x}^{\infty } {\text{erfc}\left( \xi \right)\text{d}\xi } .$$

(12.237)

The temperature at x = 0 is given by

$$T = \frac{{2\dot{Q}_{0} }}{k}\left( {\frac{\alpha t}{\pi }} \right)^{{\frac{1}{2}}} .$$

(12.238)

The above discussion can be extended to the infinite composite solid. Suppose the region x > 0 is that of one substance with the properties k₁, ρ₁ and α₁ while the region x < 0 is that of another substance with the properties k₂, ρ₂ and α₂. The boundary conditions on the plane of separation x = 0 are

$$\begin{aligned} T_{1} & = T_{2} \quad x = 0,\quad t > 0 \\ k_{1} \frac{{\partial T_{1} }}{\partial x} & = k_{2} \frac{{\partial T_{2} }}{\partial x}\quad x = 0,\quad t > 0, \\ \end{aligned}$$

(12.239)

where we write T₁ for the temperature in the region x > 0 and T₂ for the temperature in the region x < 0.

Suppose that heat is supplied for t > 0 at the constant rate $\dot{Q}_{0}$ per unit time per unit area on the plane x = 0. Similar to Eq. (12.235), we assume

$$\begin{aligned} T_{1} & = \frac{{2\dot{Q}_{01} \sqrt {\alpha_{1} t} }}{{k_{1} }}\text{ierfc}\left( {\frac{x}{{2\sqrt {\alpha_{1} t} }}} \right)\quad x > 0 \\ T_{2} & = \frac{{2\dot{Q}_{02} \sqrt {\alpha_{2} t} }}{{k_{2} }}\text{ierfc}\left( {\frac{\left| x \right|}{{2\sqrt {\alpha_{2} t} }}} \right)\quad x < 0, \\ \end{aligned}$$

(12.240)

where the unknown constants T₁ and T₂ are to be found from the boundary conditions at x = 0, which gives

$$\frac{{\dot{Q}_{01} \sqrt {\alpha_{1} } }}{{k_{1} }} = \frac{{\dot{Q}_{02} \sqrt {\alpha_{2} } }}{{k_{2} }},\quad \text{and}\quad \dot{Q}_{01} + \dot{Q}_{02} = \dot{Q}_{0} .$$

(12.241)

Therefore,

$$\begin{aligned} T_{1} & = \frac{{2\dot{Q}_{0} \sqrt {\alpha_{1} \alpha_{2} t} }}{{k_{1} \sqrt {\alpha_{2} } + k_{2} \sqrt {\alpha_{1} } }}\text{ierfc}\left( {\frac{x}{{2\sqrt {\alpha_{1} t} }}} \right)\quad x > 0 \\ T_{2} & = \frac{{2\dot{Q}_{0} \sqrt {\alpha_{1} \alpha_{2} t} }}{{k_{1} \sqrt {\alpha_{2} } + k_{2} \sqrt {\alpha_{1} } }}\text{ierfc}\left( {\frac{\left| x \right|}{{2\sqrt {\alpha_{2} t} }}} \right)\quad x < 0. \\ \end{aligned}$$

(12.242)

Considering $\text{ierfc}(0) = 1/\sqrt \pi$ and T₁|_x=0 = ΔT, Eq. (12.242) yields

$$\Delta T = \frac{{2\dot{Q}_{0} \sqrt {\alpha_{1} \alpha_{2} t} }}{{k_{1} \sqrt {\alpha_{2} } + k_{2} \sqrt {\alpha_{1} } }}\frac{1}{\sqrt \pi } = \frac{{\frac{{2\dot{Q}_{0} }}{{k_{1} }}\left( {\frac{{\alpha_{1} t}}{\pi }} \right)^{{\frac{1}{2}}} }}{{\frac{{k_{2} }}{{k_{1} }}\left( {\frac{{\alpha_{1} }}{{\alpha_{2} }}} \right)^{{\frac{1}{2}}} + 1}}.$$

(12.243)

Note 12.8 Eqs. (12.125) and (12.134)

Equation (12.125)

The Navier–Stokes equation is expressed as Eq. (12.67). In the thin-film lubrication problem, terms of $\partial^{2}/\partial z^{2}$ are dominant and the modified Reynolds number Re_mod is used instead of the Reynolds number Re:

$$Re_{\bmod } = \frac{\text{inertia term}}{\text{viscosity term}} = \frac{{\rho U^{2} /l}}{{\eta U/h^{2} }} = \frac{\rho Ul}{\eta }\left( {\frac{h}{l}} \right)^{2} = Re \left( {\frac{h}{l}} \right)^{2} ,$$

