CN103303157A - 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

A kind of four-wheel driving electric vehicle torque distribution method

Technical field

The invention belongs to electric vehicle motor control technology field, more specifically say, relate to a kind of four-wheel driving electric vehicle torque distribution method, namely when four-wheel driven electric vehicle moves, the problem how torque distributes to four-wheel.

Background technology

The problem that becomes increasingly conspicuous in order to tackle energy shortage, environmental pollution etc., battery-driven car has become the focus of research at present with its anti-emission carburetor, low noise and other advantages.Four-wheel driving electric vehicle because its control and energy-conservation above advantage, inside the field of research of electronlmobil, occupy increasing proportion.As the Torque-sharing strategy of one of four-wheel driven electric vehicle electronic control technology, be subjected to the attention of numerous scientific research institutions and enterprise.

Four-wheel driving electric vehicle torque distribution method mainly is based on the distribution method of rule and the distribution method that based target is optimized at present, mostly the torque distribution algorithm is take economy and stability as optimization aim, consider the impact of longitudinal force or side force, torque distributes to four-wheel.In the drive torque allocation strategy for traditional automobile and most of rule-based approach, only be to consider that the large tire of certain dynamic test of control carries out brake-power control, thereby reach corresponding control target, such control does not take into full account yaw moment that each tire can provide and the adhesion value state of each tire, easily cause the complete locking of certain wheel, affect vehicle stability.

Summary of the invention

The object of the invention is to overcome the deficiencies in the prior art, a kind of four-wheel driving electric vehicle torque distribution method is provided, with assurance vehicle body stability and controllability, and utilize to greatest extent frictional ground force.

For realizing above purpose, four-wheel driving electric vehicle torque distribution method of the present invention is characterized in that, may further comprise the steps:

(1), determine objective function:

$J=\frac{c_{x1}{F}_{x1}^2+c_{y1}{\mathrm{<figure-callout\; id="1"\; label="F\; y"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_y^1}{{\left(u_1{F}_{z1}\right)}^2}+\frac{c_{x2}{F}_{x2}^2+c_{y2}{\mathrm{<figure-callout\; id="2"\; label="F\; y"\; filenames="HDA00003367001500012.png,HDA00003367001500061.png"\; state="\{\{state\}\}">F</figure-callout>}}_y^2}{{(u_2+F_{z2})}^2}+\frac{c_{x3}{F}_{x3}^2}{\min ({\left(u_3{F}_{z3}\right)}^2,{({\mathrm{<figure-callout\; id="3"\; label="T"\; filenames="HDA00003367001500012.png,HDA00003367001500062.png"\; state="\{\{state\}\}">T</figure-callout>}}_3^{\max }/r_3)}^2)}+\frac{c_{x4}{F}_{x4}^2}{\min ({\left(u_4{F}_{z4}\right)}^2,{(T_4^{\max }/r_4)}^2)}---\left(1\right)$

Wherein, c _XiThe longitudinal force weight coefficient that represents i tire, c _YiThe side force weight coefficient that represents i tire, F _XiBe the longitudinal force of tire of i tire, F _YiBe the side force of tire of i tire, u _iBe the adhesion value of i tire, F _ZiBe the tyre load of i tire,

Be the wheel hub motor maximum drive moment of i tire, r _iBe the radius of wheel of i tire, x is longitudinal direction, and y is lateral, and z is perpendicular to xy in-plane, min(*, *) represent to get the minimum value in the two, i=1,2,3,4 represent respectively the near front wheel, off front wheel, left rear wheel, off hind wheel;

(2), select the side force F of front left wheel _Y1With two longitudinal force F of left and right sides front-wheel _X1, F _X2Be unknown parameter, then:

$F_{x3}=-F_{x1}+\frac{l_f(u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+u_2{F}_{z2})}{B*u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z}*{\mathrm{<figure-callout\; id="1"\; label="F\; y"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_y+\frac{F_{x\_\mathrm{des}}}{2}-\frac{M_z}{B}---\left(2\right)$

$F_{x4}=-F_{x2}-\frac{l_f(u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+u_2{F}_{z2})}{B*u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z}*{\mathrm{<figure-callout\; id="1"\; label="F\; y"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_y+\frac{F_{x\_\mathrm{des}}}{2}+\frac{M_z}{B}---\left(3\right)$

${\mathrm{<figure-callout\; id="2"\; label="F\; y"\; filenames="HDA00003367001500012.png,HDA00003367001500061.png"\; state="\{\{state\}\}">F</figure-callout>}}_y=\frac{u_2{F}_{z2}}{u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z}*F_{y1}---\left(4\right)$

Wherein, B represents wheel front-wheel two-wheeled spacing;

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}The vertically clean power of expression expectation, M _zThe needed yaw moment of expression vehicle stabilization, it is determined by the yaw moment control module;

(3), formula (2), (3), (4) are updated in the objective function (1), find the solution the minimal value of present objective function, respectively to the side force F of front-wheel left side wheel _Y1With two longitudinal force F of left and right sides front-wheel _X1, F _X2Ask local derviation and make it equal 0, can get:

