JP3381993B2 - Center line calculator - Google Patents

Description

Detailed Description of the Invention

[0001]

BACKGROUND OF THE INVENTION 1. Field of the Invention The present invention relates to a centerline calculating device for calculating a centerline of a road.

[0002]

2. Description of the Related Art Generally, in a mapping system, map data such as road boundaries is line data (vector data).
And is displayed and used as appropriate on a display or the like in response to an operator's instruction.

[0003]

By the way, in such a mapping system, not only road boundaries but also
The center line of the road was also input from the digitizer. This is because the road center line is used for road width calculation (road decoration) when calculating road distances and drawing small-scale drawings.

However, in order to input the road center line, the road data is once output, the road center line is written in the road data to create the following figure, and the following figure is pasted on the digitizer to input the road center line. Therefore, the work was troublesome.

Therefore, the present applicant has developed a system for automatically calculating the center line of a road, and has already applied for a patent for this system (Japanese Patent Application No. 5-238758).

However, the center line calculation device of Japanese Patent Application No. 5-238758 does not disclose processing when one road intersects another road.

The present invention has been made in view of such a demand, and its object is to automatically calculate the center line of a road even when one road intersects with another road. It is to provide a center line calculation device capable of performing the above.

[0008]

In order to achieve the above-mentioned object, the present invention provides a database for storing closed line data constituting a polygon, a means for searching an end portion of the line data of the database, Based on the line data, the means for extracting the data of the left side portion and the right side portion as a pair of opposing line data when viewed from the end portion, and the coordinates of the corresponding points of the pair of opposing line data are The corresponding point calculating means includes means for calculating corresponding point calculating means, means for calculating coordinates of respective middle points of the corresponding points, and means for storing the calculated coordinates of the middle point in a database. If the line data of the other party is a straight line, draw a perpendicular line from the point on the line data to the line data of the other party, and use the coordinates of the intersection as the coordinates of the corresponding point. It determined, when a point on the line data is bending point is the center line calculation device according to the coordinates of the point corresponding to the coordinates of the bending points of the line data of the other party.

[0009]

According to the present invention, the end portion of the line data of the database is searched, and the data of the left side portion and the right side portion are extracted as a pair of opposing line data when viewed from the end portion based on the line data, The coordinates of the corresponding points of the pair of opposing line data are calculated, the coordinates of the middle points of the corresponding points are calculated, and the calculated coordinates of the middle points are stored in the database.

[0010]

Embodiments of the present invention will now be described in detail with reference to the drawings. FIG. 1 is a block diagram showing a hardware configuration of a center line calculation device according to one embodiment of the present invention.

The database 1 stores line data such as road boundaries. The computer 3 uses the line data stored in the database 1 to calculate the center line, and the calculated center line data is used again in the database 1
To store.

FIG. 2 is a diagram showing an example of line data indicating a road boundary, and FIG. 3 is a diagram showing a record 11 of the line data stored in the database 1. This line data is a closed coordinate sequence for expressing a polygon. The polygon is used to extract the area of the road and the equipment on the road. Also, this polygon is clockwise.

As shown in FIG. 2, the line data has a point [1].
To [9]. And it intersects with other roads at points [2] and [3]. In FIG. 3, “1
"3" indicates the number of data. That is, 13 pieces of data are written after "13" with X and Y as a pair. Coordinate data of point [1] and point [2] is stored after the number of data. "-2" and "0" are flags indicating hidden lines, and these flags are inserted at the end of the road or a portion intersecting with another road. The hidden line is not displayed on the display or is displayed as a broken line. After that, the coordinate values from the point [3] to the point [5] are stored, and "-2" indicating a hidden line,
The coordinate value from the point [6] to the point [9] is stored after "0", and the coordinate value of the point [10] is stored after "-2" and "0" indicating a hidden line. The point [10] and the point [1] represent the same point. Steps 401 and 402 FIG. 4 is a flowchart showing the processing of the computer 3. The computer 3 first searches for the end portion of the road (step 401), and generates line data 13 indicating the road boundary on the left side and line data 15 indicating the road boundary on the right side as viewed from the end portion (step 402).

FIG. 5 is a flowchart showing the processing of step 401. First, the parameter i is cleared (step 501) and i is incremented by "1" (step 50).
2) Take out the coordinate data X _i , Y _i (step 50)
3). It is determined whether X _i is equal to “−2”, and “−
If not equal to 2 ”, then return to step 502.

