CN103303157B - Torque distribution method of four-wheel drive electric vehicle - Google Patents

Abstract

The invention provides a torque distribution method of a four-wheel drive electric vehicle, which comprises the following steps: considering parameters such as front wheel longitudinal force and transverse force, back wheel tyre longitudinal force, tyre friction coefficient, tyre load, tyre radius and maximal driving torque of a wheel hub motor, optimizing weight coefficient of the parameters, meanwhile considering the driving force limit condition of a drive motor, adopting a linear analytic method, providing a stable distribution and optimization method, guaranteeing the stability and controllability of a vehicle body, and utilizing ground friction force to the maximal limit.

Description

Torque distribution method for four-wheel drive electric automobile

Technical Field

The invention belongs to the technical field of electric vehicle motor control, and particularly relates to a four-wheel drive electric vehicle torque distribution method, namely a problem of how to distribute four-wheel torque when a four-wheel drive electric vehicle runs.

Background

In order to deal with the increasingly prominent problems of energy shortage, environmental pollution and the like, electric vehicles have been the hot spot of research at present with the advantages of low emission, low noise and the like. Four-wheel drive electric vehicles have been increasingly used in the research field of electric vehicles because of their advantages in control and energy saving. The torque distribution strategy as one of the electric control technologies of the four-wheel drive electric vehicle is emphasized by a plurality of scientific research institutions and enterprises.

At present, a four-wheel drive electric vehicle torque distribution method is mainly based on a rule distribution method and a target optimization distribution method, and most of torque distribution algorithms are based on economy and stability as optimization targets, and four-wheel torque is distributed by considering the influence of longitudinal force or lateral force. In the conventional automobile and most of driving torque distribution strategies based on the rule method, only tires with large influence factors are controlled to perform braking force control so as to achieve corresponding control targets, and the control does not fully consider the yaw moment which can be provided by each tire and the adhesion coefficient state of each tire, so that a certain wheel is easily locked completely, and the stability of the automobile is influenced.

Disclosure of Invention

The invention aims to overcome the defects of the prior art and provide a torque distribution method of a four-wheel drive electric automobile, so as to ensure the stability and controllability of an automobile body and utilize the ground friction force to the maximum extent.

In order to achieve the above object, the present invention provides a torque distribution method for a four-wheel drive electric vehicle, comprising the steps of:

(1) determining an objective function:

$J=\frac{c_{x1}{F}_{x1}^2+c_{y1}{F}_{y1}^2}{{\left(u_1{F}_{z1}\right)}^2}+\frac{c_{x2}{F}_{x2}^2+c_{y2}{F}_{y2}^2}{{(u_2+F_{z2})}^2}+\frac{c_{x3}{F}_{x3}^2}{\min ({\left(u_3{F}_{z3}\right)}^2,{(T_3^{\max }/r_3)}^2)}+\frac{c_{x4}{F}_{x4}^2}{\min ({\left(u_4{F}_{z4}\right)}^2,{(T_4^{\max }/r_4)}^2)}---\left(1\right)$

wherein, c_xiRepresents the longitudinal force weight coefficient of the i-th tire, c_yiRepresents the lateral force weight coefficient, F, of the ith tire_xiIs the tire longitudinal force of the ith tire, F_yiIs the tire lateral force of the ith tire, u_iIs the adhesion coefficient of the i-th tire, F_ziFor the tire load of the ith tire,maximum driving torque of hub motor of ith tire_iTaking the minimum value of the radius x, the radius y, the radius z and the minimum value min (×) of the radius x, the radius y, the radius z and the minimum value min (×) of the radius x, the radius z of the ith tire represent a left front wheel, a right front wheel, a left rear wheel and a right rear wheel respectively;

(2) selecting the lateral force F of the left front wheel_y1And two longitudinal forces F of the left and right front wheels_x1、F_x2Unknown parameters, then:

$F_{x3}=-F_{x1}+\frac{l_f(u_1{F}_{z1}+u_2{F}_{z2})}{B*u_1{F}_{z1}}*F_{y1}+\frac{F_{x\_\mathrm{des}}}{2}-\frac{M_z}{B}---\left(2\right)$

$F_{x4}=-F_{x2}-\frac{l_f(u_1{F}_{z1}+u_2{F}_{z2})}{B*u_1{F}_{z1}}*F_{y1}+\frac{F_{x\_\mathrm{des}}}{2}+\frac{M_z}{B}---\left(3\right)$

$F_{y2}=\frac{u_2{F}_{z2}}{u_1{F}_{z1}}*F_{y1}---\left(4\right)$

wherein, B represents the distance between two wheels of the front wheel;

F_{x_des}=F_x1+F_x2+F_x3+F_x4

M_z=l_f(F_y1+F_y2)+B(|F_x1-F_x2|+|F_x3-F_x4|)/2

F_{x_des}indicating the desired net longitudinal force, M_zA yaw moment representing a vehicle stability requirement, determined by a yaw moment control module;

(3) substituting the formulas (2), (3) and (4) into the objective function (1), solving the minimum value of the existing objective function, and respectively applying the minimum value to the lateral force F of the left wheel of the front wheel_y1And two longitudinal forces F of the left and right front wheels_x1、F_x2Calculating the partial derivative and making it equal to 0, we can obtain:

<math> <mrow> <mfrac> <mrow> <mo>&PartialD;</mo> <mi>J</mi> </mrow> <msub> <mrow> <mo>&PartialD;</mo> <mi>F</mi> </mrow> <mrow> <mi>x</mi> <mn>1</mn> </mrow> </msub> </mfrac> <mo>=</mo> <mn>2</mn> <mo>*</mo> <mrow> <mo>(</mo> <mfrac> <msub> <mi>c</mi> <mrow> <mi>x</mi> <mn>1</mn> </mrow> </msub> <msup> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>1</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mfrac> <mo>+</mo> <mfrac> <msub> <mi>c</mi> <mrow> <mi>x</mi> <mn>3</mn> </mrow> </msub> <mrow> <mi>min</mi> <mrow> <mo>(</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>3</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>3</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>,</mo> <msup> <mrow> <mo>(</mo> <msubsup> <mi>T</mi> <mn>3</mn> <mi>max</mi> </msubsup> <mo>/</mo> <msub> <mi>r</mi> <mn>3</mn> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>)</mo> </mrow> <mo>*</mo> <msub> <mi>F</mi> <mrow> <mi>x</mi> <mn>1</mn> </mrow> </msub> <mo>-</mo> </mrow> </math>

$2*\frac{l_f(u_1{F}_{z1}+u_2{F}_{z2})}{B*u_1{F}_{z1}}*\frac{c_{x3}}{\min ({\left(u_3{F}_{z3}\right)}^2,{(T_3^{\max }/r_3)}^2)}*F_{y1}----\left(5\right)$

$2*(\frac{F_{x\_\mathrm{des}}}{2}-\frac{M_z}{B})*\frac{c_{x3}}{\min ({\left(u_3{F}_{z3}\right)}^2,{(T_3^{\max }/r_3)}^2)}=0;$

<math> <mrow> <mfrac> <mrow> <mo>&PartialD;</mo> <mi>J</mi> </mrow> <msub> <mrow> <mo>&PartialD;</mo> <mi>F</mi> </mrow> <mrow> <mi>x</mi> <mn>2</mn> </mrow> </msub> </mfrac> <mo>=</mo> <mn>2</mn> <mo>*</mo> <mrow> <mo>(</mo> <mfrac> <msub> <mi>c</mi> <mrow> <mi>x</mi> <mn>2</mn> </mrow> </msub> <msup> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>2</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>2</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mfrac> <mo>+</mo> <mfrac> <msub> <mi>c</mi> <mrow> <mi>x</mi> <mn>4</mn> </mrow> </msub> <mrow> <mi>min</mi> <mrow> <mo>(</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>4</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>4</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>,</mo> <msup> <mrow> <mo>(</mo> <msubsup> <mi>T</mi> <mn>4</mn> <mi>max</mi> </msubsup> <mo>/</mo> <msub> <mi>r</mi> <mn>4</mn> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>)</mo> </mrow> <mo>*</mo> <msub> <mi>F</mi> <mrow> <mi>x</mi> <mn>2</mn> </mrow> </msub> <mo>+</mo> </mrow> </math>