$\frac{\&PartialD;J}{{\&PartialD;F}_{x1}}=2*(\frac{c_{x1}}{{\left(u_1{F}_{z1}\right)}^2}+\frac{c_{x3}}{\min ({\left(u_3{F}_{z3}\right)}^2,{({\mathrm{<figure-callout\; id="3"\; label="T"\; filenames="HDA00003367001500012.png,HDA00003367001500062.png"\; state="\{\{state\}\}">T</figure-callout>}}_3^{\max }/r_3)}^2)})*F_{x1}-$

$2*\frac{l_f(u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+u_2{F}_{z2})}{B*u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z}*\frac{c_{x3}}{\min ({\left(u_3{F}_{z3}\right)}^2,{({\mathrm{<figure-callout\; id="3"\; label="T"\; filenames="HDA00003367001500012.png,HDA00003367001500062.png"\; state="\{\{state\}\}">T</figure-callout>}}_3^{\max }/r_3)}^2)}*F_{y1}----\left(5\right)$

$2*(\frac{F_{x\_\mathrm{des}}}{2}-\frac{M_z}{B})*\frac{c_{x3}}{\min ({\left(u_3{F}_{z3}\right)}^2,{({\mathrm{<figure-callout\; id="3"\; label="T"\; filenames="HDA00003367001500012.png,HDA00003367001500062.png"\; state="\{\{state\}\}">T</figure-callout>}}_3^{\max }/r_3)}^2)}=0;$

$\frac{\&PartialD;J}{{\&PartialD;F}_{x2}}=2*(\frac{c_{x2}}{{\left(u_2{F}_{z2}\right)}^2}+\frac{c_{x4}}{\min ({\left(u_4{F}_{z4}\right)}^2,{(T_4^{\max }/r_4)}^2)})*F_{x2}+$

$2*\frac{l_f(u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+u_2{F}_{z2})}{B*u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z}*\frac{c_{x4}}{\min ({\left(u_4{F}_{z4}\right)}^2,{(T_4^{\max }/r_4)}^2)}*F_{y1}----\left(6\right)$

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

$\frac{\&PartialD;J}{{\&PartialD;\mathrm{<figure-callout\; id="1"\; label="F\; y"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_y}=-2*\frac{l_f(u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+u_2{F}_{z2})}{B*u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z}*\frac{c_{x3}}{\min ({\left(u_3{F}_{z3}\right)}^2,{({\mathrm{<figure-callout\; id="3"\; label="T"\; filenames="HDA00003367001500012.png,HDA00003367001500062.png"\; state="\{\{state\}\}">T</figure-callout>}}_3^{\max }/r_3)}^2)}*F_{x1}+$

$2*\frac{l_f(u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+u_2{F}_{z2})}{B*u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z}*\frac{c_{x4}}{\min ({\left(u_4{F}_{z4}\right)}^2,{(T_4^{\max }/r_4)}^2)}*F_{x2}+$

$2*\left(\begin{array}{c}\frac{c_{y1}}{{\left(u_1{F}_{z1}\right)}^2}+{\left(\frac{u_2{F}_{z2}}{u_1{F}_{z1}}\right)}^2*\frac{c_{y2}}{{\left(u_2{F}_{z2}\right)}^2}+{\left(\frac{l_f(u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+u_2{F}_{z2})}{B*u_1{F}_{z1}}\right)}^2*\\ (\frac{c_{x3}}{\min ({\left(u_3{F}_{z3}\right)}^2,{({\mathrm{<figure-callout\; id="3"\; label="T"\; filenames="HDA00003367001500012.png,HDA00003367001500062.png"\; state="\{\{state\}\}">T</figure-callout>}}_3^{\max }/r_3)}^2)}+\frac{c_{x4}}{\min ({\left(u_4{F}_{z4}\right)}^2,{(T_4^{\max }/r_4)}^2)})\end{array}\right)*F_{y1}+---\left(7\right)$

$2*\frac{l_f(u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+u_2{F}_{z2})}{B*u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z}*\left(\begin{array}{c}(\frac{F_{x\_\mathrm{des}}}{2}-\frac{M_z}{B})*\frac{c_{x3}}{\min ({\left(u_3{F}_{z3}\right)}^2,{({\mathrm{<figure-callout\; id="3"\; label="T"\; filenames="HDA00003367001500012.png,HDA00003367001500062.png"\; state="\{\{state\}\}">T</figure-callout>}}_3^{\max }/r_3)}^2)}-\\ (\frac{F_{x\_\mathrm{des}}}{2}+\frac{M_z}{B})*\frac{c_{x4}}{\min ({\left(u_4{F}_{z4}\right)}^2,{(T_4^{\max }/r_4)}^2)}\end{array}\right)=0;$

Solving equation (5), (6), (7) obtain the side force F of the front left wheel of required distribution _Y1With two longitudinal force F of left and right sides front-wheel _X1, F _X2, calculated again the side force F of the right front wheel of all the other required distribution by formula (2), (3), (4) _Y2With two longitudinal force F of left and right sides trailing wheel _X3, F _X4

(4), according to the side force F of front left wheel _Y1, right front wheel side force F _Y2And two longitudinal force F of left and right sides front-wheel _X1, F _X2Calculate driving torque and the steering wheel angle correction of left and right sides front-wheel, with two longitudinal force F of left and right sides trailing wheel _X3, F _X4Calculate the driving torque of left and right sides trailing wheel, then, with four driving torque control four-wheel driving electric vehicles.

