JPH09138641A - Electronic map display method and device therefor - Google Patents

Abstract

PROBLEM TO BE SOLVED: To shorten a processing time till a display by segmenting the paper map of FD of a map preparing maker into a mesh size smaller than a page unit. SOLUTION: The FD is segmented into the page unit of the paper map (like in pages i, i+1, j, j+1). At the time of performing a display, the FD is provide with a virtual area shown by a broken line as a plotting area and a display area (b) is made in the virtual area and the map data of the display area (b) are displayed. Hence, the management of the plotting area and the display area is performed in the mesh unit segmented smaller than the page. Since the FD is moved to memories in an HDD, the management of the memories is also performed in the mesh unit.

Description

Detailed Description of the Invention

[0001]

BACKGROUND OF THE INVENTION 1. Field of the Invention The present invention relates to an electronic map display method and device, and more particularly to a display method and device suitable for high speed display of an electronic map such as a commercially available FD.

[0002]

2. Description of the Related Art An electronic map is one in which map data is stored in an external memory such as a floppy disk or an optical disk instead of a paper map and is used for display. It is being widely used for car navigation systems, transportation systems, and various emergency deployment systems (ambulances, fire engines, police cars), etc. The external memory is generally created by a map company and used by an electronic device manufacturer as a database. In many cases, electronic maps created by map companies are based on commercially available paper maps. Examples of electronic maps include image map examples in which a paper map is read as an image (conversely, there is also a method of making image data and printing it on a paper map at the time of printing), and vector map examples in which straight lines and circles are vectorized as constituents. is there.

Management on the external memory is performed on a paper map page basis. For example, one page is horizontal x vertical = 3400
The pixel size is set to 2000 pixels.

[0004]

When the map data in the external memory is loaded into various map display systems, it takes time from reading to displaying if the map data is handled in page units. Further, a high-speed buffer memory is provided between the external memory and the display (for example, CRT), and the map data is temporarily stored in the high-speed buffer memory (usually called drawing), and then the display is displayed. There are many examples of how to display. In this case, the capacity of the high-speed buffer memory is several times to several tens times larger than the display area (synonymous with the visible area), and is usually called a virtual area.
At the time of display, the display area is cut out from the virtual area and displayed through the image memory. Scrolling is realized by moving the display area in the virtual area. This method of using the virtual area has a problem that the external memory is handled in page units, which further increases the series of processing time from reading to display. This is illustrated in FIG.

FIG. 2 is a diagram showing page division as a storage unit of the external memory and straddling the display area. As pages, four adjacent pages (i, i + 1, j,
j + 1) example is shown. Each page shows one map. In the figure, map numbers i, i + 1, j, and j + 1 are attached to each page. The four states a, b, c, and d are shown as the display area.
The display area is an area taken in the virtual area, and the map data in this display area is displayed on the display unit. As a matter of course, the size of the display area is sufficiently smaller than the page size.

The state a is a case where the display area exists within one page (i), and there is no span between pages. State b has four pages (i, i + 1, j, j +
This is an example in which the display area extends over 1). States c and d are two pages (i + 1 and j + 1, j and j + 1)
Indicates straddle. In this way, even if the external memory is divided into pages, the display area is set regardless of the page division. Since the virtual area is set around the display area as shown in FIG. 3, the virtual area also crosses pages.

It is convenient to manage the map data on a page-by-page basis as shown in FIG. 2 because the map data provided on the external memory can be handled as it is. However, since there is a large area that needs to be read as map data (usually stored in the form of map DB) in the external memory such as FD, it is not displayed in the display area. There is a problem that wasteful time is required and the processing time until display becomes long. Since the FD is generally read in the hard disk drive (HDD), the above external memory includes the memory in the HDD in addition to the FD.

An object of the present invention is to provide an electronic map display method and apparatus which can shorten the processing time until display.

[0009]

The present invention divides the page data into page units of the external memory into smaller mesh units and manages the map data in the mesh units. An electronic map display method for displaying is disclosed.

Further, according to the present invention, when a map is displayed, a virtual area is set in a high-speed memory and a display area is set in this virtual area, and the display area is moved on the virtual area to display the map on the display screen. In the electronic map display method that performs scroll display, the map data divided into page units of the external memory is divided into smaller mesh units, and the divided mesh units are divided into virtual regions and display regions. Disclosed is an electronic map display method for managing the map data of the above.

