From: Ian Primus
Sent: Wednesday, February 23, 2011 8:53 AM
According to sources online, a DecTape can hold 144k
of data on a PDP-10,
using 18 bit words. Assuming here that they're going by the base 10
definition of "K", as most media manufacturers are want to do, then the
following must be true:
"144K" is misleading for anything other than a PDP-11. Nobody else does
things in bytes, and even on -11's DECtape use is thought of in words.
DECtape as used on the PDP-10 (and earlier the PDP-6) stores 578 blocks of
128 36-bit words. DECtape as used on the PDP-4/7/9/15 stores 578 blocks of
256 18-bit words; on the PDP-11, the same format stores 256 16-bit words
per block, with hard zeroes in the high-order bits.
On the PDP-5, PDP-8 family, and PDP-12 in PDP-8 mode, DECtape stores 1474
blocks of 129 12-bit words (with 1 word unused in some operating systems).
129 12-bit words maps to 86 18-bit words, and the overhead frames (10 per
block, 5 on each end) are identical between 12- and 18-bit formats.
144,000 * 18 = 2,592,000 bits.
578 * 128 * 36 = 2,663,424 data bits, so you're close.
1474 * 129 * 12 = 2,281,752 data bits, so your figure is in the middle.
This of course, assumes, that the figure of 144k
accounts for the error
correcting nature of DecTape, and that the actual bit capacity of the
spool would therefore be about double - but since only 144k 18bit words
are usable from PDP-10, then that's all the data that needs to be stored
to image the tape
DECtape is not an "error correcting" medium. It is highly resistant to
data loss, which is not the same thing. Each frame of the tape contains
10 bits, laid out as
T M 1 2 3 1' 2' 3' M' T'
T is the timing track. M is the Mark track, which runs linearly and
defines the contents of each 6 frames by unique codes which were chosen
because they are complement-obverse pairs (that is, they read the same
forwards and backwards).
18 bits of data is written in 6 frames. The head assignments marked with
primes record the same data as the corresponding unmarked heads, so 3 bits
of data per frame.
The 10 overhead frames per block contribute 34,680 more bits to 18-bit
format tapes, and 88,440 to 12-bit format tapes.
Then there are the "landing zones", stretches of tape before block 0 and
after block MAX, of a couple of thousand (I'm not going to go look up the
exact number) frames to account for the mechanics of tape, and the several
thousand frames marked as unusuable so that the controller can find the front
end zone to get started looking for data.
So, then we have to calculate the size of a 1TB hard
disk. This, as we
know, is once again using the base 10 definition of all units, (also
known as HDD maker's "lying gigabytes"):
1TB = 1,000GB = 1,000,000MB = 1,000,000,000KB =
1,000,000,000,000bytes =
8,000,000,000,000 bits.
So, 8,000,000,000,000 / 2,592,000 = 3086419.75ish
So, I'm going to guess 3,086,419 DecTapes can be
completely stored on a
1TB hard disk, assuming no file system. And quite a few more if hard disk
manufacturers didn't blatently lie about the capacity of their products.
I already knew the answer to Guy's riddle (someone made the same joke a few
months ago on one of the relevant Usenet newsgroups), but I'd not worked
out the actual numbers. Thanks for doing the work.
$DEITY, I know too much about the internals of DECtape! Forensics, you know?
Rich Alderson
Vintage Computing Sr. Server Engineer
Vulcan, Inc.
505 5th Avenue S, Suite 900
Seattle, WA 98104
mailto:RichA at
vulcan.com
mailto:RichA at
LivingComputerMuseum.org
http://www.LivingComputerMuseum.org/