That is, since the flag is "-2" at the end of the road and the portion where it intersects with another road, "-"
The end of the road is detected by detecting the "2" flag. If X _i is equal to "-2" (step 50)
4), the value of (i-1) is assigned to j, the value of (i + 1) is assigned to k (step 505), X _{j is set} to X _L0 , Y _{j is set} to Y _L0 , and X _{k is set} to X _R0. , Y _k into Y _R0 (step 506).

Next, j is decreased by "1" (step 5
07), take out X _j , Y _j (step 508),
^{_{[(X j -X L0) 2}} + (Y j -Y L0) 2] ^1/2 values L
To the calculated L and threshold L _M (step 509)
Are compared with each other (step 510), and if L <L _M , the process returns to step 507, and if L ≧ L _M , the process proceeds to the next process.

Next, k is increased by "1" (step 5
11), take out X _k , Y _k (step 512),
^{_{[(X k -X R0) 2}} + (Y k -Y R0) 2] ^1/2 values L
To the calculated L and threshold L _M (step 513)
Are compared (step 514), and if L <L _M , the process returns to step 511, and if L ≧ L _M , the process proceeds to the next process.

Since there are many short vectors near the intersection in the processes of steps 510 and 514, the threshold L _{M is set} to, for example, 50 m to remove a peculiar shape near the intersection.

Next, (X _j −X _L0 ) · (X _k −X _R0 ).
And "0" are compared in magnitude (step 515),
If (X _j −X _L0 ) · (X _k −X _R0 ) <0, step 5
Return to 02, proceed to _{(X j -X L0) · (} X k -X R0)> 0 if the next step.

FIG. 6 is an explanatory diagram of the processing of step 515. As shown in FIG. _6A , X _j , X _L0 , X _R0 , X
_{When k} are arranged in this order on a substantially straight line, (X _j −X _L0 ) · (X _k −X _R0 ) <0. Further, as shown in FIG. 6B, when X _L0 and X _R0 are the ends of the road, (X _j −X _L0 ) · (X _k −X _R0 )> 0. Therefore, (X _j −X _L0 ) · (X _k −X _R0 ).
The end of the road can be detected under the condition of> 0.

At step 515, (X _j −X _L0 ) · (X _k
If -X _R0 )> 0, "-10" is substituted for X _i (step 516).

FIG. 7 shows the record 20 after the processing of step 401 is performed.

FIG. 8 is a flowchart showing the process of step 402. First, the parameters i and j and the parameter M indicating the number of data are cleared (step 80
1), i and j are incremented by "1" (step 802), and X _i and Y _i are extracted from the record 20 (step 80).
3), it is determined whether X _i is equal to "-10" (step 804). If X _i is not equal to “−10”, the values of X _i and Y _i are substituted into X _j and Y _j , respectively (step 805) and M is increased by “1” (step 80).
6) and returns to step 802.

In step 804, X _i is "-10".
If equal the substitutes "-5" to X _j (step 807), the process proceeds to the next step.

The processing from step 801 to step 807 is processing for extracting line data indicating the left boundary of the road when viewed from the end portion of the road. FIG. 9A shows
The line data (X _j , Y _j ) (j = 1 to 7) subjected to the processing from step 801 to step 807 is shown. In the first part of the line data 13, the number of data M
"7" is stored as, followed by (X _j , Y _j )
Is stored.

In FIG. 8, the parameters i, j and M are cleared again (step 808), i and j are incremented by “1” (step 809), and X _Ni and Y _Ni from the record 20 are _read.
Is taken out (step 810) and it is determined whether X _Ni is equal to “−10” (step 811). If X _Ni is not equal to “−10”, the values of X _Ni and Y _Ni are substituted into X _k and Y _k , respectively (step 812), M is increased by “1” (step 813), and step 809 is performed. Return.

At step 811, X _Ni becomes "-1".
If equal to 0 "assigns" -5 "to X _k (step 814).

