On Tue, 28 Mar 2023, Ali via cctalk wrote:
https://auctions.c.yimg.jp/images.auctions.yahoo.co.jp/image/dr000/auc0
312/users/4f92de2852282d0c4055f15836cd43f760275f36/i-img1200x675-
1672137143s8ehmb271085.jpg
Looking at this picture it indicates the disk is 406TPI which is
significantly higher than the 96TPI used with 1.2MB HD drives.
Also from Wiki: 2TD drive in NEC PC-88 VA3 3 1⁄2-inch TD 1988 ca. 9.3
MB 12.5 MB[13] (13 MB unformatted)
3.5" disks for "720K" and "1.4M" are 135TPI. 80 tracks.
An exception was the Epson Geneva PX-8, which was 67.5TPI, with 40 tracks.
YES, that disk has about three times the track to track density of a
"normal" 3.5" disk. With three times as many tracks, using otherwise
identical parameters of "2.9M", it would have three times the capaciy.
They could also have tracks further in and/or further out, to have
slightly more than 24 tracks.
Also, of course, beware of how many bytes are in a "megabyte".
A "MebiByte" is 1,048,576 bytes. (2^20, or 1024*1024)
Non-computer people often call 1,000,000 a "megabyte" (10^6)
"1.4M" disks (1,474,560 bytes of data / 1.474 SI Megabytes / 1.40625
Mebibytes) are often called "1.44M", because that number is derived from
1,024,000 bytes per "megabyte" (2^10 * 10^6, 1000 * 1024), giving 1.44.
I can find no defensible reason for that corrupted size for a "megabyte".
Therefore, I call them "1.4M" (1.40625), NOT "1.44M"
--
Grumpy Ol' Fred ci...@xenosoft.com
