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JPH0821084B2 - Incomplete circle center detection method - Google Patents

Incomplete circle center detection method

Info

Publication number
JPH0821084B2
JPH0821084B2 JP63026636A JP2663688A JPH0821084B2 JP H0821084 B2 JPH0821084 B2 JP H0821084B2 JP 63026636 A JP63026636 A JP 63026636A JP 2663688 A JP2663688 A JP 2663688A JP H0821084 B2 JPH0821084 B2 JP H0821084B2
Authority
JP
Japan
Prior art keywords
center
straight line
incomplete
circle
gravity
Prior art date
Legal status (The legal status is an assumption and is not a legal conclusion. Google has not performed a legal analysis and makes no representation as to the accuracy of the status listed.)
Expired - Lifetime
Application number
JP63026636A
Other languages
Japanese (ja)
Other versions
JPH01204187A (en
Inventor
修二 佐々木
Current Assignee (The listed assignees may be inaccurate. Google has not performed a legal analysis and makes no representation or warranty as to the accuracy of the list.)
Yaskawa Electric Corp
Original Assignee
Yaskawa Electric Corp
Priority date (The priority date is an assumption and is not a legal conclusion. Google has not performed a legal analysis and makes no representation as to the accuracy of the date listed.)
Filing date
Publication date
Application filed by Yaskawa Electric Corp filed Critical Yaskawa Electric Corp
Priority to JP63026636A priority Critical patent/JPH0821084B2/en
Publication of JPH01204187A publication Critical patent/JPH01204187A/en
Publication of JPH0821084B2 publication Critical patent/JPH0821084B2/en
Anticipated expiration legal-status Critical
Expired - Lifetime legal-status Critical Current

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Description

【発明の詳細な説明】 [産業上の利用分野] 本発明は、部品ハンドリング用のロボットの目とし
て、あるいは計測用として適用される視覚装置の不完全
な円形対象物(例えばキーミゾを有する軸等)の中心検
出方法に関する。
DETAILED DESCRIPTION OF THE INVENTION [Industrial application] The present invention relates to an imperfect circular object (for example, an axis having a key groove, etc.) of a visual device applied as a robot eye for part handling or for measurement. ) Center detection method.

[従来の技術] 視覚装置で一部に突起あるいは欠けがあるような不完
全な円形対象物の中心(不完全部が無いと仮定した場合
の重心)を求めることは計測あるいはロボットによるハ
ンドリングにおいて重要である。しかしながら、不完全
円の場合、円のように単純に対象の重心を求めたのでは
誤差が大きい。このため、不完全部が小さく、あらかじ
めその不完全部の位置が想定できるような場合は、不完
全部を避けてウインドウをかけ、ウインドウ内画像を処
理することで中心を求めることができる(特願昭61−50
499「不完全円の中心検出方法」および特願昭61−46238
「円の中心検出方法」)。これらの手法はウインドウ内
の画像の重心位置または平均エッヂ位置を求め、それら
から中心のx座標を求め、次に別のウインドウをかけ、
上記と同様な処理にて中心のy座標を求めるものであ
る。
[Prior Art] It is important for measurement or robot handling to find the center (center of gravity when assuming that there is no incomplete part) of an incomplete circular object with some projections or chips on the visual device. Is. However, in the case of an incomplete circle, there is a large error if the center of gravity of the object is simply calculated like a circle. Therefore, when the incomplete portion is small and the position of the incomplete portion can be assumed in advance, the window can be applied while avoiding the incomplete portion and the center of the image can be obtained by processing the image in the window. Wish sho 61-50
499 "Method of detecting center of incomplete circle" and Japanese Patent Application No. 61-46238
"Circle center detection method"). These methods find the barycentric position or the average edge position of the image in the window, find the x-coordinate of the center from them, and then apply another window,
The y coordinate of the center is obtained by the same processing as described above.

[発明が解決しようとする課題] しかしながら、上述した従来の方法は、第5図に示す
ような不完全部が大きい場合、あるいはその位置が不定
の場合、適用できない。
[Problems to be Solved by the Invention] However, the conventional method described above cannot be applied when the incomplete portion as shown in FIG. 5 is large or when the position is indefinite.

