Tire Noise

Abstract

Tire/road noise consists of noise due to tire surface vibration and noise related to aerodynamics. The former noise is usually explained by three elements, namely the external force applied to a tire, vibration properties of the tire and the acoustic field relating to the tire surface and road surface. The external force includes the tread impact associated with the lateral grooves and road roughness. The surface vibration of a tire is calculated by multiplying external forces by the tire’s vibration properties expressed by a transfer function. Tire/road radiation noise is then calculated by surface vibration via the Helmholtz equation [i.e., employing the boundary element method (BEM)]. The most important element for tire noise is the external force because other elements may not be controlled by tire design without deteriorating other tire performances. The external force due to lateral grooves is estimated by using a phenomenological model that uses the contact shape, pattern geometry and contact pressure or by conducting FEA. Meanwhile, the external force due to road roughness is estimated by measuring the spindle force of a tire rolling over a simple roughness and conducting contact analysis employing a Winker model with nonlinear contact stiffness or FEA. The vibration properties of a tire can be predicted by conducting FEA or using an elastic ring model. In pattern design, the phenomenological model is used as a design tool to determine the pattern geometry, and the pitch sequence is optimized by a GA. The Helmholtz resonator may be added to the circumferential grooves to reduce the pipe resonance noise. Furthermore, the special wheel or sound-absorbing material may be used to reduce the acoustic cavity noise.

Notes

Note 10.1 Eq. (10.5)

Suppose that the acceleration is constant and the initial conditions are given by h(0) = 0, dh/dt|_t=0 = 0. Then, h(t) is given by h(t) = at², where a is a constant. When the average speed is dh/dt_touch during dt_touch, a is given by

$$a = {{\frac{{{\text{d}}h}}{{{\text{d}}t_{\text{touch}} }}} \mathord{\left/ {\vphantom {{\frac{{{\text{d}}h}}{{{\text{d}}t_{\text{touch}} }}} {\frac{{{\text{d}}t_{\text{touch}} }}{2}}}} \right. \kern-0pt} {\frac{{{\text{d}}t_{\text{touch}} }}{2}}}.$$

Because the acceleration is given by 2a, it is expressed as

$$\frac{{{\text{d}}^{2} h}}{{{\text{d}}t^{2} }} = 2{{\frac{{{\text{d}}h}}{{{\text{d}}t_{\text{touch}} }}} \mathord{\left/ {\vphantom {{\frac{{{\text{d}}h}}{{{\text{d}}t_{\text{touch}} }}} {\frac{{{\text{d}}t_{\text{touch}} }}{2}}}} \right. \kern-0pt} {\frac{{{\text{d}}t_{\text{touch}} }}{2}}}.$$

Note 10.2 Cutoff Frequency (Cut-on Frequency)

The bending wave equation for a thin plate lying on the x–z plane in rectangular Cartesian coordinates is given by

$$D\left( {\frac{{\partial^{4} w}}{{\partial x^{4} }} + 2\frac{{\partial^{4} w}}{{\partial x^{2} \partial z^{2} }} + \frac{{\partial^{4} w}}{{\partial z^{4} }}} \right) = - m\frac{{\partial^{2} w}}{{\partial t^{2} }}.$$

(10.124)

Consider a simple harmonic plane wave described by

$$w(x,z,t) = \tilde{w}\exp \left[ {j(\omega t - k_{x} x - k_{z} z)} \right].$$

(10.125)

The substitution of Eq. (10.125) into Eq. (10.124) yields

$$\left[ {D\left( {k_{x}^{4} + 2k_{x}^{2} k_{z}^{2} + k_{z}^{4} } \right) - m\omega^{2} } \right]\tilde{w} = 0$$

(10.126)

or

$$D(k_{x}^{2} + k_{z}^{2} )^{2} - m\omega^{2} = 0.$$

(10.127)

If we write $k_{b}^{2} = k_{x}^{2} + k_{z}^{2}$, we obtain

$$Dk_{b}^{4} - m\omega^{2} = 0.$$

(10.128)

The flat-plate waveguide shown in Fig. 10.120 takes the form of an infinitely long strip of uniform width l located between boundaries that provide simple support. w is expressed by

$$w = \tilde{w}_{0} \sin (p\pi x/l)\exp \left\{ {j(\omega t - k_{zp} z)} \right\}.$$

(10.129)

