views:

92

answers:

1

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?

+6  A: 

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).

schnaader
beat me to it :)
BlueRaja - Danny Pflughoeft
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 ;)
schnaader
+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.
Andrew
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.
schnaader
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.
schnaader