JPH04214945A - Output controlling device for vehicle - Google Patents

Abstract

PURPOSE:To stop an extreme output reduction during operation of a transmission, and prevent stalling of an engine by prohibiting a forced control to a first torque reducing means during operation by an automatic transmission in case a delay angle quantity for an ignition timing by a second torque reducing means is more than a predetermined value. CONSTITUTION:A torque control unit 75 controls actuation of a first torque reducing means to set a drive torque of an engine 11 to be a desired drive torque, and controls actuation of a second torque reducing means based on a change quantity of slip. In case a delay angle quantity for an ignition timing of the engine 11 by the second torque reducing means is more than a predetermined value, actuation of the first torque reducing means is controlled to reduce the drive torque of the engine 11 to the maximum regardless of the desired drive torque. In the meanwhile, during operation of an automatic transmission 13, this forced control is prohibited. Possibilities of operations by both reducing means to be conflicting to each other are thus eliminated, and an extreme reduction of an output during the operation of the transmission is stopped, thereby the engine 11 can be prevented from stalling.

Description

Detailed Description of the Invention [0001] [Industrial Application Field] The present invention is directed to
The engine drive torque is quickly adjusted according to the slip amount of the drive wheels.
of vehicles that have been reduced to allow for safe driving
This invention relates to an output control device. [Prior Art] Road surface conditions change rapidly while a vehicle is running.
or slippery surfaces with low friction coefficients, such as snowy or icy roads.
When a vehicle runs on a road surface such as a road, the drive wheels may spin
The vehicle becomes inoperable and becomes extremely dangerous. [0003] In such cases, it is necessary to prevent the drive wheels from spinning.
The driver adjusts the amount of accelerator pedal depression and the engine
It is extremely difficult even for an expert to delicately control the output of
It is very difficult. [0004] For this reason, it is necessary to check the idle state of the drive wheels.
is detected, and if the spin of the drive wheels exceeds a certain level, the
Regardless of how much the driver presses the accelerator pedal,
An output control device that forcibly reduces engine output.
The driver can control this output control device as needed.
Corresponds to driving using and the amount of accelerator pedal depression
You can choose between normal driving and controlling engine output.
The results have been announced. Vehicle output control based on this viewpoint
Of those related to the rotation of the drive wheel, the conventionally known
Detects the rotation speed and the rotation speed of the driven wheel, and calculates the difference between these rotation speeds.
is regarded as the slip amount of the drive wheel, and according to this slip amount
The drive torque of the engine is controlled by the engine. In other words, conventional output control devices control changes in the rotational state of the driven wheels.
Estimates the road surface condition based on and responds to this road surface condition.
Set the standard driving torque for the engine that has been
The rotational speed of the driving and driven wheels is determined by the reference driving torque.
The engine drive torque was corrected based on the difference between
The torque control means is controlled to achieve this driving torque.
ing. On the other hand, automatic transmissions for vehicles use clutches and brakes.
By supplying hydraulic pressure to frictional engagement elements such as keys etc.
Speed ratio switching by selecting rotating elements such as wheels, gears, etc.
(speed change) automatically according to the driving condition of the vehicle
In order to protect equipment and equipment and maintain a comfortable ride.
Pressure oil is supplied to this frictional engagement element after the shift start signal is sent.
gradually from the initial hydraulic pressure supplied to the
It will be held in In the case of such an automatic transmission, the driver
Arcel pedal regardless of shift operation by automatic transmission
continues to step on. Problem to be Solved by the Invention: This conventional output control device
In this system, the road surface condition is determined based on changes in the rotational state of the driven wheels.
However, the actual road surface conditions vary widely.
A method that accurately estimates the coefficient of friction of the road surface against the tires.
As a practical matter, it is extremely difficult to determine
. [0008] For this reason, in reality there are a large number of correction factors.
The control conditions must be defined by adopting
The program may become complicated and cause control delays, or
There was a risk that problems such as increased cost of the device would occur. [0009] In vehicles equipped with automatic transmissions, power reduction operation
When the vehicle is running, the vehicle speed will not increase, so the driver should
You may try to accelerate by pressing the accelerator pedal.
. At this time, it becomes easier to shift gears, and gear shifts are carried out.
and engine output is reduced, resulting in a large shift shock.
or engine stall may occur.
there were. [Means for solving the problem] When the vehicle is running at other than extremely low speeds,
When driving, the drive wheels are more or less relative to the road surface.
It has slipped. However, the relationship between the road surface and the driving wheels
When a driving torque greater than the frictional force is applied, the driving wheel
The amount of slip increases rapidly, making it difficult to maneuver the vehicle.
It is well known from experience that this is difficult. [0011] Because of this, the drive generated by the engine
Makes it difficult to maneuver the vehicle while making effective use of torque.
To prevent slippage of the drive wheels, the engine drive torque must be
The torque does not exceed the maximum value of the frictional force between the road surface and the drive wheels.
Therefore, it is desirable to control the driving torque of this engine.
stomach. [0012] In other words, the driving torque generated by the engine can be used effectively.
In order to make it work, the tire slip rate S and this
Figure 10 shows the relationship between the friction coefficient between the ear and the road surface.
In other words, the slip rate S of the tires of the driving wheels during driving is
Maximum value of friction coefficient between tire and road surface and corresponding target slippage
slip rate SO or near this target slip rate SO
The slip amount of the drive wheels is set to a value smaller than this.
to avoid energy loss and improve vehicle operation.
It is desirable not to impair longitudinal performance or acceleration performance.
stomach. [0013] Here, V is the speed of the vehicle (hereinafter referred to as the vehicle speed).
), and VD is the circumferential speed of the driving wheel.
The slip rate S of the tire is S = (VD - V)/V, and this slip rate S is the friction coefficient between the tire and the road surface.
The target slip ratio corresponding to the maximum value of SO or its vicinity
The drive of the engine 11 is set so that the value is smaller than this.
Just set the torque. [0014] The vehicle output control device according to the present invention
This was done in light of the knowledge that
The first torque low which reduces the driving torque of the engine independently of
torque reduction means, driver's operation, and first torque reduction means.
The amount of retardation of engine ignition timing independent of drive torque reduction operation
Second torque reduction that increases the engine drive torque and reduces the engine drive torque
Based on the means and driving speed of vehicles with automatic transmission
Target drive wheel speed setting method for setting the target circumferential speed of the drive wheels
speed and the target set by this target drive wheel speed setting means.
The driving torque that serves as the reference for the engine is determined in accordance with the driving wheel speed.
The reference drive torque setting means to be set and this reference drive torque.
from the reference drive torque set by the torque setting means.
Target drive of the engine based on the amount of slip of the driving wheels
target drive torque setting means for setting torque; and the engine;
The drive torque of is set by this target drive torque setting means.
The first torque is reduced so that the target drive torque is achieved.
Based on the amount of change in slip as well as controlling the operation of the means
a torque control unit that controls the operation of the second torque reduction means;
engine ignition timing retardation by the second torque reduction means.
When the angle amount is greater than a predetermined value, the machine is activated regardless of the target drive torque.
The first torque is set to the state where the drive torque of the engine is reduced to the maximum.
A forced control means for forcibly controlling the operation of the reduction means and an automatic change
During gear shifting, the first torque is reduced by forced control means.
characterized by comprising a prohibition means for prohibiting the operation of the means
That is. [Operation] The target drive wheel speed setting means sets the target driving wheel speed based on the vehicle speed.
The drive wheel speed is set and the reference drive torque setting means is based on this target.
A reference drive torque is set based on the drive wheel speed. and
, the target drive torque setting means is based on the slip amount of the drive wheels.
setting a target drive torque from the reference drive torque;
This is output to the torque control unit. [0016] Torque control is performed from the target drive torque setting means.
When the engine's target drive torque is output to the control unit, the engine
The torque control unit adjusts the engine drive torque to this target drive torque.
The operation of the first torque reducing means is controlled so that the
and a second torque reduction based on the amount of change in slip.
controlling the actuation of the means and controlling the second torque by the forced control means;
The amount of retardation of the engine's ignition timing due to the risk reduction means exceeds a predetermined value.
In this case, the engine drive torque is maximized regardless of the target drive torque.
The operation of the first torque reduction means is controlled to a state where the torque is reduced to the maximum extent.
Forced control during gear shifting of automatic transmission by means of control and prohibition.
The operation of the first torque reduction means by the means is prohibited, and the driver
Reduces engine drive torque regardless of operation by
. On the other hand, when the slip amount of the driving wheels is small,
Of course, the torque control unit will reduce the torque