(12.244)

where U is the speed of water relative to the tire, h is the thickness of the water layer and l is the contact length. Because inertia terms are negligible comparing to viscosity terms at small Re_mod, Eq. (12.67) is simplified as

$$\begin{aligned} \frac{{\partial p_{\text{f}} }}{\partial x} & = \eta \frac{{\partial^{2} u}}{{\partial z^{2} }} \\ \frac{{\partial p_{\text{f}} }}{\partial y} & = \eta \frac{{\partial^{2} v}}{{\partial z^{2} }} \\ \frac{{\partial p_{\text{f}} }}{\partial z} & = 0. \\ \end{aligned}$$

(12.245)

Equation (12.245) is called the Stokes equation. Suppose boundary conditions are given by

$$\begin{aligned} u & = U_{1} ,\quad v = V_{1} ,\quad w = W_{1} \quad \text{at}\,z = 0\quad (\text{on the pavement surface}) \\ u & = U_{2} ,\quad v = V_{2} ,\quad w = W_{2} \quad \text{at}\,z = h\quad (\text{on the tire surface}), \\ \end{aligned}$$

(12.246)

where U₂, V₂ and W₂ are the velocities of water at z = h in x-, y- and z-directions, respectively. W₂ = 0 is satisfied if the tire is assumed to have a blank tread. Furthermore, V₁ = 0 is satisfied on a road surface (z = 0) when lateral sliding does not occur. W₁ = 0 is also satisfied on a road surface (z = 0), because there is no water flow in the z-direction if water flow is not disturbed on the smooth road surface.

The boundary condition for a free surface is given by

$$p_{\text{f}} = 0.$$

(12.247)

Hence, the fundamental equations for hydroplaning are Eq. (12.245) with the boundary conditions of Eqs. (12.246) and (12.247). Solutions to Eq. (12.245) satisfying the boundary condition of Eq. (12.246) are given by

$$\begin{aligned} u & = - \frac{1}{2\eta }\frac{{\partial p_{\text{f}} }}{\partial x}z\left( {h - z} \right) + U_{1} \left( {1 - \frac{z}{h}} \right) + U_{2} \frac{z}{h} \\ v & = - \frac{1}{2\eta }\frac{{\partial p_{\text{f}} }}{\partial y}z\left( {h - z} \right) + V_{1} \left( {1 - \frac{z}{h}} \right) + V_{2} \frac{z}{h}, \\ \end{aligned}$$

(12.248)

where the average speeds of the fluid $\bar{u}$ and $\bar{v}$ are given by

$$\begin{aligned} h\bar{u} & = \int\limits_{0}^{h} {u\text{d}z = } - \frac{1}{12\mu }\frac{{\partial p_{\text{f}} }}{\partial x}h^{3} + \frac{1}{2}\left( {U_{1} + U_{2} } \right)h \\ h\bar{v} & = \int\limits_{0}^{h} {v\text{d}z = } - \frac{1}{12\mu }\frac{{\partial p_{\text{f}} }}{\partial y}h^{3} + \frac{1}{2}\left( {V_{1} + V_{2} } \right)h. \\ \end{aligned}$$

(12.249)

Assuming that the continuity equation is satisfied in the sense of the average water thickness h, we obtain

$$\int\limits_{0}^{h} {\left( {\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z}} \right)\text{d}z = 0} .$$

(12.250)

Rewriting the above equation yields²⁰

$$0 = \frac{\partial }{\partial x}\int\limits_{0}^{h} {u\text{d}z + } \frac{\partial }{\partial y}\int\limits_{0}^{h} {v\text{d}z + } \left[ { - \frac{\partial h}{\partial x}\left( u \right)_{z = h} - \frac{\partial h}{\partial y}\left( v \right)_{z = h} + \left( w \right)_{z = 0}^{z = h} } \right],$$

(12.251)

where the term in [·] is zero because it is the vertical component of the water speed on the tire and road surface.

Using the definition of the average speed of fluid $\bar{u}$ and $\bar{v}$ in Eq. (12.249), Eq. (12.251) is rewritten as

$$\frac{{\partial \left( {h\bar{u}} \right)}}{\partial x} + \frac{{\partial \left( {h\bar{v}} \right)}}{\partial y} = 0.$$

(12.252)

The substitution of Eq. (12.249) into Eq. (12.252) yields

$$\frac{\partial }{\partial x}\left( {h^{3} \frac{{\partial p_{\text{f}} }}{\partial x}} \right) + \frac{\partial }{\partial y}\left( {h^{3} \frac{{\partial p_{\text{f}} }}{\partial y}} \right) = F(x,y).$$

(12.253)

