How much disk space, in bytes, does each File Allocation Table copy consume?

Question

I am looking at a disk image and trying to determine how many bytes each FAT copy consumes I know that I can't just find it in the data I have to figure it out mathematically but I am not sure how to do this. Can someone explain to me the process please?

Answer 1

The actual FAT table is 4 bytes per entry. FAT32/8=4

The number of entries depends on the cluster size 512,1k,2k,4k,8k,16k,32k, and 64k may or may not be allowed.

((Total size /cluster size) + (3 reserved codes)) * 4 = 1 copy of FAT in bytes. Sectors are usually 512 bytes,sometimes 4k, and it is round up to the nearest 512/4k byte marker. Now multiply that by 2 as there is a backup FAT table.

Note: FAT tries to keep a copy in memory at all times, therefore microsoft has artificially limited its size to prevent memory exhaustion. 4 billion entries (theoretically possible) *4 bytes =16 billion bytes or 16gb of RAM just for the FAT table.