I see that dividing by 2352 results in a whole number, therefore the sector size is 2352. For example, I have a BIN image that is 509,713,680 bytes in size. The divisor used that results in a whole number is the sector size. If you just have the BIN image without a cue sheet, then take the file size (in bytes) of the BIN image and divide by 2048, 2336, and 2352, respectively. If you have a BIN+CUE, the cue sheet should tell you. However, you need to know the sector size of your BIN image, whether it's 2048, 2336, or 2352.
If you're working with a BIN image of Shining Force III Scenario 3, you can use a hex editor to extract the first 47 sectors of data out and save to a new file.
#SHINING FORCE 2 ROM BIN FILE ISO#
You may open it like any ISO with IsoBuster or CDmage and you'll be able to view the file system structure. SF3SCN3.ISO is exactly (47 sectors) x (2048 bytes/sector) = 96,256 bytes = 94 KB in size. Knowing that, I used CDRWin to extract sector 0 to sector 46 from the CD and saved it as MODE1/2048 data to a file which I called SF3SCN3.ISO. After sorting by LBA order in IsoBuster, I see that the first file starts at LBA 47.
To find out, you can use IsoBuster to view the BIN image or CD, then sort from smallest LBA to highest LBA by clicking on the LBA column.Ĭurrently, I have the Shining Force III Scenario 3 CD in my CD-ROM drive.
First you need to know where the file system ends and the actual first file data begins. I'm not sure if this is what you're looking for, but you can just extract the first x number of sectors out of the BIN image or from the actual CD to get the file system structure.
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