$2*\frac{l_f(u_1{F}_{z1}+u_2{F}_{z2})}{B*u_1{F}_{z1}}*\frac{c_{x4}}{\min ({\left(u_4{F}_{z4}\right)}^2,{(T_4^{\max }/r_4)}^2)}*F_{y1}----\left(6\right)$

$2*(\frac{F_{x\_\mathrm{des}}}{2}+\frac{M_z}{B})*\frac{c_{x4}}{\min ({\left(u_4{F}_{z4}\right)}^2,{(T_4^{\max }/r_4)}^2)}=0;$

<math> <mrow> <mfrac> <mrow> <mo>&PartialD;</mo> <mi>J</mi> </mrow> <msub> <mrow> <mo>&PartialD;</mo> <mi>F</mi> </mrow> <mrow> <mi>y</mi> <mn>1</mn> </mrow> </msub> </mfrac> <mo>=</mo> <mo>-</mo> <mn>2</mn> <mo>*</mo> <mfrac> <mrow> <msub> <mi>l</mi> <mi>f</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>1</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mi>u</mi> <mn>2</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>2</mn> </mrow> </msub> <mo>)</mo> </mrow> </mrow> <mrow> <mi>B</mi> <mo>*</mo> <msub> <mi>u</mi> <mn>1</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>1</mn> </mrow> </msub> </mrow> </mfrac> <mo>*</mo> <mfrac> <msub> <mi>c</mi> <mrow> <mi>x</mi> <mn>3</mn> </mrow> </msub> <mrow> <mi>min</mi> <mrow> <mo>(</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>3</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>3</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>,</mo> <msup> <mrow> <mo>(</mo> <msubsup> <mi>T</mi> <mn>3</mn> <mi>max</mi> </msubsup> <mo>/</mo> <msub> <mi>r</mi> <mn>3</mn> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>*</mo> <msub> <mi>F</mi> <mrow> <mi>x</mi> <mn>1</mn> </mrow> </msub> <mo>+</mo> </mrow> </math>

$2*\frac{l_f(u_1{F}_{z1}+u_2{F}_{z2})}{B*u_1{F}_{z1}}*\frac{c_{x4}}{\min ({\left(u_4{F}_{z4}\right)}^2,{(T_4^{\max }/r_4)}^2)}*F_{x2}+$

$2*\left(\begin{array}{c}\frac{c_{y1}}{{\left(u_1{F}_{z1}\right)}^2}+{\left(\frac{u_2{F}_{z2}}{u_1{F}_{z1}}\right)}^2*\frac{c_{y2}}{{\left(u_2{F}_{z2}\right)}^2}+{\left(\frac{l_f(u_1{F}_{z1}+u_2{F}_{z2})}{B*u_1{F}_{z1}}\right)}^2*\\ (\frac{c_{x3}}{\min ({\left(u_3{F}_{z3}\right)}^2,{(T_3^{\max }/r_3)}^2)}+\frac{c_{x4}}{\min ({\left(u_4{F}_{z4}\right)}^2,{(T_4^{\max }/r_4)}^2)})\end{array}\right)*F_{y1}+---\left(7\right)$

$2*\frac{l_f(u_1{F}_{z1}+u_2{F}_{z2})}{B*u_1{F}_{z1}}*\left(\begin{array}{c}(\frac{F_{x\_\mathrm{des}}}{2}-\frac{M_z}{B})*\frac{c_{x3}}{\min ({\left(u_3{F}_{z3}\right)}^2,{(T_3^{\max }/r_3)}^2)}-\\ (\frac{F_{x\_\mathrm{des}}}{2}+\frac{M_z}{B})*\frac{c_{x4}}{\min ({\left(u_4{F}_{z4}\right)}^2,{(T_4^{\max }/r_4)}^2)}\end{array}\right)=0;$

solving the equations (5), (6) and (7) to obtain the lateral force F of the left front wheel required to be distributed_y1And two longitudinal forces F of the left and right front wheels_x1、F_x2Then, the lateral force F of the left front right wheel to be distributed is calculated by the formulas (2), (3) and (4)_y2And two longitudinal forces F of the left and right rear wheels_x3、F_x4；

(4) According to the lateral force F of the left front wheel_y1Side force F of the right front wheel_y2And two longitudinal forces F of the left and right front wheels_x1、F_x2Calculating the driving torque and the steering wheel angle correction value of the left and right front wheels, and applying two longitudinal forces F to the left and right rear wheels_x3、F_x4And calculating the driving torques of the left and right rear wheels, and then controlling the four-wheel drive electric automobile by using the four driving torques.

(5) The required moment of the four wheels can be calculated by the lateral force and the longitudinal force of the wheels as follows:

$M_i=\sqrt{M_{\mathrm{xi}}^2+M_{\mathrm{yi}}^2},$ wherein M is_xi=l_f·F_xiM_yi=l_f·F_yi i=1,2，

M_i=M_xi=l_f·F_xi i=3,4，

Wherein M is_iRepresenting the moment of four wheels.

As a further improvement of the invention, the weight coefficient c of the longitudinal force of the ith tire in the step (1)_xiLateral force weight coefficient c_yiComprises the following steps:

<math> <mrow> <msub> <mi>C</mi> <mi>xi</mi> </msub> <mo>=</mo> <mfenced open='{' close=''> <mtable> <mtr> <mtd> <mn>0.5</mn> <mo>,</mo> <mi>m</mi> <mo>&GreaterEqual;</mo> <mn>0.6</mn> </mtd> </mtr> <mtr> <mtd> <mn>0.3</mn> <mo>+</mo> <mn>0.1</mn> <mo>*</mo> <mrow> <mo>(</mo> <mfrac> <mi>m</mi> <mn>0.6</mn> </mfrac> <mo>+</mo> <mfrac> <mrow> <mo>|</mo> <mi>n</mi> <mo>|</mo> </mrow> <mn>0.8</mn> </mfrac> <mo>)</mo> </mrow> <mo>,</mo> <mn>0</mn> <mo>&le;</mo> <mi>m</mi> <mo>&lt;</mo> <mn>0.6</mn> </mtd> </mtr> <mtr> <mtd> <mn>0.3</mn> <mo>+</mo> <mn>0.2</mn> <mo>*</mo> <mrow> <mo>(</mo> <mfrac> <mrow> <mo>|</mo> <mi>m</mi> <mo>|</mo> </mrow> <mn>0.7</mn> </mfrac> <mo>-</mo> <mfrac> <mrow> <mo>|</mo> <mi>n</mi> <mo>|</mo> </mrow> <mn>0.8</mn> </mfrac> <mo>)</mo> </mrow> <mo>,</mo> <mo>-</mo> <mn>0.7</mn> <mo>&lt;</mo> <mi>m</mi> <mo>&lt;</mo> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <mn>0.5</mn> <mo>,</mo> <mi>m</mi> <mo>&le;</mo> <mo>-</mo> <mn>0.7</mn> </mtd> </mtr> </mtable> </mfenced> </mrow> </math>

C_yi=1-C_xi；

the longitudinal gravity center transfer rate m and the lateral gravity center transfer rate n of the tire are as follows:

$m=\frac{l_f(u_1*F_{z1}+u_2*F_{z2})-l_r(u_3*F_{z3}+u_4*F_{z4})}{u_1*F_{z1}+u_2*F_{z2}+u_3*F_{z3}+u_4*F_{z4}}$

$n=\frac{\frac{B}{2}(-u_1*F_{z1}+u_2*F_{z2}-u_3*F_{z3}+u_4*F_{z4})}{u_1*F_{z1}+u_2*F_{z2}+u_3*F_{z3}+u_4*F_{z4}}$

wherein l_fRepresenting the distance between the front wheel and the centre of mass of the vehicle, B representing the distance between the front wheels of the vehicle and the two wheels, l_rRepresenting the distance between the rear wheel and the center of mass of the vehicle;

the larger the positive direction of m is, the larger the deceleration of the automobile is, the braking force is provided by the automobile, the load of the automobile is transferred forwards, the larger the negative direction of m is, the larger the acceleration of the automobile is, the driving force is provided by the automobile, and the load of the automobile is transferred backwards; the larger the positive direction of n is, the larger the left turning degree of the automobile is, the load is transferred to the right, and the larger the negative direction of n is, the larger the right turning degree of the automobile is, the load is transferred to the left; the longitudinal center of gravity transfer rate m and the lateral center of gravity transfer rate n of the vehicle are defined to be between-0.8 and 0.8, and if the calculation is out of the range, a boundary value is taken.

