Weekly Classic Computer Trivia Question (20141205)

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