(5), by the side force of wheel and longitudinal force can calculate four wheels respectively required moment be:

$M_i=\sqrt{{\mathrm{<figure-callout\; id="2"\; label="M\; xi"\; filenames="HDA00003367001500012.png,HDA00003367001500061.png"\; state="\{\{state\}\}">M</figure-callout>}}_{\mathrm{xi}}^{}+M_{\mathrm{yi}}^2},$ Wherein, M _Xi=l _fF _XiM _Yi=l _fF _YiI=1,2,

M _i=M _xi=l _f·F _xi i=3,4，

Wherein, M _iThe moment of expression four wheels.

As a further improvement on the present invention, the longitudinal force weight coefficient c of i tire described in the step (1) _Xi, side force weight coefficient c _YiFor:

$C_{\mathrm{xi}}=\left\{\begin{array}{c}0.5,m\&GreaterEqual;0.6\\ 0.3+0.1*(\frac{m}{0.6}+\frac{\left|n\right|}{0.8}),0\&le;m<0.6\\ 0.3+0.2*(\frac{\left|m\right|}{0.7}-\frac{\left|n\right|}{0.8}),-0.7<m<0\\ 0.5,m\&le;-0.7\end{array}\right.$

C _yi=1-C _xi；

Tire longitudinal center of gravity rate of transform m and side direction center of gravity rate of transform n are:

$m=\frac{l_f(u_1*{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+u_2*F_{z2})-l_r(u_3*{\mathrm{<figure-callout\; id="3"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500062.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+u_4*F_{z4})}{u_1*{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+u_2*{\mathrm{<figure-callout\; id="2"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500061.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+u_3*{\mathrm{<figure-callout\; id="3"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500062.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+u_4*F_{z4}}$

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

Wherein, l _fThe expression front-wheel is to the distance between the vehicle barycenter, and B represents wheel front-wheel two-wheeled spacing, l _rThe expression trailing wheel is to the distance between the vehicle barycenter;

The m positive dirction is larger, and expression car deceleration degree is larger, and automobile provides braking force, and automobile load shifts forward, and the m negative direction is larger, and the expression pickup is larger, and automobile provides propulsive effort, and automobile load shifts backward; The n positive dirction is larger, and the expression automobile degree of turning left is larger, and load shifts to the right, and the n negative direction is larger, and the expression automobile degree that bends to right is larger, and load shifts left; Automobile longitudinal center of gravity rate of transform m and side direction center of gravity rate of transform n are defined between-0.8 to 0.8, exceed this scope if calculate, and then get boundary value.

The object of the present invention is achieved like this:

Four-wheel driving electric vehicle torque distribution method of the present invention, the parameters such as maximum drive moment of the friction coefficient of consideration front-wheel longitudinal force and transverse force, rear tire longitudinal force, tire, load, tire radius and the wheel hub motor of tire, and optimized the weight coefficient of parameter, consider simultaneously the propulsive effort limiting condition of drive motor, adopt the linear solution analysis method, torque distribution optimization method with stability has been proposed, guarantee vehicle body stability and controllability, and maximally utilised frictional ground force.

Description of drawings

Fig. 1 is four-wheel driving electric vehicle torque distribution method of the present invention;

Fig. 2 is steering wheel angle input curve figure;

Fig. 3 is acceleration pedal depth curve figure;

Fig. 4 is weight coefficient figure;

Fig. 5 is the front wheel torque diagram of curves;

Fig. 6 is the rear wheel torque diagram of curves;

Fig. 7 is the vehicle roll angle diagram of curves;

Fig. 8 is the vehicular longitudinal velocity diagram of curves;

Fig. 9 is that vehicle side is to speed curve diagram;

Figure 10 is four-wheel driving electric vehicle torque distribution method of the present invention and mean allocation method contrast track of vehicle curve.

The specific embodiment

Below in conjunction with accompanying drawing the specific embodiment of the present invention is described, so that those skilled in the art understands the present invention better.What need to point out especially is that in the following description, when perhaps the detailed description of known function and design can desalinate main contents of the present invention, these were described in here and will be left in the basket.

The longitudinal force that ground applies in the face of tire and side force are all produced by frictional ground force, so according to the situation of ground friction coefficient, the maximum longitudinal force that ground can provide must satisfy following formula:

$F_{\mathrm{xi}}\&le;\sqrt{{\left(u_i{F}_{\mathrm{zi}}\right)}^2-F_{\mathrm{yi}}^2}=F_i^{\mathrm{limit}},i=\mathrm{1,2,3,4};$

Wherein, F _XiBe the longitudinal force of i tire, F _YiBe the side force of i tire, u _iBe the adhesion value of i tire, F _ZiBe the tyre load of i tire,

Be illustrated under the condition of actual adhesion value, ground can offer the maximum longitudinal force of i tire; X is longitudinal direction, and y is lateral, and z is perpendicular to the xy in-plane, i=1, and 2,3,4 represent respectively the near front wheel, off front wheel, left rear wheel, off hind wheel.