The purpose of the invention is realized as follows:

the invention relates to a four-wheel drive electric vehicle torque distribution method, which considers parameters such as front wheel longitudinal force and transverse force, rear wheel tire longitudinal force, tire friction coefficient, tire load, tire radius, maximum driving torque of a hub motor and the like, optimizes the weight coefficient of the parameters, considers the driving force limiting conditions of the driving motor, adopts a linear analysis method, provides a torque distribution optimization method with stability, ensures the stability and controllability of a vehicle body, and utilizes the ground friction force to the maximum extent.

Drawings

FIG. 1 illustrates a torque distribution method for a four-wheel drive electric vehicle according to the present invention;

FIG. 2 is a graph of steering wheel angle input;

FIG. 3 is a depth profile of an accelerator pedal;

FIG. 4 is a graph of weight coefficients;

FIG. 5 is a front wheel torque graph;

FIG. 6 is a rear wheel torque graph;

FIG. 7 is a graph of body roll angle;

FIG. 8 is a vehicle longitudinal speed graph;

FIG. 9 is a vehicle lateral speed graph;

FIG. 10 is a graph of torque distribution versus average distribution for a four-wheel drive electric vehicle according to the present invention.

Detailed Description

The following description of the embodiments of the present invention is provided in order to better understand the present invention for those skilled in the art with reference to the accompanying drawings. It is to be expressly noted that in the following description, a detailed description of known functions and designs will be omitted when it may obscure the subject matter of the present invention.

The longitudinal and lateral forces exerted by the ground on the tyre are both generated by the ground friction, so depending on the condition of the ground friction coefficient, the maximum longitudinal force that the ground can provide must satisfy the following formula:

<math> <mrow> <msub> <mi>F</mi> <mi>xi</mi> </msub> <mo>&le;</mo> <msqrt> <msup> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>F</mi> <mi>zi</mi> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>-</mo> <msubsup> <mi>F</mi> <mi>yi</mi> <mn>2</mn> </msubsup> </msqrt> <mo>=</mo> <msubsup> <mi>F</mi> <mi>i</mi> <mi>limit</mi> </msubsup> <mo>,</mo> <mi>i</mi> <mo>=</mo> <mn>1,2,3,4</mn> <mo>;</mo> </mrow> </math>

wherein, F_xiIs the longitudinal force of the ith tire, F_yiIs the lateral force of the ith tire, u_iIs the adhesion coefficient of the i-th tire, F_ziFor the tire load of the i tires,represents the maximum longitudinal force that the ground can provide to the ith tire under the condition of the actual adhesion coefficient; x is the longitudinal direction, y is the lateral direction, z is the direction perpendicular to the xy plane, i =1, 2, 3, 4 respectively represent the left front wheel, the right front wheel, the left rear wheel, the right rear wheel.

Therefore, when the wheel driving force distribution is performed in order to generate a sufficient yaw moment, it is considered that the driving force generated by each wheel cannot exceed the maximum longitudinal force that can be provided by the ground, and the vehicle can be in a stable state. If the torque distributed to a certain wheel exceeds the limit of the ground adhesion limit, the wheel can generate slippage, and when the wheel slippage reaches a certain threshold value, the adhesion coefficient provided by the ground can be reduced in a large range, so that the instability trend of the wheel is further aggravated.

The four-wheel drive electric vehicle is equipped with 4 in-wheel motors capable of independently distributing different driving force torques and braking force pressures. The maximum driving torque of the in-wheel motor is determined based on the characteristics of the motor itself and by the angular acceleration of the tire, so the longitudinal force provided by the tire must satisfy the following formula:

<math> <mrow> <msub> <mi>F</mi> <mi>xi</mi> </msub> <mo>&le;</mo> <msubsup> <mi>T</mi> <mi>i</mi> <mi>max</mi> </msubsup> <mo>/</mo> <mi>r</mi> <mo>,</mo> <mi>i</mi> <mo>=</mo> <mn>1,2,3,4</mn> <mo>;</mo> </mrow> </math>

wherein,the maximum driving torque of the hub motor of the ith tire is shown, and r is the radius of the wheel.

For a four-wheel drive electric vehicle with front wheels steered, the lateral force is mainly provided by the steered wheels, so that only the lateral force of the front two wheels is considered, and the longitudinal force of 4 tires is considered, and finally the longitudinal force of the front wheels, the lateral force and the longitudinal force of the rear wheels are taken as distribution objects according to 6 parameters. The objective function giving the optimization method of this paper is shown in equation (3) taking into account the road adhesion coefficient and the torque limit of the drive motor.

$J=\frac{c_{x1}{F}_{x1}^2+c_{y1}{F}_{y1}^2}{{\left(u_1{F}_{z1}\right)}^2}+\frac{c_{x2}{F}_{x2}^2+c_{y2}{F}_{y2}^2}{{(u_2+F_{z2})}^2}+\frac{c_{x3}{F}_{x3}^2}{\min ({\left(u_3{F}_{z3}\right)}^2,{(T_3^{\max }/r_3)}^2)}+\frac{c_{x4}{F}_{x4}^2}{\min ({\left(u_4{F}_{z4}\right)}^2,{(T_4^{\max }/r_4)}^2)}---\left(1\right)$

Wherein, c_xiRepresents the longitudinal force weight coefficient of the i-th tire, c_yiThe lateral force weight coefficient of the i-th tire is expressed, and min (,) represents the minimum value of the two.

In order to ensure that the four-wheel drive electric automobile meets the vehicle speed expected by the driver, the longitudinal force provided by each tire must meet the longitudinal net force output by the vehicle speed control module:

F_{x_des}=F_x1+F_x2+F_x3+F_x4； （8）

wherein, F_{x_des}Indicating a desired net longitudinal force.

The longitudinal net force required by the tires is the sum of the longitudinal forces of all the tires, so that the driving capability of the automobile is ensured.

Meanwhile, in order to ensure that the automobile can run safely and stably, the longitudinal force and the lateral force provided by each wheel must meet the yaw moment value required by the stability of the automobile:

M_z=l_f(F_y1+F_y2)+B(|F_x1-F_x2|+|F_x3-F_x4|)/2 （9）

wherein l_fRepresenting the distance between the front wheel to the vehicle's center of mass; b represents the distance between two wheels of the front wheel; m_zA yaw moment, which is required for vehicle stability, is indicated, which is determined by the yaw moment control module.

For an electric vehicle that steers the front wheels, the steering wheel angles of the two tires of the front wheels are therefore identical.

$\frac{F_{y1}}{u_1{F}_{z1}}=\frac{F_{y2}}{u_2{F}_{z2}}---\left(10\right)$

The above equation represents the relationship between the additional tire forces for the left and right front wheels. The additional tire force of the left front wheel and the right reel is in proportion to the friction circle size of the corresponding ground, namely the larger the road friction circle of the corresponding tire is, the larger the lateral force provided by the corresponding tire is.

The formulas (8), (9) and (10) are for 6 independent variables F in the objective function_x1、F_x2、F_x3、F_x4、F_y1、F_y2The 3 constraint conditions not only ensure that the vehicle meets the vehicle running state expected by the driver, but also consider the stability and controllability of the vehicle, and ensure that the final optimization result can achieve optimal control.