Here, k_zp is the wavenumber corresponding to the propagation of the waveguide modes:

$$k_{zp}^{2} = k_{b}^{2} - k_{x}^{2} = k_{b}^{2} - \left( {\frac{p\pi }{l}} \right)^{2} = \omega \left( {\frac{m}{D}} \right)^{1/2} - \left( {\frac{p\pi }{l}} \right)^{2} .$$

(10.130)

This relation is represented qualitatively in Fig. 10.120. Equation (10.130) shows that real (propagating) solutions exist for each value of p only at frequencies that satisfy the condition ω > (pπ/l)²(D/m)^1/2. The frequencies at which k_zp = 0 are resonance frequencies of a simply supported beam of length l, and they are known as the cutoff frequencies of the waveguide modes of order p. Below its cutoff frequencies, a mode cannot effectively propagate wave energy and its amplitude decays exponentially away from a point of excitation. At the modal cutoff frequencies, the modal phase velocities c_ph = ω/k_zp are infinite and the modal group velocities c_g = ∂ω/∂k_zp are zero.

Fig. 10.120

Wave behavior of a simply supported flat-plate strip [162]

Note 10.3 Eqs. (10.77), (10.81), (10.86) and (10.88)

Equation (10.77)

Considering the relation p = ρcV, where V is the particle velocity, the volume velocity U is expressed by U = VS = Sp/(ρc).

Equation (10.81)

Equation (10.81) is obtained by solving

$$A_{2} + B_{2} = A_{3} + B_{3} ,\quad \frac{S}{\rho c}(A_{2} - B_{2} ) = \frac{{S_{cp} }}{\rho c}(A_{3} - B_{3} ) \Rightarrow A_{2} - B_{2} = m(A_{3} - B_{3} ).$$

Equation (10.86)

Substituting Eq. (10.77) into Eq. (10.81) and then substituting Eq. (10.81) into Eq. (10.83), we obtain

$$\begin{aligned} A_{4} & = A_{1} {\text{e}}^{{ - jkl_{2} }} \left( {\frac{m + 1}{2m}{\text{e}}^{{ - jkl_{1} }} - \frac{m - 1}{2m}{\text{e}}^{{jkl_{1} }} } \right) \\ B_{4} & = A_{1} {\text{e}}^{{jkl_{2} }} \left( {\frac{m - 1}{2m}{\text{e}}^{{ - jkl_{1} }} - \frac{m + 1}{2m}{\text{e}}^{{jkl_{1} }} } \right). \\ \end{aligned}$$

The substitution of the above equations into Eq. (10.85) yields

$$(m + 1)\left\{ {{\text{e}}^{{ - jk(l_{1} + l_{2} )}} - {\text{e}}^{{jk(l_{1} + l_{2} )}} } \right\} + (m - 1)\left\{ {{\text{e}}^{{ - jk(l_{1} - l_{2} )}} - {\text{e}}^{{jk(l_{1} - l_{2} )}} } \right\} = 0.$$

Equation (10.88)

Equation (10.87) can be rewritten as

$$m\sin \left( {k\frac{{L_{c} - l_{cp} }}{2}} \right)\cos \left( {k\frac{{l_{cp} }}{2}} \right) + \cos \left( {k\frac{{L_{c} - l_{cp} }}{2}} \right)\sin \left( {k\frac{{l_{cp} }}{2}} \right) = 0.$$

Transforming the above equation gives

$$\left\{ {m(\cos kl_{cp} + 1) + 1 - \cos kl_{cp} } \right\}\tan k\frac{{L_{c} }}{2} - (m - 1)\sin kl_{cp} = 0,$$

$$\tan k\frac{{L_{c} }}{2} = \frac{{(m - 1)\sin kl_{cp} }}{{m + 1 + (m - 1)\cos kl_{cp} }} = \frac{{\sin kl_{cp} }}{{\frac{m + 1}{m - 1} + \cos kl_{cp} }}.$$

Note 10.4 Ray Theory [70] and Eq. (10.113)

Ray theory

Suppose that a straight ray has time dependence e^−iωt and wavenumber k. The ray can be expressed as p(s) = A(s)e^iks, where s is the distance measured along the ray, with associated amplitude A(s) and phase ks. According to ray theory, energy flux is conserved along any ray tube as shown in Fig. 10.121. A²(s)dΣ(s)/ρc is therefore constant, where dΣ(s) is the cross-sectional area of the ray tube. The elemental area dΣ(s) has two Gaussian principal directions, say 1 and 2, and principal radii of curvature σ₁ + s and σ₂ + s, respectively. Physically, σ₁ and σ₂ are the distances from s = 0 to the caustics in the two principal directions. The ratio of the area change is therefore dΣ(s)/dΣ(0) = (σ₁ + s)(σ₂ + s)/(σ₁σ₂), which leads to the ray amplitude at s:

$$A(s) = A(0)\left( {\frac{{\sigma_{1} \sigma_{2} }}{{(\sigma_{1} + s)(\sigma_{2} + s)}}} \right)^{1/2} .$$

