JPH0471742B2 - - Google Patents

Description

DETAILED DESCRIPTION OF THE INVENTION The present invention relates to a straight-line compensation device for a hydraulically driven vehicle having independent left and right hydraulic transmissions.

As shown in FIG. 1, the hydraulic transmission of a hydraulically driven vehicle includes variable displacement hydraulic pumps (swash plate pumps) 3, 3', which are driven from an engine 1 via input shafts 2, 2'.
Variable displacement hydraulic motor (swash plate motor) to which operating oil is added via hydraulic piping from the swash plate pumps 3, 3'
4, 4', swash plate pumps 3, 3' and swash plate motors 4, 4'; The rotational speed of the swash plate motors 4, 4' is appropriately controlled so that the rotational force thereof is transmitted to the tracks or wheels via the output shafts 7, 7'.

However, this type of hydraulically driven vehicle suffers from variations in the volumetric efficiency of the swash plate pumps 3, 3' and swash plate motors 4, 4', and variations in the characteristics of the speed ratio control devices 5, 5' and 6, 6'. There is a problem in that there is a difference in the rotational speed of the left and right motors, making it impossible to guarantee the straight-line performance required for construction vehicles.

In order to solve this problem, (1) reduce the above variation, (2) connect the left and right hydraulic piping via an orifice, (3) detect the speed difference between the left and right, and based on this, determine the speed of one side. Methods have been proposed to correct the direction of travel by reducing the amount.

However, in the case of (1) above, the adjustment man-hours increase and the cost is high, in the case of (2) above, the compensation range cannot be widened because the steering becomes difficult to turn, and in the case of (3) above, the direction of travel is corrected. Therefore, the amount of deviation from the desired driving path is not actively corrected, and the automatic transmission (which is almost indispensable for hydraulically driven vehicles) often causes vehicle speed fluctuations because it reduces one speed. There were other problems.

The present invention has been made to solve the above problems, and provides a straight-line compensating device for a hydraulically driven vehicle that compensates for deviation from a desired running path and does not change the engine torque due to the compensating operation. The purpose is to provide.

In order to achieve this object, the present invention detects the output speeds of the left and right hydraulic transmissions, calculates the amount of deviation of the current position from the desired travel route based on the output speeds, and calculates the amount of deviation of the current position from the desired traveling route. The gear ratio of each hydraulic transmission is adjusted accordingly. Also,
In this modification, the gear ratio of one hydraulic transmission is increased and the gear ratio of the other hydraulic transmission is decreased.

The present invention will now be described in detail with reference to the accompanying drawings.

First, a method for calculating the amount of deviation between the desired traveling route and the vehicle will be explained using a schematic diagram of the vehicle traveling state shown in FIG. 2. If the distance between the tracks of the vehicle is 2C, the speed of the right track is V _R , the speed of the left track is V _L , and the angle between the target direction and the traveling direction is θ and the vehicle travels for △t time, it will deviate from the target direction (desired travel route). The moving speed ε・ is
The following formula, ε=V _M θ (θ≪1 rad)...(1) However, it is expressed as V _M =V _R +V _L /2...(2). On the other hand, the angular velocity θ〓, if the turning radius at that time is r, becomes θ〓=V _M /r...(3), and the turning radius r becomes r=2CV _M /V _R -V _L ...( 4) becomes. Next, by integrating the above equation (3), we get θ=∫V _M /rdt (5), and by substituting equation (4) into equation (5), we get θ=∫V _R −V _L /2Cdt …(6). Next, by substituting equations (2) and (6) into equation (1), ε==V _R +V _L /2·∫V _R −V _L /2Cdt (7). Therefore, the amount of deviation ε from the target direction at any point in time can be calculated by integrating equation (7) as follows: ε=1/4C∫[(V _R +V _L )∫(V _R −V _L )dt]dt... (8) Therefore, if the deviation amount ε is set to 0, the vehicle will travel straight.

Here, with reference to FIG. 5, a description will be given of how the vehicle moves straight along the target direction according to the above calculation.

Now, in FIG. 5, it is assumed that the vehicle 100 starts moving straight from the starting point p0 toward the right in the paper. Here, the target direction A of the vehicle 100 coincides with the longitudinal center direction of the vehicle 100 at the starting point t0. It is assumed that the vehicle 100 is located at positions p1, p2, p3, . . . at each time t1, t2, t3, . . . for each Δt. Then, similar suffixes 1, 2, 3, etc. are assigned to the average speed vM of the vehicle 100, the angle θ between the target direction A and the traveling direction, the deviation speed ε〓, and the deviation amount ε, and Let us express the average speed etc. at t1, t2, t3... Further, the speed difference VR-VL is expressed as ΔV, and a suffix is also added to this value.

