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Sectors in disk

  1. May 12, 2009 #1
    Hi,

    This is probably a simple question, but not for me.

    Seagate ST3750640AS 750GB can provide maximum data flow from disk 624 Mbit/s. How many sectors are in outer zone tracks if average rotation latency is 4.16ms?

    I've no idea how to calculete this
     
  2. jcsd
  3. May 12, 2009 #2
    I think you'll need some more info:

    how much can be stored in a sector? (1bit, 10bit, 100Mbit??)

    Then, if you have 624 Mbit/s, how much could you read in one rotation (of the outer track) which lasts 4.16ms?
     
  4. May 12, 2009 #3
    Thank you for reply!
    So 1 sector= 512 bytes

    624:4160=0.15 Mbit - can read in one rotation

    0.15:8*1000=18.75KB

    18.75:0.512=36.61 - sectors in outer track
     
  5. May 13, 2009 #4
    hmmm, I get something different (try to explain what you are doing more to make things more clear)

    Ok, so,

    624Mbit/s = 624x10^6 bit/s
    4.16ms = 4.16x10^-3 seconds
    Therefore in one rotation the harddisk reads:
    624x10^6 bit/s x 4.16x10^-3 seconds = 2595840 bits (in one rotation)

    Now to convert bits to bytes...
    How many bits in a byte? (8 is normal, but i think in modern computing it is 10 bits (maybe 11, can't remember) in a byte if you include the trailing and initial bit. something like [info][01010101][trailing bit])

    Either way:
    2595840/10 = 259584 bytes
    or
    2595840/8 = 324480 bytes

    Then divide the number of bytes by your cluster size (512bytes) and you should get either:
    a)507 clusters (for a byte of 10bits)
    b) 633.75 clusters (for a byte of 8bits)

    Seeing that the clusters should be round numbers, I would round off answer b) to 634 clusters. What do you think?
     
  6. May 13, 2009 #5
    Since the old times, the number of sectors per track was constant. I just updated myself and found out that using ZBR (zone bit recording), the bit density remains more or less optimal throughout the disk by assigning different number of sectors/cylinder to different zones. The following link presents a detailed description of the techniques, and hence a better understanding of what information that is required to do the calculation.

    http://www.pcguide.com/ref/hdd/geom/tracks.htm

    Also, rotational latency refers to the delay required by the arm to switch from one track to the adjacent one. For the 7200 RPM drive, the time for one complete rotation is 60/7200=8.33 ms.
    The company claimes a SUSTAINED speed of 78 MB/s, equivalent to 624 Mbit/sec for an 8-bit byte. So the sustained read time per track is 8.33 ms - 4.16 ms = 4.17 ms.
    The rest of the calculations are essentially the same as Redargon had done, since 4.17 ms is almost identical to 4.16 ms used in the calculations. However, if the rotational latency had not been exactly half of the rotational time, the results could be different.
     
    Last edited: May 13, 2009
  7. May 13, 2009 #6
    @ Redargon
    I think your answers are more correct than my and both are valid (in my case).

    My problem is I just don't understand how to calculate data flow from disk (media transfer rate) or how to determine number of sectors in disk, ect.

    Coud it be fomula:[tex]w=\frac{n*512}{t}[/tex] where n - number of sectors, t- one rotation time, w- data flow from disk.

    I know this wouldn't work in Zoned Bit Recording as data flow from outer tracks is faster than from inner but I assume that in this task data flow is constant.

    Another similar task I can't calculate:

    Disk HITACHI GST ULTRASTAR 15K147 147GB provide maximum data flow from disk 1.128 Gbit/s. How many sectors are in outer zone tracks if rotation speed is 1 500 turn pre min?

    I guess:

    1500:60 = 250 - rotation pre sec
    1.128:250= 0.004512 Gbit can be read after one turn. It 's also 4.512 Mbit = 0.564 Mbytes(4.512:8=0.564)= 564 Kbytes in track

    1 sector= 512 bytes
    564:0.512=(approximately) 1102 sectors in outer track.

    Let me know if I am doing something awfully wrong.
     
  8. May 14, 2009 #7
    ok, you're on the right track. just one calc error:
    1500/60=25 rotations per second, not 250.
    then it would be 1.128/25=0.04512Gbit per rotation.
    0.04512Gbit= 45.12Mbit= 5.64MB (assuming 8bits per byte)= 5.64x10^6 Bytes (per rotation that can be read)
    so, then 5.64x10^6/512 = 11015.625 sectors, so around 11016 sectors

    oh, and I think your formula looks ok too. (just remember to keep your units consistent) If you use t, rotation time in seconds, then you will get an answer in bytes/s with that formula, as you are multiplying by 512, which is bytes per sector
     
  9. May 14, 2009 #8
    Last edited by a moderator: Apr 24, 2017
  10. May 14, 2009 #9
    yip, sounds more reasonable especially since it is a 15K147
     
  11. May 14, 2009 #10
    I'm very sorry for mistyping here, it is actually 15 000.
    15 000:60=250


    Thanks for helping me out!
     
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