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Physical sector location



 
 
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  #1  
Old November 5th 03, 02:49 PM
Nuno Magalhaes
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Default Physical sector location

If you have a given sector, say 5000th sector (logical), how can you
get the physical location in: platters, tracks and sectors?
  #2  
Old November 5th 03, 04:58 PM
Maxim S. Shatskih
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This severely depends on particular hard drive vendor, and is usually the
non-disclosed information.

--
Maxim Shatskih, Windows DDK MVP
StorageCraft Corporation

http://www.storagecraft.com


"Nuno Magalhaes" wrote in message
om...
If you have a given sector, say 5000th sector (logical), how can you
get the physical location in: platters, tracks and sectors?



  #3  
Old November 5th 03, 08:21 PM
Arne Joris
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Nuno Magalhaes wrote:
If you have a given sector, say 5000th sector (logical), how can you
get the physical location in: platters, tracks and sectors?


Most scsi devices support the Translate Address diagnostic page (0x40).

You can access this by using scu for example
(http://www.bit-net.com/~rmiller/kits/scu/scu.html) :

scu show diagnostic pages

Diagnostic Pages Supported by Device [1/1/0] (RZ26L):

Supported Diagnostics Page (Code = 0x00)
Translate Address Page (Code = 0x40)

scu translate address block lba 9999 to bfi summary
Logical Block 9999 - Cylinder 38, Head 4, Bfi 2254

scu translate address block lba 9999 to physical

Translate Address Diagnostic (Page 0x40):

Page Code: 0x40
Page Length: 10
Supplied Format: Logical Block
Translate Format: Physical Sector
Alternate Track of Medium: No
Alternate Sector of Medium: No
Reserved Area of Medium: No
Cylinder Number: 9
Head Number: 4
Sector Number: 47

Arne Joris

  #4  
Old November 5th 03, 10:57 PM
Malcolm Weir
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On Wed, 05 Nov 2003 19:21:20 GMT, Arne Joris
wrote:

Nuno Magalhaes wrote:
If you have a given sector, say 5000th sector (logical), how can you
get the physical location in: platters, tracks and sectors?


Most scsi devices support the Translate Address diagnostic page (0x40).

[ Snip ]

scu translate address block lba 9999 to physical

Translate Address Diagnostic (Page 0x40):

Page Code: 0x40
Page Length: 10
Supplied Format: Logical Block
Translate Format: Physical Sector
Alternate Track of Medium: No
Alternate Sector of Medium: No
Reserved Area of Medium: No
Cylinder Number: 9
Head Number: 4
Sector Number: 47


Note that some drives don't bother to provide "real" values, just CHS
values that approximate reality, e.g. by using an average number of
sectors per track across the whole drive. The results are thus
indicative, not perfect!

Arne Joris


Malc.
  #5  
Old November 6th 03, 03:21 PM
Nuno Magalhaes
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(Nuno Magalhaes) wrote in message . com...
If you have a given sector, say 5000th sector (logical), how can you
get the physical location in: platters, tracks and sectors?


I've measured disk access times from sector 0 (track 0) to the start
of other tracks and got this results:

from to access time (ms)
0..126 - 1.3568 (track 3)
0..189 - 2.0449 (track 4)
0..252 - 2.7442 (track 5)
0..315 - 3.4373 (track 6)
0..378 - 4.1318 (track 7)
0..441 - 4.8222 (track 8)
0..504 - 5.5197 (track 9)
0..567 - 5.3046 (track 10)
0..630 - 4.6257 (track 11)
0..693 - 5.3169 (track 12)
0..756 - 6.0181 (track 13)
0..819 - 8.9897 (track 14)
0..882 - 9.6792 (track 15)
0..945 - 10.3711 (track 16)
0..1008 - 1.8192 (track 17)
0..1071 - 2.5135 (track 18)
0..1134 - 3.1956 (track 19)
0..1197 - 3.8969 (track 20)
0..1260 - 4.5939 (track 21)
0..1323 - 5.2856 (track 22)
0..1386 - 5.9845 (track 23)
0..1449 - 6.6760 (track 24)
0..1512 - 7.3691 (track 25)
0..1575 - 8.0571 (track 26)
0..1638 - 8.7536 (track 27)
0..1701 - 9.4473 (track 28)
0..1764 - 10.1306 (track 29)
0..1827 - 10.8335 (track 30)
0..1890 - 11.5330 (track 31)
0..1953 - 1.1275 (sometimes 12 ms) (track 32)
0..2016 - 3.6675 (track 33)

What should this mean? That there are 16 platters and the sectors are
enumerated from the inner cylinder to the outter cylinder?