Operation of the engine based on the operation of the driver without activating the means
will be held. [Embodiment] The vehicle output control device according to the present invention has four forward speeds.
A front-wheel drive vehicle equipped with an automatic transmission with one reverse gear.
Figure 1 showing a schematic mechanism of an embodiment applied to and its concept
As shown in FIG. 2, the output shaft 12 of the engine 11 is equipped with oil.
An input shaft 14 of a pressure type automatic transmission 13 is connected thereto. this
The hydraulic automatic transmission 13 is operated by a driver (not shown).
engine depending on the selected position of the lever and the operating condition of the vehicle.
Electronic control unit (hereinafter referred to as “electronic control unit”) that controls the operating status of 11
Based on commands from 15 (hereinafter referred to as ECU), hydraulic pressure
Automatically select a predetermined gear via the control device 16
It looks like this. Specific details of this hydraulic automatic transmission 13
Regarding the structure and operation, etc., see, for example, Japanese Patent Application Laid-Open No. 58-542.
It has already been published in Publication No. 70 and Japanese Patent Application Laid-Open No. 61-31749.
As is well known, there is a hydraulic automatic changer in the hydraulic control device 16.
Engagement operation of a plurality of frictional engagement elements forming a part of the speed gear 13
A pair of shift systems (not shown) for performing opening and opening operations
Standard solenoid valves are incorporated, and these solenoid valves for shift control
The ECU 15 controls the on/off operation of energization.
By doing so, you can shift to any gear among 4 forward gears and 1 reverse gear.
This achieves smooth gear shifting operations. 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. 3, 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. [0020] 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. [0022] 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 against the claw part 34 of the accelerator lever 23 and
Torsion coil spring 3 biasing the torque valve 20 in the direction to open it
6 is a pair of cylindrical cylinders fitted on the throttle shaft 22.
Through the spring receivers 37 and 38, the throttle shaft 22 and the
It is installed in the shape of a shaft. Also, throttle body 2
Stopper pin 39 protruding from 1 and accelerator lever 23
Also, insert the claw part 34 of the Arcel lever 23 into the slot between
the throttle valve by pressing it against the stopper 35 of the lever 24.
20 in the closing direction, and press it against the accelerator pedal 31.
Torsion coil spring 40 for providing a detent feeling
is connected to the cylindrical portion 2 of the accelerator lever 23 via the collar 33.
5 coaxially with the throttle shaft 22.
. At the tip of the throttle lever 24,
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 Arcel lever 23 and
Compression coil spring 45 biasing the throttle valve 20 in the direction to open it
is included. And these two springs 36,
45 is greater than the sum of the spring forces of the torsion coil spring 40.
The spring force is set higher, which causes the accelerator pedal to
The throttle valve 20 will not open unless the dial 31 is depressed.
There is no such thing. 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. [0025] Inside these vacuum tanks 48 and the above-mentioned
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. [0026] 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. [0027] 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.
In this example, all of these are considered to be the first aspect of the present invention.
It constitutes the torque control means. 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. Furthermore, 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.
is 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 [0031] The ECU 15 has a
Crank angle sensor 6 for detecting engine speed
2 and the rotation speed of the output shaft 63 of the hydraulic automatic transmission 13
Detects the flatness of the pair of left and right front wheels 64 and 65, which are the driving wheels.
A front wheel rotation sensor 66 for calculating the uniform peripheral speed and a slot
The throttle lever 24 is attached to the torque body 21.
A throttle opening sensor 67 detects the opening of the throttle opening.
Idle switch 68 that detects the fully closed state of the torque valve 20
In addition, it is assembled into the air cleaner 69 at the tip of the intake pipe 18.
Detects the amount of air flowing into the combustion chamber 17 of the engine 11
An air flow sensor 70 such as a Karman vortex flow meter and a machine
It is attached to engine 11 to detect the cooling water temperature of this engine 11.
The water temperature sensor 71 is assembled in the middle of the exhaust pipe 72.
Exhaust for detecting the temperature of exhaust gas flowing in the exhaust passage 73
A temperature sensor 74 is connected. [0032] These crank angle sensors 62 and
Front wheel rotation sensor 66 and throttle opening sensor 67 and
Idle switch 68 and air flow sensor 70 and water
The output signals from the temperature sensor 71 and the exhaust temperature sensor 74 are
The signals are sent to the ECU 15, respectively. [0033] Also, calculate the target drive torque of the engine 11.
Torque calculation unit (hereinafter referred to as TCL)
75, the throttle opening sensor 67 and the idler
It is attached to the throttle body 21 together with the switch 68.
Accelerator opening to detect the opening of the accelerator lever 23
A sensor 76 and a pair of left and right rear wheels 77 and 78 which are driven wheels.
Rear wheel rotation sensors 79 and 8 detect the rotational speed of
0 and when turning with reference to the straight-ahead state of the vehicle 81.
A steering angle sensor 83 that detects the turning angle of the steering shaft 82 is connected.
Next, the outputs from these sensors 76, 79, 80, 83
signals are sent respectively. [0034] The ECU 15 and TCL 75 are connected via a communication cable.
The engine speed is connected to the ECU 15 via a
The number of revolutions, the number of revolutions of the output shaft 63 of the hydraulic automatic transmission 13, and
The operation of the engine 11 such as the detection signal from the idle switch 68
Information on the rotation status is sent to the TCL 75. On the contrary, TCL7
From 5 onwards, the target drive torque calculated by this TCL75
and information regarding the ignition timing retard rate is sent to the ECU 15.
It will be done. In this embodiment, when slip control is performed
The target drive torque of the engine 11 and when swing control is performed.
The target drive torque of engine 11 is calculated by TCL75.
, the optimal final target drive from these two target drive torques.
Select the torque and adjust the driving torque of the engine 11 as necessary.
We are trying to reduce this. Also, the actuator 41
The engine 1 can also be fully closed by fully closing the throttle valve 20 through the
The ignition timing has been adjusted in consideration of the case where the output reduction in step 1 is not in time.
Set the target retardation amount and quickly reduce the driving torque of the engine 11.
We are trying to reduce it. The rough outline of the control according to this embodiment is as follows.
As shown in Figure 4, which shows the flow, the ignition switch (not shown)
The control program of this example is started by turning on the key.
At M1, the initial value δm(o
), reset various flags, or control this
The main timer counter every 15 milliseconds, which is the sampling period.
Initial settings such as starting the process are performed. [0037] Then, M2 receives detection signals from various sensors.
Based on the above, TCL75 calculates the vehicle speed V and the turning angle of the steering shaft 82.
δH etc. are calculated, and then the neutral position of the steering shaft 82 is calculated.
δM is learned and corrected using M3. The steering shaft of this vehicle 81
The neutral position δM of 82 is shown in the ECU15 and TCL75.
The ignition
The initial value δm(o) is read every time the function key is turned on.
In rare cases, when the vehicle 81 satisfies the straight running conditions described below
The learning correction is performed only when the ignition key is off and
This initial value δm(o) is corrected by learning until
It has become. Next, TCL75 uses M4 to adjust the front wheel rotation sensor.
Detection signals from sensor 66 and rear wheel rotation sensors 79 and 80
The drive torque of the engine 11 is regulated based on the detection signal of
To calculate the target drive torque TOS when performing slip control
Detection signals from rear wheel rotation sensors 79 and 80 are detected using M5.
engine based on the number and the detection signal from the steering angle sensor 83.
The mechanism when performing swing control to regulate the driving torque of 11
The target drive torque TOC of function 11 is calculated. [0039] In M6, TCL75 has these eyes.
From the standard drive torque TOS, TOC, it is small considering safety.
The value that is the best value is selected as the final target drive torque TO. Furthermore, when starting suddenly or when the road surface changes from a normal dry road to an icy road,
If there is a sudden change, the actuator 41
The output of the engine 11 is also reduced by fully closing the throttle valve 20.
There is a risk that the reduction will not be done in time, so the front wheels of M7 are reduced to 64,
The basic retard angle is based on the rate of change Gs of the slip amount s of 65.
Select the retardation ratio for correcting the amount pB, and use these
Retardation of final target drive torque TO and basic retardation amount pB
Data regarding the ratio is output to the ECU 15 via M8. [0040]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.
If necessary, delay by the target retard amount pO, thereby
to ensure that vehicle 81 runs smoothly and safely.
. [0041] 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 81 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 way, the driving torque of the engine 11 is adjusted to M9.
every 15 milliseconds, which is the sampling period of the main timer.
Control until the countdown ends, and from then on M2
The steps from to M10 are executed by the ignition key.
Repeat until it turns off. [0043] By the way, the turning control is performed in step M5.
When calculating the target drive torque TOC of the engine 11 by
, TCL75 detects a pair of rear wheel rotation sensors 79 and 80.
Based on the signal, calculate the vehicle speed V using the following equation 1, and
Based on the detection signal from the steering angle sensor 83, the front wheels 64,
The steering angle δ of 65 is calculated from the following equation 2, and the vehicle 81 at this time is
Find the target lateral acceleration GYO from the following equation 3.
Ru. [Equation 1] [Equation 1] [Equation 2] [Equation 3] [Equation 3] However, VRL and VRR are each a pair of left and right
The circumferential speed of the rear wheels 77 and 78 (hereinafter referred to as rear wheel speed)
), ρH is the steering gear gear ratio, and l is the wheel drive ratio of the vehicle 81.
