28-bit LBA limit question
 

Win98SE is limitted to 28-bit LBA drives. I understand that the
28-bit LBA limit is 137.4Gb.

My hard drive is a 160Gb Samsung Spinpoint 1614N which is a bit over
149Gb binary.

Question: Is the 137.4Gb limit of 28-bit LBA metric or binary? I
suspect it is metric. Am I right? Trying to work out exactly how
much to leave unallocated at the end of the drive.

Re: 28-bit LBA limit question
 

On Tue, 28 Jun 2005 21:03:44 +1200, GraB wrote:

> Win98SE is limitted to 28-bit LBA drives. I understand that the
> 28-bit LBA limit is 137.4Gb.
>
> My hard drive is a 160Gb Samsung Spinpoint 1614N which is a bit over
> 149Gb binary.
>
> Question: Is the 137.4Gb limit of 28-bit LBA metric or binary? I
> suspect it is metric. Am I right? Trying to work out exactly how
> much to leave unallocated at the end of the drive.

(2 ^ 28) * 512 bytes per sector = 137,438,953,472 bytes

Does that help?

Re: 28-bit LBA limit question
 

On Tue, 28 Jun 2005 21:03:44 +1200, GraB <grab@whatever.co.nz> wrote:

>Win98SE is limitted to 28-bit LBA drives. I understand that the
>28-bit LBA limit is 137.4Gb.
>
>My hard drive is a 160Gb Samsung Spinpoint 1614N which is a bit over
>149Gb binary.
>
>Question: Is the 137.4Gb limit of 28-bit LBA metric or binary? I
>suspect it is metric. Am I right? Trying to work out exactly how
>much to leave unallocated at the end of the drive.

137 GB = 128 GiB