Furthermore, the signs of the above θ, ε〓, ε are changed to the target direction A
The left area AL side with respect to the target direction A is defined as +, and the right area AR side with respect to the target direction A is defined as -.

Then, if time Δt is 1 and C is 1/2, then
The above equation (6) can be expressed as θn= _o 〓 ⁱ⁼¹ ΔV _i (9).

Similarly, the above equation (7) can be expressed as ε〓 _o =VMnθn (10).

Similarly, the above equation (8) can be expressed as ε _o = _o 〓 ⁱ⁼¹ ε〓 _i (11).

Therefore, if we assume that the vehicle 100 shifts to the left from time t0 with the right track speed VR being greater than the left track speed VL, then at time t1,
At position p1, as is clear from equations (9), (10), and (11) above, θ1=ΔV1(>0) ε〓1=VM1ΔV1(>0) ε1=ε〓1...(12) Become.

Here, as will be described later, the gear ratio is corrected according to the deviation amount ε1 in equation (12), and the left crawler speed VL is increased and the right crawler speed VR is decreased. As a result, the sign of the speed difference ΔV2 is reversed from positive to negative between t1 and t2. Here, the sign is reversed and ΔV2
If the absolute value of becomes larger than the absolute value of ΔV1, θ2=ΔV1+ΔV2 (<0), and the sign of θ2 becomes negative. Therefore, ε〓2=VM2(ΔV1+ΔV2) (<0), and ε2=ε〓1+ε〓2 (＜ε1)...(13). According to the above equation (13), although the deviation amount ε2 is smaller in absolute value than the deviation amount ε1, it still remains positive.

Further, in the next period t2 to t3, the gear ratio is corrected according to the deviation amount ε2, and the left crawler speed VL is increased and the right crawler speed VR is decreased.
Note that since ε2 is smaller than ε1, the absolute value of ΔV3 is smaller than the absolute value of ΔV2. Negative ΔV3 is θ2
is added to θ3=ΔV1+ΔV2+ΔV3 (<0), and the sign of θ3 remains negative even during the next period from t2 to t3, and the absolute value becomes larger than θ2. Therefore, ε〓3=VM3(ΔV1+ΔV2+,V3) (<0) ε3=ε〓1+ε〓2+ε〓3 (＜ε2)...(14).

According to the above equation (14), the deviation amount ε3 becomes smaller than the deviation amount ε2, but the sign still remains positive.

Further, during the next period t3 to t4, the speed ratio is corrected according to the deviation amount ε3, and the left crawler speed VL is increased and the right crawler speed VR is decreased.
In this case as well, the absolute value of ΔV4 becomes smaller than the absolute value of ΔV3, corresponding to the smaller deviation amount ε3. Then, the negative ΔV4 is added to θ3, resulting in θ4=ΔV1+ΔV2+ΔV3+ΔV4 (<0), and the sign of θ4 remains negative even between t3 and t4, and its absolute value becomes larger than θ3.

Therefore, ε〓4=VM4(∆V1+∆V2+∆V3+∆V4) (<
0) ε4=ε〓1+ε〓2+ε〓3+ε〓4...(15) As is clear from the above equation (15), as a result of the successive addition of the negative deviation velocity ε〓, the sign of the deviation amount ε4 reverses from positive to negative at time t4. As a result, during the next period from t4 to t5, the gear ratio is corrected according to the negative deviation amount ε4, and the left crawler track speed VL is decreased and the right crawler track speed VR is increased. That is,
By reversing the sign of ΔV5 and adding it to θ4, θ5=ΔV1+ΔV2+ΔV3+ΔV4+ΔV5 (<0), and although the sign is negative, the absolute value is smaller than θ4. As a result, ε〓5=VM5(ΔV1+ΔV2+ΔV3+ΔV4+ΔV5)
(<0), and the negative shear velocity ε〓5 is added to ε4, resulting in ε5=ε〓1+ε〓2+ε〓3+ε〓4+ε〓5...(16). In the above equation (16), the sign of the deviation amount ε5 remains negative, and its absolute value becomes even larger than ε4. Therefore, in the next period from t5 to t6, as ε5 becomes larger in absolute value than ε4, the left crawler track speed VL is further decreased and the right crawler track speed VR is further increased. As a result, the speed difference ΔV6 (positive sign) becomes even larger, and by adding this to θ5, the sign of θ6 is reversed to positive.

