JPH03137870A - Video reproducing device - Google Patents

Abstract

PURPOSE:To access one picture at high speed by using plural compact disk-ROM driving devices jointly and reading a picture element data recorded in dividing one screen into plural disks. CONSTITUTION:The picture element data is stored in plural CD-ROM disks by dividing one screen respectively. When one block of the data is read by the CD-ROM driving devices 6A-6D respectively, the picture element data relative to the individual divided screen of the same screen from the devices 6A-6D is received in turn by a computer 5 via a sequence switch 8 one by one block respectively within a prescribed time standardized for transferring one block, so as to be stored in a memory 10. By this method, the access time to one picture is shortened.

Description

DETAILED DESCRIPTION OF THE INVENTION (A) Field of Industrial Application The present invention relates to an apparatus for reproducing image information recorded on a CD-ROM disc.

(b) Conventional technology CD-ROM discs have a data capacity of up to 540 Mbytes, so by taking advantage of this large capacity, they can record not only text information, but also image information, music information, etc. as computer data. applied in the field.

By the way, regarding the data transfer speed from a CD-ROM drive to a computer, CD-ROM drives generally have a maximum capacity of 2M bytes/5ec(7), but the fundamental factor that determines that speed is the laser. The speed at which the pickup reads data from the disc. Therefore, the data transfer speed must depend on the rotational speed of the disk, but according to the currently standardized rotational speed, the reading speed is 75 blocks /
see, and in this case, one block corresponds to 2048 bytes, so the speed at which a CD-ROM drive reads data from the disk with a laser pickup is 15
If the data transfer rate is 0 bytes/see, the data transfer rate will also be 150 bytes/see. This is a unified standard that includes CD-ROM and CD audio, and faster playback is not possible.

(c) Problems to be Solved by the Invention Therefore, CD-ROM drives have a high data transfer rate, and in line with this, computers that receive data also have to read data at a high speed of, for example, IM bytes/see. Even if you set the data transfer time to be fast, it takes 150 bytes/sec for a CD-ROM drive to read data from a CD-ROM disc.
There is a problem that the actual data transfer takes a long time because it takes a long time.

This will be explained in detail with reference to FIGS. 2, 3, and 3A. FIG. 2 is a diagram showing the exchange of signals between the computer and the CD-ROM drive, and FIGS.
The figure is a timing chart.

The computer 1 reads the CD-ROM on the condition that the data enable signal DE of the CD-ROM drive device 2 is “L”.
The operation of reading one block of data from the OM drive device 2 is started (FIG. 3).

During this reading operation period T, the computer 1 has 204
Eight read signals RD are output in sequence, and in response, C
One byte of data output from the D-ROM drive device 2 to the data bus DB is read (FIG. 3A). Therefore, 2048 bytes (2 bytes) of data are transferred from the CD-ROM drive 2 to the computer 1 during the reading operation period T, but if the data reading speed of the computer 1 is IM bytes/sec, this The certificate collection operation period T is 'l m5ec. On the other hand, CD-R
For OM drive 2 to read 2 bytes of data from disk in the next block, the data read speed must be 1.
50 bytes/sea so 13゜3 m5ec
, and the CD-ROM drive device 2 is de-enabled once.
Once the tie enable signal DE is output, the next output will be 13.3 rnsec later. Therefore, the computer 1 finishes receiving one block of data from the CD-ROM drive device 2 2 m5ec after the data enable signal DE becomes "L", but it continues to receive the next block of data. To receive the data of 1
You have to wait for 1,3 m5ec.

Therefore, when such a CD-ROM is used to store image information, the amount of image information is so large that even one image consists of many blocks of data.
It takes 11 for computer 1 to receive one block of data.
．． If a waiting time of 3 seconds is required, there is a drawback that access to one image becomes slow.

For this reason, the CD-ROM drive device 2 has a buffer memory that stores several blocks of data, and the data can be read in advance, and block data specified by a command can be stored in the buffer memory. There is a method in which the data is read from the CD-ROM drive device 2 at some point. However, this method apparently increases the data transfer speed, but cannot increase the transfer speed when reading an amount of data that exceeds the capacity of the buffer memory.