The processing from step 808 to step 814 is processing for extracting line data indicating the right boundary of the road when viewed from the end portion of the road. FIG. 9B shows
The line data (X _j , Y _j ) (j = 1 to 7) subjected to the processing from step 808 to step 814 is shown. In the first part of the line data 17, the number of data M
Is stored as "5", followed by (X _j , Y _j )
Is stored. Step 403 Next, the distance and gradient between each point are calculated (Step 40
3). FIG. 10 shows the distance and the gradient calculated for the record 13. Here, the distance is a straight line distance between the points, and for example, the distance between the point [1] and the point [2] is L ₂
Is. If the gradient a is, for example, the gradient between the point [I] (X _i , Y _i ) and the point [J] (X _j , Y _j ), then a =
Obtained in _{(Y i -Y j) / (} X i -X j). Figure 11
FIG. 6 is a diagram showing a distance and a gradient calculated for a record 17. Step 404 Next, the corresponding coordinate points on the opposing line data are calculated (step 404). That is, as shown in FIG. 12, the point [2 ′] (X _B , Y _B ) on the line data 17 corresponding to the point [2] on the line data 13 is calculated, and similarly, the point [4] on the line data 17 is calculated. The point [4] (X _A , Y _A ) on the line data 13 corresponding to “′” is calculated.

FIG. 13 is a flow chart showing in detail the processing of step 404. In the figure, SA is the full extension from the starting point [1] in the line data 13, SB is the full extension from the starting point [1 ′] in the line data 17, iA
Indicates the data number on the line data 13, and iB indicates the data number on the line data 17.

First, the parameters SA and SB are cleared (step 1001), and "2" is substituted into the parameters iA and iB (step 1002). Then, new SA and SB are obtained by adding L _iA and L _iB to SA and SB, respectively (step 1003). Here, L _iA and L _iB are shown in FIG.
It is the distance between each point shown in FIG.

Next, it is judged whether or not both X _iA and X _iB are "-5" (step 1004). If they are "-5", it means that the data has ended and the process is ended.
If both X _iA and X _iB are not “−5”, the full extension SA and the full extension SB are compared (step 1005), and SA is S
When it is considerably smaller than B, a perpendicular is drawn between the point iA of the line data 13 and the point (iB-1) of the line data 17 and the point iB to obtain the intersection (X, Y) (step 1006).

FIG. 14 shows the processing at this time.
A perpendicular line is drawn from the point K to the points I and J, and the intersection point M (X,
It is explanatory drawing when Y) is calculated | required.

The intersection M (X, Y) is, is on a line segment _{IJ, Y = (Y j -Y} i) / (X j -X i) · (X-X i) +
Y _i (1), and since the line segment KM is orthogonal to the line segment IJ, its gradient is − (X _j −X _i ) / (Y _j −Y _i ), and the intersection point M is
Since (X, Y) is on the line segment KM, Y = − (X _j −X _i ) / (Y _j −Y _i ) · (X−X _k ).
+ Y _k (2) is established.

[0034] Formula 1 (1) and when organizing the formula _{(2), (X j -X} i) Y- (Y j -Y i) X + Y i X j -Y j
X _i = 0 (3) (Y _j −Y _i ) Y + (X _j −
X _i ) X + Y _k (Y _j −Y _j ) + (X _j −X _i ) X _k = 0
(4), and the intersection point M (X, Y) is obtained by solving the simultaneous equations of the equations (3) and (4).

Next, the coordinate data of the line data 17 after the point iB is shifted backward by one data (step 10).
07). Then, the coordinates X and Y of the intersection are respectively _added to X _iB and Y _iB.
(Step 1008), iA is incremented by 1 (step 1009), SA is incremented by L _iA (step 1010), and the process returns to step 1004.

In step 1005, SA is SB
Is much smaller than point i of line data 17
A perpendicular line is drawn from B to the point (iA-1) of the line data 13 and the point iA to obtain the intersection point (X, Y) (step 1).
011).

Next, the coordinate data after the point iA of the line data 13 is shifted backward by one data (step 10).
12). Then, the coordinates X and Y of the intersection are respectively _added to X _iA and Y _iA.
Is substituted (step 1013), iB is incremented by 1 (step 1014), SB is incremented by L _iB (step 1015), and the process returns to step 1004.

If SA and SB are approximately equal in step 1005, it is judged whether or not the gradients a _iA , a _{iA + 1} , a _iB , and a _{iB + 1} are approximately equal (step 1016). In this case, the intersection flags F _iA and F _iB of the line data 13 and 17 are set to “1” (step 1017). Next, iA and iB are each increased by "1" (step 10
18) and increase SA by L _iA (step 101
9) increase SB by _LiB (step 1020),
Returning to step 1004.