Equation (12.253) is called the Reynolds equation. F(x, y) is expressed as

$$\begin{aligned}F(x,y) &= 12\eta h\frac{\partial }{\partial x}\left( {\frac{{U_{1} + U_{2} }}{2}} \right) + 12\eta \frac{\partial h}{\partial x}\left( {\frac{{U_{1} + U_{2} }}{2}} \right) \\ &\quad+\, 12\eta h\frac{\partial }{\partial y}\left( {\frac{{V_{1} + V_{2} }}{2}} \right) + 12\eta \frac{\partial h}{\partial y}\left( {\frac{{V_{1} + V_{2} }}{2}} \right).\end{aligned}$$

(12.254)

If flow in the x-direction is neglected, Eq. (12.254) can be simplified as

$$F(x,y) = 12\eta h\frac{\partial }{\partial y}\left( {\frac{{V_{1} + V_{2} }}{2}} \right) + 12\eta \frac{\partial h}{\partial y}\left( {\frac{{V_{1} + V_{2} }}{2}} \right) = 12\eta \frac{\partial }{\partial y}\left( {h\frac{{V_{1} + V_{2} }}{2}} \right).$$

(12.255)

• Equation (12.134)

Using Eq. (12.121), we obtain

$$\int\limits_{{x_{\text{m}} }}^{l} {\dot{Q}_{\text{c}} \text{d}t} = \frac{1}{V}\int\limits_{{x_{\text{m}} }}^{l} {\dot{Q}_{\text{c}} \text{d}x} = \frac{1}{V}\int\limits_{{x_{\text{m}} }}^{l} {k_{\text{i}} T_{\text{b}} } \sqrt {\frac{V}{{\pi \alpha_{\text{i}} \left( {x - x_{\text{m}} } \right)}}} \text{d}x = 2k_{\text{i}} T_{\text{b}} \left\{ {\frac{{l - x_{\text{m}} }}{{\pi \alpha_{\text{t}} V}}} \right\}^{{\frac{1}{2}}} .$$

Note 12.9 Eq. (12.149) [115]

The amount of heat conducted into ice during the time interval l/V is expressed by

$$Q_{c1} = \frac{{k_{1} bl\Delta T_{1} }}{\delta }\frac{l}{V},$$

(12.256)

where l is the contact length of the tire, b is the contact width of the tire, V is the axle velocity, k₁ is the thermal conductivity of ice, ΔT₁ is the temperature difference between the contact surface and ice, and δ is the thickness of the layer into which heat is conducted. This amount of heat is equal to the energy stored in the heated layer expressed by

$$Q_{{c1\_\text{stored}}} = \rho_{1} bl\delta c_{1} \Delta T_{1} /2,$$

(12.257)

where ρ₁ is the density of ice and c₁ is the specific heat capacity of ice.

Using Eqs. (12.256) and (12.257), the thickness δ can be eliminated:

$$Q_{c1} = bl\Delta T_{1} \left( {\frac{l}{2V}} \right)^{{\frac{1}{2}}} \left( {k_{1} c_{1} \rho_{1} } \right)^{{\frac{1}{2}}} .$$

(12.258)

Considering heat conduction from the interface to the tire, the total heat flow due to conduction is

$$Q_{\text{c}} = bl\left( {\frac{l}{2V}} \right)^{{\frac{1}{2}}} \left\{ {\Delta T_{1} \left( {k_{1} c_{1} \rho_{1} } \right)^{{\frac{1}{2}}} + \Delta T_{2} \left( {k_{2} c_{2} \rho_{2} } \right)^{{\frac{1}{2}}} } \right\},$$

(12.259)

where k₂ is the thermal conductivity of the tire, ΔT₂ is the temperature difference between the contact surface and the tire, ρ₂ is the density of the tire tread rubber, and c₂ is the specific heat capacity of the tire tread rubber.

The melting of an ice layer of thickness d requires energy Q_m:

$$Q_{\text{m}} = bl\text{d}L\rho_{1} ,$$

(12.260)

where L is the latent heat of melting for ice. The frictional energy produced during a time interval l/V is expressed by

$$Q_{\text{f}} = \mu F_{z} V_{\text{s}} \cdot l/V,$$

(12.261)

where F_z is the load applied to the tire, μ is the friction coefficient of the tire on ice, and V_s is the sliding velocity.

