# A Turning Brake System for Motorcycles via an Autoregulative Optimal Slip Ratio
## 2 Optimization Model for the Optimum Slip Ratio  
If a motorcycle brakes during a turn, the body leans at an angle. the body roll angle is equal to the camber angle, i.e., \(\phi=\gamma\). The gravity generates torque to balance the torque that is exerted by the centrifugal force \(F_{\rm C}\) on the body. The centrifugal force creates a side reaction force from the ground on the tire. In other words, tires are subject to longitudinal and lateral forces. According to [PERSON] tire model, the longitudinal force \(F_{x}\) can be expressed as follows:  
\[F_{x}(\kappa,\alpha,\gamma,F_{z},\mu_{r})=G_{xx}\cdot F_{x0} \tag{1}\]  
The parameters in Equation (1) are shown as follows:  
\[\left\{ \begin{array}{l}F_{x0}=\mu_{r}D_{x}sin[C_{x}arctan\{B_{x}\kappa_{x}- E_{x}(B_{x}\kappa_{x}-arctan(B_{x}\kappa_{x}))\}]+S_{Vx}\\ \kappa_{x}=\kappa+S_{Hx};\quad C_{x}=p_{Cx1};\quad\quad f_{z}=(F_{z}-F_{z0})/ F_{z0};\\ D_{x}=(p_{Dx1}+p_{Dx2}d{f_{z}})\big{(}1-p_{Dx3}\gamma^{2}\big{)}F_{z}\\ E_{x}=\Big{(}p_{Ex1}+p_{Ex2}d{f_{z}}+p_{Ex3}d{f_{z}}^{2}\Big{)}[1-p_{Ex4}sgn( \kappa_{x})]\\ K_{x}=F_{z}(p_{Ex1}+p_{Kx2}d{f_{z}})\exp(p_{Ex3}d{f_{z}});\quad B_{x}=K_{x}/( C_{x}D_{x})\\ S_{Hx}=p_{Hx1}+p_{Hx2}d{f_{z}};\quad S_{Vx}=F_{z}(p_{Vx1}+p_{Vx2}d{f_{z}})\\ S_{Hxa}=r_{Hx1};\quad\quad\alpha_{s}=\alpha+S_{Hxa};\quad B_{xx}=r_{Bx1}\cos[ arctan({f_{Bx2}\kappa})]\\ C_{xx}=r_{Cx1};\quad\quad E_{xx}=r_{Ex1}+r_{Ex2}d{f_{z}}\\ G_{xx}=\frac{cos[C_{x}arctan\{B_{xx}\kappa_{x}-E_{x}(B_{xx}\kappa_{x}-arctan (B_{xx}\kappa_{x}))]\}}{cos[C_{x}arctan\{B_{xx}S_{Hxa}-E_{x}(B_{xx}\kappa_{x} -arctan(B_{xx}S_{Hxa}))]\}}]\end{array} \right.\]  
Similarly, the lateral force \(F_{y}\) can be expressed as follows:  
\[F_{y}(\kappa,\alpha,\gamma,F_{z},\mu_{r})=G_{y\kappa}\cdot F_{y0}+S_{Vy\kappa} \tag{2}\]  
The parameters in Equation (2) are shown as follows:\[\left\{ \begin{array}{c}F_{y0}=\mu_{r}\big{\{}D_{y}sin[C_{y}arctan\{B_{y}\alpha_{y} -E_{y}\big{(}B_{y}\alpha_{y}-arctan(B_{y}\alpha_{y})\big{)}\}]+S_{vy}\big{\}}\\ \alpha_{y}=\alpha+S_{Hy};\quad C_{y}=p_{Cy1};\quad df_{z}=(F_{z}-F_{z0})/F_{z0} ;\\ D_{y}=\big{(}p_{Dy1}+p_{Dy2}df_{z}\big{)}\big{(}1-p_{Dy3}\gamma^{2}\big{)}F_{z} \\ E_{y}=\big{(}p_{Ey1}+p_{Ey2}df_{z}\big{)}\big{[}1-\big{(}p_{Ex3}+p_{Ex4}\gamma \big{)}sgn\big{(}\alpha_{y}\big{)}\big{]}\\ K_{y}=p_{Ky1}F_{z0}sin\big{\{}2\arctan\big{[}F_{z}/\big{(}p_{ky2}F_{z0}\big{)} \big{]}\big{\}}\big{(}1-p_{ky3}|\gamma\big{)}\\ B_{y}=K_{y}/\big{(}C_{y}D_{y}\big{)};\quad S_{Hy}=p_{Hy1}+p_{Hy2}df_{z}+p_{Hy 3}\gamma\ ;\quad S_{Hy}=r_{Hy1}+r_{Hy2}df_{z}\\ S_{yy}=F_{z}[Py\gamma_{1}+p_{y2}df_{z}+\big{(}p_{y3}+p_{vy4}df_{z}\big{)} \gamma\big{]}\\ B_{yx}=r_{By1}\cos\big{\{}arctan\big{[}r_{yB_{y2}}(\alpha-r_{By3})\big{]}; \quad C_{yx}=r_{Cy1};\quad E_{yx}=r_{Ey1}+r_{Ey2}df_{z}\\ D_{vyx}=\big{(}p_{Dy1}+p_{Dy2}df_{z}\big{)}F_{z}\big{(}r_{vy1}+r_{y2}df_{z}+r_ {y3}\gamma\big{)}\cos\big{[}\arctan\big{(}r_{vy4}\alpha\big{)}\big{]}\\ S_{yyx}=D_{yyx}\sin\big{[}r_{yy5}\text{arctan}\big{(}r_{vy4}\alpha\big{)} \big{]}\\ G_{yx}=\frac{\cos[C_{y}arctan[B_{y}\alpha_{x}-E_{y}\big{(}B_{yx}\alpha_{x}- arctan(B_{y}\alpha_{x})\big{)}]]}{\alpha[C_{y}arctan[B_{y}\alpha_{y}-E_{yx}\big{(}B_{yx} \alpha_{Hyx}-arctan(B_{yx}\alpha_{Hyx})\big{)}]]}\end{array} \right.\]  
Taking the tire (mc150/55R17) as the research object, the parameters are shown in Table 1.  
These parameters are inserted into Equations (1) and (2) by MATLAB, and, given the slip ratio, sideslip angle, roll angle, vertical force and road friction coefficient, the longitudinal and lateral force can be obtained. The influence of longitudinal force on the lateral force can be demonstrated by friction ellipse, as shown in Figure 1.  
The friction ellipse is composed of longitudinal force and lateral force; in the case of a certain sideslip angle, with the increase in longitudinal force, due to the change in the lateral elasticity of the tire, the limit lateral force provided by the ground decreases.