The present invention further discloses an electronic map display method in which the subdivided mesh unit is determined from the display area size, the virtual area size, and the map page size.

Further, according to the present invention, the external memory is divided into page units, the map data corresponding to the page is stored in each section, and the page unit, which is a division unit, is subdivided into a plurality of meshes. And a virtual area, and a display area that is allocated within the virtual area and has a size smaller than the virtual area. Disclosed is an electronic map display device comprising an internal memory for reading the map data of 1. and a display unit for displaying the map data of the display area on the internal memory.

The present invention further discloses an electronic map display device in which the subdivided mesh unit is determined from the display area size, the virtual area size, and the map page size.

Further, the present invention discloses an electronic map display device in which means for moving the display region on the virtual region is added to the electronic map display device.

Furthermore, the present invention is an electronic map display method in which map data from an external memory is moved onto a virtual area and the map data of a display area within this virtual area is displayed. A method of displaying an electronic map that minimizes the processing time until the display of the map data from the display area is minimized is disclosed.

[0016]

FIG. 1 shows an embodiment of the present invention. FIG. 1 is a diagram illustrating an example of division in a memory area. The external memory is divided into page units (page i, i +
4 adjacent maps of 1, j, j + 1 are shown).
For this division, each page is divided into 8 divisions of vertical × horizontal = 2 × 4 (this is called a mesh to distinguish from pages). # 1 to # 8 are mesh numbers. An example of the size of these page section, mesh section, virtual area, and display area is shown below. Data size of map for one page Horizontal × vertical = 3400 pixels × 2000 pixels Size of virtual area Horizontal × vertical = 2740 pixels × 2070 pixels Display area size Horizontal × vertical = 1100 pixels × 1000 pixels 1 mesh data size Horizontal × Vertical = 850 pixels x 1000 pixels

FIG. 4 shows an example in which one page is meshed into 16 pieces. 16 meshes are vertical x horizontal = 4 x 4
It is. The mesh numbers are indicated by # 1 to # 16. The size example in FIG. 4 is as follows. Data size of 1 mesh Horizontal x vertical = 850 pixels x 500 pixels

The 1-page 8-meshing of FIG. 1 is a preferable example, and the 1-page 16-meshing of FIG. 4 is a less preferable example. A method of setting the number of meshes and an example of setting the optimum number of meshes will be described below. The smaller the mesh size, the closer the amount of data required to draw the display area will be to the area of the display area, and there will be no waste, but the read time will increase because the number of times of reading from the HDD increases, and Time increases. That is,
There is an optimal mesh size. Hereinafter, this relationship will be quantitatively analyzed.

In the case of 8-mesh division, the size of 1 mesh is 850 × 1000. To cover the display area (1100 × 1000) with this mesh, a minimum of 2 meshes (state e) and a maximum of 6 meshes (state a) It is clear from FIG. 1 that () is required. State b is an example of 4 meshes. On the other hand, in the case of 16 mesh division, the size of one mesh is 850 × 500, and a minimum of 6 meshes (state b) and a maximum of 9 meshes (state a) are required to cover the display area with this mesh. Is clear from FIG.

Hereinafter, in order to evaluate in the worst case, the case of the maximum number of meshes will be evaluated. Consider the following three cases. Case 1: No mesh division. Let the processing time be T1 (conventional method). Case 2: 8 mesh division. Let the processing time be T2 (a mode of the present invention). Case 3: 16 mesh division. The processing time is T3. Processing time = time proportional to the amount of data (α) + time proportional to the number of data items (β). Here, time proportional to the amount of data: data transfer time + data calculation time + data display time Time proportional to the number of data items: read time from the HDD. The processing time per unit is α and β, respectively. Considering Case 1, the amount of data is one map and the maximum is four maps. Therefore, the time proportional to the amount of data is 4 × 1 × α = 4α. HDD as data
Therefore, T1 is expressed as follows: Time proportional to the number of data = 4 × β = 4β

## EQU1 ## T1 = 4 × 1 × α + 4 × β = 4 (α + β)

Considering Case 2, the amount of data is ⅛ of the amount of one map, and the maximum is 6 meshes. Therefore, time proportional to the amount of data = (6 × 1) / (8 × α In addition, since the number of data cases is one-eighth mesh of one map managed as one data in the HDD, time proportional to the number of data cases = 6 × β Therefore, T2 is expressed as follows.

