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Partition problem how to solve it?


zillah

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This one works, sorry:

http://www.win.tue.nl/~aeb/partitions/

About the multiple systems, it is perfectly safe IF it is accurately planned beforehand.

Read here how I like to setup systems:

http://www.msfn.org/board/index.php?showtopic=34761

http://www.msfn.org/board/index.php?showtopic=33964

The above is just my personal opimion, the only thing I won't recommend is having more than one Primary partitions viewable by DOS/Win9x systems, as it is known to create problems.

jaclaz

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Thanks jaclaz

When I read the thread that you involved in it, I felt that you encourage people to use Ranish Partition,,,Am I right?

Is there any guide line to show me how to use it ?

My second query:

now I want to do backup,,,by using Ghost,,,,ghost was unable to do backup, because it was complaining there was error in partition,,,this is what happend to me more than once(not only one time).

What should I do?

Thanks

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Well, yes, I like Ranish's because:

1) it is freeware

2) lets you SEE what you are doing

3) lets you FINE TUNE every aspect of partition table

It has it's drawbacks too:

1) it is not "EASY", you have to know what you are doing

2) you need to boot from a DOS floppy to use it

In my opinion, the BEST utility is ACRONIS.

I don't like Partition Magic as in some configurations can mess up things.

More precisely:

PM2 worked like a charm

PM3 was the BEST edition (no drives larger than 8Gb, though)

PM4 was CRAP

PM5 was so and so (incompatibilities with some systems, namely ME)

PM6 was CRAP

PM7 was so and so

PM8 cannot say as I don't use it anymore

I never tried it, but there are good reports about BootIt NG.

The guide of Ranish's is self contained, there is a "Partition Primer" and some Faq's linked from the homepage:

http://www.ranish.com/part/primer.htm

http://www.ranish.com/part/faq.htm

and you could get to the discussion group:

http://groups.yahoo.com/group/partman

Yes, LOTS of reading.

About "repairing" a Partition that has errors, best option is to use the

Checkdisk tool that comes with every Operating System.

You could also try to test the partition with FSUTIL:

http://www.microsoft.com/resources/documen...-us/fsutil.mspx

You should anyway backup the data BEFORE.

I suggest you to backup the data TWICE, 1st by simply copying it to CD's from your Operating System, and 2nd making a digital image of that partition, if GHOST does not work, use another utility, you can try with dd under LINUX or it's Win32 ports:

http://uranus.it.swin.edu.au/~jn/linux/rawwrite/dd.htm

or

http://gnuwin32.sourceforge.net/packages/fileutils.htm

or

http://users.erols.com/gmgarner/forensics/

Once you have an IMAGE file, try doing all the repairs on the IMAGE, not on the original.

You can use this program:

http://chitchat.at.infoseek.co.jp/vmware/vdk.html

to mount the RAW image

(eventually using my little GUI to it):

http://home.graffiti.net/jaclaz:graffiti.n...ts/VDM/vdm.html

jaclaz

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Thanks for these details

In my opinion, the BEST utility is ACRONIS

I used Acronis Bootable CD to check partitions on my laptop,,,it shows me red cross on Linux partitions,,,,as I told you before these parttions have been created by using magic partition,,,,

Now do you thing this error on these partitons because I used Magic Parttition to create them? As I told you before I experienced same problem many times!!!!!

Do I need to worry about this problem, or I can leave it as its?

With my Acronis CD there is no option for delete,,is this normal ?

Thanks

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Now do you thing this error on these partitons because I used Magic Parttition to create them?
Yes, it is possible, as I said it happened to me that PM dealt with partitions in some "non-standard" way.

Can't say though if this applies to you.

It does not say "what" it has found as bad?

Try having a look at them with Ranish, and post results.

Do I need to worry about this problem, or I can leave it as its?

Cannot say, usually the good old "if ain't broken don't fix it" applies.

jaclaz

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@zillah

Here is what #116 error means.

The hard-disk partition table contains two inconsistent descriptions of the
partition's starting sector. This error can occur if the operating system reports a
hard-disk geometry that is different than the geometry in use when the partition
table was written. Possible causes of the hard-disk geometry changing are:
• Different operating systems (for example, DOS and OS/2) report different
hard-disk geometries.
• You boot from a diskette that loads a different driver than is loaded when you
boot from the hard drive.
• Upgrading the operating system (for example, from OS/2 2.x to OS/2 Warp)
causes a different driver to be used.
• The hard drive or controller has been changed.
• The BIOS has been upgraded.
• The BIOS LBA setting has been changed.
• There is a partition table virus present on the hard drive.

