JPS5984110A - Method and device for measuring position of body - Google Patents

Abstract

PURPOSE:To improve measurement precision by setting an object point whose position is to be measured at the center points of the visual fields of two television cameras, and measuring the distance between the cameras and object point. CONSTITUTION:Observation points ER and EL are set on the left and right sides of an origin 0. Mirrors are provided rotatably at the observation points ER and EL and the object point P is positioned in the centers of visual fields of the camera. Then, the object point P(X, Y) is calculated from angles thetaL and thetaR between a (y) axis and the mirrors at points ER and EL and the distance L between the mirrors. This principle applies to object points P in a three- dimensional space.

Description

DETAILED DESCRIPTION OF THE INVENTION The present invention relates to a method and apparatus for measuring the position of a target object.

(a) Automated machines that combine conventional shape recognition devices industrial robots and visual (image processing) devices are being put into practical use.

In these attempts, most of the visual functions required are ``shape recognition and object identification.'' Measurement of the position of the target object cannot be determined. This is because the position is usually determined in advance.

For example, as shown in FIG. 10, the object shape recognition device
It is assumed that the television camera C recognizes all target objects placed on the workbench T.

In this case, the television camera C is installed almost directly above the object 0, the vertical distance between the camera C and the worktable T is 2, and the height of the object is h.
Give it in advance.

Then, regarding the plane coordinates (X, Y) on the workbench, the object 0
It can be detected by Mk television camera C.

This is because the distance (z-h) between the camera and the object is known, so the position of object 0 can be known. It is possible to recognize not only the center of gravity and center position of an object, but also its shape.

However, in such a visual device, the camera is fixed in an upper space, so the field of vision is narrow.

It is not possible to search for the target object over a wide range and measure its position.

If the television camera C is moved in parallel, the measurement area will be expanded.

For this purpose, a drive device such as a rail or a servo motor is required to fully move the camera in the X-axis and Y-axis directions.

Since a heavy camera is moved in parallel, the rails, motors, reducers, control circuits, etc. must be large. Visual equipment becomes large and expensive. Installing the visual equipment in the space above the workbench is also not easy.

However, the biggest drawback of this method is that the vertical distance 2 between the camera and the platform and the height h of the object must be determined in advance.

[ is completely useless if the vertical distance is unknown.

Rather, there are many cases where you want to know the distance itself.

(b) Binocular stereoscopic viewing device There are already known methods for measuring the distance to a target object.

One method is to take advantage of the camera's defocus.

This is simple but has poor accuracy.

Another method is to irradiate a target with laser light or the like and find the distance from the reflected light. However, this method is limited in scope and environment. -Poorly lethal.

All of these had one camera that received the light.

This is a monocular method.

In contrast, a stereoscopic viewing device using two eyes has also been proposed.

Having two cameras improves measurement accuracy.

``Binocular Stereoscopic Vision of Dagumi's Object Recognition'' (Yasueri-1 Yoshiaki Shirai, published December 1972), Vol. 37, No. 12, p. 49 of the Spiritual Report of the Electronic Technology Research Institute, describes how to recognize objects with two eyes. A distance measuring device is disclosed.

It consists of two fixed side mirrors, a switching mirror in between, and a camera. The switching mirror selectively guides the light from the left mirror and the light from the right mirror to the camera.

The light from the left and right mirrors is then captured by a camera and divided into pixels. All the divided parts are stored in the computer. Although there is only one camera, the reflected light from the left and right mirrors is selectively imaged, so it is effectively the same as having two cameras.

The line segment connecting the two mirrors is called the baseline. In the above device, the mirrors always move symmetrically. In other words, the base line and the normal of the two trees erected at the center of the mirror always form an isosceles triangle.

Therefore, by knowing the base angle α and multiplying it by the length Vcpnα of the base line length 2a, the distance from the center of the base line to the center of the workbench on which the object is placed can be found.

The object is on the workbench, but it is not always centered.

The pixels of the image from the left mirror and the pixels from the image of the right mirror are related to each other so as to correspond to a certain point P on the object.

For example, one image is divided into 256 horizontal by 256 vertical pixels. The number of pixels is 256×256.

