Weekly Classic Computer Trivia Question (20141205)

[Chuck Guzis]

On 12/06/2014 10:48 AM, Johnny Billquist wrote:

> Actually, the 1.44MB floppy holds 1,474,560 bytes.
> 512 bytes per sector, 18 sectors per track, 80 tracks per side, and two
> sides... Where did you pull the 1440*1024 from???

These are primarily PC distinctions.  The HD 3.5" disk is sold with a 
nominal raw capacity of 2.0MB.  In other words, if you assume a 300 RPM 
spindle speed and 500KHz clock, a track will hold about 12,500 bytes 
using MFM.  Drives are specified for 80 (double-sided) cylidners, so you 
get exactly 2,000,000 unformatted bytes on a disk.  So far so good.

But a "360K" 5.25" disk is formatted in PC terms as tracks of 9 sectors 
of 512 bytes or 720 sectors total or exactly 360 * 1024 bytes.  So far 
so good.

Now, if we extend this to the "1.2MB" disk with 15 sectors of 512 bytes 
per track, we run into a problem.  Counting only one side, we have 
15*512*80 = 600,000 bytes.  Making this double-sided is 1,200 * 1024, 
which was 1.2 * 1,000 * 1024--a mongrel.  I prefer to call this 
1200K--and it's 3.5" cousing 1440K--and indeed some Microsoft utilities 
access this--but the world is what it is.

The drive, btw, is no more 1.44MB than is the format.  Various 
manufacturers have placed more or less on the floppies, dependent on 
format and spindle speed.

I take hard drive capacities as stated as more-or-less; not exact.  It's 
been the case since the manufacturers started putting printed defect 
maps on the drives.

FWIW, sectors sizes of other than powers-of-two have been well-known for 
a long time.  CDC Cyber used 644 6-bit characters   The Zilog 
development system employed 132 bytes per sector on 8" floppies.  Older 
decimal-based machines often used sector lengths that were a multiple of 10.

--Chuck