Approximating Performance For Different Environments - IBM C1B 1.12 Brick On Sled Carrier 128-pin HPC Specification Sheet

1.12/2.25 gb - 1.0" high 4.51 gb - 1.6" high 3.5 form factor disk drive version 5.0
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O E M FUNCTIONAL SPECIFICATION ULTRASTAR XP (DFHC) SSA MODELS 1.12/2.25 GB - 1.0" H I G H
3.3.2.1 Theoretical Data Sector Transfer Rate
This Rate does not account for time required for error recovery or defective sectors (the Typical Data Sector
Transfer Rate described in 3.3.1, "Basic Component Descriptions" on page 40 does include those effects).
Each group of cylinders with a different number of gross sectors per track is called a notch. The following
shows values for notch #1 of C4x models. The "Average" values used in this specification are sums of the
individual notch values weighted by the number of LBAs in the associated notches. For the other notches
and block lengths use values that correspond to those notches and block lengths.
Data Sector Transfer Rate
Bytes/cylinder
time for 1 cyl of data
time for track skews
time for 1 cyl skew
Data Sector Transfer Rate
Note: See 2.0, "Specifications" on page 11 for the descriptions of
tracks/cyl (trk/cyl)
gross sectors/track (gs/trk)
spares/cyl (b1spr/cyl and b2spr/cyl)
user bytes/sector (ub/sct)
gross bytes/sector (gb/sct)
See 3.5, "Skew" on page 46 for the descriptions of
track skew (tss)
cylinder skew (css)
Average sector times per notch can be calculated as follows:
average sector time (ast) =
1 sec
120.045 × gs/trk

3.4 Approximating Performance for Different Environments

Source filename=PERFORM
USER RESPONSIBLE F O R VERIFYING VERSION A N D COMPLETENESS
=
Bytes/cylinder
time for 1 cyl + track skews + 1 cyl skew
= {(tracks/cyl)(gross sectors/track) - spares/cyl}(user bytes/sector)
= {(16)(135) - 40}(512)
= 1,085,440 Bytes/cyl
= {(tracks/cyl)(gross sectors/track) - spares/cyl}(avg. sector time)
= {(16)(135) - 40}(.061705)
= 130.815 msec/cyl
= (tracks/cyl - 1)(track skew)(avg. sector time)
= (16-1)(13)(.061700)
= 12.032 msec/cyl
= (cylinder skew)(avg. sector time)
= (25)(.061705)
= 1.543 msec/cyl
=
1,085,440 Bytes
130.815 msec + 12.032 msec + 1.543 msec
= 7.517 MB/sec (Notch #1)
IBM Corporation
Page 43 of 87

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