JP3081763B2 - Control method for autonomous driving robot - Google Patents

Description

DETAILED DESCRIPTION OF THE INVENTION

[0001]

BACKGROUND OF THE INVENTION 1. Field of the Invention The present invention relates to a vehicle driving simulation in which a driving wheel is mounted on a rotating drum of a chassis dynamometer to drive a vehicle and a dynamic driving performance test of the vehicle is performed indoors. The present invention relates to a method for controlling an automatic driving robot for an automobile that automatically drives a vehicle.

[0002]

2. Description of the Related Art Conventionally, in order to test the dynamic running performance of an automobile, a real vehicle running simulation operation has been performed by a chassis dynamometer. An autonomous driving robot (hereinafter referred to as an autonomous driving robot) in which a plurality of actuators are individually driven by a motor or the like, and these actuators can be used to operate an accelerator pedal, a brake pedal, a clutch pedal, and the like, and to shift gears. ) Is being used.

[0003] In the actual vehicle running simulation driving, it is necessary to drive the vehicle according to a predetermined driving program (running pattern), and the accelerator pedal depression amount (hereinafter referred to as accelerator depression amount) F
_A is the sum of a term proportional to the target speed of the automobile and a term proportional to the target acceleration (time derivative of the target speed). That is, if the target speed is V _NOM , the target acceleration is dV
Is expressed as _NOM / dt, the F _A is represented as _{F A = K 1 V NOM +} K 2 dV NOM / dt, is compared with the K ₁ V _NOM and K ₂ dV _NOM / dt, the acceleration term, K ₂ dV _NOM / dt is larger.

In the prior art, in order to set the target speed V _NOM and the target acceleration dV _NOM / dt, as shown in FIG. _Is interpolated by a straight line and the target speed V _NOM
(Shown by curve 1), and the target velocity V _NOM is differentiated with respect to time to obtain a target acceleration dV _NOM / dt (shown by curve 2).

[0005]

The target speed V _NOM is set as described above, and the target acceleration dV _NOM /
When dt is obtained, as shown in FIG. 7, the target acceleration dV _NOM / dt changes stepwise before and after the speed regulation point every second and becomes discontinuous, and as a result, the accelerator depression amount becomes smaller. This is disadvantageous in the exhaust gas certification test because the concentrations of CO and HC in the exhaust gas increase and the fuel efficiency deteriorates. FIG.
(A) is a part of the operation program (time 163.0 seconds-
22B), and FIG. 17B is an enlarged view of the time 175.0 seconds to 190.0 seconds.
The same applies to FIGS. 2, 4 and 6 described below.

SUMMARY OF THE INVENTION The present invention has been made in consideration of the above-mentioned problems, and is intended to prevent a target acceleration of a vehicle from becoming discontinuous before and after a speed regulation point, to smooth an accelerator pedal depression amount, and to further improve driving performed by a human driver. It is an object of the present invention to provide a control method of an automatic driving robot for an automobile, which can drive the vehicle close to the vehicle.

[0007]

In order to achieve the above object, a method of controlling an automatic driving robot for an automobile according to the present invention comprises:
Calculate the target speed by interpolating between the two speed specified points with a straight line
Time derivative of this target speed
The target acceleration is obtained by using the spline function,
The accelerator depression amount is obtained using the target acceleration and the target speed .

[0008]

According to the above-mentioned control method, the acceleration does not change stepwise, the change in the accelerator depression amount becomes smooth, and the driving can be performed more like a human driver.

