GTHEALL@PENNDRLS.UPENN.EDU (George A. Theall) (03/28/91)
[ The following comes from Warner Losh. I expect it is of general interest so I'm posting it here, GAT ] I wondered that for a long time as well. My assumption was that there was less than 1" to record on. Until you asked the question again, I didn't even think about how there could be only 1" on a 5+" disk. So I got out a ruler just now and measured the disk to come up with the following crude diagram of a 5.25" floppy: +-----------------------------+ -+- -+- --- | --- | | -+- | --- .25 | | | | | | | | | | | | ~1.0 *** 2.625 | --- | | -+- | --- .25 | +-----+ | | -+- | --- | | | | 5.25 | -+- | | | | | 1.5 | +-----+ | | -+- | | | | | | +-----------------------------+ -+- If you take a ruler and measure the size of the opening for the disk, you will find that it about an inch long (measured from the ends of the rounding bevel on the opening). Since there is probably some slop at each end for the heads to seat on the disk, etc, 80 tracks seems about right if there are 96 tpi. This represents about .08 inches on each side. Now for space efficancy of this scheme. We have an inner radius of about 1" and an outer raduis of about 2", so the area that is recorded is about 9.4 sq. in. This is out of about 27.5 sq. in. the entire disk occupies. So only about 1/3 of the area of the disk is actually used to store data. Warner P.S. Feel free to post this if you like.