A is the stability factor of vehicle 81, which will be described later.
It's Kuta. As is clear from this number 3, the vehicle 81
If front wheels 64 and 65 are toe-in adjusted during maintenance,
Due to age-related changes such as wear of the steering gear (not shown),
If the neutral position δM of the steering shaft 82 changes, the steering shaft
The actual turning position of 82 and the front wheels 64 and 65, which are the steering wheels.
A deviation occurs between the steering angle and the steering angle δ. As a result, the vehicle 81
Target lateral acceleration GYO cannot be calculated accurately.
This may make it difficult to properly control the turning. Moreover, in this embodiment, the slip control at the M4 step
In this case, this target lateral acceleration GYO is the front wheel acceleration described later.
Used to calculate the correction amount GKC and slip correction amount VKC.
Because of this, there is a risk that slip control may not be performed properly.
There is. For this reason, the neutral position δ of the steering shaft 82
It is necessary to perform learning correction on M in step M3. The neutral position δM of the steering shaft 82 is determined by learning compensation.
As shown in FIGS. 5 and 6 showing the correcting procedure, TCL75
is based on the detection signals from the rear wheel rotation sensors 79 and 80,
In H1, this steering shaft neutral position δM is learned and will be described later.
The vehicle speed V for turning control is calculated using Equation 1 above. Next, TCL75 adjusts the rear wheel speed VRL in H2.
, VRR difference (hereinafter referred to as rear wheel speed difference) |V
Calculate RL−VRR| [0051] After that, TCL75 increases the vehicle speed V at H3.
Determine whether or not is greater than a preset threshold value VA
. This operation requires the vehicle 81 to reach a certain high speed.
Rear wheel speed difference due to steering |VRL-VRR|, etc. cannot be detected.
The threshold value VA is necessary for the vehicle 81
For example, based on the running characteristics of
It is set as appropriate, such as km. [0052] Then, if the vehicle speed V is equal to or higher than the threshold value VA,
If it is determined, TCL75 will change the rear wheel speed difference |V at H4.
RL-VRR｜preset, for example, 0.3km/hour
Whether or not the vehicle 81 is smaller than a threshold value VB such as
Determine whether or not the vehicle is traveling straight. Here, the threshold V
B is not 0km/h because the left and right rear wheels 77, 78
If the air pressures of the tires are not equal, the vehicle 81 is traveling straight.
The circumference of the pair of left and right rear wheels 77 and 78
When the speeds VRL and VRR are different and the vehicle 81 is traveling straight,
This is to avoid determining that there is no such information. [0053] At this H4 step, rear wheel speed difference | VRL
-VRR| is determined to be less than or equal to the threshold value VB,
TCL75 is the current steering shaft turning position δm(n) at H5.
The previous steering shaft turning position detected by the steering angle sensor 83
It is determined whether the position is the same as δm(n-1). child
To avoid being affected by the driver's hand shake, etc.,
The turning detection resolution of the steering shaft 82 by the steering angle sensor 83 is
For example, it is desirable to set it to around 5 degrees. [0054] At this step H5, the current steering axis rotation
The position δm(n) is the previous steering shaft turning position angle δm(n-1
), TCL75 changes to H6.
It is determined that the current vehicle 81 is traveling straight, and this TC
Counting of the learning timer (not shown) built into the L75
and continues this for, for example, 0.5 seconds. Next, TCL75 sets the learning timer at H7.
Whether 0.5 seconds have passed since the start of counting, that is, the vehicle
Determine whether 81 continues to go straight for 0.5 seconds
. In this case, at the beginning of the journey of the vehicle 81, the learning tie
Since 0.5 seconds have not elapsed since the car started counting, the car
At the beginning of the run of both 81, steps from H1 to H7 are repeated.
It will be returned. [0056] Then, from the start of counting of the learning timer,
When determining that 0.5 seconds have passed, TCL75 goes H.
At step 8, the steering angle neutral position learned flag FH is set.
In other words, whether this learning control is the first time or not.
judge. [0057] In this step H8, the steering angle neutral position is learned.
If it is determined that the completed flag FH is not set,
In H9, the current steering shaft turning position δm(n) is changed to a new
This is regarded as the neutral position δM(n) of the steering shaft 82 and is set as TC.
Load it into the memory in L75 and read the steering angle neutral position learning flag.
Set FH. In this way, the new neutral position of the steering shaft 82 is achieved.
After setting the position δM, the neutral position of the steering shaft 82 is set.
The turning angle δH of the steering shaft 82 is calculated based on δM.
On the other hand, the learning timer count was cleared in H10.
Then, the steering angle neutral position learning is performed again. [0059] Steering angle neutral position learning in step H8
completed flag FH is set, that is, the steering angle is in neutral position.
If it is determined that this is the second or subsequent time, TCL7
5 is H11, and the current steering shaft turning position δm(n) is the previous time.
is equal to the neutral position δM(n-1) of the steering shaft 82, i.e.
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 82.
-1), the H10 step is as is.
The next steering angle neutral position learning is performed once again. [0060] At step H11, the current steering shaft turning position
position δM(n) Due to play in the steering system, etc., the previous steering
It is determined that it is not equal to the neutral position δM(n-1) of the axis 82.
In this case, the current steering shaft turning position δm(n) is changed as is.
This is not determined to be the neutral position δM(n) of the steering shaft 82.
The absolute value of the difference between them differs by more than the preset correction limit amount Δδ
, the previous neutral position δM of the steering shaft 82 (
Subtract or add this correction limit amount Δδ to n-1)
This is the new neutral position δM(n) of the steering shaft 82.
, I tried to read this into the memory inside TCL75.
Ru. [0061] In other words, TCL75 changed its current operation in H12.
From the rudder shaft turning position δm(n) to the previous neutral position of the steering shaft 82
The value obtained by subtracting the value δM(n-1) is the preset negative correction.
It is determined whether the control amount is smaller than the control amount -Δδ. and,
The value subtracted in this H12 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 82 is set to the previous steering shaft.
82 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. [0062] This causes the steering angle to change due to some reason.
Even if an abnormal detection signal is output from the sensor 83,
The neutral position δM of the steering shaft 82 does not change suddenly;
It is possible to quickly respond to abnormalities. On the other hand, the value subtracted in step H12 is
If it is determined that the negative correction limit amount - Δδ is larger than
, from the current steering shaft turning position δm(n) in H14 to the previous time
The value obtained by subtracting the neutral position δM(n-1) of the steering shaft 82 is
It is determined whether or not it is larger than the positive correction limit amount Δδ. So
Then, the value subtracted in this step of H14 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 82 to the previous steering position.
Neutral position δM(n-1) of shaft 82 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.
We are taking care not to make this happen. [0064] This causes the steering angle to change due to some reason.
Even if an abnormal detection signal is output from the sensor 83,
The neutral position δM of the steering shaft 82 does not change suddenly;
It is possible to quickly respond to abnormalities. [0065] However, the value subtracted in step H14 is
If it is determined that it is smaller than the positive correction limit amount Δδ,
In H16, the current steering shaft turning position δm(n) was
Read it out as it is. In this way, the neutral position δM of the steering shaft 82
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 79 and 80.
performs slip control to regulate the driving torque of the engine 11.
Calculate the target drive torque TOS for the case. In this embodiment, rear wheel rotation sensors 79, 80
of the target front wheels 64, 65 based on the detection signal from the
Peripheral speed (hereinafter referred to as target front wheel speed for standard torque calculation)
) Calculate the VFO and set the target front wheel for this reference torque calculation.
Rate of change of speed VFO (hereinafter referred to as target front wheel acceleration)
) Find GFO, and calculate the magnitude of this target front wheel acceleration GFO.
The standard drive torque TB of engine 11 corresponding to
Based on the actual front wheel speed VF detected from the rotation sensor 66
Deviation from target front wheel speed VFO for semi-torque calculation (hereinafter referred to as
is called the slip amount), and the target drive is
The dynamic torque TOS is calculated. [0068] Calculate the target drive torque TOS of this engine 11.
Figures 7, 8, and 9 show the calculation blocks for
First, the TCL75 detects the vehicle speed V using the rear wheel rotation sensor 7.
Calculated based on the detection signals from 9 and 80, but in this implementation
In the example, the low vehicle speed selection unit 101 selects two rear wheel speeds VRL and V.
The smaller value of RR is selected as the vehicle speed V, and the high vehicle speed
The selection unit 102 selects the higher of the two rear wheel speeds VRL and VRR.
Select the higher value as the vehicle speed V, and then press the switch switch.
The output of either selection section 101 or 102 is determined by the switch 103.
You can now choose whether to take in power. In other words, rear wheel speed slip control actually
In a state where the drive torque of connection 11 is reduced, that is, when the
When the top control flag FS is set, the switch
switch 103, the lower of the two rear wheel speeds VRL and VRR is set.
The driver selects the lower value as the vehicle speed V and applies slip control.
Even if you wish to control the engine, the driving torque of engine 11 will be reduced.
In other words, the slip control flag FS is reset.
In the set state, the higher of the two rear wheel speeds VRL and VRR
The higher value is selected as the vehicle speed V. [0070] 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 81 is turning
The smaller value of the two rear wheel speeds VRL and VRR at
is selected as the vehicle speed V, there will be slippage on the front wheels 64 and 65.
A slip has occurred even though no slip has occurred.
Therefore, the driving torque of the engine 11 is reduced.
In order to avoid such problems and to improve the running safety of vehicle 81.
Taking into consideration, once the driving torque of the engine 11 is reduced,