本発明の目的は、不完全部が円の半分を占めるように
大きくても、あるいは不完全部の位置が不定でも、不完
全円の中心を求めることができる方法を提供することで
ある。
An object of the present invention is to provide a method capable of obtaining the center of an incomplete circle even if the incomplete portion is large so as to occupy half of a circle or the position of the incomplete portion is indefinite.

[課題を解決するための手段] 本発明の不完全円の中心検出方法は、対象物の重心の
位置、対象物と慣性主軸および慣性副軸まわりの慣性モ
ーメントが等しい楕円である等価楕円の長軸の長さ、短
軸の長さを求め、 前記重心を通り、前記慣性主軸と重なる第1の直線、
前記重心を通り、第1の直線に対してπ/2の角度をなす
第2の直線、前記重心を通り、第1の直線に対してπの
角度をなす第3の直線と、前記対象物の円周部分の交点
を、各直線上に前記重心からの距離が(前記長軸の長さ
+前記短軸の長さ)/4の点を中心とする所定の大きさの
エリアを設定し、該エリア内にあって前記対象物の円周
上にある点のうち、当該直線との距離が一定値以下の点
を求めることにより求め、 前記3つの交点を通る円の中心を求める。
[Means for Solving the Problems] The method for detecting the center of an incomplete circle according to the present invention is the length of an equivalent ellipse which is an ellipse in which the position of the center of gravity of an object and the moment of inertia about the object and the principal inertial axis and the minor inertial axis are equal. The length of the axis and the length of the minor axis are obtained, and a first straight line passing through the center of gravity and overlapping with the principal axis of inertia,
A second straight line passing through the center of gravity and forming an angle of π / 2 with respect to the first straight line; a third straight line passing through the center of gravity and forming an angle of π with respect to the first straight line; and the object Set an area of a predetermined size centered on the point where the distance from the center of gravity is (the length of the long axis + the length of the short axis) / 4 on each straight line at the intersections of the circumferential portions of Of the points in the area and on the circumference of the object, the points whose distance to the straight line is a fixed value or less are obtained, and the center of the circle passing through the three intersections is obtained.

[作用] 第2の直線が不完全円部の方に設定された場合、エリ
アの点で、かつ当該直線との距離が一定値以下の円周上
の点は存在しない。このときは、第2の直線を180°反
対方向に設定し直せば、第2の直線と対象物の円周上の
交点が求められる。以上のようにして対象物の円周上の
3点が求まれば、これら3点を通る円の中心が不完全円
の中心となる。抽出する3点は隣接する2点のなす円弧
角が60°〜90°程度あれば3点を通る円の中心検出精度
が確保されるので、完全な円弧部は半円程度で良く、従
って、不完全部の大きさは半円程度になっても差しつか
えない。
[Operation] When the second straight line is set to the incomplete circle portion, there is no point on the circumference of the area and the distance to the straight line is equal to or less than the fixed value. At this time, if the second straight line is set again in the opposite direction of 180 °, the intersection point of the second straight line and the circumference of the object can be obtained. If three points on the circumference of the object are obtained as described above, the center of the circle passing through these three points becomes the center of the incomplete circle. As for the three points to be extracted, the accuracy of detecting the center of a circle passing through the three points can be secured if the arc angle formed by the two adjacent points is about 60 ° to 90 °, so a complete arc part can be about a half circle, and therefore, The size of the incomplete part can be about a half circle.

[実施例] 次に、本発明の実施例について図面を参照して説明す
る。
Example Next, an example of the present invention will be described with reference to the drawings.

第1図から第4図は本発明の不完全円の中心検出方法
の一実施例の説明図で、第1図は対象1の等価楕円2を
示す図、第2図は直線l1,l2,l3と対象1の交点PK
PL,PMを示す図、第3図は交点PK,PL,PMを求める方法
の説明図、第4図は3点PK,PL,PMを通る円を示す図で
ある。
1 to 4 are explanatory views of an embodiment of the method for detecting the center of an incomplete circle according to the present invention. FIG. 1 shows an equivalent ellipse 2 of an object 1, and FIG. 2 shows straight lines l 1 and l. The intersection P K of 2 and l 3 with object 1
P L, illustrates the P M, Figure 3 is the intersection point P K, P L, explanatory diagram of a method for obtaining the P M, Figure 4 is a view showing a circle passing through three points P K, P L, the P M is there.

