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Re: Disk Size

Re: Disk Size

2003-10-13 by Laurent Lemaire

Hi Peter,

Let me help you with this...

Your HD has got after (being formated) 433118 sectors of 256 Bytes.

433118 * 256 = 110878208 Bytes

110878208 / 1024 = 108279 Kilo-Bytes

108279 / 1024 = 105 Mega-Bytes

Your disk is the same as mines. It's a Maxtor XT1140
It's a 140MB unformatted hard disk.

Hope it helps.

	Laurent.

Re: Disk Size

2003-10-13 by Laurent Lemaire

Hi again Peter,

I forgot... You may have checked the capacity of the data
partition using the "FREE" command. But your HD has at least
4 partitions: 3 systems partitions (K0, K1, K2) of 1MB each
and the "C0" partition were your data are stored. To find out
your HD entire capacity, you need to check the "SC00" capacity
which represent the total of the four partitions. To do so :
# FREE /SC00

Please consult the kmi9000 web site for more details.

Hope it helps.

	Laurent.

Disk Size

2003-10-13 by Peter Connelly (Core Design Ltd.)

Hi guys,

I've exited to OS9 and typed FREE /C0 (or just FREE) and it mentions I have
a capacity of 433,118 sectors and 135,960 Free sectors. It mentions in the
manual each sector is 256 bytes and to divide by 4 to find number of
kilobytes. Ok, I've multiplied 433,118 by 256 then divided by 4 (27,719,552k
= 27, 069.875 Meg). Is this right or am I missing something? I thought I had
a 190 Meg HD???

Regards,
Peter

Re: [Fairlight-CMI] Disk Size

2003-10-13 by Colin Ross

Hi Peter
I've work it out as 110,878kb, it looks like its a 120 mb drive. !!!!!!
Regards
Colin


----- Original Message ----- 
From: "Peter Connelly (Core Design Ltd.)" <PeterC@Core-Design.com>
To: "Fairlight CMI - Yahoo Groups (E-mail)" <Fairlight-CMI@yahoogroups.com>
Sent: Monday, October 13, 2003 12:24 PM
Subject: [Fairlight-CMI] Disk Size


> Hi guys,
>
> I've exited to OS9 and typed FREE /C0 (or just FREE) and it mentions I
have
> a capacity of 433,118 sectors and 135,960 Free sectors. It mentions in the
> manual each sector is 256 bytes and to divide by 4 to find number of
> kilobytes. Ok, I've multiplied 433,118 by 256 then divided by 4
(27,719,552k
> = 27, 069.875 Meg). Is this right or am I missing something? I thought I
had
Show quoted textHide quoted text
> a 190 Meg HD???
>
> Regards,
> Peter
>
>
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