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09/11/2006, 12:23 AM
#19
 Originally Posted by +Kardboard+
Uh...in plain Engrish? hahah
Honestly, I'm not sure how plain this is. The first thing I wrote was an analogy using pie, but it made me hungry and concerned I might come off condescending. You'd be amazed how many people fly off the handle when they hear, "Well, it's like pie."
Anyway, file allocation blocks are the smallest size a file can be, and each takes some overhead for dealing with: the more chunks have to be moved, the longer it'll take, so one 32k chunk moves faster than four 8k chunks; additionally, the File Allocation Table from which FAT32 gets its name also has to account for every chunk, and sets aside space as a sort of table of contents. The more chunks you have, the more space you lose to the file allocation table.
(That's one reason your "120GB" hard drive might actually show up as "100GB," for example; the other is a marketting trick that has to do with the definition of "megabyte." 1/6th of your space would be a huge FAT (heh-heh) but I'm just making up numbers to serve as an example.)
However, there's a trade off with large chunks, because if you have 4k of data on 32k chunks, your 4k file actually occupies the full chunk, wasting 28k. If you have a 35k file, it takes two chunks, wasting 33k, etc. With larger drives a little waste is no big deal, but with smaller ones things get increasingly cramped. I wonder if there's an upper limit to chunk size as it relates to the total size ... but this is stuff I learned a long time ago and I'm rather fuzzy on some of the details.
Is there a way to check cluster size? I formatted it on a "default" setting in Windows, but no idea what that is.
Also from the command line, type CHKDSK DRIVE: (and add /F to do error-correction as well). It'll say "XXXX bytes in each allocation unit" near the bottom of the report.
This actually does reveal something useful, maybe: my 40GB drives have 4k allocation chunks, so 32k over 4GB is enormously wasteful. I'm not surprised that it doesn't even allow it for smaller disks.
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