This is because consideration was given to ensuring that this situation continues in case of
. Next, based on this vehicle speed V, the standard front wheel acceleration
The degree GFB is calculated, but when the vehicle 81 is running, it
The front wheels 64, which are the driving wheels, always have a slip amount of about 3%.
, 65, so this slip
The target front wheel speed VFO is lowered by the multiplier 104 in consideration of the amount of
Set as in number 4. ##EQU00004## Then, the group at the n-th sampling is
The quasi-front wheel acceleration GFB(n) is calculated by the differential calculation unit 105.
Calculate according to number 5. [Equation 5] [0075] However, Δt is the sampling period of the main timer.
15 milliseconds, where g is the gravitational acceleration. Slip control is performed when the vehicle 81 accelerates.
is the basic principle, so the base calculated using this number 5 is
Of the semi-front wheel acceleration GFS, less than 0 is sent to the clip part 106.
is truncated, and further filtered by the filter section 107 to remove noise.
Perform filter processing to obtain the corrected reference front wheel acceleration GFF.
Ru. By the way, when the vehicle 81 accelerates, the engine 11
In order to make the generated drive torque work effectively, please refer to Figure 10.
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 81, the handling performance and acceleration performance
It is desirable to avoid damage. [0078] Here, the target slip ratio SO depends on the condition of the road surface.
It can fluctuate between 0.1 and 0.25 depending on the situation.
It is known that the coefficient of friction between the tires and the road surface is
It can be considered to be equivalent to the longitudinal acceleration GX of 81.
Therefore, the longitudinal acceleration GX applied to this vehicle 81 is detected.
and convert the corrected reference front wheel acceleration GFF to this longitudinal acceleration.
It is sufficient to correct it based on GX. [0079] First, the vehicle speed V(n) calculated this time is
From the previously calculated vehicle speed V(n-1), the current
The longitudinal acceleration GX(n) of the vehicle 81 is sent to the differential calculation unit 108.
Calculate as shown in equation 6 below. [Equation 6] [0081] 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 109, the front and back should not exceed 0.6g.
Clip the acceleration GX(n) to 0.6g. In addition,
The filter section 110 performs filter processing as detailed below.
Calculate the corrected longitudinal acceleration GXF. [0082] This filter processing is performed by
The degree GX(n) is equivalent to the coefficient of friction between the tire and the road surface.
Therefore, the longitudinal acceleration of the vehicle 81
The maximum value of degree GX(n) changes and the tire slip rate S
is the target speed corresponding to the maximum value of the friction coefficient between the tire and the road surface.
If the rip rate is about to deviate from SO or its vicinity,
Even if the tire slip rate S is the friction between the tire and the road surface,
The target slip ratio SO corresponding to the maximum value of the coefficient or its
around it so that it is maintained at a smaller value than this.
This is to correct the acceleration GX(n), and the specific
is carried out as follows. [0083] The current longitudinal acceleration GX(n) is filtered.
If the corrected longitudinal acceleration GXF or higher, that is, the vehicle
When 81 continues to accelerate, GXF=(28/256)・Σ(GX(n)-GXF)
Remove noise by delay processing and accelerate before and after correction.
The degree GXF follows the longitudinal acceleration GX(n) relatively quickly.
Go. [0084] The longitudinal acceleration GX(n) is the corrected longitudinal acceleration G
If it is less than XF, that is, vehicle 81 is not accelerating much.
Sometimes the following processing is performed every sampling period Δt of the main timer.
I do. [0085] 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 81 is decelerating.
Therefore, by setting GXF = GXF - 0.002, the decrease in the corrected longitudinal acceleration GXF is suppressed, and the driver
This ensures responsiveness to acceleration requests from the vehicle 81.
. [0086] 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
Both 81s have a low degree of acceleration, so there are no safety issues.
Therefore, by setting GXF = GXF - 0.002, the decrease in the corrected longitudinal acceleration GXF is suppressed, and the driver
This ensures responsiveness to acceleration requests from the vehicle 81.
. Furthermore, driving the engine 11 by slip control
Front wheels 64 and 65 slip while reducing torque
When the amount s is negative, that is, the vehicle 81 is decelerating, the correction
The maximum value of longitudinal acceleration GXF is maintained, and the vehicle is controlled by the driver.
81 to ensure responsiveness to acceleration requests. Corrected longitudinal acceleration obtained in this way
Based on GXF, this corrected longitudinal acceleration corresponds to GXF.
The front wheel acceleration correction amount GKF set in advance is
The main section 111 reads out a map as shown in FIG.
This is added to the correction reference front wheel acceleration GFF in the addition unit 112.
Add to. This front wheel acceleration correction amount GKF is before and after correction.
It increases in stages as the value of acceleration GXF increases.
However, in this example, the driving test
This map is created based on the following. On the other hand, the friction between the tires and the road surface during turning
Figure 10 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
Target slip rate of the tire that 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
It is considerably smaller than the target slip rate SO of the tire.
I understand that. Therefore, while the vehicle 81 is turning, this vehicle 8
In order for 1 to turn smoothly, the corrected standard front wheel acceleration GF
It is desirable to set F to be smaller than when traveling straight. [0090] Therefore, the vehicle 81 calculated using the above equation 3
Front wheel acceleration correction amount GK based on the target lateral acceleration GYO of
C from the map shown in FIG. 12 in the turning correction unit 113.
The subtraction unit 114 reads out this front wheel acceleration correction amount GKC.
Subtract it from the corrected reference front wheel acceleration GFF at
As a result, the target front wheel acceleration GFO shown in the following formula is obtained. GFO=GFF+GKF-GKC 0091]Then, this target front wheel acceleration GFO is -0
．． If it is less than 6g or more than 0.6g, there is a calculation error.
In consideration of safety against
Clip it to a range of -0.6g to 0.6g, and then
After converting this into torque in the torque conversion unit 116, the running resistance is
The running resistance TR calculated by the resistance calculating unit 117 is added to the adding unit.
118, and the standard drive torque TB shown in the formula below is obtained.
calculate. TB = GFO・Wb r+TR [0092] Here, Wb is the vehicle weight, r is the front wheel 64
, 65 effective radius. Also, the running resistance TR is the vehicle
It can be calculated as a function of speed V, but in this example,
is obtained from a map as shown in FIG. in this case,
Since running resistance TR differs between flat roads and uphill roads,
In the diagram, the flat road version is shown as a solid line, and the one for flat roads is shown as a two-dot chain line.
This is a diagram of the vehicle 81 with "for use on uphill roads" written on it.
Based on the detection signal from the tilt sensor not shown,
I try to choose one or the other, but it also includes downhill etc.
It is also possible to set the running resistance more precisely. On the other hand, TCL75 is the front wheel rotation sensor 66.
The actual front wheel speed VF is calculated based on the detection signals from the
As mentioned above, this actual front wheel speed VF and the reference torque mentioned above
The correction torque is set based on the target front wheel speed VFO for calculation.
Slip amount, which is the deviation from the target front wheel speed VFS for torque calculation
Feedback control of the reference torque TB using s
By performing the following, the target drive torque TOS of the engine 11 is
The details are as follows. In this example, the target front wheel acceleration GFO is calculated.
When starting, the acceleration correction section 111 and the turning correction section 113
Since the correction standard front wheel acceleration GFF is corrected with
, make similar corrections to the target front wheel speed VFO for calculating the reference torque.
Then, the target front wheel speed VFS for calculating the correction torque is calculated. [0095] That is, the TCL 75
From the map shown in Figure 14, the above-mentioned corrected longitudinal acceleration can be calculated.
Read out the slip correction amount VK corresponding to GXF and
The adder 119 calculates the target front wheel speed VF for calculating the reference torque.
Add to O. As a result, the target front wheel for calculating the correction torque
The speed VFS increases, and the slip rate S during acceleration increases as shown in Figure 1.
Target slip ratio SO shown by the solid line in 0 or its vicinity
is set to a value smaller than this. Similarly, the turning correction section 113 performs the rotation correction as shown in FIG.
From the map, the speed corresponding to the target lateral acceleration GYO is determined.
Read out the lip correction amount VKC and send it to the subtraction unit 120.
and subtract it from the target front wheel speed VFO for calculating the reference torque. child
As a result, the target front wheel speed VFO for calculating the correction torque decreases.
, the slip rate S when turning is the target slip when going straight.
The acceleration performance of the vehicle 81 becomes smaller than the lip ratio SO.
Although the speed decreases slightly, good turning performance is ensured. [0097] Furthermore, after turning on the ignition key,
The initial learning of the neutral position δM of the steering shaft 82 performed in
Until this is done, the reliability of the turning angle δH of the steering shaft 82 will be
Since there is no front wheel acceleration correction amount GKC and slip compensation
Multiply the correct amount VKC by 0 and set the steering angle neutral position learned flag FH
After this is set, these front wheel acceleration correction amounts GKC
and the slip correction amount VKC is multiplied by 1. As a result of the above, the target front wheel speed for calculating the correction torque
The degree VFS is as shown in the formula below. VFS=VFO+VK-VKC 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 121 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
Then, -2.5km/hour is clipped by the clip part 122.
This prevents the engine 11 from running out of control due to a calculation error. [0100] Also, the slip amount s after this clipping process is
Perform proportional correction and differential correction as described later, and then integrate
Final torque TP is obtained by performing integral correction using correction ΔTi.