Therefore, when carrying out the distribution of wheel propulsive effort in order to produce enough yaw moments when, will be understood that, the propulsive effort that each wheel produces all can not surpass the maximum longitudinal force that ground can provide, just this moment, vehicle can be in stabilized conditions.If the torque that certain wheel distributes surpasses ground limit of adhesion restriction, just wheel can produce slippage, when wheelslip reached certain threshold, the adhesion value that ground provides just can descend on a large scale, had more increased the weight of the unsettled trend of wheel.

Four-wheel driving electric vehicle is equipped with 4 can independent allocation different propulsive effort torques and the wheel hub motor of braking force pressure.The wheel hub motor maximum driving torque is based on the basis of motor self character and by the angle of tire and accelerates to determine, so the longitudinal force that tire provides must satisfy following formula:

$F_{\mathrm{xi}}\&le;T_i^{\max }/r,i=\mathrm{1,2,3,4};$

Wherein, Be the wheel hub motor maximum drive moment of i tire, r is radius of wheel.

Four-wheel driving electric vehicle for front-wheel steering, side force mainly is to be provided by wheel flutter, so the side force of two-wheeled before only considering is considered the longitudinal force of 4 tires simultaneously, finally with 6 parameters of longitudinal force of longitudinal force, side force and the trailing wheel of front-wheel as distribution object.Consider the torque limitation of coefficient of road adhesion and drive motor, provide this paper optimization method objective function as shown in Equation (3).

$J=\frac{c_{x1}{F}_{x1}^2+c_{y1}{\mathrm{<figure-callout\; id="1"\; label="F\; y"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_y^1}{{\left(u_1{F}_{z1}\right)}^2}+\frac{c_{x2}{F}_{x2}^2+c_{y2}{\mathrm{<figure-callout\; id="2"\; label="F\; y"\; filenames="HDA00003367001500012.png,HDA00003367001500061.png"\; state="\{\{state\}\}">F</figure-callout>}}_y^2}{{({\mathrm{<figure-callout\; id="2"\; label="u"\; filenames="HDA00003367001500012.png,HDA00003367001500061.png"\; state="\{\{state\}\}">u</figure-callout>}}_2+F_{z2})}^2}+\frac{c_{x3}{F}_{x3}^2}{\min ({\left(u_3{F}_{z3}\right)}^2,{({\mathrm{<figure-callout\; id="3"\; label="T"\; filenames="HDA00003367001500012.png,HDA00003367001500062.png"\; state="\{\{state\}\}">T</figure-callout>}}_3^{\max }/r_3)}^2)}+\frac{c_{x4}{F}_{x4}^2}{\min ({\left(u_4{F}_{z4}\right)}^2,{(T_4^{\max }/r_4)}^2)}---\left(1\right)$

Wherein, c _XiThe longitudinal force weight coefficient that represents i tire, c _YiThe side force weight coefficient that represents i tire, min(*, *) expression gets the minimum value in the two.

In order to guarantee that four-wheel driving electric vehicle satisfies the speed of a motor vehicle of chaufeur expectation, the longitudinal force that each tire provides must satisfy the vertically clean power of speed of a motor vehicle control module output:

F _{x_des}=F _x1+F _x2+F _x3+F _x4； （8）

Wherein, F _{X_des}The vertically clean power of expression expectation.

The needed vertically clean power of tire is the longitudinal force sum of all tire, has guaranteed the power-handling capability of automobile.

Simultaneously for travelling of guaranteeing that automobile can security and stability, the longitudinal force that each wheel provides and side force must satisfy the needed yaw moment value of vehicle stabilization:

M _z=l _f(F _y1+F _y2)+B(|F _x1-F _x2|+|F _x3-F _x4|)/2 （9）

Wherein, l _fThe expression front-wheel is to the distance between the vehicle barycenter; B represents wheel front-wheel two-wheeled spacing; M _zThe needed yaw moment of expression vehicle stabilization, it is determined by the yaw moment control module.

For the electronlmobil of front-wheel steering, so the steering wheel angle of two tires of front-wheel is consistent.

$\frac{F_{y1}}{u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z}=\frac{F_{y2}}{u_2{F}_{z2}}---\left(10\right)$

Relation on the left of following formula represents between the extra tire force of front-wheel and right side front wheel.It is that friction circle size with corresponding ground is mutually ratio that left side front-wheel and right-hand line are taken turns extra tire force, and namely the pavement friction of corresponding tire circle is larger, and then its side force that provides is corresponding just larger.

Formula (8), (9), (10) are to 6 independent variable F in the objective function _X1, F _X2, F _X3, F _X4, F _Y1, F _Y23 constraint conditions, these 3 constraint conditions have guaranteed that not only vehicle satisfies the vehicle running state of chaufeur expectation, have also considered stability and the controllability of vehicle simultaneously, guarantee that final optimum results can reach the control optimum.

