JPH04252736A - Differential limit device for drive wheel - Google Patents

Abstract

PURPOSE:To provide a differential limit device for drive wheels to produce consistently proper differential restraining torque without altering the turning performance of a vehicle. CONSTITUTION:A differential limit device consists of a restraining torque adjusting clutch which is assembled in a differential gear 90 to selectively adjust differential restraining torque, integrated wheel speed sensors 66, 97 to detect the speed of drive wheels 64, 65, a wheel speed difference computing section to compute the absolute value of a peripheral speed difference between the drive wheels 64, 65 in accordance with detected signals from these wheel speed sensors 66, 97, a differential restraining torque computing section to compute the differential restraining torque of a restraining torque adjusting clutch in accordance with the computed absolute value of the peripheral speed difference between the drive wheels 64, 65, and an electron control unit 15 to properly control the operation of the restraining torque adjusting clutch corresponding to output from the differential restraining torque computing section.

Description

Detailed Description of the Invention [0001] [Industrial Field of Application] The present invention is directed to
According to the difference in circumferential speed between a pair of drive wheels,
Differential restriction of drive wheels that allows you to arbitrarily change differential restriction torque
Regarding equipment. [0002] Generally, a type of vehicle that travels on the road via wheels
In type vehicles, there is a gap between the engine and the pair of left and right drive wheels.
A differential device is used to reduce the circumference of the drive wheels when the vehicle turns.
Allows for speed differences. [0003] For this reason, the road surface conditions differ greatly between the left and right sides.
For example, if there is a low-friction area such as an icy road on one side,
A road surface with a friction coefficient, and the other is a road surface such as dry asphalt.
When a vehicle is driving on a road surface with a high coefficient of friction,
The driving force of the transmission is exclusively reduced by the reduction of road resistance by the action of the differential.
As a result, the power is transmitted to the other drive wheel.
There is a risk that the wheels will spin and the vehicle will be unable to drive. [0004] In order to prevent such problems, conventional
restricts the function of the differential and directly connects the left and right drive wheels.
Differential restraints that hold the differential in place, or
The peripheral speed difference between the pair of left and right drive wheels is kept constant by providing bundle torque.
Assemble a differential limiting device, etc. that suppresses the following into the differential gear.
There is a lot of work going on. Problem to be Solved by the Invention: Incorporating a differential restraint device
In vehicles with
or, conversely, to activate the function of the differential.
It is necessary to operate the dynamic restraint device appropriately, which is cumbersome to operate.
Therefore, a certain degree of skill is required to operate this differential restraint device.
required. On the other hand, in the case of a vehicle incorporating a differential limiting device,
In this case, the differential locking torque of the conventional differential limiting device depends on the road surface condition.
The differential is constant regardless of the road surface condition, so it may be inappropriate depending on the road surface condition.
This may result in a restraining torque. In particular, road surfaces with a low coefficient of friction
A large differential constraint is installed in the differential limiting device in consideration of the fact that the
Having bundle torque makes it difficult for the vehicle to turn when turning on general road surfaces.
Problems such as difficulty in bending occur. [0007]Object of the invention: The present invention provides a pair of left and right
A differential is applied to these drive wheels according to the difference in circumferential speed of the drive wheels.
Vehicle turning performance can be improved by arbitrarily changing the restraint torque.
Always generates appropriate differential locking torque without compromising
The purpose is to provide a drive wheel differential limiting device that obtains
. [Means for Solving the Problems] Differential driving wheels according to the present invention
The motion limiter is installed between the engine and the pair of left and right drive wheels.
The difference between these drive wheels is
A locking torque adjustment clamp that allows you to arbitrarily adjust the dynamic locking torque.
and detect the speeds of the pair of left and right drive wheels.
A pair of wheel speed sensors and a pair of wheel speed sensors.
The absolute value of the circumferential speed difference of the drive wheels is determined based on the detection signals of the drive wheels.
In this wheel speed difference calculation section, the wheel speed difference calculation section calculates
Based on the absolute value of the calculated circumferential speed difference of the driving wheels.
Calculate the differential locking torque of the locking torque adjustment clutch.
A differential restraint torque calculation unit, and a differential restraint torque calculation unit
The movement of the locking torque adjusting clutch is controlled according to the output from the
It is equipped with an electronic control unit that controls the operation. [Operation] A pair of wheel speed sensors detect the speed of a pair of left and right drive wheels.
When each of these is detected, the wheel speed difference calculation section calculates the
Circumferential speed of the drive wheel based on the detection signal from the wheel speed sensor
The absolute value of the difference is calculated, and then this wheel speed difference calculating section calculates the difference.
Differential restraint is applied based on the absolute value of the circumferential speed difference of the driven wheels.
Differential restraint of clutch for restraint torque adjustment in torque calculation unit
Torque is calculated. Then, calculate this differential restraint torque
The electronic control unit operates according to the differential restraint torque calculated by the
The clutch controls the operation of the clutch for adjusting the restraining torque and
The differential restraint torque on the wheels is adjusted appropriately. [Embodiment] The drive wheel differential limiting device according to the present invention is applied to an engine.
Applied to front-wheel drive vehicles equipped with an output control device.
Figure 1 shows the concept of one embodiment and the schematic structure of the vehicle.
As shown in FIG. 2, the output shaft 12 of the engine 11 has hydraulic pressure.
An input shaft 14 of a type automatic transmission 13 is connected thereto. this oil
The pressure type automatic transmission 13 is operated by a selection (not shown) by the driver.
Engine 1 depending on Trevor's selected position and vehicle operating condition.
1. Electronic control unit (hereinafter referred to as “electronic control unit”) that controls the operating status of
(referred to as ECU) Based on the command from 15, the hydraulic control
A predetermined gear is automatically selected via the control device 16.
It's becoming a sea urchin. Specific details of this hydraulic automatic transmission 13
Regarding the structure, action, etc., see, for example, Japanese Patent Application Laid-Open No. 58-5427.
Already known in Publication No. 0 and Japanese Patent Application Laid-Open No. 61-31749, etc.
The hydraulic control device 16 includes a hydraulic automatic transmission.
Engagement operation of a plurality of frictional engagement elements constituting a part of the machine 13
A pair of shift controls (not shown) for performing the and opening operations
A solenoid valve for shift control is incorporated, and the solenoid valve for shift control is
The ECU 15 controls the on/off operation of energization.
Therefore, in this embodiment, any one of four forward speeds and one reverse speed can be selected.
This enables smooth gear shifting operations. The output shaft 63 of the hydraulic automatic transmission 13 and
There is a difference between the front wheels 64 and 65, which are a pair of left and right drive wheels.
A clamp for adjusting the restraint torque that can arbitrarily change the dynamic restraint torque.
A differential gear 90 incorporating a switch 89 is interposed. Book
In the embodiment, this restraint torque adjustment clutch 89
Adopts an electromagnetic clutch, and energizes this electromagnetic clutch.
Clutch coupling force, i.e. differential restraint, by changing the amount
The torque can be changed arbitrarily, but in addition to this,
A clutch that can arbitrarily change the coupling force using fluid pressure, etc.
Of course, it is also possible to adopt the following. [0012] Therefore, the restraint torque is adjusted according to the road surface condition.
Continuously changing the amount of current applied to the clutch 89
Expresses the relationship between the amount of energization and the differential locking torque at this time.
As shown in FIG. 3, the differential restraint torque of the front wheels 64 and 65
is continuously changed to optimize the rotation difference between front wheels 64 and 65.
can be controlled. In this embodiment, the steering force of the driver is reduced.
Therefore, a power steering device is incorporated into the steering mechanism.
Figure 4 shows the concept of this power steering device.
As shown, the pair of left and right front wheels 64, 65 are operated by a steering wheel.
a rack and pinion mechanism (not shown) connected to the handle 85;
, a power actuator connected to this rack and pinion mechanism.
A power steering device 92 composed of a motor 91 and
They are connected via tie rods 93, respectively. Said
The power actuator 91 includes a steering wheel 85 that is operated by a steering wheel 85.
Pressure oil for this power actuator 91 is
Hydraulic pump 95 via steering valve 94 that switches the flow of
is connected. Also, this engine driven by the engine 11
The hydraulic pump 95 and the power actuator 91 include
, a reservoir tank 96 for storing pressure oil is connected thereto. Therefore, the steering wheel 85 is controlled by the driver.
When the turning operation is performed, the power actuator is activated via the steering valve 94.
Flow of pressure oil from hydraulic pump 95 to tuator 91
is switched to correspond to the steering direction of the steering handle 85.
The steering force is applied to the rack and pinwheel via the power actuator 91.
As a result of being transmitted to the on-mechanism, the front wheels 64, 6 are turned on with light steering force.
5 is designed to be steered. Intake pipe connected to combustion chamber 17 of engine 11
In the middle of the intake pipe 18, there is an intake pipe formed by the intake pipe 18.
By changing the opening degree of the air passage 19, air is supplied into the combustion chamber 17.
A throttle valve 20 is incorporated to adjust the amount of intake air.
A throttle body 21 is interposed. Figure 1 and cylindrical
An enlarged cross-sectional structure of the part of the throttle body 21 that forms the
As shown in FIG. 5, the throttle body 21 has a
Both sides of the throttle shaft 22 to which the throttle valve 20 is fixed integrally
The end portion is rotatably supported. into the intake passage 19.
At one end of this throttle shaft 22, there is an accelerator lever.
The bar 23 and the throttle lever 24 are fitted coaxially.
are combined. The throttle shaft 22 and the accelerator lever 2
A bush 26 and a spacer 27 are provided between the cylindrical portion 25 of No. 3 and
is interposed, thereby causing the accelerator lever 23 to slide into the slot.
It is rotatable about the torque shaft 22. In addition, slot
Washer 28 and nut attached to one end of the torque shaft 22
29 from the throttle shaft 22 to the accelerator lever 23
This prevents it from falling off. Also, this accessory
The cable receiver 30, which is integrated with the lever 23, has a
Therefore, the operated accelerator pedal 31 connects the cable 32.
The amount of depression of the accelerator pedal 31
The accelerator lever 23 moves relative to the throttle shaft 22 according to the
It is designed to rotate. On the other hand, the throttle lever 24 is
It is fixed integrally with the torque shaft 22, so this slot
By operating the torque lever 24, the throttle valve 2
0 rotates together with the throttle shaft 22. Also, accelerator
A collar 33 is coaxially integrated with the cylindrical portion 25 of the bar 23.
It is fitted into the tip of the throttle lever 24.
is engaged with a claw portion 34 formed on a part of this collar 33.
A stopper 35 is formed. These claw portions 34 and
The stopper 35 refers to a stopper 35 that is a throttle in the direction in which the throttle valve 20 opens.
Rotate the throttle lever 24 or use the throttle valve
When the accelerator lever 23 is rotated in the direction in which the
They are set in such a positional relationship that they lock with each other when they are connected. [0018] The throttle body 21 and the throttle lever
A stopper of the throttle lever 24 is provided between the bar 24 and the stopper of the throttle lever 24.
35 to the claw portion 3 of the collar 33 that is integrated with the accelerator lever 23.
4 to bias the throttle valve 20 in the direction of opening.
The torsion coil spring 36 is fitted onto the throttle shaft 22.
This spring is connected via a pair of cylindrical spring receivers 37 and 38.
It is mounted coaxially with the throttle shaft 22. or,
A stopper pin 39 protruding from the throttle body 21 and
The claw portion of the collar 33 is also located between the accelerator lever 23 and the accelerator lever 23.
34 against the stopper 35 of the throttle lever 24.
to bias the throttle valve 20 in the direction of closing, and press the accelerator pedal.
Twisting to give detent feeling to dull 31
The coil spring 40 is connected to the accelerator lever via the collar 33.
- 23 has a cylindrical portion 25 coaxial with the throttle shaft 22.
It is installed. At the tip of the throttle lever 24, there is a
The base end is fixed to the diaphragm 42 of the actuator 41.
The distal end portions of the control rods 43 are connected to each other. This actu
The pressure chamber 44 formed in the ether 41 has the torsion
The stopper of the throttle lever 24 together with the coil spring 36
35 against the claw part 34 of the collar 33 and remove the throttle valve.
A compression coil spring 45 is incorporated to bias 20 in the opening direction.
It is rare. And the springs of these two springs 36 and 45
The spring force of the torsion coil spring 40 is greater than the sum of the spring forces.
This causes the accelerator pedal 31 to be set larger.
Make sure that the throttle valve 20 does not open unless the
It has become. Connected to the downstream side of the throttle body 21
surge tank 46 that forms part of the intake passage 19
A vacuum tank 48 is connected via a connecting pipe 47.
This vacuum tank 48 and connecting piping 47
Between the vacuum tank 48 and the surge tank 4
A check valve 49 that only allows air to move to 6 is interposed.
ing. As a result, the pressure inside the vacuum tank 48 is
Set to a negative pressure almost equal to the lowest pressure in the surge tank 46
be done. [0021] The inside of these vacuum tanks 48 and the above-mentioned AC
The pressure chamber 44 of the tuator 41 is connected via piping 50.
It is in a energized state, and there is a part in the middle of this piping 50 when it is not energized.
A closed-type first torque control solenoid valve 51 is provided.
Ru. In other words, this torque control solenoid valve 51 has piping 50.
A spring biases the plunger 52 against the valve seat 53 so as to close the valve seat 53.
54 is included. [0022] Furthermore, the first torque control solenoid valve 51 and
A throttle valve is connected to the piping 50 between the actuator 41 and the actuator 41.
Piping 55 communicating with the intake passage 19 on the upstream side of the valve 20
is connected. There is no communication in the middle of this piping 55.
A second torque control solenoid valve 56 of an electrically open type is provided.
ing. In other words, this torque control solenoid valve 56 has piping.
Spring 5 biasing plunger 57 to open 55
8 is included. [0023] The two torque control solenoid valves 51 and 56
The ECU 15 is connected to each of the ECU 1.
Torque control solenoid valves 51, 56 based on commands from 5
The on/off of energization is controlled by duty.
It has become. For example, the torque control solenoid valves 51 and 56
When the duty rate is 0%, the pressure of the actuator 41
The chamber 44 is located in the intake passage 19 on the upstream side of the throttle valve 20.
The atmospheric pressure becomes almost equal to the pressure inside the throttle valve 20.
The opening degree corresponds one-to-one to the amount of depression of the accelerator pedal 31.
respond. Conversely, the duty of the torque control solenoid valves 51 and 56 is
When the tee rate is 100%, the pressure chamber of the actuator 41
44 is a negative pressure approximately equal to the pressure inside the vacuum tank 48
As a result, the control rod 43 is pulled up diagonally to the left in Figure 1.
As a result, the throttle valve 20 is depressed when the accelerator pedal 31 is depressed.
It is closed regardless of the amount of penetration, and the driving torque of the engine 11 is
It will be in a state where it is forcibly reduced. In this way,
Adjust the duty rate of the torque control solenoid valves 51 and 56.
By this, it is related to the amount of depression of the accelerator pedal 31.
The opening degree of the throttle valve 20 is changed without changing the opening degree of the throttle valve 20, and the engine 11 is driven.
The dynamic torque can be adjusted arbitrarily. In addition, in this embodiment, the opening degree of the throttle valve 20
simultaneously with the accelerator pedal 31 and actuator 41
However, there are two slots in the intake passage 19.
Arrange throttle valves in series and use one throttle valve to accelerate
Connect only to the pedal 31 and connect the other throttle valve.
Connect only to actuator 41 and connect these two slots.
It is also possible to control each torque valve independently. On the other hand, on the downstream end side of the intake pipe 18, there is a
Fuel (not shown) is injected into the combustion chamber 17 of the engine 11
The fuel injection nozzle 59 of the injection device is connected to each cylinder (main cylinder) of the engine 11.
In the example, a four-cylinder internal combustion engine is assumed)
The duty control is controlled by the ECU 15.
Fuel is supplied to the fuel injection nozzle 59 via the controlled solenoid valve 60.
supplied to In other words, the opening time of the solenoid valve 60 is controlled.
By this, the amount of fuel supplied to the combustion chamber 17 is adjusted.
, the spark plug 61 in the combustion chamber 17 reaches a predetermined air-fuel ratio.
It is designed to be ignited by [0027] The ECU 15 has a
Crank angle setting for detecting the engine speed NE
sensor 62 and a pair of left and right front wheels 64 and 65 which are driving wheels.
Circumferential speed (hereinafter referred to as front wheel speed) VFL, VF
A pair of front wheel rotation sensors 66 and 97 that respectively detect R
and the throttle is attached to the throttle body 21.
The opening degree of the lever 24 (hereinafter referred to as the throttle opening degree)
) a throttle opening sensor 67 that detects θT;
An idle switch that detects the fully closed state of the throttle valve 20
In addition to the air cleaner 69 at the tip of the intake pipe 18,
The amount of air that is assembled into the combustion chamber 17 of the engine 11 and flows into the combustion chamber 17 of the engine 11
An air flow sensor 70 such as a Karman vortex flowmeter that detects
, it is assembled into the engine 11 and the cooling water temperature of this engine 11 is controlled.
Assemble the water temperature sensor 71 to detect and the middle of the exhaust pipe 72
detects the temperature of the exhaust gas flowing through the exhaust passage 73.
exhaust temperature sensor 74 and ignition key switch 7
5, and the left and right sides (not shown) of the power actuator 91
The power steering wheel is installed in a pair of pressure chambers, respectively.
The operating pressure of the steering device 92 (hereinafter referred to as power steering pressure)
) a pair of pressure sensors 98 for detecting PS;
99 is connected. [0028] These crank angle sensors 62 and
Front wheel rotation sensors 66, 97 and throttle opening sensor 6
7 and idle switch 68 and air flow sensor 70
and water temperature sensor 71, exhaust temperature sensor 74 and ignition
from the function key switch 75 and pressure sensors 98 and 99.
The output signals of each are now sent to the ECU 15.
ing. [0029] Also, calculate the target drive torque of the engine 11.
Driving force control unit (hereinafter referred to as TCL)
76 includes the throttle opening sensor 67 and the idler
It is attached to the throttle body 21 together with the switch 68.
the opening of the accelerator lever 23 (hereinafter referred to as the accelerator)
Accelerator opening sensor that detects θA (referred to as opening)
77 and the circumferential speed of the pair of left and right rear wheels 78 and 79 which are driven wheels.
degree (hereinafter referred to as rear wheel speed) VRL, VRR
Rear wheel rotation sensors 80 and 81 detecting each, and vehicle 8
The steering shaft 83 at the time of turning is based on the straight-ahead state of 2.
The turning angle (hereinafter referred to as the steering shaft turning angle) δH is
Steering angle sensor 84 to detect and steering integrated with steering shaft 83
The normal phase of the steering wheel 85 every 360 degrees (when the vehicle 82 is approximately
(This includes the phase that causes a straight traveling state) δN
The steering shaft reference position sensor 86 to be detected is connected, and these
The output signals from sensors 77, 80, 81, 84, 86 are
Each is sent to the TCL 76. [0030] The ECU 15 and TCL 76 are connected via a communication cable.
The engine speed is connected to the ECU 15 via a
The rotation number NE and the detection signal from the idle switch 68, etc.
In addition to information on the operating state of the engine 11, front wheel speeds VFL, VF
R and road surface friction coefficient (hereinafter referred to as road surface μ)
etc. is sent to the TCL 76. On the other hand, TCL76?