Therefore, the present invention divides one screen into a plurality of blocks, records the pixel data of each divided screen on a plurality of disks, and also records a plurality of blocks corresponding to each disk.
A D-ROM drive device is provided, and when pixel data is read from one drive device and the drive device enters a transfer suspension period, pixel data is read from the next drive device, thereby speeding up access to one image. The present invention provides an image reproducing device as described above.

(d) Means for Solving the Problems In order to solve the above problems, the image reproducing apparatus according to the present invention includes a plurality of CD-ROM drive devices, and a CD-ROM from which each of the CD-ROM drive devices reads data.
- Pixel data for each divided screen of one screen is recorded on the ROM disk, and multiple pixel data are recorded within a predetermined time standardized for the CD-ROM drive to transfer one block of data. The above-mentioned apparatus is equipped with a processing device that sequentially inputs pixel data for each divided screen of the same screen from the ffD-ROM drive device.

(e) The processing device receives one block of data regarding one split screen from one CD-ROM drive, and when this CD-ROM drive goes into a transfer suspension period, it then transfers data to the next CD-ROM. Data regarding the next divided screen of the same screen is received from the ROM drive device. When the processing device receives each pixel data for each of the divided screens constituting the same screen in this manner, the image reproducing device reproduces the screen based on these data.

(F) Embodiment FIG. 1 is a block diagram showing a schematic configuration of an image reproducing apparatus according to an embodiment of the present invention, which includes a computer 5 as a processing device and four CD-ROM drive devices 6A, 6B, 6C. ，
6D is daisy-chained by a pass line 7. This pass line 7 has eight data buses and a CD
- Four selection signal lines for selecting a ROM drive device, and a computer 5 for each CD-ROM drive device 6A, 6B.
, 6C, 6D, and a read signal line for outputting the aforementioned read signal RD to the CD-ROM drive devices 6A, 6.
B, 6C, and 6D include data enable signal lines for outputting a data enable signal DE to the computer 5. Each CD-ROM drive device 6A, 6B, 6C, 6
D is equipped with a sequence switch 8 in which a predetermined number is set using a binary code, so that the CD-ROM drive device whose number data indicated by the aforementioned selection signal line matches the set number is driven. It has become.

The CRT 9 reproduces images under the control of the computer 5, and has a screen configuration of 400 dots vertically and 640 dots horizontally. In this example, one screen is divided into 4 screens as shown in Figure 7.
It is divided into CD-ROMg movement f5l6 A,
Each CD-ROM disk that performs 6B, 6C, and 6D Si removal operations has 4
Each pixel data of the divided screens a to d is recorded. One split screen consists of 64,000 dots, and each dot has 1 byte of R and G.
, B are required. Therefore, each CD-ROM
Each CD-R corresponding to drive device 6A, 6B, 6C, 6D
On the OM disk, 64 bytes each of R, G, and B data are recorded for the same screen. Therefore, the computer 5 drives each CD-ROM drive device 6A, 6B.
, 6C, and 6D (64 bytes x 3) allows one screen to be reproduced on the CRT9.

FIG. 4 shows that the computer 5 has each CD-ROM drive device 6A.
, 6B, 6C, and 8D. As mentioned above, each CD-
ROM drive unit 6A, 6B, 6C, 6D is 13.3m5
One block of data is transferred in one data transfer period of EC, but the actual data transfer time is 2fflSeC.
It only requires a Therefore, CD-ROM drives are standardized to transfer one block of data.
Four data reading operation periods are provided within a time of 3m5ec, and in the first reading operation period T1, the CD-ROM
The driving device 6A is driven to read one block (2048 bytes) of pixel data for the divided screen a, and then a second reading operation period T begins.

Then, the CD-ROM drive device 6B is driven to read one block of pixel data for the divided screen, and in the next third reading operation period T, the CD-ROM drive device 6B is driven.
C is driven to read one block of pixel data for the divided screen C, and then the next fourth reading operation period T
4, the CD-ROM drive device 6D is driven to read one block of pixel data for the divided screen d. Therefore, by processing each of the four reading operation periods once, data for 2048 dots for each of the divided screens a, b, c, and d is transferred to the computer 5. When the 13°3 m5ec period ends, the computer 5 starts the reading operation period TI in the next 13.3 nisec period.

By sequentially processing T,,T,,T,, a,b.