Select the side force F of front left wheel _Y1With two longitudinal force F of left and right sides front-wheel _X1, F _X2Be unknown parameter.Then concrete equation is as follows:

$F_{x3}=-{\mathrm{<figure-callout\; id="1"\; label="F\; x"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_x+\frac{l_f(u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+u_2{F}_{z2})}{B*u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z}*{\mathrm{<figure-callout\; id="1"\; label="F\; y"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_y+\frac{F_{x\_\mathrm{des}}}{2}-\frac{M_z}{B}---\left(2\right)$

$F_{x4}=-F_{x2}-\frac{l_f(u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+u_2{F}_{z2})}{B*u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z}*{\mathrm{<figure-callout\; id="1"\; label="F\; y"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_y+\frac{F_{x\_\mathrm{des}}}{2}+\frac{M_z}{B}---\left(3\right)$

${\mathrm{<figure-callout\; id="2"\; label="F\; y"\; filenames="HDA00003367001500012.png,HDA00003367001500061.png"\; state="\{\{state\}\}">F</figure-callout>}}_y=\frac{u_2{F}_{z2}}{u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z}*F_{y1}---\left(4\right)$

Wherein, B represents wheel front-wheel two-wheeled spacing.

Formula (2), (3), (4) are updated in the objective function (1) as necessary condition, find the solution the minimal value of present objective function, respectively to the side force F of front-wheel left side wheel _Y1With two longitudinal force F of left and right sides front-wheel _X1, F _X2Ask local derviation and make it equal 0, can get:

$\frac{\&PartialD;J}{{\&PartialD;F}_{x1}}=2*(\frac{c_{x1}}{{\left(u_1{F}_{z1}\right)}^2}+\frac{c_{x3}}{\min ({\left(u_3{F}_{z3}\right)}^2,{({\mathrm{<figure-callout\; id="3"\; label="T"\; filenames="HDA00003367001500012.png,HDA00003367001500062.png"\; state="\{\{state\}\}">T</figure-callout>}}_3^{\max }/r_3)}^2)})*F_{x1}-$

$2*\frac{l_f(u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+u_2{F}_{z2})}{B*u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z}*\frac{c_{x3}}{\min ({\left(u_3{F}_{z3}\right)}^2,{({\mathrm{<figure-callout\; id="3"\; label="T"\; filenames="HDA00003367001500012.png,HDA00003367001500062.png"\; state="\{\{state\}\}">T</figure-callout>}}_3^{\max }/r_3)}^2)}*F_{y1}----\left(5\right)$

$2*(\frac{F_{x\_\mathrm{des}}}{2}-\frac{M_z}{B})*\frac{c_{x3}}{\min ({\left(u_3{F}_{z3}\right)}^2,{({\mathrm{<figure-callout\; id="3"\; label="T"\; filenames="HDA00003367001500012.png,HDA00003367001500062.png"\; state="\{\{state\}\}">T</figure-callout>}}_3^{\max }/r_3)}^2)}=0;$

$\frac{\&PartialD;J}{{\&PartialD;F}_{x2}}=2*(\frac{c_{x2}}{{\left(u_2{F}_{z2}\right)}^2}+\frac{c_{x4}}{\min ({\left(u_4{F}_{z4}\right)}^2,{(T_4^{\max }/r_4)}^2)})*F_{x2}+$

$2*\frac{l_f(u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+u_2{F}_{z2})}{B*u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z}*\frac{c_{x4}}{\min ({\left(u_4{F}_{z4}\right)}^2,{(T_4^{\max }/r_4)}^2)}*F_{y1}----\left(6\right)$

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

$\frac{\&PartialD;J}{\&PartialD;{\mathrm{<figure-callout\; id="1"\; label="F\; y"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_y}=-2*\frac{l_f(u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+u_2{F}_{z2})}{B*u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z}*\frac{c_{x3}}{\min ({\left(u_3{F}_{z3}\right)}^2,{({\mathrm{<figure-callout\; id="3"\; label="T"\; filenames="HDA00003367001500012.png,HDA00003367001500062.png"\; state="\{\{state\}\}">T</figure-callout>}}_3^{\max }/r_3)}^2)}*F_{x1}+$

$2*\frac{l_f(u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+u_2{F}_{z2})}{B*u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z}*\frac{c_{x4}}{\min ({\left(u_4{F}_{z4}\right)}^2,{(T_4^{\max }/r_4)}^2)}*F_{x2}+$

$2*\left(\begin{array}{c}\frac{c_{y1}}{{\left(u_1{F}_{z1}\right)}^2}+{\left(\frac{u_2{F}_{z2}}{u_1{F}_{z1}}\right)}^2*\frac{c_{y2}}{{\left(u_2{F}_{z2}\right)}^2}+{\left(\frac{l_f(u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+u_2{F}_{z2})}{B*u_1{F}_{z1}}\right)}^2*\\ (\frac{c_{x3}}{\min ({\left(u_3{F}_{z3}\right)}^2,{({\mathrm{<figure-callout\; id="3"\; label="T"\; filenames="HDA00003367001500012.png,HDA00003367001500062.png"\; state="\{\{state\}\}">T</figure-callout>}}_3^{\max }/r_3)}^2)}+\frac{c_{x4}}{\min ({\left(u_4{F}_{z4}\right)}^2,{(T_4^{\max }/r_4)}^2)})\end{array}\right)*{\mathrm{<figure-callout\; id="1"\; label="F\; y"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_y+---\left(7\right)$