[0009]

DETAILED DESCRIPTION OF THE PREFERRED EMBODIMENTS First, an approximation by a cubic equation that passes through two points and where the differential coefficients at the two points coincide with each other As shown in FIG. 3, four velocity defining points P ₁ (t ₁ ,
y ₁ ), P ₂ (t ₂ , y ₂ ), P ₃ (t ₃ , y ₃ ), P
₄ (t ₄ , y ₄ ), and a cubic expression for interpolating between the two central velocity defining points P ₂ , P ₃ (t ₂ ≦ t ≦ t ₃ ) is determined. In that case, an attempt was made to express the target speed V _NOM between the _two speed specified points P ₂ and P ₃ with a cubic function of time satisfying the following conditions. Passing through the _two speed regulation points P ₂ and P _3, and the differential coefficient at the left speed regulation point P ₂ being the point P ₂
Derivative in that - the right speed prescribed point P ₃ to match the derivative of the cubic function passing between the further and the speed prescribed point P ₁ on the left when the, the point P ₃
And the differential coefficient of a cubic function that passes between the speed stipulating point P ₄ and the speed stipulating points P ₂ , P 4
The value of the differential coefficient in ₃ is given the average slope between the speed specified points sandwiching that point.

Now, the interpolation function is given by y (t) = at ³ + bt ² + ct + d (1)

Two speed defining points P ₂ (t ₂ , y ₂ ), P
₃ (t _3, y ₃₎ by passing _{through, y (t 2) = at} 2 3 + bt 2 2 + ct 2 + d = y 2 ...... (2) y (t 3) = at 3 3 + bt 3 2 + ct 3 + D = y ₃ (3)

When the inclinations at the speed specified points P ₂ and P ₃ are y ₂ ′ and y ₃ ′, y ₂ ′ = (y ₃ −y ₁ ) / (t ₃ −t ₁ ) (4) y ₃ ′ = (y ₄ −y ₂ ) / (t ₄ −t ₂ ) (5)

[0013] By decided slope at two speeds defined point _{P 2, P 3, y '} (t 2) = 3at 2 2 + 2bt 2 + c = y 2' ...... (6) y '(t 3) = ^{_{3at 3 2 + 2bt 3 + c}} = y 3 ' becomes ... (7).

The above four equations (2), (3), (6),
From (7), four unknowns a, b, c, and d are obtained. Here, replacing t = t−t ₂ and T = t ₃ −t ₂ ,
The above equations (2), (3), (6) and (7) are expressed as follows: y (0) = d = y ₂ ... (2) ′ y (T) = aT ³ + bT ² + cT + d = y ₃ ( 3) 'y' (0) = c = y ₂ '(6)' y '(T) = 3aT ² + 2bT + c = y ₃ ' (7) '

From the above equations (2) 'and (6)', c and d are determined. From equations (3) ′ and (7) ′, aT ³ + bT ² = y ₃ −cT−d (8) 3aT ² + 2bT = y ₃ ′ −c (9) is obtained.

By performing (9) × T− (8) × 2, aT ³ = y ₃ 'T-cT-2y ₃ + 2cT + 2d = y ₃ ' T-2y ₃ + cT + 2d (10) is obtained. By transforming 8), bT ² = y ₃ −cT−d−aT ³ (11) is returned.

From the above, four unknowns a, b, c,
Since d is obtained, t ₂
Y (t) = a (t−t ₂ ) ³ + b (t−t ₂ ) ² + c (t−t ₂ ) + d for an arbitrary t at ≦ t ≦ t ₃ (12) I do.

When the target speed V _NOM and the target acceleration dV _NOM / dt are obtained by the cubic function expressed by the above equation (12), a curve 1 representing an operation program as shown in FIG.
2 is obtained.

According to the method I, the target speed V _NOM is continuous before and after the speed specified point, but there are two problems as follows. That is, one is the target acceleration dV _NOM
/ Dt is not continuous. Other is Oite the target acceleration dV _NOM / d t, it is that it is extra horns. The most prominent example of this horn is shown in FIG.
In (A), as indicated by reference numeral 3, it appears before the point where the target speed V _NOM starts increasing from zero. Target speed V
Target acceleration dV _{NOM at the} point where _NOM starts to increase from zero /
dt is a positive value according to the above condition. That means
The target speed V _NOM immediately before that becomes a negative value, and the target acceleration dV _NOM / dt once increases after becoming negative, and does not match the actual operation.