are the target drive torque TO calculated by this TCL76.
and the traveling speed of the vehicle 82 (hereinafter referred to as vehicle speed)
) In addition to V, the steering shaft turning angle δH and the ignition timing retardation ratio
Information regarding the failure is sent to the ECU 15. In this embodiment, the front wheels 66, 9, which are the driving wheels,
Absolute value of circumferential speed difference (front wheel speed difference) of 7 |VFL-VFR|
The differential restraint torque of the restraint torque adjustment clutch 89 is
Control for setting torque TF (hereinafter, this is referred to as differential restraint torque
control of the front wheels 64, 65).
When the slip amount s in the direction becomes larger than the preset amount,
In this case, the drive torque of the engine 11 is reduced to improve maneuverability.
control to ensure energy security and prevent energy loss (hereinafter referred to as
This is called slip control) Engine 1
1 target drive torque TOS and the target drive torque generated in the vehicle during turning.
Lateral acceleration (hereinafter referred to as lateral acceleration) G
When Y is greater than or equal to a preset value, engine 11
The vehicle 82 deviates from the turning path by reducing the driving torque of
(hereinafter referred to as swing control)
), the target drive torque TOC of the engine 11 is
The TCL76 calculates these two target drives.
Optimal final target drive torque T from torques TOS and TOC
Select O and reduce the driving torque of engine 11 as necessary.
We are trying to reduce it. Also, via the actuator 41
Even when the throttle valve 20 is fully closed, the engine 11
The ignition timing should be set in consideration of the case where the output cannot be reduced in time.
Set the target retardation amount po to quickly increase the drive torque of the engine 11.
We are making it possible to reduce this to The rough outline of the control according to this embodiment is as follows.
As shown in Figure 6, which shows the flow, in this example, slip control is
The target drive torque TOS of the engine 11 when the control is performed,
Target drive torque TO of engine 11 when turning control is performed
C is always calculated in parallel with TCL76, and these two
The optimal final target drive is determined from the target drive torques TOS and TOC.
Select the dynamic torque TO and determine the required driving torque of engine 11.
It is reduced depending on the driving condition of the vehicle 82 and the road surface μ.
The differential locking torque of the locking torque adjusting clutch 89 is adjusted accordingly.
I am trying to change TF. Specifically, the ignition key switch
The control program of this embodiment is started by the ON operation of 75.
In M1, first, the initial value δm(0) of the steering shaft turning position is determined.
Read, reset various flags, or sample this control.
Main timer count every 15 milliseconds which is the pulling cycle
Initial settings such as starting are performed. [0034] Then, M2 receives detection signals from various sensors.
Based on the number, TCL76 calculates vehicle speed V, etc.
The neutral position δM of the steering shaft 83 is learned and corrected using M3.
Ru. The neutral position δM of the steering shaft 83 of this vehicle 82 is E
It is stored in a memory (not shown) in the CU15 or TCL76.
Since the ignition key switch 75 is not
The initial value δm(0) is read each time the ON operation is performed, and the vehicle 8
The learning supplement will only be applied if 2 satisfies the straight running conditions described later.
corrected, and the ignition key switch 75 is turned off.
This initial value δm(0) is corrected by learning until
It has become. Next, TCL76 uses M4 to adjust the front wheel rotation sensor.
Detection signals from sensors 66, 97 and rear wheel rotation sensors 80, 8
The driving torque of the engine 11 is determined based on the detection signal from 1.
Target drive torque TO when performing regulating slip control
S is calculated, and M5 is used to calculate the signal from the rear wheel rotation sensors 80 and 81.
Based on the detection signal and the detection signal from the steering angle sensor 84
When the swing control is performed to regulate the drive torque of the engine 11,
The target driving torque TOC of the engine 11 is calculated. [0036] Then, in M6, TCL76
The optimal final target drive torque is determined from the target drive torques TOS and TOC.
The method described below is mainly concerned with safety.
Choose from more. Furthermore, when starting suddenly or when the road surface is normally dry,
If there is a sudden change from a road to an icy road, the actuator
Even when the throttle valve 20 is fully closed via the throttle valve 41,
There is a possibility that the output reduction of engine 11 will not be done in time, so M7
Based on the rate of change Gs of the slip amount s of the front wheels 64 and 65 at
Based on this, the retardation ratio for correcting the basic retardation amount pB is
select. Next, at M8, the ECU 15 controls the front wheel rotation setting.
The restraining torque is determined based on the detection signals from the sensors 66 and 97.
Setting the differential restraint torque TF of the adjustment clutch 89
After performing differential restraint torque control, these final steps are performed using M9.
The retardation ratio of the target drive torque TO and the basic retardation amount pB
While outputting related data to the ECU 15, the differential restraint torque
The amount of current corresponding to the torque TF is read out. [0038]The driver then presses a manual switch (not shown).
If you wish to perform slip control or turning control by operating
In this case, the ECU 15 determines that the driving torque of the engine 11 is at this final stage.
A pair of torque controllers are used for torque control so that the target driving torque TO
Controls the duty rate of the solenoid valves 51 and 56, and also controls the basic delay.
Based on the data regarding the retardation ratio of the angle amount pB, this E
Calculate the target retard amount pO in the CU15 and set the ignition timing P.
While delaying by the target retard amount pO as necessary, the differential
The amount of energization corresponding to the restraining torque TF is restricted from the ECU 15.
The bundle torque is applied to the clutch 89 for adjusting the bundle torque, thereby increasing the torque of the vehicle.
I try to make the 82 run smoothly and safely. [0039] It should be noted that the driver presses a manual switch (not shown).
If you do not want to perform slip control or turning control by operating
, the ECU 15 has a pair of torque control solenoid valves 51 and 5.
As a result of setting the duty rate of 6 to the 0% side, the vehicle 82 is
The notification corresponds to the amount of depression of the accelerator pedal 31 by the driver.
It will be in normal operating condition. In this case, in this example, differential restraint is applied.
Torque control is linked to slip control.
Therefore, the differential restraint torque of the restraint torque adjustment clutch 89
TF becomes 0 and the differential device 90 continues to function.
It happens. In this way, the driving torque of the engine 11 is set to M1.
At 0, every 15 milliseconds is the main timer's sampling period.
control until the countdown ends, and from then on M
Steps from 2 to M11 to the ignition key
This is repeated until the switch 75 is turned off. [0041] By the way, the turning control is performed in step M5.
When calculating the target drive torque TOC of the engine 11 by
, TCL76 detects a pair of rear wheel rotation sensors 80 and 81.
If the vehicle speed V is calculated based on the signal using the formula (1) below,
Both the front wheels 6
Calculate the steering angle δ of 4,65 using the formula (2) below, and at this time
The target lateral acceleration GYO of the vehicle 82 is calculated from the formula (3) below.
I'm looking for each. V=(VRL+VRR)/2
　　　　　　　　　　　　　　　　　　　　　　　　　
...(1) δ=δH /ρH
　　　　　　　　　　　　　　　　　　　　　　　　　
...(2) G
YO=δ/ω・(A+1/V2)
...(3) However
where ρH is the steering gear gear ratio, and ω is the wheel of the vehicle 82.
base, A is the stability factor of the vehicle 82, which will be described later.
It is. As is clear from this equation (3), the vehicle
Toe-in adjustment of front wheels 64 and 65 was performed during maintenance of 82.
due to changes over time such as wear of the steering gear (not shown).
If the neutral position δM of the steering shaft 83 changes,
The turning position δm of the steering shaft 83 and the front wheels 64, which are the steered wheels,
A deviation occurs between the actual steering angle δ of 65 and the actual steering angle δ. As a result
, to accurately calculate the target lateral acceleration GYO of the vehicle 82.
There is a risk that it will not be possible to perform the turning control properly.
It becomes difficult. Moreover, in the present invention, the step of M4 is
Cornering drag correction (described later) during lip control
The means controls the engine 11 based on the turning angle δH of the steering shaft 83.
The standard drive torque of
μ estimating means determines the road surface based on the turning angle δH of the steering shaft 83.
Since μ is estimated, slip control and differential restraint are
There is also a possibility that torque control may not be performed well. like this
Therefore, the neutral position δM of the steering shaft 83 is set by the step M3.
It is necessary to perform learning correction in the program. The neutral position δM of the steering shaft 83 is determined by learning compensation.
As shown in FIG. 7, FIG. 8, and FIG. 9 showing the correcting procedure,
In TCL76, the turning control flag FC is set at H1.
Determine whether or not. And this H1 stage
When it is determined that the vehicle 82 is under turning control,
The output of the engine 11 learns the neutral position δM of the steering shaft 83.
There is a risk that a sudden change may occur due to the correction, worsening the ride comfort.
etc., the learning correction of the neutral position δM of the steering shaft 83
Do not do this. On the other hand, the vehicle 82 turns at step H1.
If it is determined that the control is not in progress, the steering shaft 83 is set to neutral.
Even if the learning correction of the position δM is performed, no problem will occur, so
, TCL76 receives detection signals from rear wheel rotation sensors 80 and 81.
Based on the above, the neutral position δM is learned in H2 and will be described later.
Calculate vehicle speed V for turning control using equation (1) above.
do. Next, TCL76 rear wheel speed VRL, VR in H3
R difference (hereinafter referred to as rear wheel speed difference) | VRL-
After calculating VRR｜, TCL76 adjusts the steering axis base at H4.
The reference position δN of the steering shaft 83 is determined by the quasi-position sensor 86.
Learning correction of the neutral position δM was performed in the detected state.
Whether or not the reference position δN of the steering shaft 83 is detected.
The steering angle neutral position learned flag FHN is set when
Determine whether or not it is true. Turning on the ignition key switch 75
Immediately after operation, the steering angle neutral position learned flag FHN is set.
In other words, the neutral position δM is learned for the first time.
Then, the steering shaft turning position δm(n) calculated this time in H5
is equal to the previously calculated steering shaft turning position δm(n-1).
Determine whether or not. At this time, due to driver's hand shake etc.
The steering axis is determined by the steering angle sensor 84 so as not to be affected by the steering angle sensor 84.
For example, set the rotation detection resolution of 83 to around 5 degrees.
This is desirable. [0046] The operation calculated this time in this step H5
The rudder shaft turning position δm(n) is the previously calculated steering shaft turning position
If it is determined that the position is equal to δm(n-1), then in H6
to determine whether the vehicle speed V is greater than a preset threshold value VA.
judge. This operation allows the vehicle 82 to reach a certain high speed.
Otherwise, the rear wheel speed difference due to steering |VRL-VRR| etc.
This is necessary because it cannot be detected, and the threshold value VA
is based on the running characteristics of the vehicle 82, etc., and is determined by experiment etc.
For example, the speed is set as appropriate, such as 10 km/h. [0047] Then, in step H6, the vehicle speed V reaches the threshold value.
If it is determined that it is equal to or higher than VA, the TCL76
Example of rear wheel speed difference |VRL-VRR| set in advance in 7.
Is it smaller than a threshold value VX, such as 0.3 km/h?
In other words, it is determined whether the vehicle 82 is traveling straight.
Ru. Here, the reason why the threshold value VX is not set to 0 km/h is that
If the tire pressures of the right rear wheels 78 and 79 are not equal
, even though the vehicle 82 is traveling straight, the pair of left and right
The circumferential speeds VRL and VRR of the rear wheels 78 and 79 are different and the car
Avoid determining that both 82 are not traveling straight.
It's for a reason. Note that the left and right rear wheels 78 and 79 are empty.
If the air pressures are not equal, the rear wheel speed difference | VRL - VRR
| tends to increase in proportion to vehicle speed V, so this
For example, the threshold value VX is mapped as shown in Figure 10.
Then, read the threshold value VX based on the vehicle speed V from this map.
You can also put it out. [0049] At this H7 step, rear wheel speed difference | VRL
If it is determined that -VRR| is less than or equal to the threshold VX,
At H8, the steering shaft reference position sensor 86 detects the reference position of the steering shaft 83.
It is determined whether the position δN is detected. and,
In this step H8, the steering shaft reference position sensor 86
The reference position δN of the rudder shaft 83 is detected, that is, the vehicle 8
If it is determined that 2 is traveling straight, TC is applied at H9.
The first learning timer (not shown) built into L76
Start counting. [0050] Next, TCL76 completed this first process in H10.
Whether 0.5 seconds have passed since the learning timer started counting.
In other words, whether the vehicle 82 continued to go straight for 0.5 seconds.
The first learning timer starts counting.
If 0.5 seconds have not elapsed since then, the vehicle speed V is set at H11.
It is determined whether or not is larger than the threshold value VA. This H
In step 11, it is determined that the vehicle speed V is greater than the threshold value VA.
In case of disconnection, rear wheel speed difference | VRL - VRR at H12
Whether | is less than a threshold value VB such as 0.1 km/h
Determine. Rear wheel speed difference at this H12 step | VRL - VRR |
is below the threshold value VB, that is, the vehicle 82 is traveling straight
If it is determined that
Open the stored second learning timer (not shown).
start [0051] Then, in H14, this second learning tie was created.
Whether or not 5 seconds have elapsed since the start of the count of vehicle 8.
Determine whether the straight-ahead state of 2 continues for 5 seconds, and
If 5 seconds have not passed since the learning timer started counting,
If so, return to step H2 and repeat this step H2.
The operations from step H14 to step H14 are repeated. [0052] At step H8 in the middle of this iterative operation,
The steering shaft reference position sensor 86 detects the reference position δN of the steering shaft 83.
is detected, and in step H9 the above-mentioned
The first learning timer starts counting, and in H10 this timer starts counting.
0.5 seconds have passed since the first learning timer started counting.
In other words, it was determined that the vehicle 82 was traveling straight for 0.5 seconds.
In this case, the reference position δN of the steering shaft 83 is changed at H15.
The steering angle neutral position learned flag FHN in the detected state
the reference position δN of the steering shaft 83 at H16.
Steering angle neutral position learned flag FH when is not detected
Determine whether or not is set. Also, the above H1
Does the second learning timer start counting in step 4?
Even if it is determined that 5 seconds have passed since the
to the top. [0053] With the above operations, the reference position of the steering shaft 83 is still
Steering angle neutral position learned frame when position δN is not detected
Since the lug FH is not set, this H16
The reference position δN of the steering shaft 83 is not detected at step
The steering angle neutral position learned flag FH is set in
In other words, the reference position δN of the steering shaft 83 is not detected.
It is determined that the learning of the neutral position δM in this state is the first time.
, in H17, the current steering shaft turning position δm(n) was changed to a new one.
This is regarded as the neutral position δM(n) of the steering shaft 83, and this is the TC
Read it into the memory in L76 and set the reference position of the steering shaft 83.
Steering angle neutral position learned frame when position δN is not detected
Set the lug FH. [0054] In this way, the new neutral position of the steering shaft 83 is achieved.
After setting the position δM(n), the neutral position of this steering shaft 83
Calculate the turning angle δH of the steering shaft 83 based on the position δM
On the other hand, the count of the learning timer was cleared in H18.
Then, the steering angle neutral position learning is performed again. [0055] Note that the current value calculated in step H5 above is
The steering axis turning position δm(n) is the steering axis that was calculated last time.
If it is determined that the turning position is not equal to δm (n-1),
, in step H11, the vehicle speed V is not greater than the threshold value VA.
In other words, the rear wheel speed difference |V calculated in step H12
If it is determined that RL-VRR｜ is unreliable, or
At step H12, the rear wheel speed difference |VRL-VRR| is the threshold
If it is determined that the value is larger than VB, both cars
Since both 82 are not moving straight, the step of H18 is
to the top. [0056] Also, in step H7, the rear wheel speed difference |V
If it is determined that RL-VRR| is larger than the threshold VX
In this case, the steering shaft reference position sensor 86 is detected in step H8.
It is determined that the reference position δN of the steering shaft 83 has not been detected.
If so, in H19 the counter of the first learning timer
Clear the list and move on to step H11, but the previous step
If the vehicle speed V is below the threshold value VA in step H6,
Even when it is determined that the vehicle 82 is traveling straight,
Since this is not possible, the process moves to step H11. On the other hand, in step H4, the steering shaft 83
Rudder angle neutral position with the reference position δN detected
The completed flag FHN is set, that is, the neutral position δ
If it is determined that learning M is for the second time or later, H
At 20, the steering shaft reference position sensor 86 detects the reference position of the steering shaft 83.
It is determined whether the position δN is detected. and,
In this step H20, the steering shaft reference position sensor 86 is
It was determined that the reference position δN of the steering shaft 83 was detected.
In this case, the vehicle speed V is the preset threshold value VA in H21.
Determine whether the value is greater than or not. [0058] In this step H21, the vehicle speed V reaches the threshold value V.
If it is determined that the value is A or higher, TCL76 will be set to H2.
2, the rear wheel speed difference |VRL-VRR| is equal to the threshold value VX.
In other words, is the vehicle 82 traveling straight?
judge whether And in this H22 step
Rear wheel speed difference |VRL-VRR| is smaller than threshold value VX
If it is determined that the steering axis rotation calculated this time in H23
The turning position δm(n) is the previously calculated steering shaft turning position δm
(n-1) is determined. This H23 star
The steering shaft turning position δm(n) calculated this time at step
Equal to the previously calculated steering shaft turning position δm(n-1)
If it is determined, the first learning timer will be set in H24.
Start counting. [0059] Next, TCL76 completed this first process in H25.
Whether 0.5 seconds have passed since the learning timer started counting.
In other words, whether the vehicle 82 continued to go straight for 0.5 seconds.
0 from the start of the first learning timer count.
．． If 5 seconds have not passed, proceed to step H2 above.
Go back and repeat steps H2-H4 and H20-H25.
Go back. On the contrary, in this H25 step, the first learning
It was determined that 0.5 seconds had passed since the timer started counting.
If so, proceed to step H16. [0060] Note that in step H20, the steering axis base
The quasi-position sensor 86 detects the reference position δN of the steering shaft 83
If it is determined that the car has not been
The speed V is not greater than the threshold value VA, that is, step H22.
The reliability of the rear wheel speed difference |VRL-VRR| calculated by
If it is determined that there is no
It is determined that the speed difference |VRL-VRR| is larger than the threshold value VX.
In the case of disconnection or the operation calculated this time in the step of
The rudder shaft turning position δm(n) is the previously calculated steering shaft turning position
If it is determined that the position is not equal to δm(n-1), then
If there is a difference, the process also proceeds to step H18. [0061] In step H16, the steering angle neutral position is determined.
The completed flag FH is set, that is, the neutral position
If it is determined that δM is learned for the second time or later, TC
L76 has the current steering shaft turning position δm(n) in H26.
Equal to the previous neutral position δM(n-1) of the steering shaft 83,
That is, it is determined whether δm(n)=δM(n-1). And the current steering axis rotation
The position δm(n) is the previous neutral position δM(n) of the steering shaft 83.
-1), if it is determined that it is equal to
The next steering angle neutral position learning is performed. [0062] In step H26, the current steering axis rotation is
The rotation position δm(n) was changed from the previous time due to play in the steering system, etc.
is not equal to the neutral position δM(n-1) of the steering shaft 83 of
If it is determined, in this embodiment, the current steering shaft turning position δm
(n) is changed to the new neutral position δM(n) of the steering shaft 83.
), and the absolute value of these differences is a preset correction.
If the difference is more than the limit amount Δδ, the previous steering axis rotation
Reduce this correction limit amount Δδ for rotation position δm (n-1).