## EQU2 ## T2 = (6 × 1) / (8 × α) + 6 × β = 6
(Α / 8 + β)

Considering Case 3, the amount of data is 1/16 of the amount of one map, and the maximum mesh is 9 meshes. Therefore, time proportional to the amount of data = (9 × 1) / (16 × α In addition, since the number of data cases is 1/16 mesh of one map is managed in the HDD as one data, time proportional to the number of data cases = 16 × β Therefore, T3 is expressed as follows.

## EQU3 ## T3 = (9 × 1) / (16 × α) + 9 × β = 9
(Α / 16 + β) From the above equations 1 to 3,

## EQU00004 ## Solving the inequality under the condition of T3.ltoreq.T2.ltoreq.T1,

## EQU5 ## 8 / 13β ≦ α ≦ 16β is obtained. This indicates that the unit processing time proportional to the amount of data is almost equal to or longer than the unit processing time proportional to the number of times.

Or vice versa

## EQU6 ## (α × 1) / 16 ≦ β ≦ (α × 13) / 8 where α = 16

1 ≦ β ≦ 26 As an example, let α = β = 16, and T1 (α, β) = T1 (1
6, 16), T1 (16,16) = 4 (16 + 16) = 4 × 32 = 128 T2 (16,16) = 6 (2 + 16) = 6 × 18 = 108 T3 (16,16) −9 (1 + 16) = 9 × 17 = 153, and T2 is certainly optimal. As an example, α = 16,
If β = 2, T1 (16,2) = 4 (16 + 2) = 4 × 18 = 72 T2 (16,2) = 6 (2 + 2) = 6 × 4 = 24 T3 (16,2) = 9 (1 + 2 ) = 9 × 3 = 27, and T2 is optimal. T2 is 24 / compared to T1
It takes about 1/3 the processing time from 72. This indicates that the present invention is more effective when the unit processing time proportional to the amount of data takes much longer than the unit processing time proportional to the number of times.

In the above example, the case 2 that is, the case where one map is divided into 8 meshes is optimum, but this is because of the relationship between the size of one map and the size of the display area. Is. That is, as mentioned above,
Various conditions are as follows as a typical mesh division example of a map. Data size of map 1 3400 × 2000 Size of virtual area 2740 × 2070 Size of display area 1100 × 1000 (Unit: number of screen pixels) In this case, in the case of 8 mesh division of one map, 1 mesh is 850 × The width is 1000, and the width of one mesh is slightly smaller than the width of the display area. In this case, the worst case is 6 meshes. That is, it is generally as follows.
Compare the size of each side of the mesh and the vertical and horizontal sides of the display area. Case 1 Maximum number of spans for each side if mesh ≥ display area = 4 Case 2 Maximum number of spans for one side if display area ≥ mesh ≥ display area / 2 = 6 Case 3 One side is a display area / 2 ≧ mesh ≧ display area /
If it is 3, the maximum number of strides = 9 Case 4 Two sides are display area / 2 ≧ mesh ≧ display area /
If it is 3, the maximum number of strides = 12 ...

The general expression of processing time is as follows.