II Next, in order to solve the first drawback of the cubic function described in the above I, that is, the drawback that the rate of change of the target acceleration dV _NOM / dt is not continuous,
An attempt was made to express a target speed V _NOM between two speed specified points by a _quintic function of time satisfying the following conditions. That is,
As shown in FIG. 5, six speed regulation points P ₁ (t ₁ ,
y ₁ ), P ₂ (t ₂ , y ₂ ), P ₃ (t ₃ , y ₃ ), P
₄ (t ₄ , y ₄ ), P ₅ (t ₅ , y ₅ ), P ₆ (t ₆ ,
y ₆ ), and the two central velocity defining points P ₃ and P ₄ (t
_A quintic equation for interpolating between ₃ ≦ t ≦ t ₄ ) is determined. The conditions are as follows. Passing through the two speed defining points P ₃ and P _4, and differentiating a quintic function whose differential coefficient at the left speed defining point P ₃ passes between the point P ₃ and the left speed defining point P ₂ The second order differential coefficient of the quintic function passing between the point P ₃ and the further left speed defining point P ₂ is the second order differential coefficient at the left speed defining point P ₃ . derivative in that - the right speed prescribed point P ₄ match the found-that match the derivative of the quintic function passing between the further right speed prescribed point P ₅ and the point P ₄ The second derivative at the right speed defining point P ₄ matches the second order derivative of the quintic function passing between that point P ₄ and the further right speed defining point P ₅ ; Speed regulation points P ₃ , P
The value of the differential coefficient in ₄ was given the average slope between the speed specified points sandwiching that point. Also, the speed regulation points P ₃ , P ₄
For the value of the second-order differential coefficient in the above, the differential coefficient at a speed specified point sandwiching that point was similarly obtained, and the average was used as the slope.

[0021] Now, the interpolation function, and ^{y (t) = at 5 +} bt 4 + ct 3 + dt 2 + et + f ...... (21).

The two speed regulation points P ₃ (t ₃ , y ₃ ), P
₄ (t _4, y ₄₎ by passing _{through, y (t 3) = at} 3 5 + bt 3 4 + ct 3 3 + dt 3 2 + et 3 + f = y 3 ...... (22) y (t 4) = at 4 ^{_{5 + bt 4 4 + ct 4}} 3 + dt 4 2 + et 4 + f = y 4 ...... (23) and represented.

Then, assuming that the inclinations at the speed specified points P ₃ and P ₄ are y ₃ ′ and y ₄ ′, respectively, y ₃ ′ = (y ₄ −y ₂ ) / (t ₄ −t ₂ ) 24) y ₄ ′ = (y ₅ −y ₃ ) / (t ₅ −t ₃ ) (25) is obtained.

By determining the slopes at the two speed defining points P ₃ and P ₄ , y ′ (t ₃ ) = 5 at ₃ ⁴ +4 bt ₃ ³ + 3ct ₃ ² + 2dt ₃ + e = y ₃ ′ (26) y _{'(t 3) = 5at 4} 4 + 4bt 4 3 + 3ct 4 2 + 2dt 4 + e = y 4' becomes ... (27).

Assuming that the inclinations at the two speed specified points P ₃ and P ₄ are respectively y ₃ ″ and y ₄ ^″ , y ₃ ″ = (y ₄ ^′ −y ₂ ^′ ) / (t ₄ −t ₂ ) ...... (28) y ₄ "= a ^{_{(y 5 '-y 3')}} / (t 5 -t 3) ...... (29). Here, y ₂ ^′ and y ₅ ^′ are the slopes at points P ₂ (t ₂ , y ₂ ) and P ₅ (t ₅ , y ₅ ) obtained in the same manner as in equation (24).

By determining the inclination of the inclination at the two velocity defining points P ₃ and P ₄ , y ″ (t ₃ ) = 20 at ₃ ³ +12 bt ₃ ² +6 ct ₃ + 2d = y ₃ ″ (30) y ″ ^{_{(t 4) = 20at 4 3}} + 12bt 4 2 + 6ct 4 + 2d = y 4 " becomes a ... (31).