(10.131)

Fig. 10.121

A ray tube

When an incident acoustic ray from the caustic S is reflected by a rigid surface, the reflected ray does not change phase at the reflection point. The ray is emitted from the image caustic S′ with an amplitude that leads to p(0) at point O, as shown in Fig. 10.122. Therefore, incorporating the above equation, the ray at s is

$$p(s) = p(0)\left( {\frac{{\sigma_{1} \sigma_{2} }}{{(\sigma_{1} + s)(\sigma_{2} + s)}}} \right)^{1/2} {\text{e}}^{{{\text{i}}ks}} ,$$

(10.132)

where σ₂ is the distance from O to the caustic of the ray in the second principal direction. From elementary geometrical relations, we obtain

$$\sigma_{1} = \frac{{\rho_{1} R\cos \theta }}{{2\rho_{1} + R\cos \theta }}.$$

(10.133)

For this cylindrical tire, the radius of curvature is infinitely large across the tire belt. Substituting R = ∞ into Eq. (10.133) and changing the suffix from 1 to 2, we obtain

$$\sigma_{2} = \rho_{2} .$$

(10.134)

Equation (10.132) is applied recursively to calculate the ray amplitude for each reflection on the tire surface.

Fig. 10.122

Ray reflection from a smoothly curved surface

Derivation of ( 10.133 )

Using the sine theorem illustrated in Fig. 10.122, we obtain

$$\begin{aligned} & \frac{{R{\text{d}}\phi }}{{\sin (2{\text{d}}\phi + {\text{d}}\psi )}} = \frac{{\sigma_{1} }}{{\sin \left( {\frac{\pi }{2} - \frac{3}{2}{\text{d}}\phi - \theta - {\text{d}}\psi } \right)}} \\ & \frac{{R{\text{d}}\phi }}{{\sin ({\text{d}}\psi )}} = \frac{{\rho_{1} }}{{\sin \left( {\frac{\pi }{2} - \frac{{{\text{d}}\phi }}{2} - \theta - {\text{d}}\psi } \right)}}. \\ \end{aligned}$$

From the above equations, we obtain

$$\begin{aligned} & R\left\{ {\cos \theta - \sin \theta \left( {{\text{d}}\psi + \frac{3}{2}{\text{d}}\phi } \right)} \right\}{\text{d}}\phi = \sigma_{1} (2{\text{d}}\phi + {\text{d}}\psi ) \\ & R\left\{ {\cos \theta - \sin \theta \left( {{\text{d}}\psi + \frac{{{\text{d}}\phi }}{2}} \right)} \right\}{\text{d}}\phi = \rho_{1} {\text{d}}\psi . \\ \end{aligned}$$

Neglecting the small terms, we obtain

$$\begin{aligned} R\cos \theta \frac{{{\text{d}}\phi }}{{{\text{d}}\psi }} & = \sigma_{1} \left( {2\frac{{{\text{d}}\phi }}{{{\text{d}}\psi }} + 1} \right) \\ R\cos \theta \frac{{{\text{d}}\phi }}{{{\text{d}}\psi }} & = \rho_{1} \\ \end{aligned}$$

and eliminating dϕ/dψ, we obtain σ₁ = ρ₁R cos θ/(2ρ₁ + R cos θ).

Equation (10.113)

$$\begin{aligned} & \frac{{d_{2M} }}{{\sin \left\{ {\frac{\pi }{2} - (\phi + \theta_{2M} )} \right\}}} = \frac{L}{{\sin \left\{ {\frac{\pi }{2} + \theta_{2M} } \right\}}} \\ & \frac{{d_{2M} }}{{\cos (\phi + \theta_{2M} )}} = \frac{{d_{0} \cos \theta_{0} }}{{\cos (\phi + \theta_{2M} )\cos \theta_{2M} }} = \frac{L}{{\cos \theta_{2M} }} \\ & \frac{{l_{2M} }}{\sin \phi } = \frac{L}{{\sin \left\{ {\frac{\pi }{2} + \theta_{2M} } \right\}}} \\ & \theta_{2M} = \theta_{0} + 2M\alpha \\ \end{aligned}$$