Information about the brightness of a certain point is divided into 64 levels. The intensity is defined as an integer from 0 to 63. The brightness of each pixel is measured and stored in the memory corresponding to each pixel.

This method effectively uses two cameras. A mirror can be considered a camera. Therefore, the following description assumes that left and right cameras are attached to the positions of the left and right mirrors.

The left and right cameras are in the center of the workbench. VC@ is kicked and fixed. Cameras A and B and the center 0 of the workbench form an isosceles triangle.

Center A of the image of camera A, and center B of the image of camera day
oii, both correspond to the center 0 of the workbench. Center M of baseline AB connecting the camera. The distance i was a-α.

Point A9 of the PH1A camera at a point other than the center 0 of the workbench, 8
It is assumed that this corresponds to one point on the camera day. Here A9, B,)
is one of the pixels. Ap and Bp generally do not match. The distance between the camera base line center M and the point P can be determined from the X, Y coordinates of the pixel point A9, day, and α, a.

Such binocular stereopsis has the advantage that the positions of all points on the workbench can be determined by calculation without moving the camera and therefore the mirror. Once the two images of the left camera and right camera are determined, all subsequent processing is determined by calculation.

However, this device requires, for example, 64 levels (6 bits) of information for 256 x 256 pixels. It requires a large-scale information processing device that detects brightness and stores memory.

This contains a lot of unnecessary processing and information.

For example, the corresponding point A of the target point P in the left and right camera images
9. To know B, to find the brightness level using a local correlation method, etc., compare the brightness information of A, the neighborhood, day, and the neighborhood, and if the local correlations of both places match, A9 point B9
This means that it is a point. Data for more distant parts are not needed.

In this way, there is a lot of unnecessary processing, so the equipment is also large and
The processing time is also long, and it cannot be said to be a simple device.

Also, accuracy is limited by pixel size. Decreasing the pixel size increases accuracy, but requires increased computer storage capacity.

The biggest problem lies in something else. Since the left and right cameras are looking at the center o of the workbench, the distance calculation results for objects placed far away from the 0 point will contain larger errors. This is mainly due to optical reasons. The left and right cameras are focused on the 0 point, but the straight line connecting cameras A and B and the 0 point is not perpendicular to the workbench. Therefore, other points on the workbench will not be correctly imaged. Therefore, the brightness level is also affected by the degree of brightness of adjacent pixels.

(ri) Principle of the position measuring method of the present invention In the present invention, a mirror is moved and a target point P is observed. In other words, two cameras are effectively used to track the target point Pt with the two cameras and align the target point P' with the center of the field of view.

Then, the distance between the camera and the target point P is calculated based on the direction of the camera.

In the method described in (a), the target point P is seen at one corner of the field of view of the two images, but in the present invention, the camera is moved so that the target point P is brought to the center of the field of view.

Actually, instead of moving the camera, you move the left and right mirrors. But mirror? Moving it is effectively the same thing as moving the camera.

The basic principle of the present invention will be explained with reference to FIG.

The distance is determined by a simple application of the law of sine.

Think on a plane. Take the horizontal y-axis and the vertical KX-axis.

Observation points E r, E where mirrors should be placed on the left and right of the origin 0
Take 1.

There is a distance between the two.

Mirrors are rotatably provided at observation points E2 and El.

This is effectively the same as placing the camera so that it can rotate freely. In other words, E2 and El are eyes, and it can be said that these are moving eyes. The eyes mentioned in the previous section do not move.

Since the eyes move, the angle between pi = 1, PE, and the y axis is θ1,
θ, can be directly known. Here, P(x, y) is an arbitrary point to be measured. Let PEI=11+.

According to the law of sine, g, 5in(θ1+θ, ) −L sinθ, (1)
It is. The position of P(x, y) is x=/7, sin θ1(2) y2−/! Isen θH+ L/2 (31. The angle of the eyes can be changed, and the line of sight PE, , PE,
It is important to be able to directly know the angle formed between and the Y axis (axis along the base line).

Although FIG. 1 is a diagram of the basic principle in a two-dimensional space, it is easy to extend it to a three-dimensional space.