本実施例では、第1図に示すような、図面右下部に不
完全部のある不完全円1を対象とする。
In this embodiment, an incomplete circle 1 having an incomplete portion in the lower right part of the drawing as shown in FIG. 1 is targeted.

まず、画像を取込み、対象1の重心3の位置、等価楕
円2(対象物1と慣性主軸および副軸まわりの慣性モー
メントが等しい楕円)の長軸5の長さLmajor,短軸6の
長さLminor,慣性主軸角度θ(長軸5とx軸のなす角
度)を求める(第1図参照)。
First, the image is taken in, the position of the center of gravity 3 of the object 1, the length Lmajor of the major axis 5 and the length of the minor axis 6 of the equivalent ellipse 2 (ellipse having the same inertia moment about the principal axis of inertia and the minor axis as the object 1) Lminor, the angle of principal axis of inertia θ (the angle formed by the long axis 5 and the x axis) is obtained (see FIG. 1).

次に、重心3を通り、傾きが慣性主軸角度θである直
線l1,重心3を通り、傾きがθ+π/2である直線l2およ
び重心3を通り、傾きがθ+πである直線l3と対象1の
円周との交点PK(xK,yK),PL(xL,yL),PM(xM
yM)(第2図参照)を一点ずつ求める。
Then, through the center of gravity 3, linear l 1 slope of the principal axis of inertia angle theta, the street centroid 3, through a linear l 2 and the center of gravity 3 inclination is θ + π / 2, the straight line l 3 is inclination theta + [pi Intersection points P K (x K , y K ), P L (x L , y L ), P M (x M ,
y M ) (see Fig. 2) is calculated point by point.

この場合の各点PK,PL,PMの求め方は以下の通りであ
る。
In this case, the points P K , P L , and P M are calculated as follows.

第3図に示すように直線l(=直線l1,l2,l3)上に
重心3からの距離disがdis=(Lmajor+Lminor)/4であ
る点P(=P1,P2,P3)を求め、次に点Pを中心とする
一辺が2w(w=max(Lmajor−Lminor,const.)なる正方
形のエリア(第2図の11,12,13)を求める。ここで、co
nstは定数で、不完全部が小さな場合、Lmajor−Lminor
が小さくなるため、上記エリアに最小値を設定するため
である。
As shown in FIG. 3, a point P (= P 1 , P 2 , P on the straight line l (= straight lines l 1 , l 2 , l 3 ) where the distance dis from the center of gravity 3 is dis = (Lmajor + Lminor) / 4. 3 ), and then a square area (11, 12, 13 in FIG. 2) whose one side is 2w (w = max (Lmajor-Lminor, const.)) Centered on the point P. Here, co
nst is a constant, Lmajor−Lminor when the imperfections are small
Is to set the minimum value in the above area.

次に、対象1の円周上の点Pj(xj,yj)を取り出し、
その点が上記エリア内であるか否かをチェックする。も
しエリア内の点であれば直線lとの距離rを計算する。
重心3の座標を(xz,yz)とするとき、点Pj(xj,yj
と直線l:ax+by+c=0(ただし、a=−sinα,b=cos
α,c=xgsinα−ygcosα)とのなす距離rは となる。
Next, the point P j (x j , y j ) on the circumference of the object 1 is extracted,
Check if the point is within the above area. If it is a point within the area, the distance r to the straight line 1 is calculated.
When the coordinates of the center of gravity 3 are (x z , y z ), the point P j (x j , y j )
And the straight line l: ax + by + c = 0 (where a = -sinα, b = cos
α, c = x g sin α−y g cos α) Becomes

この距離rがΔ(定数)以下となればこの点Pjは求め
る円弧上の一点としてこの処理を終える。Δより大であ
るか、点Pjがエリア外の点であれば次の円周上の点Pj+1
を取り出し、上記処理を繰り返す。
If this distance r becomes equal to or less than Δ (constant), this point P j is regarded as one point on the arc to be obtained, and this processing is ended. If it is larger than Δ or if the point P j is outside the area, then the next point P j + 1 on the circumference
Is taken out and the above process is repeated.