The calculated or added value is the new neutral position δ of the steering shaft 83.
M(n) and read it into the memory in TCL76
That's what I do. [0063] In other words, TCL76 changed its current operation in H27.
From the rudder shaft turning position δm(n) to the previous neutral position of the steering shaft 83
The value obtained by subtracting the value δM(n-1) is the preset negative correction.
It is determined whether the limit amount is smaller than the limit amount -Δδ. and,
The value subtracted in this H27 step is the negative correction limit amount
- If it is determined that it is smaller than Δδ, the new
The neutral position δM(n) of the steering shaft 83 is set to the previous steering shaft.
83 neutral position δM (n-1) and negative correction limit amount -Δδ
From this, change δM(n) = δM(n-1) - Δδ, and the learning correction amount per time will unconditionally increase to the negative side.
I take care not to make it too loud. [0064] This causes the steering angle to change due to some reason.
Even if an abnormal detection signal is output from the sensor 84,
The neutral position δM of the steering shaft 83 does not change suddenly;
It is possible to quickly respond to abnormalities. On the other hand, the value subtracted in step H27 is
If it is determined that the negative correction limit amount - Δδ is larger than
, from the current steering shaft turning position δm(n) to the previous time in H29.
The value obtained by subtracting the neutral position δM(n-1) of the steering shaft 83 is
It is determined whether or not it is larger than the positive correction limit amount Δδ. So
Then, the value subtracted in this H29 step is a positive correction.
If it is determined that the limit amount Δδ is larger than the limit amount Δδ, the
to set the new neutral position δM(n) of the steering shaft 83 to the previous steering position.
Neutral position δM(n-1) of shaft 83 and positive correction limit amount Δδ
From this, change δM(n) = δM(n-1) + Δδ, and the learning correction amount per time will unconditionally increase to the positive side.
I take care not to make it too loud. [0066] This causes the steering angle to change due to some reason.
Even if an abnormal detection signal is output from the sensor 84,
The neutral position δM of the steering shaft 83 does not change suddenly;
It is possible to quickly respond to abnormalities. [0067] However, the value subtracted in step H29 is
If it is determined that it is smaller than the positive correction limit amount Δδ,
In H31, the current steering shaft turning position δm(n) was changed to a new steering position.
Read it as is as the neutral position δM(n) of the rudder shaft 83
. In this way, in this embodiment, the inside of the steering shaft 83 is
When learning and correcting the standing position δM, the rear wheel speed difference |VRL-V
In addition to using only RR|, the steering shaft reference position sensor 86
A method is adopted in which detection signals from the vehicle 82 are also used.
The neutral position δ of the steering shaft 83 is reached relatively quickly after the vehicle starts moving.
M can be learned and corrected, and the steering axis reference position
Rear wheel speed difference even if sensor 86 fails for some reason | VRL
−VRR｜Learning and supplementing the neutral position δM of the steering shaft 83
can be corrected and has excellent safety. Therefore, the front wheels 64 and 65 remain in the turning state.
When the stopped vehicle 82 starts, the steering at this time
FIG. 11 shows an example of the changing state of the neutral position δM of the shaft 83
As shown in FIG. 2, learning control of the neutral position δM of the steering shaft 83
When is the first time, the steering axis rotation in step M1 mentioned above
The amount of correction from the initial value δm(0) of the rotation position is very large.
However, the neutral position δM of the steering shaft 83 from the second time onwards
By the operations in steps H17 and H19,
It becomes suppressed. In this way, the neutral position δM of the steering shaft 83
After learning and correcting, the detection signal from the front wheel rotation sensor 66
based on the number and the detection signals from the rear wheel rotation sensors 80 and 81.
performs slip control to regulate the driving torque of the engine 11.
Calculate the target drive torque TOS for the case. By the way, the coefficient of friction between the tires and the road surface is
The rate of change in vehicle speed V applied to both 82 (hereinafter referred to as longitudinal acceleration)
can be considered equivalent to GX (referred to as degree).
Therefore, in this example, this longitudinal acceleration GX is expressed as rear wheel rotation.
This is calculated based on the detection signals from the sensors 80 and 81.
Engine 11 standard corresponding to the maximum value of longitudinal acceleration GX of
The drive torque TB is detected from the front wheel rotation sensor 66.
front wheel speed VF corresponding to the vehicle speed V and the target front wheel speed VF corresponding to the vehicle speed V.
The deviation from O (hereinafter referred to as slip amount) s
The target drive torque TOS is calculated based on the correction. [0072] Calculate the target drive torque TOS of this engine 11.
Figures 12 and 13 show the calculation blocks for
First, TCL76 controls the vehicle speed VS for slip control.
is based on the detection signals from the rear wheel rotation sensors 80 and 81.
However, in this embodiment, the low vehicle speed selection unit 101 calculates the second
The smaller of the two rear wheel speeds VRL and VRR is set to
High vehicle speed selection
In part 102, the higher of the two rear wheel speeds VRL and VRR is determined.
Select this value as the second vehicle speed VS for slip control.
Then, use the selector switch 103 to select between the two options.
Change which output from parts 101 and 102 is to be imported.
It is now possible to select. [0073] In this embodiment, the low vehicle speed selection section 101
The first vehicle speed VS selected is the two rear wheel speeds VRL.
, VRR, the smaller value VL is given by equation (1) above.
Weighting coefficient KV corresponding to vehicle speed V calculated from
is multiplied by the multiplier 104, and this and the two rear wheel speeds VRL
, VRR to the larger value VH (1-KV)
is multiplied by the multiplication unit 105 and added.
I'm looking for more. That is, the slip control actually causes the engine 11 to
The state in which the driving torque of the
When the service flag FS is set, the changeover switch is
103, the smaller of the two rear wheel speeds VRL and VRR is determined.
The driver selects the higher value as the vehicle speed VS.
Even if control is desired, the driving torque of engine 11 is reduced.
In other words, the slip control flag FS is reset.
In the set state, one of the two rear wheel speeds VRL and VRR is
The larger value is now selected as the vehicle speed VS.
There is. [0075] This is because the driving torque of the engine 11 is reduced.
The driving torque of the engine 11 is reduced from the state where the
At the same time as making it difficult to transition to the state, the transition in the opposite case
This is to make it more difficult. For example, while the vehicle 82 is turning
The smaller value of the two rear wheel speeds VRL and VRR at
If you select VS as the vehicle speed, the front wheels 64 and 65 will be
Slip occurs even though lip does not occur.
The driving torque of engine 11 is reduced.
To avoid such malfunctions and to ensure the running safety of the vehicle 82.
Considering the performance, the driving torque of the engine 11 is temporarily reduced.
This is because we have taken care to ensure that this situation continues in the event that
be. [0076] Also, the low vehicle speed selection section 101 selects the vehicle speed VS.
When calculating, the smaller of the two rear wheel speeds VRL and VRR.
The weighting coefficient KV is added to the value VL of the lower value in the multiplier 104.
Multiply this by the greater of the two rear wheel speeds VRL and VRR.
Multiplying unit 105 adds (1-KV) to the listening value VH.
For example, adding the product multiplied by
When driving on a turning path with a small radius of curvature, such as a right or left turn,
Average value of front wheel speed VFL, VFR (VFL+VFR)/2
and the smaller value VL of the two rear wheel speeds VRL and VRR.
As a result, the feedback
The correction amount of the drive torque is too large, and the vehicle 82
This is because there is a possibility that acceleration performance may be impaired. In this embodiment, the weighting coefficient K
(1) above, where V is the average value of rear wheel speeds VRL and VRR
Based on the vehicle speed V in the equation, read from a map as shown in Figure 14.
I try to let it out. [0078] For slip control calculated in this way
The longitudinal acceleration GX is calculated based on the vehicle speed VS of
, First, the vehicle speed VS(n) calculated this time and the vehicle speed calculated previously
From the vehicle speed VS (n-1), the current longitudinal acceleration of the vehicle 82
The degree GX(n) is calculated by the differential calculation unit 106 as shown in the following formula.
do. GX(n)={Vs(n)-Vs(n-1)}/3.6
・Δt・g However, Δt is the sampling period of this control.
15 milliseconds, g is the gravitational acceleration. [0079] Then, the calculated longitudinal acceleration GX(n)
If the value exceeds 0.6g, please take precautions against calculation errors, etc.
Considering safety, the maximum value of this longitudinal acceleration GX(n) is
At the clip part 107, the front and back should not exceed 0.6g.
Clip the acceleration GX(n) to 0.6g. In addition,
The filter unit 108 performs filter processing to remove noise.
and calculate the corrected longitudinal acceleration GXF. [0080] This filter processing is performed by
The degree GX(n) is equivalent to the coefficient of friction between the tire and the road surface.
Since it can be considered that the longitudinal acceleration of the vehicle 82 is
The maximum value of GX(n) changes and the tire slip rate S increases.
Target slippage corresponding to the maximum value of the friction coefficient between the tire and the road surface
If you are about to deviate from the top rate SO or its vicinity
However, the tire slip rate S is the coefficient of friction between the tire and the road surface.
target slip ratio SO corresponding to the maximum value of the number or its
To keep the value smaller than this in the neighborhood, add before and after.
This is to correct the speed GX(n), specifically
is performed as follows. [0081] The current longitudinal acceleration GX(n) is filtered.
Previously corrected longitudinal acceleration GXF(n-1) or more
In the case of , that is, when the vehicle 82 continues to accelerate,
GXF(n) = 28・Σ{GX(n)−GXF(n−
1) Remove noise by delay processing as }/256
The corrected longitudinal acceleration GXF(n) can be adjusted longitudinally relatively quickly.
It follows the speed GX(n). [0082] The current longitudinal acceleration GX(n) is the previous correction
If the longitudinal acceleration is less than GXF (n-1), that is, the vehicle
When 82 is not accelerating much, main timer sampling
The following processing is performed for each programming period Δt. [0083] The slip control flag FS is set.
In other words, the drive torque of the engine 11 is not controlled by slip control.
If the torque is not reduced, the vehicle 82 is decelerating.
Therefore, GXF(n) = GXF(n-1) -0.002
to suppress the decrease in the corrected longitudinal acceleration GXF(n) and improve driving performance.
ensuring responsiveness to a request for acceleration of the vehicle 82 by a person.
There is. [0084] Also, the drive torque of the engine 11 is controlled by slip control.
When the slip amount s is positive while reducing the torque, that is, the slip amount s is positive.
Even when wheels 64 and 65 are slipping to some extent, the car
Since both 82 are decelerating, there is no problem with safety.
, GXF(n) = GXF(n-1) -0.002
This suppresses the decline in the corrected longitudinal acceleration GXF, allowing the driver to
Responsiveness to requests for acceleration of the vehicle 82 is ensured. Furthermore, the engine 11 is driven by slip control.
Front wheels 64 and 65 slip while reducing torque
When the amount s is negative, that is, the vehicle 82 is decelerating, the correction
The maximum value of longitudinal acceleration GXF is maintained, and the vehicle is controlled by the driver.
82 to ensure responsiveness to acceleration requests. Similarly, the engine 11 is driven by slip control.
by the hydraulic control device 16 while the dynamic torque is being reduced.
During the upshift of the hydraulic automatic transmission 13, the driver
Due to the need to ensure a sense of acceleration, the corrected front and rear acceleration GXF
Keep the maximum value of . [0087] Then, the filter section 108 removes noise.
The corrected longitudinal acceleration GXF is sent to the torque conversion unit 109.
This is converted into torque using the torque conversion unit 109.
Naturally, the calculated value should be a positive value.
Therefore, calculation errors are prevented in the clip section 110.
After clipping this to 0 or more for the purpose, running resistance calculation section
The running resistance TR calculated in step 111 is sent to adding section 112.
and further based on the detection signal from the steering angle sensor 84.
Calculated by the cornering drag correction amount calculation unit 113
Adding section adds cornering drag correction torque TC
The standard drive torque is added at 114 and shown in the formula (4) below.
Calculate TB. TB = GFO・Wb ・r+TR +TC
・
...(4) However, Wb is the vehicle weight, r is the front wheel 6
It has an effective radius of 4.65. The running resistance TR is as a function of the vehicle speed V.
However, in this example, it is calculated as shown in FIG.
I'm looking for it from the map. In this case, flat road and uphill road
Since the running resistance TR is different, the actual map is not shown in the map.
For flat roads shown by lines and for uphill roads shown by two-dot chain lines.
A tilt sensor (not shown) installed in the vehicle 82
Select one based on the detection signal from the sensor.
However, the running resistance, including downhill slopes, etc.
It is also possible to set up anti-TR. Furthermore, in this embodiment, the cornering drag
Calculate the correction torque TC from the map shown in Figure 16.
This allows engine 1 to approximate the actual running condition.
1 standard drive torque TB can be set, and turning
The standard drive torque TB of engine 11 immediately after becomes large.
Therefore, the acceleration rate of the vehicle 82 after exiting the turning path is
Feeling improves. Note that the group calculated by the above formula (4)
For the semi-driving torque TB, in this example, the variable clip
By setting the lower limit value in section 115, this reference drive is
The final correction torque TPID, which will be described later, is obtained from the dynamic torque TB.
The value subtracted by the subtraction unit 116 will be negative.
This prevents such problems. This standard drive torque TB
As shown in the map shown in Figure 17, the lower limit of
in stages according to the elapsed time from the start of lip control.
I'm trying to lower it. On the other hand, TCL76 is a front wheel rotation sensor 66,
Based on the detection signal from 97, the actual front wheel speed VF is determined.
Average value of left and right front wheel speeds VFL and VFR (VFL+VF
R)/2, and as mentioned earlier, this front wheel speed VF
and vehicle speed VS for slip control.
Correction torque calculation set based on target front wheel speed VFO
Using the slip amount s, which is the deviation from the target front wheel speed VFS,
, performs feedback control of the reference drive torque TB.
The target drive torque TOS of the engine 11 is calculated by
put out By the way, when the vehicle 82 accelerates, the engine 11
In order to make the generated drive torque work effectively, it is necessary to
As shown by the solid line in the middle, the tires of the front wheels 64 and 65 are
The slip rate S of the tire is the coefficient of friction between the tire and the road surface.
Target slip ratio SO corresponding to the maximum value or its vicinity
Adjust the value to be smaller than this, and set the energy
In addition to avoiding losses in vehicle 82, the handling performance and acceleration performance
It is desirable to avoid damage. Here, the slip rate S of the tire is S=(
VF - V)/V, and the target slip ratio SO is 0 depending on the road surface condition.
．． It is known that it can vary in the range of 1 to 0.25.
. Therefore, while the vehicle 82 is running, it is approximately 10% of the road surface.
A slip amount s is generated in the front wheels 64 and 65, which are the driving wheels.
It is desirable to Considering the above points, set the target front wheel speed.
The VFO is set in the multiplier 117 according to the following formula. VFO=1.1·V [0094]Then, the TCL76 is the acceleration correction section 118.
From the map shown in Figure 19, the acceleration before and after the above correction is calculated.
Read the slip correction amount VK corresponding to degree GXF,
The adder 119 adds this to the target front wheel speed V for calculating the reference torque.
Add to FO. This slip correction amount VK is before correction
Increases gradually as the value of rear acceleration GXF increases
However, in this example, we conducted a driving test.
This map has been created based on experiments, etc. [0095] As a result, the target front wheel speed for calculating the correction torque
As VFS increases, the slip rate S during acceleration decreases as shown in Figure 18.
At or near the target slip ratio SO shown by the solid line in the middle
It is set to a value smaller than this. On the other hand, the friction between the tires and the road surface during turning
Figure 18 shows the relationship between the friction coefficient and the slip rate S of this tire.
As shown by the dash-dotted line in the middle, the relationship between the tires and the road during turning is
The tire slip rate, which is the maximum value of the coefficient of friction with the surface, is
The maximum value of the coefficient of friction between the tires and the road surface while driving straight.
be considerably smaller than the target slip rate SO of the tire.
I understand. Therefore, while the vehicle 82 is turning, this vehicle 82
To ensure smooth turning, the target front wheel speed VFO is adjusted from when going straight.
It is desirable to set the value to a small value. Therefore, the turning correction section 120 performs the actual operation shown in FIG.
Corresponding to the target lateral acceleration GYO from the map shown by the line
The slip correction amount VKC is read out and subtracted by the subtraction unit 1.
21, subtract from the target front wheel speed VFO for calculating the reference torque.
Ru. However, the ON operation of the ignition key switch 75
The first neutral position δM of the steering shaft 83 is determined after
Until the learning is carried out, the turning angle δH of the steering shaft 83 is not reliable.
Since there is no
The broken line in Figure 20 is based on the lateral acceleration GY that actually acts.
Read the slip correction amount VKC from the map shown in
put out. By the way, the target lateral acceleration GYO is the steering
Based on the detection signal from the angle sensor 84, the above formula (2) is calculated.
Calculate the rudder angle δ, and use this rudder angle δ to calculate the above (3).
In addition to finding the neutral position δM of the steering shaft 83 using the formula,
Xi is making corrections. Therefore, the steering angle sensor 84 or the steering axis reference
When an abnormality occurs in the position sensor 86, the target lateral acceleration GY
It is possible that O will be a completely incorrect value. Therefore, the operation
If an abnormality occurs in the steering angle sensor 84, etc., the rear wheel speed difference
The actual value generated in the vehicle 82 using |VRL-VRR|
Calculate the lateral acceleration GY and use this as the target lateral acceleration GYO.
Use instead. Specifically, this actual lateral acceleration GY is
TCL76 from rear wheel speed difference | VRL - VRR | and vehicle speed V
The lateral acceleration calculation unit 122 built into the unit calculates the following formula (5).
The filter section 123 filters the noise
The corrected lateral acceleration GYF that has been subjected to noise removal processing is used. GY=｜VRL-VRR｜・V/3.62
・b・g ...(5)
However, b is the tread of the rear wheels 78 and 79, and
In the router section 123, the lateral acceleration GY(n) calculated this time and the front
From the corrected lateral acceleration GYF(n-1) calculated twice, this time
The corrected lateral acceleration GYF(n) of
Low-pass processing is performed by calculation. GYF(n) =Σ20・{GY(n)−GYF(n−
1) }/256 [0101] The steering angle sensor 84 or
Or whether an abnormality has occurred in the steering shaft reference position sensor 86.
For example, TCL7 is detected by the disconnection detection circuit shown in FIG.
It can be detected at 6. In other words, the steering angle sensor 8
4 and the output of the steering shaft reference position sensor 86 are connected to the resistor R.
and ground it with capacitor C.
Input the output as it is to the A0 terminal of TCL76 and use various controls.
At the same time, it is connected to the A1 terminal through the comparator 88.
I am inputting it. The negative terminal of this comparator 88 has a reference voltage of 4.
A specified value of 5 volts is applied, and the steering angle sensor 84 is
If the wire is disconnected, the input voltage of the A0 terminal will exceed the specified value and the controller will be disconnected.
The parator 88 turns on and the input voltage at the A1 terminal continues.
The signal becomes high level H. Therefore, the input voltage of A1 terminal is
If is at a high level H for a certain period of time, for example 2 seconds, the wire will be disconnected.
It is determined that these steering angle sensors 84 or the steering shaft reference position
The TCL 76 is configured to detect the occurrence of an abnormality in the position sensor 86.
The program has been set. [0102] In the embodiment described above, the operation is performed by hardware.
Although it was designed to detect abnormalities in the steering angle sensor 84, etc., the software
Of course, it is also possible to detect the abnormality using software.
. For example, an example of this abnormality detection procedure is shown below.
As shown in FIG. 22, the TCL 76 first performs the
An abnormality is determined by the disconnection detection shown in 1.
