JPS5851305A - Path interpolation system for robot hand - Google Patents

Abstract

PURPOSE:To obtain polygonal continuation and to perform real-time path interpolation by making the path of a robot, given as a polygonal line, continuous so that the hand in equal-acceleration motion near the polygonal line. CONSTITUTION:A processor 1 couples through a bus line 2 with a memory 3, a table 4 of trigonometric and inverse trigonometric functions, etc., a multiplier and divider 5, an output board 6 for a robot body 9, an input board 7 for signals from a robot detector, and a teaching box 8. Firstly, the teaching box 8 drives the robot to a prescribed position to read the value of the displacement detector of an actuator, and the position of the hand in the space is calculated from the read value and then stored in a memory. Then, the hand is driven along straight lines connecting the obtained points. Its speeds of movement along the straight lines are specified by the teaching box 8.

Description

DETAILED DESCRIPTION OF THE INVENTION The present invention relates to a method for determining the route that the robot hand should take based on predetermined position data.

When making a path that was conventionally given by two broken lines continuous,
Two broken lines are made continuous using a circle with a specified radius, but this method complicates the calculations], which is difficult to do when driving the robot! The disadvantages are that it is difficult to calculate the route in terms of time and that teaching becomes very complicated.

SUMMARY OF THE INVENTION The present invention provides a path interpolation method for a robot hand that eliminates the above-mentioned drawbacks of the prior art, makes it possible to make the break line continuous by a simple calculation, and allows path interpolation to be performed in real time.

That is, the present invention uses an interpolation calculation that equally divides the alert positions determined to determine the robot's path, so that the robot moves along the robot's path given by the broken line with constant acceleration near the broken line. It is characterized by making the diameter W & continuous.

The present invention will be specifically described below based on embodiments shown in the drawings.

#1! Nine examples are shown in J with interpolation at equal intervals along two broken lines. In Figure 1, five points A, B, and C are taught to the robot and indicate the positions of nine hands.

Fast customs clearance! -9R with fast WLν easy hands AB, B
It is assumed that the vehicle is driven along C. At this time, the control circuit determines the position through which the hand passes on the line segment at each sampling time T, and controls the robot using this position as the target point at each sampling point. Therefore, customs clearance is n = [, tB/ν, 7')
(11 is %h-CEC/lJ]
It passes with the sampling number (2). For example, in equation (1), K() is A and ν, T≦, and <A and sr,
Indicates that it is a natural number in the range of 7'+1. Therefore, the target point at each sampling point is a
3. Equal division, tt--point of equal division. However, if the robot's hand is driven using this method, the acceleration will become infinite at point B, and the robot's hand will actually become undriveable. Therefore, the direction is changed from the line segment Kf1 before sampling from point β, and the route is corrected using a parabola so that the line segment BCIIC is reached after sampling from point B.

As shown in Figure 2, for the three points ABC, point B is the origin and the y-axis is in the direction that bisects the angle ABC. x to li
@ Once established, the equation of the radiation I#IJ line is expressed in the form %Formula % (31), and the tangent to this parabola is expressed by the differential equation of y=-Hz+c(9+d-(4)).Equation ( By differentiating 4: with respect to X and finding its singular solution, we obtain the equation (5). Substituting this into equation (4) with K and sorting it out, we get 1 y = -a, + d
俤) Tonapu, adult) and (6) from a Yaw 1/4 c + b = d
ffl, 181 relationships are obtained.

By the way, if the lengths of line segments DB and BE are both t,
Point and tangent of parabola at E#C! ! passes through the origin, so if the angle AECt-.theta., the slope of the tangent at the point and .theta. is ±14-, so we obtain the relationship. On the other hand, from equation (3), we obtain the slope of the tangent line. Therefore, by calculating the angle from the trout and substituting it into equation (6), the equation of the parabola becomes:

By the way, since the robot is under sampling control, the movement from point to point E must be performed by the printer 1i#X times. Is this sampling number t-2? If the position of the hand on the parabola at the 10,000th sampling from the start is (J,, 311B), then it is a uniform motion in the two axis directions or 0 (2)! , = (--1) t-l. Substituting this into equation (2) and rearranging, tθ 'lm=-((-1)" + 1) ?IB->
(2) Obtain le. Also, the slope of the tangent to the parabola at point (J,, y,) is -(m/1-1)bei (goods 2). The intersection point (,r,', y,') between this tangent and the line segment before is m θ m
θ (J,', y,') = (-noodles-1--mu)
a◆21% 2 2rL2, these points are 1.2, 3. . . . When changed to 9, the lines are plotted at equal intervals on the father line.

Also, the points ' ``*-1+ 'I s-1) and (x, ',
distance between y,') and point (xs-1+L+a-1)
The ratio of the distance between and (J, , y,) is from the component in the X direction. That is, Je is a point ("s −f+ 3's−
1) and (J,'.

The value of s'1wa is determined by dividing the line segment connecting y,') into equal parts a+1-++a,/2.If the difference between y, and y,-1 is small, the value of s'1wa is approximately the same as - It can be taken on a line segment.

Applying this result to the interpolation of the hand's path along the parabola, the algorithm is as follows: First, divide the line segment 1i into 25 equal parts, and start from the point closest to point B (ro',
5), (there'* 3')...--* (%'-t
3'z-). The last point coincides with E. Next, set the score to (% t %), and first select Coo @ lo
) and (he'*l) is divided into +14 equal parts, and (qey
Let the next point of o) be ('1 + yt).

Next, divide the line segment (Hei * yr) (#, yl') into equal parts, and now 9h), (West * 3/m) (匈t
'ls') line segment t-1172 equally divided 's*y,)
......, the second point is point E.
K-interval, ('s t 3'l) (%,'/2)
, ...-(M-, M-) is a parabola. As mentioned above, this parabolic interpolation calculation can be performed only by linear interpolation calculation that divides the line segment into intervals.
It has the advantage of not requiring complicated calculations.