なお、直線l2が不完全円部の方に設定された場合、正
方形のエリアの点で、かつ直線l2との距離rがΔ以下の
円周上の点は存在しないが、このときは、直線l2を180
°反対方向に設定し直せば、直線l2と対象1の円周上の
交点が前述したように求められる。
When the straight line l 2 is set toward the incomplete circle part, there is no point on the circumference of the square area and the distance r to the straight line l 2 is Δ or less, but at this time, , Straight line l 2 180
° If set in the opposite direction, the intersection of the straight line l 2 and the object 1 on the circumference can be obtained as described above.

以上のようにして3点PK,PL,PMが求まれば、これら
3点PK,PL,PMを通る円の中心が不完全円1の中心4と
なる。
When the three points P K , P L , and P M are obtained as described above, the center of the circle passing through these three points P K , P L , and P M becomes the center 4 of the incomplete circle 1.

この時の計算式は、第4図において、求める中心4の
座標を(xc,yc)とすると、 ただし xLK=xL−xK xMK=xM−xK yLK=yL−yK yMK=yM−yK として求められる。
The calculation formula at this time is, assuming that the coordinates of the center 4 to be obtained in FIG. 4 are (x c , y c ), However obtained as x LK = x L -x K x MK = x M -x K y LK = y L -y K y MK = y M -y K.

なお、あらかじめ不完全部の位置が分っている場合
は、本実施例のような対象の姿勢情報(慣性主軸および
副軸)を抽出して不完全部の位置を検出する手間がはぶ
ける。すなわち、不完全部の位置があらかじめわかって
いる場合は、不完全部と重心を結ぶ直線を引き、不完全
部の反対側の円弧とその直線の交点をPLとし、またその
直線に直交し、重心を通る直線と円弧の交点をPM,PK
することで、円弧上の3点を求めることができる。
When the position of the incomplete portion is known in advance, it takes time and effort to detect the position of the incomplete portion by extracting the target posture information (main axis of inertia and secondary axis) as in the present embodiment. That is, if the position of the incomplete portion is known in advance, draw a straight line that connects the incomplete portion and the center of gravity, set the intersection point of the arc on the opposite side of the incomplete portion and that straight line as P L, and intersect the line at right angles. , 3 points on the circular arc can be obtained by setting the intersection points of the straight line passing through the center of gravity and the circular arc as P M and P K.

[発明の効果] 以上説明したように本発明は、対象の慣性主軸を用い
て不完全部を避け、完全な円弧上に互いに離れた3点を
抽出し、これら3点を通る円の中心を、不完全円の中心
として求めることにより、不完全部が大きくても、また
不完全部がどこに位置していても正しく中心を検出で
き、計測、ハンドリングにおける視覚装置の応用範囲を
広める等の効果がある。
EFFECTS OF THE INVENTION As described above, the present invention avoids imperfections by using the principal axis of inertia of the object, extracts three points separated from each other on a complete arc, and determines the center of the circle passing through these three points. By calculating the center of an incomplete circle, the center can be correctly detected regardless of where the incomplete part is large and where the incomplete part is located. This has the effect of expanding the range of applications for visual devices in measurement and handling. There is.

【図面の簡単な説明】[Brief description of drawings]

第1図から第4図は本発明の不完全円の検出方法の一実
施例の説明図で、第1図は対象1の等価楕円2を示す
図、第2図は直線l1,l2,l3と対象1の交点PK,PL,PM
を示す図、第3図は交点PK,PL,PMを求める方法の説明
図、第4図は3点PK,PL,PMを通る円を示す図、第5図
は従来の方法では中心の求められない不完全円の例を示
す図である。 1……対象(不完全円)、2……等価楕円、3……重
心、4……中心、5……長軸、6……短軸、11,12,13…
…正方形のエリア。
1 to 4 are explanatory views of an embodiment of the method for detecting an incomplete circle according to the present invention. FIG. 1 shows an equivalent ellipse 2 of an object 1, and FIG. 2 shows straight lines l 1 and l 2. , the intersection of l 3 and the target 1 P K, P L, P M
FIG. 3, FIG. 3 is an explanatory diagram of a method for obtaining the intersection points P K , P L , P M , FIG. 4 is a diagram showing a circle passing through the three points P K , P L , P M, and FIG. It is a figure which shows the example of the incomplete circle whose center is not calculated | required by the method of. 1 ... Target (incomplete circle), 2 ... Equivalent ellipse, 3 ... Center of gravity, 4 ... Center, 5 ... Long axis, 6 ... Short axis, 11, 12, 13 ...
… Square area.