If it is determined that there is no
and determines whether there is an abnormality in the rear wheel rotation sensors 80, 81.
to be determined. In this step W2, each rotation sensor 66, 8
If it is determined that there is no abnormality in 0 and 81, operate it in W3.
The rudder shaft 83 rotates more than one turn in the same direction, for example, more than 400 degrees.
Determine whether or not the vehicle has been steered. This step of W3
It was determined that the rudder shaft 83 was steered more than 400 degrees in the same direction.
In this case, the steering is performed from the steering shaft reference position sensor 86 at W4.
Whether there was a signal informing the reference position δN of the axis 83
to judge. [0104] Then, in this step W4, the steering shaft 8
If it is determined that there is no signal informing the reference position δN of 3.
If the steering shaft reference position sensor 86 is normal, the steering shaft reference position sensor 86 is normal.
The signal informing the reference position δN of the rudder shaft 83 is at least
It must have happened once, so the steering angle sensor 84 is abnormal at W4.
It is determined that the abnormality is occurring and sets the abnormality flag FW.
Ru. [0105] In step W3, the steering shafts 83 are the same.
If it is determined that the vehicle is not being steered more than 400 degrees in the direction,
Alternatively, the reference position δN of the steering shaft 83 is determined in step W4.
There was a signal from the steering shaft reference position sensor 86 indicating that
If it is determined that the steering shaft neutral position δM is
Whether learning has been completed or not, i.e. two steering angle neutral position learning
At least one of completed flags FHN and FH is set.
Determine whether or not. [0106] At this step W6, the steering shaft 83
When it is determined that the neutral position δM of has been learned,
For example, in W7, the rear wheel speed difference |VRL-VRR| is
exceeds 1.5 km, and the vehicle speed V is 20 km/h at W8.
m and 60km/h, and at this time in W9
For example, the absolute value of the turning angle δH of the steering shaft 83 is less than 10 degrees.
, that is, the vehicle 82 is turning at a certain speed.
If it is determined that the steering angle sensor 84 is not functioning normally.
, the absolute value of the turning angle δH is 10 degrees or more.
Therefore, the steering angle sensor 84 is set at W10.
It is determined that there is an abnormality. [0107] Note that the above corresponding to the target lateral acceleration GYO
The slip correction amount VKC is determined by the driver's steering wheel 85.
Since additional cuts can be considered, this target lateral acceleration GYO is
In small areas, the slit corresponding to the modified lateral acceleration GYF
This is set to be smaller than the drop correction amount VKC. Also, vehicle speed
Ensure acceleration of the vehicle 82 in a region where V is small.
is desirable, and conversely, if this vehicle speed V exceeds a certain speed,
, it is necessary to consider the ease of turning, so from Fig. 20
The read slip correction amount VKC corresponds to the vehicle speed V.
Read the correction coefficient from the map shown in Figure 23 and multiply it.
By this, the corrected slip correction amount VKF is calculated.
. [0108] As a result, the target front wheel speed for calculating the correction torque
VFO decreases and slip ratio S when turning becomes lower when going straight
becomes smaller than the target slip ratio SO at
Although the acceleration performance of 82 is slightly reduced, it has good turning performance.
Secured. [0109] These target lateral acceleration GYO and actual lateral acceleration
As shown in FIG. 24 showing the selection procedure of the speed GY,
L76 is for calculating the slip correction amount VKC at T1.
The corrected lateral force from the filter section 123 is expressed as the lateral acceleration of
Adopts speed GYF and sets slip control flag F at T2
Determine whether S is set. [0110] In this step T2, the slip control
If it is determined that the lag FS is set, the above
The corrected lateral acceleration GYF is adopted as is. This is a pickpocket
This is the standard for determining the slip correction amount VKC during slip control.
Convert the lateral acceleration from the corrected lateral acceleration GYF to the target lateral acceleration GY
When changing to O, the slip correction amount VKC changes greatly.
This is because there is a possibility that the behavior of the vehicle 82 may be disturbed. [0111] In step T2, the slip control
If it is determined that the lag FS is not set, T
3, two steering angle neutral position learned flags FHN, FH
Determine whether one of the
Ru. Here, two steering angle neutral position learned flags FHN,
If it is determined that neither FH is set,
Again, the above-mentioned modified lateral acceleration GYF is adopted as is.
. Also, in this step T3, two steering angle neutral positions are learned.
Either completed flag FHN or FH is set.
If it is determined that there is a slip correction amount VKC at T4.
The target lateral acceleration GY is used as the lateral acceleration for calculating
Adopt O. [0112] As a result of the above, the target front wheel speed for calculating the correction torque
VFS is as shown in the formula below. VFS=VFO+VK-VKF Next, from the detection signal of the front wheel rotation sensor 66,
The results obtained by filtering for the purpose of noise removal etc.
Front wheel speed VF and the target front wheel speed VF for calculating the correction torque
The subtraction unit 124 calculates the slip amount s, which is the deviation from S.
do. If this slip amount s is less than or equal to a negative set value, e.g.
For example, if the speed is -2.5 km/h or less, the slip amount s
and -2.5km/hour is clipped by the clip part 125.
The slip amount s after this clipping process will be described later.
Performs proportional correction to prevent overcontrol in this proportional correction.
to prevent output hunting from occurring. [0114] Also, the slip amount s before this clipping process is
Then, perform integral correction using the integral constant ΔTi, which will be described later.
Then, perform differential correction to obtain the final correction torque TPID.
calculate. [0115] As the proportional correction, the multiplier 126
The basic correction amount is obtained by multiplying the slip amount s by the proportional coefficient KP.
Then, the multiplier 127 calculates the value of the hydraulic automatic transmission 13.
The correction coefficient ρKP set in advance by the gear ratio ρm is
The proportional correction torque TP is obtained by multiplication. In addition, proportional
The coefficient KP depends on the slip amount s after clipping.
The data is read from the map shown in 25. [0116] Also, as the integral correction, the slip amount s is
Integral calculation is performed to achieve correction that corresponds to gradual changes.
A basic correction amount is calculated in the section 128, and a calculation is made for this correction amount.
Then, the multiplier 129 calculates the gear ratio ρ of the hydraulic automatic transmission 13.
Multiply by a preset correction coefficient ρKI based on m
, an integral correction torque TI is obtained. In this case, this implementation
In the example, a constant minute integral correction torque ΔTI is integrated.
Therefore, the slip amount s is calculated every 15 ms sampling period.
If it is positive, add the minute integral correction torque ΔTI.
, conversely, if the slip amount s is negative, the minute integral correction torque
ΔTI is subtracted. [0117] However, this integral correction torque TI depends on the vehicle speed.
The lower limit value TI as shown in the map of FIG. 26 is variable depending on V.
L is set, and by this clipping process, the vehicle 82 is
Large integral correction torque when starting, especially when starting on an uphill slope.
TI is activated to secure the driving force of the engine 11, and the vehicle 8
On the contrary, after the vehicle speed V increases after the start of step 2, the correction becomes larger.
If it is too high, control stability will be lost, so the integral correction torque TI
is made smaller. It also improves control convergence.
In order to
gm, and the integral correction torque is set by this clipping process.
Luk TI changes as shown in FIG. [0118] Proportional correction torque calculated in this way
TP and the integral correction torque TI are added by the adding section 130.
Then, calculate the proportional-integral correction torque TPI. [0119] Note that the correction coefficients ρKP and ρKI are hydraulic pressure
Preset in association with the gear ratio ρm of the automatic transmission 13
Read from the map shown in Figure 28.
There is. [0120] Also, in this embodiment, the differential operation section 131 performs step
Calculate the rate of change Gs of the lip amount s, and apply the differential coefficient K to this.
D is multiplied by the multiplier 132 to calculate the sudden change in the slip amount s.
Calculate the basic correction amount for And to this
Set upper and lower limits for each value obtained.
Make sure that the differential correction torque TD does not become an extremely large value.
The clipping unit 133 performs clipping processing to
The corrected torque TD is obtained. This clip part 133
are wheel speeds VF, VRL, VRR while the vehicle 82 is running.
may change momentarily depending on the road surface condition, the driving condition of the vehicle 82, etc.
This may cause the machine to idle or become locked.
When the rate of change Gs of the slip amount s becomes extremely positive or negative,
If the value becomes large, there is a risk that the control will diverge and the responsiveness will decrease.
For example, clip the lower limit to -55kgm.
At the same time, the upper limit is clipped to 55 kgm, and differential correction is performed.
To prevent torque TD from becoming an extremely large value,
It's a special thing. [0121] Then, in the adding section 134, these proportional
Add integral correction torque TPI and differential correction torque TD
The final correction torque TPID obtained by this is reduced.
The calculation unit 116 subtracts it from the above-mentioned reference drive torque TB,
Furthermore, in the multiplier 135, the engine 11 and the axles of the front wheels 64, 65 are
By multiplying the reciprocal of the total reduction ratio between 89 and 90,
The target drive torque for slip control is shown in equation (6) below.
Calculate the total TOS. TOS=(TB-TPID)/ρm・ρ
d・ρT...
(6) However, ρd is the differential gear reduction ratio, and ρT is
This is the torque converter ratio, and the hydraulic automatic transmission 13 is
When performing a top shift operation, after the shift is completed,
The gear ratio ρm on the high speed gear side is output.
. In other words, the upshift speed change of the hydraulic automatic transmission 13
In the case of operation, the gear shift on the high gear side is performed at the time the gear shift signal is output.
If the ratio ρm is adopted, it is clear from equation (6) above that
As the target drive torque TOS increases during gear shifting, the engine
Since 11 is blown up, a signal to start shifting is output.
After the shift operation is completed, for example, 1.5 seconds,
Shifting to the lower gear side that can reduce the target drive torque TOS
The ratio ρm is held, and after outputting the signal to start shifting, 1
．． After 5 seconds, the gear ratio ρm on the high speed side is adopted. similar
For this reason, the downshift change of the hydraulic automatic transmission 13
In the case of speed operation, the lower gear side is changed at the time the gear shift signal is output.
The speed ratio ρm is immediately adopted. [0122] Target drive torque calculated using the above formula (6)
Should TOS be a positive value as a matter of course?
In order to prevent calculation errors in the clip section 136,
Clip the target drive torque TOS to 0 or more to control slippage.
Start/end determination unit for determining the start or end of the control
According to the determination process at step 137, this target drive torque TO
Information regarding S is output to the ECU 15. [0123] The start/end determination section 137 performs the following (a) ~
(e) Slip control is applied when all the conditions shown in
control is judged to have started, and sets the slip control flag FS.
and slips the output from the low vehicle speed selection section 101.
Switch to select vehicle speed VS for control.
switch 103 to determine the target drive torque TOS.
Outputs information to ECU 15 to determine end of slip control
until the slip control flag FS is reset.
continues this process. (a) The driver operates a manual switch (not shown) to
I want lip control. (b) The driving torque Td requested by the driver is
The minimum driving torque required to run the 82, e.g.
It is 4 kgm or more. [0124] In this embodiment, this required driving torque T
d is calculated based on the detection signal from the crank angle sensor 62.
The engine speed NE and the accelerator opening sensor 76
The accelerator opening degree θA calculated from the detection signals of
29, which is preset based on the map shown in Figure 29.
It's protruding. (c) The slip amount s is 2 km/h or more. (d) The rate of change Gs of the slip amount s is 0.2g or more.
be. (e) The actual front wheel speed VF is time differentiated by the differential calculation unit 138.
The calculated actual front wheel acceleration GF is 0.2g or more. On the other hand, the start/end determination section 137
After determining the start of drop control, perform the following (f) and (g).
If any of the conditions shown in
Determines that the slip control has ended and resets the slip control flag FS.
Set the target drive torque TOS for the ECU 15.
The transmission is stopped and the output from the high vehicle speed selection section 102 is
Switch to select vehicle speed VS for slip control.
Activate the changeover switch 103. (f) Target drive torque TOS is required drive torque Td
above, and the slip amount s is a constant value, for example, −
2km or less for a certain period of time, e.g. 0.5 seconds or more
continuing. (g) Idle switch 68 turns from off to on.
state, that is, the driver released the accelerator pedal 31.
The state continues for a certain period of time, for example, 0.5 seconds or more. [0126] In the vehicle 82, slip control is performed by the driver.
A manual switch (not shown) is provided to select the
The driver operates this manual switch to control the slippage.
If you select Slip Control, operate the slip control as described below.
I do. In addition, in this example, part of this slip control
The differential restraint torque control of the present invention is performed by
Therefore, the differential restraint torque described later is applied only when slip control is in progress.
It is now possible to perform torque control. FIG. 3 shows the flow of this slip control process.
As shown in 0, the TCL75 has the various devices described above in S1.
Target drive torque TOS is determined by data detection and calculation processing.
is calculated, but this calculation operation is based on the operation of the manual switch mentioned above.
is done regardless of. [0128] Next, in S2, first the slip control flag is set.
It is determined whether FS is set or not, but initially
Since the slip control flag FS is not set.
, TCL76 has the slip amount s of front wheels 64 and 65 in S3.
is greater than a preset threshold, e.g. 2 km/h?
Determine whether or not. [0129] In this step S3, the slip amount s is
If it is determined that the speed is greater than 2km, TCL76 will move to S4.
The rate of change Gs of the slip amount s is greater than 0.2g at
Determine whether or not. [0130] In this step S4, the slip amount change rate
If Gs is determined to be larger than 0.2g, TCL7
6, the driver's requested drive torque Td is 82 at S5.
The minimum driving torque required to run the vehicle, e.g. 4k
gm , that is, whether the driver drives the vehicle 82 or not.
It is determined whether there is an intention to run the vehicle. [0131] In this step S5, the required drive torque T
d is greater than 4 kgm, i.e. the driver is driving the vehicle 82
If it is determined that there is an intention to run the
Set the slip control flag FS and set slip control in S7.
Determine again whether or not the serving flag FS is set.
do. [0132] In this step S7, the slip control
If it is determined that the lug FS is being set, S8
(6) above as the target drive torque TOS of the engine 11.
Target drive torque for slip control calculated in advance using the formula
Adopt TOS. [0133] Also, slip control is performed in step S7.
When it is determined that the medium flag FS has been reset
In S9, TCL76 is set as the target drive torque TOS.
Outputs the maximum torque of the engine 11, which causes the ECU 15 to
sets the duty rate of the torque control solenoid valves 51 and 56 to 0%.
As a result of lowering the engine 11 to the side, the engine 11 is
Generates driving torque according to the amount of depression of the pedal 31
. [0134] Furthermore, in step S3, the front wheels 64, 65
If it is determined that the slip amount s is smaller than 2 km/h,
or in step S4, the slip amount change rate Gs
is smaller than 0.2g, or if S5
At the step, the required drive torque Td is less than 4 kgm.
If it is determined that the size is small, proceed directly to step S7 above.
In step S9, the TCL76 starts the target drive.
The maximum torque of the engine 11 is output as the torque TOS, and this
This causes the ECU 15 to control the torque control solenoid valves 51 and 56.
As a result of reducing the duty rate to 0% side, the engine 11 is operated.
Driving according to the amount of depression of the accelerator pedal 31 by the driver.
Generates dynamic torque. [0135] On the other hand, in step S2, the slip control is
When it is determined that the service flag FS is set,
In S10, the slip amount s of the front wheels 64 and 65 is as described above.
-2km/h or less, which is the threshold value, and the required drive torque Td
is below the target drive torque TOS calculated in S1
It is determined whether or not the period continues for 0.5 seconds or more. [0136] In this step S10, the slip amount s is
The target driving torque is less than 2 km/h and the required driving torque Td is
The condition below the target drive torque TOS continues for 0.5 seconds or more.
i.e. the driver has already requested acceleration of the vehicle 82.
If it is determined that this is not the case, the slip control flag FS is set in S11.
is reset, and the process moves to step S7. [0137] In step S10, the slip amount s is
Greater than 2km/h or required drive torque Td
remains below the target drive torque TOS for more than 0.5 seconds.
The driver does not want to accelerate the vehicle 82.
If it is determined that there is, TCL76 will idle in S12.
The switch 68 is on, that is, the throttle valve 20 is fully closed.
It is determined whether it continues for 0.5 seconds or more. [0138] In this step S12, the idle switch is
If the driver determines that the switch 68 is on, the driver presses the accelerator.
Since the pedal 31 is not depressed, the step S11 is
and reset the slip control flag FS.
do. Conversely, it is determined that the idle switch 68 is off.
In this case, the driver must depress the accelerator pedal 31.
Therefore, the process returns to step S7. [0139] It should be noted that the driver's method of selecting slip control is
If the automatic switch is not operated, the TCL76 will operate as described above.
Calculate the target drive torque TOS for slip control as follows.
Target drive of engine 11 when turning control is performed after
Calculate torque. By the way, the lateral acceleration GY of the vehicle 82 is
Using the wheel speed difference |VRL-VRR|, use the formula (5) above.
Although it can be more practically calculated, the steering shaft turning angle δH
The lateral acceleration acting on the vehicle 82 by utilizing
Since it is possible to predict the value of degree GY, quick control can be performed.
It has the advantage of being able to [0141] Therefore, when controlling the turning of this vehicle 82,
TCL76 determines the vehicle speed based on the steering shaft turning angle δH and the vehicle speed V.
Calculate the target lateral acceleration GYO of 82 using the formula (3) above.
However, the vehicle 82 does not undergo extreme understeering.
The acceleration in the longitudinal direction of the vehicle body, that is, the target longitudinal acceleration G
XO is set based on this target lateral acceleration GYO. So
Then, the engine 11 corresponding to this target longitudinal acceleration GXO
Calculate target drive torque TOC. FIG. 31 shows the calculation block of this turning control.
And as shown in FIG.
At 0, the vehicle speed is determined from the output of the pair of rear wheel rotation sensors 80 and 81.
V is calculated using equation (1) above, and the steering angle sensor
The steering angle δ of the front wheels 64 and 65 is determined based on the detection signal from 84.
is calculated from equation (2) above, and the target lateral acceleration calculation unit 14
1, the target lateral acceleration GYO of the vehicle 82 at this time is expressed as (
3) Calculate from the formula. In this case, the area where the vehicle speed V is small
For example, when the speed is below 23 km/h, turning control is performed.
It is better to prohibit turning control, for example, at intersections with heavy traffic.
Sufficient acceleration can be obtained when turning left or right at points, making it safer.
In this example, since it is often convenient in terms of gender,
The correction coefficient multiplier 142 calculates the correction coefficient K as shown in FIG.
Y is multiplied by the target lateral acceleration GYO according to the vehicle speed V.
Ru. By the way, the learning of the steering shaft neutral position δM is
If not, the target lateral acceleration is determined based on the steering angle δ.
From the point of view of reliability, it is important to calculate the degree GYO using formula (3).
Since there is a problem, learning of the steering shaft neutral position δM is performed.
It is desirable not to start turning control until the deer
Immediately after the vehicle 82 starts traveling, it starts traveling on a curved road.
In this case, the vehicle 82 will be in a state requiring turning control.
, the learning start condition for the steering shaft neutral position δM is easily met.
This causes a problem in which the rotation control does not start.
There is a possibility that Therefore, in this embodiment, the steering shaft neutral position δ
Until the learning of M is performed, the changeover switch 143