θ6=ΔV1+ΔV2+ΔV3+ΔV4ΔV5+ΔV6
(>0) As a result, ε〓6=VM6(ΔV1+ΔV2+ΔV3+ΔV4 +ΔV5+ΔV6) (>0). And, ε6=ε〓1+ε〓2+ε〓3+ε〓4+ε〓5+ε〓6…(17
), the positive deviation speed ε〓6 is added to the negative deviation amount ε5, so that ε6 remains negative and its absolute value becomes smaller than ε5.

Thereafter, similar calculations are performed, and the vehicle 100 moves straight along the target direction A.

Note that a trajectory L indicated by a dashed line on the left side of the drawing is a trajectory of a left turn of the vehicle 100, and if the operator returns the steering wheel to a straight-ahead state after turning along such a trajectory, as will be described later. The above integration is started at time t0 when the steering wheel is turned back. Therefore, vehicle 10
0 means that the vehicle 100 travels straight along the longitudinal direction A when the steering wheel is returned, as expected by the operator. As described above, the present invention calculates the deviation amount ε from the above equation (8), and uses this to cause the vehicle to move straight.

FIG. 3 schematically shows the drive system and control system of the hydraulically driven vehicle according to the present invention. The calculation device 10 controls the gear ratio control devices 5, 5' and 6, 6', and outputs left and right steering signals S _L ,
S _R , a vehicle speed setting signal V _E , and speed signals V _R and V _L from speed detectors 8 and 8' for detecting the rotational speeds of the output shafts 7 and 7' are added. The calculation device 10 operates the gear ratio control device 5 based on the vehicle speed setting signal V _E and manual operation signals such as steering signals S _L and S _R.
5' and 6,6' as well as speed signal
The amount of deviation of the vehicle is determined by calculating the equation (8) from V _R and V _L , and the manual operation signal is corrected in accordance with this amount of deviation.

FIG. 4 is a block diagram showing details of the arithmetic unit 10. The vehicle speed setting signal V _E is a signal corresponding to the position of the vehicle speed setting lever, throttle lever, etc., and this signal is applied to multipliers 11 and 12 and a zero cross detector 13. Note that this vehicle speed setting signal V _E changes, for example, from -Vcc to +Vcc, and when the signal V _E is 0, it corresponds to a vehicle speed of 0. Multipliers 11 and 12 have steering signals S _L and S _R added to their other inputs, respectively, and multiplier 11 multiplies signal V _E and signal S _L and outputs this to multiplier 14 to perform multiplication. The divider 12 multiplies the signal V _E and the signal S _R and sends this to the divider 1.
Output to 5. Note that the steering signals S _L and S _R are signals in the range of "0" to "1", and both signals are "1" when the vehicle is not being steered (when the vehicle is traveling straight).

Multiplier 14 and divider 15 each have other inputs to which a speed ratio correction signal is applied from sample hold circuit 16. This gear ratio correction signal is a signal that changes within the range of "0.5" to "1.5" corresponding to the amount of deviation between the desired traveling route and the vehicle, and when the amount of deviation is 0, the signal becomes "1". . Note that details of this gear ratio correction signal will be described later. Therefore, when the vehicle is traveling straight and the amount of deviation between the desired travel route and the vehicle is 0, the vehicle speed setting signal
V _E is directly applied to limiters 17 and 18.

Limiters 17 and 18 apply different limiters to the input signals for the swash plate pump and swash plate motor (see the characteristic graph in the figure), and apply these to the speed ratio control devices 5' and 6' (left pump, left motor side). ) and speed ratio control devices 5 and 6 (right pump, right motor side).

Now, referring to FIGS. 6a and 6b, the limiter 17,
18 will be explained in more detail.

The limiters 17 and 18 function as function generators that make the relationship between the vehicle speed signal (signal input to the limiters) sp and the rotational speed nM of the left and right motors 4' and 4 linear.