FIG. 2 is an explanatory perspective view showing a method for determining the position of point P (x, y% Z) in a three-dimensional space using both eyes placed at left and right observation points E1 and E.

The X axis and Y1iIl] are taken in the horizontal direction, and the two axes are taken vertically downward. Left and right observation points E1 and E are provided on the Y-axis at the same position as the origin 0. The distance between two points iI'i: is given.

Let km be a plane containing points P, E, , El. Let the angle between the plane m and the vertical Z-axis be θ2.

A straight line included in the plane m, passing through the origin O, and perpendicular to the Y axis is o
Let it be x'. From point P(x, y, z), straight line ox
Let the foot of the perpendicular line drawn to ' be X'.

Regarding the diagonal coordinate system OX'Y[, equations (2) and (3) hold true.

In equation (2), X may be replaced with X'.

x = x's in terms of coordinate systems OX'Y and 0XYZ
in ll z(4) z == z'oO5θz(5).

fil ~ From formula (5), the value of each coordinate of the target point P (x, y, z) is L sinθrSfnθ1 sinθ2L sin (
θ1−θr) L sinθrIitnθ1oosθ2.

The distance X' between the point P and the Y axis connecting the surface observation points E1 and Er is given by:

The distance from the origin 0 to P is given by OpH''j:.

In the example of Fig. 2, two observation points El and Er1d are located on the horizontal axis (Y axis, and two axes are vertically downward).

However, the observation points E1 and Er are not necessarily required to be on the horizon. The straight line EIEr may be inclined with respect to the horizontal.

Instead, the center of the observation points E1 and Er should be set as the origin 0, the straight Y axis passing through El and Er should be set, and the X and Z axes should be determined to be perpendicular to the Y axis. And P, El, Er
The angle θ2 formed by the plane m including the two axes and the line of sight PEI, PE
The important point is that the position of point P can be calculated by knowing the angles θ1 and θrf that r makes with the Y axis.

The method for measuring the position of an object according to the present invention is to
The straight line passing through observation points E1 and Er on the right is the Y axis, and the midpoint of EIEr is 0, and the angle θ1 that the straight line that fits the orthogonal coordinates xYz and connects the target point P and observation points E1 and Er with the Y axis is θr and
The X, Y, and Z coordinates of point P are determined by knowing the angle θ2 that the plane m including P, El, and Er makes with the two axes.

Set up a direct camera at observation point E1, Er VCtris, move the connecting member (tilt axis) along the Y axis to adjust θ2, and rotate the camera about the support shafts set at El and Er to obtain θ1 and θ. ``It may also be possible to adjust.

Alternatively, a rotatable mirror may be provided at the observation point and light reflected by the mirror may be guided to the camera.

In these structures, the camera is connected to the Y-axis connecting member (tilt axis).
must be moved with it.

Therefore, especially in the apparatus of the present invention, mirrors are placed at the observation points E1 and E', and the reflected light from the mirrors is transmitted to the camera via an image fiber.

(1) Object position measuring device according to the present invention The object position measuring device according to the present invention will be explained with reference to the embodiment shown in FIG.

The support column 1 is a vertical bar member that supports the mirror and the drive unit, and does not move.

A horizontal horizontal bar 2 is fixed above the column 1. Support rods 3.3 are fixed diagonally to both ends of the horizontal rod 20.

A tilt support 4 is rotatably provided parallel to the horizontal bar 2. The tilt support 4 has a bent shape at both ends, and is rotatably supported by the support rod 3 at the bent portion. In this example, one is supported by the pin 5 and the other by the tilt drive unit 6. The tilt drive unit 6 includes a drive mechanism such as a servomorph or a decelerator, and a tilt angle detection mechanism that detects the angle of the tilt rod 4.

A left mirror 9 and a right mirror 10 are rotatably provided at equal positions from the center of the tilt keyaki 4 by a swing shaft 11.

The swing beam 11 is perpendicular to the tilt beam 4 and is included in a plane that includes the pin 5, the tilt drive section 6, and the center of the tilt beam 4.

