Path: utzoo!utgpu!jarvis.csri.toronto.edu!cs.utexas.edu!mailrus!accuvax.nwu.edu!tank!eecae!cps3xx!usenet
From: usenet@cps3xx.UUCP (Usenet file owner)
Newsgroups: comp.binaries.ibm.pc.d
Subject: Re: Formatting 720K disks to 1.44 Megs HELP!!
Keywords: disks, media
Message-ID: <5592@cps3xx.UUCP>
Date: 30 Nov 89 03:30:24 GMT
References: <1989Nov27.212809.7241@ux1.cso.uiuc.edu> <1114@bridge2.ESD.3Com.COM> <89Nov28.223437est.19733@me.utoronto.ca>
Reply-To: hendrick@frith.UUCP (Kenneth J. Hendrickson)
Organization: Michigan State University
Lines: 68

In article <89Nov28.223437est.19733@me.utoronto.ca> yap@me.utoronto.ca (Davin Yap) writes:
>FYI, the difference between 1.44 and 720K disks is much less than that
>between 1.2 and 360K disks.  The coercivity of the media is as follows:
>Disk Type      Coercivity (oersteds (sp?))
>360K             300
>720K             600
>1.2 Meg          600
>1.44 Meg         700

I am not familiar with the units of oersteds, or the idea of coercivity.
I do know, however, that a very important parameter is #bits/area.

Consider: we are all familiar with the increase in quality of audio or
video tape when the tape travels at a faster speed.  The reason for this
is that a larger AREA is used to store the information.  Higher
signal/noise ratios, and wider frequency response are achieved using
this technique.

Now, lets compare areas.

The area of a 5+1/4" disk has an area of 21.65 sq. in., with a center
of radius 1.1" that is not available for use.  The area of this center
section is 3.80 sq. in.  Thus the total area of a 5+1/4" disk which is
available for use is 17.85 sq. in.

The area of a 3.5" disk is 9.62 sq. in., with a center of radius 0.8"
that is not available for use.  The area of this center section is 2.01
sq. in.  Thus the total area of a 3.5" disk which is available for use
is 7.61 sq. in.

Some thoughts:

Size    Area(sq.in.)    Bytes/sq.in.    $/byte (mail order)
----    ------------    ------------    -------------------
360k       17.85         20.2 * 10^3    .694 * 10^-6 (@$.25)
1.2M       17.85         67.2 * 10^3    .417 * 10^-6 (@$.50)
720k        7.61         94.6 * 10^3    1.04 * 10^-6 (@$.75)
1.44M       7.61        189.2 * 10^3    1.28 * 10^-6 (@$1.85)

The 360k media is the most mature technology.
The 1.2M media is the most affordable.
The 1.44M media is the most expensive.

The minimum quality of a 1.2M disk should be 3.33x better than the minimum
quality of a 360k disk, since there is 3.33x the data stored on the same
area.

The minimum quality of a 1.2M disk only needs to be 0.71x as good as the
minimum quality of a 720k disk, since there is 1.67x the data stored on
2.34x the area.

The minimum quality of a 1.44M disk should be 9.38x as good as the
minimum quality of a 360k disk, since there is 4x the data stored on
0.43x the area.

The minimum quality of a 1.44M disk should be 2x better than the minimum
quality of a 720k disk, since there is 2x the data stored on the same
area.

The minimum quality of a 1.44M disk should be 2.79x better than the
minimum quality of a 1.2M disk, since there is 1.2x the data stored on
0.43x the area.

You can figure out the rest.  I think at this point I might be rambling.

In the rare case that original ideas   Kenneth J. Hendrickson    N8DGN
are found here, I am responsible.      Owen W328, E. Lansing, MI 48825
Internet: kjh@pollux.usc.edu           UUCP: ...!uunet!pollux!kjh