Generally, the function of a hydraulic motor is expressed by equation (18) below.

nM=np・(Dp/DM) …(18) nM: Motor rotation speed (rev/s) np: Pump rotation speed (rev/s) Dp: Pump displacement (cc/rev) (Discharge amount per rotation ) DM: Motor displacement (cc/rev) (suction amount required for one rotation) The above Dp is the gear ratio control device 5 shown in Fig.
5', and the DM is controlled by the gear ratio control devices 6, 6'.

The hydraulic drive device consisting of a variable displacement pump and a variable displacement motor shown in Fig. 3 generally uses the following control in order to ensure output torque at low speeds and to enable high speed rotation. method is adopted.

(a) In a region where the vehicle speed sp is small (sp≦sp0): - Dp is increased as the vehicle speed increases (until Dp reaches the maximum value DpMAX) - DM is maintained at the maximum value DMMAX.

(b) In the region where the vehicle speed sp is high (sp>sp0), - Dp is kept at the maximum value (DpMAX).

・DM decreases as vehicle speed increases.

Generally, the gear ratio control device 5', 6', 5, 6
input signal (output signal of limiters 17 and 18) and
The following relationship holds between Dp and DM.

Dp＝Kp・inp …(19) DM＝DMMAX−KM・inM …(20) Kp: constant inp: limiter output for pump DMMAX: constant (maximum displacement of motor) KM: constant inM: limiter output for motor Then, the vehicle speed signal SP (limiter input signal)
In order to make the relationship between and motor rotation speed nM linear, the following relationship between vehicle speed sp and limiter output can be obtained.

Regarding the pump, the above equations (18) and (19) and conditions
It is clear from (a) and (b) that it is sufficient if there is a relationship between the vehicle speed sp and the limiter output inp as shown in FIG. 6a. Here, when the vehicle speed sp is sp0, Dp becomes the maximum value DpMAX, and DM becomes the maximum value D MMAX.

On the other hand, regarding the motor, it is clear from the above conditions (a) and equation (20) that the limiter output inM is zero in a region where the vehicle speed sp is less than or equal to sp0.

Therefore, let us consider the region where sp>sp0.

Substituting equation (20) into equation (18) (from condition (b), Dp=
DpMAX), nM=np・{DpMAX/(DMMAX−KM・
inM)} …(21) is obtained. On the other hand, assuming that the vehicle speed signal sp and the motor rotation speed nM are linear, there is a relationship where nM=np・Ksp・sp (22) Ksp: constant. Here, from the above condition (a), nM=np(DpMAX/DMMAX)=np・KSP・
sp0…(23) Therefore, Ksp=(DpMAX/DMMAX)・(1/
sp0) …(24) is obtained. Limiter output from equations (21), (22), and (24) above
Solving for inM, inM={DMMAX−DpMAX/Ksp・sp)}/
KM = {DMMAX (1-sp0/sp)}/KM (25). From the above, the relationship between the vehicle speed sp regarding the motor and the limiter output inM is as shown in FIG. 6b.

Figure 6 a and b are when the vehicle speed sp is positive (vehicle moving forward)
It is. Therefore, when considering the case where the vehicle speed sp is negative (vehicle moving backward), the characteristic graph shown in FIG. 4 is obtained which is symmetrical with respect to the vehicle speed sp of zero.

On the other hand, the right crawler belt speed V _R and the left crawler belt speed V _L detected by speed detectors 8 and 8' are added to a subtracter 19 and an adder 20, respectively. Subtractor 19 subtracts velocity V _L from velocity V _R and outputs this to integrator 21 . The integrator 21 calculates the above equation (6), θ=∫{(VR−VL)/2C}dt. In other words, the input speed difference VR−VL
is integrated over time (1/s), and the distance between tracks is
By dividing by 2C, the angle θ between the vehicle's target direction and the traveling direction is calculated. The calculated angle θ is output to the multiplier 23.

The adder 20 and the divider 22 perform the calculation of equation (2) above, VM=(VR+VL)/2. That is, the adder 20 adds the speed VR and the speed VL, and the sum of these speeds is added to the divider 2.
2 is output. The divider 22 calculates the velocity sum VR+
VL is divided by 2, and the average speed of the left and right track speeds
VM is calculated and output to the multiplier 23.

The multiplier 23 calculates the above equation (7), ε〓=VM・θ (VM=(VR+VL)/2, θ=∫{(VR−
VL)/2C}dt). That is, the input angle θ and the input average velocity VM are multiplied to obtain the velocity ε〓 of deviation from the target direction, and is added to the integrator 24.