I wouldn't using Acronis or other tools in your case, because latter it can bring you other problems.

And remember, it's bad idia to mix different tools to create partition.

One more, mostly you have had all these errors from PM because you not fixed main error.

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  • 3 weeks later...

Hi jaclaz

If you look to this:

Note that MBR occupies one sector at cylinder 0, side 0, sector 1 and partition starts on the cylinder 0, side 1, sector 1. The 62 sector gap between them was left unused, because we want all partitions to start at the cylinder boundary or, at least, on the side boundary. This is not required with LBA, but we need to follow this rule in order to make happy old software (MS-DOS for example).
from :

]http://www.ranish.com/part/primer.htm]

My query:

He mentioned side 0 (Heads 0), and side 1 (Head 1), he did not mention how many tracks on each head (side),,,why did he say the 62 sector gap? this will be true if the partition starts on cylinder 0, side 0 (instead of 1), sector 63,,,Am I right?

Any clue

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Well, when talking about this matter it is VERY easy to get confused.

CHS is the old way (but still simpler) to address sectors.

I'll try to explain it as best as I can:

1) Track 0 (zero) is the 1st sector in the hard disk (512 bytes long)

2) It's address is CHS 0 0 1 or LBA 1

3) In track 0 is stored the MBR (or Master Boot Record)

4) The MBR is composed of two main parts, see here for reference:

http://therdcom.com/asm/mbr/MBR_in_detail.htm

the first part is the boot (more properly IPL Initial Program Loading) code, whilst the second (the last 66 bytes) are the 64 byte Partition table plus the 55AA "signature bytes".

5) All FDISK tools will default the first partition entry to CHS 0 1 1

6) supposing that the disk (like almost all of them) has 63 sectors per head, this leaves the 62 sectors gap you mention:

Total of sectors on Cilynder 0 Head 0 Sector 1 = 63 - 1 sector (taken by MBR) leaves 62 of them.

Track as used in point one above is actually not a proper term, unless it is used as Track0 , which it is the only one that we know about.

No one, apart from disk manufacturers, actually know how many tracks are actually on the disk, to say it better, track is a name for a physical thing, let's call it a "magnetic groove", of which there is a certain number on the disk's surfaces.

This unknown number of physical tracks are identified and numbered, by the disk on-board controller, in an unknown way, then this physical address is "translated" to the "outside" (the BIOS) as consecutive 512 bytes SECTORS.

Finally this address can be interpreted either as CHS or LBA.

Actually on a disk there are quite a few more "tracks" than the ones reported from the controller.

This is because on ALL disks there is a certain number of defective tracks, that are marked as bad by the manufacturer "LOW-LEVEL" utilities.

I hope that the above does help.

jaclaz

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Thanks

Please correct me if I am wrong:

In each side(head) of Disk there are numbers of Tracks, within each tarck there is number of sectors (normaly 63 sectors),,,,Am I right?

Now the URL that I gave you he mentioned that the MBR on side0 (head0) and the partition table on side1 (head1), in this case he can not say gap 62 sectors because both of them not on the same side (head),,,correct me if I misunderstood.

Second:

1) Track 0 (zero) is the 1st sector in the hard disk (512 bytes long)
Why did not you say that Track 0 or Track 1 or Track 20 each of them has 63 sectors, the 1st sector in Track 0 is for MBR? Correct my understanding if I am wrong

Third:

6) supposing that the disk (like almost all of them) has 63 sectors per head,

Why did you assume that the disk consist of sectors only? what about the tracks?why did not you say : that the one head of disk instead of disk only, has n number of tracks and in each track there is 63 sectors?

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In each side(head) of Disk there are numbers of Tracks, within each tarck there is number of sectors (normaly 63 sectors),,,,Am I right?
No, sorry,

let's see if I can explain it better:

on each side (head) there is a determinate area (physical space) where data can be stored.

PHYSICALLY this area is made of TRACKS (i.e. magnetic grooves).