Calculate ID. [0101] As the proportional correction, the multiplier 123
The basic correction amount is obtained by multiplying the slip amount s by the proportional coefficient KP.
is calculated, and then the multiplier 124 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. [0102] Also, as the differential correction, the differential calculation section 12
5, calculate the rate of change GS of the slip amount s, and add a small amount to this.
The multiplier 126 multiplies the coefficient KD to detect sudden slip.
A basic correction amount for a change in the amount s is calculated. and
, the gear ratio ρm of the hydraulic automatic transmission 13 in the multiplier 127
Multiply by a preset correction coefficient ρKD based on
The resulting values are constrained by upper and lower limits, respectively.
A limit is set to prevent the differential correction torque TD from reaching an extremely large value.
Clip processing is performed in the clip unit 128 to prevent
The differential correction torque TD is obtained. This clip part
128 indicates wheel speeds VF, VRL,
The VRR may change momentarily depending on the road surface condition, the driving condition of the vehicle 81, etc.
It may become idle or locked for a while, and if this happens,
In such a case, the rate of change Gs of the slip amount s is positive or negative.
If the value becomes extremely large, control diverges and responsiveness decreases.
For example, click the lower limit value to -55 kgm.
At the same time as clipping the upper limit to 55 kgm, perform differential compensation.
Prevent positive torque TD from becoming an extremely large value
It is for. [0103] Proportional correction torque calculated in this way
TP and the differential correction torque TD are added by an adding section 129.
Then, the reference proportional differential correction torque TPD is calculated. stop
This standard proportional difference is calculated according to the friction coefficient between the tire and the road surface.
In order to change the correction width of the correction torque TPD, the multiplier 13
Corrected longitudinal acceleration GX from the map shown in Figure 16 at 0
Read out the slip correction coefficient Km corresponding to F, and
Multiply the reference proportional differential correction torque TPD to obtain the final proportional differential
Calculate the correction torque TPDm. This slip correction
Several km is small when the corrected longitudinal acceleration GXF is small.
It has a tendency to have a value of
Final proportional differential correction torque when driving on easy roads.
Narrow the TPDm correction range to avoid overcontrol.
I am considerate. [0104] Also, in this embodiment, the slip amount s is gentle.
In order to realize correction corresponding to such changes, the integral calculation unit 13
Calculate the basic correction amount in step 1, and multiply this correction amount by
The calculation unit 132 calculates the gear ratio ρm of the hydraulic automatic transmission 13.
Multiply by a preset correction coefficient ρK1 based on
A correction torque T1 is obtained. In this case, in this example
Constant minute correction torque ΔT to improve control convergence
I is integrated every 15 ms sampling period.
When the slip amount s is positive, the minute correction torque ΔT
I, and conversely, if the slip amount s is negative, a small compensation is applied.
Positive torque ΔTI is subtracted. However, actual front wheel speed VF
If the time that VFS exceeds the target front wheel speed is the opposite
The longer the length, the better the acceleration, so the integral correction torque T
Set the upper limit for I, for example 0 kgm, and perform clip processing.
I am rational. In addition, in order to prevent calculation errors, etc., integral correction is performed.
Set the lower limit value for torque TI, for example -100kgm.
clip processing. These clip processing
Therefore, the integral correction torque TI changes as shown in Fig. 17.
Ru. [0105] Note that the correction coefficients ρKP, ρKD, ρK
I is related to the gear ratio ρm of the hydraulic automatic transmission 13.
Read from the preset map shown in Figure 18.
I'm doing it. [0106] Thereafter, in the adding section 133, these final
Proportional differential correction torque TPDm and integral correction torque TI
The final correction torque TPI obtained by adding
The subtraction unit 134 calculates D by the above-mentioned reference drive torque TB.
The engine 11 and the front wheels 64,
Multiply the reciprocal of the total reduction ratio between 65 and axles 85 and 86.
By doing so, the target drive for slip control shown in Equation 7 below can be obtained.
Calculate the dynamic torque TOS. [Equation 7] [0108] However, ρd is the differential gear reduction ratio.
, the hydraulic automatic transmission 13 performs an upshift gear change operation.
When the gear shift is completed, the gear ratio ρm on the high gear side becomes
It is now output. In other words, hydraulic automatic transmission
In the case of a gear change operation for upshifting machine 13, the gear change signal
If the gear ratio ρm on the high gear side is adopted at the time of output, the upper
As is clear from number 7, the target drive torque is
Because TOS increases and engine 11 blows up,
Example of completing the shift operation after outputting the shift start signal
For example, for 1.5 seconds, the target drive torque TOS is made smaller.
The gear ratio ρm on the lower gear side that can be maintained is maintained, and the shift start time is
1.5 seconds after outputting the signal, the gear ratio ρm on the high speed side
will be adopted. For similar reasons, hydraulic automatic transmission 1
In the case of downshifting (3), the gear shift signal is output.
At the moment of power, the gear ratio ρm on the lower gear side is immediately adopted. [0109] The target drive torque TOS calculated by equation 7 is
Naturally, it should be a positive value, so click
In order to prevent calculation errors, the target drive torque is
Clip TOS above 0 and start slip control.
or in the start/end determination unit 137 for determining the end.
According to the determination process, the target drive torque TOS is
Information is output to the ECU 15. [0110] 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 the lower speed side of the two rear wheel speeds VRL and VRR.
is selected as the vehicle speed V by the changeover switch 103.
The information regarding the target drive torque TOS is sent to the EC.
Outputs to U15 to determine the end of slip control and
This process will continue until the under control flag FS is reset.
continue the 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 vehicle, e.g. 4k
gm or more. Note that in this embodiment, this required drive torque Td is
The engine speed calculated from the detection signal from the angle sensor 62
Rotation number NE and detection signal from accelerator opening sensor 76
Based on the accelerator opening θA calculated by
Reading from the set map shown in Figure 19
. (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. [0111] The start/end determination section 137 performs a slip control.
After determining the start of control, as shown in (f) and (g) below.
If any of the conditions are met, slip control
Determine the end and reset the slip control flag FS
and transmits the target drive torque TOS to the ECU 15.
At the same time as stopping, the higher speed of the two rear wheel speeds VRL and VRR
Select the changeover switch 10 to select the side as the vehicle speed V.
Activate 3. (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.
In other words, the driver released the accelerator pedal 31.
The state continues for a certain period of time, for example, 0.5 seconds or more. [0112] In the vehicle 81, slip control is selected by the driver.
A manual switch (not shown) is provided to select the
, the driver operates this manual switch to control slip.
If selected, perform the slip control operations described below.
cormorant. FIG. 2 shows the flow of this slip control process.
0, As shown in Figure 21, TCL75 is as described above in S1.
The target drive torque is determined by detecting and calculating various data.
This calculation operation is performed by the manual switch.
This is done independently of the operation of the key. 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.
, TCL75 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. [0115] In this step S3, the slip amount s is
If it is determined that the speed is greater than 2km, TCL75 will move to S4.
The rate of change Gs of the slip amount s is greater than 0.2g at
Determine whether or not. [0116] In this step S4, the slip amount change rate
If Gs is determined to be larger than 0.2g, TCL7
5, the driver's requested drive torque Td is 81 at S5.
The minimum driving torque required to run the vehicle, e.g. 4k
gm, that is, whether the driver is driving the vehicle 81.
Determine whether or not there is an intention to do so. [0117] In this step S5, the required drive torque T
d is greater than 4 kgm, that is, the driver is driving the vehicle 81.
If it is determined that there is an intention to run the car, it will slip at S6.
Set the control flag FS and perform slip control in S7.
Determine again whether the medium flag FS is set.
Ru. [0118] In this step S7, the slip control
If it is determined that the lug FS is being set, S8
As the target drive torque TOS of the engine 11, use the above equation 7.
Target drive torque TOS for slip control calculated in advance
Adopt. [0119] Also, slip control is performed in step S7.
When it is determined that the medium flag FS has been reset
In S9, TCL75 is 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
. [0120] 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
The required drive torque Td is smaller than 4kgm at the step.
If it is determined that the
Then, in step S9, the TCL75 sets the target drive torque.
The maximum torque of engine 11 is output as 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, engine 11 starts operating.
Drive according to the amount of depression of the accelerator pedal 31 by the person
Generates torque. [0121] On the other hand, in step S1, the slip control
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. [0122] In this step S10, the slip amount s is
-2km/h or less and the required drive 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 vehicle 81.
If it is determined that there is no slip control flag F in S11,
S is reset and the process moves to step S7. [0123] In step S10, the slip amount s is
Greater than -2km/h or required drive torque T
d is below the target drive torque TOS for 0.5 seconds or more
not continuing, i.e. the driver wishes to accelerate the vehicle 81.
If it is determined that the
The switch 68 is on, that is, the throttle valve 20 is fully closed.
It is determined whether or not the period continues for 0.5 seconds or more. [0124] 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. [0125] When controlling the turning of the vehicle 81, the TCL 75
From the steering shaft turning angle δH and the vehicle speed V, the target lateral direction of the vehicle 81 is determined.
Calculate the acceleration GYO, and if the vehicle 81 has extreme understeering.