Selecting the lateral force F of the left front wheel_y1And two longitudinal forces F of the left and right front wheels_x1、F_x2Are unknown parameters. The specific equation is as follows:

$F_{x3}=-F_{x1}+\frac{l_f(u_1{F}_{z1}+u_2{F}_{z2})}{B*u_1{F}_{z1}}*F_{y1}+\frac{F_{x\_\mathrm{des}}}{2}-\frac{M_z}{B}---\left(2\right)$

$F_{x4}=-F_{x2}-\frac{l_f(u_1{F}_{z1}+u_2{F}_{z2})}{B*u_1{F}_{z1}}*F_{y1}+\frac{F_{x\_\mathrm{des}}}{2}+\frac{M_z}{B}---\left(3\right)$

$F_{y2}=\frac{u_2{F}_{z2}}{u_1{F}_{z1}}*F_{y1}---\left(4\right)$

wherein B represents the distance between the front wheels and the two wheels.

Substituting the formulas (2), (3) and (4) into the objective function (1) as necessary conditions, solving the minimum value of the existing objective function, and respectively applying the minimum value to the lateral force F of the left wheel of the front wheel_y1And two longitudinal forces F of the left and right front wheels_x1、F_x2Calculating the partial derivative and making it equal to 0, we can obtain:

<math> <mrow> <mfrac> <mrow> <mo>&PartialD;</mo> <mi>J</mi> </mrow> <msub> <mrow> <mo>&PartialD;</mo> <mi>F</mi> </mrow> <mrow> <mi>x</mi> <mn>1</mn> </mrow> </msub> </mfrac> <mo>=</mo> <mn>2</mn> <mo>*</mo> <mrow> <mo>(</mo> <mfrac> <msub> <mi>c</mi> <mrow> <mi>x</mi> <mn>1</mn> </mrow> </msub> <msup> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>1</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mfrac> <mo>+</mo> <mfrac> <msub> <mi>c</mi> <mrow> <mi>x</mi> <mn>3</mn> </mrow> </msub> <mrow> <mi>min</mi> <mrow> <mo>(</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>3</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>3</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>,</mo> <msup> <mrow> <mo>(</mo> <msubsup> <mi>T</mi> <mn>3</mn> <mi>max</mi> </msubsup> <mo>/</mo> <msub> <mi>r</mi> <mn>3</mn> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>)</mo> </mrow> <mo>*</mo> <msub> <mi>F</mi> <mrow> <mi>x</mi> <mn>1</mn> </mrow> </msub> <mo>-</mo> </mrow> </math>

$2*\frac{l_f(u_1{F}_{z1}+u_2{F}_{z2})}{B*u_1{F}_{z1}}*\frac{c_{x3}}{\min ({\left(u_3{F}_{z3}\right)}^2,{(T_3^{\max }/r_3)}^2)}*F_{y1}----\left(5\right)$

$2*(\frac{F_{x\_\mathrm{des}}}{2}-\frac{M_z}{B})*\frac{c_{x3}}{\min ({\left(u_3{F}_{z3}\right)}^2,{(T_3^{\max }/r_3)}^2)}=0;$

<math> <mrow> <mfrac> <mrow> <mo>&PartialD;</mo> <mi>J</mi> </mrow> <msub> <mrow> <mo>&PartialD;</mo> <mi>F</mi> </mrow> <mrow> <mi>x</mi> <mn>2</mn> </mrow> </msub> </mfrac> <mo>=</mo> <mn>2</mn> <mo>*</mo> <mrow> <mo>(</mo> <mfrac> <msub> <mi>c</mi> <mrow> <mi>x</mi> <mn>2</mn> </mrow> </msub> <msup> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>2</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>2</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mfrac> <mo>+</mo> <mfrac> <msub> <mi>c</mi> <mrow> <mi>x</mi> <mn>4</mn> </mrow> </msub> <mrow> <mi>min</mi> <mrow> <mo>(</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>4</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>4</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>,</mo> <msup> <mrow> <mo>(</mo> <msubsup> <mi>T</mi> <mn>4</mn> <mi>max</mi> </msubsup> <mo>/</mo> <msub> <mi>r</mi> <mn>4</mn> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>)</mo> </mrow> <mo>*</mo> <msub> <mi>F</mi> <mrow> <mi>x</mi> <mn>2</mn> </mrow> </msub> <mo>+</mo> </mrow> </math>

$2*\frac{l_f(u_1{F}_{z1}+u_2{F}_{z2})}{B*u_1{F}_{z1}}*\frac{c_{x4}}{\min ({\left(u_4{F}_{z4}\right)}^2,{(T_4^{\max }/r_4)}^2)}*F_{y1}----\left(6\right)$

$2*(\frac{F_{x\_\mathrm{des}}}{2}+\frac{M_z}{B})*\frac{c_{x4}}{\min ({\left(u_4{F}_{z4}\right)}^2,{(T_4^{\max }/r_4)}^2)}=0;$

<math> <mrow> <mfrac> <mrow> <mo>&PartialD;</mo> <mi>J</mi> </mrow> <mrow> <mo>&PartialD;</mo> <msub> <mi>F</mi> <mrow> <mi>y</mi> <mn>1</mn> </mrow> </msub> </mrow> </mfrac> <mo>=</mo> <mo>-</mo> <mn>2</mn> <mo>*</mo> <mfrac> <mrow> <msub> <mi>l</mi> <mi>f</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>1</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mi>u</mi> <mn>2</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>2</mn> </mrow> </msub> <mo>)</mo> </mrow> </mrow> <mrow> <mi>B</mi> <mo>*</mo> <msub> <mi>u</mi> <mn>1</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>1</mn> </mrow> </msub> </mrow> </mfrac> <mo>*</mo> <mfrac> <msub> <mi>c</mi> <mrow> <mi>x</mi> <mn>3</mn> </mrow> </msub> <mrow> <mi>min</mi> <mrow> <mo>(</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>3</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>3</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>,</mo> <msup> <mrow> <mo>(</mo> <msubsup> <mi>T</mi> <mn>3</mn> <mi>max</mi> </msubsup> <mo>/</mo> <msub> <mi>r</mi> <mn>3</mn> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>*</mo> <msub> <mi>F</mi> <mrow> <mi>x</mi> <mn>1</mn> </mrow> </msub> <mo>+</mo> </mrow> </math>

$2*\frac{l_f(u_1{F}_{z1}+u_2{F}_{z2})}{B*u_1{F}_{z1}}*\frac{c_{x4}}{\min ({\left(u_4{F}_{z4}\right)}^2,{(T_4^{\max }/r_4)}^2)}*F_{x2}+$

$2*\left(\begin{array}{c}\frac{c_{y1}}{{\left(u_1{F}_{z1}\right)}^2}+{\left(\frac{u_2{F}_{z2}}{u_1{F}_{z1}}\right)}^2*\frac{c_{y2}}{{\left(u_2{F}_{z2}\right)}^2}+{\left(\frac{l_f(u_1{F}_{z1}+u_2{F}_{z2})}{B*u_1{F}_{z1}}\right)}^2*\\ (\frac{c_{x3}}{\min ({\left(u_3{F}_{z3}\right)}^2,{(T_3^{\max }/r_3)}^2)}+\frac{c_{x4}}{\min ({\left(u_4{F}_{z4}\right)}^2,{(T_4^{\max }/r_4)}^2)})\end{array}\right)*F_{y1}+---\left(7\right)$

$2*\frac{l_f(u_1{F}_{z1}+u_2{F}_{z2})}{B*u_1{F}_{z1}}*\left(\begin{array}{c}(\frac{F_{x\_\mathrm{des}}}{2}-\frac{M_z}{B})*\frac{c_{x3}}{\min ({\left(u_3{F}_{z3}\right)}^2,{(T_3^{\max }/r_3)}^2)}-\\ (\frac{F_{x\_\mathrm{des}}}{2}+\frac{M_z}{B})*\frac{c_{x4}}{\min ({\left(u_4{F}_{z4}\right)}^2,{(T_4^{\max }/r_4)}^2)}\end{array}\right)=0;$

solving the equations (5), (6) and (7) to obtain the lateral force F of the left front wheel required to be distributed_y1And two longitudinal forces F of the left and right front wheels_x1、F_x2Then, the lateral force F of the left front right wheel to be distributed is calculated by the formulas (2), (3) and (4)_y2And two longitudinal forces F of the left and right rear wheels_x3、F_x4；

