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Re: Re: [Qemu-devel] kqemu 0.6.2 and XP
From: |
Andreas Bollhalder |
Subject: |
Re: Re: [Qemu-devel] kqemu 0.6.2 and XP |
Date: |
Mon, 18 Apr 2005 23:32:33 +0200 |
I tried the math on my WinXP host with Win2k guest.
When using the following math with my 8192MB QCOW image, QEMU refuse
to start:
1 Block = 512 Bytes
Cylinders = Blocks / (Heads * Sectors)
(kBytes * 2) / (16 * 63)
(8388608 kBytes * 2) / (16 * 63)
For 8192MB Image: -hdachs 16644,16,63
I brute forced the cylinder value, QEMU will start with a lower or
equal value of 16383 cylinders.
I made my own math based on this:
Cylinders = Blocks / (Heads * (Sectors + 1)) - 1
(kBytes * 2) / (16 * 64) - 1
(8388608 kBytes * 2) / (16 * 64) - 1
For 8192MB Image: -hdachs 16383,16,63
I'm realy confused about what's right or not. I created the image with
the QEMU Manager. Can anyone explain this ???
Andreas
Ben Taylor wrote:
> I forgot one very important part. The calculation needs
> to be <img size in bytes> / (heads * sectors per cylinder
> * 512 bytes) = cylinders. so in your case, the calculation is close
> well, the default sector size is 512, and you've done
> your sizing in number of 1k blocks, your calculations was
> close. It should be:
>
> blocks secs/block hds sec/cyl
>
> (2 000 000 * 2 ) / ( 16 * 63 ) = 3968
> The first time I did the math, my 2g partition went
> to 8g partition when I used the -hdachs parameters
> cause I really screwed up the math. (it appears if
> you mess your c-h-s values, your image since may
> change. :-)
> HTH, and sorry for the confusion.
> Ben