New data for 2048 dots for each split screen of C and d is introduced.

In this example, the computer 5 introduces pixel data in the order of R, G, and H. Therefore, the computer 5 first inputs 2 bytes of R pixel data from each CD-ROM drive device 6A, 6B, 6C, 6D every 13.3 m5ec in order to introduce 64 bytes of R pixel data in each divided screen a to d. The above transfer process that introduces data one by one is performed in 32 steps.
Repeat times. By repeating this operation for 0-pixel data and 8-pixel data and performing the transfer process a total of 96 times, all the data necessary to reproduce one image on the screen of the CRT 9 is transferred to the computer 5.

Therefore, the time required to access one image is approximately 1
, 3 seconds, which reduces the data transfer time to 1/4 compared to the conventional method that receives and receives data one block at a time within a standardized time of 13.3m5ec. Access time is faster.

As shown in FIG. 5, the memory 10 that is controlled by the computer 5 and stores image information for one screen includes an image data memory M1 that stores pixel data for split screen a, and pixel data for split screen A. an image data memory M that stores pixel data for the split screen C, an image data memory M that stores pixel data for the split screen C, and a split screen d.
Image data memory M4 that stores pixel data about
Each of the memories M+, Mt, Ms, and M4 has 96 storage areas. And memory M is 1
It is equipped with storage areas indicated by codes 00 to 195, in which R pixel data is stored in storage areas 100 to 131, 0 pixel data is stored in storage areas 132 to 163, and storage areas 164 to 195 are stored. By now, 8 pixel data can be stored. Similarly, memory M*,
M3. M4 is also provided with storage areas indicated by codes 200 to 295.300 to 395.400 to 495, respectively.
31.400-431 stores R pixel data, and each storage area 232-263, 332-363.432-463
stores 0 pixel data, and each storage area 264-295.
364-395 and 464-495 are designed to store 8 pixel data.

FIG. 6 shows a counter that the computer 5 has, where API is an address pointer indicating the storage location of the image data memory M, AP2 is an address pointer indicating the storage location of the image data memory M, and AP3 is an address pointer indicating the storage location of the image data memory M. This is an address pointer indicating a storage location in data memory M. C1 is a transfer block counter that counts the number of data transfers, and C2 provides numerical information for specifying a storage area to each address pointer when each address pointer specifies a storage location within a predetermined storage area. Basic address counter, C3 is a drive select counter that specifies the CD-ROM drives 6A, 6B, and 6C, C4 is an RGB counter that specifies whether RlG or B pixel data is to be read, and C5 is one block transfer at a time. This is a 1 block counter that counts the number of bytes (204B).

The computer 5 is a CD-ROM drive device 6A.

R and G for one screen from 6B, 6C, and 6D.

The operation of introducing each pixel data of B will be explained based on the flowchart of FIG.

First, in step S1, the computer 5 sets the RGB counter C4 to 3 in order to introduce data from the R pixel.

In the next step S2, the computer 5 sets the transfer block counter C1 to 32.

In the next step S3, the computer 5 initializes the basic address counter C2.

In the next step S4, the computer 5 initializes the drive select counter C3.

Then, in the next step S5, the computer 5 determines whether the drive select counter C3 is "0", and if it is '0', the process proceeds to step S6.
6, the computer 5 has an image data memory M,
The value obtained by adding the contents of the basic address counter C2 to the value representing the start address of the address pointer AP
Set to I. In this case, the contents of the counter C2 are 10. due to the initialization in step S3. Because it is,
The first address of the storage area 100 of the image data memory M1 is designated by the address pointer API.

After setting the address pointer API in step S6, the process proceeds to step S12.

In this step S12, the computer 5 drives the CD-ROM drives 6A, 6B, and 6C.

Select 6p. In this case, since the drive select counter C3 is "0", the computer 5 outputs data indicating "0." to the selection signal line of the bus line, thereby indicating that the CD-ROM drive device 6A is designated. Each CD-ROM drive device 6A, 6B, 6C, 6D
Normally, the data enable signal line of the pass line 7 is set to H, but the computer 5 is set to one C on the selection signal line.
When the D-ROM drive device is identified, the identified CD-R
The OM drive device sets the data enable signal line to "L".

In the next step S13, the computer 5 sets the one block counter C5 to '2048'.

