Where is sector 0?

[crysis]

Hi All,
after reading your very interesting articles, I would like ask 3 questions if I can:

- Where is located the sector 0?
- why sector 0 (low LBAs) has best performances, min access time and max transfer rate?
- Performing Read surface test, int the graph, these sudden vertical falls down of KB/Sec are the "Zone" starting?

I found another hd documentation where its written that the sector 0 (and lowest LBAs) is located on the outer track of the hd platter, because this:

"Data transfer rates for data that is on the outer tracks is generally 180 to 240% that of data that is on the inner tracks. This is due to zoned-bit recording (more densely packed sectors) and faster angular velocity of the disk in that region"

Can someone explain this?
I guess I understood that this position is exactly the OD "Zone 0" where the hd performs worst - highest BER and smallest TPI, right?
So, as for my understanding, sector 0 can NOT be in OD, because data located there have the worst performances, right?

Thanks for the attention

[fzabkar]

- Where is located the sector 0?
- why sector 0 (low LBAs) has best performances, min access time and max transfer rate?
- Performing Read surface test, int the graph, these sudden vertical falls down of KB/Sec are the "Zone" starting?

I found another hd documentation where its written that the sector 0 (and lowest LBAs) is located on the outer track of the hd platter, because this:

"Data transfer rates for data that is on the outer tracks is generally 180 to 240% that of data that is on the inner tracks. This is due to zoned-bit recording (more densely packed sectors) and faster angular velocity of the disk in that region"

Can someone explain this?

The angular velocity is constant. Zoned-Bit Recording (ZBR) divides the platter into zones in an attempt to maintain a constant linear bit density (BPI, bits per inch).

See this article:
http://hddscan.com/doc/HDD_Tracks_and_Zones.html

The discontinuities in the read performance curve do in fact correspond to the zone boundaries.

There are many examples here:
http://www.hdtune.com/testresults.html

I guess I understood that this position is exactly the OD "Zone 0" where the hd performs worst - highest BER and smallest TPI, right?
So, as for my understanding, sector 0 can NOT be in OD, because data located there have the worst performances, right?

AFAIK, the TPI (Tracks Per Inch) is a constant value for any particular head.

The data transfer rate at the outer zones is twice that of the inner zones. Why do you say that the performance is worst at the OD?

[BlackST]

Doesn't make sense, then, putting service areas at id , od or scattered on the surface... ? And that's what happens in real hdds. You find both situations.

[crysis]

Hi People
yes, Im trying my way out in hd understanding.. thanks for your answer!

So if I well understood your notes, sector 0 (the boot sector where is partition table) is really in the OD=outer track=Zone 0
Right?

Yes, I read the article HDD_Tracks_and_Zones.html, and it seems to me its written that the "Zone 0" has the highest BER (problem) and smallest TPI, the less Tracks Per Inch..
so few tracks there, and I imagine that few tracks means when the head reads in Zone 0, it reads few tracks..
so in my opinion thats was illogical, it doesnt make sense.. unless someone can tell me how possible reading fewest tracks results in "max transfer rate".. it should result in the MIN transfer rate, isnt it?

So, why "data transfer rate at the outer zones is twice that of the inner zones", why data transfer at OD (=Zone 0=physical outer tracks) is the max?

Again, anybody knows why sector 0 has the "min access time"?

Thanks

[drc]

Constant angular velocity = Higher linear velocity at larger radius, lower linear velocity at smaller radius

[Vulcan]

Yes, I read the article HDD_Tracks_and_Zones.html, and it seems to me its written that the "Zone 0" has the highest BER (problem)

No, that is not exactly what the article says. I suggest you read it carefully again - the BER is managed within acceptable limits, even in the OD e.g. by varying the write (and therefore read) frequency. That area does not have a "problem".

I imagine that few tracks means when the head reads in Zone 0, it reads few tracks..

That does not make sense in English - can you try to explain your thoughts again?

so in my opinion thats was illogical, it doesnt make sense.. unless someone can tell me how possible reading fewest tracks results in "max transfer rate".. it should result in the MIN transfer rate, isnt it?

You seem to be confusing number of tracks with transfer rate, but your description is unclear in English (e.g. mentioning "fewest tracks" makes no sense in relation to transfer rate), so it is not very clear what you are suggesting

So, why "data transfer rate at the outer zones is twice that of the inner zones", why data transfer at OD (=Zone 0=physical outer tracks) is the max?

Forum members drc and fzabkar have already explained this.

Again, anybody knows why sector 0 has the "min access time"?

I know what I believe that term means, but what exactly do you mean by "min access time"? Are you meaning transfer rate? Or seek time? Or what, exactly? Notice that this term is not in the article which you are quoting, so if you are asking for an explanation, you should explain exactly what question you are trying to answer.

[fzabkar]

AFAIK, the TPI (Tracks Per Inch) is a constant value for any particular head.

Sorry, you are right, I'm wrong.

