Vehicle Dynamics Simulation Using Elliptical Combined-Slip Tire Model

Abstract

This paper constitutes the second part of the publication series on introducing a new simplified combined-slip tire model for vehicle dynamics. The effectiveness of the proposed model (Elliptic Model) is validated using a comparative dynamic simulation. A five degree of freedom vehicle model is presented and used for simulation. Tire forces are calculated based on both the Elliptic Model and Magic Formula, and simulation results are compared. It has been shown that the simplified tire model shows promising performance and may be used for estimating the tire force capacity, specially up to the saturation point of tire slips.

Appendix—Magic Formula Reference Parameters

The following parameters are taken from Appendix 3 of [1] and used as reference values for simulations and comparisons:

$$ C_{x} = p_{Cx1} = 1.579 $$

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$$ D_{x} = \left( {p_{Dx1} + p_{Dx2} df_{z} } \right)F_{z} = \left( {1.0422 - 0.08285df_{z} } \right)F_{z} $$

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$$ \begin{aligned} E_{x} & = \left( {p_{Ex1} + p_{Ex2} df_{z} + p_{Ex3} df_{z}^{2} } \right)\left[ {1 - p_{Ex4} {\text{sgn}}\left(\upkappa \right)} \right] \\ & = \left( {0.111113 + 0.3143df_{z} - 0df_{z}^{2} } \right)\left[ {1 - 0.001719{\text{sgn}}\left(\upkappa \right)} \right] \\ \end{aligned} $$

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$$ \begin{aligned} B_{x} & = F_{z} \left( {p_{Kx1} + p_{Kx2} df_{z} } \right).\exp \left( {p_{Kx3} df_{z} } \right)/\left( {C_{x} D_{x} } \right) \\ & = F_{z} \left( {21.687 + 13.728df_{z} } \right).\exp \left( { - 0.4089df_{z} } \right)/\left( {C_{x} D_{x} } \right) \\ \end{aligned} $$

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$$ C_{y} = p_{Cy1} = 1.338 $$

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$$ D_{y} = \left( {p_{Dy1} + p_{Dy2} df_{z} } \right)F_{z} = \left( {0.8785 - 0.06452df_{z} } \right)F_{z} $$

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$$ E_{y} = \left( {p_{Ey1} + p_{Ey2} df_{z} } \right)\left[ {1 - p_{Ey3} {\text{sgn}}\left(\upalpha \right)} \right] = \left( { - 0.8057 - 0.6046df_{z} } \right)\left[ {1 - 0.09854\,{\text{sgn}}\left(\upalpha \right)} \right] $$

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$$ \begin{aligned} B_{y} & = p_{Ky1} F_{z0}^{{\prime }} \sin \left\{ {2\arctan \left[ {\frac{{F_{z} }}{{p_{Ky2} F_{z0}^{{\prime }} }}} \right]} \right\}/\left( {C_{y} D_{y} } \right) \\ & = \left( { - 15.324} \right)\left( {4000} \right)\sin \left\{ {2\arctan \left[ {\frac{{F_{z} }}{{\left( {1.715} \right)\left( {4000} \right)}}} \right]} \right\}/\left( {C_{y} D_{y} } \right) \\ \end{aligned} $$

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$$ {\text{B}}_{{{\text{x}}\upalpha}} = {\text{r}}_{{{\text{Bx}}1}} \cos \left[ {\arctan \left( {{\text{r}}_{{{\text{Bx}}2}}\upkappa} \right)} \right] = 13.046\cos \left[ {\arctan \left( {9.718\upkappa} \right)} \right] $$

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$$ C_{{x\upalpha}} = r_{Cx1} = 0.9995 $$

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$$ B_{{y\upkappa}} = r_{By1} \cos \left[ {\arctan \left( {r_{By2}\upalpha} \right)} \right] = 10.622\cos \left[ {\arctan \left( {7.82\upalpha} \right)} \right] $$

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$$ C_{{y\upkappa}} = r_{Cy1} = 1.0587 $$

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where

$$ df_{z} = \frac{{F_{z} - F_{z0}^{{\prime }} }}{{F_{z0}^{{\prime }} }} = \frac{{F_{z} - 4000}}{4000} $$

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