The above six equations (22), (23), (2)
6) From (27), (30), and (31), six unknowns a, b, c, d, e, and f are obtained. Here, t = t−
t ₃ , T = t ₄ −t ₃ , the above equation (2)
2), (23), (26), (27), (30), (3)
1), y (0) = f = y 3 ...... (22) 'y (T) = aT 5 + bT 4 + cT 3 + dT 2 + eT + f = y 4 ...... (23)' y '(0) = e = _{y 3 '...... (26)'} y '(T) = 5aT 4 + 4bT 3 + 3cT 2 + 2dT + e = y 4' ...... (27) 'y "(0) = 2d = y 3" ...... (30)' y ^{"(T) = 20aT 3 +} 12bT 2 + 6cT + 2d = y 4" becomes a ... (31) '.

The above equations (22) ', (26)', (3
0) ', d, e, and f were determined. Equation (2)
3) ', (27)', ( ^{'from, aT 5 + bT 4 + cT} 3 = y 4 -dT 2 -eT-f ...... (32) 5aT 4 + 4bT 3 + 3cT 2 = y 4' 31) -2dT-e ^{...... (33) 20aT 3 + 12bT} 4 2 + 6cT 4 = y 4 "-2d ...... (34) is obtained.

Here, (34) × T ² − (33) × 2T
^{Doing, 10aT 5 + 4bT 4 = y} 4 "T 2 -2y 4 'T + 2dT 2 + 2eT ...... (35) is obtained, when a (33) × T- (32) × 3, 2aT 5 + b T _{^{4 = y 4 'T-3y}} 4 + dT 3 + 2eT + 3f ...... (36) is obtained, and, (35) - (36) When performing _{^{× 4, 2aT 5 = y 4}} "T 2 -6y 4' T + 12y 4 2dT ² -6eT-12f (37) is obtained, and the equation (36) is transformed to obtain bT ⁴ = y ₄ ′ T-3y ₄ + dT ³ + 2eT + 3f-2aT ⁵ (38) Then, the expression (32) is transformed to obtain cT ³ = y ₄ -dT ² -eT-f-aT ⁵ -bT ⁴ (39).

As described above, the six unknowns a, b, c,
Since d, e, and f have been obtained, using these, for any t at t ₃ ≦ t ≦ t ₄ , y (t) = a (t−t ₃ ) ⁵ + b ( (t−t ₃ ) ⁴ + c (t−t ₃ ) ³ + d (t−t ₃ ) ² + e (t−t ₃ ) + f (40)

When the target speed V _NOM and the target acceleration dV _NOM / dt are obtained by the _quintic function represented by the above equation (40), a curve 1 representing an operation program as shown in FIG.
2 is obtained. As can be understood from FIG. 6, according to the method II, the target acceleration dV _NOM is obtained before and after the speed regulation point.
/ Dt and its rate of change are continuous, and the first disadvantage of the method I has been solved.

However, according to the above-mentioned method II, the horn, which has been a problem in the above-mentioned method I, becomes sharper than in the method I, as indicated by reference numeral 4 in FIG. ing. This is probably because the degree of the function used for the interpolation increased, and the curve was wavy, and the increase and decrease of the fine acceleration increased.

III So, finally, the first floor with two points, 2
Approximation by a cubic equation in which the differential coefficients of the order match (the target velocity V between two velocity stipulated points by a cubic base spline function)
_NOM ), that is, as shown in FIG. 1, four speed regulation points P ₁ (t ₁ , y ₁ ), P ₂ (t ₂ , y ₂ ),
There are P ₃ (t ₃ , y ₃ ) and P ₄ (t ₄ , y ₄ ), and a cubic function of time satisfying the following conditions is obtained.・
Derivative in the left of the velocity defined point P ₂ is, the point P ₂ further speed specified point that, left to match the derivative of the cubic function passing between the speed prescribed point P ₁ on the left and P ₂ second order differential coefficient at the, in the second-order differential coefficient and that the right and the speed prescribed point P ₃ match cubic function passing between the speed prescribed point P ₁ of the further left and the point P ₂ The differential coefficient coincides with the differential coefficient of a cubic function passing between the point P ₃ and the further right speed defining point P _4.
Second order differential coefficient at _3, and it is consistent with the second-order differential coefficient of the cubic function passing between the point P ₃ and further to the right of the speed prescribed point P _4, the rate prescribed point P _2, As the value of the derivative at P _3, the value of the derivative at that point of the quadratic function passing through the three points, that point and the left and right points, was given.
The value of the second-order differential coefficient at that point of the same quadratic function was given to the value of the second-order differential coefficient at the speed regulation points P ₂ and P ₃ .