Acceleration in the longitudinal direction of the vehicle body,
Change the target longitudinal acceleration GXO to this target lateral acceleration GYO.
Set based on. And this target longitudinal acceleration GXO
Calculate the target drive torque TOC of the engine 11 corresponding to
. By the way, the lateral acceleration GY of the vehicle 81 is
Can be actually calculated using wheel speed difference |VRL-VRR|
However, by using the steering shaft turning angle δH,
Therefore, the prediction of the value of the lateral acceleration GY acting on the vehicle 81 is
This allows for the advantage of rapid control.
have [0127] Note that the actual lateral force generated on this vehicle 81
The speed GY is given by b as the tread of the rear wheels 77 and 78.
, rear wheel speed difference |VRL-VRR| and vehicle speed V, as shown in the formula below.
It is calculated as follows. GY = (|VRL-VRR|・V)/3.62・b
・g0128] Here, the machine set every 15 milliseconds
The increase/decrease in the target drive torque TOC of Seki 11 is very large.
In this case, a shock occurs as the vehicle 81 accelerates or decelerates,
The target drive of engine 11 is
The increase or decrease in the dynamic torque TOC causes a decrease in the ride comfort of the vehicle 81.
If the target drive torque becomes large enough to cause
It is also necessary to regulate the increase and decrease of TOC. [0129] Operation of turning control taking into consideration the above knowledge.
As shown in FIG. 22 representing the calculation block, the TCL75 has one
The vehicle speed V is determined from the output of the pair of rear wheel rotation sensors 79 and 80.
Calculate using equation 1 and detect from steering angle sensor 83
Based on the signal, the steering angle δ of the front wheels 64 and 65 is calculated from the above equation 2.
The target lateral acceleration GYO of the vehicle 81 at this time is calculated as described above.
Calculate using equation 3. [0130] By the way, the stability factor mentioned above
As is well known, A indicates the configuration of the suspension system of the vehicle 81 and the tires.
This value is determined by the characteristics of the vehicle, road surface conditions, etc. concrete
Specifically, the actual lateral movement that occurs on the vehicle 81 during a steady circular turn
Acceleration GY and steering angle ratio δH of the steering shaft 82 at this time
/δHO (lateral with reference to the neutral position δM of the steering shaft 82)
Steering axis during extremely low speed driving where acceleration GY is close to 0
Steering shaft 82 during acceleration for turning angle δHO of 82
For example, Fig. 23 shows the relationship between the turning angle δH and
expressed as the slope of the tangent line in the graph shown
. In other words, the lateral acceleration GY is small and the vehicle speed V is too high.
In the region where the stability factor A is almost constant (
A=0.002), but the lateral acceleration GY is 0
．． When the weight exceeds 6g, the stability factor A increases rapidly,
Vehicle 81 shows extremely strong understeering tendency
It becomes like this. [0131] Based on the above, based on FIG.
In this case, stability factor A is set to 0.002.
The target lateral acceleration GY of the vehicle 81 is calculated using equation 3.
The driving torque of the engine 11 is adjusted such that O is less than 0.6 g.
control. [0132] In addition, on slippery roads such as frozen roads, etc.
In this case, the stability factor A is set to 0.005, for example.
You can set it before or after. [0133] 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 81 set according to the speed V
The XO is stored in the TCL75 in advance as shown in FIG.
engine 1 using this target longitudinal acceleration GXO.
1's target drive torque TOC is calculated using the following equation 8. [Equation 8] [0135] However, TL is the lateral acceleration GY of the vehicle 81
road resistance, which is the resistance of the road surface found as a function of
(Road-Load) torque, and in this example,
It is obtained from a map as shown in FIG. [0136] In the vehicle 81, the turning control is selected by the driver.
A manual switch (not shown) is provided to
The driver operated this manual switch to select rotation control.
If the
ing. [0137] This target drive torque TOC for turning control is
Figures 26 and 27 show the control flow for determining
Detection and processing of various data mentioned above in C1
The target drive torque TOC is calculated by this method.
The operation is performed independently of the operation of the manual switch. Next, at C2, the vehicle 81 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.
The target drive torque TOC is set in advance in C3.
It is determined whether or not it is less than or equal to a threshold value, for example (Td -2). In other words, even when the vehicle 81 is traveling straight, the target drive torque TOC
can be calculated, but its value is driven by the driver's request.
It is usually larger than the torque Td. However, this
When the vehicle 81 is turning, the required driving torque Td is generally
Since the target drive torque TOC becomes smaller than the threshold value (Td
-2) Determine the following as the start condition for turning control.
I'm trying to set it up. [0139] Note that this threshold value is set as (Td -2).
The hysteresis was added to prevent control hunting.
As a standard. [0140] 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. [0141] 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, at C6, the steering angle
Whether the neutral position learned flag FH is set or not
, that is, the steering angle δ detected by the steering angle sensor 83 is
Authenticity is determined. [0142] At step C6, the steering angle neutral position learned flag is
If it is determined that the lag FH is set, at C7
Check whether the turning control flag FC is set or not.
and determined. [0143] 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,
, since the steering angle δ calculated by Equation 2 is unreliable, Equation 8
Without using the target drive torque TOC calculated in TC
L75 is the maximum torque of the engine 11 as the target drive torque TOC.
The torque is output at C9, and this causes the ECU 15 to output the torque.
Reduce the duty rate of control solenoid valves 51 and 56 to 0% side
As a result, the engine 11 is activated by the driver's accelerator pedal 3.
A driving torque is generated according to the amount of depression. [0145] Also, in step C3, the target drive torque
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, TCL75 sets 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. 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,
L75 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. [0147] 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 81 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 81 is suppressed to 0.1 g per second.
If you want, use Equation 8 above to calculate TK = 0.1・(Wb・r)/(ρm・ρd)
・It becomes Δt. [0148] The eyes 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 . [0149] At step C11, the target drive torque TO
C and the target drive torque TOC (n-1) calculated last time.
It is determined that the difference ΔT is larger than the negative increase/decrease tolerance TK.
Then, the target drive torque TOC calculated this time and the target drive torque TOC calculated last time are
Absolute value of the difference between target drive torque TOC(n-1) |
Since ΔT| is smaller than the allowable increase/decrease amount TK, it is not calculated.
The calculated value in step C8 this time is used as is to drive the target.
Adopted as torque TOC. [0150] 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(n-1) is negative increase/decrease tolerance TK.
If it is determined that the current target drive is not large, C12
Correct the torque TOC using the formula below and apply it to the C8 step.
adopted as the calculated value in the process. TOC=TOC(n-1) -TK [0151] 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. [0152] 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 [0153] In other words, the target drive torque TO that was calculated this time
C 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
This reduces the acceleration shock that accompanies the increase. [0154] As described above, the target drive torque TOC is
Once set, TCL75 sets this target drive torque in C14.
torque TOC is larger than the driver's required driving torque Td
Determine whether or not. [0155] 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. [0156] In 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 Then, at this step C6, the steering angle neutral position
It is determined that the learned flag FH is set, and further
The turning control flag FC is set in step C7.
If it is determined that C1 or C12 or C13 is
The calculated value adopted in step is the target drive torque for swing control.
Selected as the TOC. [0157] Also, in step C14, the target drive torque is set.
torque TOC is larger than the driver's required driving torque Td
If it is determined that the vehicle 81 has finished turning,
This means that TCL75 is free during turning control with 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 turn the flag during turning control.
Reset the FC. [0158] At this C16, the turning control flag FC is set.
When reset, the TCL75 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. [0159] The target drive torque TOC for this turning control is
After calculating, TCL75 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 81, the target drive is for the smaller number.
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. [0160] As shown in Figure 28, which shows the flow of this process,
, M11 is the target drive torque TOS for slip control.
After calculating the target drive torque TOC for turning control, M
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 by M13 and output to the ECU 15. [0161] On the other hand, if there is a slip in step M12,
It was determined that the control flag FS was not set.
If so, the turning control flag FC is set at M14.
The turning control flag FC is set.
If it is determined that the target drive torque is
M1 is the target drive torque TOC for turning control as TO.
5 and outputs it to the ECU 15. [0162] Also, during the turning control in step M14,
If it is determined that flag FC is not set,
TCL75 reaches the final maximum torque of engine 11 at M16.
It is output to the ECU 15 as the target driving torque TO. [0163] As described above, the final target drive torque TO
while selecting the slot via the actuator 41.
The output of the engine 11 is also reduced by fully closing the torque valve 20.
When starting suddenly or when the road surface changes from a normal dry road to ice.
If there is a sudden change in the road, TCL75 will change to ECU15.
The retardation of the ignition timing P with respect to the basic retardation amount pB set by
The angle ratio is set and outputted to the ECU 15. One
Therefore, the retardation ratio of the ignition timing P to the basic retardation amount pB
The means for setting the torque is the second torque reduction means. [0164] 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. Moreover, as the retardation ratio, in this embodiment, the basic retardation amount pB
0 level that sets the value to 0, and the basic retard amount pB to 2/3