Claims (1)

【特許請求の範囲】[Claims] 【請求項1】画像処理装置を用いて、一部に欠けや突起
がある不完全な円形対象物に対して、欠けや突起を除い
た完全円の中心を求める不完全円の中心検出方法であっ
て、 対象物の重心の位置、対象物と慣性主軸および慣性副軸
まわりの慣性モーメントが等しい楕円である等価楕円の
長軸の長さ、短軸の長さを求め、 前記重心を通り、前記慣性主軸と重なる第1の直線、前
記重心を通り、第1の直線に対してπ/2の角度をなす第
2の直線、前記重心を通り、第1の直線に対してπの角
度をなす第3の直線と、前記対象物の円周部分の交点
を、各直線上に前記重心からの距離が(前記長軸の長さ
+前記短軸の長さ)/4の点を中心とする所定の大きさの
エリアを設定し、該エリア内にあって前記対象物の円周
上にある点のうち、当該直線との距離が一定値以下の点
を求めることにより求め、 前記3つの交点を通る円の中心を求める、不完全円の中
心検出方法。
1. A method of detecting an incomplete circle center for an incomplete circular object partially lacking or protruding by using an image processing device to find the center of a perfect circle excluding the missing or protruding. Then, the position of the center of gravity of the object, the length of the major axis and the length of the minor axis of the equivalent ellipse, which is an ellipse having the same inertia moment about the principal axis of inertia and the minor axis of inertia as the object, are obtained, A first straight line overlapping the principal axis of inertia, a second straight line passing through the center of gravity and forming an angle of π / 2 with respect to the first straight line, passing through the center of gravity, and forming an angle of π with respect to the first straight line. The intersection of the third straight line and the circumference of the object is centered on each straight line at a distance of (the length of the major axis + the length of the minor axis) / 4 from the center of gravity. Area of a predetermined size to be set, among the points in the area on the circumference of the object, the distance to the straight line is A method for detecting the center of an incomplete circle, wherein the center of a circle passing through the three intersections is found by finding points below a certain value.
JP63026636A 1988-02-09 1988-02-09 Incomplete circle center detection method Expired - Lifetime JPH0821084B2 (en)

Priority Applications (1)

Application Number Priority Date Filing Date Title
JP63026636A JPH0821084B2 (en) 1988-02-09 1988-02-09 Incomplete circle center detection method

Applications Claiming Priority (1)

Application Number Priority Date Filing Date Title
JP63026636A JPH0821084B2 (en) 1988-02-09 1988-02-09 Incomplete circle center detection method

Publications (2)

Publication Number Publication Date
JPH01204187A JPH01204187A (en) 1989-08-16
JPH0821084B2 true JPH0821084B2 (en) 1996-03-04

Family

ID=12198936

Family Applications (1)

Application Number Title Priority Date Filing Date
JP63026636A Expired - Lifetime JPH0821084B2 (en) 1988-02-09 1988-02-09 Incomplete circle center detection method

Country Status (1)

Country Link
JP (1) JPH0821084B2 (en)

Families Citing this family (1)

* Cited by examiner, † Cited by third party
Publication number Priority date Publication date Assignee Title
EP0727760A3 (en) 1995-02-17 1997-01-29 Ibm Produce size recognition system

Family Cites Families (1)

* Cited by examiner, † Cited by third party
Publication number Priority date Publication date Assignee Title
JPS62208180A (en) * 1986-03-10 1987-09-12 Yaskawa Electric Mfg Co Ltd How to detect the center of an incomplete circle

Also Published As

Publication number Publication date
JPH01204187A (en) 1989-08-16

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