$2*\frac{l_f(u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+u_2{F}_{z2})}{B*u_1{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z}*\left(\begin{array}{c}(\frac{F_{x\_\mathrm{des}}}{2}-\frac{M_z}{B})*\frac{c_{x3}}{\min ({\left(u_3{F}_{z3}\right)}^2,{({\mathrm{<figure-callout\; id="3"\; label="T"\; filenames="HDA00003367001500012.png,HDA00003367001500062.png"\; state="\{\{state\}\}">T</figure-callout>}}_3^{\max }/r_3)}^2)}-\\ (\frac{F_{x\_\mathrm{des}}}{2}+\frac{M_z}{B})*\frac{c_{x4}}{\min ({\left(u_4{F}_{z4}\right)}^2,{(T_4^{\max }/r_4)}^2)}\end{array}\right)=0;$

Solving equation (5), (6), (7) obtain the side force F of the front left wheel of required distribution _Y1With two longitudinal force F of left and right sides front-wheel _X1, F _X2, calculated again the side force F of the right front wheel of all the other required distribution by formula (2), (3), (4) _Y2With two longitudinal force F of left and right sides trailing wheel _X3, F _X4

Different state of kinematic motions, the weight factor that needs is not identical yet, so determine that the motion state parameters of automobile is most important.Select the longitudinal center of gravity rate of transform of battery-driven car to represent the degree that automobile accelerates and slows down, the side direction center of gravity rate of transform of selection battery-driven car represents the degree of automobile turning, then calculates the weight coefficient of motor tire power with two center of gravity rates of transform of vertical and horizontal.Tire longitudinal center of gravity rate of transform m and side direction center of gravity rate of transform n computing formula are as follows:

$m=\frac{l_f({\mathrm{<figure-callout\; id="1"\; label="u"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">u</figure-callout>}}_1*{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+{\mathrm{<figure-callout\; id="2"\; label="u"\; filenames="HDA00003367001500012.png,HDA00003367001500061.png"\; state="\{\{state\}\}">u</figure-callout>}}_2*F_{z2})-l_r({\mathrm{<figure-callout\; id="3"\; label="u"\; filenames="HDA00003367001500012.png,HDA00003367001500062.png"\; state="\{\{state\}\}">u</figure-callout>}}_3*{\mathrm{<figure-callout\; id="3"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500062.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+u_4*F_{z4})}{{\mathrm{<figure-callout\; id="1"\; label="u"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">u</figure-callout>}}_1*{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+{\mathrm{<figure-callout\; id="2"\; label="u"\; filenames="HDA00003367001500012.png,HDA00003367001500061.png"\; state="\{\{state\}\}">u</figure-callout>}}_2*{\mathrm{<figure-callout\; id="2"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500061.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+{\mathrm{<figure-callout\; id="3"\; label="u"\; filenames="HDA00003367001500012.png,HDA00003367001500062.png"\; state="\{\{state\}\}">u</figure-callout>}}_3*{\mathrm{<figure-callout\; id="3"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500062.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+u_4*F_{z4}}$

$n=\frac{\frac{B}{2}(-{\mathrm{<figure-callout\; id="1"\; label="u"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">u</figure-callout>}}_1*{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+{\mathrm{<figure-callout\; id="2"\; label="u"\; filenames="HDA00003367001500012.png,HDA00003367001500061.png"\; state="\{\{state\}\}">u</figure-callout>}}_2*F_{z2}-{\mathrm{<figure-callout\; id="3"\; label="u"\; filenames="HDA00003367001500012.png,HDA00003367001500062.png"\; state="\{\{state\}\}">u</figure-callout>}}_3*{\mathrm{<figure-callout\; id="3"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500062.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+u_4*F_{z4})}{{\mathrm{<figure-callout\; id="1"\; label="u"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">u</figure-callout>}}_1*{\mathrm{<figure-callout\; id="1"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+{\mathrm{<figure-callout\; id="2"\; label="u"\; filenames="HDA00003367001500012.png,HDA00003367001500061.png"\; state="\{\{state\}\}">u</figure-callout>}}_2*{\mathrm{<figure-callout\; id="2"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500061.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+{\mathrm{<figure-callout\; id="3"\; label="u"\; filenames="HDA00003367001500012.png,HDA00003367001500062.png"\; state="\{\{state\}\}">u</figure-callout>}}_3*{\mathrm{<figure-callout\; id="3"\; label="F\; z"\; filenames="HDA00003367001500012.png,HDA00003367001500062.png"\; state="\{\{state\}\}">F</figure-callout>}}_z+u_4*F_{z4}}$