The left and right mirrors 9,10 are rotated about the swing center by the left swing drive 7 and the right swing drive 8. In this case, the swing shaft 11 of the mirror 9.10 has a pulley 12, and the left and right swing drive parts γ, n
A pulley 14 is fixed to the output shaft of the motor, and a belt 13 connects both pulleys 12 and 14.

The tilt drive section 6, left swing drive section 7, and right swing drive section 8 are connected to drive section control signal lines 15, 16, and 17.

The control signals provide forward and reverse rotation signals to these drives. The drive section is composed of, for example, a step motor, and provides pulses as control signals. At this time, the angular displacement in the forward and reverse directions can be found by counting pulses.

In such a case, the tilt, left/right swing drive unit 6.7.
8 only needs to include a motor, a speed reducer, etc.

However, since it is necessary to know the accurate angular displacement of the tilt axis and swing axis, it is necessary to provide an angular displacement detection mechanism within the tilt drive unit 6 and left/right swing drive unit 7.8, or as a separate body. You can also do it.

The center of the left and right mirrors 9 and 10 is the rotation axis of the tilt rod 4,
In other words, the pin 5 and the output end of the tilt drive unit 6 are located on the straight line of the book deposit.

A left imaging section 18 and a right imaging section 19 made of image fibers are provided diagonally below the left and right mirrors 9 and 10. The reflected images of the left and right mirrors are incident on the end face of the image fiber.

Mirror 9.1 is placed on the extension of the center line of the image fibers.
This is done so that the center of 0 always exists. Fixed piece 20.
Reference numeral 21 attaches the left and right imaging sections 18 and 19 to the horizontal bar 2.

(e) Swing angle, tilt angle and inclination angle 62 Inclusion angle θr
, θ1 As already mentioned, the line of sight PE+, PEr when looking at the object from the left and right observation points E1, Er, the viewing angles θr, θI with the Chilean axis (Y axis), and the triangle PElEr. The position of the object P can be determined by knowing the inclination angle θzk that the plane m including k forms with the two axes.

If the object placed at the observation point is a camera, the rotational displacement of the camera directly determines θ11θr, θz). Therefore, the position of point P can be immediately calculated based on the rotational displacement of the camera.

However, if the object placed at the observation point is a mirror, the swing angles α, β, and tilt angle γ of the mirror are immediately changed to θ1,
θr and θzK are not equal.

This is because the positions of the end faces of the left and right imaging units 18 and 19 are related to the center of the mirror.

FIG. 4 is a perspective view showing the relationship between the tilt axis, the swing axis, the mirror, and the imaging section.

Tilt axis f% (axis, two axes are taken vertically downward from the center of the mirror at 0.0 point.

The X-axis is taken in the direction perpendicular to the Y-axis and the Z-axis. The X and Y axes are horizontal.

When the mirror is in the XY plane, the tilt angle γ and the swing angle α are defined as 0.

The angle at which the tilt axis (and therefore the mirror) is tilted is the tilt angle γ. Take the Ox' coordinate that forms an angle γ with Ox.

The mirror is on the x'y plane.

Furthermore, assume that the mirror is tilted by α with respect to the x'o axis.

The normal OM to the mirror is α in the Y direction from the O2' axis.
It is in a tilted position.

It is assumed that the end l of the image fiber (imaging section) is within the O2 plane and forms an angle φ with the Z axis.

When the tilt angle is -0 and the swing angle is -00, the CO force direction ray is reflected at the 0 point and enters the end 1 of the image filer, then the inclination angle θp (C) and the line of sight angle for co θ+(C) (also 1d-(lr(C)) is (1
1) θ, (C) = φ (θ1(C)=72.

When the mirror is tilted toward the Y axis by a tilt angle γ, the tilt angle becomes γ and the swing angle becomes IO. Assume that the AO force direction light is reflected at the zero point of the mirror and enters the image fiber.

Inclination angle θ about straight line AO, (A) line of sight angle θ1 (A)
is θp(”A) = φ−2γ (13
)θ, (A) = Of141. Here, the swing axis is Ox'. Suppose that the mirror is tilted by α instead of Ox'. It is assumed that a beam in a horizontal direction having a tilt angle of γ and a swing angle of α is reflected by a mirror and enters the image fiber end 1.