Because I want to measure disk rotation times (to generate random
numbers according to a theorem) and this seems strange.
  #6  
Old November 6th 03, 04:44 PM
Arne Joris
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Nuno Magalhaes wrote:
(Nuno Magalhaes) wrote in message . com...

If you have a given sector, say 5000th sector (logical), how can you
get the physical location in: platters, tracks and sectors?



I've measured disk access times from sector 0 (track 0) to the start
of other tracks and got this results:

from to access time (ms)
0..126 - 1.3568 (track 3)
0..189 - 2.0449 (track 4)
0..252 - 2.7442 (track 5)
0..315 - 3.4373 (track 6)
0..378 - 4.1318 (track 7)
0..441 - 4.8222 (track 8)
0..504 - 5.5197 (track 9)
0..567 - 5.3046 (track 10)
0..630 - 4.6257 (track 11)
0..693 - 5.3169 (track 12)
0..756 - 6.0181 (track 13)
0..819 - 8.9897 (track 14)
0..882 - 9.6792 (track 15)
0..945 - 10.3711 (track 16)
0..1008 - 1.8192 (track 17)
0..1071 - 2.5135 (track 18)
0..1134 - 3.1956 (track 19)
0..1197 - 3.8969 (track 20)
0..1260 - 4.5939 (track 21)
0..1323 - 5.2856 (track 22)
0..1386 - 5.9845 (track 23)
0..1449 - 6.6760 (track 24)
0..1512 - 7.3691 (track 25)
0..1575 - 8.0571 (track 26)
0..1638 - 8.7536 (track 27)
0..1701 - 9.4473 (track 28)
0..1764 - 10.1306 (track 29)
0..1827 - 10.8335 (track 30)
0..1890 - 11.5330 (track 31)
0..1953 - 1.1275 (sometimes 12 ms) (track 32)
0..2016 - 3.6675 (track 33)

What should this mean? That there are 16 platters and the sectors are
enumerated from the inner cylinder to the outter cylinder?


Have you averaged these access times ? Rotational latency can create some pretty
bad one-shot access times : let's say you want to start reading block 0 and it
just passed by the head on the platter, the disk will have to wait for the
platter to spin 359 degrees before it can start reading block zero.
So you should average at least 5 attempts per block range.
Also, were you doing reads or writes or both ?

Because I want to measure disk rotation times (to generate random
numbers according to a theorem) and this seems strange.


Isn't the disk rotation constant and determined by the model, eg. 10000 RPM ?
I don't know if you can use disk access times as input for a theoretical model;
the disk's firmware usually has all kinds of exotic optimisations to cut corners
on access times wherever it can, which can cause wildly varying (and
unpredictible) access times


Arne Joris

  #7  
Old November 6th 03, 05:26 PM
Bill Todd
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"Nuno Magalhaes" wrote in message
om...
(Nuno Magalhaes) wrote in message

. com...
If you have a given sector, say 5000th sector (logical), how can you
get the physical location in: platters, tracks and sectors?


I've measured disk access times from sector 0 (track 0) to the start
of other tracks and got this results:

from to access time (ms)
0..126 - 1.3568 (track 3)
0..189 - 2.0449 (track 4)
0..252 - 2.7442 (track 5)
0..315 - 3.4373 (track 6)
0..378 - 4.1318 (track 7)
0..441 - 4.8222 (track 8)
0..504 - 5.5197 (track 9)
0..567 - 5.3046 (track 10)
0..630 - 4.6257 (track 11)
0..693 - 5.3169 (track 12)
0..756 - 6.0181 (track 13)
0..819 - 8.9897 (track 14)
0..882 - 9.6792 (track 15)
0..945 - 10.3711 (track 16)
0..1008 - 1.8192 (track 17)
0..1071 - 2.5135 (track 18)
0..1134 - 3.1956 (track 19)
0..1197 - 3.8969 (track 20)
0..1260 - 4.5939 (track 21)
0..1323 - 5.2856 (track 22)
0..1386 - 5.9845 (track 23)
0..1449 - 6.6760 (track 24)
0..1512 - 7.3691 (track 25)
0..1575 - 8.0571 (track 26)
0..1638 - 8.7536 (track 27)
0..1701 - 9.4473 (track 28)
0..1764 - 10.1306 (track 29)
0..1827 - 10.8335 (track 30)
0..1890 - 11.5330 (track 31)
0..1953 - 1.1275 (sometimes 12 ms) (track 32)
0..2016 - 3.6675 (track 33)

What should this mean?


Ignoring the slightly anomalous figures for 'tracks' 11 through 13, it
appears to mean that there is one recording surface, that a single-track
seek takes around 0.5 ms., that your disk rotates at 5400 rpm, and that
track 0 on the disk contains a bit under 1000 sectors (which kind of sounds
like it might be a Seagate Barracuda with track 0, as usual, the outermost
track).

- bill



 




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