[geeks] Seagate buckles to math ignorant consumers

[John Francini]

Do a "df -B 512".

On a true 1-gigabyte unit, the capacity should read 2,097,152 blocks.
On a true 2-gigabyte unit, the capacity should read 4,194,304 blocks.
On a true 4-gigabyte unit, the capacity should read 8,388,608 blocks.

Etc.

John





On 2 Nov 2007, at 15:50, Francois Dion wrote:

> On 11/2/07, Shannon Hendrix <shannon at widomaker.com> wrote:
>> On Nov 2, 2007, at 10:53 AM, Francois Dion wrote:
>>
>>> On flash, they already are. Check df -h on a 256MB or 2GB usb stick,
>>> you will be unpleasantly surprised. The 2GB usb stick I'm giving  
>>> away
>>> on my blog, with Belenix pre-loaded is showing:
>>> /dev/dsk/c5t0d0s0      1.8G   833M   961M    47%    /media/JD  
>>> FIREFLY
>>
>> Oh please, think for a moment.
>>
>> That's due to filesystem overhead.
>>
>> You should know better.
>>
>> This is a geeks list after all.
>
> That's why I also listed rmformat's output.  This is raw, without
> filesystem overhead. One might expect 5% overhead, but i'm showing
> clearly 10% with df of which half is due to the manufacturer calling a
> key 2GB but raw space is 1.9GB.
>
> Similarly, my 320GB usb hard disk shows as:
>  1. Logical Node: /dev/rdsk/c1t0d0p0
>         Physical Node: /pci at 0,0/pci1028,1c1 at 1d,7/storage at 1/disk at 0,0
>         Connected Device: MAXTOR S         9QF00FYG D
>         Device Type: Removable
>         Bus: USB
>         Size: 305.2 GB
>         Label: <Unknown>
>         Access permissions: Medium is not write protected.
>
> And my 4GB flash as:
>      2. Logical Node: /dev/rdsk/c5t0d0p0
>         Physical Node: /pci at 0,0/pci1028,1c1 at 1d,7/storage at 7/disk at 0,0
>         Connected Device: LEXAR    JD FIREFLY       1100
>         Device Type: Removable
>         Bus: USB
>         Size: 3.8 GB
>         Label: <None>
>         Access permissions: Medium is not write protected.
>
> The loss is even more impressive on my 500.
>
> As far as the size, in rmformat it is calculated based on the struct
> smmedium_prop_t, specifically:
>
> int32_t sm_blocksize;	/* Medium block size in bytes */
> int32_t sm_capacity;	/* Medium capacity in no. of blocks */
>
> And then displayed in human readable format (/1024 for MB and one more
> time for GB). So 320,000,000 ends up as 305GB and 2,000,000 ends up as
> 1.9GB.
>
> Francois
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