On Tue, Dec 24, 2002 at 09:21:23AM -0800, jw schultz wrote:
> On Tue, Dec 24, 2002 at 10:18:52AM +0100, Roy Sigurd Karlsbakk wrote:
> > keep in mind that only around half of the seek time is because of the
> > partition! Taking an IBM 120GXP as an example:
> >
> > Average seek: 8.5ms
> > Full stroke seek: 15.0ms
> > Time to rotate disk one round: 1/(7200/60)*1000 = 8.3ms
>
> I'm afraid your math is off.
>
> The rotational frequency should be 7200*60/sec which makes
> for 2.31 us which would produce an average rotational
> latency of 1.16us if such a condition even still applies.
> My expectation is that the whole track is buffered starting
> from the first sector that syncs thereby making the time
> rotfreq + rotfreq/nsect or something similar. In any case
> the rotational latency or frequency is orders of magnitude
> smaller than the seek time, even between adjacent
> tracks/cylinders.
>
> If the the stated average seek is 50% of full stroke and not
> based on reality then 76% of the cost of an average seek is
> attributed to distance and likewise 87% of the cost of a
> full. Based on that i'd say the seek distance is a much
> bigger player than you are assuming. If it weren't the
> value of elevators would be much less.
No. Your math is correct. Mine is upside down. Don't know
where that came from. Apologies for the bad smell.
-- ________________________________________________________________ J.W. Schultz Pegasystems Technologies email address: jw@pegasys.wsRemember Cernan and Schmitt - To unsubscribe from this list: send the line "unsubscribe linux-kernel" in the body of a message to majordomo@vger.kernel.org More majordomo info at http://vger.kernel.org/majordomo-info.html Please read the FAQ at http://www.tux.org/lkml/
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