Draw a sphere o of unit radius centered at point 0. Let the intersections of sphere 0 and rays 01, oc, O8, OA be kl, C, B, and A.

FIG. 5 shows rays of light extending from the tilt axis direction. 02', O
M appears to overlap and forms an angle γ with oz. Three points 0, /, R
1 always includes the normal line of the mirror. The mirror is
This is because it rotates around the Ox' axis. plane P and circle 0
The shared point group of forms a great circle. This is called the great circle P. When the mirror is rotated around the swing axis, the locus of the intersection of the mirror normal and circle 0 is a great circle P.

AO is the incident ray when the swing angle is 0.

That is, A and I are at symmetrical positions with respect to the plane P.

The incident ray 80 and the reflected ray o1 are symmetrical with respect to the normal OM of the mirror.

Therefore, the locus of the intersection of the incident ray BO and the circle 0 passes through the point A' and must be on the plane q parallel to the plane P. Define the small circle q as a group of common points between the plane q and the circle 0. do. When the mirror is rotated around the swing axis, let BO be the incident ray that should produce the reflected ray o1, then the straight line BO and the circle 0
The locus of the intersection day is a small circle q.

FIG. 6 is a relationship diagram of light rays viewed from the swing axis direction. In this figure, a large circle P and a small circle q (a perfect circle centered on l-io) appear.

In FIG. 5, let H and G be the legs of a perpendicular line drawn from point A%B to plane P.

H and G appear to overlap with point A and 8 in Fig. 6 when viewed from the swing axis.

What we want to find is not the angle that the incident light angle 0 makes with the two axes. We want to find the inclination angle θzk between the two axes of the plane containing Y[ll11] and the incident ray EIO.

FIG. 5 shows the orthogonal projection onto the xz plane for the incident light edge date 0. Therefore, as shown in FIG. 5, /BO2 is equal to the inclination angle θ2. This is because the angle of inclination θ2 is a projection of an included angle between the incident light beam BO and the two axes onto the XZ plane.

Consider ΔBGO by placing the positive 1 of points B, a, and o as a shadow K (onto the 'xz plane). Here, Δday GO is triangle BG
It is an orthogonal projection onto the Oo'xz plane, not the triangle GO itself.

OGm(θz + 1) in △BGO, = B G
(15). In this formula, OG is the orthogonal projection of the line segment OG onto the xz plane, and is not the true length of the line segment OG itself. Since BG is parallel to the xz plane, it can be said that it is the true length of the line segment BG and the length of the orthogonal projection.

On the other hand, at 0AHK, OHm (tlp+7") = A H (16). Here, O, A, and H are on the xz plane, so 0)
-1,AH is the true length and the length of the orthogonal projection. θp is the inclination angle between OA and the Z axis, and θp in equation (13)
(A) means.

Points A and B are on plane q, and points G and H are on plane p.

BG, AH are perpendicular to plane p. Therefore, four points B, A
, H, and G form a rectangle. Therefore, 8G=AH(+7).

(15) ~From 0no It is.

Since this is an orthogonal projection of the 0GH1 line segment OG onto the xz plane, if a perpendicular line is drawn from point G to the z'X plane in FIG. 6 and its leg is J, then the length is equal to OJ.

On the other hand, OH appears as it is in FIG. 6 as the radius of the projected image of the small circle q onto the Z'Y plane. Figure 6 is viewed from the swing axis, so the mirror has a swing angle α with respect to the Y axis.
It's only tilted. Incident ray B.

The projection of BOid onto the z'Y plane is tilted by 2α with respect to the oz' axis.

Projection triangle △GJO of three points G and Jlo onto the z′Y plane
In this case, OG deposit 2α = OJ Q9).

Since OG in the equation (18) is the orthogonal projection of the line segment OG onto the Xz plane, it is equal to OJ in the equation (+9) shown in FIG. 6.

(]+8 formula oH is H, G 7'fi in Figure 6.
Since it is located at the second point of the small circle 'Q', it is 0Hc90G (20) in Figure 6.

