How do I compute the capacity of a hard disk?

Question

I have a worked example of how to compute the capacity of a hard disk, could anyone explain where the BOLD figures came out of?

RPM: 7200

no of sectors: 400

no of platters: 6

no of heads: 12

cylinders: 17000

avg seek time: 10millisecs

time to move between adj cylinders: 1millisec

the first line of the answer given to me is:

12 x 17 x 4 x 512 x 10^5

I just want to know where the parts in bold came from.The 512 I dont know. The 10 is from the seek time but its power 5?

Solution

The answer is

heads x cylinder x sectors x 512 (typical size of one sector in bytes)

so this is

12 x 17000 x 400 x 512

which is the same as

12 x 17 x 1000 x 4 x 100 x 512

and

100 = 10^2
1000 = 10^3
10^2 x 10^3 = 10^5

As you want the capacity, you don't need any times here.

A reference for the 512 bytes can be found at Wikipedia, for example (and it also has a similar example with the same formula a bit below).

Comments

[Dijar]

Was pretty lucky that I saw the question almost as it appeared - I'll wait a bit on the next one so you can take it ;)

[Andrew]

+1 with caveat, I think heads should really be the platters x 2 - although normally you would expect 1 head per side of a platter it isn't guarenteed.

[Dijar]

See the Wikpedia link: "Naturally, a platter has 2 sides and thus 2 surfaces on which data could be manipulated; usually there are 2 heads per platter--one on each side, but not always." So I guess we can assume that and as the number of heads is given, I chose to use that one.

[Dijar]

But of course you're right that it's still an assumption, as well as the sector size of 512 bytes which will perhaps change to 4096 bytes in the near future.