Different motion states require different weighting factors, so that the determination of the motion state parameters of the automobile is of great importance. The longitudinal gravity center transfer rate of the electric vehicle is selected to represent the degree of acceleration and deceleration of the vehicle, the lateral gravity center transfer rate of the electric vehicle is selected to represent the degree of turning of the vehicle, and then the weight coefficient of the vehicle tire force is calculated by using the longitudinal and transverse gravity center transfer rates. The calculation formulas of the longitudinal gravity center transfer rate m and the lateral gravity center transfer rate n of the tire are as follows:

$m=\frac{l_f(u_1*F_{z1}+u_2*F_{z2})-l_r(u_3*F_{z3}+u_4*F_{z4})}{u_1*F_{z1}+u_2*F_{z2}+u_3*F_{z3}+u_4*F_{z4}}$

$n=\frac{\frac{B}{2}(-u_1*F_{z1}+u_2*F_{z2}-u_3*F_{z3}+u_4*F_{z4})}{u_1*F_{z1}+u_2*F_{z2}+u_3*F_{z3}+u_4*F_{z4}}$

the larger the positive direction of m is, the larger the deceleration of the automobile is, the braking force is provided by the automobile, the load of the automobile is transferred forwards, the larger the negative direction of m is, the larger the acceleration of the automobile is, the driving force is provided by the automobilePower, the vehicle load shifts to the right; the larger the positive direction of n is, the larger the left turning degree of the automobile is, the load is transferred to the right, and the larger the negative direction of n is, the larger the right turning degree of the automobile is, the load is transferred to the left. The longitudinal center of gravity transfer rate m and the lateral center of gravity transfer rate n of the vehicle are defined to be between-0.8 and 0.8, and if the calculation is out of the range, a boundary value is taken. Thereby the weight coefficient C_xiAnd C_yiThe method is obtained by calculating the longitudinal gravity center transfer rate m and the lateral gravity center transfer rate n of the automobile. (i =1, 2, 3, 4 respectively represent the left front wheel, the right front wheel, the left rear wheel, the right rear wheel)

<math> <mrow> <msub> <mi>C</mi> <mi>xi</mi> </msub> <mo>=</mo> <mfenced open='{' close=''> <mtable> <mtr> <mtd> <mn>0</mn> <mo>.</mo> <mn>5</mn> <mo>,</mo> <mi>m</mi> <mo>&GreaterEqual;</mo> <mn>0.6</mn> </mtd> </mtr> <mtr> <mtd> <mn>0.3</mn> <mo>+</mo> <mn>0.1</mn> <mo>*</mo> <mrow> <mo>(</mo> <mfrac> <mi>m</mi> <mn>0.6</mn> </mfrac> <mo>+</mo> <mfrac> <mrow> <mo>|</mo> <mi>n</mi> <mo>|</mo> </mrow> <mrow> <mn>0</mn> <mo>.</mo> <mn>8</mn> </mrow> </mfrac> <mo>)</mo> </mrow> <mo>,</mo> <mn>0</mn> <mo>&le;</mo> <mi>m</mi> <mo>&lt;</mo> <mn>0.6</mn> </mtd> </mtr> <mtr> <mtd> <mn>0.3</mn> <mo>+</mo> <mn>0.2</mn> <mo>*</mo> <mrow> <mo>(</mo> <mfrac> <mrow> <mo>|</mo> <mi>m</mi> <mo>|</mo> </mrow> <mn>0.7</mn> </mfrac> <mo>-</mo> <mfrac> <mrow> <mo>|</mo> <mi>n</mi> <mo>|</mo> </mrow> <mn>0.8</mn> </mfrac> <mo>)</mo> </mrow> <mo>,</mo> <mo>-</mo> <mn>0.7</mn> <mo>&lt;</mo> <mi>m</mi> <mo>&lt;</mo> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <mn>0.5</mn> <mo>,</mo> <mi>m</mi> <mo>&le;</mo> <mo>-</mo> <mn>0.7</mn> </mtd> </mtr> </mtable> </mfenced> <mo>,</mo> </mrow> </math>

C_yi=1-C_xi

The invention has the advantages that:

1. and a torque distribution algorithm based on target optimization adopts the minimum tire load rate as a control target, and performs torque distribution on the premise of ensuring the stability and controllability of a vehicle body.

2. In the control algorithm, the distribution proportion of longitudinal force and lateral force of the electric automobile under different working conditions is ensured by correcting the weight coefficient, and the ground friction is ensured to be utilized to the maximum extent.

Examples of the invention

The verification is carried out on the four-wheel-drive electric vehicle with the total mass of 1296kg, the rotating inertia of 1750kgm2 around the Z axis, the wheel base of 2.57m, the distance from the center of mass to the front shaft of 1.25m, the distance from the center of mass to the rear shaft of 1.32m, the front wheel base of 1.405m, the rear wheel base of 1.399m, the height of the center of mass of 0.45m and the wheel radius of 0.326 m. The speed of the vehicle is increased from 0km/h to 70km/h, 7 obstacles are bypassed within 8-25 s, the snakelike motion working condition is simulated, the road surface friction coefficient is set to be u =0.7, and the road surface friction coefficient is the normal friction coefficient of a dry asphalt road surface.

In the torque distribution method of the four-wheel drive electric automobile shown in fig. 1, firstly, a real-time driving torque (T) of each hub motor of the automobile is acquired by an automobile sensor system_i) And yaw moment value (M)_z) (ii) a The longitudinal force weighting factor (C) is then determined by a weighting factor calculation module_xi) And a lateral force weighting factor (C)_yi) (ii) a Finally, the optimal longitudinal force and the optimal lateral force F of the tire are calculated by a torque distribution method_xi、F_yiAnd then the parameters are converted into driving torque and steering wheel angle correction values of each tire by the parameter correction module, and finally the purpose of controlling the four-wheel drive electric automobile to enable the four-wheel drive electric automobile to stably run is achieved.

As shown in fig. 2 and 3, the driver's expected motion state, obviously, the vehicle keeps a large acceleration when running in a straight line, and reaches a vehicle speed of 70km/h around 7s, and then after 8s, the driver turns the steering wheel according to a snake-shaped working condition, so that the vehicle avoids the front 7 obstacles.

As can be seen from fig. 4, 5, and 6, the control amount of the vehicle torque distribution and the weight coefficient are corrected such that the weight coefficient C is increased during the sudden and abrupt steering wheel operation_xiThe weight coefficient C is reduced_yiAnd in the process of automobile steering, continuously correcting C according to the steering direction and the positive and negative of the required torque_xiA value of (d); in the correction curve of the steering wheel of the automobile, the steering angle is corrected according to different steering directions of the steering wheel, and in the process of distributing the driving torque of the left wheel and the right wheel, the driving torque of the outer wheels is larger than that of the inner wheels, so that extra yaw moment is provided, and the steering stability is ensured.

Fig. 7 shows the vehicle stability index, obviously, when the roll angle of the vehicle is small and the vehicle is in a stable state, fig. 8 and 9 show the moving state of the vehicle, the longitudinal speed of which starts to increase rapidly, the vehicle basically keeps unchanged when turning, and the transverse speed changes in real time according to the turning direction.

Fig. 10 is a comparison curve of the torque distribution method of the four-wheel drive electric vehicle according to the present invention and the average torque distribution method, the average torque distribution method is a control method simulating the torque of the conventional vehicle, when the average torque distribution method avoids the last two obstacles, the vehicle is already unstable, the motion curve completely does not meet the driver's expectation, and the motion trajectory of the four-wheel torque distribution algorithm always meets the driver's expectation motion trajectory and is in a stable state.

Although illustrative embodiments of the present invention have been described above to facilitate the understanding of the present invention by those skilled in the art, it should be understood that the present invention is not limited to the scope of the embodiments, and various changes may be made apparent to those skilled in the art as long as they are within the spirit and scope of the present invention as defined and defined by the appended claims, and all matters of the invention which utilize the inventive concepts are protected.