In other words, OG and OJ in equation (19) are OH and O in equation (1B).
Corresponds to G.

Therefore It is.

In this way, the inclination angle θ2 formed by the two axes of the plane including the incident light beam BO having the swing angle α and the tilt angle γ and the Y axis was determined. By fully substituting equation (13) into θp, we get cos 2α. θ2 becomes. φ is the angle formed between oz and the image fiber end 01, and is a constant. For example, it can be about 74.

Next, the angle of view θl (or θr) is determined. This is the angle between the incident light beam BO and the Y axis.

FIG. 7 is an explanatory perspective view for determining the viewing angle.

OY is on the side and oz is on the bottom. Incident light 800
Rectangle F3 with o-4 angle and one side on the Yz plane
Thinking about JVWGSOU.

From the day, the foot of the perpendicular line drawn on the ZY plane, xz plane, and XY plane is G.
, J, and W. It is assumed that when a perpendicular line is drawn from J, W, and V to the yz plane, it coincides with S, U, and O. line segment ou
, ov, oSu Y axis, X axis, Z axis nial.

/BLJG-θ2. The angle of view is θIH/UOB.

/GO5 is 2α.

UG = UO嬶2α (
24) For △SOU, BU = (JQtanθ1(20) For △BGtJ, BU = UG See Oz
(26) holds true. From (24) to (26), the theory θ1 = cot 2αki θz (27
) As mentioned above, from the swing angles α, β and tilt angle γ of the mirror, the viewing angles θ1, θ' and the inclination angle θ2 are oz −
tan ' ((φ-γ)/ω1+2(E)-γ(c)θH-tan I (cot 2(Z secθz
) (29) θr-tan' (cat2
βqiθz) (30).

The X, Y, Z (7) values of the object P (x, y, χ) are (6)
~ (8) Calculated by formula.

(Force) Control device of the present invention A control device for controlling the mirror drive unit @ shown in FIG. 3 will be explained.

FIG. 8 is a system diagram of the control device.

The tilt drive circuit 22, left swing drive circuit 23, and right swing drive circuit 24 provide control signals to the tilt drive section 6, left swing drive section 7, and right swing drive section 8 in FIG.

If the prime mover of the drive unit is a step motor, these drive circuits 22 to 24 generate pulses until the command values of tilt angle γ, left swing angle α, and right swing angle β are reached, and provide them to the drive unit.

The command values of γ, α, and β are instructed to the drive circuits 22 to 24 by the position calculation circuit 40.

The left imaging section 18 and the right imaging section 19 are configured by the left imaging circuit 2.
5. Connected to the right imaging circuit 26. This is a two-dimensional reproduction of the reflected image of the mirror.

The outputs of the left and right imaging circuits 25.26 are further processed by left and right image processing circuits 30.31.

This is the center P0 of the target object and the center A of the image.
There is a function to measure the difference between . Image processing circuit 30.31
are the x and y components of the positional deviation (po-AO), △X, △y
Output.

The position calculation circuit 40 moves in the direction of decreasing △X and △yi.
Command values γ, α, and β are outputted and given to tilt and swing drive circuits 22 to 24. Then, the tilt and swing drive unit 6
~8 rotates the mirror. The positional deviations △X and △y will be smaller than last time. Furthermore, a command value is given. Repeat below.

In this way, the positional deviation gradually decreases, and ΔX and Δy become zero. This means that the center of the image corresponds to the center of the object. This is done for the left and right imaging circuits.

At this time, the mirror tilt angle γ, left and right swing angles α, β
The position calculation circuit 40 knows this, and as a result, (6) to (8)
), glance ~ Using equation (30), calculate the X, y,
Calculate the value of the z coordinate.

The image processing circuits 30 and 31 detect the deviation between the object center P0 and the image center Ao, as shown in FIG. In left and right image processing, the same object center Po must be viewed.

This is because, in FIG. 2, the left and right line of sight lines PE+ and PEr must match.

For this reason, the center of gravity or the center is adopted as the center of the object.