The m positive dirction is larger, and expression car deceleration degree is larger, and automobile provides braking force, and automobile load shifts forward, and the m negative direction is larger, and the expression pickup is larger, and automobile provides propulsive effort, and automobile load shifts to the right; The n positive dirction is larger, and the expression automobile degree of turning left is larger, and load shifts to the right, and the n negative direction is larger, and the expression automobile degree that bends to right is larger, and load shifts left.Automobile longitudinal center of gravity rate of transform m and side direction center of gravity rate of transform n are defined between-0.8 to 0.8, exceed this scope if calculate, and then get boundary value.Weight coefficient C thus _XiAnd C _YiTo be calculated by automobile longitudinal center of gravity rate of transform m and side direction center of gravity rate of transform n.(i=1,2,3,4 represent respectively the near front wheel, off front wheel, left rear wheel, off hind wheel)

$C_{\mathrm{xi}}=\left\{\begin{array}{c}0.5,m\&GreaterEqual;0.6\\ 0.3+0.1*(\frac{m}{0.6}+\frac{\left|n\right|}{0.8}),0\&le;m<0.6\\ 0.3+0.2*(\frac{\left|m\right|}{0.7}-\frac{\left|n\right|}{0.8}),-0.7<m<0\\ 0.5,m\&le;-0.7\end{array}\right.,$

C _yi=1-C _xi

Advantage of the present invention is:

1, the torque distribution algorithm of based target optimization adopts the loading of tire rate minimum for controlling target, and assurance vehicle body stability and controllability are prerequisite, carry out torque distribution.

2, the allocation proportion that guarantees that by revising weight coefficient electronlmobil longitudinal force and side force under different operating modes are occupied in the control algorithm guarantees to utilize to greatest extent frictional ground force.

Example

For total mass 1296kg, around Z axis rotor inertia 1750kgm2, wheelbase 2.57m, barycenter to front axle apart from 1.25m, barycenter to rear axle apart from 1.32m, track front 1.405m, track rear 1.399m, height of center of mass 0.45m, the four-wheel driven electric vehicle of radius of wheel 0.326m is verified.The speed of a motor vehicle is increased to 70km/h by 0km/h, walks around 7 obstacles simultaneously in 8～25s, simulation serpentine locomotion operating mode, and surface friction coefficient is made as u=0.7, and this surface friction coefficient is the normal friction coefficient of dry asphalt surface.

Four-wheel driving electric vehicle torque distribution method shown in Figure 1 is at first by the Real Time Drive moment (T of each wheel hub motor of automobile sensor system collection vehicle _i) and yaw moment value (M _z); Then determine longitudinal force weight factor (C by the weight coefficient computing module _Xi) and side force weight factor (C _Yi); Calculate optimum longitudinal force of tire and side force F by the torque distribution method at last _Xi, F _Yi, then convert thereof into driving torque and steering wheel angle correction into each tire by the parameter correcting module, finally reach the control four-wheel driving electric vehicle, make the purpose of its smooth operation.

Fig. 2 and shown in Figure 3, the state of kinematic motion of chaufeur expectation, obviously vehicle keeps larger acceleration/accel when straight-line travelling, and about 7s, reach the speed of a motor vehicle of 70km/h, then after 8s, chaufeur makes automobile avoid 7 obstacles in the place ahead according to snakelike operating mode steering wheel rotation.

Can find out that from Fig. 4, Fig. 5 and Fig. 6 the controlling quantity of automobile torque distribution in the weight coefficient correction, in the anxious process rapidly of bearing circle, has improved weight coefficient C _Xi, reduced weight coefficient C _Yi, and in the process of motor turning, according to the positive and negative continuous correction C of steering direction and demand moment _XiValue; In the vehicle steering correction curve, different according to the wheel steering direction, all steering angle is revised, and in the process that the drive torque left and right wheels is distributed, all be that the drive torque of outboard wheels is greater than the drive torque of inboard wheel, extra yaw moment is provided, has guaranteed the stability that turns to.

Fig. 7 has represented the vehicle stability index, and obviously, this moment, the angle of roll of vehicle was little, vehicle is in stabilized conditions, and Fig. 8 and Fig. 9 have shown the state of kinematic motion of vehicle, and its longitudinal velocity begins quick increase, substantially remain unchanged during turning, cross velocity is according to the difference of turn direction, real-time change.

Figure 10 is the correlation curve of four-wheel driving electric vehicle torque distribution method of the present invention and average torque distribution method, all the torque distribution method is the mode of simulation orthodox car moment, the average torque distribution method was when keeping away latter two obstacle, vehicle is unstability, curve movement does not meet the chaufeur expectation fully, and the path of motion of four-wheel torque distribution algorithm satisfies chaufeur desired motion track always, is in stabilized conditions.

Although the above is described the illustrative specific embodiment of the present invention; so that those skilled in the art understand the present invention; but should be clear; the invention is not restricted to the scope of the specific embodiment; to those skilled in the art; as long as various variations appended claim limit and the spirit and scope of the present invention determined in, these variations are apparent, all utilize innovation and creation that the present invention conceives all at the row of protection.

