partition sizes on flash drive


N

nutso fasst

New PNY 1GB USB2 flash drive: 979MB FAT partition.
Repartitioned/formatted in Win2K Disk Management: 972MB FAT partition.
Repartitioned/formatted in Win98 w/DOS fdisk/format: 977MB FAT partition.

I can use SYS.COM in Win98 to make any of the above bootable as a USB ZIP
drive.

In WinXP the option to remove the existing partition is not available. I
tried the HP tool for creating bootable flash drives (google HP
SP27213.exe). This resulted in a 979MB FAT partition, same as a new drive.
(The tool did NOT create a bootable drive, however - I could not get it to
boot even after reformatting it and copying Win98 boot files with SYS.)

Anyway, that's a 7MB variation in capacities with different partitioning
tools. Why?

nf
 
Ad

Advertisements

A

Arno Wagner

Previously nutso fasst said:
New PNY 1GB USB2 flash drive: 979MB FAT partition.
Repartitioned/formatted in Win2K Disk Management: 972MB FAT partition.
Repartitioned/formatted in Win98 w/DOS fdisk/format: 977MB FAT partition.
I can use SYS.COM in Win98 to make any of the above bootable as a USB ZIP
drive.
In WinXP the option to remove the existing partition is not available. I
tried the HP tool for creating bootable flash drives (google HP
SP27213.exe). This resulted in a 979MB FAT partition, same as a new drive.
(The tool did NOT create a bootable drive, however - I could not get it to
boot even after reformatting it and copying Win98 boot files with SYS.)
Anyway, that's a 7MB variation in capacities with different partitioning
tools. Why?

Probably some C/H/S emulation differences and the number of
wasted sectors (because they do not make up a full cylinder)
varies.

Arno
 
N

nutso fasst

Arno Wagner said:
Probably some C/H/S emulation differences and the number of
wasted sectors (because they do not make up a full cylinder)

Seems reasonable for 979 and 977MB partitions, but only 972MB (1,019,625,472
bytes) from W2K is hard to figure. Thx for reply.

nf
 
E

Eric Gisin

nutso fasst said:
Seems reasonable for 979 and 977MB partitions, but only 972MB (1,019,625,472
bytes) from W2K is hard to figure. Thx for reply.
The default CHS geometry for 2K/XP is heads=255 sectors=63, or 8MB per cylinder.
 
N

nutso fasst

Eric Gisin said:
The default CHS geometry for 2K/XP is heads=255 sectors=63, or 8MB per
cylinder.

That doesn't appear to be the case here. W2K partitioning gives
1,019,625,472 bytes. Sector size = 512 bytes. 1,019,625,472 / 512 =
1,991,456 sectors, which is not evenly divisible by 63 or 255.

All 3 partition sizes - the others being 2,001,600 and 2,006,464 512-byte
sectors - are divisible by 32. If all are 32 heads then CHS geometry doesn't
explain the size differences.

thx,
nf
 
N

nutso fasst

nutso fasst said:
All 3 partition sizes - the others being 2,001,600 and 2,006,464 512-byte

Oops, sorry, I wrote the wrong number for the FDISK partitioned drive. It's
2002368, not 2001600.

For the sake of further confusion, I just ran TeraByte's PARTINFW.EXE, which
reported 2002912 sectors for the FDISKed drive and 2007008 for the
'stocker'. LBA: 32; heads: 64; hidden sectors: 32. Even subtracting hidden
sectors, sector count > opsys reports. Total sectors are greater than opsys
numbers by 544, so 1991456 + 544 = 1992000, which IS divisible by 255 and
(63 * 32) - W2K default CHS.

Moral: capacity numbers reported by Windows are not exact.

thx,
nf
 
Ad

Advertisements

F

Folkert Rienstra

That doesn't appear to be the case here.
( No? Any reason why you cant look it up and say for sure?)

Not necessarily:
He didn't say anything about only full cylinders being used.
W2K partitioning gives 1,019,625,472 bytes.
Sector size = 512 bytes. 1,019,625,472 / 512 = 1,991,456 sectors,
which is not evenly divisible by 63 or 255.

Which makes the babblebot wrong, as always.
You can use the full capacity by using the max LBA value in the LBA parameter sections of the MBR. It depends on the apps doing the
partitioning and formatting whether those values will be put there in
the first place and whether they are subsequently used for formatting.
All 3 partition sizes - the others being 2,001,600 and 2,006,464 512-byte
sectors - are divisible by 32.

Which is significant how?
If all are 32 heads then CHS geometry doesn't explain the size differences.

Or whatever. Geometry *difference* is not the only cause.
Manufacturers wouldn't make drives with incomplete (logical) cylinders
if those cylinders could never be used. Thats what you have LBA for,
so the full capacity can be used independent of the CHS geometry values.
 

Ask a Question

Want to reply to this thread or ask your own question?

You'll need to choose a username for the site, which only take a couple of moments. After that, you can post your question and our members will help you out.

Ask a Question

Top