In the above, the modification from the filter section 123 based on equation (5) is performed.
Enables turning control using normal lateral acceleration GYF
There is. In other words, the two steering angle neutral position learned flags FHN
, FH are reset, the switch is turned off.
Modified lateral acceleration GYF is adopted by switching switch 143
Then, the two steering angle neutral position learned flags FHN and FH are set.
If at least one of the
The target side from the correction coefficient multiplier 142 is set by the switch 143.
Acceleration GYO is selected. [0144] Furthermore, the stability factor A mentioned above is
As is well known, the structure of the suspension system of the vehicle 82 and the characteristics of the tires
Alternatively, it is a value determined by road surface conditions, etc. in particular
, the actual lateral acceleration that occurs in the vehicle 82 during a steady circular turn.
GY and the steering angle ratio δH /δH of the steering shaft 83 at this time
O (lateral acceleration with reference to the neutral position δM of the steering shaft 83
The steering shaft 83 in an extremely low speed running state where GY is close to 0.
Turning of the steering shaft 83 during acceleration with respect to the turning angle δHO
For example, as shown in FIG.
It is expressed as the slope of the tangent line in a graph. wife
, a region where the lateral acceleration GY is small and the vehicle speed V is not very high.
In this case, the stability factor A is approximately a constant value (A=0.
002), but if the lateral acceleration GY is 0.6g
When the stability factor A increases rapidly, the vehicle becomes 82
shows an extremely strong understeering tendency.
Ru. [0145] From the above, dry paved roads
FIG. 34 corresponding to the surface (hereinafter referred to as a high μ road)
, the stability factor A is set to 0.0.
02, and the vehicle 82 calculated by equation (3).
engine so that the target lateral acceleration GYO is less than 0.6g.
11 drive torque is controlled. [0146] Note that slippery roads such as frozen roads (
(hereinafter referred to as a low μ road), the stability
For example, it is sufficient to set the tifactor A to around 0.005.
stomach. In this case, on a low μ road, the actual lateral acceleration GY is
Since the target lateral acceleration GYO has a larger value, the target lateral acceleration
Acceleration GYO has a preset threshold value, for example (GYF-2
), and determine whether the target lateral acceleration GYO is greater than
If it is larger than this threshold, the vehicle 82 is running on a low μ road.
Determines that the vehicle is currently underway, and controls turning for low μ roads as necessary.
All you have to do is Specifically, based on equation (5) above,
Add 0.05g to the calculated corrected lateral acceleration GYF.
As a result, the target lateral acceleration GYO is higher than the preset threshold value.
Whether it is large or not, that is, the actual lateral acceleration GY on low μ roads
Since the target lateral acceleration GYO has a larger value than
Determine whether the target lateral acceleration GYO is greater than this threshold.
and if the target lateral acceleration GYO is larger than the threshold,
, it is determined that the vehicle 82 is traveling on a low μ road.
. [0147] In this way, the target lateral acceleration GYO is calculated.
If so, determine the size of this target lateral acceleration GYO and the vehicle in advance.
Target longitudinal acceleration G of the vehicle 82 set according to the speed V
The target longitudinal acceleration calculation unit 144 predicts XO to the TCL 76.
Read from the stored map as shown in FIG.
Reference drive of engine 11 corresponding to target longitudinal acceleration GXO of
The torque TB is calculated using the following formula (
7) Calculate by. TB = (GXO・Wb・r+TL)/ρ
m・ρd・ρT...(7)
However, TL is a function of the lateral acceleration GY of the vehicle 82.
Road resistance (Roa), which is the resistance of the road surface required for
d-Load) torque, and in this example, it is shown in Fig. 36.
I'm looking for it from the map shown. [0148] Here, depending on the steering shaft turning angle δH and the vehicle speed V,
Therefore, simply finding the target drive torque of the engine 11 will not work.
The driver's will is not reflected at all, and the maneuverability of the vehicle 82 is affected.
There is a risk that drivers may remain dissatisfied. For this reason, drivers rarely
Accelerate the desired driving torque Td of the engine 11.
This required drive torque is determined from the amount of depression of the pedal 31.
Set the target drive torque of the engine 11 by taking Td into consideration.
This is desirable. Therefore, in this embodiment, the reference drive torque TB
In order to determine the adoption ratio of
The correction standard is calculated by multiplying the dynamic torque TB by the weighting coefficient α.
Find the driving torque. This weighting coefficient α is the vehicle 8
2 is set empirically by turning, but on high μ roads it is set to 0.
．． Adopt a value around 6. On the other hand, the crank angle sensor 55 detects
Detected by the engine speed NE and the accelerator opening sensor 77.
The driver desires based on the output accelerator opening θA.
Is the required driving torque Td a map as shown in FIG. 35?
Then, in the multiplier 147, the weighting coefficient α
The corrected required driving torque corresponding to the required driving torque Td
Calculated by multiplying by (1-α). for example,
When α is set to 0.6, the reference drive torque TB
The adoption ratio of the required driving torque Td is 6:4. [0151] Therefore, the target driving torque TOC of the engine 11
is calculated by the adding section 148 using the following equation (8). TOC=α・TB +(1−α)・Td
...
(8) 0152] By the way, it is set every 15 milliseconds.
The increase/decrease in the target drive torque TOC of engine 11 is very large.
If it is large, a shock occurs as the vehicle 82 accelerates or decelerates.
Engine 11
The increase or decrease in the target drive torque TOC affects the ride comfort of the vehicle 82.
If it becomes large enough to cause a decline, this target drive
It is desirable to regulate the increase/decrease in torque TOC. Therefore, in this embodiment, the amount of change clip section 1
The target drive torque TOC(n) calculated this time in 49 and
Difference from the previously calculated target drive torque TOC (n-1)
If the absolute value |ΔT| is smaller than the allowable increase/decrease amount TK
is the calculated current target drive torque TOC(n)
is adopted as is, but the target drive torque T calculated this time
OC(n) and the previously calculated target drive torque TOC(n
-1) The difference ΔT is larger than the negative increase/decrease tolerance TK
If not, the current target drive torque TOC(n)
is set by the following formula. TOC(n) =TOC(n-1) -TK 015
4] In other words, the target drive torque TOC(n-
1) The amount of decrease to
Reduces deceleration shock due to reduction in drive torque of Seki 11.
Ru. Also, the target drive torque TOC(n) calculated this time and
Difference from the previously calculated target drive torque TOC (n-1)
If ΔT is greater than or equal to the allowable increase/decrease amount TK, the current target drive
Set the dynamic torque TOC(n) using the formula below. TOC(n) =TOC(n-1) +TK 015
5] In other words, the target drive torque TOC(n) calculated this time
and the target drive torque TOC (n-1) calculated last time.
If the difference ΔT exceeds the allowable increase/decrease amount TK, the previous calculation
Increase in target drive torque TOC (n-1)
The width is regulated by the allowable increase/decrease amount TK, and the driving torque of the engine 11 is
Reduce acceleration shock due to increase. [0156] Then, the start or end of turning control is determined.
According to the determination process in the start/end determination unit 150 to
Therefore, information regarding this target drive torque TOC is sent to the ECU1.
5 is output. [0157] The start/end determination unit 150 performs the following (a)
~(d) Turning control is performed when all conditions shown in (d) are satisfied.
It is determined that the turning control is being started and the turning control flag FC is set.
In addition, information regarding the target drive torque TOC is sent to the ECU 15.
and determines the end of swing control and sets the swing control flag.
This process continues until the FC is reset. (a) The target drive torque TOC is the required drive torque Td
less than the threshold value, for example, 2 kgm. (b) The driver operates a manual switch (not shown) to turn the
I would like to control the times. (c) The idle switch 68 is in the off state. (d) The control system for turning is normal. [0158] On the other hand, the start/end determining section 150 rotates.
After determining the start of control, proceed to (e) and (f) below.
If any of the conditions shown are satisfied, the rotation control
Deciding that it is finished, reset the turning control flag FC,
Stop sending target drive torque TOC to ECU15
do. (e) Target drive torque TOS is required drive torque Td
That's all. (f) There is an abnormality such as a failure or disconnection in the control system for turning.
Ru. By the way, the output of the accelerator opening sensor 77
Naturally, there is a difference between the voltage and the accelerator opening θA.
There is a certain proportional relationship between the two and the accelerator opening θA is fully closed.
In this case, the output voltage of the accelerator opening sensor 77 is, for example, 0.
to the throttle body 21 so that it is 6 bolts.
The accelerator opening sensor 77 is assembled. However, the vehicle
Accelerator from throttle body 21 during inspection and maintenance of 82
This happens when the opening sensor 77 is removed and reassembled.
Accurately return the accelerator opening sensor 77 to its original installation condition.
It is virtually impossible to restore it, and due to changes over time, etc.
Accelerator opening sensor 77 for throttle body 21
There is also a risk that the position of the Therefore, in this embodiment, the accelerator opening sensor
77's fully closed position is learned and corrected, and this
Therefore, based on the detection signal from the accelerator opening sensor 77
The reliability of the accelerator opening θA calculated by
Ru. [0161] When the accelerator opening sensor 77 is in the fully closed position,
As shown in Figure 37 representing the learning procedure, the idle switch
68 is on and the ignition key switch 75 is
After changing from on to off state, for a certain period of time, e.g. 2 seconds
The output of the accelerator opening sensor 77 is monitored, and the
The minimum value of the output of the accelerator opening sensor 77 is set as the accelerator opening θ.
A is taken as the fully closed position and incorporated into the ECU15.
Store it in a RAM with backup (not shown).
, the output of this accelerator opening sensor 77 until the next learning.
The accelerator opening degree θA is corrected using the lowest value as a reference. [0162] However, a power storage device (not shown) mounted on the vehicle 82
If the pond is removed, the memory in the RAM will be erased.
Therefore, in such a case, see Figure 38 and Figure 39.
The learning procedure shown will be adopted. [0163] In other words, TCL76 opens the accelerator at A1.
The fully closed value θAC of degree θA is stored in the RAM.
It is determined whether or not, and the accelerator opening degree is determined in this step A1.
It is determined that the fully closed value θAC of θA is not stored in the RAM.
If disconnected, save the initial value θA(0) to RAM at A2.
Make me remember. On the other hand, at this step A1, the accelerator is opened.
It is determined that the fully closed value θAC of degree θA is stored in the RAM.
If it is disconnected, turn the ignition key switch on A3.
75 is in the on state. This A3 space
Ignition key switch 75 is on at step
If it is determined that the state has changed from
A learning timer (not shown) starts counting. stop
After the learning timer starts counting, press the button A5.
It is determined whether the dollar switch 68 is in the on state. [0165] At this step A5, the idle switch
If it is determined that 68 is in the off state, the above-mentioned operation is performed at A6.
The count of the learning timer has reached the set value, for example 2 seconds.
It is determined whether or not, and the process returns to step A5 again. or,
At step A5, the idle switch 68 is in the on state.
If it is determined that there is, the accelerator opening sensor at A7
Read the output of 77 at a predetermined period, and read the current address using A8.
The accelerator opening θA(n) is the previous accelerator opening θA.
It is determined whether or not it is smaller than the minimum value θAL. [0166] Here, the current accelerator opening θA(n) is
Greater than the previous minimum value θAL of accelerator opening θA
If it is determined that the current accelerator opening θA is
Keep the minimum value θAL as it is, and conversely, open the accelerator this time.
The degree θA(n) is the minimum value θ of the accelerator opening degree θA so far.
If it is determined that it is smaller than AL, use A9 for this time.
Update accelerator opening θA(n) as new minimum value θAL.
New. This operation is performed on the learning tie in step A6.
Repeat until the machine count reaches the set value, e.g. 2 seconds.
vinegar. [0167] The count of the learning timer has reached the set value.
Then, the minimum value θAL of the accelerator opening θA at A10
is a preset clip value, e.g. 0.3 volts and 0.3 volts.
Determine whether the voltage is between 9 volts or not. And this
The minimum value θAL of the accelerator opening θA is the preset threshold.
If it is determined that the value is within the range of A1
1, the initial value θA (0) of the accelerator opening θA or the full
The closing value θAC is set to a constant value in the direction of the minimum value θAL, e.g.
Acceleration based on this learning for objects that are 0.1 volts closer
The fully closed value θAC(n) of the opening degree θA is assumed. In other words, a
Initial value θA(0) or fully closed value θA of throttle opening θA
If C is larger than its minimum value θAL, set θAC(n) = θAC(0) -0.1 or θAC(n) = θAC(n-1) -0.1.
, conversely, the initial value θA (0) of the accelerator opening θA or the full
If the closing value θAC is larger than its minimum value θAL, then
Set θAC(n) = θAC(0) +0.1 or θAC(n) = θAC(n-1) +0.1.
Ru. [0168] In step A10, the accelerator opening degree θ
Is the minimum value θAL of A within the preset clip value range?
If it is determined that it is off, check A12.
The clip value for the accelerator opening θA is the minimum value θAL
and move to step A11 above to activate
Learn and correct the fully closed value θAC of the cell opening degree θA. [0169] In this way, the minimum value of the accelerator opening θA
By setting an upper limit value and a lower limit value to θAL, the
Even if the cell opening sensor 77 fails, incorrect learning will not occur.
There is no risk and the learning correction amount per time is set to a constant value.
This prevents incorrect learning from occurring due to disturbances such as noise.
There will be nothing to do. [0170] In the embodiment described above, the accelerator opening sensor
77's fully closed value θAC is determined by the ignition key.
-When the switch 75 changes from the on state to the off state
based on the seating control system built into the seat (not shown).
Turn on the ignition key switch 75 using
The seat sensor detects when the driver leaves the seat even when the driver is in the seat.
Detection is made using changes in seat pressure and positional displacement,
You may start the learning process after the step.
. Additionally, a door lock device (not shown) can be operated from the outside of the vehicle 82.
key entry system.
detected that the door lock device was operated by the system.
Learning the fully closed value θAC of the accelerator opening sensor 77 at the time
It is also possible to start. In addition, hydraulic automatic change
The shift lever (not shown) of the speed gear 13 is in the neutral position.
position or parking position (if equipped with a manual transmission)
(in the case of a vehicle, the neutral position) and the manual brake
key is operated and the air conditioner is turned off.
, that is, the learning process is performed when the idle state is not up.
You can do it as well. [0171] In the vehicle 82, turning control is selected by the driver.
A manual switch (not shown) is provided to select the
, the driver operates this manual switch to select turning control.
If so, perform the rotation control operation described below.
It has become. [0172] The target drive torque TOC for this turning control is
Figures 40 and 41 show the control flow for determining
Detection and calculation of various data mentioned above in C1
The process calculates the target drive torque TOC, but this
The operation is performed independently of the operation of the manual switch.
. Next, at C2, the vehicle 82 is under turning control.
In other words, whether the turning control flag FC is set.
Determine whether the At first, the turning control is not in progress.
Therefore, the turning control flag FC is in the reset state.
For example, in C3, determine whether it is less than (Td -2) or not.
judge. In other words, even when the vehicle 82 is traveling straight, the target drive torque is
It is possible to calculate the torque TOC, but the value is
Normally, the required driving torque Td is larger than the required driving torque Td. However, when the vehicle 82 turns, this required drive torque Td
Since the target drive torque TOC is generally smaller than
The turning control start condition is set when the threshold value (Td -2) or less is reached.
I am trying to judge it as a case. [0174] Note that this threshold value is set as (Td -2).
The hysteresis was added to prevent control hunting.
As a standard. [0175] At step C3, target drive torque TOC
is less than the threshold (Td −2), TCL
76 indicates whether the idle switch 68 is in the off state at C4.
Determine. [0176] At this step C4, the idle switch
68 is in the off state, that is, the accelerator pedal 31 is not pressed by the driver.
If it is determined that the driver is stepping on the vehicle, control the turn using C5.
The control flag FC is set. Next, two at C6
The steering angle neutral position learned flag FHN, FH.
Whether or not at least one is set, that is, the steering angle sensor
The authenticity of the steering angle δ detected by the sensor 84 is determined.
. [0177] Two steering angle neutral positions at step C6
At least one of the completed flags FHN and FH is set.
If it is determined that the rotation is being controlled, the turning control flag is set at C7.
It is determined again whether FC is set. [0178] In the above procedure, turning is performed at step C5.
Since the control flag FC is set, the C7 step
At step, the turning control flag FC is set.
It is determined that, in C8, the previously calculated value, that is, the step of C1
The target drive torque TOC at is used as is. On the other hand, in step C6, the steering angle is set to the neutral position.
If it is determined that the setting learned flag FH is not set,
, (2) Since the rudder angle δ calculated by formula is unreliable,
, (8) The target drive torque TOC calculated using the formula is adopted.
TCL76 is used as the target drive torque TOC for the engine.
The maximum torque of 11 is output at C9, which causes the ECU
15 is the duty rate of the torque control solenoid valves 51 and 56.
As a result, the engine 11 is activated by the driver.
Generates driving torque according to the amount of depression of the cell pedal 31
do. [0180] Also, in step C3, the target drive torque is
If it is determined that the TOC is not below the threshold (Td -2),
, from step C6 or C7 without transitioning to turning control.
Move to step C9, TCL76 sets the target drive torque.
Outputs the maximum torque of engine 11 as TOC, and
The ECU 15 controls the duty of the torque control solenoid valves 51 and 56.
As a result of reducing the tee rate to the 0% side, engine 11
The drive torque according to the amount of depression of the accelerator pedal 31
cause a problem. [0181] Similarly, at step C4, the idle switch is
switch 68 is on, that is, the accelerator pedal 31 is
Even if it is determined that the TC has not been taken into account,
L76 is the maximum torque of the engine 11 as the target drive torque TOC.
This outputs the torque, which causes the ECU 15 to control the torque control voltage.
As a result, the duty ratio of the magnetic valves 51 and 56 is reduced to the 0% side.
As a result, the engine 11 is activated when the driver presses the accelerator pedal 31.
Generates drive torque according to the amount of intrusion and shifts to swing control
do not. [0182] At step C2, the turning control flag is set.
If it is determined that FC is set, C10
The target drive torque TOC calculated this time and the previously calculated
The difference ΔT from the target drive torque TOC (n-1) is set in advance.
Determine whether it is larger than the specified increase/decrease tolerance TK.
. This allowable increase/decrease amount TK provides the acceleration/deceleration shock of the vehicle 82 to the occupants.
The amount of torque change is such that you do not feel any shock, for example.
The target longitudinal acceleration GXO of vehicle 82 is suppressed to 0.1 g/s
If you want, use equation (7) above to calculate TK =0.1・Wb・r・Δt/ρm・ρd・
It becomes ρT. [0183] The number calculated this time in step C10 above
Target drive torque TOC and previously calculated target drive torque TO
The difference ΔT from C(n-1) is the preset increase/decrease tolerance T
If it is determined that it is not larger than K, then in C11
is the target drive torque TOC and the previously calculated target drive torque
TOC (n-1) difference ΔT is negative increase/decrease tolerance TK
Determine whether it is larger than . [0184] The target drive calculated this time in step C11
dynamic torque TOC and the previously calculated target drive torque TOC (
n-1) is larger than the negative increase/decrease tolerance TK
If it is determined that the target drive torque TOC calculated this time is
and the target drive torque TOC (n-1) calculated last time.
The absolute value of the difference |ΔT| is smaller than the allowable increase/decrease amount TK.
In this case, the calculated current target drive torque TO
Use C as it is as the calculated value in step C8.
. [0185] Also, the eyes calculated this time in step C11
Target drive torque TOC and previously calculated target drive torque TO
The difference ΔT from C is not larger than the negative increase/decrease tolerance TK
If it is determined, the current target drive torque TOC is determined at C12.