Using the equation Q_f = Q_c + Q_m for Eqs. (12.259), (12.260) and (12.261), we obtain

$$\mu F_{z} V_{\text{s}} \frac{l}{V} = bl\left( {\frac{l}{2V}} \right)^{{\frac{1}{2}}} \left\{ {\Delta T_{1} \left( {k_{1} c_{1} \rho_{1} } \right)^{{\frac{1}{2}}} + \Delta T_{2} \left( {k_{2} c_{2} \rho_{2} } \right)^{{\frac{1}{2}}} } \right\} + bl\text{d}L\rho_{1} ,$$

(12.262)

from which the thickness of the water layer d can be obtained as

$$d = \frac{1}{{L\rho_{1} }}\left[ {\frac{{\mu F_{z} V_{\text{s}} }}{bV} - \left( {\frac{l}{2V}} \right)^{{\frac{1}{2}}} \left\{ {\Delta T_{1} \left( {k_{1} c_{1} \rho_{1} } \right)^{{\frac{1}{2}}} + \Delta T_{2} \left( {k_{2} c_{2} \rho_{2} } \right)^{{\frac{1}{2}}} } \right\}} \right].$$

(12.263)

Neglecting the frictional force in the dry contact region, the frictional force F_μ generated by the viscous shear in the water layer between the ice and tire is expressed as

$$F_{\mu } = \tau bl = \eta \frac{{\text{d}v}}{{\text{d}y}}bl = \eta \frac{{V_{\text{s}} }}{\text{d}}bl,$$

(12.264)

where τ is the shear stress and η is the viscosity of water.

Considering that F_μ = μF_z and substituting d of Eq. (12.263) into Eq. (12.264) yields

$$\mu^{2} - \frac{b}{{F_{z} }}\frac{V}{{V_{\text{s}} }}\left( {\frac{l}{2V}} \right)^{{\frac{1}{2}}} \left\{ {\Delta T_{1} \left( {\lambda_{1} c_{1} \rho_{1} } \right)^{{\frac{1}{2}}} + \Delta T_{2} \left( {\lambda_{2} c_{2} \rho_{2} } \right)^{{\frac{1}{2}}} } \right\}\mu - \frac{{\eta VblL\rho_{1} b}}{{F_{z}^{2} }} = 0.$$

(12.265)

When a tire is locked, V_s = V is satisfied. Substituting this relation into Eq. (12.265), μ is obtained as

$$\begin{aligned} \mu &= \frac{b}{{2F_{z} }}\left( {\frac{l}{{2V_{\text{s}} }}} \right)^{{\frac{1}{2}}} \left\{ {\Delta T_{1} \left( {k_{1} c_{1} \rho_{1} } \right)^{{\frac{1}{2}}} + \Delta T_{2} \left( {k_{2} c_{2} \rho_{2} } \right)^{{\frac{1}{2}}} } \right\} \\ &\quad +\, \left[ {\frac{{b^{2} l}}{{8F_{z} 2V_{\text{s}} }}\left\{ {\Delta T_{1} \left( {k_{1} c_{1} \rho_{1} } \right)^{{\frac{1}{2}}} + \Delta T_{2} \left( {k_{2} c_{2} \rho_{2} } \right)^{{\frac{1}{2}}} } \right\}^{2} + \frac{{\eta_{0} V_{\text{s}} blL\rho_{1} b}}{{F_{z}^{2} }}} \right]^{{\frac{1}{2}}} . \end{aligned}$$

(12.266)

Note 12.10 Eq. (12.251)

$$\begin{aligned}& \int {u(x,y,z)\text{d}z} = U(x,y,z),\quad \int\limits_{0}^{h(x,y)} {u(x,y,z)\text{d}z} = U(x,y,h(x,y)) - U(x,y,0) \\ &\frac{\partial }{\partial x}\int\limits_{0}^{h(x,y)} {u(x,y,z)\text{d}z} = \frac{\partial }{\partial x}U(x,y,h(x,y)) + \frac{\partial }{\partial h}U(x,y,h(x,y))\frac{\partial h}{\partial x} - \frac{\partial }{\partial x}U(x,y,0) \\ & \int\limits_{0}^{h(x,y)} {\frac{\partial }{\partial x}u(x,y,z)\text{d}z} = \frac{\partial }{\partial x}U(x,y,h(x,y)) - \frac{\partial }{\partial x}U(x,y,0) \\ &\int\limits_{0}^{h(x,y)} {\frac{\partial }{\partial x}u(x,y,z)\text{d}z} = \frac{\partial }{\partial x}\int\limits_{0}^{h(x,y)} {u(x,y,z)\text{d}z} - \frac{\partial }{\partial h}U(x,y,h(x,y))\frac{\partial h}{\partial x} \\ &\quad = \frac{\partial }{\partial x}\int\limits_{0}^{h(x)} {u(x,y,z)\text{d}z} - u(x,y,h(x,y))\frac{\partial h}{\partial x} \\ \end{aligned}$$