In the next step S14, the computer 5 checks whether the data enable signal line of the pass line 7 is "Ln", confirms that the CD-ROM drive device 6A is in a state where it can send image data, and proceeds to step 915. .

In this step S15, the computer 5 outputs the read signal RD to the read signal line of the pass line 7, and in response, the CD-ROM drive 6A outputs 1-byte data to the data bus of the pass line 7. read.

In the next step 516, the 1-byte data read is saved at the address of the memory M1 indicated by the address pointer AP1.

In the next step S17, 1 is added to the address pointer, and in this case, 1 is added to API.

In the next step 818, 1 is subtracted from the one block counter C5, and in the next step S19, it is determined whether this counter C5 is "0", and if it is not "O", the process returns to step S15. Therefore step S1
The processes from Step 5 to Step S1B are repeated until the value of counter C5 becomes "0", that is, 2048 times.

Then, the computer 5 receives 2048 read signals RD.
By sequentially outputting , the 2048-byte R pixel data for the split screen a is introduced from the CD-ROM drive 6A one byte at a time, and this data is stored in the 2048 storage units in the storage area 100 of the memory M. When the data is saved one by one, the processing of the computer 5 proceeds to step S20.

In this step S20, the computer 5 adds 1 to the drive select counter C3, and proceeds to the next step 82.
In step S1, it is determined whether the counter C3 exceeds 3 degrees, and if it does not, the process returns to step S5.

In step S5, the computer 5 determines whether the drive select counter C3 is r□, and if it is other than "0", the process proceeds to step S7 and determines whether the counter C3 is "1". If the value is "1" as a result of the addition operation to the counter C3 in step 820, the computer 5 proceeds from step S7 to step S8 in order to introduce R pixel data for the next divided screen.

In step S8, the computer 5 adds the contents of the basic address counter C2 to the value representing the start address of the image data memory M, and sets the value in the address pointer AP2. In this case, since the content of the counter C2 is "rO", this means that the first address of the storage area 200 in the memory M has been designated.

Then, in the next step S12, the computer 5 outputs data indicating "14" to the selection signal line of the pass line 7 according to the contents of the drive select counter C3, and thereby the CD-ROM drive device 6B is activated. Subsequently, the computer 5 sets '2048.' in the one block counter C4 in step S13, and further checks the data enable signal line in step S14, and then proceeds to step S12. By repeating the above-described series of processes from step S15 to step 81B for the ROM drive device 6B, 2048 bytes of 3-pixel data for the split screen are introduced one byte at a time from the CD-ROM drive device 6B, and the memory M2 The data is saved one by one in 2048 storage units in the storage area 300 of .

After introducing 2048 bytes of 3-pixel data for the split screen, the computer 5 adds "1" to the drive select counter C3 and proceeds to step S21. In this case, the value of the counter C3 becomes "2" due to the addition, but since it does not exceed 3, the processing of the computer 5 returns to step S5.

In step S5, the computer 5 determines whether the drive select counter C3 is l"0", but here, since the value is "r2" due to the addition to the counter C3 in step S20, From step 87, the process further advances from step S9 to step S10.

In step 810, the computer 5 adds the contents of the basic address counter C2 to the value representing the start address of the image data memory M, and sets the value in the address pointer AP3. In this case, since the content of the counter C2 is "O2," this means that the first address of the storage area 300 in the memory M has been designated.

Then, in the next step S12, the computer 5 connects the drive select counter C to the selection signal line of the pass line 7.
According to the contents of 3, data indicating ``2'' is outputted, thereby specifying the CD-ROM drive device 6C. Subsequently, the computer 5 sets '2048.' to the one block counter C5 in step S13. J is set, and after checking the data enable signal line in step S14, the process advances to step S15. Then, the computer 5 repeats the above-described series of processes from step S15 to step S19 for the CD-ROM drive device 6C, thereby obtaining 2048 bytes of 3-pixel data for the split screen C from the CD-ROM drive device 6C. The data is introduced byte by byte and saved one by one in 2048 storage units in the storage area 300 of the memory M.

After introducing 2048 bytes of 3-pixel data for the split screen C, when the process proceeds to step S20, the computer 5 adds rl to the drive select counter C3 and proceeds to step S21. In this case, the value of the counter C3 becomes -13'' due to the addition, but since it has not yet exceeded 3, the processing of the computer 5 returns to step S5.