## EQU8 ## T = M × (α / n + β) where M: number of straddles, n: number of mesh divisions of one map Therefore, the optimum mesh of the present invention is calculated by calculating T according to the environment used. It becomes possible to divide. Hereinafter, some cases will be shown. The processing time in the case of n division is Tn
As (α, β), T4 (8,2) = 4 (2 + 2) = 16 α: β = 4: 1 T8 (8,2) = 6 (1 + 2) = 18 T4 (16,2) = 4 (4 + 2 ) = 24 α: β = 8: 1 T8 (16,2) = 6 (2 + 2) = 24 T4 (24,2) = 4 (6 + 2) = 32 α: β = 12: 1 T8 (24,2) = 6 (3 + 2) = 30 T4 (32,2) = 4 (8 + 2) = 40 α: β = 16: 1 T8 (32,2) = 6 (4 + 2) = 36 T4 (8,3) = 4 (2 + 3) = 20 α: β = 8: 3 = 2.7: 1 T8 (8,3) = 6 (1 + 3) = 24 T4 (16,3) = 4 (4 + 3) = 28 α: β = 16: 3 = 5 .3: 1 T8 (16,3) = 6 (2 + 3) = 30 T4 (24,3) = 4 (6 + 3) = 36 α: β = 24: 3 = 8: 1 T8 (24,3) = 6 ( 3 + 3) = 36 T4 32,3) = 4 (8 + 3 ) = 44 α: β = 32: 3 = 10.7: 1 T8 (32,3) = 6 (4 + 3) = 42 T4 (8,4) = 4 (2 + 4) = 24 α: β = 8: 4 = 2: 1 T8 (8,4) = 6 (1 + 4) = 30 T4 (16,4) = 4 (4 + 4) = 32 α: β = 16: 4 = 4: 1 T8 ( 16, 4) = 6 (2 + 4) = 36 T4 (24,4) = 4 (6 + 4) = 40 α: β = 24: 4 = 6: 1 T8 (24,4) = 6 (3 + 4) = 42 T4 ( 32, 4) = 4 (8 + 4) = 48 α: β = 32: 4 = 8: 1 T8 (32,4) = 6 (4 + 4) = 48 T4 (8,5) = 4 (2 + 5) = 28 α: β = 8: 5 = 1.6: 1 T8 (8,5) = 6 (1 + 5) = 36 T4 (16,5) = 4 (4 + 5) = 36 α: β = 16: 5 = 3.2: 1 T8 (16,5) = 6 ( +5) = 42 T4 (24,5) = 4 (6 + 5) = 44 α: β = 24: 5 = 4.8: 1 T8 (24,5) = 6 (3 + 5) = 48 T4 (32,5) = 4 (8 + 5) = 52 α: β = 32: 5 = 6.4: 1 T8 (32,5) = 6 (4 + 5) = 54 As described above, the optimal mesh division is based on the ratio of the speeds of α and β. Fluctuates in various ways. In this case α: β = 8: 1
The result changes with the boundary.

The average access time of the current standard 5-inch HDD is set to about 50 ms. Therefore β = 50 (m
s). If α = 1000 ms, α: β = 20:
Since it is 1, it is better to divide it into 8 meshes. At this time, the processing time is 6 × (1000/8 + 5
0) = 1050 ms = 1.05 sec. If this is processed without dividing into meshes, 4 × (1000 + 50)
= 4200 ms = 4.2 sec. That is, dividing into 8 meshes is 4 times faster. On the other hand, in a 4-mesh division, 4 × (1000/4 + 50) = 1200 ms. That is, the 8-mesh division according to the present invention is optimal. In this embodiment, even divisions such as 4 divisions and 8 divisions are adopted. On the other hand, 6-division and 9-division are conceivable, but since one side is an odd-division, inconvenience is likely to occur in actual processing, which is not realistic. Note that, from the form of Equation 8, α is the time required to process one piece of data.

FIG. 5 shows a procedure flow until display in the conventional division by page unit shown in FIG.
FIG. 6 shows a procedure flow until display when the page is further meshed as shown in FIG. 1 of the present invention. In FIG. 5, in the worst case, four pages are overlapped with each other in FIG. 2 (state b). Therefore, the map data of the related four pages of maps (four maps) in the flow 1 is read from the HDD to the high-speed buffer memory. In Flow 2, this map data is subjected to conversion calculation necessary for display. For example, in the case of vector map data, a numerical value is converted into image data in the image memory. In the flow 3, the map data is sent to the high speed buffer memory so as to fill the entire virtual area. Display on the screen in Flow 4. Flows 1 to 3 are realized by loop processing for the number of blocks (number of data) of data.

In FIG. 6, in flow 10, map data for the display area in the virtual area is first read out from the HDD to the high speed buffer memory. In the flow 20, conversion operation necessary for displaying the map data is performed. In flow 30, map data that fills the entire display area is sent to the display memory. In the flow 40, display on the screen is performed. Further, in a flow 50, map data other than the display area is read from the HDD to the high speed buffer memory. This prepares for scrolling. In a flow 60, conversion calculation necessary for displaying the map data is performed. In flow 70, map data that fills the entire virtual area other than the display area is transferred to the display memory for display. Incidentally, the flow 10-40, 50
All of .about.70 actually depend on the loop processing for the block (number of cases) of data.