FIG. 3 shows an example of route interpolation when path=5.

Mu, E1. &, ..., & are the points where the line segment Ω is divided into six equal parts. First, divide the line turtle into 3.5 equal parts and determine the points.

Divide the hen into S fathoms and make 8 pieces.Similarly, make stripes, hem...1 item each 2.5.2. ..., divide into 1 equal parts, 8, A, ...
, R is the bird).

FIG. 4 shows the relationship between the interpolation points obtained as described above and the parabola of equation (2). The circles in the figure are interpolation points, and the broken lines indicate parabolas.The two provinces almost match. The difference between the two is due to the fact that the interpolation point is set on the tangent to the parabola, as shown in FIG. In addition, in Fig. 4, the connection = 10. θ=60
°.

The method described above is based on the release 1 [! This method can be used as an interpolation method for general acceleration/deceleration control.

Assume that the previous method is applied when moving at a constant speed along the drum and tray as shown in FIG.

In this case, let us consider nine motion components along ' in the line segment that includes the line segment. Figure 5 shows the case of Satoru 5.

Project the point Ei (j=1 to 6) onto 2 and set the dead center to Ecp. At this time, if the velocity of the hand between points A and D and between the bird and C is l'jj, the velocity component between r along line segment 2' is vBc""vmaθ, and the previous method is If applied along E6', the hand will accelerate or practice at a constant acceleration from point to point E6'. In other words, t of the parabolic path interpolation result at 1ijAEc
In the projection to AC', the opening of the interpolation points is kept proportional, and the interpolation method is also linear, so the interpolation method 4 along AC'
*It is the same as the tl method. In this way, No.
i Moves along AC', and during 9 the speed is ν41 + E
When driving with acceleration or deceleration near point B at speed ν1C between jC' and speed νjj + ''' between n
The distance between the hooks is divided equally to correspond to the speed vBσ, the distance between the hooks is further divided into two, and E1', E; , . . . s 4 are determined. Next, the DEj function is divided into 15 equal parts as shown in equation (2), and the Pj - anti interval is divided into 5 equal parts to obtain Pi.

You can go as...

In this method, especially when 1IB-JQ, the no.nd corresponds to a motion that accelerates with uniform acceleration during 24 samplings from a stopped state, and conversely corresponds to a motion that reaches a stop and stops when 1.nu.IC=0.

FIG. 6 shows the interpolation results when the vehicle is accelerated from a stop according to the above method. The circles in the figure are the interpolation results, and the solid lines indicate the parabolas when the motion is accelerated. The two are in good agreement. Note that in this example, turtle=10.

Fig. 97 shows an example of the distance when the speeds along the broken line are different. Here the speed between the two is half the speed between the two and the path=10.

As described above, the θ position of the hand can be parabolically delayed and interpolated. Regarding the interpolation of the hand posture, the hand posture is usually expressed by five independent parameters, and these parameters can be interpolated using similar means.

-m 1+In the interpolation operation, the length of each line segment is divided into n+7 equal parts to determine each interpolation point. ,, n+2, and it is also possible to make corrections and adjust acceleration/deceleration.

FIG. 9 shows an example of a robot control circuit that implements the present invention.

Pro Seven Tuna 1 is connected to bus 2 in 2'ft through memory 3, table 41 for trigonometric functions, inverse trigonometric functions, etc. Multiplier/divider 5. Output port to robot body 9.

A robot detector signal input port door and a teaching box 8 are connected.

First, the robot is driven to a predetermined position using the teaching box 8, the value of the displacement detector (not shown) of the actuator at this position is read 9, and the position of the hand in space is calculated from this value. Store in memory. The hand is driven along i[lI, which connects the points thus obtained. Speed 1 along Im at this time is also designated by the teaching box.

When implementing this patent, the positions of the nine hands stored in the memory as described above and the speed along the line when consecutive positions are connected with a straight line are used as shown in Figure 5. interpolates the given nine position data in a simple manner.

The interpolation operation according to the present invention can be performed using a multiplier/divider S, and the interpolation operation shown in the present invention is performed from the position data stored in the memory 5, and the resulting position data of the friend hand is It is converted into a pairwise displacement of the robot and outputted as a target value to the robot via the output port 1P.

As described above, according to the present invention, it is possible to easily perform interpolation calculations on the movements of the four bot hands, including acceleration and deceleration, and the path interpolation of the robot hands can be performed in real time.

[Brief explanation of the drawing]

FIG. 1 is a diagram showing a broken line path, FIG. 2 is a diagram for explaining the method of continuous YrFi of a path using a parabola according to the present invention, and FIGS. 93 to 7 are diagrams showing interpolation results according to the present invention. , FIG. 8 is a diagram for explaining the division interpolation correction method, and FIG. 9 is a diagram showing -Ij! of a control circuit that embodies the present invention. It is a figure showing an example. 1... Processor 2... Pass line 3...
Memory 4...Table 5...Multiplier/divider 6...Output port door...Input port
8...Teaching box 9...Patent attorney representing the robot body Usui 1) Toshiyuki 17 Figure 3 Rate 40,! I-6 Figure 7 Figure 80

Claims (1)

[Claims] To determine the route of the tcl bot, we use an interpolation calculation that divides the distance between the emergency positions, and the path is determined by # of the robot given by a broken line, and /1 is near the break point. This is a fighting robot's 1nd route interpolation method, which is characterized by making the route continuous so as to perform continuous motion. 2. A path interpolation method for a robot node according to claim 1, characterized in that the hand is accelerated and decelerated along a straight line with respect to the movement of the hand along a straight line.