Claims (2)

1. A four-wheel drive electric vehicle torque distribution method is characterized by comprising the following steps:

(1) and an objective function:

$J=\frac{c_{x1}{F}_{x1}^2+c_{y1}{F}_{y1}^2}{{\left(u_1{F}_{z1}\right)}^2}+\frac{c_{x2}{F}_{x2}^2+c_{y2}{F}_{y2}^2}{{\left(u_2{F}_{z2}\right)}^2}+\frac{c_{x3}{F}_{x3}^2}{\min ({\left(u_3{f}_{z3}\right)}^2,{(T_3^{\max }/r_3)}^2)}+\frac{c_{x4}{F}_{x4}^2}{\min ({\left(u_4{f}_{z4}\right)}^2,{(T_4^{\max }/r_4)}^2)}---\left(1\right)$

wherein, c_xiRepresents the longitudinal force weight coefficient of the i-th tire, c_yiRepresents the lateral force weight coefficient, F, of the ith tire_xiIs the tire longitudinal force of the ith tire, F_yiIs the tire lateral force of the ith tire, u_iIs the adhesion coefficient of the i-th tire, F_ziFor the tire load of the ith tire,maximum driving torque of hub motor of ith tire_iThe radius of the wheel of the ith tire is x, the longitudinal direction is x, the lateral direction is y, z is a direction perpendicular to the xy plane, min (—) represents the minimum value of the x, the i is 1, 2, 3 and 4 respectively represent a left front wheel, a right front wheel, a left rear wheel and a right rear wheel;

(2) selecting the lateral force F of the left front wheel_y1And two longitudinal forces F of the left and right front wheels_x1、F_x2Unknown parameters, then:

$F_{x3}=-F_{z1}+\frac{l_f(u_1{F}_{x1}+u_2{F}_{z2})}{B*u_1{F}_{z1}}*F_{y1}+\frac{F_{x\_\mathrm{drs}}}{2}-\frac{M_z}{B}---\left(2\right)$

$F_{x4}=-F_{x2}-\frac{l_f(u_1{F}_{z1}+u_2{F}_{z2})}{B*u_1{F}_{z1}}*F_{y1}+\frac{F_{x\_\mathrm{dds}}}{2}-\frac{M_z}{B}---\left(3\right)$

$F_{y2}=\frac{u_2{F}_{z2}}{u_1{F}_{z1}}*F_{y1}---\left(4\right)$

wherein l_fThe distance between the front wheel and the mass center of the vehicle is shown, and B represents the distance between the front wheel and the front wheel of the vehicle;

F_{x_des}＝F_x1+F_x2+F_x3+F_x4

M_z＝l_f(F_y1+F_y2)+B(|F_x1-F_x2|+|F_x3-F_x4|)/2

F_{x_des}indicating the desired net longitudinal force, M_zA yaw moment representing a vehicle stability requirement, determined by a yaw moment control module;

(3) substituting the formulas (2), (3) and (4) into the objective function (1), solving the minimum value of the existing objective function, and respectively applying the minimum value to the lateral force F of the left wheel of the front wheel_y1And two longitudinal forces F of the left and right front wheels_x1、F_x2Calculating the partial derivative and making it equal to 0, we can obtain:

<math> <mrow> <mfenced open='' close='' separators=' '> <mtable> <mtr> <mtd> <mfrac> <mrow> <mo>&PartialD;</mo> <mi>J</mi> </mrow> <msub> <mrow> <mo>&PartialD;</mo> <mi>F</mi> </mrow> <mrow> <mi>x</mi> <mn>1</mn> </mrow> </msub> </mfrac> <mo>=</mo> <mn>2</mn> <mo>*</mo> <mrow> <mo>(</mo> <mfrac> <msub> <mi>c</mi> <mrow> <mi>x</mi> <mn>1</mn> </mrow> </msub> <msup> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>1</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mfrac> <mo>+</mo> <mfrac> <msub> <mi>c</mi> <mrow> <mi>x</mi> <mn>3</mn> </mrow> </msub> <mrow> <mi>min</mi> <mrow> <mo>(</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>3</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>3</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>,</mo> <msup> <mrow> <mo>(</mo> <msubsup> <mi>T</mi> <mn>3</mn> <mi>max</mi> </msubsup> <mo>/</mo> <msub> <mi>r</mi> <mn>3</mn> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>)</mo> </mrow> <mo>*</mo> <msub> <mi>F</mi> <mrow> <mi>x</mi> <mn>1</mn> </mrow> </msub> <mo>-</mo> </mtd> </mtr> <mtr> <mtd> <mn>2</mn> <mo>*</mo> <mfrac> <mrow> <msub> <mi>l</mi> <mi>f</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>1</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mi>u</mi> <mn>2</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>2</mn> </mrow> </msub> <mo>)</mo> </mrow> </mrow> <mrow> <mi>b</mi> <mo>*</mo> <msub> <mi>u</mi> <mn>1</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>1</mn> </mrow> </msub> </mrow> </mfrac> <mo>*</mo> <mfrac> <msub> <mi>c</mi> <mrow> <mi>x</mi> <mn>3</mn> </mrow> </msub> <mrow> <mi>min</mi> <mrow> <mo>(</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>3</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>3</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>,</mo> <msup> <mrow> <mo>(</mo> <msubsup> <mi>T</mi> <mn>3</mn> <mi>max</mi> </msubsup> <mo>/</mo> <msub> <mi>r</mi> <mn>3</mn> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>*</mo> <msub> <mi>F</mi> <mrow> <mi>y</mi> <mn>1</mn> </mrow> </msub> <mo>-</mo> </mtd> </mtr> <mtr> <mtd> <mn>2</mn> <mo>*</mo> <mrow> <mo>(</mo> <mfrac> <msub> <mi>F</mi> <mrow> <mi>x</mi> <mo>_</mo> <mi>des</mi> </mrow> </msub> <mn>2</mn> </mfrac> <mo>-</mo> <mfrac> <msub> <mi>M</mi> <mi>z</mi> </msub> <mi>B</mi> </mfrac> <mo>)</mo> </mrow> <mo>*</mo> <mfrac> <msub> <mi>cx</mi> <mn>3</mn> </msub> <mrow> <mi>min</mi> <mrow> <mo>(</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>3</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>3</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>,</mo> <msup> <mrow> <mo>(</mo> <msubsup> <mi>T</mi> <mn>3</mn> <mi>max</mi> </msubsup> <mo>/</mo> <msub> <mi>r</mi> <mn>3</mn> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>=</mo> <mn>0</mn> </mtd> </mtr> </mtable> <mo>;</mo> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>5</mn> <mo>)</mo> </mrow> </mrow> </math>

