A typical LBA size is (going to be) 4KB, a multiplier of 32 is not universal. You could use ceil(LBA_size_in_bytes/16), but it sometimes leads to odd multipliers, and still there is no contiguous LBA processing with other ugly sizes. -Laszlo
> -------- Original Message -------- > Subject: LBA to I mapping > From: "Colin Sinclair" <[EMAIL PROTECTED]> > Date: Wed, December 14, 2005 12:09 pm > To: <[EMAIL PROTECTED]> > > There is still some confusion on this mapping: > > The minutes (and Rob just now) said "offset by 64K" which to me means > > I = 2^16 + LBA*32 + (0..31) for 512 byte sectors > and I = 2^16 + LBA*33 + (0..32) for 520 byte sectors > > but Laszlo said: > > "C. I also agree with the proposal to count > the cipher blocks as I = LBA*2^16+(0,1,2...)." > > which is clearly different, and contiguous sector processing is not possible. > > I'm guessing the former was the intention. > > > Rob said: > > I would prefer that I increment without knowledge of the underlying > > block size; let the unaligned devices take the hit of a full adder, > > since they already have to stall to do the CTS (as you mentioned). > > The LBA size is required to be known with whatever scheme is chosen, because > to get the initial I > value for a random access sector you have to do a multiply of LBA * > sector_size_in_blocks. My > mapping approach which works for any sector size avoids an awkward multiply > by 33 for odd-sized > sectors. It allows a contiguous increment for consecutive nice-sized sectors > as you desire, but > makes odd-sized sectors only slightly more painful (reset bottom n bits to > zero and add 2^n). > > > LBAs in both SCSI logical units and ATA devices start at 0. > > This is important. I am slightly worried that I values of 0 are not allowed > (since T is then 0) and > I think this is perhaps why the current spec defines a key scope starting at > 1. The offset by 64K > proposal eliminates this problem; with my mapping scheme I would just set the > MS bit of I (add > 2^127), which shouldn't interfere with the 30 or so LBA bits at all: > > I = 2^127 + LBA<<n + (0,1,2...j-1), where n = ceil(log2(j)) > > Colin.