FIG. 7 shows the map system of the present invention.
The common bus 15, the CPU 11, the main memory 12, the HDD 1
3, buffer 12A, display memory (image memory) 16,
A keyboard 18, a mouse 19, and a printer 21 are commonly connected. The HDD 13 stores paged and database (DB) map data read from the FD. The CRT 17 displays the map data temporarily stored in the display memory 16 as a map on the screen.
The buffer 12A is an expansion memory for the main memory 12 and forms a virtual area in this example.

Although the FD is divided into pages, the reading unit 14 transfers the contents read from the FD to the memory in the HDD 13 via the common bus 15.
New allocation from the page section to the mesh section is performed for the memory in the HDD 13. The positional relationship between the page and the mesh at this time is associated as management data and latched in the main memory 12 or the like and used for conversion from the mesh to the page or the like. New allocation from the page section to the mesh section may be performed every time the size of the display area is determined according to the above-mentioned various formulas, or the optimum number of meshes is made into a table and read and used. Good. This new allocation is CPU11
Alternatively, the operation is performed by operating the keyboard 18 or the mouse 19 according to the operator's instruction. Data is transferred between the HDD 13 and the common bus 15 under this new mesh division. The CPU 11 can access the HDD 13 by mesh division without being aware of the page, and the buffer 12A is also managed by this mesh division.
Of course, the mesh is further subdivided into block units, and this block unit becomes a processing unit. In addition, it is naturally possible to move the display area on the virtual area (left and right, up and down, diagonally) to perform horizontal, vertical and diagonal scrolling display of the map. Although the example of the FD has been described as the external memory, there is an example of another external memory such as an optical disk memory.

[0031]

According to the present invention, one map is divided into meshes, which are smaller units than page units.
The wasteful processing time can be reduced. Also, the optimum mesh size can be determined from the relationship between the display area and the page size, which enables higher speed display.

[Brief description of the drawings]

FIG. 1 is a diagram showing an example of mesh allocation according to the present invention.

FIG. 2 is a diagram showing conventional page allocation.

FIG. 3 is a diagram showing a relationship between a virtual (storage) area and a display area.

FIG. 4 is a diagram showing another example of mesh allocation.

FIG. 5 is a diagram showing a conventional processing procedure.

FIG. 6 is a diagram showing a processing procedure of the present invention.

FIG. 7 is a diagram showing an example of a display system of the present invention.

[Explanation of symbols]

 11 CPU 12 Main Memory (MM) 12A Buffer 13 HDD 14 FD Reading Section 15 Common Bus 16 Image Memory (Display Memory) 17 CRT 18 Keyboard 19 Mouse 21 Printer

Claims (7)

[Claims] 1. An electronic map display wherein map data is divided into page units of an external memory, the page units are divided into smaller mesh units, and the map data is managed and displayed in mesh units. Method. 2. When displaying a map, a virtual area is set on a high-speed memory, and a display area is set in this virtual area.
In the electronic map display method of scrolling the map on the display screen by moving the display area on the virtual area, the map data divided into page units of the external memory is divided into smaller mesh units. An electronic map display method in which the map data in the virtual area and the display area are managed in this divided mesh unit. 3. The electronic map display method according to claim 2, wherein the subdivided mesh unit is determined from a display area size, a virtual area size, and a map page size. 4. An external memory that is divided into page units and stores map data corresponding to each page, and a page unit that is a division unit is subdivided into a plurality of meshes, and the external memory is divided into mesh units. And a virtual area and a display area which is allocated in the virtual area and has a size smaller than the virtual area. The map data in the external memory is included in the virtual area including the display area in the mesh unit. An electronic map display device comprising: an internal memory for reading out the data and display means for displaying map data in a display area on the internal memory. 5. The electronic map display device according to claim 3, wherein the subdivided mesh unit is determined from a display area size, a virtual area size, and a map page size. 6. The electronic map display device according to claim 4, further comprising means for moving the display area on the virtual area. 7. An electronic map display method in which map data from an external memory is moved onto a virtual area and the map data in a display area within this virtual area is displayed, and the amount of drawing data in the virtual area is minimized. By doing so, an electronic map display method that minimizes the processing time until the display of the map data from the display area is completed.