In any case, AFAICS, if the tracks were more tightly spaced in one zone than in another, this would not affect the transfer rate since the track-to-track seek time would not be measurably different.

[Doomer]

Yes, I read the article HDD_Tracks_and_Zones.html, and it seems to me its written that the "Zone 0" has the highest BER (problem)

I already told you that you are mistaken
Why you keep insisting on this wrong guess?

[crysis]

Well
again thanks for you answers
Ok, my questions come from wikipedia, where I found:
"For hard drives, disk access time is determined by a sum of the spin-up time, seek time, rotational delay, and transfer time."
So, my special case is the OD (sector 0) vs. all other sectors on HD.
I guess the spin-up is almost irrelevant in the equation, constant.
About seek and rotational, I dont know, maybe you can tell me, are them constant or not?
About transfer time, that one looks very changeable.. in the special case of OD, its minimum, why?

Hope this time I asked better..
Thanks

[Vulcan]

my questions come from wikipedia

Thanks, so now it becomes clear...

where I found:
"For hard drives, disk access time is determined by a sum of the spin-up time, seek time, rotational delay, and transfer time."

I don't fully agree with that list - which is another reason why you shouldn't believe everything on Wikipedia, and why IMHO it's not reasonable to expect long explanations from other people, for information that you read elsewhere.

For example, spin-up time only applies in situations where the drive is not already at full rotational-speed (e.g. spun-down or slow RPM mode). For most data accesses, this is not a factor, since the drive is already up-to-speed & ready. There are also other factors which they have not included in that list. I could write lots on this subject, but since I believe that this whole thread is becoming increasingly pointless, I'll just reply to your questions below.

So, my special case is the OD (sector 0) vs. all other sectors on HD.

Sector 0 is not different than "all other sectors"! Its performance is no different than sector 1, for example. Your ideas about sector 0 seem to be rather confused

If you didn't mean sector 0, but instead the OD (zone 0) - then the OD (zone 0) is not much different from the next inner zone (zone 1). The performance changes are small (gradual) between zones. It is not that zone 0 = good performance; all others = bad performance.

Anyway, continuing the comparison that you were asking about:

I guess the spin-up is almost irrelevant in the equation, constant.

Yes, but see above - it's usually not relevant, except when the drive is spun-down or in power-saving slow RPM mode.

About seek and rotational, I dont know, maybe you can tell me, are them constant or not?

Both are effectively constant, on a given disk drive model, when averaged between I/Os to 2 random sector numbers. Obviously there are max & min values, for specific tests for both parameters e.g. sequential I/O is very different to random I/O & track-to-track seek time is much lower than full-sweep seeks etc. etc.

Since you did not understand this yourself, then IMHO you really do need to learn more by reading reference information, and not by asking questions. To use an analogy: What you are doing is equivalent to knowing nothing about surgery, and then asking a surgeon how the human body works, via a web forum - it is inefficient and frustrating, as you will never get a full, clear understanding that way! There are simply too many questions that you could ask I suggest that you research disk benchmarking information, to start your learning about the topic of disk performance and its measurement.

About transfer time, that one looks very changeable.

Ignoring effects of read caching etc., transfer time (i.e. time to read/write when the drive is actually transferring data from media to drive electronics) is not what I would call very changeable, as it is related exactly to the media data transfer rate. Therefore it only varies by a factor of approx 2, between ID and OD. You have already quoted that in your earlier postings, where you said:

Data transfer rates for data that is on the outer tracks is generally 180 to 240% that of data that is on the inner tracks.

So therefore it is not clear to me why you are describing this as "very changeable".

in the special case of OD, its minimum, why?

This has already been explained to you by drc and fzabkar, and in the webpage you quoted by doomer - shorter transfer time for sectors closer to the OD (compared to sectors closer to the ID), is the effect of ZBR (linear media speed under the head is faster at the OD).

However, in real life consumer situations, the increased transfer rate of the sectors near the OD, is less noticed, except for sequential I/O. The use of read & write caching (and some filesystem behaviours) helps to mask the performance of the underlying media. In random I/O, factors like disk seek times (which can also be affected by filesystem algorithms) are much more likely to have a noticeable effect on performance. That is why SSDs are becoming more popular.

Hope this time I asked better

I'm glad that you stopped saying zone 0 has a "problem". It is still not clear to me, why you seem to have this obsession with sector 0 or the performance of the OD of a disk - when in normal consumer situations (e.g. no short-stroking of heads), your data is all over the disk (or at least a partition), and you have relatively little control about exactly where it is written. Sector 0 itself, is very rarely read from.

I think you've had enough explanation from me. Good luck with your research.

[Doomer]

About transfer time, that one looks very changeable.. in the special case of OD, its minimum, why?

I guess you missed this - http://en.rlab.ru/forum/index.php?topic=1657.0

[crysis]

Yes Doomer, I saw it thanks Thanks Vulcan, I guess dummies like me always prefer to ask hints to the professionals, because if not they would hung around internet a while without understand nothing, if they are convinced about the wrong self guessings..