Here, the center of gravity is the center of gravity when equal load density is distributed on the image. The center of the image and the border are the x of the object,
Find the expansion in the y direction, and refer to its center.

However, more generally, the @ point of origin of other objects is defined as the object center P. You can close it.

(g) Effects of the present invention (1) Since two eyes are used to move the eyes so as to gaze at the object, objects that are not on the perpendicular trisecting plane of the two eye positions E1 and Er can also be included in the visual field. It can be captured in the center. The eyes have three degrees of freedom, α, β, and γ, which are necessary and sufficient to direct both eyes to the same object.

The technology described as a known technique has one degree of freedom for the eye (
There are only two constraint conditions: α-β, γ-0). The position of objects in the corners of the field of vision was calculated primarily through calculations. The calculation data becomes huge. The accuracy is poor for that reason.

In the present invention, since the eyes are moved and the object is gazed at, there is almost no need for a memory circuit. A simple servo control system is sufficient.

The calculation is also easy.

The measurement accuracy depends on the base line length EI Er, but since it is not divided into pixels as described as a known technique, the accuracy does not decrease depending on the size of the pixel.

Experimental data are shown.

As a drive unit, a pulse motor with 48 pulses per revolution and a lever are used.
A reduction gear was used. Approximately 1. , ll11 object 04 The position accuracy of the body was about 1%. Baseline lengths were arranged in column order.

(2) To carry out the method of the present invention, a camera or a mirror may be placed as an eye at the observation points E1 and Er.

The device described as the device of the present invention uses a mirror.

When a mirror is used, the mirror is significantly lighter than a camera, so all small parts such as the morph and gears of the drive unit can be used.

Since the inertial force against the drive is small, high-speed and highly accurate measurements are possible.

(3) When an image fiber is used as the imaging section, the movable section becomes lightweight. Although it is small, it has a wide field of view.

(4) The control circuit can be simplified by using the center of gravity or the center of the object as the corresponding point between the left and right images. The conventional method for finding corresponding points in images is to divide the two images into pixels,
A correlation function of the degree of shading between the pixels was calculated, and when the values matched, the points were determined to be corresponding points.

In other words, shading information of the entire image was required.

In the present invention, since the center of gravity or the center is used, calculation can be made more easily.

(i) Applications of the present invention The present invention can be applied to three-dimensional position measuring devices, visual devices for robots, and the like.

[Brief explanation of drawings]

