We've been through that already. His question is, why can't we use it with
2068? (The reason he asked being that more space is "wasted" with 512-byte
sectors due to sector mark overhead.)
The short answer is that most of the kernel assumes that disk sector sizes
are a power of 2 and therefore can be represented via bit shifting 2068
isn't a power of 2, so isn't eligible. 2048 might well be doable, though,
if you really wanted it....
| optimization to select the proper head, etc. Therefore I doubt
| that the "real" block-size is 2068. In practice, it's usually
| 512 or a multiple thereof.
+--->8
It has been suggested that they chose a format that makes the disk look as
big as possible. That it's not *usable* in that format (to my knowledge, no
other OS does 2068-byte sectors either) isn't significant to marketroids
pushed to "demonstrate" their numbers....
-- brandon s. allbery [os/2][linux][solaris][japh] allbery@kf8nh.apk.net system administrator [WAY too many hats] allbery@ece.cmu.edu electrical and computer engineering KF8NH carnegie mellon university
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