The II level compressed to
Three levels are set, including the output level III.
Basically, as the rate of change GS of the slip amount s increases,
Therefore, the map of the retardation ratio that results in a large amount of retardation is T.
It is stored in the CL75. [0165] Therefore, the second torque reducing means
Created based on the rate of change GS (amount of change in slip) of the amount s.
The movement is controlled and output reduction operations can be performed with good responsiveness. FIG. 29 shows the procedure for reading out this retardation ratio.
, As shown in Figure 30, the TCL75 is first
Based on the map shown in 31, the rate of change in the slip amount s
Level of retardation ratio based on GS (0, II, III
). Next, at P1, the retardation ratio reaches the III level.
It is determined whether or not the retardation ratio is at the III level.
If it is determined that the hydraulic control device 16 is
It is determined whether the hydraulic automatic transmission 13 is changing gears by
. [0167] At this step P2, the hydraulic automatic transmission
If it is determined that 13 is changing gears, the output will be extremely low.
In order to prevent engine 11 from stalling due to the
In step 3, the retardation ratio is reset to II level. I'll explain later
If the retard rate is set to III level,
, the throttle valve 20 is fully closed by the forced control means.
Become. When the hydraulic automatic transmission 13 is changing gears, the retardation ratio is set to II.
Reset the throttle valve by forced control means to the level
The fully closed operation of 20 is prevented from occurring. This is the original
It has become a prohibited method in the Ming Dynasty. Therefore, during gear shifting, throttle
There is no longer a risk that the valve 20 will be fully closed, resulting in an extreme reduction in output.
There will be no more trouble. [0168] At P3, set the retard ratio to II level again.
After that, at P4, is this retard ratio again at II level?
Determine whether or not. In this case, the retardation ratio is at II level.
Therefore, move on to the next P5 and calculate the slip amount of front wheels 64 and 65.
Determine whether or not s is less than 5 km/h, and calculate this P5.
If the slip amount s is less than 5 km/h at the step, immediately
It was determined that front wheels 64 and 65 were not slipping much.
In this case, change rate GS of slip amount s is 0g in P6.
Determine whether or not the value is less than or equal to the value. [0169] In step P6, the slip amount s is changed.
If it is determined that the conversion rate GS is less than 0g, proceed to P7.
Set the retard ratio to 0 level and set this to ECU15.
After outputting, the process returns to step P1. On the contrary, this P
In step 6, the rate of change GS of the slip amount s is 0 g or more.
If it is determined that the
Determine whether the slip amount s is less than 0 km/h
. [0170] In this step P8, the slip amount s is
If it is determined that the speed is less than 0km, retard at P9.
Reset the ratio to 0 level and output this to ECU15.
After pressing the button, return to step P1. On the other hand, this P8
The slip amount s is 0 km/h or more in step ,
In other words, it was determined that there was some slip in front wheels 64 and 65.
In this case, the retardation ratio was maintained at the II level at P10.
After outputting this to the ECU 15, the step of P1
Return to top. [0171] In step P5, the slip amount s is
5 km or more, that is, front wheels 64 and 65 are slipping.
If it is determined that the
After outputting the ratio as it is to the ECU 15, the step of P1
Back to Tep. Also, in this P1 step, the retardation ratio is
If you decide that it is not level III, proceed to step P4.
If the retard ratio is not at level II,
If it is disconnected, change the 0 level retardation ratio to ECU1 at P12.
5, the process returns to step P1. On the other hand, in step P2, the hydraulic automatic
If it is determined that the transmission 13 is not shifting, P1
3, the slip amount s of front wheels 64 and 65 is less than 12 km/h.
If the speed is less than 12km/h,
If it is determined that the slip amount s changes in P14,
It is determined whether the rate GS is smaller than -0.05g. Then, the rate of change GS of the slip amount s is -0.05g.
If it is determined in step P14 that the
15, reset the retardation ratio to II level and
Move on to step 4. [0173] In step P13, the front wheels 64 and 65
The slip amount s is 12 km/h or more, that is, the slip
If it is determined that a large drop is occurring, proceed to P16.
to maintain the retardation ratio at the III level, and set this to ECU1.
5, the process returns to step P1. Also, P1
In step 4, the rate of change GS of the slip amount s is -0.
05g or more, in other words, it is judged that the slip is increasing.
Then, in P17, the slip amount s is 5k/hour.
It is determined whether it is less than m. [0174] In this step P17, the slip amount s is
5 km/h or more, that is, the slip amount s is 5 km/h
If it is determined that the speed is above 12km/h or less, P1
8, the retardation ratio is maintained at the III level, and this
After outputting to the ECU 15, return to step P1.
. However, it was determined that the slip amount s was less than 5 km/h.
Then, in P19, the change rate GS of the slip amount s is
is less than 0g, that is, whether the vehicle 81 is in a deceleration state.
Determine whether [0175] In step P19, the slip amount s is
If it is determined that the rate of change GS is less than 0g, the vehicle 81
is not in acceleration state, so set the retardation ratio to 0 level at P20.
After setting it again and outputting this to ECU15,
Return to step P1. Also, in step P19
The rate of change GS of the lip amount s is 0 g or more, that is, the vehicle 81 is
If it is determined that it is in an accelerating state, it will slip at P21.
It is determined whether or not the drop amount s is less than 0 km/hour. [0176] Then, in step P21, the slip
The amount of slip s is less than 0 km/h, that is, slip occurs.
If it is determined that it is not, set the retard ratio to 0 in P22.
After resetting the bell and outputting this to ECU15,
Return to step P1. Also, the slip amount s is 0 per hour.
km or more, if it is determined in step P21 that the
III level slowness may increase
The angle ratio was maintained as it was and outputted to the ECU 15.
After that, the process returns to step P1. [0177] 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 5 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. [0178] 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. [0179] The target retard amount pO in this retard control is
As shown in FIGS. 32 and 33 showing the calculation procedure, first EC
U15 is the slip control flag FS mentioned above in Q1.
This slip control determines whether or not is set.
If it is determined that the medium flag FS is set, Q2
Is the retardation ratio set to III level?
Determine. In other words, in Q2, is the retard amount greater than or equal to the predetermined value?
, whether the retardation ratio is set to III level or not.
It is decided by [0180] 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 value of target drive torque TOS
Q so that the throttle valve 20 is fully closed regardless of
In step 4, set the duty rate of the torque control solenoid valves 51 and 56.
Set to 100% and force engine 11 to idle
Realize the state of affairs. Forced control means performs this control.
Ru. [0181] Therefore, the ignition timing is controlled by the forced control means.
When the retardation amount of P is equal to or greater than a predetermined value, the target drive torque TOS
Regardless of the situation, the throttle valve 20 is fully closed and the engine 11 is
The driving torque is reduced to the maximum, and the engine 11
In this case, the operations by the two torque reduction means become contradictory.
There is no risk of it happening. [0182] 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. [0183] Then, in this step Q5, the retardation ratio
If it is determined that the
Set the retard amount pO as shown in the formula below, and set the ignition timing P to the target retard.
The angle is retarded by the angle amount pO. pO = pB ・(2/3) [0184] Furthermore, the ECU 15 sets the target drive torque TOS
The duty of the torque control solenoid valves 51 and 56 is adjusted according to the value of
The acceleration rate is set in Q7, and the driver presses the accelerator pedal 3.
The driving torque of the engine 11 is determined regardless of the amount of depression of the engine 11.
reduce [0185] 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. [0186] 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 [0187] 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.
. On the other hand, in step Q5, the retardation ratio is II.
If it is determined that the level is not reached, the target retard angle is set in Q8.
Determine whether the quantity pO is 0 or not, and if it is 0,
If it is determined, move on to step Q7 and adjust the ignition timing.
Torque is adjusted according to the target drive torque TOS without retarding P.
Set the duty rate of the solenoid valves 51 and 56 for brake control, and
It has nothing to do with the amount of depression of the accelerator pedal 31 by the driver.
Therefore, the driving torque of the engine 11 is reduced. [0189] Also, in step Q8, the target retardation amount p
If it is determined that O is not 0, the target retardation amount p is immediately set.
If O is set to 0, the amount of fluctuation in the driving torque of the engine 11
becomes large and a shock occurs, so in Q9
The target retardation amount pO is calculated every current sampling period Δt.
For example, by pump control, the pO is reduced by 1° until pO = 0.
Then, proceed to step Q7. [0190] 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 Q10 set pO = 0 and set the ignition timing P.
Torque control solenoid valves 51 and 56 at Q11 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. Effects of the Invention According to the vehicle output control device of the present invention,
Based on the vehicle running speed by the target drive wheel speed setting means.
Set the target drive wheel speed, and adjust the forward speed according to this target drive wheel speed.
The reference drive torque of the reference engine is set by the reference drive torque setting means.
Set the drive wheel slip amount based on this reference drive torque.
Set target drive torque of engine based on
The drive torque of the engine is set by the target drive torque.
The torque control unit reduces the torque so that
Since the operation of the reducing means is controlled, it is better than the conventional one.
It is possible to reduce the correction factor for control of
There may be problems such as delays or increased cost of computing equipment.
There is almost no energy loss, making the vehicle safer while avoiding energy loss.
It can be run completely. Also, with automatic transmission
During gear shifting, the driving torque by the first torque reduction means is maximized.
Since the operation that reduces the limit is prohibited, the
This eliminates the possibility of extreme output reduction and reduces engine stall.
can be prevented from occurring.