is corrected using the formula below, and this is the calculated value in step C8.
Adopted as. TOC=TOC(n-1) -TK [0186] In other words, the previously calculated target drive torque TO
The amount of decrease relative to C(n-1) is defined by the allowable increase/decrease amount TK.
control and reduce the deceleration shock caused by the reduction of the driving torque of engine 11.
Make it less. [0187] On the other hand, in step C10 above, the current calculation
The calculated target drive torque TOC and the previously calculated target drive torque
The difference ΔT from TOC(n-1) is the allowable increase/decrease amount TK.
If it is determined that the current target drive torque is
Correct the TOC using the formula below, and use this in step C8.
Adopted as the calculated value. TOC=TOC(n-1) +TK [0188] In other words, in the case of an increase in driving torque, the above-mentioned
As in the case of drive torque reduction, the target drive calculated this time
Torque TOC and previously calculated target drive torque TOC (n
-1) If the difference ΔT exceeds the allowable increase/decrease amount TK
is the target drive torque TOC (n-1) calculated last time.
The increase amount for engine 11 is regulated by the allowable increase/decrease amount TK.
This reduces acceleration shock caused by increased drive torque.
Ru. [0189] As described above, the target drive torque TOC is
Once set, TCL76 sets this target drive torque at C14.
torque TOC is larger than the driver's required driving torque Td
Determine whether or not. [0190] Here, the turning control flag FC is set.
If the target drive torque TOC is
Since it is not larger than the driving torque Td, the
It is determined whether the idle switch 68 is in the on state. [0191] At this step C15, the idle switch
If it is determined that the switch 68 is not in the ON state, turning control is required.
Since this is the required state, move on to step C6 above.
go [0192] Also, in step C14, the target drive torque is
torque TOC is larger than the driver's required driving torque Td
If it is determined that the vehicle 82 has finished turning,
This means that TCL76 is used for turning control at C16.
Reset the lag FC. Similarly, step C15
It is determined that the idle switch 68 is in the on state.
Then, the accelerator pedal 31 is not depressed.
Therefore, move to step C16 and set the flag during turning control.
Reset the FC. [0193] At this C16, the turning control flag FC is set.
When reset, TCL76 sets the target drive torque TOC.
output the maximum torque of engine 11 at C9, and
Therefore, the ECU 15 controls the torque control solenoid valves 51 and 56.
As a result of reducing the party rate to 0%, engine 11
The drive torque corresponds to the amount of depression of the accelerator pedal 31.
generates a certain amount of energy. [0194] In order to simplify the above-mentioned turning control procedure,
Ignore the driver's requested drive torque Td in order to
Of course, it is also possible, and in this case, the target drive torque is
Reference drive torque TB that can be calculated using the above formula (7)
You should adopt. In addition, as in this example, the driver's request
Even when considering the driving torque Td, the weighting coefficient
Instead of setting α to a fixed value, the time elapsed after the start of control
At the same time, the value of the coefficient α is gradually decreased, or the vehicle speed V is increased.
The driver's required drive torque Td is gradually decreased accordingly.
The adoption ratio may be gradually increased. Similarly,
For a while after starting control, the value of coefficient α is kept constant.
set and gradually decrease after a predetermined period of time, or
As the rudder shaft turning amount δH increases, the value of the coefficient α is increased.
, especially for turning paths where the radius of curvature gradually becomes smaller.
, it is also possible to make the vehicle 82 run safely.
be. [0195] In the embodiment described above, the target drive for high μ road
I tried to calculate the torque, but this high μ road and low μ road
Calculate the target drive torque for swing control corresponding to each
The final target drive torque is calculated from these target drive torques.
It is also possible to select one. In addition, the above-mentioned arithmetic processing
In this method, the addition due to sudden fluctuations in the driving torque of the engine 11 is
To prevent deceleration shock, target drive torque TOC
When calculating, this target drive is determined by the allowable increase/decrease amount TK.
We are trying to regulate torque TOC, but this regulation is not yet reached the target.
The process may be performed for the rear acceleration GXO. [0196] The target drive torque TOC for this turning control is
After calculating, TCL76 calculates these two target drive torques.
Optimal final target drive torque TO from torque TOS and TOC
is selected and output to the ECU 15. In this case, the car
Considering the driving safety of both 82, the target drive is set for a smaller value.
Outputs dynamic torque with priority. However, in general, slip
The target drive torque TOS for swing control is the same as the target drive torque TOS for swing control.
Because it is always smaller than the dynamic torque TOC, the slip control
Set the final target drive torque TO in the order of control use and swing control use.
All you have to do is choose. As shown in FIG. 42, which shows the flow of this process,
, B11 is the target drive torque TOS for slip control.
After calculating the target drive torque TOC for turning control, B
12, the slip control flag FS is set.
It is determined whether or not the slip control flag FS is
If it is determined that the final target drive torque is set,
Target drive torque TOS for slip control
is selected at B13 and output to the ECU 15. On the other hand, if there is a slip in step B12,
It was determined that the control flag FS was not set.
If so, the turning control flag FC is set at B14.
The turning control flag FC is set.
If it is determined that the target drive torque is
B1 is the target drive torque TOC for turning control as TO.
5 and outputs it to the ECU 15. [0199] Also, during the turning control in step B14,
If it is determined that flag FC is not set,
TCL76 reaches the final maximum torque of engine 11 at B16.
It is output to the ECU 15 as the target driving torque TO. [0200] As described above, the final target drive torque TO
while selecting the slot via the actuator 41.
The output of the engine 11 can also be reduced by operating the torque valve 20.
When starting suddenly or when the road surface changes from a normal dry road to an icy road.
If there is a sudden change in the temperature, TCL76 should be set in ECU15.
Retard ratio for the basic retard amount pB of the ignition timing P determined
and outputs it to the ECU 15. [0201] The basic retard amount pB is determined by the operation of the engine 11.
This is the maximum value of the retard angle that does not interfere with engine 11.
It is set based on the intake air amount and engine speed NE.
, basically the rate of change Gs of the slip amount s increases.
According to
ing. As this retardation ratio, in this example, the basic retardation amount is
0 level to set pB to 0 and basic retard amount pB to 3 minutes
The I level to be compressed to 2 and the basic retardation amount pB are kept as they are.
The II level to be output and the basic retardation amount pB remain as they are.
III to fully close the throttle valve 20 while outputting the output.
There are four levels set. That is, this III
The above-mentioned fully closing operation of the throttle valve 20 at
In combination with the retard operation, the engine can be activated extremely quickly.
11 drive torque to reduce the slippage of the front wheels 64 and 65.
can converge. FIG. 41 shows the procedure for reading out this retardation ratio.
As shown in FIG. 42, TCL76 first turns on at P1.
Reset the fire timing control flag FP and set the slider at P2.
Determine whether the drop control flag FS is set or not.
to be determined. At this P2 step, the slip control flag is set.
If it is determined that FS is set, ignition occurs at P3.
Set the timing control flag FP and slip at P4
It is determined whether the amount s is less than 0 km/hour. Also, the above P2
The slip control flag FS is set in step
If it is determined that the
Ru. [0203] In this step P4, the slip amount s is
time is less than 0 km, that is, increase the driving torque of the engine 11.
If it is determined that there is no problem, set the retard ratio to 0 level in P5.
and outputs this to the ECU15. vice versa
, in this step P4, the slip amount s is 0 km/h or more.
If it is determined that the slip amount is above, change the slip amount in P6.
Determine whether the rate GS is 2.5g or less, and
In step 6, the slip amount change rate GS is 2.5g or more.
If it is determined that the retardation ratio is
Determine whether it is at I level. [0204] Also, in step P6, the slip amount is changed.
The conversion rate GS exceeds 2.5g, that is, the front wheel 64,
If it is determined that 65 is slipping, at P8
Is the final target drive torque TO less than 4 kgm?
This final target drive torque TO is 4kgm.
In other words, the driving torque of the engine 11 is sharply suppressed.
If it is determined that it is necessary to
Set to level III and move to step P7 above.
do. Conversely, in step P8, the final target drive torque T
If it is determined that O is 4 kgm or more,
Proceed to step P7. [0205] In this step P7, the retardation ratio is I
If it is determined that the level is
It is determined whether the drop amount change rate GS exceeds 0 g. Here, the slip amount change rate GS exceeds 0g,
In other words, when it is determined that the slip amount s tends to increase
The ignition timing control flag FP is set at P11.
It is determined whether or not the
The slip amount change rate GS is 0g or less, that is, the slip
If it is determined that the amount of drop s is trending toward a phenomenon, P12
Check whether the slip amount s of the lever exceeds 8 km/h.
judge. [0206] In step P12, the slip amount s is
If it is determined that the speed exceeds 8km/h, the above P1
Shifting to step 1, the slip amount s becomes 8 km/h.
If it is determined that the
Switch from III level to II level, slip on P14
Determine whether the drop amount change rate GS is 0.5g or less.
do. Similarly, in step P7, the retardation ratio is set to II.
Even if it is determined that you are not at the I level, this P14
Move to step. [0207] The slip amount changes in this step P14.
The rate GS is 0.5g or less, that is, the change in the slip amount s
If it is determined that the change is not too rapid, then on P15
It is determined whether the retardation ratio is at level II. Also, P
At step 14, the slip amount change rate GS is 0.5g.
If it is determined that it is not below, the retardation rate is determined in P16.
is set to the II level and moves to step P15. [0208] Then, in step P15, the delay angle division is
If it is determined that the situation is at level II, please proceed to P16.
Determine whether the slip amount change rate GS exceeds 0g.
, conversely, if it is determined that the retardation ratio is not at II level,
The slip amount change rate GS is 0.3g or less on P17.
Determine whether or not. In step P16 above,
The lip amount change rate GS does not exceed 0g, that is, the slip
If it is determined that the drop amount s is on a decreasing trend, P18
Does this slip amount s exceed 8 km/h?
Determine whether or not. Then, in step P18,
If it is determined that the lip amount s is less than 8 km/h,
, Switch the retardation ratio from II level to I level at P19.
Yes, proceed to step P17. Also, the above P16
If the slip amount change rate GS is 0g or more in step
In other words, when it is determined that the slip amount s is on an increasing trend.
, and the slip amount s is 8 km/h at step P18.
If it is determined that the amount of slip exceeds s, that is, the amount of slip s is large,
If so, the process moves to step P11. [0209] Slip amount change in step P17
The ratio GS is 0.3g or less, that is, the slip amount s is almost
If it is determined that there is no increasing trend, retard the rate at P20.
It is determined whether or not the current is at I level. On the contrary, P17
Slip amount change rate GS exceeds 0.3g at step
In other words, the slip amount s is increasing to some extent.
If it is determined that
Set to . [0210] Then, at P20, the retardation ratio is at I level.
If it is determined that there is, check the slip amount change rate in P22.
Determine whether GS exceeds 0g, and determine if this is 0g.
or less, that is, it is determined that the slip amount s is on a decreasing trend.
If the slip amount s is less than 5 km/h in P23,
Determine whether the conditions are satisfied. In this step of P23
The slip amount s is less than 5 km/h, that is, the front wheels 64,
If it is determined that 65 is not slipping, P2
4, set the retardation ratio to 0 level, and set this to ECU15.
Output to. Also, in step P20, the retardation ratio is set to I level.
If it is determined that it is not a bell, or in step P22.
The slip amount change rate GS exceeds 0g, that is, the slip
If it is determined that the drop amount s is on an increasing trend, or P2
In step 3, the slip amount s is 5 km/h or more.
, that is, when it is determined that the slip amount s is relatively large,
The process then proceeds to step P11. [0211] On the other hand, in this step P11, the ignition timing
If it is determined that the control flag FP is set
For example, in P25, the final target drive torque TO is 10 kgm.
Determine whether or not the value is less than or equal to the value. Also, step P11
Ignition timing control flag FP is not set in
If it is determined that
After setting, proceed to step P25. [0212] Then, in this P25, the final target drive torque is set.
engine 11 is more than 10 kgm.
If it is determined that a slightly larger driving force is being generated,
At P27, it is determined whether the retardation ratio is at II level or not.
, if this retardation ratio is determined to be at II level,
, at P28, the retardation ratio is reduced to I level, and this is set to EC.
Output to U15. [0213] In step P25, the final target drive torque is set.
If it is determined that the TO is less than 10 kgm,
, the retardation ratio is not at II level in step P27.
If it is determined that the hydraulic automatic transmission 13
It is determined whether or not the gear is being changed. And hydraulic automatic transmission
If it is determined that 13 is in the process of shifting, the speed is slowed down at P30.
Determine whether the angular ratio is at the III level, and
If the retardation ratio is at the III level at step 30,
If it is determined, set the retard ratio to II level on P31.
and outputs this to the ECU 15. Also, the page 29
It was determined that the hydraulic automatic transmission 13 was not in the process of shifting.
If it is disconnected, or if the retardation ratio is
If it is determined that it is not at the II level, P
32, the retardation ratio set earlier is used as is in the ECU15.
Output to. [0214] For example, in step P9,
When the retardation ratio of the slip angle is set, the slip amount change rate GS
exceeds 0g and the slip amount s is 8km/h.
, that is, the rate of increase in the slip amount s is rapid.
and the final target drive torque TO is less than 10 kgm.
If you just retard the ignition timing, the front wheels 64 and 65 will slip.
If it is determined that it is difficult to sufficiently suppress the
is the throttle with the III level retard ratio selected.
The opening of the valve 20 is forced to the fully closed state to prevent slippage.
We are trying to efficiently suppress this in its early stages. [0215] The ECU 15 controls the engine speed NE and the engine speed.
The ignition timing P is preset based on the intake air amount of function 11.
and a map (not shown) regarding the basic retardation amount pB
From these, the ignition timing P and basic retardation amount pB are clamped.
Detection signal from angle sensor 62 and air flow sensor 7
Based on the detection signal from 0, this is read out based on the detection signal from TCL7.
Correct based on the retardation ratio sent from 6 and set the target retardation amount.
I am trying to calculate pO. In this case,
Upper limit of exhaust gas that does not damage the exhaust gas purification catalyst
The upper limit of the target retardation amount pO is set according to the temperature.
The temperature of this exhaust gas is detected by the exhaust temperature sensor 74.
Detected by the output signal. [0216] Note that the temperature detected by the water temperature sensor 71 is
When the cooling water temperature of Seki 11 is lower than the preset value
, retarding the ignition timing P prevents engine 11 from knocking.
When igniting as shown below,
The retardation operation for period P is canceled. [0217] The target retard amount pO in this retard control is
As shown in FIGS. 45 and 46 showing the calculation procedure, first, E
CU15 sets the slip control flag FS mentioned above in Q1.
This slip control is determined whether or not is set.
When it is determined that the service flag FS is set, Q
Whether the retardation ratio is set to III level in 2.
Determine whether [0218] Then, in this step Q2, the retardation ratio
If it is determined that the
Target the basic retardation amount pB read from the map as is.
The ignition timing P is used as the retard amount pO, and the ignition timing P is set to the target retard amount p
Delay by O. Furthermore, the final target drive torque TO
So that the throttle valve 20 is fully closed regardless of the value.
, the duty of the torque control solenoid valves 51 and 56 at Q4.
Set the rate to 100% and force the throttle valve 20 to fully open.
Achieve a closed state. As a result, the slip amount change rate Gs
Even if there is a sudden increase in
can be effectively suppressed at the initial stage. [0219] Also, in step Q2, the retardation ratio is
If it is determined that the
It is determined whether or not the level is set to II level. So
Then, in this step Q5, the retardation ratio is at II level.
If it is determined that there is, follow the steps in Q3 above.
In Q6, the target retardation amount pO is read out from the map.
Using the angle amount pB as it is as the target retard amount pO,
The ignition timing P is retarded by the target retardation amount pO. Furthermore, Q
At step 7, the ECU 15 operates according to the value of the target drive torque TOS.
The duty rate of the torque control solenoid valves 51 and 56 is set to Q7.
When the driver depresses the accelerator pedal 31,
The driving torque of the engine 11 is reduced regardless of the amount. [0220] Here, the ECU 15 has the engine speed NE and
Throttle using the drive torque of engine 11 as a parameter
A map for determining the opening degree θT is stored, and E
CU15 uses this map to determine the current engine speed NE
and the target slot corresponding to this target drive torque TOS.
Read out the opening degree θTO. [0221] Next, the ECU 15 selects this target throttle.
Opening θTO and output from throttle opening sensor 67
Find the deviation from the actual throttle opening θT, and
The duty rate of the solenoid valves 51 and 56 for torque control is set to the above deviation.
Each torque control solenoid valve 51, 56 is set to a value commensurate with
Apply current to the solenoids of the plungers 52 and 57, and the
The actual throttle opening θ is determined by the actuation of the actuator 41.
T is controlled so that it falls to the target throttle opening θTO.
control [0222] Furthermore, as the target drive torque TOS, engine 1
If the maximum torque of 1 is output to ECU15, ECU
15 is the duty rate of the torque control solenoid valves 51 and 56.
0% side, and the driver presses the accelerator pedal 31.
Generates a driving torque in the engine 11 according to the amount of depression.
. [0223] In step Q5, the retardation ratio is set to II level.
If it is determined that it is not a bell, the retard rate is determined in Q8.
Determine whether or not it is set to I level. This Q8
If the retardation ratio is set to I level in step
If determined, set the target retardation amount pO as shown in the formula below.
to retard the ignition timing P by the target retard amount pO, and further advance the ignition timing P.
Proceed to step Q7. pO = pB ・2/3 [0224] On the other hand, in step Q8, the retardation ratio is
If it is determined that it is not at the I level, the goal will be determined in Q10.
Determine whether or not the retardation amount pO is 0, and determine if this is 0.
If it is determined that the
without retarding the timing P, according to the value of the target drive torque TOS.
Set the duty rate of the torque control solenoid valves 51 and 56.
, is not related to the amount of depression of the accelerator pedal 31 by the driver.
Regardless, the driving torque of the engine 11 is reduced. [0225] Also, in step Q10, the target retardation amount
If it is determined that pO is not 0, the main
Set the target retard amount pO every timer sampling period Δt.
By ramp control, for example, once at a time until pO = 0.
By subtracting the
After reducing the shock, move on to step Q7. [0226] It should be noted that the slip control is applied in step Q1 above.
If it is determined that the service flag FS has been reset
In this case, the driving torque of the engine 11 is not reduced.
control, and in Q12 set pO = 0 and set the ignition timing P.
Torque control solenoid valves 51 and 56 at Q13 without retarding
By setting the duty rate of engine 11 to 0%,
depends on the amount of depression of the accelerator pedal 31 by the driver.
generates driving torque. [0227] In this way, the eye in ignition timing control can be adjusted.
After calculating the target retardation amount pO, the ECU 15 calculates the restraint torque.
Adjust the differential restraint torque of the adjustment clutch 89 to the absolute value of the front wheel speed difference.
Set based on the value |VFR−VFL|. [0228] Calculation procedure for this differential restraint torque control
As shown in FIGS. 47 and 48, the ECU 15 is
The front wheel is rotated based on the detection signals from the wheel rotation sensors 66 and 97.
The absolute value of the speed difference |VFR−VFL| is calculated by the circumferential speed difference calculation unit 15
Calculated using 1. At this time, if the vehicle 82 is turning
Inevitably, the front wheels 64 and 65 have a circumferential speed difference ΔVF due to turning.
occurs, so the turning correction calculation unit 152 corrects this turning.
The circumferential speed difference ΔVF between the front wheels 64 and 65 is the absolute difference in front wheel speed.
It is necessary to subtract from the versus value |VFR−VFL|. here
Now, if the turning radius of the vehicle 82 is Cr, then ΔVF = b·V/Cr, but Cr = ω·(1+A·V2)/δ. [0229] However, in this embodiment, the unit is
