China DOS Union

To All：

I remember seeing articles discussing low-density disks (360K) before, but they left out the most important part: the calculation process?

Can some expert give both the result and the calculation process?

This requires a very deep understanding of the physical structure of floppy disks, the partition format of floppy disks, the file storage mechanisms of DOS and Windows, as well as the sector allocation mechanism. I advise DOS newbies not to discuss this thankless and difficult problem.

You bastard! Even though I'm a newbie, I still want to reply!
Not trying to make a low-effort post, I just want to know the answer too, so I'll help you bump the thread!

9494
Everyone has the right to reply

Hehe, I really want to know too

Depends on the file size... If they're 0-byte files, then...

For a 1.44MB floppy, theoretically it is 1474560 bytes, a total of 2880 sectors. According to the allocation convention of the FAT12 file system, each cluster is 2 sectors. Even the smallest file has to occupy one cluster number, and once that cluster is allocated, other files cannot use it. That means each file occupies at least two sectors, so theoretically a 1.44 floppy can store at most 1440 files.

It's been so long that I've forgotten: the DOS boot record reserved area, the FAT tables, and the FDT table also take up the first few sectors :）

I remember that on a 360K floppy, the data area started from track 0, head 1, sector 4. For a 1.44M floppy, it's more than that.

I'm getting old, and can't even tune the strings right anymore

7680 single-byte filenames