The integrator 24 calculates the above equation (8), ε=∫ε〓dt (ε〓=VM·θ). That is, the input speed ε〓 is integrated (1/s) over time. The amount of deviation ε is determined in this way, but here the amount of deviation ε is further amplified with a gain K to adjust the correction sensitivity, and this is applied to the speed ratio correction signal generator 25.

The gear ratio correction signal generator 25 generates a gear ratio correction signal in the range of "0.5" to "1.5" according to the signal Kε corresponding to the amount of deviation. Here, the gear ratio correction signal becomes a signal larger than "1" when the signal Kε is positive (when the vehicle deviates to the left from the target direction), and when the signal Kε is negative (when the vehicle deviates to the right from the target direction). ), the signal will be smaller than "1". Of course, when the signal Kε is "0", the gear ratio correction signal is "1".

This gear ratio correction signal is sent to the sample hold circuit 1.
6 through multiplier 14 and divider 15 respectively.
added to. Therefore, when the gear ratio correction signal is larger than "1" (when the vehicle deviates to the left from the target direction), the signal applied to the limiter 17 increases and the gear ratio of the left pump and motor increases. At the same time, the signal applied to the limiter 18 becomes smaller and the speed ratio of the right pump and motor decreases. Further, when the gear ratio correction signal is smaller than "1", the increase/decrease in the gear ratio is reversed from the above case.

In this way, the amount of deviation of the vehicle from the desired travel route is corrected. In addition, when correcting the amount of deviation, it is possible to decelerate one motor and simultaneously accelerate the other motor so that the total engine torque does not change.
The straight-line compensation operation can be made less likely to disturb the automatic gear shift in the automatic transmission device.

Next, a case where the steering wheel is operated will be explained. Steering signals S _L and S _R are applied to multipliers 11 and 12, respectively, as described above, and also to comparators 26 and 27, respectively. Here, when the vehicle turns left, the steering signal S _L becomes smaller than the signal "1", and when the vehicle turns right, the steering signal S _R
becomes smaller than the signal "1".

The comparators 26 and 27 respectively determine whether or not the steering signals S _L and S _R are below a predetermined threshold. When the steering signals S L and S R become below the predetermined threshold, a high level signal (hereinafter referred to as H level) is detected. signal). Each output of the comparators 26 and 27 is applied to an exclusive OR circuit 28,
The output of 8 is applied to a sample and hold circuit 16. Therefore, when the steering is operated and an H level signal is output from one of the comparators, the exclusive OR circuit 28 outputs an H level signal to the sample hold circuit 16.

The sample hold circuit 16 is an exclusive OR circuit 28
When an H level signal is input from the transmission gear ratio correction signal generator 25, the transmission ratio correction signal applied from the transmission ratio correction signal generator 25 at that time is held while the H level signal is inputted.

Further, each output of the comparators 26 and 27 is applied to an exclusive OR circuit 29. Therefore, when the steering wheel is operated and an H level signal is output from one of the comparators, the exclusive OR circuit 29
is the H level signal through the delay circuit 30 and the OR circuit 3.
1 to integrators 21 and 24, respectively. Integrators 21 and 24 are reset to 0 when an H level signal is input.

Thereby, when the steering lever (or pedal) is released, the vehicle can move straight in the direction set by the operator through the steering operation. Furthermore, by holding the speed ratio correction signal by the sample hold circuit 16, it becomes easier to turn slowly. The zero cross detector 13 detects when the vehicle speed setting signal V _E reaches 0, and outputs an H level signal to the integrators 21 and 24 via the OR circuit 31 at this time. Therefore, for example, when the vehicle speed setting signal _VE switches from forward to reverse, the integrators 21 and 24 are reset, and then the amount of deviation of the vehicle is calculated anew.

As explained above, according to the present invention, the amount of deviation of the vehicle from the desired traveling route is calculated and the vehicle is controlled to reduce this amount of deviation, so that accurate straight-ahead compensation can be performed. can. In addition, when performing straight-line compensation operation, it is possible to decelerate one output speed while simultaneously accelerating the other output speed so that the total engine torque remains unchanged. Disturbances can be made less likely to occur, and a great effect can be expected in vehicles equipped with automatic transmissions.