<math> <mrow> <mfenced open='' close='' separators=''> <mtable> <mtr> <mtd> <mfrac> <mrow> <mo>&PartialD;</mo> <mi>J</mi> </mrow> <msub> <mrow> <mo>&PartialD;</mo> <mi>F</mi> </mrow> <mrow> <mi>x</mi> <mn>2</mn> </mrow> </msub> </mfrac> <mo>=</mo> <mn>2</mn> <mo>*</mo> <mrow> <mo>(</mo> <mfrac> <msub> <mi>c</mi> <mrow> <mi>x</mi> <mn>2</mn> </mrow> </msub> <msup> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>2</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>2</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mfrac> <mo>+</mo> <mfrac> <msub> <mi>c</mi> <mrow> <mi>x</mi> <mn>4</mn> </mrow> </msub> <mrow> <mi>min</mi> <mrow> <mo>(</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>4</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>4</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>,</mo> <msup> <mrow> <mo>(</mo> <msubsup> <mi>T</mi> <mn>4</mn> <mi>max</mi> </msubsup> <mo>/</mo> <msub> <mi>r</mi> <mn>4</mn> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>)</mo> </mrow> <mo>*</mo> <msub> <mi>F</mi> <mrow> <mi>x</mi> <mn>2</mn> </mrow> </msub> <mo>+</mo> </mtd> </mtr> <mtr> <mtd> <mn>2</mn> <mo>*</mo> <mfrac> <mrow> <msub> <mi>l</mi> <mi>f</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>1</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mi>u</mi> <mn>2</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>2</mn> </mrow> </msub> <mo>)</mo> </mrow> </mrow> <mrow> <mi>b</mi> <mo>*</mo> <msub> <mi>u</mi> <mn>1</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>1</mn> </mrow> </msub> </mrow> </mfrac> <mo>*</mo> <mfrac> <msub> <mi>c</mi> <mrow> <mi>x</mi> <mn>4</mn> </mrow> </msub> <mrow> <mi>min</mi> <mrow> <mo>(</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>4</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>4</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>,</mo> <msup> <mrow> <mo>(</mo> <msubsup> <mi>T</mi> <mn>4</mn> <mi>max</mi> </msubsup> <mo>/</mo> <msub> <mi>r</mi> <mn>4</mn> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>*</mo> <msub> <mi>F</mi> <mrow> <mi>y</mi> <mn>1</mn> </mrow> </msub> <mo>-</mo> </mtd> </mtr> <mtr> <mtd> <mn>2</mn> <mo>*</mo> <mrow> <mo>(</mo> <mfrac> <msub> <mi>F</mi> <mrow> <mi>x</mi> <mo>_</mo> <mi>des</mi> </mrow> </msub> <mn>2</mn> </mfrac> <mo>+</mo> <mfrac> <msub> <mi>M</mi> <mi>z</mi> </msub> <mi>B</mi> </mfrac> <mo>)</mo> </mrow> <mo>*</mo> <mfrac> <msub> <mi>cx</mi> <mn>4</mn> </msub> <mrow> <mi>min</mi> <mrow> <mo>(</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>4</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>4</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>,</mo> <msup> <mrow> <mo>(</mo> <msubsup> <mi>T</mi> <mn>4</mn> <mi>max</mi> </msubsup> <mo>/</mo> <msub> <mi>r</mi> <mn>4</mn> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>=</mo> <mn>0</mn> </mtd> </mtr> </mtable> <mo>;</mo> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>6</mn> <mo>)</mo> </mrow> </mrow> </math>

<math> <mrow> <mfenced open='' close=''> <mtable> <mtr> <mtd> <mfrac> <mrow> <mo>&PartialD;</mo> <mi>J</mi> </mrow> <msub> <mrow> <mo>&PartialD;</mo> <mi>F</mi> </mrow> <mrow> <mi>y</mi> <mn>1</mn> </mrow> </msub> </mfrac> <mo>=</mo> <mo>-</mo> <mn>2</mn> <mo>*</mo> <mfrac> <mrow> <msub> <mi>l</mi> <mi>f</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>1</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mi>u</mi> <mn>2</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>2</mn> </mrow> </msub> <mo>)</mo> </mrow> </mrow> <mrow> <mi>b</mi> <mo>*</mo> <msub> <mi>u</mi> <mn>1</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>1</mn> </mrow> </msub> </mrow> </mfrac> <mo>*</mo> <mfrac> <msub> <mi>c</mi> <mrow> <mi>x</mi> <mn>3</mn> </mrow> </msub> <mrow> <mi>min</mi> <mrow> <mo>(</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>3</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>3</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>,</mo> <msup> <mrow> <mo>(</mo> <msubsup> <mi>T</mi> <mn>3</mn> <mi>max</mi> </msubsup> <mo>/</mo> <msub> <mi>r</mi> <mn>3</mn> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>*</mo> <msub> <mi>F</mi> <mrow> <mi>x</mi> <mn>1</mn> </mrow> </msub> <mo>+</mo> </mtd> </mtr> <mtr> <mtd> <mn>2</mn> <mo>*</mo> <mfrac> <mrow> <msub> <mi>l</mi> <mi>f</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>1</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mi>u</mi> <mn>2</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>2</mn> </mrow> </msub> <mo>)</mo> </mrow> </mrow> <mrow> <mi>b</mi> <mo>*</mo> <msub> <mi>u</mi> <mn>1</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>1</mn> </mrow> </msub> </mrow> </mfrac> <mo>*</mo> <mfrac> <msub> <mi>c</mi> <mrow> <mi>x</mi> <mn>4</mn> </mrow> </msub> <mrow> <mi>min</mi> <mrow> <mo>(</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>4</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>4</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>,</mo> <msup> <mrow> <mo>(</mo> <msubsup> <mi>T</mi> <mn>4</mn> <mi>max</mi> </msubsup> <mo>/</mo> <msub> <mi>r</mi> <mn>4</mn> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>*</mo> <msub> <mi>F</mi> <mrow> <mi>x</mi> <mn>2</mn> </mrow> </msub> <mo>+</mo> </mtd> </mtr> <mtr> <mtd> <mn>2</mn> <mo>*</mo> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <mi></mi> <mfrac> <msub> <mi>c</mi> <mrow> <mi>y</mi> <mn>1</mn> </mrow> </msub> <msup> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>1</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mfrac> <mi></mi> <mo>+</mo> <mi></mi> <msup> <mrow> <mi></mi> <mrow> <mo>(</mo> <mfrac> <mrow> <msub> <mi>u</mi> <mn>2</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>2</mn> </mrow> </msub> </mrow> <mrow> <msub> <mi>u</mi> <mn>1</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>1</mn> </mrow> </msub> </mrow> </mfrac> <mo>)</mo> </mrow> <mo>*</mo> </mrow> <mn>2</mn> </msup> <mi></mi> <mfrac> <msub> <mi>cy</mi> <mn>2</mn> </msub> <msup> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>2</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>2</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mfrac> <mi></mi> <mo>+</mo> <mi></mi> <msup> <mrow> <mo>(</mo> <mfrac> <mrow> <msub> <mi>l</mi> <mi>f</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>1</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mi>u</mi> <mn>2</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>2</mn> </mrow> </msub> <mo>)</mo> </mrow> </mrow> <mrow> <mi>B</mi> <mo>*</mo> <msub> <mi>u</mi> <mn>1</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>1</mn> </mrow> </msub> </mrow> </mfrac> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>*</mo> </mtd> </mtr> <mtr> <mtd> <mo>(</mo> <mfrac> <msub> <mi>c</mi> <mrow> <mi>x</mi> <mn>3</mn> </mrow> </msub> <mrow> <mi>min</mi> <mrow> <mo>(</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>3</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>3</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>,</mo> <msup> <mrow> <mo>(</mo> <msubsup> <mi>T</mi> <mn>3</mn> <mi>max</mi> </msubsup> <mo>*</mo> <msub> <mi>r</mi> <mn>3</mn> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>+</mo> <mo></mo> <mo></mo> <mo></mo> <mo></mo> <mo></mo> <mo></mo> <mo></mo> <mo></mo> <mo></mo> <mo></mo> <mo></mo> <mo></mo> <mo></mo> <mo></mo> <mo></mo> <mo></mo> <mo></mo> <mo></mo> <mo></mo> <mo></mo> <mo></mo> <mo></mo> <mo></mo> <mo></mo> <mo></mo> <mo></mo> <mo></mo> <mo></mo> <mo></mo> <mo></mo> <mfrac> <msub> <mi>c</mi> <mrow> <mi>x</mi> <mn>4</mn> </mrow> </msub> <mrow> <mi>min</mi> <mrow> <mo>(</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>4</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>4</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>,</mo> <msup> <mrow> <mo>(</mo> <msubsup> <mi>T</mi> <mn>4</mn> <mi>max</mi> </msubsup> <mo>*</mo> <msub> <mi>r</mi> <mn>4</mn> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>)</mo> </mtd> </mtr> </mtable> </mfenced> <mo>*</mo> <msub> <mi>F</mi> <mrow> <mi>y</mi> <mn>1</mn> </mrow> </msub> <mo>+</mo> </mtd> </mtr> <mtr> <mtd> <mn>2</mn> <mo>*</mo> <mfrac> <mrow> <msub> <mi>l</mi> <mi>f</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>1</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mi>u</mi> <mn>2</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>2</mn> </mrow> </msub> <mo>)</mo> </mrow> </mrow> <mrow> <mi>b</mi> <mo>*</mo> <msub> <mi>u</mi> <mn>1</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>1</mn> </mrow> </msub> </mrow> </mfrac> <mi></mi> <mo>*</mo> <mi></mi> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <mrow> <mo>(</mo> <mfrac> <msub> <mi>F</mi> <mrow> <mi>x</mi> <mo>_</mo> <mi>des</mi> </mrow> </msub> <mn>2</mn> </mfrac> <mo>-</mo> <mfrac> <msub> <mi>M</mi> <mi>z</mi> </msub> <mi>B</mi> </mfrac> <mo>)</mo> </mrow> <mi></mi> <mo>*</mo> <mfrac> <msub> <mi>c</mi> <mrow> <mi>x</mi> <mn>3</mn> </mrow> </msub> <mrow> <mi>min</mi> <mrow> <mo>(</mo> <msup> <mrow> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mrow> <mi></mi> <mn>3</mn> </mrow> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>3</mn> </mrow> </msub> <mo>)</mo> </mrow> <mi></mi> </mrow> <mn>2</mn> </msup> <mo>,</mo> <msup> <mrow> <mo>(</mo> <msubsup> <mi>T</mi> <mn>3</mn> <mi>max</mi> </msubsup> <mo>/</mo> <msub> <mi>r</mi> <mn>3</mn> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>)</mo> </mrow> </mrow> </mfrac> </mtd> </mtr> <mtr> <mtd> <mo>(</mo> <mfrac> <msub> <mi>F</mi> <mrow> <mi>x</mi> <mo>_</mo> <mi>des</mi> </mrow> </msub> <mn>2</mn> </mfrac> <mi></mi> <mo>+</mo> <mfrac> <msub> <mi>M</mi> <mi>z</mi> </msub> <mi>B</mi> </mfrac> <mo>)</mo> <mo>*</mo> <mfrac> <msub> <mi>c</mi> <mrow> <mi>x</mi> <mn>4</mn> </mrow> </msub> <mrow> <mi>min</mi> <mrow> <mo>(</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>u</mi> <mn>4</mn> </msub> <msub> <mi>F</mi> <mrow> <mi>z</mi> <mn>4</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>,</mo> <msup> <mrow> <mo>(</mo> <msubsup> <mi>T</mi> <mn>4</mn> <mi>max</mi> </msubsup> <mo>/</mo> <msub> <mi>r</mi> <mn>4</mn> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>)</mo> </mrow> </mrow> </mfrac> </mtd> </mtr> </mtable> </mfenced> <mo>=</mo> <mn>0</mn> </mtd> </mtr> </mtable> </mfenced> <mo>;</mo> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>7</mn> <mo>)</mo> </mrow> </mrow> </math>

