> > Quick simple puzzle: why would a drive that steps both disks at the same
> > time be significantly SLOWER when DISKCOPY'ing a disk than using separate
> > drives?
> > Hint: MUCH faster when copying if you write software specifically for
> it.
On Thu, 23 Aug 2001, Pete Turnbull wrote:
> Because it will read some number of tracks, stepping inward as it does so,
> then have to step back to the correct track to start writing. Example:
> read tracks 0 to 19 (20 in total, 19 steps), step back 19 tracks, write 20
> tracks (19 steps again); repeat. Total number of steps to copy n tracks is
> just less than 3n. If you want to read the tracks back to verify them,
> life is even worse, as the total is almost 5n.
>
> On the other hand, if you do it one track at a time: read a track, write
> that track, then step forward one track, the total number of steps to copy
> n tracks is n-1 steps. And still only n-1 if you verify each track as you
> go.
It is further compounded by the single stepper! When it switches to the
"other" drive, it finds out that the drive is NOT on the track that it
remembers it being on. So, ... It has to "recalibrate" by going to track
0 and counting in again EVERY time it switches "drive".
That can, of course be fixed by doing your own software that is optimized
for that kind of drive.
--
Grumpy Ol' Fred cisin_at_xenosoft.com
Received on Thu Aug 23 2001 - 20:00:25 BST