I keep reading conflicting things about this topic.
I wanted to make a dban disk wiping floppy. Some
people say use:
Code:
dd if=dban.IMG of=/dev/fd0 bs=512
others say
Code:
dd if=dban.IMG of=/dev/fd0 bs=18k
or even:
Code:
dd if=dban.IMG of=/dev/fd0 bs=1440
I would think the latter two would be much faster
because the 512 bytes would need about 2800
write events to do the whole 1.44 megs
but doing it in 18k or 1440 chunks would be faster.
But someone said that if the bs is too big
there might be a problem if the image is
really crammed to fit the floppy, you might lose
that last part? The reason I was wanting to not
use 512 is the drive seems to take forever using
it that way. If the exact byte size of the image
is:
Code:
ls -l
1474560 dban-beta.2006031900_i386.IMA
could I do this:
Code:
dd if=dban-beta.2006031900_i386.IMA of=/dev/fd0
bs=1474560 count=1
just write the whole thing in one big stream?
Mark