[Brief explanation of drawings]

Fig. 1 is a drive system diagram of a hydraulically driven vehicle having two systems of hydraulic transmissions, Fig. 2 is a schematic diagram of the running state of the vehicle used to explain the calculation according to the present invention, and Fig. 3 is a diagram of the driving state of the vehicle according to the present invention. FIG. 4 is a block diagram showing details of the arithmetic unit shown in FIG. 3, and FIG. 5 shows how the vehicle of the embodiment moves straight along the target direction. Figures 6a and 6b used to explain the above are graphs showing the characteristics of the limiter of the embodiment when the vehicle speed is positive. 1... Engine, 2, 2'... Input shaft, 3, 3'... Variable capacity hydraulic pump, 4, 4'... Variable capacity hydraulic motor, 5, 5', 6, 6'... Gear ratio control device, 7.
7'...Output shaft, 10...Arithmetic unit, 11, 12, 1
4, 23... Multiplier, 13... Zero cross detector, 1
5, 22... Divider, 16... Sample hold circuit, 17, 18... Limiter, 21, 24... Integrator, 25... Gear ratio correction signal generator, 26, 27...
Comparator, 30...delay circuit.

Claims (1)

[Claims] 1. A system for steering and driving a hydraulically driven vehicle by changing the gear ratios of left and right hydraulic transmissions provided for each of left and right tracks or left and right wheels in response to a manual operation signal. A straight-line compensating device for a hydraulically driven vehicle, comprising: speed detecting means for detecting the output speeds of the left and right hydraulic transmissions, and integrating the difference between the detected output speeds of the speed detecting means to determine the traveling direction of the hydraulically driven vehicle. The angle that the hydraulic drive vehicle makes with respect to the target direction is determined, and based on the determined angle and the average speed of both detected output speeds of the speed detection means, the hydraulically driven vehicle deviates laterally with respect to the target direction. calculating means for calculating the amount of lateral positional deviation of the hydraulically driven vehicle with respect to the target direction by integrating the calculated deviation speed; and a gear ratio correcting means for correcting a gear ratio corresponding to the manual operation signal in accordance with the amount of lateral positional deviation. 2. The straight-line compensating device for a hydraulically driven vehicle according to claim 1, wherein the manual operation signal is a vehicle speed signal according to a set vehicle speed of the hydraulically driven vehicle and a steering signal according to a steering operation. 3. The hydraulically driven vehicle according to claim 2, wherein the calculation means resets the calculated lateral positional deviation amount to zero when the vehicle speed signal changes to a signal corresponding to zero vehicle speed. straight line compensator. 4. The gear ratio correction means corrects the gear ratio by increasing the gear ratio of one hydraulic transmission and decreasing the gear ratio of the other hydraulic transmission at the same time. Straight line compensator for hydraulically driven vehicles. 5. A hydraulically driven vehicle that steers the hydraulically driven vehicle by changing the gear ratio of the left and right hydraulic transmissions provided for each of the left and right tracks or the left and right wheels in response to a steering signal corresponding to a steering operation. In the straight-line compensating device, speed detection means detects the output speeds of the left and right hydraulic transmissions, and a difference between the detected output speeds of the speed detection means is integrated to determine whether the traveling direction of the hydraulically driven vehicle is relative to the target direction. determining the angle at which the hydraulically driven vehicle deviates in a lateral direction with respect to the target direction based on the determined angle and the average speed of both detected output speeds of the speed detecting means; a calculating means for calculating the amount of lateral positional deviation of the hydraulically driven vehicle with respect to the target direction by integrating the deviation speed; and resetting the integral calculation value in the calculating means to zero in response to input of a reset signal, and canceling the reset. calculation control means for starting an integral calculation in the calculation means in response to a signal input; and when the steering signal has a predetermined magnitude that instructs the hydraulically driven vehicle to move straight, the lateral The gear ratio corresponding to the steering signal is corrected according to the amount of lateral positional deviation so that the amount of directional positional deviation becomes zero, and the hydraulically driven vehicle is turned left or When the steering signal changes to a predetermined magnitude that instructs a right turn, the gear ratio corresponding to the steering signal is corrected according to the amount of lateral positional deviation calculated by the calculation means at the time of the change, and outputting the reset signal, and when the steering signal changes from a magnitude instructing the left turn or the right turn to a magnitude instructing the straight ahead, outputting the reset release signal at the time of the change; A straight-line compensating device for a hydraulically driven vehicle, comprising a gear ratio correcting means.