[Brief explanation of the drawing]

FIG. 1 is a conceptual diagram of an embodiment in which a vehicle output control device according to the present invention is applied to a front-wheel drive vehicle incorporating a hydraulic automatic transmission with four forward speeds and one reverse speed.

FIG. 2 is a schematic configuration diagram of a vehicle incorporating an output control device according to the present invention.

FIG. 3 is a sectional view showing a throttle valve drive mechanism of a vehicle incorporating the output control device according to the present invention.

FIG. 4 is a flowchart showing the overall flow of control of a vehicle incorporating the output control device according to the present invention.

FIG. 5 is a flowchart showing the flow of learning and correcting the neutral position of the steering shaft.

FIG. 6 is a flowchart showing the flow of learning and correcting the neutral position of the steering shaft.

FIG. 7 is a block diagram showing a calculation procedure for a target drive torque for slip control.

FIG. 8 is a block diagram showing a calculation procedure for a target drive torque for slip control.

FIG. 9 is a block diagram showing a calculation procedure for a target drive torque for slip control.

FIG. 10 is a graph showing the relationship between the coefficient of friction between a tire and a road surface and the slip rate of this tire.

FIG. 11 is a map showing the relationship between corrected longitudinal acceleration and front wheel acceleration correction amount.

FIG. 12 is a map showing the relationship between lateral acceleration and front wheel acceleration correction amount.

FIG. 13 is a map showing the relationship between vehicle speed and running resistance.

FIG. 14 is a map showing the relationship between corrected longitudinal acceleration and slip correction amount accompanying acceleration.

FIG. 15 is a map showing the relationship between lateral acceleration and slip correction amount associated with turning.

FIG. 16 is a map showing the relationship between corrected longitudinal acceleration and road surface condition correction coefficient.

FIG. 17 is a graph showing areas of increase and decrease in integral correction torque.

FIG. 18 is a map showing the relationship between each gear stage of a hydraulic automatic transmission and a correction coefficient corresponding to each correction torque.

FIG. 19 is a map showing the relationship between engine speed, required drive torque, and accelerator opening.

FIG. 20 is a flowchart showing the flow of slip control.

FIG. 21 is a flowchart showing the flow of slip control.

FIG. 22 is a block diagram showing a procedure for calculating a target drive torque for turning control.

FIG. 23 is a graph showing the relationship between lateral acceleration and steering angle ratio to explain the stability factor.

FIG. 24 is a map showing the relationship between target lateral acceleration, target longitudinal acceleration, and vehicle speed.

FIG. 25 is a map showing the relationship between lateral acceleration and load torque.

FIG. 26 is a flowchart showing the flow of turning control.

FIG. 27 is a flowchart showing the flow of turning control.

FIG. 28 is a flowchart showing the flow of final target torque selection operation.

FIG. 29 is a flowchart showing the flow of a retard ratio selection operation.

FIG. 30 is a flowchart showing the flow of a retard ratio selection operation.

FIG. 31 is a map showing the level of change rate and retardation rate.

FIG. 32 is a flowchart showing a procedure for engine output control.

FIG. 33 is a flowchart showing a procedure for engine output control.

[Explanation of symbols]

11 Engine 13 Hydraulic automatic transmission 15 ECU 16 Hydraulic control device 20 Throttle valve 23 Accelerator lever 24 Throttle lever 31 Accelerator pedal 32 Cable 34 Claw 35 Stopper 41 Actuator 43 Control rod 47 Connection piping, 48 Vacuum tank 49 Check valve 50 , 55 Piping 51, 56 Torque control solenoid valve 60 Solenoid valve 61 Spark plug 62 Crank angle sensor 64, 65 Front wheel 66 Front wheel rotation sensor 67 Throttle opening sensor 68 Idle switch 70 Air flow sensor 71 Water temperature sensor 74 Exhaust temperature sensor 75 TCL 76 Accelerator opening sensor 77, 78 Rear wheel 79, 80 Rear wheel rotation sensor 81 Vehicle 82 Steering shaft 83 Steering angle sensor 84 Communication cable 104, 135 Multiplier 105 Differential calculation section 111 Acceleration correction section 112, 118 Addition section 115 Clip section 116 Torque conversion unit 117 Running resistance calculation unit 134 Subtraction unit A Stability factor FH Steering neutral position learned flag FS Slip control flag GF Actual front wheel acceleration GFB Reference front wheel acceleration GFF Modified reference front wheel acceleration, GFO Target front wheel acceleration GKC, GKF Front wheel acceleration Correction amount GS Slip amount change rate GX Longitudinal acceleration GXF Corrected longitudinal acceleration GXO Target longitudinal acceleration GYO Target lateral acceleration g Gravity acceleration NE Engine speed P Ignition timing pB Basic retardation amount pO Target retardation amount r Wheel effective radius SO Target slip rate s Slip amount TB Standard drive torque TD Differential correction torque Td Requested drive torque T1 Integral correction torque TO Final target drive torque TOS Target drive torque for slip control TP Proportional correction torque TPID Final correction torque TR Running resistance Δt Sampling period V Vehicle speed VF Actual front wheel Speed VFO, VFS Target front wheel speed VK, VKC Slip correction amount VRL Left rear wheel speed RRR Right rear wheel speed Wb Vehicle weight δ Front wheel steering angle δH Turning angle of steering angle ρd Operating gear reduction ratio ρKD Differential correction coefficient ρKI Integral correction coefficient ρKP Proportional correction coefficient ρm Transmission gear ratio

Claims (1)

[Claims] 1. A first torque reducing means for reducing the driving torque of the engine independently of the operation by the driver; and a first torque reducing means for reducing the driving torque of the engine independently of the operation by the driver; a second torque reduction means that increases the retardation amount of the engine to reduce the driving torque of the engine; and a target driving wheel speed setting that sets a target circumferential speed of the driving wheels based on the traveling speed of a vehicle equipped with an automatic transmission. a reference drive torque setting means for setting a reference drive torque of the engine in accordance with the target drive wheel speed set by the target drive wheel speed setting means; target drive torque setting means for setting a target drive torque of the engine based on a slip amount of the drive wheels from a reference drive torque; and a target drive torque in which the drive torque of the engine is set by the target drive torque setting means. a torque control unit that controls the operation of the first torque reduction means and also controls the operation of the second torque reduction means based on the amount of change in slip; and a torque control unit that controls the operation of the second torque reduction means based on the amount of change in slip; A forced control means for forcibly controlling the operation of the first torque reducing means to a state where the driving torque of the engine is reduced to the maximum regardless of the target driving torque when the angle amount is equal to or greater than a predetermined value; An output control device for a vehicle, comprising prohibition means for prohibiting the operation of the first torque reduction means by the forced control means.