LOGICALLY this same area is addressed as SECTORS.

You cannot say how many physical tracks correspond to how many logical sectors, as this is managed by the Internal Disk Controller.

The only thing that you can say is that each side(head) has in the example 63 sectors.

Why did not you say that Track 0 or Track 1 or Track 20 each of them has 63 sectors, the 1st sector in Track 0 is for MBR? Correct my understanding if I am wrong

Sorry again, there is no known (fixed) relationship whatsoever between Tracks and Sectors, see above, the only thing you know is that on the first Track (track 0) there must start the first Sector (sector 1).

Why did you assume that the disk consist of sectors only? what about the tracks?why did not you say : that the one head of disk instead of disk only, has n number of tracks and in each track there is 63 sectors?
See previous answers.

I don't want to complicate things further, but you must understand that the internal PHYSICAL geometry of the hard disk can be very different from the LOGICAL (addressing) one(which is the one you manage).

From the Ranish's Partition Primer:

When IDE (Integrated Drive Electronics) disks came out they had a little processor on each drive. This helped to free up CPU time by implementing more sophisticated set of commands. The disk space was also used more efficiently. Engineers had placed more sectors on the outer tracks, but still provided software writers with a convenient "cubical" look of the disk by doing internal translation of CHS (cylinders, heads, sectors). For example my old 340M disk has only two platters = 4 heads (sides), but it reports 665 cylinders, 16 heads (sides), and 63 sectors. In reality it, probably, has more then 4*63 sectors on each outer track and a little less than 4*63 on the most inner tracks, but we could never know for sure.

jaclaz

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Thanks jaclaz, the reason for saying this in my previous response:

In each side(head) of Disk there are numbers of Tracks, within each tarck there is number of sectors (normaly 63 sectors),,,,Am I right?.
link:

http://biology.ncsa.uiuc.edu/library/SGI_b...#LE21428-PARENT

As shown in Figure 1-2, a ring on one surface is called a track. Each track is divided into disk blocks. Sometimes called sectors, these physical blocks on a disk are different from filesystem blocks.

And in the book "Operating System concept" by Silberschatz page 33:

The surface of platter is logically diveded into circular tracks, which are subdivede into secotrs."

also see this picture:

http://www.pctechguide.com/04disks_Construction.htm

I am looking to your comment just to clarify me.

Thanks

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Yes, if you have a look at your second link:

http://www.pctechguide.com/04disks_Construction.htm

When a disk undergoes a low-level format, it is divided it into tracks and sectors. The tracks are concentric circles around the central spindle on either side of each platter. Tracks physically above each other on the platters are grouped together into cylinders which are then further subdivided into sectors of 512 bytes apiece. The concept of cylinders is important, since cross-platter information in the same cylinder can be accessed without having to move the heads. The sector is a disk's smallest accessible unit. Drives use a technique called zoned-bit recording in which tracks on the outside of the disk contain more sectors than those on the inside.

It better explains what I am trying to.

Let's try with a fictional (very simplified) hard disk platter (single platter with just one side).

Imagine it like a shooting range target (circular concentric grooves or tracks).

Let's say the disk is 10 cm diameter and each groove is 1 cm thick, and let us assume that the central axle needs a 2 cm in diameter spacer to hold the plate.

This leaves space for (10-2)/2=4 cm = 4 "tracks", let us number them as track 0 to 3 from the outer to the inner.

The outer track (track 0) will have a mean radius of (10-1)/2=4.5 cm, track 1 will have 4.5-1=3.5 cm, track 2 will have 3.5-1=2.5 cm, the inner track 3 will have 2.5-1=1.5 cm radius.

Consequently, mean length of each track (let's simplify Pi to 3.1416) will be:

Track 0 -> 2*4.5*Pi= 28.27 cm

Track 1 -> 2*3.5*Pi= 21.99 cm

Track 2 -> 2*2.5*Pi= 15.71 cm

Track 3 -> 2*1.5*Pi= 09.42 cm

------------

Total linear length = 75.39 cm

Supposing (due to magnetic density of the media) each 512 byte needs exactly 1 mm (0.1 cm) of linear length, maximum theoretical capacity of the disk "should" be 75.3/0.1=753 sectors * 512 bytes= 385536 bytes.