Standard of front wheels 64, 65 per steering angle (e.g. 120 degrees)
The circumferential speed difference ΔVBF is ECed by the reference circumferential speed difference calculation unit 153.
Read from the map shown in Figure 49 stored in U15.
This is multiplied by the steering angle (δH /120
), the circumferential speed of the front wheels 64, 65 during turning
The difference ΔVF is calculated. Then, the turning correction calculation section 1
In step 52, calculate the corrected front wheel speed difference ΔVFF as shown in the formula below.
, based on this corrected front wheel speed difference ΔVFF, the reference differential restraint torque is set.
50 stored in the ECU 15 by the torque calculation unit 155
Calculate the reference differential restraint torque TBF from the map shown in
do. ΔVFF=|VFR−VFL|−ΔVBF・δH /1
20 [0230] In this case, there is no relation between high μ road and low μ road.
Changing the differential restraint torque of the bundle torque adjustment clutch 89
, that is, on high μ roads, the differential restraint torque is set to be strong.
, it is desirable to set the differential locking torque weakly on low μ roads.
Delicious. Therefore, the standard differential restraint torque is adjusted according to the road surface μ.
It is necessary to correct the TBF, and then the road surface μ estimation means 15
6 to estimate the road surface μ. By the way, the steering status of the right front wheel 65 is shown below.
As shown in FIG.
The steering force DF is as shown in the formula (9) below.
Ru. DF ∝δF ・μ
　　　　　　　　　　　　　　　　　　　　　　　　　
...(9) However, δF is the progress of the vehicle 82.
row direction (the longitudinal direction of the vehicle 82 corresponds to the vertical direction in the figure)
The sideslip angle of the front wheel 65 with respect to
It is. [0232] Here, sideslip angle δF and cornering
As shown in Figure 52, which shows the relationship with force DF, sideslip
Even if the angle δF is a constant value, cornering may occur depending on the road surface condition.
DF is very different and common
As the road surface μ increases, the sideslip angle δF increases.
It becomes a large value. Also, cornering force DF and pa
As is clear from Figure 50, waste pressure PS is a mechanical
Since there is an almost proportional relationship based on the relationship, C1 can be compared.
As an example constant, we can transform equation (9) to the following equation (10):
It can be expressed as follows. PS = C1 ・δF ・μ
　　　　　　　　　　　　　　　　　　　　　　　　　
...(10) 0233 On the other hand, sideslip angle
Since δF can be expressed by the following formula (11), (10
) and this equation (11), power steering pressure PS and steering shaft
The ratio to the turning angle δH, that is, PS /δH, is calculated using the following formula (12
). δF = C2 ・V2 ・δH / (μ+C3
・V2) ...(1
1) PS/δH = μ・C1・C2・V
2/(μ+C3・V2)...(12)
However, C2 and C3 are each constants. [0234] Therefore, it is output to the road surface μ estimating means 156.
Power steering pressure PS, steering shaft turning angle δH, and vehicle speed V
Based on the equation (12) above, the road surface μ can be calculated.
I can do that. [0235] Calculation procedure by this road surface μ estimating means 156
As shown in FIG. 53, the pressure sensors 98 and 99
The power steering pressure PS calculated based on the detection signal of
The power output detected by these pressure sensors 98 and 99
Difference between the pressures PLS and PRS in the pressure chamber of the actuator 91
The subtraction unit 157 calculates the power steering pressure PS, which is the absolute value of the pressure.
After the calculation, the road surface μ is calculated via the phase compensation filter 158.
It is output to section 159. In addition, the detection from the steering angle sensor 84
The steering shaft turning angle δH calculated based on the output signal and the rear
Calculated based on detection signals from wheel rotation sensors 80 and 81
The vehicle speed V is transmitted from the TCL 76 via the communication cable 87.
Then, it is output to this road surface μ calculating section 159.
There is. [0236] The phase compensation filter 158 includes the subtractor 1
During the signal corresponding to the power steering pressure PS output from 57
In addition to removing the noise of the steering wheel 85,
Power steering pressure PS versus steering shaft turning angle δH during transfer period
This is to compensate for the phase lead of. In other words, steering
Change in steering shaft turning angle δH and power steering pressure PS
As shown in Figure 54, which shows the relationship between the phase compensation
If the filter 158 is not used, the characteristics of the steering valve 94
As shown by the solid line in the figure, the steering handle 8
Regarding the change in the steering shaft turning angle δH due to the notch of 5.
The power steering pressure PS increases quickly and greatly, and the steering
Change in steering shaft turning angle δH due to turning back of the handle 85
There is a tendency for power steering pressure PS to fall earlier than
have However, using the phase compensation filter 158
By doing so, the steering shaft turning angle can be adjusted as shown by the broken line in the figure.
The pattern can be adjusted without causing a phase shift with respect to changes in δH.
The steering wheel 85 is adjusted to follow the change in waste pressure PS.
Eliminate phase advance of power steering pressure PS during steering transition period
can do. [0237] Road surface μ calculated by road surface μ calculating section 159
stabilizes the μ fluctuation limiter 160 and the value of the road surface μ.
47 and 48 through the stabilizing filter 161 for
The road surface correction coefficient calculation unit 162 shown in FIG. here,
The μ fluctuation limiter 160 controls the rate of change of road surface μ within a predetermined range.
In some cases, the road surface μ calculated by the road surface μ calculation unit 159 is
Because it is output to the stabilizing filter 161, extreme fluctuations are avoided.
The stable road surface μ is corrected by the stabilization filter 161.
It is output to the coefficient calculating section 162. [0238] Figure 55 shows the flow of this road surface μ estimation operation.
As shown in FIG. 56, first, the rear wheel speed sensor 8 is
0,81 and steering angle sensor 84 and pressure sensor 98,9
Each vehicle is calculated based on the detected detection signal from 9.
speed V, steering shaft turning angle δH, and power actuator
The pressure PLS and PRS in the pressure chamber of 91 are read respectively.
Then, in J2, this power actuator 91 is
Differential pressure between the pressures PLS and PRS in the pressure chamber, i.e. power steering pressure
PS is calculated. And for power steering pressure PS
Processing by the phase compensation filter 158 described above in J3
is applied, and the steering shaft turning angle δH is 0 at J4.
or whether there is a steering shaft turning angle δH calculated this time.
(n) is the previously calculated steering shaft turning angle δH(n-1)
It is determined whether they are the same or not. [0239] At this step J4, the steering shaft turning angle δH
is 0, or the currently calculated steering shaft turning angle δH
(n) is the previously calculated steering shaft turning angle δH(n-1)
If it is determined that they are not the same, return to step J1.
However, in this step J4, the steering shaft turning angle δH becomes 0.
, or the steering shaft turning angle δH(n
) is the same as the previously calculated steering shaft turning angle δH(n-1)
If it is determined that the steering shaft turning angle δH is
The absolute value of is a preset value δH1 (for example, 10 degrees
) or more is determined. [0240] At this step J5, the steering shaft turning angle δH
When it is determined that the absolute value of is less than the predetermined value δH1
returns to step J1, but in step J5
The absolute value of the steering shaft turning angle δH is greater than or equal to the predetermined value δH1.
If it is determined that the power steering pressure PS and steering are
The ratio of the shaft rotation angle δH, that is, PS /δH, is
12). [0241] After that, the power steering pressure PS is adjusted to be positive at J7.
Whether the negative and the positive and negative of the steering shaft turning angle δH are the same, that is,
It is determined whether the sign of PS /δH is positive or not. This J
In step 7, it is determined that the sign of PS /δH is negative.
If disconnected, use the phase compensation filter at step J3.
Due to processing, power steering pressure PS and steering shaft turning angle δH
It is determined that there is a phase inversion between the
Return to step. Also, in this J7 step PS /
If it is determined that the sign of δH is positive, in J8
The multiplication coefficient Km for calculating road surface μ is as shown in Fig. 57.
Read from the map. This map corresponds to vehicle speed V.
The multiplication coefficient Km is defined in advance.
The data is stored in a memory (not shown) in 5. [0242] Here, if we transform the above equation (12),
μ=PS ・{1+C3 ・V2 /C1 ・C2
・V2 }/δH, but the multiplication coefficient Km is Km = 1+C3 ・V2 /C1 ・C2 ・V2
This corresponds to [0243] Therefore, the road surface μ can be expressed by the following formula
. μ=PS ・Km /δH 0244] Next, in J9, read in step J8
The calculated multiplication coefficient Km and the parameter calculated in step J6
Ratio between waste pressure PS and steering shaft turning angle δH PS/δ
The road surface μ is calculated by multiplying by H. [0245] After this, at J10, the rate of change of road surface μ dμ/
The absolute value of dt is a predetermined value Δμ (for example, every second
It is determined whether or not it is within 0.2μ). This J10
In step , the absolute value of the rate of change dμ/dt of the road surface μ is found.
If it is determined that it exceeds the fixed value Δμ, the J1
Returning to the step, at this J10 step the road surface μ
It is determined that the absolute value of the rate of change dμ/dt is within a predetermined value Δμ.
In the case of disconnection, the road surface μ calculated in step J9
In order to stabilize the value of
After the processing is performed, the road surface μ is output at J12. [0246] In this embodiment, the operation is performed in step J5.
Is the absolute value of the rudder shaft turning angle δH greater than or equal to the predetermined value δH1?
By determining whether or not the steering shaft turning angle δH is
If the value δH1 or more, that is, the front wheels 64 and 65 are being steered.
The power steering pressure PS has increased substantially, and J7
The positive/negative of the power steering pressure PS and the steering shaft turning angle are determined at the step of
By determining whether the positive and negative signs of δH are the same, the
The directions of waste pressure PS and steering shaft turning angle δH are the same.
Since the road surface μ is calculated only when the road surface
μ can be estimated accurately. That is, the steering valve 94
Eliminates the effects of inertia associated with characteristics and steering of front wheels 64 and 65
The road surface μ can be calculated accurately. On the other hand, J4
, J5, and J7, if any
If the result is negative, the processing from step J8 onwards is executed.
In this case, the previously calculated road surface μ is
It will be output as is. Furthermore, in this embodiment, the steps after step J8
Even if the process is performed and the road surface μ is calculated, the road surface μ
If the rate of change dμ/dt is larger than the predetermined value Δμ,
The value of road surface μ is updated by the judgment operation in step J10.
It is set so that it does not update, and the judgment in step S10 is
Even if the
Since the surface μ is output, the output road surface μ
There was no sudden change in the value, and its value was stable.
Become something. [0248] In this example, the power steering pressure PS is detected.
When releasing, the pressure on the left and right sides of the power actuator 91
The pressure in the chamber is detected by a pair of pressure sensors 98 and 99.
The differential pressure in the pressure chamber is calculated as the power steering pressure PS.
However, this power steering pressure PS is discharged from the hydraulic pump 95.
Based on the output from a single pressure sensor built into the side
It is also possible to detect. In addition, in this example, the road surface μ
The information is used for differential restraint torque control, but
It can be used to determine road surface μ for turning control as explained in
Of course, it is also possible to
The road surface μ may be estimated using the method described in
. [0249] In this way, the road surface μ is estimated, and the ECU1
From the map shown in Fig. 58 stored in 5 to the road surface μ.
The corresponding road surface correction coefficient KR is calculated by the road surface correction coefficient calculation unit 16.
2, this road surface correction coefficient KR and the reference difference
The multiplier 163 multiplies the dynamic restraint torque TBS and corrects it.
Calculate differential restraint torque TBF. [0250] By the way, the front wheel speed difference |VFL-VFR|
In the method of setting the differential restraint torque based on the front wheel 64
, 65 only when slip occurs on at least one side.
Increase the dynamic restraint torque and restart the system when this slip is resolved.
In order to weaken the differential restraint torque, especially when starting the vehicle 82,
There is a risk of control hunting occurring. For this reason, the vehicle
82 is the time of starting, using the accelerator opening θA and the vehicle speed V.
The restraining torque is adjusted when the vehicle 82 starts.
Hold the differential locking torque of clutch 89 slightly stronger.
It is desirable to leave it there. [0251] Therefore, the output from the accelerator opening sensor 77
Based on the force signal, the ECU
The reference calculation is made from a map as shown in FIG.
Reads the calculation restraint torque TSS and calculates the vehicle speed correction coefficient 1
65, the map as shown in FIG. 60 stored in the ECU 15 is
Read the vehicle speed correction coefficient KS corresponding to the vehicle speed V from the
Then, the multiplier 166 calculates the reference addition restraint torque TSS.
and further based on the detection signal from the steering angle sensor 84.
Then, the turning angle correction coefficient calculation unit 167 calculates the
Read the turning angle correction coefficient KC from the map and multiply it by
The calculation unit 168 multiplies the value calculated by the previous multiplication unit 166,
Obtain differential restraint torque TSP for addition. [0252] In this way, in this embodiment, when the vehicle 82 starts
The state is estimated from the accelerator opening θA and the vehicle speed V, and the vehicle
of the clutch 89 for adjusting the restraint torque when starting the vehicle 82.
Although the differential locking torque was maintained slightly stronger, other
The starting state of the vehicle 82 can also be estimated using a well-known method.
Of course it is possible. [0253] Then, in the differential restraint torque calculation section 169,
The modified differential restraint torque TBF and the additional differential restraint torque
Add TSP and lower the final differential locking torque TF.
Calculate according to the formula. TF = TBF + TSP 0254] After that, in consideration of control safety etc.
Differential restraint torque calculated by restraint torque calculation unit 169
Apart from TF, the maximum differential restraint torque T according to the road surface μ
SL based on the output from the road surface correction coefficient calculating section 162.
The maximum differential restraint torque calculation unit 170 calculates the
The differential restraint torque calculated by the differential restraint torque calculation unit 169
The clip processing unit 171 calculates the maximum difference between the maximum value of
The dynamic restraint torque is regulated to TSL. However, in this example, the maximum
Clutch 8 for adjusting the large differential locking torque TSL
9 is set to a value that does not result in a direct connection state. [0255] This allows the spare tire to be installed.
If this spare tire becomes flat, there will be an abnormal front wheel speed difference.
Excessive differential locking torque TF based on |VFL-VFR|
is calculated by the differential restraint torque calculation unit 169, but the
The locking torque adjustment clutch is controlled by the lip processing section 171.
The differential locking torque is suppressed so that 89 is not directly connected.
available. [0256] Processing flow for this differential restraint torque control
As shown in FIGS. 62 and 63, the front wheel
The absolute value of the speed difference |VFR−VFL| is calculated. Then,
In D2, the front wheel speed difference ΔVF due to the turning of the vehicle 82 is calculated as follows.
Subtract this from the absolute value of front wheel speed difference |VFR-VFL|
Based on the corrected front wheel speed difference ΔVFF calculated by D
In step 3, the modified differential restraint torque TBF is calculated. and,
Modified differential restraint torque TBF(n) calculated this time in D4
is the modified differential restraint torque TBF (n-1) calculated last time.
Determine whether it is larger than . [0257] Correction calculated this time in step D4
The differential restraint torque TBF(n) is the differential restraint calculated last time.
If it is determined that the torque is greater than TBF (n-1)
This time, the modified differential restraint torque TBF was calculated at D5.
Set the corrected differential restraint torque TBF(n) and adjust the control response.
The driving safety of the vehicle 82 is ensured by increasing responsiveness. and
, D6, the differential restraint torque TSP for addition is calculated. or
, the modified differential restraint torque calculated this time in step D4.
TBF(n) is the modified differential restraint torque TB calculated last time
If it is judged to be smaller than F(n-1), make corrections.
Differential restraint calculated this time using differential restraint torque TBF as is
By setting the torque TBF(n), the correction difference
Low running stability due to sudden drop in dynamic restraint torque TBF
Since there is a possibility that the
The revised differential restraint torque TBF (
n-1) A preset constant value ΔTBF was subtracted from
D while ensuring the running stability of the vehicle 82.
Move to step 6. [0258] Addition differential restraint torque at step D6
After calculating TSP, modify differential restraint torque TB at D8.
Differential restraint is achieved by adding F and the additional differential restraint torque TSP.
Calculate the torque TF, then use D9 to calculate the torque TF according to the road surface μ.
Calculate the maximum differential locking torque TSL. And D10
The differential locking torque TF is this maximum differential locking torque T
It is determined whether it is smaller than SL. [0259] At this step D10, the differential restraint torque
It is determined that TF is larger than the maximum differential locking torque TSL.
In this case, set the differential locking torque TF to the maximum with D11.
Corrected the differential locking torque TSL and controlled the slip with D12.
Determine whether the service flag FS is set or not.
. Also, at step D10, the differential locking torque TF reaches its maximum value.
If it is determined that the large differential locking torque is less than TSL,
, there is no problem with the set differential restraint torque TF, so
This set differential restraint torque TF should not be modified.
The process then moves to step D12. [0260] Slip control is in progress at step D12.
If flag FS is determined to be set,
Tires mounted on vehicle 82 at D13 during braking
(not shown) to maintain the slip ratio between the road surface and the road surface at an optimal value.
The skid control device (hereinafter referred to as ABS)
Determine whether it is functioning or not. Also, step D12
The slip control flag FS is not set in
If it is determined that
is also stopped, so the differential locking torque TF is set at D14.
is reset to 0, and the process moves to step D13. [0261] If ABS is operating at step D13,
If it is determined that there is no
The amount of current corresponding to torque TF is determined from the map shown in Figure 3.
The ECU 15 reads this differential restraint torque at D15.
The current corresponding to TF is applied to the restraint torque adjustment clutch 89.
Turn on electricity. Also, ABS operates at this step D13.
If it is determined that the
Since all rotations of 4, 65, 78, and 79 are controlled,
In order to prevent control interference with this ABS, D16
to reset the differential locking torque TF to 0, and set the D15 step again.
State where the differential gear 90 is fully functional after shifting to step
to hold. [0262] Furthermore, if the vehicle 82 is not equipped with ABS,
In this case, steps D12 and D14 to step D15
to the top. In addition, in this example, a front-wheel drive type vehicle is used.
82, but for rear-wheel drive vehicles
Of course, it can also be applied. [0263] According to the drive wheel differential limiting device of the present invention,
, a pair of cars that detect the speed of the left and right drive wheels, respectively.
Wheel speed sensor and detection signals from these pair of wheel speed sensors
A wheel that calculates the absolute value of the circumferential speed difference of the driving wheels based on the
Speed difference calculation unit and the drive calculated by this wheel speed difference calculation unit
A clamp for adjusting the restraining torque is based on the absolute value of the circumferential speed difference between the wheels.
Differential restraint torque calculation unit that calculates the differential restraint torque of the
and the differential restraint torque for the drive wheels can be adjusted arbitrarily.
This differential locking torque controls the operation of the locking torque adjustment clutch.
Control according to the differential restraint torque calculated by the calculation unit
An electronic control unit is installed, and a pair of left and right controls are installed based on the road surface conditions.
to these driving wheels according to the absolute value of the difference in circumferential speed of the driving wheels.
The differential locking torque of the locking torque adjustment clutch is assigned to
Since it can be changed at will, it can be used on any road surface.
Always maintain appropriate differential restraint torque without compromising vehicle turning performance.
can be applied to the locking torque adjustment clutch.
.

