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Size of DOS 3.3 800k volumes?



I know they split into "drive 1" and "drive 2" each 400k.

but is it really exactly 400k?  by my calculations it is: 368640
bytes.

256 (bytes/sect) * 16 (sect/track)  * 97 (tracks/disk) = 368640
(bytes/disk)

do i have that wrong?  does this leave a gap between volumes or are
they separated by the "sides" of the disk?  if the disk is separated
by "sides" that would mean, since i must access it as a block device,
when I read block 0, i'm getting 2 sectors, side 0 sector 0 and side 1
sector 0 right?

anyone got any more format info on this?