Then, the processing of the computer 5 is such that the value of the counter C3 is r2.
”, the process proceeds from step S5 to step 87 and further from step S9 to step S11.

In step S11, the computer 5 adds the contents of the basic address counter C2 to the value representing the start address of the image data memory M4 and sets the value in the address pointer AP4. In this case, since the content of the counter C2 is "0.", this means that the first address of the storage area 400 in the memory M has been designated.

Then, in the next step S12, the computer 5 connects the drive select counter C to the selection signal line of the pass line 7.
According to the contents of ``3'', data indicating ``3'' is outputted, thereby specifying the CD-ROM drive device 6D. Subsequently, the computer 5 sets '2048.' to the one block counter C5 in step S13. is set, and after checking the data enable signal line in step S14, the process advances to step S15. And the computer 5 is as follows:
By repeating a series of processes from step S15 to step S19 for the D-ROM drive device 6D, the CD
- 204 for split screen d from ROM drive 6D
The 8-byte 3-pixel data is introduced one byte at a time, and is saved one by one in 2048 storage units in the storage area 400 of the memory M.

After introducing 2048 bytes of R pixel data for the split screen d, when the process proceeds to step S20, the computer 5 adds "1" to the drive select counter C3 and proceeds to step S21. In this case, as a result of the addition, the value of the counter C3 becomes 14'', which exceeds 3, and the process then proceeds to step S22.

In this step S22, the computer 5 indicates 800 in hexadecimal in the basic address counter C2.
Add 0H.

In the next step S23, the transfer block counter C1
``1.'' is subtracted from the value, and in the next step S24, it is determined whether the value of the counter C1 is rO.

If the counter C1 is not "0.", the processing of the computer 5 returns to step S4. Therefore, by initializing the drive select counter C3, each CD-ROM drive device 6A, 6B, 6C.

Reading of pixel data from 6p is resumed.

In this case, the basic address counter C2 is
19, the computer 5 uses the address pointer API to sequentially address the 2048 storage units in the storage area 101 of the memory M, and adds the data from the CD-ROM drive 6A to 1. A total of 2048 bytes of R pixel data for split screen a, which is transferred byte by byte, is saved. Also, when reading the next 02048 bytes of R pixel data for the split screen from the CD-ROM drive 6B, the address pointer AP2 sequentially addresses the 2048 storage units in the storage area 201 of the memory M. The R pixel data transferred one byte at a time is saved one by one.

Similarly, when reading the next 2048 bytes of R pixel data for the split screen C from the CD-ROM drive 6C, the 2048 storage units in the storage area 301 of the memory M are sequentially addressed using the address pointer AP3. , here one byte at a time CD-ROM drive device 6C
Save the data sent from the CD-R
The following 20 for split screen d from OM drive device 6D
When reading 48 bytes of R pixel data, the address pointer AP4 sequentially addresses 2048 storage units in the storage area 401 of the memory M, and data is sent there one byte at a time from the CD-ROM drive device 6D. Save the data one by one.

When the transfer block counter C1 reaches "rO", that is, each R pixel data of the divided screens a, b, c, and d is read out in two bytes from each CD-ROM drive device 6A, 6B, 6C, and 6D, and is stored in the memory M1.

When the process of sequentially saving data in each of the 32 storage areas Ms, Ms, and M4 is completed, the computer 5 has introduced all R pixel data for one screen.

If it is determined in step S24 that the transfer block counter C1 is 10'', the process of the computer 5 proceeds to step S25. In this step S25, RG
"11" is subtracted from B counter C4.

Then, in the next step S25, "rl" is subtracted from the RGB counter C4, and in the next step S26, it is determined whether the counter C4 is 10". In this case, if 11'' is subtracted from the counter C4, r2. Therefore, the process advances to step S27, and the computer 5 next starts introducing 0 pixel data.

In this step S27, the computer 5 adds '40000H' indicating 40000 in hexadecimal to the basic address counter C2.