Now, the interpolation function is given by y (t) = at ³ + bt ² + ct + d (41)

The inclination at the speed specified point P ₂ (t ₂ , y ₂ ) is y ₂ ′, and the inclination of the inclination is y ₂ ″. Y ₂ ′ and y
₂ ″ are three points P ₁ (t ₁ , y ₁ ), P ₂ (t ₂ ,
y ₂ ), t = t _{2 of a} quadratic function passing through P ₃ (t ₃ , y ₃ )
In the first and second order.

A quadratic function passing through the three points P ₁ , P ₂ , and P ₃ is represented by Y (t) = At ² + Bt + C (42).

[0037] By passing through the point quadratic function _{3 P 1, P 2, P} 3, Y (t 1) = At 1 2 + Bt 1 + C = y 1 ...... (43) Y (t 1) = At a ^{_{2 2 + Bt 2 + C =}} y 2 ...... (44) Y (t 1) = At 3 2 + Bt 3 + C = y 3 ...... (45).

Here, t = t−t ₂ , t ₁ ′ = t ₁ −t
₂ , t ₃ ′ = t ₃ −t ₂ , Y (t ₁ ′) = At ₁ ′ ² + Bt ₁ ′ + C = y ₁ (46) Y (0) = C = y ₂ (47) ) Y (t ₃ ′) = At ₃ ′ ² + Bt ₃ ′ + C = y ₃ (48)

From the expressions (46) and (48), At ₁ ' ² + Bt ₁ ' = y ₁ -y ₂ (49) At ₃ ' ² + Bt ₃ ' = y ₃ -y ₂ (50) From these equations (49) and (50), A = {(y ₁ −y ₂ ) t ₃ ′ − (y ₃ −y ₂ ) t ₁ ′} / (t ₁ ′ ² t ₃ ′ − ^{_{t 1 't 3' 2)}} ...... (51) B = {t 1 '2 (y 3 -y 2) -t 3' 2 (y 1 -y 2)} / (t 1 '2 t 3' - t ₁ 't ₃ ' ² ) (52) is obtained.

From the equations (47), (51), and (52), unknown coefficients A, B, and C are obtained, and three points P ₁ ,
A quadratic function passing through P ₂ and P ₃ was determined. And
y ₂ ′, y ₂ ″ is 1 at t = t ₂ of this quadratic function.
Since it is the second-order differential coefficient, y ₂ ′ = Y ′ (0) = 2A × 0 + B = B y ₂ ″ = Y ″ (0) = 2A.

Similarly, three points P ₂ (t ₂ ,
y ₂ ), a quadratic function passing through P ₃ (t ₃ , y ₃ ) and P ₄ (t ₄ , y ₄ ) is obtained, and the first-order and second-order differential coefficients at t = t ₃ are calculated as y ₃ ′ , Y ₃ ″.

From the above equation (41), y ′ (t) = 3 at ² +2 bt + c (53) y ″ (t) = 6 at + 2b (54) is obtained.

[0043] than the conditions of _{interpolation, y '(t 2) =} 3at 2 2 + 2bt 2 + c = y 2' ...... (55) y '(t 3) = 3at 3 2 + 2bt 3 + c = y 3' ...... ( 56) y ″ (t ₂ ) = 6 at ₂ +2 b = y ₂ ″ (57) y ″ (t ₃ ) = 6 at ₃ +2 b = y ₃ ″ (58)

Here, if t = t−t ₂ and T = t ₃ −t ₂ are replaced, the above equations (55), (56), (57),
(58), y '(0) = c = y 2' ...... (55) 'y' (T) = 3aT 2 + 2bT + c = y 3 '...... (56)' y "(0) = 2b = y ₂ is represented as "...... (57) 'y" (T) = 6aT + 2b = y 3 "...... (58)'.

