Edited and amended from a previously posted reply, a mere 40 messages back....
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For those of you that "appear" to have the two-platter version of this drive, does the placard on your harddrive
match the placard that is shown in the
Tom's Hardware review of the WD2000JB harddrive - better secondary view of the
WD2000JB harddrive....
or does it more
closely match one of the following placards in these 'views' of the WD1200JB harddrive????
Placard type 1
Placard type 2
You will note in the second picture of the WD2000JB, that the serial number is
WMACK1032153. I've noticed that the serial number in all the WD1200JBs I've looked at, have a serial number that includes one numerical digit (eg.
WMA8C264xxxx). Now this is purely speculation, but what if the serial number decodes as follows:
W - Western Digital
MA - Malaysia
CK - production facility
1 - production run
032153 - individual unit number
If Western Digital is producing the JB Special Edition series harddrives at several locations, it's likely they would only shutdown
one production facility at a time, to retool - thus resulting in an overlap in manufacture dates between the two versions of the same drive. Again, this is just speculation, but possibly the serial number may in some way indicate which harddrives are the two-platter versions of the WD2000JB drive. And then again, I could simply be full of.... However, the current "focus" seems to be on using the LBA listed on the harddrive's placard, to determine 40GB versus 60GB platter versions of this drive....
Yet when you review earlier reviews of the WD1000JB and WD1200JB harddrives, both versions of the "placard" have shown-up - in reviews written well before the (speculated) release of the 60GB platter drives. Or were they....????
Here's an example of the placard believed to have been used on the 40GB platter drives (review dated 05 Mar 2002):
Western Digital WD1200JB With 8 MB Cache: Outperforms SCSI Drives - image 1 (appears to have LBA: 23437500; manufacture date 2001);
Western Digital WD1200JB With 8 MB Cache: Outperforms SCSI Drives - image 2 (LBA and manufacture date unreadable).
Yet this earlier review of the [then] new WD1000JB and WD1200JB hardrives have placards thought to be in use with the newest 60GB platters drives, which shows jumper settings for configuring the drive (note review dated 20 Nov 2001):
New Performers From Western Digital: 100 GB Hard Disk With 8 MB Cache, Plus A New 120 GB Drive - image 1 (LBA and manufacture date unreadable);
New Performers From Western Digital: 100 GB Hard Disk With 8 MB Cache, Plus A New 120 GB Drive - image 2 (LBA unreadable; manufacture date 2001).
But what truly makes me question whether LBA 234441648
actually designates the newer 60GB platter harddrives is the following image from this same November 2001 review (review dated 20 Nov 2001):
New Performers From Western Digital: 100 GB Hard Disk With 8 MB Cache, Plus A New 120 GB Drive - image 3. In this image you can clearly see the LBA number "
234441648", as well as the
12 Sep 2001 date of manufacture!. The placard on this harddrive
also appears to have the Western Digital model number
WD1200JB - 00CRA1 - although I am looking at these images on an Inspiron 15" LCD screen. Furthermore, in looking at an even older review of the WD1000BB, it too has the placard type that is being assocaited with the 60GB platter harddrives (review dated 26 Sep 01):
Western Digital WD1000BB - Taking Grand Prize! - image 1 (LBA for WD1000BB; 14 Jul 2002 date of manufacture).
Regardless, there are clearly examples of both placard types
*AND* LBA numbers appearing
WELL BEFORE Sep 2002.
And let's also look at this from a "logical" perspective. LBA refers to the physical locations on the harddrive's platter(s) or what is commonly referred to as "addresses" (see definitions below). So does it stand to reason that with fewer platters there should be
more address locations, or
fewer address locations? Not knowing "how" Western Digital defines the addresses on their harddrives, if they simply cataloged all their addresses (initially) by platter (eg. P1-0, P1-1, P2-0, P2-1, P3-0, P3-1), changing from a three-platter design to two-platter design would reduce the base addresses by 1/3 - which is a fairly large number. Such an addressing system is relatively inefficient when using very large harddrives, due to the unique cylinder-head-sector addresses requiring a larger number of parameters to be defined for each block address on the harddrive.
Now if we assume that the addresses are cataloged simply with some sort of numerical code (eg. 00000001, 00000002, 00000003, etc) corresponding to a section/sector of the harddrive/platter, the relative reduction in the number of physical addresses the harddrive would have to define would still be reduced, but not by near the amount resulting from the previous example using an inefficient addressing system. So if you're not using this "ineffificient" addressing method, why would there be any change at all - I mean 120GB is 120GB, right?
Well, yes - 120GB is 120GB, but the amount of physical space required to store 120GB is based on
both the space to store the data, and space used to separate the tracks of data. When a harddrive is "formatted", you lose a portion of the harddrive's capacity to two (2) primary things: 1) the space required to record how the drive is formatted; and 2) the space required to write the tracks the data will be written
within.