Then, the computer 5 repeats the above-described processing from step S4 in order to introduce 0 pixel data for the divided screens a, b, c, and d from the CD-ROM drives 6A, 6B, 6C, and 6D. At this time, in step S6,
In order to add '40000 HJ, which is the content of the counter C2, to the address pointer API of the computer 5, the 2048-byte 0 pixel data for the split screen a introduced from the CD-ROM drive 6A is stored in the memory M1. The data is saved one by one in 2048 storage units within the storage area 131. Further, when the process of step S8 is reached, the computer 5 subsequently adds '40000 HJ to the address pointer AP2.
The 2048 bytes of 0 pixel data for the divided screen introduced from the ROM drive device 6B are saved one by one in 2048 storage units in the storage area 231 of the memory M. Similarly, when it comes to the process of step S10 or step S11, the computer 5 sets the address pointer AP3 or the address pointer AP4 to '', respectively.
In order to add 40000H, the 0 pixel data of 2048 bytes for each split screen C and split screen d introduced from the CD-ROM drive device 6C or CD-ROM drive device 6D is stored in the storage area of the memory M. 331 or the storage area 431 of the memory M4.

Then, 2048 bytes of 0 pixel data for each of the divided screens a, b, c, and d are stored in the CD-ROM drives 6A and 6.
After reading from B, 6C, and 6D, the computer 5 again performs the process from step S4, and stores the next 2048 bytes of 0 pixel data for each of the divided screens a, b, c, and d in the storage area 132 of the memory M+. , memory M
, the storage area 232 of the memory M, the storage area 332 of the memory M, and the storage area 432 of the memory M, respectively.

In this way, the 0 pixel data of split screens a, b, c, and d are
Read 48 bytes at a time to memory Ml.

When the operation of saving each of M, ,M, ,M, is repeated 32 times, the computer 5 has introduced all 0 pixel data for one screen.

Then, the computer 5 performs steps S24 to S24.
25, and further from step S26 to step S27.
In step 327, '40000H' is added to the basic address counter C2. This starts the process from step S4, and then displays the split screen a.
, b, c, and d when CD-ROM drives 6A, 6B, 6C, and 6D are installed, 32 types of 8 pixel data of 2048 bytes each for split screen a are stored in the storage area of memory M1. 163-195, 32 of 2048 bytes each for split screen
The 8 pixel data is stored in memory areas 263 to 2 of memory M.
95, 32 types of 8 pixel data of 2048 bytes each for split screen C are saved in storage areas 363 to 395 of memory M, and 32 types of 8 pixel data of 2048 bytes each for split screen d. are saved in storage areas 463-495 of memory M4.

In this way, each R, G, B of the split screen a, b, c, d
When all pixel data of is stored in memories M and -M4,
The RGB counter C4 becomes "r□" and the operation of introducing one screen of pixel data from the CD-ROM drives 6A, 6B, 6C, and 6D of the computer 5 is completed.

(G) Effects of the Invention According to the present invention, pixel data for the same screen can be read from another CD-ROM drive device using the transfer suspension period of one CD-ROM drive device. Therefore, the time required to access one screen can be shortened.

[Brief explanation of the drawing]

FIG. 1 is a block diagram showing a schematic configuration of an embodiment of the present invention, FIG. 2 is a diagram illustrating the exchange of signals between a CD-ROM drive and a computer, and FIGS. 3 and 3A are A chart showing the timing at which a computer reads data from one CD-ROM drive; FIG. 4 is a chart showing the timing at which the computer reads data in the present invention; and FIG. 5 schematically shows memory storage. 6 is a diagram for explaining a counter included in the computer, FIG. 7 is a diagram for explaining division of one screen, and FIG. 8 is a flowchart for explaining the operation. 5... Funputer (processing device), 6A, 6B,
6C, 6D...CD-ROM drive device.

Claims (1)

[Scope of Claims] 1. In an image playback device in which a CD-ROM drive device plays back images based on image information read from a CD-ROM disk, a plurality of the CD-ROM drive devices are installed together, and each CD-ROM drive device plays back an image based on image information read from a CD-ROM disk. - A CD from which the ROM drive reads data.
- Pixel data for each divided screen of one screen is recorded on the ROM disk, and multiple pixel data are recorded within a predetermined time standardized for the CD-ROM drive to transfer one block of data. An image reproducing device comprising: a processing device for sequentially introducing pixel data for each divided screen of the same screen from the CD-ROM drive device.