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How to calculate the capacity of a disk

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  1. Mar 23, 2017 #1
    A 2.5 inch diameter disk plate has 6 plates , 512 bytes per sector , 256 sectors , 5268 tracks per surface. What is the capacity of the disk in terms of Giga bytes ?

    In the solution of my book ..written:

    6 plates = 6x2-2 = 10 recording surfaces // I dont understand this part.

    Could you please explain this ?

    Ans: 6.43 GB
     
  2. jcsd
  3. Mar 23, 2017 #2

    jedishrfu

    Staff: Mentor

    The topmost and bottommost surfaces aren't used so for 6 plates you have 12 surfaces which amounts to 10 usable surfaces.
     
  4. Mar 23, 2017 #3
    answer given in the book :

    Disk capacity = 10x5268x256x512/2^30 = 6.43 GB

    what is the relation here in sector and track. ... I see they are multiplied ....I could not get this part
     
  5. Mar 23, 2017 #4
    The tracks are concentric circles on the surface of the disk, in this case there are 5268 of them.
    Each track contains discrete chunks of data, these are the sectors, there are 256 sectors per track in this case.
    The sectors each have a capacity of 512 bytes; read or write operations are from or to 1 whole such sector at a time,
    (even if only part of the 512 bytes contains relevant data).
     
    Last edited: Mar 23, 2017
  6. Mar 23, 2017 #5
    not clear. Could you please show me an image to visualize this concept ?

    Is not sectors looks like arc in circle ? or sectors lay in concentric circle i.e track ? unable to visualize this part.
     
  7. Mar 23, 2017 #6
    Think of it as a pages in a book.
    Each page has a lines, and you can only have so many lines on one page.
    and only so many words on one line
     
  8. Mar 23, 2017 #7
    sectors are like words in your example ? :smile:

    and lines are tracks ?

    Is that correct understanding ?
     
  9. Mar 23, 2017 #8
    More like sentences, but you got the right idea
     
  10. Mar 23, 2017 #9
    okay ..
    I dont see this being used in solution . Is this a useless information in this problem ?
     
  11. Mar 23, 2017 #10
    Yes, the physical size of the storage medium is not relevant, It only matters what is their capacity to store data.
    The earliest hard disk drives were very heavy devices (tens of Kg), they needed to be installed in a whole room by themselves with air cooling,
     
  12. Mar 23, 2017 #11
    awesome.

    thanks for your time.
     
  13. Mar 23, 2017 #12

    rbelli1

    User Avatar
    Gold Member

  14. Mar 23, 2017 #13
  15. Mar 25, 2017 #14
    The track is established by the mechanics. Once the read/write heads have been positioned, each head has access to all the information on the track - all 128Kbytes of it.

    But 128Kbytes is a pretty large block of storage. To use it efficiently, you want to allow it to be shared among many small files - and you want to be able to write data to one of those files without changing data for the other files. So the track is divided into sectors of a more convenient size. This division is done during disk formatting when special disk instructions are used to write whole tracks with data patterns. Those data patterns are then read during normal read and write operations so that the sectors can be properly recognized and targeted.
     
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