FIGS. 15 and 16 are diagrams showing the line data 13 and the line data 17 which have undergone such processing.
Here, the intersection point flag is used, for example, when one bending point of one line data corresponds to a plurality of bending points of the other line data. Step 405 When the corresponding coordinate points on the opposing line data are calculated in this way, the coordinates of the middle point of the corresponding coordinate points are calculated (step 405).

FIG. 17 is an explanatory diagram for calculating the coordinates of the midpoint. The coordinate value from the point [1] to the point [5] is calculated on the line data 13 by the above-described procedure, and the coordinate value from the point [1] to the point [5] is calculated on the line data 17. . Then, the midpoint (1) to the midpoint (5) of the corresponding points (for example, the point [1] and the point [1 ′]) are obtained.

That is, if the midpoint (X, Y) of the point [I] (X _i , Y _i ) and the point [J] (X _j , Y _j ), then X = (X _i + X _j ) / 2. The coordinates of the midpoint are calculated according to Y = (Y _i + Y _j ) / 2. Step 406 Then, the midpoint (1) to the midpoint (5) are connected to form the center line data, which is stored in the database 1 (step 40).
6).

As described above, according to the present embodiment, the center line of the line data facing each other such as a road can be easily calculated.

In the case where the road 33 intersects with the road 31 as shown in FIG. 18, the mounting portion is constituted by the points [2], [3] and [4], [5]. In such a case, point [3] and point [4] may be ignored and processed.

If the data is arranged in the order shown in FIG. 19, the start point and the end point are obtained by repeating the process of step 401 twice, and the line data of step 402 is obtained from the start point to the end point. .

It should be noted that in the present invention, the two opposing polygons are not polygons.
It can also be applied to the case of finding the center line of one line data (for example, two line data facing each other such as a railway).

..

As described above in detail, according to the present invention, even when one road intersects with another road,
The center line of the road can be calculated automatically.

[Brief description of drawings]

FIG. 1 is a block diagram of a system for carrying out an embodiment of the present invention.

FIG. 2 is a diagram showing closed line data forming a polygon.

FIG. 3 is a diagram showing line data stored in a database 1.

FIG. 4 is a flowchart showing the processing of the computer 3.

FIG. 5 is a flowchart showing the processing of step 401.

FIG. 6 is an explanatory diagram of detection of a terminal portion.

FIG. 7 is a diagram showing converted line data.

FIG. 8 is a flowchart showing the process of step 402.

FIG. 9 is a diagram showing line data 13 and 17

FIG. 10 is a diagram showing a table regarding the line data 13 in which the distance and the gradient are calculated.

FIG. 11 is a diagram showing a table regarding the line data 17 in which the distance and the gradient are calculated.

FIG. 12 is a diagram showing line data in which corresponding points are calculated.

FIG. 13 is a flowchart showing the process of step 404 in detail.

FIG. 14 is an explanatory diagram of calculation of intersections of perpendicular lines.

FIG. 15 is a diagram showing a table related to line data 13.

FIG. 16 is a diagram showing a table relating to line data 17.

FIG. 17 is a diagram showing line data 13 and 17 and a center line.

FIG. 18 is a diagram showing another line data.

FIG. 19 is a diagram showing another line data.

[Explanation of symbols]

1 ... database 3 ... Computer 13, 15, 17 ... Line data

Front page continuation (58) Fields surveyed (Int.Cl. ⁷ , DB name) G06T 1/00 G06T ⁷ /00-7/60 G06T 11/60 G06F 17/50

Claims (2)

(57) [Claims] 1. A database for storing closed line data forming a polygon, a means for searching an end portion of the line data of the database, and a portion on the left side of the line data based on the end portion. Means for extracting the data of the right part as a pair of opposing line data, corresponding point calculating means for calculating the coordinates of the corresponding points of the pair of opposing line data, and each of the corresponding points. And a means for storing the coordinates of the calculated midpoint in a database , wherein the corresponding point calculation means is such that the line data of the other party is a straight line.
If there is a line on the other side,
A perpendicular line is drawn on the data and the coordinates of the intersection are
The coordinates on the line data are the bending points.
Sometimes, the coordinates of the bending point of the line data of the other side are
A center line calculation device characterized by using coordinates of corresponding points . 2. The corresponding point calculating means includes means for determining whether or not the corresponding point is a mounting portion of another road, and means for ignoring the corresponding coordinate point when the corresponding point is a mounting portion. The centerline calculation device according to claim 1, wherein