solving the equations (5), (6) and (7) to obtain the lateral force F of the left front wheel required to be distributed_y1And two longitudinal forces F of the left and right front wheels_x1、F_x2Then, the lateral force F of the left front right wheel to be distributed is calculated by the formulas (2), (3) and (4)_y2And left and right rear wheelsForce F_x3、F_x4；

(4) According to the lateral force F of the left front wheel_y1Side force F of the right front wheel_y2And two longitudinal forces F of the left and right front wheels_x1、F_x2Calculating the driving torque and the steering wheel angle correction value of the left and right front wheels according to the two longitudinal forces F of the left and right rear wheels_x3、F_x4Calculating the driving torques of the left and right rear wheels, and then controlling the four-wheel drive electric automobile by using the four driving torques;

(5) the required moment of the four wheels can be calculated by the lateral force and the longitudinal force of the wheels as follows:

<math> <mrow> <msub> <mi>M</mi> <mi>i</mi> </msub> <mo>=</mo> <msqrt> <msup> <mrow> <mo>(</mo> <msub> <mi>l</mi> <mi>f</mi> </msub> <mo>&CenterDot;</mo> <msub> <mi>F</mi> <mi>xi</mi> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>l</mi> <mi>f</mi> </msub> <mo>&CenterDot;</mo> <msub> <mi>F</mi> <mi>i</mi> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> </msqrt> <mo>,</mo> <mi>i</mi> <mo>=</mo> <mn>1,2</mn> <mo>,</mo> </mrow> </math>

M_i＝l_f·F_xi i＝3,4，

wherein M is_iRepresenting the moment of four wheels.

2. The torque distribution method for four-wheel drive electric vehicle according to claim 1, wherein the weight coefficient c of longitudinal force of the i-th tire_xiLateral force weight coefficient c_yiComprises the following steps:

<math> <mrow> <msub> <mi>C</mi> <mi>xi</mi> </msub> <mo>=</mo> <mfenced open='{' close=''> <mtable> <mtr> <mtd> <mn>0.5</mn> </mtd> <mtd> <mi>m</mi> <mo>&GreaterEqual;</mo> <mn>0.6</mn> </mtd> </mtr> <mtr> <mtd> <mn>0.3</mn> <mo>+</mo> <mn>0.1</mn> <mo>*</mo> <mrow> <mo>(</mo> <mfrac> <mi>m</mi> <mn>0.6</mn> </mfrac> <mo>+</mo> <mfrac> <mrow> <mo>|</mo> <mi>n</mi> <mo>|</mo> </mrow> <mn>0.8</mn> </mfrac> <mo>)</mo> </mrow> </mtd> <mtd> <mn>0</mn> <mo>&le;</mo> <mi>m</mi> <mo>&lt;</mo> <mn>0.6</mn> </mtd> </mtr> <mtr> <mtd> <mn>0.3</mn> <mo>+</mo> <mn>0.2</mn> <mo>*</mo> <mo>(</mo> <mfrac> <mrow> <mo>|</mo> <mi>m</mi> <mo>|</mo> </mrow> <mn>0.7</mn> </mfrac> <mo>+</mo> <mfrac> <mrow> <mo>|</mo> <mi>n</mi> <mo>|</mo> </mrow> <mn>0.8</mn> </mfrac> <mo>)</mo> </mtd> <mtd> <mo>-</mo> <mn>0.7</mn> <mo>&lt;</mo> <mi>m</mi> <mo>&lt;</mo> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <mn>0.5</mn> </mtd> <mtd> <mi>m</mi> <mo>&le;</mo> <mo>-</mo> <mn>0.7</mn> </mtd> </mtr> </mtable> </mfenced> </mrow> </math>

C_yi＝1-C_xi；

the longitudinal gravity center transfer rate m and the lateral gravity center transfer rate n of the tire are as follows:

$m=\frac{l_f(u_1*F_{z1}+u_2*F_{z2})-l_r(u_3*F_{z3}+u_4*F_{z4})}{u_1*F_{z1}+u_2*F_{z2}+u_3*F_{z3}+u_4*F_{z4}}$

$n=\frac{\frac{B}{2}({-u}_1*F_{z1}+u_2*F_{z2}-u_3*F_{z3}+u_4*F_{z4})}{u_1*F_{z1}+u_2*F_{z2}+u_3*F_{z3}+u_4*F_{z4}}$

wherein l_fRepresenting the distance between the front wheel and the centre of mass of the vehicle, B representing the distance between the front wheels of the vehicle and the two wheels, l_rRepresenting the distance between the rear wheel and the center of mass of the vehicle;

the larger the positive direction of m is, the larger the deceleration of the automobile is, the braking force is provided by the automobile, the load of the automobile is transferred forwards, the larger the negative direction of m is, the larger the acceleration of the automobile is, the driving force is provided by the automobile, and the load of the automobile is transferred backwards; the larger the positive direction of n is, the larger the left turning degree of the automobile is, the load is transferred to the right, and the larger the negative direction of n is, the larger the right turning degree of the automobile is, the load is transferred to the left; the longitudinal center of gravity transfer rate m and the lateral center of gravity transfer rate n of the vehicle are defined to be between-0.8 and 0.8, and if the calculation is out of the range, a boundary value is taken.