🚀 go-pugleaf

[ 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.