Claims (2)

1. a four-wheel driving electric vehicle torque distribution method is characterized in that, may further comprise the steps:

(1), 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 _XiThe longitudinal force weight coefficient that represents i tire, c _YiThe side force weight coefficient that represents i tire, F _XiBe the longitudinal force of tire of i tire, F _YiBe the side force of tire of i tire, u _iBe the adhesion value of i tire, F _ZiBe the tyre load of i tire,

Be the wheel hub motor maximum drive moment of i tire, x is longitudinal direction, and y is lateral, and z is perpendicular to xy in-plane, min(*, *) represent to get the minimum value in the two, i=1,2,3,4 represent respectively the near front wheel, off front wheel, left rear wheel, off hind wheel;

(2), select the side force F of front left wheel _Y1With two longitudinal force F of left and right sides front-wheel _X1, F _X2Be unknown parameter, 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 wheel front-wheel two-wheeled spacing;

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}The vertically clean power of expression expectation, M _zThe needed yaw moment of expression vehicle stabilization, it is determined by the yaw moment control module;

(3), formula (2), (3), (4) are updated in the objective function (1), find the solution the minimal value of present objective function, respectively to the side force F of front-wheel left side wheel _Y1With two longitudinal force F of left and right sides front-wheel _X1, F _X2Ask local derviation and make it equal 0, can get:

$\frac{\&PartialD;J}{{\&PartialD;F}_{x1}}=2*(\frac{c_{x1}}{{\left(u_1{F}_{z1}\right)}^2}+\frac{c_{x3}}{\min \left({\left(u_3{F}_{z3}\right)}^2\right),{(T_3^{\max }/r_3)}^2)})*F_{x1}-$

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

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

$\frac{\&PartialD;J}{\&PartialD;F_{x2}}=2*(\frac{c_{x2}}{{\left(u_2{F}_{z2}\right)}^2}+\frac{c_{x4}}{\min ({\left(u_4{F}_{z4}\right)}^2,{(T_4^{\max }/r_4)}^2)})*F_{x2}+$

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

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

$\frac{\&PartialD;J}{{\&PartialD;F}_{y1}}=-2*\frac{l_f(u_1{F}_{z1}+u_2{F}_{z2})}{B*u_1{F}_{z1}}*\frac{c_{x3}}{\min ({\left(u_3{F}_{z3}\right)}^2,{(T_3^{\max }/r_3)}^2)}*F_{x1}+$

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

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

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

Solving equation (5), (6), (7) obtain the side force F of the front left wheel of required distribution _Y1With two longitudinal force F of left and right sides front-wheel _X1, F _X2, calculated again the side force F of the right front wheel of all the other required distribution by formula (2), (3), (4) _Y2With two longitudinal force F of left and right sides trailing wheel _X3, F _X4

(4), according to the side force F of front left wheel _Y1, right front wheel side force F _Y2And two longitudinal force F of left and right sides front-wheel _X1, F _X2Calculate driving torque and the steering wheel angle correction of left and right sides front-wheel, with two longitudinal force F of left and right sides trailing wheel _X3, F _X4Calculate the driving torque of left and right sides trailing wheel, then, with four driving torque control four-wheel driving electric vehicles.

(5), by the side force of wheel and longitudinal force can calculate four wheels respectively required moment be:

$M_i=\sqrt{M_{\mathrm{xi}}^2+M_{\mathrm{yi}}^2},$ Wherein, M _Xi=l _fF _XiM _Yi=l _fF _YiI=1,2,

M _i=M _xi=l _f·F _xi i=3,4，

Wherein, M _iThe moment of expression four wheels.

2. a four-wheel driving electric vehicle torque distribution method is characterized in that, the longitudinal force weight coefficient c of described i tire _Xi, side force weight coefficient c _YiFor:

$C_{\mathrm{xi}}=\left\{\begin{array}{c}0.5,m\&GreaterEqual;0.6\\ 0.3+0.1*(\frac{m}{0.6}+\frac{\left|n\right|}{0.8}),0\&le;m<0.6\\ 0.3+0.2*(\frac{\left|m\right|}{0.7}-\frac{\left|n\right|}{0.8}),-0.7<m<0\\ 0.5,m\&le;-0.7\end{array}\right.$

C _yi=1-C _xi；

Tire longitudinal center of gravity rate of transform m and side direction center of gravity rate of transform n are:

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

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

Wherein, l _fThe expression front-wheel is to the distance between the vehicle barycenter, and B represents wheel front-wheel two-wheeled spacing, l _rThe expression trailing wheel is to the distance between the vehicle barycenter;

The m positive dirction is larger, and expression car deceleration degree is larger, and automobile provides braking force, and automobile load shifts forward, and the m negative direction is larger, and the expression pickup is larger, and automobile provides propulsive effort, and automobile load shifts backward; The n positive dirction is larger, and the expression automobile degree of turning left is larger, and load shifts to the right, and the n negative direction is larger, and the expression automobile degree that bends to right is larger, and load shifts left; Automobile longitudinal center of gravity rate of transform m and side direction center of gravity rate of transform n are defined between-0.8 to 0.8, exceed this scope if calculate, and then get boundary value.

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