Let's say you take a piece of blank paper. Since you're horrible at writing in a straight-line when no lines are present, you take your #2 pencil and draw nice, thick, dark lines to write between and keep your lines looking straight. As you get to the end of filling-in that paper with information, you realize you're going to be about one line too short. So you resort to writing on-top of one or more of the last few nice, thick dark lines - knowing that even if part of a letter is unreadable because the cross-line on the "e" super-imposes on-top of the line (making it look like either an "e" or "c"), you will be able to figure out which letter it is, based on the adjacent letter(s) which combine to form the word (eg. "the" versus "thc"). Unfortunately you cannot do this with computer data, since the data is stored electronically as binary code of "1s" and "0" ("ONs"and "OFFs"). So a portion of the platter surface must be dedicated to defining the tracks within which the data are stored.
Basically, small electrical fields on the harddrive's surface arrange the magnetic particles on the platters surface to create this binary code which represents the written letters, numbers, and digits that make-up the words, text, and graphics stored on the harddrive.
So, if we were to try to graphically show how the tracks (exclamation points) and data (numbers and letters) are arranged, it would look something like this, with the "0s" representing track "zero", the "1s" representing track "one", the "2s" representing track "two", etc., and the exclamation points representing the "lines" between the tracks of data - something like this:
!0!1!2!3!4!5!6!7!8!9!
!0!1!2!3!4!5!6!7!8!9!
!0!1!2!3!4!5!6!7!8!9!
!0!1!2!3!4!5!6!7!8!9!
!0!1!2!3!4!5!6!7!8!9!
!0!1!2!3!4!5!6!7!8!9!
!0!1!2!3!4!5!6!7!8!9!
!0!1!2!3!4!5!6!7!8!9!
!0!1!2!3!4!5!6!7!8!9!
!0!1!2!3!4!5!6!7!8!9!
If we now take binary data from each of the three platters, where the binary data 0-9 are from platter number 1, A-J are from platter number 2, and a-j are from platter number 3, any given track could look like this:
!0!1!2!3!4!5!6!7!8!9!
!A!B!C!D!E!F!G!H!I!J!
!a!b!c!d!e!f!g!h!i!j!
If we count the number of exclamation points (which represent the track "boundaries"), we get 11 spaces lost on each platter to record 10 binary digits - or a total of 33 spaces lost on each platter to record 30 binary digits (52.38-percent lost). When we increase areal density to the two-platter level, we get the following:
!0!1!2!3!4!5!6!7!8!9!A!B!C!D!E!
!F!G!H!I!J!a!b!c!d!e!f!g!h!i!j!
Now if we count the number of exclamation points (which represent the tracks), we get 16 spaces lost on each platter to record 15 binary digits - or a total of 32 spaces lost on each platter to record 30 binary digits (51.62-percent lost). Thus increasing the areal density to the two-platter level results in an efficiency gain of 0.768-percent (0.00768) for the same amount of stored data.
The difference between LBA 234441648 and LBA 234375000 is 0.0000284283, or 0.00284283-percent. Not having *any* clue to the number of tracks, sectors, etc., (or how many address locations - "LBAs) would be gained through this efficiency, it should be expected that there would be a small, but measureable reduction in the LBA addresses that results from increasing areal density and the resulting reduction from three- to two-platter design. Hence it is my proposal, that the two-platter 60GB versions of the Western Digital WD1200JB are encoded with
LBA 23437500 - not 234441648 as has been reported.
Thus this should result in a relative insignificant - but none-the-less measureable - reduction in address locations. Therefore, there should be
fewer NOT
more LBA adresses with the reduction in the number of platters.
And I still don't buy the arguement that there is no performance difference between the two- and three-platters versions of this harddive.
It seems logical, that the higher areal density of the two-platter drives would equate to faster read/write speeds since the heads have less distance to travel in reading/writing each data bit, as well as in traveling between read/write tasks on separate areas of the disks. It also stands to reason, that mixing two- and three-platter versions within a raid-array, would create a performance hit, since the two drives could not read/write at the same relative rate due to the different relative positions of the read/write [heads, as well as the difference relative read/write] locations on the two different platters.... [edited text from prior post]
Definitions:
CHS (Cylinder, Head, Sector) Addressing ? An inefficient method of referencing the sectors on a harddrive as a collection of unique cylinder, head and sector addresses. Each block on the drive will have a unique cylinder, head and sector address.
LBA (Logical Block Addressing) ? A method of addressing the sectors on a harddrive. Addresses the sectors on the drive/platter as a single group of logical block numbers instead of cylinder, head and sector addresses. It allows for accessing larger drives than is normally possible with CHS addressing.
Reply question posted by: K-squared
Hey
KavMan - can you look to see if there is a major difference in the serial number (as listed externally on the retail box) for the four (4) harddrives? I'm curious as to whether the serial number listed on the bottom of the box would also identify the two- versus three-platter harddrives.....
Originally posted by: KavMan
I have 4 WD 1200JB's
1 - 24 Feb 2002 - LBA: 234375000 - Rev A00
2 - 09 May 2002 - LBA: 234375000 - Rev A00
3 - 16 Aug 2002 - LBA: 234441648 - Can't find a Rev # - This drive has pictures of the power connector and jumper settings that the other 3 doesn't have and also it has WD1200JB - 00CRA1 instead of ther other three that has WD1200JB - 75CRA0
4 - This is the drive Dell sent - 15 Oct 2002 - LBA: 234375000 - Rev A00