| As far as the bits go:
1 tape inch 1600 bits 1 byte 1 block 1600
1 tape inch = X --------- X ------ X --------- = ------- blocks
tape inch 8 bits 512 bytes 8 X 512
which is approximately 200/500 = 0.4 blocks.
Unfortunately, the gaps on the tape are larger than the data bits,
so we must also factor in the number of record gaps and file gaps.
Does anyone know how long a record gap is (in inches)?
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| Assuming one uses BACKUP to write the tape (e.g., for software distribution),
and one sticks to rather conventional defaults, then at 1600 bpi:
1 foot ~= 32 blocks
For 6250 bpi the density is approximately 85 blocks/foot.
If your data consists primarily of a large number of small files, then
the density will be somewhat less, but that isn't a concern for distribution
purposes. There's already some fudge (0.7) in the factors above which you'll
see if you do the same arithmetic I did.
Some notes of interest --
"bpi" is a linear measure. For tapes, capacity is an area and things being
what they are, 1600 bits/inch works out to be 1600 Bytes/inch of storage.
Likewise for 6250.
BACKUP writes data onto the tape with a blocking factor of 8 KBytes, regardless
of file format or file size. Inter-record gap is .4in for 6250 bpi and I think
its .75in for 1600 bpi. I could be wrong about the .75 but not by enough to
make a significant difference since 8KB is already 5+ inches.
The figures above make sense when compared against our advertising specs --
A TU77 @ 1600 bpi, 125 ips operates at a bandwidth of 200 KBytes/second and
can store 40 MBytes of data on a 2400 foot reel at 8 KBytes/block.
A TE16 @ 1600 bpi, 45 ips operates at a bandwidth of 72 KBytes/second and
can store 40 MBytes of data on a 2400 foot reel at 8 KBytes/block.
A TA78 @ 6250 bpi, 125 ips operates at a bandwidth of 781 KBytes/second and
can store 140 MBytes of data on a 2400 foot reel at 8 KBytes/block.
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| Don't forget to add the overhead for BACKUP's redundancy blocks! Every n
blocks (n=10, by default), BACKUP writes a block that is the XOR of the
n blocks just written. Hence, the available space on the tape is n/n+1 times
what you think (about 10% lower, with the default redundancy group).
-- Jerry
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| Updating my previous answer to allow for Jerry's point and the correct values
for interblock gaps (1600 --> .6in, 6250 --> .3in, thanx to Howard Kaikow),
we have:
Assuming one uses BACKUP to write the tape (e.g., for software distribution),
and one sticks to rather conventional defaults, then at 1600 bpi:
1 foot ~= 30 blocks
For 6250 bpi the density is approximately 108 blocks/foot.
If your data consists primarily of a large number of small files, then the
density will be somewhat less, but that isn't a concern for distribution
purposes. There's already some fudge in the factors above which you'll see if
you do the arithmetic below. One note of interest: "bpi" is a linear measure.
For tapes, capacity is an _area_ and things being what they are, 1600 bits/inch
works out to be 1600 Bytes/inch of storage. Likewise for 6250.
BACKUP writes data onto the tape with a blocking factor of 8 KBytes, regardless
of file format or file size. Every 10th block, BACKUP appends an 11th "XOR"
block for redundancy. We therefore have:
14.5454 user d_blocks = 16.00 data d_blocks = 8K bytes
(8K bytes) / (1600 bytes/in) = 5.12in. / t_block "data"
(5.12 in. / t_block "data") + (.6in "gap") = 5.72in. / t_block "net"
So: 14.5454 user disk blocks = 5.72 inches of tape
or: 1 inch = 2.54 blocks (1 cm = 1 block!!!)
or 1 foot = 30+ blocks
John
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