#### MORE INFORMATION

The correct size is determined by multiplying the number of tracks, sides,
sectors per track, and 512 bytes per sector, then subtracting the bytes
required to format the disk, and then dividing this figure by 1024. For a
"1.44-MB" 3.5-inch floppy disk, there are

80 tracks

18 sectors per track

512 bytes per sector

2 sides

Multiplying the above gives you 1,474,560 bytes. This is the unformatted
size.

To determine the number of bytes formatting requires, you need to know how
many bytes are used for the boot sector, file allocation table (FAT), and
root directory.

There is 1 sector used for the boot sector, which is 512 bytes; 18
sectors for the two FATs (9 sectors each), which is 9216 bytes (512 *
18 = 9216); and 14 sectors for the root directory, which is 7168
bytes.

NOTE: There are two ways to arrive at the 7168 number:

224 entries * 32 bytes per entry = 7168 bytes

512 bytes per sector (14 * 512 = 7168 bytes)

Adding these figures gives you 16,896 bytes.

Subtracting the amount used for formatting from the total unformatted size
gives you 1,457,664. (1,474,560 - 16,896 = 1,457,664 bytes)

Dividing the above figure by 1024 bytes generates 1440. (1,474,560 /
1024 = 1440 KB)

To convert to megabytes, divide by 1024. (1440 KB / 1024 = 1.406 MB)

This formula works for 1.2-MB disks as well. The only variable is the
number of sectors, which is 15, for the calculations with 1.2-MB disks.

From the calculations shown above, we can see that the 3.5-inch disk
considered to have 1.44 MB free disk space actually has 1.40 MB, and the
5.25-inch disk considered to have 1.2 MB actually has 1.17 MB.

The misunderstanding comes from the incorrect calculation below:

The calculation should be:

There are 1024 bytes in a kilobyte, not 1000.

Note that in Windows 95, the properties for a blank, formatted 3.5-inch
1.44-MB disk show that there are 1.38 MB of free disk space.