But as the sector (512 bytes) is the minimal unit, you must see how many sectors will fit in each track:

28.27/0.1=282

21.99/0.1=219

15.71/0.1=157

09.42/0.1=94

---------------

Total 752 sectors

which is the "real" maximum capacity of the disk.

The manufacturer will reserve some sectors, either for internal use or for substituting (in case) some bad sectors, let's say they keep to themselves 52 sectors.

The capacity of the drive (the one on the label) or the one seen either by BIOS or by some disk utility will be 700 sectors or 700*512=358400 bytes.

You cannot say HOW this capacity will be addressed to by the Disk INTERNAL controller:

Manufacturer A could arrange it as follows:

Track 0 260 sectors (+22 reserved)

Track 1 200 sectors (+19 reserved)

Track 2 150 sectors (+7 reserved)

Track 3 90 sectors (+ 4 reserved)

translated on the outside as 7 Cylinder by 100 sectors (Logical address)

Whilst

Manufacturer B could arrange it as follows:

Track 0 280 sectors (+2 reserved)

Track 1 210 sectors (+9 reserved)

Track 2 140 sectors (+17 reserved)

Track 3 70 sectors (+ 24 reserved)

translated on the outside as 10 Cylinder by 70 sectors (Logical address)

So you cannot say WHICH translation occurs, and you have no way to actually link a LOGICAL address to a PHYSICAL one.

Hope the above makes it clear.

jaclaz

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I like this discussion because it will rich my knowledge

I will go thorug what you said, and if I have any unclear thing , I will let you know.

But quickly I am going to ask about few things just to be clarified

Let's say the disk is 10 cm diameter and each groove is 1 cm thick
Do you mean by groove track?

If so, then I guess you have to do (10-2)/1=8 ?

Could you please let me know if you agree with this concept or not?:

And in the book "Operating System concept" by Silberschatz Fifth Edition page 33:

The surface of platter is logically diveded into circular tracks, which are subdivede into secotrs."

Yes, if you have a look at your second link:

http://www.pctechguide.com/04disks_Construction.htm

When a disk undergoes a low-level format, it is divided it into tracks and sectors. The tracks are concentric circles around the central spindle on either side of each platter. Tracks physically above each other on the platters are grouped together into cylinders which are then further subdivided into sectors of 512 bytes apiece. The concept of cylinders is important, since cross-platter information in the same cylinder can be accessed without having to move the heads. The sector is a disk's smallest accessible unit. Drives use a technique called zoned-bit recording in which tracks on the outside of the disk contain more sectors than those on the inside.

My understanding of the above is still:

That each track is consisting from sectors,but with this technique (zoned-bit recording), the outside track contain more sectors than those on the inside. But for most disk drives-may for simplicity-they use same number of sectors for inner and outer tracks.

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If so, then I guess you have to do (10-2)/1=8 ?
Well, no, inthe (absolutely hypothetical example above) 10 cm is diameter, central space diameter is 2 cm, track groove width is 1, but it must be compared with radius (i.e. diameter/2):

ASCII CROSS SECTION :

|1cm|1cm|1cm|1cm|1cm        |*|1cm        |1cm|1cm|1cm|1cm| =10cm
|Tr0|Tr1|Tr2|Tr3|half spacer|*|half spacer|Tr3|Tr2|Tr1|Tr0|
Here is the actual center   |*| of the platter (axis)

That each track is consisting from sectors,but with this technique (zoned-bit recording), the outside track contain more sectors than those on the inside. But for most disk drives-may for simplicity-they use same number of sectors for inner and outer tracks.

Yes, correct, but personally I never found ANY "non zone bit recorded" hard disks. Maybe VERY early models had constant sectors per track.

See here also:

http://www.storagereview.com/map/lm.cgi/zone

Since all modern hard disks use ZBR and don't have a single number of sectors per track across the disk, they use logical geometry for the BIOS setup. IDE hard disks up to 8.4 GB usually tell the BIOS 63 sectors per track and then translate to the real geometry internally; no modern drive uses 63 sectors on any track, much less all of them. Hard drives over 8.4 GB can't have their parameters expressed using the IDE BIOS geometry parameters anyway (because the regular BIOS limit is 8.4 GB) so these drives always have 63 sectors per track as "dummy" geometry parameters, and are accessed using logical block addressing.

jaclaz

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