FIG. 1 is a principle diagram for explaining the basic principle of the present invention in a two-dimensional space. FIG. 2 is a diagram showing the principle of the present invention in detail in a three-dimensional space. FIG. 3 is a perspective view of the visual mechanism section of the object position measuring device of the present invention. Figure 4 shows the mirror, incident light, reflected light, tilt axis (Y axis),
FIG. 3 is a perspective view showing the positional relationship of the swing axis (X' axis). FIG. 5 is a projection diagram of stray rays from the tilt axis (CY axis) direction. Figure 6 is a projection view of the rays of light from the direction of the swing axis. FIG. 7 is a perspective view for explaining how to determine the relationship among the tilt angle, swing angle, and viewing angle of the mirror. FIG. 8 is a system diagram of a control device that controls a mirror drive section. Figure 9 shows the deviation between the image center AO and the object center po, X, △y.
The figure which shows the example of the image in the image pick-up part which is not enough to show. FIG. 10 is a schematic perspective view of a conventional object shape recognition device. 1・・・・・・・・・・・・・・・・・・Support 2・・・・・・・・・・・・・・・Horizontal bar 3...
・・・・・・・・・・・・・・・Support rod 4・・・・
・・・・・・・・・・・・・・・Tilt bar 5...
・・・・・・・・・・・・・・・Pin 6・・・・
・・－・・・・・・・・・・Tilt drive section 7 ・・
・・・・・・・・・・・・・・・・・・Left swing drive unit 8 ・・・・・・・・・・・・・・・Right swing drive unit 9 ・・・・・・・・・・・・・・・・・・・・・
Left Mira -10 ・・・・・・・・・
・・・・・・－・・Right mirror -11・・
・・・・・・・・・・・・・・・Fuig axis 12・・・・
・・・・・・・・・・・・Puri 13・・・・・・
・・・・・・・・・、7,114・・・・・・・・・
...Print 15.16.17...Control signal line 18°°゛°゛°°゛゛°゛゛°°°Left imaging section (
Image fiber) 19 °°°゛゛°゛°°゛°゛°
°°゛Right imaging section (image fiber) 20.21...
......Fixed piece 22...
・・・・・・・・・・・・・Tilt drive circuit 23・
・・・・・・・・・・・・・・・・・・Left swing drive section path 24・・・・・・・・・・・・・・・・・・・・・
Right swing drive circuit 25・・・・・・・・・・・・・・・
......Left imaging circuit 26...
......Right imaging circuit 30...
−・・・・・・・・・・Left image processing circuit 31・・・・
・・・・・・・・・・・・・・・Right image processing circuit 40
・・・・・・・・・・・・・・・・・・Position calculation circuit El, Er・・・・・・・・・・・・Left and right observation points L・・・・・・・・・・・・・・・ Observation point interval (
Base line length) P......Target object θ1, θr...Left and right viewing angle θ2...・・・・・・・・・・・・・・・・・・
・Inclination angle α, β ・・・・・・・・・・・・・・・Left and right swing angle γ ・・・・・・・・・・・・・・・・・・
・・・Tilt Angle Y axis・・・・・・・・・・・・・・・
...Horizontal axis in the tilt axis direction
・・・・・・Horizontal axis X' axis perpendicular to the tilt axis...
・・・・・・・・・・・・・・・Two swing axes forming a tilt angle i with the X axis・・・・・・・・・・・・・・・・・・
・Axis plane m facing vertically downward...
・・・・・・Plane 0 including object and observation point ・・・・・・
・・・・・・・・・・・・Center of mirror, origin circle of XYZ coordinates 0・・・・・・・・・・・・・・・Reference circle great circle with center O and radius 1 P・・・・・・・・・・・・・・・
...The large circle and the small circle q including the normal OM set at the center of the mirror...... Change the swing angle a so that the travel time incident ray 80 becomes circle 0. Small circle θp made up of intersecting schools of fish・・・・・・・・・・・・・・・−) #)
Inclination angle φ when the angle is γ and spink° angular force is 0... Image phi/ ) Yutaka Figure 4 Figure 7 Figure 5 Figure 6 Figure 8 Left imaging unit 18 Right imaging unit 19 Object position x, y, z Figure 9

Claims (4)

[Claims] (1) Visual devices are installed on the left and right observation points El and Er, which are separated by a distance r@L, and the visual devices are directed toward the target object P, and the viewing angles θ1 and θ are formed by the line of sight PE+ and PEr with respect to the baseline. A method for measuring the position of an object, characterized in that the position of the object P is measured by finding the angle θ2 of inclination that a plane m including P%E1 and Er makes with an axis perpendicular to the base line. (2) Rotatably provided tilt support 4 and tilt support 4
a left mirror 9 and a right mirror 10 which are rotatably provided on the swing shaft 11 at observation points E1 and Er, which are separated by a distance on the tilt support 4, and a left mirror 9. , a left swing drive section 7 for rotationally displacing the right mirror 10, a right swing drive section 8, and a left and right imaging section 18 for capturing reflected images of the left and right mirrors 9 and 10.
．． 19, a left image processing circuit 30 and a right image processing circuit 31 that calculate the deviations ΔX and Δy between the center Po of the target object and the center A of the image, and a left swing angle α and a right swing angle corresponding to the deviations ΔX and Δy. A position calculation circuit 40 that issues command values regarding angle β and tilt angle γ, and a tilt drive unit 6, left and right swing drive units 7.8', and command values issued by the position calculation circuit 400.
f: An object position measuring device characterized by comprising a tilt drive circuit 22 and left and right swing drive circuits 23 and 24. (3) The object position measuring device according to claim (2), wherein the left and right imaging unit transmission paths are image fibers. (4) The left and right image processing circuits 30 and 31 operate at the center P of the object.
The object position measuring device according to claim (2), wherein the center of gravity or the center is adopted as o.