[Brief explanation of the drawing]

FIG. 1 is a conceptual diagram of an embodiment in which a driving wheel differential limiting device according to the present invention is applied to a front wheel drive type vehicle equipped with an engine driving force control device.

FIG. 2 is a schematic configuration diagram of this embodiment.

FIG. 3 is a graph showing the relationship between the amount of current applied to the restraint torque adjusting clutch and the differential restraint torque.

FIG. 4 is a schematic configuration diagram of a power steering device in this embodiment.

FIG. 5 is a sectional view showing a throttle valve drive mechanism according to the present embodiment.

FIG. 6 is a flowchart showing the overall flow of control according to the present embodiment.

FIG. 7 is a flowchart showing the flow of learning and correcting the neutral position of the steering shaft.

FIG. 8 is a flowchart showing the flow of learning and correcting the neutral position of the steering shaft.

FIG. 9 is a flowchart showing the flow of learning and correcting the neutral position of the steering shaft.

FIG. 10 is a map showing the relationship between vehicle speed and variable threshold value.

FIG. 11 is a graph showing an example of the amount of correction when learning and correcting the neutral position of the steering shaft.

FIG. 12 is a block diagram showing a calculation procedure for a target drive torque for slip control.

FIG. 13 is a block diagram showing a calculation procedure for a target drive torque for slip control.

FIG. 14 is a map showing the relationship between vehicle speed and correction coefficient.

FIG. 15 is a map showing the relationship between vehicle speed and running resistance.

FIG. 16 is a map showing the relationship between the steering shaft turning amount and the correction torque.

FIG. 17 is a map regulating the lower limit value of the target drive torque immediately after the start of slip control.

FIG. 18 is a graph showing the relationship between the coefficient of friction between the tire and the road surface and the slip rate of the tire.

FIG. 19 is a map showing the relationship between target lateral acceleration and slip correction amount accompanying acceleration.

FIG. 20 is a map showing the relationship between lateral acceleration and slip correction amount associated with turning.

FIG. 21 is a circuit diagram for detecting an abnormality in a steering angle sensor.

FIG. 22 is a flowchart showing the flow of abnormality detection processing of the steering angle sensor.

FIG. 23 is a map showing the relationship between vehicle speed and correction coefficient.

FIG. 24 is a flowchart showing the flow of a lateral acceleration selection procedure.

FIG. 25 is a map showing the relationship between slip amount and proportionality coefficient.

FIG. 26 is a map showing the relationship between vehicle speed and lower limit value of integral correction torque.

FIG. 27 is a graph showing an increase/decrease range of integral correction torque.

FIG. 28 is a map showing the relationship between each gear stage of the hydraulic automatic transmission and a correction coefficient corresponding to each correction torque.

FIG. 29 is a map showing the relationship between engine speed, required drive torque, and accelerator opening.

FIG. 30 is a flowchart showing the flow of slip control.

FIG. 31 is a block diagram showing a procedure for calculating a target drive torque for turning control.

FIG. 32 is a block diagram showing a procedure for calculating a target drive torque for turning control.

FIG. 33 is a map showing the relationship between vehicle speed and correction coefficient.

FIG. 34 is a graph showing the relationship between lateral acceleration and steering angle ratio for explaining the stability factor.

FIG. 35 is a map showing the relationship between target lateral acceleration, target longitudinal acceleration, and vehicle speed.

FIG. 36 is a map showing the relationship between lateral acceleration and load torque.

FIG. 37 is a graph showing an example of a procedure for learning and correcting the fully closed position of the accelerator opening sensor.

FIG. 38 is a flowchart illustrating another example of the flow of learning and correcting the fully closed position of the accelerator opening sensor.

FIG. 39 is a flowchart illustrating another example of the flow of learning and correcting the fully closed position of the accelerator opening sensor.

FIG. 40 is a flowchart showing the flow of turning control.

FIG. 41 is a flowchart showing the flow of turning control.

FIG. 42 is a flowchart showing the flow of a final target torque selection operation.

FIG. 43 is a flowchart showing the flow of a retard ratio selection operation.

FIG. 44 is a flowchart showing the flow of a retard ratio selection operation.

FIG. 45 is a flowchart showing a procedure for engine output control.

FIG. 46 is a flowchart showing a procedure for engine output control.

FIG. 47 is a block diagram showing the calculation procedure of differential restraint torque control.

FIG. 48 is a block diagram showing the calculation procedure of differential restraint torque control.

FIG. 49 is a map showing the relationship between vehicle speed and reference circumferential speed difference.

FIG. 50 is a map showing the relationship between corrected front wheel speed difference and reference differential restraint torque.

FIG. 51 is a geometric conceptual diagram showing the steering state of the right front wheel.

FIG. 52 is a graph showing the relationship between sideslip angle and cornering force.

FIG. 53 is a block diagram showing a calculation procedure by road surface μ estimating means.

FIG. 54 is a graph showing the relationship between steering shaft turning angle and power steering pressure.

FIG. 55 is a flowchart showing a road surface μ estimation procedure.

FIG. 56 is a flowchart showing a road surface μ estimation procedure.

FIG. 57 is a map showing the relationship between vehicle speed and multiplication coefficient.

FIG. 58 is a map showing the relationship between road surface μ and road surface correction coefficient.

FIG. 59 is a map showing the relationship between accelerator opening and additional restraint torque.

FIG. 60 is a map showing the relationship between vehicle speed and vehicle speed correction coefficient.

FIG. 61 is a map showing the relationship between the steering shaft turning angle and the turning correction coefficient.

FIG. 62 is a flowchart showing the procedure of differential restraint torque control.

FIG. 63 is a flowchart showing the procedure of differential restraint torque control.

[Explanation of symbols]

11 is an engine, 12 and 63 are output shafts, 13 is a hydraulic automatic transmission, 14 is an input shaft, 15 is an ECU, 16 is a hydraulic control device, 17 is a combustion chamber, 18 is an intake pipe, 19 is an intake passage, 2
0 is a throttle valve, 21 is a throttle body, 22 is a throttle shaft, 23 is an accelerator lever, 24 is a throttle lever, 25 is a cylinder part, 26 is a bush, 27 is a spacer,
28 is a washer, 29 is a nut, 30 is a cable receiver, 31
is the accelerator pedal, 32 is the cable, 33 is the collar, 3
4 is a claw portion, 35 is a stopper, 36 is a torsion coil spring,
37, 38 are spring receivers, 39 is a stopper pin, 40 is a torsion coil spring, 41 is an actuator, 42 is a diaphragm, 43 is a control rod, 44 is a pressure chamber, 45 is a compression coil spring, 46 is a surge tank, 47 is a connection Piping, 48 is a vacuum tank, 49 is a check valve, 50, 55 are piping, 51, 56 are torque control solenoid valves, 52 is a plunger, 53 is a valve seat, 54 is a spring, 57 is a plunger, 58 is a spring, 59 is a fuel injection nozzle, 60 is a solenoid valve, 61 is a spark plug, 62 is a crank angle sensor, 64, 65 are front wheels, 66, 97 are front wheel rotation sensors, 67 is a throttle opening sensor, 68 is an idle switch, 69 is Air cleaner, 70 is an air flow sensor, 71 is a water temperature sensor, 72 is an exhaust pipe, 73 is an exhaust passage, 74 is an exhaust temperature sensor, 75 is an ignition key switch, 76 is a TCL, 77 is an accelerator opening sensor, 78 and 79 are rear 80 and 81 are rear wheel rotation sensors, 82 is a vehicle, 83 is a steering shaft, 84 is a steering angle sensor, 85 is a steering handle, 86 is a steering shaft reference position sensor, 87 is a communication cable, 88 is a comparator, 8
9 is a restraint torque adjustment clutch, 90 is a differential device, 91
92 is a power steering device, 93 is a tie rod, 94 is a steering valve, 95 is a hydraulic pump, 96 is a reservoir tank, and 98 and 99 are pressure sensors. Further, 101 and 102 are selection units, 103 is a changeover switch, 104, 105, 117, 126, 127,
129,132,135,146,147,154,1
63, 166, 168 are multiplication parts, 106, 131, 13
8 is a differential operation section, 107, 110, 125, 133, 1
36 is a clip part, 108, 123 is a filter part, 10
9 is a torque conversion unit, 111 is a running resistance calculation unit, 112,
114, 119, 130, and 134 are addition units; 113 is a cornering drag correction amount calculation unit; 115 is a variable clip unit; 116, 121, 124, and 157 are subtraction units;
8 is an acceleration correction section, 120 is a turning correction section, 122 is a lateral acceleration calculation section, 128 is an integral calculation section, 137 and 150 are start/end determination sections, 140 is a vehicle speed calculation section, 141 is a target lateral acceleration calculation section, 142 143 is a changeover switch; 144 is a target longitudinal acceleration calculation unit; 145
149 is a reference drive torque calculation section, 149 is a variation clipping section,
151 is a circumferential speed difference calculation section, 152 is a turning correction calculation section, 1
53 is a reference circumferential speed difference calculation unit, 155 is a reference restraint torque calculation unit, 156 is a road surface μ estimation means, 158 is a phase compensation filter, 159 is a road surface μ calculation unit, 160 is a μ fluctuation restriction unit,
161 is a stabilization filter, 162 is a road surface correction coefficient calculation unit, 164 is an additional restraint torque calculation unit, 165 is a vehicle speed correction coefficient calculation unit, 167 is a turning correction coefficient calculation unit, 169 is a differential restraint torque calculation unit, and 170 is a maximum The differential restraint torque calculation section 171 is a clip processing section. Furthermore, A is the stability factor, b is the tread of the rear wheel, C1, C2,
C3 is a constant, Cr is the turning radius of the vehicle, DF is the cornering force, FC is the turning control flag, FH
, FHN is the steering angle neutral position learned flag, FP is the ignition timing control flag, FS is the slip control flag, FW
is the abnormality flag, GF is the actual front wheel acceleration, GKC,
GKF is front wheel acceleration correction amount, Gs is slip amount change rate, GX is longitudinal acceleration, GXF is corrected longitudinal acceleration, GX
O is target longitudinal acceleration, GY is lateral acceleration, GYF is corrected lateral acceleration, GYO is target lateral acceleration, g is gravitational acceleration, K
C is the turning angle correction coefficient, KD is the differential coefficient, Km is the multiplication coefficient, KP is the proportional coefficient, KR is the road surface correction coefficient, K
S is the vehicle speed correction coefficient, KV is the weighting coefficient, KY
is the correction coefficient, NE is the engine speed, P is the ignition timing, PL
S, PRS is the pressure in the hydraulic chamber, PS is the power steering pressure, pB
is the basic retardation amount, po is the target retardation amount, r is the front wheel effective radius, SO is the target slip ratio, s is the slip amount, TB
is the reference drive torque, TBF is the modified differential restraint torque, TC
is the cornering drag correction torque, TD is the differential correction torque, Td is the required drive torque, TF is the differential restraint torque, TI is the integral correction torque, TIL is the lower limit of the integral correction torque, TK is the allowable increase/decrease amount, TL is the road load torque , TO is the final target drive torque, TOC is the target drive torque for turning control, TOS is the target drive torque for slip control, TP is the proportional correction torque, TPI is the proportional integral correction torque, TPID is the final correction torque, TR is the running resistance, TSL is the maximum differential restraint torque, TSP is the differential restraint torque for addition, TSS is the standard addition restraint torque, ΔT is the difference between the target drive torque between this time and the previous time, ΔTBF is a constant value, ΔTI is the minute integral correction torque, ΔTi is the integral constant, Δt is the sampling period, V is the vehicle speed, VA, VB
, VX is the threshold, VF is the actual front wheel speed, VFL is the left front wheel speed, VFO is the target front wheel speed for standard torque calculation, VFR is the right front wheel speed, VFS is the target front wheel speed for corrected torque calculation, VH is the larger rear wheel. speed, VK and VKC are slip correction amounts,
VKF is the corrected slip correction amount, VL is the smaller rear wheel speed, VRL is the left rear wheel speed, VRR is the right rear wheel speed, VS is the vehicle speed for slip control, ΔVBF is the reference peripheral speed difference, ΔVF
is the front wheel speed difference due to turning, ΔVFF is the corrected front wheel speed difference, W
b is the vehicle weight, α is the weighting coefficient, δ is the front wheel steering angle, δF is the sideslip angle, δH is the steering shaft turning angle, δH1,
Δμ is a predetermined value, δM is the neutral position, δm is the steering shaft turning position, δN is the steering shaft reference position, Δδ is the correction limit amount, μ
is the friction coefficient, θA is the accelerator opening, θA (0) is the initial value of the accelerator opening, θAC is the fully closed value of the accelerator opening, θ
AL is the minimum accelerator opening, θT is the throttle opening, θTO is the target throttle opening, ρd is the differential gear reduction ratio, ρH is the steering gear gear ratio, ρKI is the integral correction coefficient,
ρKP is a proportional correction coefficient, ρm is a gear ratio of the hydraulic automatic transmission, ρT is a torque converter ratio, and ω is a wheel base.

Claims (1)

[Claims] 1. A restraint torque adjustment clutch that is assembled to a differential device interposed between an engine and a pair of left and right drive wheels, and capable of arbitrarily adjusting differential restraint torque for these drive wheels; a pair of wheel speed sensors that respectively detect the speeds of the pair of drive wheels, and a wheel speed difference calculation unit that calculates the absolute value of the circumferential speed difference of the drive wheels based on detection signals from the pair of wheel speed sensors; a differential restraint torque calculation unit that calculates a differential restraint torque of the restraint torque adjustment clutch based on the absolute value of the circumferential speed difference of the driving wheels calculated by the wheel speed difference calculation unit; A differential limiting device for drive wheels, comprising: an electronic control unit that controls operation of the restraint torque adjusting clutch according to an output from a torque calculating section.