The unknown coefficients b and c are determined from the equations (55) 'and (57)'. From equation (58), 6aT = y ₃ ″ −2b (59)

Here, even if the unknown a is obtained by substituting b and c obtained from the equations (55) ′ and (57) ′ into the equation (56) ′, t ₁ , t ₂ , t ₃ , t _{If 4} are equally spaced, the result is the same. Thus, the unknowns a, b, and c have been obtained.

The unknown d is indefinite, but t =
The value of y at t ₂ , that is, y (t ₂ ) is defined as t ₁ ≦
It can be obtained by adding a condition that the value of the interpolation function at t ≦ t ₂ is t = t ₂ .

Using a, b, c, and d obtained in this manner, for any t at t ₂ ≦ t ≦ t ₃ , y (t) = a (t−t ₂ ) ³ + b (T−t ₂ ) ² + c (t−t ₂ ) + d (60)

As described above, the cubic equation is determined by the method of III, but since the conditions are all related to the differential coefficient, the constant term is not determined. Therefore, the target speed V _NOM is not determined. When the coefficients other than the constant term are obtained and the target acceleration dV _NOM / dt is obtained, curves 1 and 2 representing the operation program as shown in FIG. 2 are obtained. Incidentally, in this figure, curve 1 representing the target speed V _NOM is is depicted as conveniently through the speed prescribed point P ₂ on the left.

According to the method III, all the disadvantages of the methods I and II have been solved. That is, the target acceleration dV _NOM / dt and the rate of change thereof are continuous before and after the speed regulation point, and there is no extra horn on the curve 2 representing the target acceleration dV _NOM / dt.

However, the speed is not continuous before and after the speed specified point, and therefore, the target speed V _NOM is not determined. For example, if the vehicle passes through the left speed specified point P ₂ , the right speed specified point P ₃ It will be discontinuous. Therefore, the target speed V _NOM is formed by an appropriate method such as, for example, a straight line interpolation between the speed specified points, as in the related art.
Preferably, V _NOM / dt is _created using the base spline function.

[0052]

As described above, the method for controlling an automatic driving robot according to the present invention provides a method for controlling the speed between two specified speed points.
The target speed is obtained by interpolating with a straight line,
Differentiate the degree over time and use the base spline function
By calculating the target acceleration,
Since the accelerator depression amount is obtained using the target speed , the acceleration does not change in a stepwise manner, and the change in the accelerator depression amount becomes smooth,
Driving closer to a human driver can be performed. Therefore,
Advantageous results are also obtained in exhaust gas certification tests.

[Brief description of the drawings]

FIG. 1 is a diagram for explaining a method for controlling an automatic driving robot for an automobile according to the present invention.

FIG. 2 is a diagram showing an example of a traveling pattern representing a target speed and a target acceleration obtained by the method.

FIG. 3 is a diagram for explaining a method tried in a process leading to the present invention.

FIG. 4 is a diagram showing an example of a traveling pattern representing a target speed and a target acceleration obtained by the trial method.

FIG. 5 is a diagram for explaining another method tried in a process leading to the present invention.

FIG. 6 is a diagram showing an example of a traveling pattern representing a target speed and a target acceleration obtained by the other trial method.

FIG. 7 is a diagram for explaining a conventional technique.

──────────────────────────────────────────────────続 き Continuation of the front page (56) References JP-A-4-249735 (JP, A) JP-A-5-312685 (JP, A) JP-A-63-61307 (JP, A) (58) Field (Int.Cl. ⁷ , DB name) G01M 17/00 G01M 17/007 G05B 19/416

Claims (1)

(57) [Claims] 1. A linear interpolation method between two specified speed points.
While obtaining the target speed, the target speed is differentiated with time,
Furthermore, by using the base spline function,
A method for controlling an automatic vehicle driving robot, wherein an acceleration is obtained, and an accelerator pedal depression amount is obtained using the target acceleration and the target speed .